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2025-03-21T14:48:29.803493
| 2020-02-07T20:09:32 |
352170
|
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"Alessandro Codenotti",
"Asaf Karagila",
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|
Stack Exchange
|
Undetermined Banach-Mazur games: beyond DC
This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; however, the situation with ZF+$\neg$DC is still open (there was an error in a claimed answer, and this question has been modified from its original form to accommodate that).
I'd like to ask what the ZF-situation is. However, in ZF alone things are rather weird so there are multiple plausible "right" versions of this question. Unfortunately this does make the question rather subjective (presumably based on the answers we somehow pin down which version was "right," but how?), but I think it's within $\epsilon$ of precise for a sufficiently small $\epsilon$.
The most obvious question to ask is whether the result, phrased exactly as previously, continues to hold:
Version 1: Does ZF prove "There is an undetermined Banach-Mazur game"?
On second thought, however, it's not obvious to me that this is the right question. Without DC determinacy is a bit weird, and it's arguably better to talk about quasistrategies. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.
A game is quasidetermined if one player or the other has a winning quasistrategy. In ZF+DC determinacy is the same as quasideterminacy, but in ZF they are distinct. So we can separately ask:
Version 2: Does ZF prove "There is an un-quasidetermined Banach-Mazur game?"
Yet on third thought, even that isn't obviously the right question: it's consistent with ZF that both players have winning quasistrategies in a given game! (Consider "Alternate choosing new elements of an amorphous set.") So we might want to land somewhere in the middle - say that a game is weakly determined if exactly one player has a winning quasistrategy.
Version 3: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?
My guess is that looking at Gitik's model would be a good starting point.
@AsafKaragila Yeah, I'm looking at it but it's really complicated.
If it was easy you'd have solved it by now...
@AsafKaragila I meant the Gitik model.
I meant the Gitik model as well.
If it helps, I have a model where every set is the image of a Dedekind-finite set...
ZF+¬DC implies that there is a BM game that fails to be weakly determined. Take a pruned tree of height $\omega$ with no paths and put the upper set topology on it. The quasistrategy of responding to any move with the cone above some element not minimal in the previous set is winning for both players.
It's actually not clear to me that DC implies that any weakly determined BM game is determined.
It's not quite a Banach-Mazur game, but if you're not aware of it already 21.4 in Kechris's classical DST (see also the long hint at the end of the book) asks to construct a non quasidetermined game in ZF, maybe that construction can be modified
|
2025-03-21T14:48:29.803720
| 2020-02-07T20:42:03 |
352172
|
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"Kate ",
"Paata Ivanishvili",
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}
|
Stack Exchange
|
Martingale polynomial functions
If $B_t$ is a Brownian motion then using Hermite polynomials one can find that
$$1, B_t, B_t^2-t, B_t^3 - 3tB_t,...$$
are martingales.
If $X_t$ is a diffusion
$dX_t = \mu(X_t,t)dt + \sigma(X_t,t)dt$
Is it possible to create sequence of polynomials that would be martingales, but now with $X_t$ instead of Brownian motion?
I will be very grateful for any resources, articles that researched this question
If $H_{n}(x)$ is n'th degree (probabilists) Hermite polynomial then $v(x,t)=t^{n/2}H_{n}(x/\sqrt{t})$ satisfies backward heat equation $v_{t}+\frac{v_{xx}}{2}=0$ (with the boundary condition $v(x,0)=x^{n}$). This is the reason that the process $v(B_{t},t)$ is the martingale. Now for the stochastic process $dX_{t} = \mu(X_{t},t)dt+\sigma(X_{t},t)dt$, the new process $u(X_{t},t)$ is martingale if and only if $u(x,t)$ satisfies PDE $u_{t}+\mu u_{x}+\frac{\sigma^{2} u_{xx}}{2}$.
So you are asking for which $\mu$ and $\sigma$ the solutions of the PDE for $u$ with boundary condition $u(x,0)=x^{n}$ are polynomials in variables $(x,t)$. I doubt that there is a nice answer in such general setting, however, for some concrete $\mu$ and $\sigma$ they can be polynomials.
For example if $\mu$ and $\sigma$ are independent of $t$, then by letting $L:=\mu \partial_{x} + \frac{\sigma^{2}}{2} \partial_{xx}^{2}$, we want to solve PDE $u_{t}+Lu=0$ with boundary data $u(x,0)=x^{n}$. Informally you can write $u(x,t)=e^{-tL} x^{n}$. So this is polynomial if $L^{k} x^{n}=0$ starting for some $k \geq N(n)$. And this is true if $\mu$ and $\sigma$ are constants.
@PaataIvanishvili thank you very much for your reply! Regarding your last comment that this will be true if $\mu$ and $\sigma$ are constants, do I understand correctly that this will be true if some $m$th order derivatives of $\mu$ and $\sigma$ are zero?
We need $\mu$ and $\sigma$ to be constant. If $\mu$ and $\sigma$ are polynomials (nonconstant) this will not help. If $\mu$ and $\sigma$ are constants then it is quite easy to recover the polynomials $u(x,t) = (\sum_{k=0}^{\infty} \frac{(-t)^{k}L^{k}}{k!})x^{n}$. For example, if $n=2$, then $u(x,t) =e^{-tL}x^{2}=(1-tL+\frac{t^{2}L^{2}}{2})x^{2} = x^{2}-2t\mu x - t\sigma^{2}+t^{2}\mu^{2}$, i.e., $X^{2}{t}-2t\mu X{t}-t\sigma^{2}+t^{2}\mu^{2}$ is martingale.
|
2025-03-21T14:48:29.803885
| 2020-02-07T20:50:51 |
352173
|
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|
Stack Exchange
|
Characters of sets of representations closed under tensor product
Let $R$ denote the set of all irreducible representations of a group $G$ over a complex vector space. Let $U \subset R$ denote a subset of representations which is closed under tensor product (i.e., if $\rho$, $\mu \in U$, then $\rho \otimes \mu = \oplus_{i} \nu_i$ with each $\nu_i \in U$). For $g \in G$, let $\chi_\rho(g) =$ Tr$\rho(g)$ denote the character given by the trace of $g$ in the $\rho$ representation.
Is the following statement true: $U$ is a proper subset of $R$ if and only if there exists some $g \neq 1 \in G$ such that $\chi_{\rho}(g) = \chi_\rho(1)$ for all $\rho \in U$? Here $\chi_\rho(1):=$ Tr$\rho(1)$ is the trace of the identity element $1 \in G$ in the $\rho$ representation.
I am quite sure this statement is true for $G$ a finite group, and I am especially interested to know if it also holds true for $G$ a Lie group.
The first question is whether such proper subsets closed under tensor products exist at all. For SU(2) one can use Clebsch-Gordan to see that the only such proper subset is the one consisting of all odd-dimensional irreducible reps. For this set, however, I wouldn‘t see why your claim should be true for some g.
The claim is true for SU(2). The only proper subset closed under tensor products is indeed the one consisting of all odd-dimensional irreducible reps, and the element g is given by -1, the nontrivial element in the Z2 center of SU(2).
Your claim is indeed true for finite groups. By a theorem of Blichfeldt/Burnside, it follows, if there is no such $g$, that any irreducible character $\mu$ of $G$ occurs as a constituent of some power of $\sum_{\chi \in U}\chi$, and then the assumed properties of $U$ force $\chi \in U$.
|
2025-03-21T14:48:29.804028
| 2020-02-07T21:03:26 |
352174
|
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|
Stack Exchange
|
Vectors that satisfy $\sum_{i=1}^n Y_i X_i^\top = I$ and $\sum_{i=1}^n \frac{1}{p_i}Y_iY_i^\top = \Sigma(P)^{-1}$
Let $X_1,\dots,X_n$ be vectors in $\mathbb{R^d}$. Assume all of the vectors are inside the unite $\ell_2$ ball. Let $P$ be a vector in the probability simplex $\Delta_n$ with $P_i>0$ for all $i$. Consider the second moment matrix $\Sigma(P) = \sum_{i=1}^n P_i X_i X_i^\top$. Assume the $X_i$s are such that $\Sigma(P)$ is full rank. We need to find vectors $Y_1,\dots, Y_n$ such that the following two conditions hold:
\begin{align}
\sum_{i=1}^n Y_i X_i^\top &= I\\
\sum_{i=1}^n \frac{1}{p_i}Y_iY_i^\top &= \Sigma(P)^{-1}
\end{align}
We observe that $Y_i=p_i \Sigma(P)^{-1} X_i$ is a solution. Are there any other solutions?
|
2025-03-21T14:48:29.804105
| 2020-02-07T21:10:34 |
352176
|
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"Zack",
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|
Stack Exchange
|
Augmentations of wrapped Floer cochains
Let $M$ be a closed, simply-connected spin manifold and let $F_b \subset T^*M$ be the cotangent fiber over a point $b \in M$. Let $CW^*(L,L)$ be the $A_{\infty}$-algebra of wrapped Floer cochains over a field k and let $C_{-*}(\Omega_b M)$ be the dg-algebra of chains on the based loop space.
Abouzaid proved that there is an equivalence of $A_{\infty}$-algebras $CW^*(L,L) \simeq C_{-*}(\Omega_b M)$.
Observe now that the collapse map $M \mapsto b$ induces an augmentation $C_{-*} (\Omega_b M) \to C_{-*} ({pt}) \simeq k$. Precomposing with Abouzaid's equivalence, we conclude that $CW^*(L,L)$ is an augmented $A_{\infty}$-algebra (i.e. it admits a morphism of $A_{\infty}$-algebras to $k$).
Question: can this augmentation be seen at the level of wrapped Floer homology, i.e. without going through Abouzaid's equivalence? A natural thing to do would be to intersect with the zero section. However, this does not appear define a morphism of $A_{\infty}$-algebras in an obvious way.
Intersecting with the zero section defines a 1-dimensional module, i.e. a $A_\infty$-homomorphism to $End_k(k)\cong k$, which is the same as an augmentation.
|
2025-03-21T14:48:29.804209
| 2020-02-07T21:13:09 |
352177
|
{
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"Student",
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"https://mathoverflow.net/users/41862",
"zibadawa timmy"
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"url": "https://mathoverflow.net/questions/352177"
}
|
Stack Exchange
|
Classification of $\operatorname{Rep}D(H)$
Question
Let $H$ be a finite dimensional complex Hopf algebra and $D(H)$ its quantum double. Can we classify the simple objects in $\operatorname{Rep}D(H)$ if the representations of $H$ are well-understood?
My impression is that this is not yet completely developed. However, the more examples the better! Would you mind sharing papers/resources that deal with the representations of quantum doubles, even those that deal with specific examples?
Example: finite group algebra
For example, if $H$ is the complex group algebra of some finite group $G$, both $\operatorname{Rep}(G)$ and $\operatorname{Rep}D(G)$ are well-enough understood (to my standard). See my previous questions about this
Representations of $D(G)$ as an object in the center of $\operatorname{Rep}(G)$
Classification of $\operatorname{Rep} D(G)$
Interestingly, irreducible representations of $D(G)$ do not restrict to irreducible ones as $G$-reps. Viewing them as objects of $Z(\operatorname{Rep}G)$ reveals hidden structures among irreps of $G$: nontrivial half-braidings arise naturally!
Example: Taft algebra
As another example, the representations of the double of Taft algebras are examined here by Chen.
In what sense does the center construction itself not answer your question? My initial impression otherwise is that the answer is "no" in a sense, because even in the case of group algebras you can have non-isomorphic, non-isocategorical groups whose doubles have equivalent representations; or in another sense the double's representations can have multiple tensor equivalence types of categorical Lagrangian Grassmanians. A result I've mentioned to you elsewhere on the site.
To any skew-pairing $\lambda:U\otimes H\rightarrow k$, one can associate a hopf algebra $D(U, H)$ (built on $U\otimes H$) which is called the generalized quantum double of $U$ and $H$.
(If $H$ is finite dimensional, $U=H^{*cop}$ and $\lambda$ is the usual evaluation map, then, this corresponds to the usual quantum double $D(H)$ introduced by Drinfeld).
In the article On the irreducible representations of generalized quantum doubles, the authors describe the irreducible representations of generalized quantum double hopf algebras $D_{\lambda_f}(U,H)$, where $\lambda_f:U\otimes H\rightarrow k$ is a skew-pairing induced by a surjective hopf algebra map $f:U\rightarrow H^{*cop}$.
One of the main results of the paper (as i understand it) is theorem 1.1 (p.2), which describes the simple objects of $\operatorname{Rep}\big(D_{\lambda_f}(U,H)\big)$, in a way which generalizes the corresponding description of the simple objects of $\operatorname{Rep}D(G)$ (discussed in the questions linked to the OP). Then the paper proceeds in a classification of the generalized quantum double, simple modules, for the case where both $U,H$ are semisimple hopf algebras and there is a surjective map $U\rightarrow H^{*cop}$.
Furthermore:
In The Representations of Quantum Double of Dihedral Groups, the finite dimensional, indecomposable, left $D(kD_n)$-modules are classified, where $D_n$ is the dihedral group of order $2n$ and $k$ is an algebraically closed field, of odd characteristic $p\ |\ 2n$.
Burciu's paper reduces the problem to the representations of smaller ones: $U \bowtie L(g)$, where $g$ is a conjugacy class of the universal grading group $G$ of $\operatorname{Rep}(H)$. In case of doing actual computations, we still need to know (1) how to compute information of $G$ (e.g. is it finite? of order?) (2) how to compute $L(g)$, whose definition is indirect in the paper? (3) The representations of $U \bowtie L(g)$ provided enough knowledge of $\operatorname{Rep}(U)$ and $\operatorname{Rep(L(g))}$.
It might be too much to ask.. perhaps I should just ask: how practical it is to compute $Rep(D(H))$ using the knowledge of $Rep(H)$?
|
2025-03-21T14:48:29.804449
| 2020-02-07T22:20:25 |
352180
|
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"Andi Bauer",
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|
Stack Exchange
|
What is a sufficient set of links in a simplicial complex to represent any PL manifold?
The link of a vertex in a $n$-dimensional simplicial complex is the $(n-1)$-dimensional simplicial complex formed by the $(n-1)$-simplices each of which, together with the vertex, spans an $n$-simplex. A combinatorial PL manifold is a simplicial complex where the links of all vertices are topologically equivalent to a PL sphere.
It is known that equivalence classes of combinatorial PL manifolds under Pachner moves are in one-to-one correspondence with equivalence classes of PL manifolds under homeomorphism.
Instead of allowing all links with PL topology of a sphere, we can restrict to a finite set of allowed links. This includes restricting to Pachner moves that transform a combinatorial PL manifold with allowed links in another one with allowed links (note that a Pachner move in the manifold acts also with Pachner moves on the links of the involved vertices).
If this set of allowed links is big enough then the equivalence in the second paragraph will still hold. Note that this includes 1) we still have to be able to represent all PL manifolds and 2) we still have to be able to transform any equivalent combinatorial manifolds into each other with Pachner moves, without going via any non-allowed links.
E.g. in 2 dimensions, it is easy to show that 5-, 6-, and 7-gons as links are enough to represent any 2-manifold, and it seems highly plausible that any two such restricted combinatorial manifolds can be transformed by Pachner moves if also 3- to, say, 10-gons are allowed.
Is there a general scheme to construct a sufficiently large set of allowed links that is still rather small? Optimally, this would work for any dimension, but I'm also fine with dimensions 2 and 3. I'm not searching for the strictly smallest possible set of allowed links, but I also would want something better than "provably, taking links with $<10000$ vertices is sufficient in $3$ dimension".
Not sure about Pachner moves, but as for links, three suffice, see the math review of:
Cooper, D.; Thurston, W. P., Triangulating 3-manifolds using 5 vertex link types, Topology 27, No. 1, 23-25 (1988). ZBL0656.57004.
Kevin Walker walks among us, so he may comment further.
Thanks, that looks promising that small sufficient sets of links exist. However, there is not a single Pachner move transforming a link in this set to into another link in the set, so one won't be able to go between equivalent combinatorial manifolds via simplicial complexes with only those links. The latter constraint seems to be the hard bit of the problem. Without it, it should not be that difficult to find small sufficient sets of links by refining and "smoothing" a given triangulation.
|
2025-03-21T14:48:29.804752
| 2020-02-08T00:13:27 |
352184
|
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|
Stack Exchange
|
fair-indexed tournament
Let's say that two tournaments $T$ and $T'$ with vertex respectivally $T_1,T_2,....T_n$ and $T'_1,T'_2,....T'_n$ are in the same score-class iff there is a $n$-permutation $s$, such that for any $i\in \left\{1,2,....n\right\}$, $T_i$ and $T'_{s(i)}$ have the same number of victory.
Let's say that a tournament whose vertex are $T_1,T_2,...,T_n$ is fair-indexed iff for any $i\in \left\{1,2,3,....,n\right\}$ there exists $j\in \left\{i,i+1,...,n\right\}$ such that for any $k>i$, $T_i$ beats $T_k$ iff $k>j$.
Is there a fair-indexed tournament in any score-class?
(I would also be glad to kwow if there is already a standard appelation for "fair-indexed", or if the question has been studied already, if it is the case i will edit)
The mathematician who presented me the problem (in quite a different form) told me that it would solve a problem of interpolation of multivalued polynomes, that has only been solved for the characteristic zero - I'll give some precisions when I will get them, if anyone is interested... anyway I find the question for itself nice enough to be asked and investigated.
Sorry about my stupid "answer". I will try to think intelligently about your question.
Thank you for your interest, I looked if the question was clear after your answer and thanks to it, I edited "iff" insted of "if"^^
From looking at very small examples I got the idea that maybe there is always a fair indexing with $s_1\ge s_2\ge\cdots\ge s_n$ where $s_i$ is the score (out-degree) of the vertex $T_i$. This was a wrong idea, as shown by the example $${6,6,6,4,4,4,4,4,4,3}.$$
@bof : indeed, if it was true, by symetry, it would be true as well for increasing sequences. The smallest example that does not work monotonally (both increasingly and decreasingly) is $\left{2,2,3,3,3,4,4\right}$
What symmetry is that? I learned long ago that it doesn't work for increasing sequences such as ${1,1,1,3}$, but only recently learned that decreasing sequences don't work either.
Sorry i made a confusion with the symetry thing! Is it a famous problem ? (because you wrote ""I lerarned long ago..."
I meant relatively long ago, i.e., a few days after you posted the question.
By the way, what happens if your "fair indexing" requirement is strengthened, so as to require both your condition "if $i\lt j\lt k$ and $T_i$ beats $T_j$ then $T_i$ beats $T_k$" and simultaneously "if $i\lt j\lt k$ and $T_k$ beats $T_i$ then $T_k$ beats $T_j$"? Do you have a counterexample to that?
If i don't mistake i think 22225555 is a counterexample...
Lets call the strengthened a bifair indexation. I thought once if score (wied as a increasing fonction) that does not increase of mire than 1 might have a bifair indexation tournament in the class that they define. (In the counterexample i gave, the score is increasing by 3 units from 2 to 5) I don't know if the statement of bifairity for the 0 or 1 increasing score is true or not... but i'm going to think about it!
by the way the tournament that are bifair indexable (lets call them BFI), are maybe intersting to study, as well as the set of 0 or 1 incresing score (lets call them ZOOI) The Zooi are intersting because 1) it seems that almost every tournament is Zooi 2) you can get all the score by iterated substitutions of Zooi in a zooi. I will be more precise in both 1) and 2) if you ask me^^
Thank you for the counterexample!
|
2025-03-21T14:48:29.805022
| 2020-02-08T00:34:17 |
352186
|
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|
Stack Exchange
|
Alternating 1D lattice sum
Are there any equivalent representations of the following (real valued) sum, in particular that are suitable for evaluation as $z\rightarrow0$ ?
$$ S=\sum_{k=-\infty}^\infty \frac{i^k(z-2ik)}{(\rho^2+(z-2ik)^2)^{3/2}} $$
I am aware that $S$ resembles a coulomb force sum which can be rearranged using Lekner summation into a seires of Bessel functions, but the factor $i^k$ seems to prohibit this transformation.
I don't understand the statement about the terms decreasing as $k^{-1/2} $. Isn't $k/(k^2)^{3/2} = k^{-2} $?
Thanks for pointing this out, I don't know how I reached that conclusion. I have reworded the question appropriately.
For $\rho$ equal to an even integer the sum $S$ diverges as $1/z^{3/2}$ when $z\rightarrow 0$. For $\rho$ unequal to an even integer and $z>0$, one has
$$S_0=\lim_{z\rightarrow 0}\sum_{k=-\infty}^\infty \frac{i^k(z-2ik)}{(\rho^2+(z-2ik)^2)^{3/2}}=-4\,\Re\sum_{k=1}^\infty\frac{ i^k k}{\left(4 k^2-\rho^2\right)^{3/2}}.$$
Yes this techincally is the limit as $z\rightarrow0$, but I would like to find an expression that isn't conditionally convergent - I have updated the question to specify this.
@MattMajic -- please clarify for me: the sum I wrote down has terms that decrease as $1/k^2$ --- why is this only "conditionally convergent" ?
you're right, I mistakenly took them to go as $k^{-1/2}$, which would be conditional.
$S$ may be expressed as a series of Bessel functions:
$$S=-\pi^2\sum_{n=1}^\infty \left(n-1/4\right) \exp[-(n-1/4)\pi z] J_0[(n-1/4)\pi\rho],$$
however this form diverges for $z=0$.
|
2025-03-21T14:48:29.805163
| 2020-02-08T00:51:23 |
352188
|
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|
Stack Exchange
|
Artin's "On isolated rational singularities of surfaces"
My question deals with Michael Artin's paper "On isolated rational singularities of surfaces"; more precisely the proof of Theorem 4 on page 133. Here the relevant excerpt:
The Setting: Let $\bar{V}=Spec(A)$ where $A$ is a local, normal $2$-dimensional ring with algebraically closed residue field $k=A/\frak{m}$.
Let $\pi:V \to \bar{V}$ be a birational proper map with $V$ regular, i.e. it "resolves" the singularity $s$ of $\bar{V}$, where the point $s$ corresponds to the maximal ideal $\frak{m}$ $ \subset A$.
Denote by $Z= \sum_i X_i$ the fundamental cycle; here the definition:
$Z$ is defined as the unique smallest cycle satisfying property
One of the intermediate steps in the proof is to show that for the ideal $I_Z \mathcal{O}_V$ that determines the fundamental cycle (that is a closed subscheme of $V$) we have
$$\mathfrak{m} \cdot \mathcal{O}_{\text{V}}= I_Z$$
Artin reduces the problem to verification of the surjectivity of
(*) $$H^0(Z, \mathcal{O}_{(n+1)Z}) \to H^0(Z, \mathcal{O}_{nZ})$$
for each $n$.
Lemma 5 proves it.
Problem/Question: Then it is claimed that "moreover, because of (*), it follows that the canonical map $A/\mathfrak{m}$ $ \to H^0(nZ, \mathcal{O}_{nZ})$ is surjective.
Why? I don't understand it. How the argument works?
What I tried: My first approach was to argue by induction but this gives an obstacle that I can't solve.
Denote $K_n:= ker(\mathcal{O}_{(n+1)Z} \to \mathcal{O}_{nZ})$. One can show that $K_n \cong I_{nZ} \otimes_V \mathcal{O}_Z$ and one obtain the diagram
$$
\require{AMScd}
\begin{CD}
\mathfrak{m}^n/\mathfrak{m}^{n+1} = \mathfrak{m}^n \otimes A/\mathfrak{m} @>{} >> A/\mathfrak{m}^{n+1} @>{} >> A/\mathfrak{m}^n \\
@VaVV @VbVV @VcVV \\
H^0(V, K_n) @>{} >> H^0(Z, \mathcal{O}_{(n+1)Z}) @>{}>> H^0(Z, \mathcal{O}_{nZ});
\end{CD}
$$
By induction hypothesis we may assume that $c$ is surjective. Question/Problem: Why is $a$ surjective? (we need surjectivity of $a$ to conclude that $b$ is surjective).
Additionally let me loose some words on inkspot's
observation about the map
$ H^0(nZ,O_{nZ}) \to A/{\mathfrak m}^n $: I think that
it is a typo and the map should be
$A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$
because considering the composition
$nZ \subset V \to \bar{V}=Spec(A)$ induces naturally
$A \to H^0(nZ,O_{nZ})$ which factorize through
$\mathfrak m^n$, thus I think the only map which can be
extracted can only go in the opposite direction
of that one in Artin's paper.
I don't know whether this will answer the question, but you write $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ while Artin's homomorphism goes in the opposite direction.
I think that it was just a typo in the paper. That is we consider indeed $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ induced canonically by $nZ \subset V \to \bar{V}=Spec(A)$ on the ring side ($A \to H^0(nZ,O_{nZ})$ factorizes through $A/{\mathfrak m}^n$)
I think I have (almost) the answer I was looking for: The problem is to show that $a$ surjective and by induction hypothesis $c_n: A/m^k \to H^0(Z,O_{kZ})$ is surjective for all $k < n+1$. Assume $A/m \to H^0(Z,O_{Z})$ is surjective (that's a seriuos problem: see below problem 2, but let at first assume we know it). We tensor it with $m^n$ and obtain the surjection $m^n/m^{n+1} \to m^n \otimes H^0(V, K_n)$. The goal would be to show $m^n \otimes H^0(Z,O_{Z})= H^0(V, K_n)$.
That is that $H^0$ and $\otimes$ commute. I'm facing two problems now: first one: to show that $m O_V$ is invertible $O_V$-module. (then it would flat and $H^0$ and $\otimes$ commute and we are done). The second problem is the initial induction step: how to show that $A/m \to H^0(Z,O_{Z})$ is surjection?
|
2025-03-21T14:48:29.805403
| 2020-02-08T01:02:33 |
352189
|
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|
Stack Exchange
|
Exhaustion of a manifold by open $n$-cells
Suppose that a differentiable manifold $M^n$ has an exhaustion $M^n=\bigcup U_i$ with $U_i \subset U_{i+1}$ such that each element $U_i$ is diffeomorphic to $\mathbb R^n$.
Can we prove that $M^n$ is diffeomorphic to $\mathbb R^n$?
Notice that if $n\ne 4$, the question is true from this paper since there is a unique smooth structure on $\mathbb R^n$. So the question only concerns four-dimensional case.
Sorry for a naïve question, but how can one deduce that $M$ is diffeomorphic to $\mathbb R^n$ from the uniqueness of smooth structure on $\mathbb R^n$? It seems you'd need first to know that $M$ is a topological $\mathbb R^n$; is that obvious?
@LSpice: You didn't follow the link.
https://math.stackexchange.com/questions/3177579/manifold-with-a-good-exhaustion?noredirect=1&lq=1 and https://math.stackexchange.com/questions/3065755/characterization-of-mathbbrn?noredirect=1&lq=1
@MoisheKohan I guess the link you gave requires all $\bar U_i \subset U_{i+1}$ and all $\bar U_i$ is diffeomorphic to a closed $n$-ball.
If $U_i$ is diffeo to $R^n$, it contains arbitrarily large smooth compact $n$-balls $V_i$; this yields an exhaustion by $int(V_i)$ such that $V_i\subset int(V_{i+1})$.
|
2025-03-21T14:48:29.805523
| 2020-02-08T01:24:57 |
352191
|
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|
Stack Exchange
|
Stalks of perverse cohomology sheaves?
For a complex of sheaves $\cal{F}^{\bullet}$ on a variety $X$, a useful fact is that the stalks of the cohomology sheaves of $\mathcal{F}^{\bullet}$ agree with the cohomology groups of the complex of stalks:
$$\mathcal{H}^{i}(\cal{F}^{\bullet})_{x} \cong \mathcal{H}^{i}(\cal{F}^{\bullet}_{x})$$
Consider now the category $D_{c}^{b}(X)$ of bounded constructible complexes on $X$, and the full abelian subcategory $\operatorname{Perv}(X)$ of perverse sheaves. Let us work with sheaves of $\mathbb{Q}$ or $\mathbb{C}$-modules. The perverse cohomology sheaves ${}^{\mathfrak{p}}\mathcal{H}^{i}(\cdot):D_{c}^{b}(X) \to \operatorname{Perv}(X)$ are themselves perverse sheaves.
I'm wondering, are there any nice results helping to compute stalks of ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathcal{F}^{\bullet})$? Like something analogous to the basic result I recall at the beginning? These stalks should be complexes of vector spaces.
The basic result above for ordinary sheaves can't work as it is. For example, if $X$ is smooth of dimension $n$, then $\mathbb{C}_{X}[n]$ is perverse. And ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathbb{C}_{X}[n])$ vanishes unless $i=0$ in which case it is just $\mathbb{C}_{X}[n]$. So at any point $q \in X$, the stalk is the skyscrapper sheaf:
$${}^{\mathfrak{p}}\mathcal{H}^{0}(\mathbb{C}_{X}[n])_{q} = \mathbb{C}_{q}[n]$$
But if we first pass to the stalk, and then take perverse cohomology ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathbb{C}_{q}[n])$, this vanishes unless $i=-n$, in which case it is just $\mathbb{C}_{q}$.
So are there any useful ways of computing stalks of perverse cohomology sheaves, or do you just have to do it directly?
Certainly none as nice as for the usual cohomology sheaves, except at a generic point, or more concretely a point where all the cohomology sheaves are lisse - at these points, the perverse filtration is the usual filtration shifted by n. If your complex of sheaves is one to which the decomposition theorem can be applied, the decomposition theorem is a big help. Other than that I don't know any big general principle.
I have been told by various experts in the area that stalks aren’t the right notion for perverse sheaves. Instead, one should be considering nearby and vanishing cycles with respect to various linear forms. These functors (appropriately shifted) do commute with perverse cohomology, as they’re exact on the category of perverse sheaves.
Sorry, what do you mean that "that stalks aren’t the right notion for perverse sheaves."? I don't understand. I think taking stalks preserve the exactness (in the triangulated category).
@zxx Taking stalks of an exact sequence of sheaves yields an exact sequnce of modules. Taking stalks of an exact sequence of perverse sheaves does not. Taking the stalk is the same as taking the inverse image $i^$ along the inclusion of a point $i$, and for perverse sheaves $i^$ is only right exact in general.
Thanks a lot! I see you argument. I mixed the perverse sheaf category and the derived category of it.
|
2025-03-21T14:48:29.805732
| 2020-02-08T02:45:29 |
352193
|
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|
Stack Exchange
|
What's the state of the art on stochastic representations of hyperbolic PDE?
I saw this paper: https://arxiv.org/abs/1306.2382
Chatterjee gives a representation of many solutions of the wave equation in terms of Brownian motion. I haven't seen much other than this. Is there any recent developments?
I am not sure about the state of the art, but there is a nice paper of Pal-Shkolnikov which shows hows hyperbolic PDE arise from intertwining relations between diffusion processes: https://arxiv.org/abs/1306.0857
Is Navier-Stokes considered Hyperbolic ? If yes, then see this:Constantin, Peter, and Gautam Iyer. "A stochastic Lagrangian representation of the three‐dimensional incompressible Navier‐Stokes equations." CPAM (2008)
|
2025-03-21T14:48:29.805814
| 2020-02-08T04:19:57 |
352195
|
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|
Stack Exchange
|
On the set $\{n>0:\ n\ \text{is a quadratic nonresidue modulo the}\ n\text{th prime}\}$
Let $S$ denote the set of positive integers $n$ with $n$ a quadratic nonresidue modulo the $n$th prime $p_n$. The first 20 elements of $S$ are
$$2,\, 3,\, 6,\, 7,\, 8,\, 10,\, 11,\, 13,\, 15,\, 18,\, 21,\, 24,\, 26, \,27,\, 28,\, 32,\, 33,\, 39, \,41,\, 44.$$
For more elements of $S$ see http://oeis.org/A332021. I have the following natural conjecture.
Conjecture. The asymptotic density $d(S)$ of the set $S$ is $1/2$, that is,
$$\lim_{N\to+\infty}\frac{|\{n\in S:\ n\le N\}|}{N}=\frac12.\tag{1}$$
QUESTION. How to prove the conjecture? Is it practical to show that the set $S$ is infinite?
It seems that $(1)$ can be strengthened as follows:
$$|\{n\in S:\ n\le x\}|=\frac x2+O(\sqrt x). \tag{2}$$
Your comments are welcome!
I suspect this conjecture is completely hopeless. The $n$-th prime function has no sensible arithmetic properties. Asking whether $n$ is a square mod $p_n$ is as good as a coin toss, so while we certainly would expect the density to be $1/2$, I doubt we have tools to prove it.
Agree, your question is weird, it suggests that you don't know "the $n$-th prime" has no direct meaning, we don't have any useful generating series for the function $n \to p_n$. Then, did you ever try to check a conjecture with the random model for the primes ? Here you'd need to check if extending the random model to take in account quadratic residues and $n\to p_{f(n)}$ has any chance to give non-contradictory results.
This conjecture appeals for studying the exact values of the prime-counting function.There are few results in this direction. For example, in my 2017 paper in Ramanujan J. , I proved that for any integer m>4 there is a positive integer n with π(mn)=m+n.
|
2025-03-21T14:48:29.805969
| 2020-02-08T06:27:23 |
352196
|
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"authors": [
"Alessandro Codenotti",
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|
Stack Exchange
|
Different locally compact metrizable second countable topologies on the same group
Let $G$ be a non-Abelian infinite group. Can $G$ admit more than one (inequivalent) non-compact locally compact metrizable second countable topologies that make it a topological group?
Thank you.
Turning my comment into an aswer: yes, it is possible for such a $G$ to have more than one topology with your requirements.
Consider $(\Bbb R,+)$ with its usual topology. As a group it is isomorphic to $(\Bbb C,+)$, so there is also a topology in which $(\Bbb R,+)$ is a topological group homeomorphic to $\Bbb R^2$, both topologies obviously satisfy your requirements.
The only issue is that $\Bbb R$ is Abelian, but that can be fixed by taking a direct product with any locally compact, second countable, metrizable, nonabelian topological group, since all of this properties are preserved under finite products.
Although "direct product with any locally compact, second countable, metrizable, nonabelian topological group" will preserve the distinctness of the two topologies, preserving non-homeomorphism seems to require being more specific about the nonabelian factor. (Any finite, discrete, nonabelian factor will do.)
MO recommends avoiding comments of gratitude, but I like thanking people for their help. Thank you, this was enlightening. In this example there exists no continuous group isomorphism (not necessarily open) between the two. Is it possible to answer the same question but with the additional requirement that there is a continuous abstract group isomorphism between the two? If you think that this makes things very different, I will set up a new question. Thanks again.
@Bedovlat I don't see an easy way to construct a pair with such an isomorphism, I'm not sure whether it's a much harder question or whether there is another neat construction
A continuous bijection from compact to Hausdorff is a homeomorphism. From locally compact to Hausdorff is a similar situation, so I agree that it is probably impossible (or at least much harder).
|
2025-03-21T14:48:29.806125
| 2020-02-08T07:49:34 |
352198
|
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|
Stack Exchange
|
Jacquet module of irreducible principal series
Let $F$ be a local field of characteristic zero and $G=\operatorname{GL}_n(F)$.
Let $B=UT$ be a Borel subgroup of $G$ and $\chi=(\chi_1,\cdots,\chi_n)$ is an unramified character of $B$.
Consider unramified principal series $\pi$ of $G$ defined by $\operatorname{Ind}_B^G \chi$. Assume that $\pi$ is irreducible.
Then for any parabolic subgroup $P_{a,n-a}=NM$ of $G$ whose Levi subgroup $M$ is $GL_{a} \times GL_{n-a}$, it is known that the any subquotient of Jacquet module $J_N(\pi)$ with respct to $N$ is also irreducible principal seires of $M$. (i.e. any subquotient of $J_N(\pi)|_{{GL_{a}}}$ and $J_N(\pi)|_{{GL_{n-a}}}$ are both principal series.)
I don't know why it holds. If it is not easy to explain, would you recommend some reference on this?
And I also wondering whether principal series representations can have finite dimensional subrepresentation.
There is a very general formula describing composition factors of Jacquet modules of an induced representation in Bernstein-Zelevinsky.
@GTA, Thank you very much for your comments. If you don't mind, would you suugest me some reference on such result? Though I refered BZ's papers, I couldn't find such formula.
Look at Theorem 5.4 in Prasad and Raghuram's notes on representations of $GL_n(F).$ It is a special case of what is sometimes called the geometric lemma. In the case of $GL_n(F)$ see Theorem 1.2 on page 170 in Zelevinsky, Induced representations of reductive p-adic groups.
@Not a grad student, Sorry for late reply. Finally I catched your point. Though the general geometric lemma can't specify every subquotients, in this case, it works because we begins from the character of the maximal torus $GL(1) \times \cdots \times GL(1)$ corresponding the partiton (1,1,...,1). Am I right? Also I think I should correct $J_N(\pi)|{GL_a}$. Instead, it would be right $(J_N(\pi)){ss}=\oplus_w Ind_{T_a}^{GL_a} (\chi^w)\otimes Ind_{T_{n-a}}^{GL_{n-a}} (\chi'^w)$ and what I mean for irr.subquo of $J_N(\pi)|{GL_a}$ is $Ind{T_a}^{GL_a} (\chi^w)$ for some $w$. Does it makes sense?
I will explicitly work out the details in Section 1.2 of Zelevinsky "Induced representations of reductive $\mathfrak p$-adic groups II." In their notation, $\beta=(1,\dots,1)$ and $\gamma=(a,n-a)$. Now $I_i=\{i\}$ for each $i=1,\dots,n$ and $\mathscr I_1=\{1,\dots,a\}$ and $\mathscr I_2=\{a+1,\dots,n\}$. Now $$W^{\beta,\gamma}=\{w\in W|w^{-1}(1)<\cdots <w^{-1}(a),w^{-1}(a+1)<\cdots<w^{-1}(n)\}$$
is a subset of $W$ of size ${n\choose a}$.
For each such permutation, we can define $\gamma':=\gamma\cap w(\beta)=(1,\dots,1)$ and $\beta':=\beta\cap w^{-1}(\gamma)=(1,\dots,1)$. Now Theorem 1.2 states that
$$\begin{align*}
J_N(\pi)^{\mathrm{s.s.}}&=\bigoplus_{w\in W^{\beta,\gamma}}i_{\gamma,\gamma'}\circ w\circ r_{\beta',\beta}(\chi_1\otimes\cdots\otimes\chi_n)\\
&=\bigoplus_{w\in W^{\beta,\gamma}}i_{\gamma,\gamma'}(\chi_{w^{-1}(1)}\otimes\cdots\otimes\chi_{w^{-1}(n)})\\
&=\bigoplus_{w\in W^{\beta,\gamma}}\mathrm{Ind}_B^{\mathrm{GL}_a}(\chi_{w^{-1}(1)}\otimes\cdots\otimes\chi_{w^{-1}(a)})\boxtimes \mathrm{Ind}_B^{\mathrm{GL}_{n-a}}(\chi_{w^{-1}(a+1)}\otimes\cdots\otimes\chi_{w^{-1}(n)}),
\end{align*}$$
where as usual $\mathrm{Ind}_B^{\mathrm{GL}_a}$ denotes normalized induction. In fact, since you assume $\pi$ is irreducible, i.e., none of the $\chi_i\chi_j^{-1}$ are of the form $\nu^{\pm1}$ for any $i\ne j$, each direct summand above is irreducible, and moreover the Ext groups must vanish, so in fact the above decomposition is true before semi-simplification.
P.S. Note that none of the above used that $\chi$ is unramified.
|
2025-03-21T14:48:29.806690
| 2020-02-08T07:56:04 |
352200
|
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|
Stack Exchange
|
Is cyclic replacement inconsistent with ZFC-Foundation?
Replacement: if $\phi(x,z)$ is a formula in which all and only symbols $``x,z,x_1,..,x_n"$ occur free, and non of them occur as bound, and in which the symbol $``B"$ never occur; then: $$\forall x_1,..,x_n \\\forall A \exists B \forall y \ [y \in B \leftrightarrow \exists x \in A \forall z (\phi(x,z) \leftrightarrow z=y)]$$; is an axiom.
The above scheme is a just a reformulation of the axiom schema of replacement.
Cyclic replacement is the above schema but with allowing the symbol $``B"$ to occur free in $\phi(x,z)$.
Is it inconsistent to replace axiom schema of replacement in ZFC-Foundation by axiom schema of cyclic replacement?
Would that be inconsistent with Aczel's anti-foundation axiom?
Unless I'm misunderstanding something here, Cyclic Replacement is inconsistent with just about any set theory. If $p$ and $q$ are two distinct sets, let $\phi(x,z,B)$ be $(z=p\land B\neq\{p\})\lor(z=q\land B=\{p\})$. Let $A$ be your favorite nonempty set. (Since you insist that $\phi$ contain $x$ free, add a clause "$\land(x=x)$".)Then cyclic replacement says there's a set $B$ such that (1) if $B=\{p\}$ then $B=\{q\}$ and (2) if $B\neq \{p\}$ then $B=\{p\}$. That's contradictory.
Possibly $$\forall x_1,..,\forall x_n \\forall A \exists B \forall y \ [ \exists x \in A \forall z (\phi(x,z) \leftrightarrow z=y) \to y \in B ]$$ might salvage it?
@ZuhairAl-Johar Let $\phi$ say that $z$ is the first ordinal not in $B$.
Thanks! I think I know now how to salvage this approach, the formula $\phi(x,z)$ must be a disjunctive formula $\psi(x,z) \lor \varphi(x,z)$ where only $B$ only occurs in $\varphi(x,z)$ which must be a positive formula (defined as in positive set theory), clearly cyclicity is anathematic to negation. a possible other way is to use acyclic (stratified) formulas.
|
2025-03-21T14:48:29.806840
| 2020-02-08T10:29:02 |
352204
|
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|
Stack Exchange
|
What is known about iterated matching as a TSP heuristic
A fairly wellknown heuristic for TSP that is based on matching is described in the 2003 paper Match twice and stitch: a new TSP tour construction heuristic by Andrew B. Kahng and Sherief Reda.
Its basic idea is to generate a vertex-covering collection of even cycles via the union of two edge-disjoint matchings and then striving to optimally combine the cycles to arrive at a short Hamilton cycle.
The performance of that heuristic is reportedly good.
My question is however related to a different idea of "directly" constructing short Hamilton cycles of a graph with $n=2^k$ vertices by starting with a minimum weight matching, defining the matching edges to be the "root-paths" having level $0$. The paths of the next level are then generated by optimally connecting pairs of paths from the previous level and to finally insert the edge that connects the ends of the final path.
That idea has already been pursued in the 1996 German phd thesis *Traveling Salesman Problem mit iteriertem Matching* (Traveling Salesman Problem via Iterated Matching) by Frank Lauxtermann.
That paper is however the only one I could find where the idea of repeatedly combining pairs of paths at minimal cost is mentionend and apart from that the algorithm in the cited paper left a fuzzy impression on me in the sense that the combination of paths is based on rather involved criteria that aim at yielding low-weight combinations.
Questions:
has the idea to repeatedly combining pairs of paths (starting with the set of edges in a matching) also been investigated by other authors?
what is known about algorithms for determining the pairs of paths whose optimal combination yields the optimal solution of reducing the number of paths from $2m$ to $m$?
For optimally merging pairs of paths, you can introduce a complete graph with one node per path and edge weight for each path pair $(i,j)$ equal to the minimum of the four possible original edge weights to join one endpoint of path $i$ to one endpoint of path $j$. Then solve a minimum-weight perfect matching problem on this graph.
|
2025-03-21T14:48:29.807137
| 2020-02-08T10:53:50 |
352206
|
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|
Stack Exchange
|
Lower bound for the chromatic number in terms of minimum feedback vertex set
Let $MFVS(G)$ denote the size of minimum feedback vertex set of $G$.
We believe we proved $\chi(G) \ge (|G| - MFVS(\overline{G}))/2$
and this bound is sharp.
Is this known or trivial result?
This is standard notation and to address comments
$|G|$ is the number of vertices of $G$, $\overline{G}$
is the complement of $G$ and feedback vertex set $S$ is subset of $V(G)$
such that deleting $S$ from $G$ leaves acyclic graph.
Is $\overline G$ the complement of $G$? Is $|G|$ the number of vertices? And what is a feedback vertex set of $G$?
@bof This is standard notation, yet I edited with the definitions.
It might be easier to think of $|G|-MFVS(\bar G)$ as the size of the largest induced subforest of $\bar G$.
Here is an easy proof: consider any proper colouring of $G$, with colour classes $n_1,\dots,n_{\chi(G)}$. Then the complement of the complete multipartite graph is a collection of cliques, from each of which we can choose at most two vertices to have an induced subforest. Since this collection of cliques is a subgraph of $\bar G$, the largest induced subforest of the latter can only be smaller.
@MartinRubey Thanks :) My proof is different. Is this known? And can your solution compute the RHS in time n^(2*chi(G)) if chi(G) is given?
I have no idea because I rarely do graphs :-) I am sceptical that my proof yields a way to compute the right hand side in sensible time, because I need a colouring to start with!
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2025-03-21T14:48:29.807283
| 2020-02-08T11:15:11 |
352207
|
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|
Stack Exchange
|
$k$-linear $\infty$ stable categories and dg categories
This question is related to this question, where I asked about the relation between the derived category of a fiber product $Y \times_Z W$ and the push out of the diagram of derived categories one obtains considering the pullback functors. It was pointed out to me that it was proven by Ben-Zvi, Francis and Nadler, arXiv:0805.0157, that if we consider the derived fiber product, then (under some assumptions) we have an equivalence $QC(Y \times_Z W) \simeq QC(Y) \otimes_{QC(Z)} QC(W)$ where $QC(-)$ denotes the $\infty$ stable category of quasi coherent sheaves and the tensor product is computed in the $\infty$ category of presentable $\infty$ categories with morphisms given by left adjoints. As I am interested in characteristic zero, the formalism of $k$ linear $\infty$ stable categories is equivalent to that of pretriangulated dg categories. I would like to restate the above result in the formalism of pretriangulated dg categories, but I am having some problems doing it. The categories need to be substituted with dg enhancements of the triangulated categories of quasi coherent sheaves and functors of $\infty$ categories need to be substituted with quasi functors, but what is the counterpart of the tensor product? I tried having a look at the paper by Cohn, arXiv:1208.2587, where the equivalence between $k$ linear $\infty$ stable and pretriangulated dg is proven, and as far as I understand one needs to replace one of the two dg categories by a flat dg category. However, I don't understand what being flat for a dg category means, as in the cited paper the definition is given for spectral categories. Would anyone be able to shed some light on this tensor product for me? Thanks.
The derived tensor product of dg-categories was explored
by Toën, see his article
The homotopy theory of dg-categories and derived Morita theory,
in particular, Section 4, where Toën explains how to derive the tensor product of dg-categories.
For a dg-category $\mathcal C$, being flat means that all enriched Hom's are flat (as k-modules). In other words, for every two objects $a,b$, one requires that the functor $\mathcal C(x,y)\otimes-$ preserve quasi-isomorphisms.
See e.g. http://www.mi-ras.ru/~akuznet/dgcat/Keller%20On%20differential%20graded%20categories.pdf
But you are interested in a relative tensor product? In this case I guess that the definition of $\mathcal C$ being flat over $\mathcal A$ (when one has a dg-functor $\mathcal A\to\mathcal C$) is that $\mathcal C$ is flat as an $\mathcal A$-module, that is $\mathcal C\otimes_{\mathcal A}-:\mathcal A-mod\to\mathcal C-mod$ preserves weak-equivalences and colimits.
|
2025-03-21T14:48:29.807492
| 2020-02-08T12:17:05 |
352209
|
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|
Stack Exchange
|
Non-self-intersecting paths on $\mathbb{C}\setminus\{0,1\}$
Let us make two small holes around points $0$ and $1$ on the complex plane and consider non-self-intersecting paths that start on the boundary of one hole and finish at the boundary of the another. It seems,
that all these paths are homotopic to the segment of the real line.
How to prove this? Is there any conceptual explanation of this fact?
Is there anything to be said about the nature of the holes or are these just tiny circular closed disks?
You have to specify the nature of the homotopy.
Just a guess but there's the winding number around each hole separately, which you can unwind by a homotopy?
The title seems in conflict with the details in the body of the question.
What do you mean precisely by "homotopic to"? If the meaning is all maps $\gamma:[0,1]\to\mathbb{C}\setminus\mathrm{smalldisks}$ are homotopic, relative to the endpoints, to the "real segment between the disks", then it's clearly false in general (if e.g. $\gamma(0)$ is not real). In other words: how are the endpoints allowed to move during the homotopy?
There are (at least) three different notions of homotopy applicable in this situation, which lead to (two) different answers. It is your responsibility to state clearly which notion of homotopy you are using. Voting to close for now. The question is also more suitable for MathStackexchange, as I do not see a research aspect to this question.
@Misha, this question seems perfect for MO.
@Wlod AA Thanks!
Supposedly "This question needs details or clarity." == But I DID supply perfect details and clarity. It is sick, that my EDIT was simply rejected without eventually supplying anything in return. It is sick that once again mathematics loses to MO administration.
Such a homotopy exists and in fact you can assume that it is an isotopy. This is a "standard fact" in the theory of mapping class groups. See Proposition 2.2 of the "Primer" by Farb and Margalit.
Thank you for the reference! It is a simple fact, indeed. My question was mostly about some kind of explanation of this fact. For example, this is not true for other surfaces, they have some non-trivial non-self-intersecting paths... Anyway, thank you for your comment!
Suppose the two holes are two small disks. Then connect their boundaries by a cylinder (homeomorphic copy of $[0,1]\times S^1$) lying out of the plane. Let us also add the complex $\infty$ to the space (meaning: let us start with the Riemann sphere in the place of complex plane). The resulting space is homeomorphic to the torus, so its fundamental group is $Z$, the integers. Any given closed path $p$ should belong to the homotopy class $n\in Z$ if and only if the total oriented number of passages of $p$ through the cylinder is $n$; I suppose this can be obvious from some particular way how the homotopy group can be calculated (but I am not strong at this presumably ?basic exercises). Your curve is to be completed to closed curve p_1 by one passage through the cylinder. The real segment is to be completed to a closed curve p_2 also by one passage through cylinder. So p_1 and p_2 are homotopic. It seems to me "obvious" that the homotopy does not do anything serious in the cylinder and can be modified to live in your complex domain (asside from trivial part that trivially transits between added parts of p_1 and p_2). I admit this "obvious" thing might require a substantial use of algebraic geometry or real analysis.
Now, the $\infty$ I added plays no role: The homotopy intersect the complex plain in a bounded region, it never reaches $\infty$.
(If one of your 'holes' is two disks itself, say $B(0,1/9) \cup B(i,1/9)$, then the curve might be non-homotopic to the segment. E.g. if it passes around above B(i,1/9) or winds it.)
Replacing two points by two disks changes the question significantly. Also, the fundamental group of a torus is not ℤ but ℤ$^2$.
Others may feel different but I don't see any positive contribution to solving the Question, not even a partial contribution.
|
2025-03-21T14:48:29.807833
| 2020-02-08T14:51:15 |
352212
|
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|
Stack Exchange
|
fundamental groups of complements to countable subsets of the plane
This question is a follow-up of this MSE post and a comment by Henno Brandsma:
Question 1. Let $S$ be the set of isomorphism classes of fundamental groups $\pi_1(E^2 - C)$, where $C$ ranges over all countably infinite subsets of the Euclidean plane $E^2$. What is the cardinality of $S$?
All what I can say is that $S$ contains at least two non-isomorphic groups:
One is the free group $F_\omega$ of countably infinite rank, the fundamental group of the complement to a closed discrete countably infinite subset of $E^2$, does not matter which one, say, $C={\mathbb Z}\subset {\mathbb R}\subset {\mathbb R}^2$.
The other is $G_{{\mathbb Q}^2}=\pi_1(E^2-C)$, where $C$ is a dense countable subset of $E^2$, again, does not matter which one, for instance, $C={\mathbb Q}^2$. This group is not free since it contains, for instance, the fundamental group of the Hawaiian Earrings. (Actually, it contains $\pi_1$ of every nowhere dense planar Peano continuum.)
The natural expectation is that $S$ is uncountable (more precisely, has the cardinality of continuum: It is clear that the cardinality of $S$ cannot be higher than that).
Edit. Following Yves' suggestion:
Question 2. Let ${\mathbb H}$ denote the Hawaiian Earrings. Is the fundamental group $\pi_1({\mathbb H})$ essentially freely indecomposable? (Here a group $G$ is essentially freely indecomposable if in every free product decomposition $G\cong G_1\star G_2$, one of the free factors $G_1, G_2$ is free of finite rank.) One can also ask for the weaker property of $G=\pi_1({\mathbb H})$, namely, that $G$ does not admit free product decompositions $G\cong G_1\star G_2$ with two uncountable factors.
The only relevant result I could find in the literature is the theorem (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $G=\pi_1({\mathbb H})$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall.
If Q2 (even in the weaker form) has positive answer, then in Q1 at least one can say that $S$ is infinite.
Note that is $C$ is closed then the $\pi_1$ is free and countable (better denoted $F_\omega$ than $F_\infty$, since uncountable free groups such as $F_{\mathfrak{c}}$ come into the game, at least as subgroup of some of these groups). Also if $C={1/n:n\ge 1}$, then $\mathbf{C}-C$ is maybe homotopy equivalent to the Hawaiian earring space.
If $D$ is a nonempty compact totally disconnected subset of the sphere $S$, one can produce a closed subset $W_D$ containing $D$, such that $D$ precisely consists of the set of accumulation points in $W_D$. Of course $W_D$ is not unique, but it might be unique up to homeomorphism of $S$ (I'm also only interested in $D$ up to homeomorphism, or [nontrivially] equivalently homeomorphism of $S$).Define $C_D=W_D\smallsetminus D$. Then one could wonder whether the isomorphism class of $\pi_1(\mathbf{R}^2-C_D)$ determines $D$ up to homeomorphism (there are continuum such $D$ up to homeomorphism).
@YCor Yes, in the first example $E^2 -C$ is h.e. to the H.E. The difficulty with the question, as I currently see it, is that I see very few invariants to distinguish isomorphism types of groups appearing as $\pi_1(E^2- C)$. For instance, I cannot even tell is $G_{{\mathbb Q}^2}$ splits nontrivially as a free product (I am quite sure, it does not) or is not isomorphic to the $\pi_1$ of the H.E (I am sure, it is not).
One can also wonder whether the HE-group can be split as free product of two uncountable groups. If the answer is no, then it's not isomorphic to the $\pi_1$ of the bouquet of two HE along a non-singular point. (This question is maybe more accessible, because this group is well-studied and quite well-understood.)
To me, Question 2 is a completely different question from Question 1 that should appear as a separate MO question.
@JeremyBrazas: OK, I will do this a bit later.
Thanks to the comments, my original cardinality bound $\aleph_1\leq |S|\leq \mathfrak{c}$ has been refined to the equality $|S|=\mathfrak{c}$ that I originally suspected.
For Question 1: $S$ has the cardinality of the continuum. It's clear that $|S|\leq \mathfrak{c}$. Below, I'll argue that $|S|$ is at least the cardinality of the set of homeomorphism types of closed nowhere dense subsets of $[0,1]$. Since this set has cardinality $\mathfrak{c}$ (using Pierre PC's comment below), we have $|S|\geq \mathfrak{c}$. For the lower bound, I'll use a construction from this paper.
Consider any infinite closed nowhere dense subset $A\subseteq[0,1]$ containing $\{0,1\}$. Let $\mathcal{I}(A)$ denote the ordered set of components of $[0,1]\backslash A$. For each $I=(a,b)\in\mathcal{I}(A)$, let $$C_I=\left\{(x,y)\in\mathbb{R}^2\mid y\geq 0,\left(x-\frac{a+b}{2}\right)^2+y^2=\left(\frac{b-a}{2}\right)^2\right\}$$ be the semicircle whose boundary is $\{(a,0),(b,0)\}$. Let $$\mathbb{W}_{A}=([0,1]\times\{0\})\cup \bigcup_{I\in\mathcal{I}(A)}C_I$$ with basepoint $(0,0)$. Here is an example where $\mathcal{I}(A)$ has the order type of the linear order sum $\omega^{\ast}+\omega+1+\omega^{\ast}$ where $\ast$ denotes reverse order.
First, notice that $\mathbb{W}_A$ is a one-dimensional Peano continuum (connected, locally path-connected, compact metric space). By picking a single point in the interior of each simple closed curve $C_I\cup (\overline{I}\times \{0\})$, we can see that $\mathbb{W}_A$ is homotopy equivalent to $E^2\backslash C$ for some countably infinite set $C$. The fundamental groups $\pi_1(\mathbb{W}_A)$ ranging over all such $A$ realize continuum-many non-isomorphic groups. Here are the heavy hitting theorems that get the job done.
Eda's homotopy classification of 1-dimensional Peano continuua: two one-dimensional Peano continua are homotopy equivalent if and only if they have isomorphic fundamental groups. I would just like to pause and emphasize how incredible and powerful this result is. When you first hear about it and realize the kind of groups and spaces it applies to, you might get the impression that you are being scammed.
Let $\mathbf{w}(X)$ denote the subspace of $X$ consisting of the points at which $X$ is not semilocally simply connected, i.e. the 1-wild set of $X$. For general spaces, the homotopy type of $\mathbf{w}(X)$ is a homotopy invariant of $X$, but monodromy actions in one-dimensional spaces have discrete graphs (see 9.13 of this paper of mine with H. Fischer). Among spaces whose monodromy actions have discrete graphs, the homeomorphism type of $\mathbf{w}(X)$ becomes a homotopy invariant of $X$ (see 9.15 of the same paper). Thus, among 1-dimensional spaces, $\mathbf{w}(X)$ is a homotopy invariant. This result is kind-of embedded in Eda's work leading up to 1. but the core idea behind what is really going on is fleshed out in Section 9 of the linked paper.
By combining 1. and 2. we have:
Corollary: If one-dimensional Peano continua $X$ and $Y$ have isomorphic fundamental groups, then $\mathbf{w}(X)\cong \mathbf{w}(Y)$. A direct consequence is that the Hawaiian earring group and the free product of the Hawaiian earring group with itself are not isomorphic because two copies of $\mathbb{H}$ adjoined by an arc has two 1-wild points.
Returning back to the spaces $\mathbb{W}_A$, notice that $\mathbf{w}(\mathbb{W}_A)$ is homeomorphic to the Cantor Bendixsion derivative of $A$, i.e. the subspace of non-isolated points of $A$. Every closed nowhere dense set $B\subseteq [0,1]$ is the Cantor Bendixsion derivative of some other $A$. Hence, the 1-wild sets of the $\mathbb{W}_A$ realize all closed nowhere dense subsets of $B\subseteq [0,1]$. By the corollary, each homeomorphism class of a nowhere dense closed subset of $[0,1]$, gives a unique isomorphism class of fundamental group $\pi_1(\mathbb{W}_A)$ and hence a unique isomorphism class of a fundamental group $\pi_1(E^2\backslash C)$ for some countably infinite set $C$. In the comments below, Pierre PC gives a construction of $\mathfrak{c}$-many non-homeomorphic closed nowhere dense subsets of $[0,1]$ (one can confirm this by analyzing the Cantor Bendixson derivatives of the neighborhoods of the described "super limit points"). Hence, $|S|=\mathfrak{c}$.
In the last paragraph, you say that there are continuum many $B$'s, but then you give as examples the countable compact ordinals. Since there are only $\aleph_1$ countable ordinals, you seem to be assuming the continuum hypothesis here to say that this is the cardinal of the continuum.
Right, I don't usually worry about this sort of thing and implicitly applied CH (I apologize to the logicians) but I guess I should either clarify or decide if there really are continuum many non-homeomorphic closed nowhere dense subsets in $[0,1]$.
Here is, I think, a construction of a different homeomorphism type of some subset $A$ of $[0,1]$ for every subset $S$ of the integers. Define a compact set $K_k$ as the gluing of (first) some copy of $\omega^k+1$, seen as a compact space with "super limit point" $x$, corresponding to the +1, and (second) a Cantor set, glued at respectively $x$ and an extremity (the maximum, say). Then fit a copy of $K_k$ in $(2^{-k-1},2^{-k})$ if and only if $k\in S$.
@JeremyBrazas: This is quite amazing. I think, you should add to your answer references to Proposition 9.13 (discreteness of monodromy) and 9.15 (discrete monodromy implies that topology of $w(X)$ is a homotopy-invariant of $X$) of your paper with Fischer.
@PierrePC nice construction! Actually this precisely answers this question (which already has an accepted answer, but the latter gives a reference and no explicit construction).
@YCor Thank you! :) I wrote an answer to that question, it is longer than a comment and maybe the readers interested by the construction may want to look at that more detailed answer.
|
2025-03-21T14:48:29.808446
| 2020-02-08T15:57:04 |
352222
|
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|
Stack Exchange
|
Journey into a strange wilderness
W. S. Anglin wrote
Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost. Rigour should be a signal to the historian that the maps have been made, and the real explorers have gone elsewhere.
What wildernesses are being explored these days and by whom?
Historically, Newton, Euler, Fourier, Heaviside, and Riemann were iconic explorers blazing trails through strange widernesses, but today ... ?
I think at present this is far too broad.
The landscape of string theory is aptly characterized as a "wilderness" for mathematicians.
"String theory is a piece of 21st-century physics that has fallen by accident into the 20th century and therefore will require 22nd-century mathematics to solve."
Daniele Amati
See also The mathematics of string theory by Robbert Dijkgraaf.
Who is Daniele Amati? A great deal of 20th century physics was solved with 20th century mathematics.. think of quantum mechanics or general relativity
Daniele Amati was referring to string theory, which seems out of reach of presently available mathematical tools.
What would it mean to "solve" string theory?
@Wojowu --- perhaps in the sense that the anti-de Sitter/conformal field theory correspondence "solves" string theory on a manifold of constant negative curvature.
@CarloBeenakker notice that we are in 2020. New insights are gained every day. It is perfectly possible that in 30 years, say, there will be a rigorous theory of the Feynman integral/infinite-dimensional integration, etc...
Certainly one of the first to spring to mind, with Witten as one of the explorers.
|
2025-03-21T14:48:29.808605
| 2020-02-08T16:37:37 |
352225
|
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|
Stack Exchange
|
Is this anti-foundation axiom consistent?
Starting with ZFC:
Replace axiom of Extensionality by weak Extensionality (nonempty sets having the same members, are identical)
Remove axiom of Foundation.
Add the anti-foundation axiom which states that: if $G$ is a graph such that there do NOT exist two distinct nodes $a,b$ of $G$ such that there exists edges of $G$ stemming from $a,b$ towards a common third node of $G$ that is distinct from $a,b$; then for every node $p$ of $G$ that reaches every other node of $G$ (i.e. for every node $n$ of $G$ other than $p$, there is a path of $G$ from $p$ to $n$); there is a set $p*$ such that the membership graph on $p*$ is isomorphic to $G$! That is: there is an isomorphic function $f$ from the transitive closure of $\{p*\}$ to the set of all nodes of $G$ such that $f(p*)=p$ and for every elements $a,b$ of the transitive closure of $\{p*\}$ we have: $$ b \in a \leftrightarrow \langle f(a) , f(b) \rangle \in G$$
A graph here can be captured by being a set of ordered pairs, a node of a graph is a projection of an element of that graph, an edge of $G$ from $a$ to $b$ can be defined as $\langle a,b \rangle \in G$, and a path of $G$ from $a$ to $b$ is defined as a subset of $G$ satisfying the usual definition of a directed finite path from $a$ to $b$.
Of course this clearly violates Aczel's anti-foundation axiom!
My question: Is the resulting theory consistent? and if so, then would that be a kind of maximal anti-foundation axiom that is compatible with weak Extensionality in the milieu of weakly extensional ZFC-Foundation?
I tried of course, yet I failed. However, I think there would be problem with Extensionality even with the weak form, but a part from that I think it would work, because the buildup of the transitive closures of added sets only copies the structure of already existing graphs.
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2025-03-21T14:48:29.808765
| 2020-02-08T18:26:26 |
352230
|
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|
Stack Exchange
|
Eigenvalue distribution of a band matrix
Let $\mathbf M_i$ be rectangular matrices of dimensions $N_{i-1}\times N_i$. We assume that their entries are random, with zero mean and variance $\sigma_i^2$.
For some positive integer $k$, I define the matrix $\mathbf{A}_k$:
$$\mathbf{A}_k = \left(\begin{array}{cccccc}
0 & \mathbf{M}_1 & 0 & \cdots & 0 & 0\\
\mathbf{M}_1^{\top} & 0 & \mathbf{M}_2 & \cdots & 0 & 0\\
0 & \mathbf{M}_2^{\top} & 0 & \ddots & 0 & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 0 & \mathbf{M}_n\\
0 & 0 & 0 & \cdots & \mathbf{M}_k^{\top} & 0
\end{array}\right)$$
Notice that $\mathbf{A}$ is symmetric. But unfortunately it is not from a rotationally invariant random ensemble, which has complicated things for me.
I have two questions.
What is the eigenvalue density of $\mathbf{A}_k$?
What happens to the eigenvalue density when I set $N_i=c_iN$, with positive constants $c_i$ and we take the limit $N\rightarrow\infty$? I presume one must perform an appropriate re-scaling here to get a meaningful result.
Even if the answer to 1. is not tractable, maybe the limiting form 2. can still be obtained.
Update: Say I have access to the singular value decomposition of $\mathbf{M}_i$. Can we say anything about the eigenvalues/eigenvectors of $\mathbf{A}_k$?
Related: https://math.stackexchange.com/questions/3538580/positive-definiteness-of-a-block-diagonal-matrix
A closed form expression for finite $N$ is most certainly not available.
A large-$N$ calculation for Gaussian distributed matrix elements has been reported in Density of eigenvalues of random band matrices. If the band width $\delta N$ increases with $N$ as $N^\beta$ for some $\beta>0$, the eigenvalue density tends in the limit $N\rightarrow\infty$ to the usual semicircle that one would obtain for the full Gaussian matrix (so for $\delta N=N$).
This is not quite the same model. The model here seems closer to the Wegner orbital model, but again not quite because of the identity on the diagonal blocks. There is some work of Peled, Shenker, Shamis and Sodin on the latter, and also work of Scherbina^2 using supersymmetry methods https://arxiv.org/abs/1802.03813. But I stress this is not exactly the same model.
I also think that it's not the same model. Please note I have updated the question, simplifying it.
|
2025-03-21T14:48:29.808954
| 2020-02-08T18:58:09 |
352231
|
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|
Stack Exchange
|
Cardinality of the set of continuous functions
Suppose $(X,\tau)$ and $(Y,\sigma)$ are topological spaces. Let $F(X,Y)$ be the set of continuous functions $X\rightarrow Y$. I want to compute the cardinality of $F(X,Y)$. It depends not only on cardinalities $|X|$ and $|Y|$ but on the topologies as well: just imagine what happens if $X$ or $Y$ is discrete.
Question: is it possible to write some "topological invariants" $|\tau|$ and $|\sigma|$ that yield the following explicit formula:
$$|F(X,Y)| = \Theta (|X|,|Y|,|\tau|,|\sigma|)$$
I am sorry for being sloppy: I really need to clarify what a formula is. Ideally, $|\tau|$ would be a collection of cardinals and the formula will be given in the cardinal arithmetic. However, I would be glad to have any other method of computing the cardinality. Thus, "the formula" may be something more general.
There is a continuum that only allow constant mappings and identity to be continuous. May be relevant
If you allow proper class invariants, then the answer is yes, because you can essentially encode the whole of F into the invariant in that case.
You can also trivially do it if you have global choice, from which you get a definable well-ordering of $V$ of order type $Ord$. For any topological space you find the lowest index $\alpha$ of a homeomorphic topological space, and then your 'cardinal invariant' is $\aleph_\alpha$.
So in particular, the answer is yes in L, where there is a definable such order.
@James Hanson. What is $V$? What is the lowest index of a homeomorphic topological space?
@BugsBunny The point is that a lot of information can be coded in cardinals and under certain common set theoretic assumptions you can actually code the homeomorphism type of $(X,\tau)$ by a unique cardinal. So if you're too permissive about what $|\tau|$ and $\Theta$ are, the answer to your question can become very sensitive to set theoretic assumptions and even trivial under certain specific assumptions.
@Matt $L$ is the constructible universe, mentioned in a comment by Joel. Instead, $V$ is the class of all sets.
OK, I understand that one can "enumerate" all topologies, say by well-ordering the set of all topologies. I guess I just want to be permissive as soon as the formula remains somewhat constructive...
Some indications of the difficulty: if $X$ is discrete, or $Y$ is indiscrete, then $F(X,Y)$ has cardinality $|Y|^{|X|}$.
Next: in Wann sind alle stetige abbildungen in $Y$ konstant? Herrlich shows that for every $T_1$-spae $Y$ there is a regular space $X$ such that every continuous map from $X$ to $Y$ is constant, so $|F(X,Y)|=|Y|$. There is no bound on the cardinality of $X$. In particular of $Y$ is the space of rationals $F(X,Y)$ is countable for very large $X$.
One more: every connected space, no matter how large, has exactly two continuous maps to the discrete two-point space.
Are there known examples, e. g. if the topological spaces are n-dim Riemann Manifolds?
@KlausLoehnert It seems to me the answer is: if $Y$ is 0-dimensional, it's $c(Y)^{c(X)}$ where $c()$ is the number of connected components; otherwise, $\mathfrak c$.
|
2025-03-21T14:48:29.809313
| 2020-02-08T20:38:58 |
352236
|
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"url": "https://mathoverflow.net/questions/352236"
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|
Stack Exchange
|
Maximizing $\iiint|(x-z)\times(y-z)|d\mu d\mu d\mu$ over probability measures on the unit sphere
This is a follow-up question to the one asked here (the unit circle case). What probability measure(s) maximize the quantity $\iiint_{\mathbb{S}^2}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z)$?
The answer may be uniform measure, by direct analogy to the unit circle problem.
|
2025-03-21T14:48:29.809371
| 2020-02-08T20:56:43 |
352237
|
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"url": "https://mathoverflow.net/questions/352237"
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|
Stack Exchange
|
Anticyclotomic extensions via ideles
Let $ K $ be an imaginary quadratic field with ring of integers $ \mathcal{O} $. Let $ \mathcal{O}_{n} = \mathbb{Z} + n \mathcal{O} $ be the order of conductor $ n $. There is an associated extension $ K_{n} $ such that $$ \mathrm{Gal}(K_{n}/K) \cong \mathrm{Cl}(\mathcal{O}_{n}) $$
which is referred to as ring class field of conductor $ n $ in literature. Adelically, this may be described as the extension associated to the compact open subgroup $ \widehat{\mathcal{O}}_{n} = \mathcal{O}_{n} \otimes \widehat{\mathbb{Z}} $ (profinite completion of $ \mathcal{O}_{n}$) of $ \mathbb{A}_{K,f} ^{\times} $, so that $$ \mathbb{A}_{K}^{\times}/K^{\times}\widehat{\mathcal{O}}_{n}C_{E} \cong \mathbb{A}_{K,f}/K^{\times} \widehat{\mathcal{O}}_{n} \cong \mathrm{Gal}(K_{n}/K) $$
where $ \mathbb{A}_{K,f} $ denotes the finite adeles of $ K$, $ C_{E} \cong \mathbb{C} $ is the connected component the idele class group. Given $ n $, $ m \geq 1 $, $ \mathcal{O}_{nm} \subset \mathcal{O}_{n}, \mathcal{O}_{m} $, whence $ K_{n}, K_{m} \subset K_{nm} $. The filtered direct limit/compositum $$ K_{\infty} = \bigcup_{n} K_{n} $$
is referred to as the anticyclotomic abelian extension of $K$.
Let $ T = \mathrm{Res}_{K/\mathbb{Q}}\mathbb{G}_{m}$, and $ U_{1} $ be norm $ 1 $ elements in $ T $ i.e. for $ \mathbb{Q}$-algebra $ R $, let $$ U_{1} (R) = \left \{ x \in ( R \otimes K ) ^ {\times} | x \sigma(x) = 1 \right \} $$
where $ \sigma \in \mathrm{Gal}(K/\mathbb{Q}) $ is the non-trivial element. There is a $ \mathbb{Q} $-map $$ \phi : T \to U_{1} \quad \quad x \mapsto \frac{\sigma(x)}{x}$$ and hence a continuous map $$ \phi : \mathrm{Gal}(K^{\mathrm{ab}}/K) \xrightarrow{\cong} \mathbb{A}_{K,f}^{\times} / E^ {\times} \to U_{1} ( \mathbb{A}_{\mathbb{Q},f})/U_{1}(\mathbb{Q}) .$$
Let $ H = \mathrm{ker} \phi $.
Question: Is $ \phi $ surjective? What is the fixed field of $ H $? Is it the anticyclotomic extension $ K_{\infty} $? If all of these are affirmative, are there compact open subgroups $ C_{n} \subset U_{1} (\mathbb{A}_{\mathbb{Q},f}) $, such that $ \mathrm{Gal}(K_{n}/K) \cong U_{1} ( \mathbb{A}_{\mathbb{Q},f}) / U_{1}(\mathbb{Q}) C_{n} $?
Surjectivity follows by Hilbert 90
The article https://arxiv.org/pdf/1709.00998.pdf, seems to answer all my questions. I'll write the answer to my post for the benefit of anyone looking for similar facts.
|
2025-03-21T14:48:29.809572
| 2020-02-08T21:58:11 |
352240
|
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|
Stack Exchange
|
Constructing a vector consisting of nonnegative entries
Consider constructing a vector $v=(a_1,a_2,\ldots,a_n)$ consisting of nonnegative integers such that $a_1=1$ and, if $a_j$'s are nonzero, then $a_j\equiv a_{n-j+2}+j-1 \pmod m\ \forall 1<j\le\frac{n}{2}$, where $m$ is the number of nonzero entries; with the additional constraint that all nonzero $a_i$'s are distinct modulo $m$. Note that the number $m$ itself appears once as a nonzero number satisfying the above congruence.
Is it always possible to construct such a vector? I think this should be possible if $m$ is odd. For example, $(1,4,2,5,3)$ and $(1,4,2,0,0,5,3)$ are such vectors. It is easy to construct if the first entries (within $\lfloor\frac{n}{2}\rfloor$) of the vector are consecutive. But, in other cases, it is not clear as to how to proceed with the construction. Any hints? Thanks beforehand.
Your two examples are nonnegative, have $a_1=1$, and have distinct nonzero entries, but the mod condition fails for both.
@RobPratt thanks! edited. see now.
Still not right. For $j=2$, $4\not\equiv 5+2-1 \pmod 5$.
@RobPratt no, we do have $4\equiv 3+2-1 \pmod 5$. In fact you misread $a_{n-j+1}$ for $n$
For the first example, $n=5$, and $j=2$ yields $a_{n-j+1}=a_{5-2+1}=a_4=5\not= 3$.
@RobPratt sorry, again edited hope its correct now.
Your second example fails for $j=6$ because $5\not\equiv 2+6-1 \pmod{5}$.
@RobPratt well observed, edited again now. Actually the condition is derived from the condition of being able to obtain a symmetric anti-circulant matrix by shifting the vector $v$.
What is given? Both $n$ and $m$?
@MaxAlekseyev yes, suppose I specify any $n$ and $m$
Yes, such construction is always possible.
Consider two sets of pairs of values:
$$\big\{ (2+t,m-t)\quad :\quad t=0\,..\,\lfloor\frac{m-1}{4}\rfloor-1\big\},$$
where differences of elements modulo $m$ are: $2,4,\dots,2\cdot\lfloor\frac{m-1}{4}\rfloor$, and
$$\big\{ (\lfloor\frac{m+1}{2}\rfloor+t+1,\lfloor\frac{m+1}{2}\rfloor-t)\quad :\quad t=0\,..\,\lfloor\frac{m+1}{4}\rfloor-1\big\},$$
where differences of elements modulo $m$ are: $1, 3, \dots, 2\cdot\lfloor\frac{m+1}{4}\rfloor-1.$ Together they give all differences from $1$ to $\lfloor\frac{m-1}{2}\rfloor$.
Notice that all elements forming pairs in these sets are $\ne0,1$ and are distinct modulo $m$.
Therefore, it's enough to assign the values from these sets to the corresponding pairs $(a_j,a_{n+2-j})$, giving $2\cdot\lfloor\frac{m-1}{2}\rfloor$ nonzero $a_j$'s. If $m$ is even we need to assign one more yet unassigned nonzero value (which is $\lfloor\frac{3m}{4}\rfloor+1$) to any of yet unassigned $a_j$ with $j>\frac{n}2$. The other unassigned $a_j$ are set to zero.
ADDED. Here is a SageMath code implementing the above construction. There is a sample call construct_a(16,11), which constructs vector $(a_1,\dots,a_n)$ for parameters $n=16$ and $m=11$.
great! how did you get such an idea?
@vidyarthi: a lot of drawing on a piece of paper ;)
you mean you drew paths of odd and even order between the numbers $a_j$ and $a_{n-j+2}$?
@vidyarthi: I drew matchings between residues modulo $m$ giving differences $1,2,\dots$
ok, you are actually a computational biologist, is number theory and combinatorics your hobby?
@vidyarthi: sort of ;)
Actually, I have another problem now. My problem requirements also state that the numbers in the individual pairs $(a_j,a_{n+2-j})$ must not be equivalent to $(j,n+2-j)$ modulo $r$. Now, is that also possible? By the above construction, this would occur at the point when $j\equiv1\pmod r$. So how to circumvent this problem?
I meant $r$ to be $m$ in my previous comment
I do not quite follow. The difference in $(j,n+2-j)$ is $2j-n-2$. If it's assigned to $(a_j,a_{n+2-j})$, the the difference is congruent to $j-1$ modulo $m$, implying $j\equiv n+1\pmod m$. Why you say $j\equiv 1\pmod m$?
I wanted the difference between $a_j$ and $a_{n+2-j}$ in the OP to be $j-1\pmod m$. Using your rule, I equate this to either $2+2t$(even difference) or $2t+1$(odd difference), obtaining $t=\frac{j-3}{2}$ and $t=\frac{j-2}{2}$ respectively. This means when $j\equiv1\pmod m$, and $j$ is odd, we would obtain $a_j=2+t= 2+\frac{j-3}{2}\equiv j\pmod m$. But, I do not want $a_j$ to be conguent to $j$. Is my method of equating $2+2t$ and $2t+1$ with $j-1$ wrong?
Yes, it is true that $j\equiv n+1\pmod m$. But, if I were to suppose that $m| n$, then I would get by conclusion. Anyways, the main thing is to ensure that the numbers $a_j$ and $a_{n+2-j}$ are not congruent to $j$ and $n+2-j$ respectively, Can this be done. Assume that $m$ divides $n$
Assume $m|n$. There are no $j\equiv 1\pmod m$ in the interval $[2,n/2]$ if $m\geq n/2$. So, the additional restriction playa role only when $m\leq n/3$. However, in this case we can simply assign all nonzero values to $a_j$'s with $j>n/2$.
sorry, I did not get your second part. Could you illustrate the part when $m\le\frac{n}{3}$. And what about the part $\frac{n}{3}<m<\frac{n}{2}$?
For $m\leq n/3$, simply set $a_{n/2+k}=k$ for $k=1..m$, and all other $a_j$ to 0. If $n/3<m<n/2$, then $m$ cannot divide $n$.
sorry, again a confusion. For $m\le\frac{n}{3}$, suppose $a_{\frac{n}{2}+3}=3$. Then, since we have $a_{\frac{n}{2}-1}=0$, therefore the difference is $-3=m-3\pmod m$.So how is the difference in this case $j-1=\frac{n}{2}-2\pmod m$? And, by the way, can I not have nonzero entries before $a_{\frac{n}{2}}$?
I do not quite follow again. According to your original question, the congruence $a_j\equiv a_{n+2-j}+j-1\pmod m$ is required only for nonzero $a_j$. In the above example, for $j=n/2-1$ we have $a_j=0$, and so the congruence may not hold.
oh yes thanks, i forgot about that. so you have completely answered the modified problem. only last thing is whether i could have nonzero entries before $a_{\frac{n}{2}}$.
How many nonzero entries you want before $a_{n/2}$? When $m\leq n/3$, it's unavoidable that some of these entries will be zero.
say, by symmetry that $\frac{n}{6}$ entries are nonzero before $a_{n/2}$
Then simply assign pairs of values as in my answer, but if such assignment violates your additional restriction, then re-assign both values to any available $a_j$ with $j>n/2$.
again a problem, the differences in the first pair $(2+t,m-t)$ are always odd if $m$ is odd. So, how can you ensure all the dfferences? Although the pairs give distinct entries, but the proof you give is not clear.
We consider differences modulo $m$. So, here it does not matter if $m$ is even or odd - in either case we get $(2+t)-(m-t)\equiv 2+2t\pmod m$.
but take $t=2, m=17$ . Then $2+t=4,m-t=15$, and the difference is odd. The problem is that the difference is $2+2t-m$, which is less than (in magnitude) $m$, whence modulo does not keep the parity unaffected
The proof i think need not mention the parity of differences. Instead, only the distinctness of the numbers in both the pairs would do to prove that all the numbers from $2$ to $m$ are covered. But anyways, the answer is right and works well!
Sorry, again a problem. Suppose I want to construct a vector like ${1,, 0, ,,,0,,0,0,0,0,0,,0,,,,0,}$, where $*$ denotes the nonnegative entries. The differences between nonnegatives in this case modulo $m=11$ in this case are $1, 3,4, 5, 7$. So isnt is asymmetric. What are the entries in this case?
@vidyarthi: This is a different problem from what you originally asked. Here is a solution obtained by brute-force: $$(1,11,0,8,7,9,0,2,0,0,0,0,0,6,0,4,3,5,0,10)$$
how is it a different problem? I think i asked the same problem. Anyways, is brute force the only way, or can we state the possibility from by a fixed algorithm; or, is it altogether unpredictable to construct such a vector
@vidyarthi: In your original problem, the positions of zeroes were not fixed. The one you just asked is more restrictive in that sense. I did not claim that brute-force is the only way - it's just the first thing that comes to mind.
yes, the positions of zeros are not fixed, but once we put a nonzero entry before $n/2$th column, we automatically put another after $n/2$, so in that sense, the nonzero entries before $n/2$th column are the real variables. Yes, brute force is the natural thought, but whether it is always possible is my only question
@vidyarthi: Since it's a different problem, I'd recommend to accurately formulate it as a new question, which would be more suitable at https://math.stackexchange.com/ I may have another algorithm for your new problem, but I cannot explain it here.
done, its here
I do not see any difference from the question here, which I answered. In the comments you imposed additional restriction on positions of zeros - why didn't you mention this restriction in the new question?
Done some slight modification. By the way, I just want that whenever I put a nonzero number, I have a complementary nonzero number determined modulo $m$. We could put the zeros in the other places. Thus, in this sense, there is no condition of zeros as such. Could you explain your dilemma a bit more?
My answer gives you a construction that answers your original question here and the new question at MSE (since it's still a duplicate). I'm not sure what else you are asking in your questions -- it seems you have some additional restrictions in mind, which you do not state explicitly.
Let us continue this discussion in chat.
Computational experiments for $1 \le m \le n \le 20$ yield feasible solutions even if you impose the upper bound $a_j \le n$. Here is an infinite family for $m \le \lceil n/2\rceil +1$:
$$(1,\underbrace{0,\dots,0}_{n-m},2,3,4,\dots,m)$$
Here are the lexicographically smallest solutions for $n=m$:
\begin{matrix}
n & a\\
\hline
1& (1)\\
2& (1, 2)\\
3& (1, 2, 3)\\
4& (1, 3, 4, 2)\\
5& (1, 3, 4, 5, 2)\\
6& (1, 3, 6, 5, 4, 2)\\
7& (1, 3, 6, 5, 7, 4, 2)\\
8& (1, 3, 6, 8, 7, 5, 4, 2)\\
9& (1, 3, 6, 8, 7, 9, 5, 4, 2)\\
10& (1, 3, 6, 10, 9, 8, 5, 7, 4, 2)\\
11& (1, 3, 6, 8, 11, 9, 10, 7, 5, 4, 2)\\
12& (1, 3, 6, 11, 9, 12, 10, 7, 5, 8, 4, 2)\\
13& (1, 3, 6, 8, 13, 12, 10, 11, 7, 9, 5, 4, 2)\\
14& (1, 3, 6, 10, 12, 14, 5, 11, 13, 9, 8, 7, 4, 2)\\
15& (1, 3, 6, 8, 14, 12, 15, 11, 13, 9, 7, 10, 5, 4, 2)\\
16& (1, 3, 6, 10, 12, 16, 15, 5, 13, 14, 9, 11, 8, 7, 4, 2)\\
17& (1, 3, 6, 8, 13, 15, 17, 14, 12, 16, 7, 11, 10, 9, 5, 4, 2)\\
18& (1, 3, 6, 8, 13, 16, 18, 17, 15, 14, 7, 10, 12, 11, 9, 5, 4, 2)\\
19& (1, 3, 6, 8, 11, 17, 19, 16, 18, 14, 15, 10, 9, 13, 12, 7, 5, 4, 2)\\
20& (1, 3, 6, 8, 13, 17, 20, 18, 15, 19, 16, 10, 7, 11, 14, 12, 9, 5, 4, 2)\\
\end{matrix}
thanks, but what about the general case? And even in the cases you have provided, what happens when I introduce zeroes in between the numbers ( thereby increasing $n$ keeping $m$ fixed) ?
|
2025-03-21T14:48:29.810220
| 2020-02-08T22:21:07 |
352241
|
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|
Stack Exchange
|
Questions about a certain sequence of naturals generated by primorials
I'm working on the following sequence of naturals (which is NOT listed in OEIS)
$$3,5,11,17,23,29,59,89,119,149,179,209,419,629,839,1049,1259,1469,1679,...$$
whose elements are generated this way
$$3=(p_1-1)+p_1\cdot(1)$$
$$5=(p_1-1)+p_1\cdot(p_2-1)$$
$$11=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(1))$$
$$17=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(2))$$
$$23=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(3))$$
$$29=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot(p_3-1))$$
$$59=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot((p_3-1)+p_3\cdot(1)))$$
$$...$$
$$209=(p_1-1)+p_1\cdot((p_2-1)+p_2\cdot((p_3-1)+p_3\cdot(p_4-1)))$$
$$...$$
where $p_i$ stands for the $i$-th prime. It's quite evident that the sequence is generated by adding the primorials $p_i\#$.
By construction, each element $\,m\,$ of the sequence has the following property:
$$m\equiv -1\mod p_i\;$$
for each $\,i\lt k\,$ where $\,k\,$ is the first index such that $\,p_k\# \gt m$.
If we truncate the sequence at the last element generated by using only the first $n$ primes (id est, for example, $209\,$ for $\,n=3$), the whole number of elements is given by
$$\sum_{i=1}^{n}(p_{i+1}-1)=(\sum_{i=2}^{n+1}p_i)-n\sim \frac12n^2\log n$$
But what about the number of primes in the sequence (3,5,11,17,23,29,59,89,149,179,419,...)?
Some numeric experiments let me suppose that the number of primes in the truncated sequence above defined could be $\sim 2\,n\log n$.
Is it the correct asymptotical behavior? Could anybody give me the sketch of a proof?
Many thanks.
It is not clear to me how you construct your sequence ("succession"). Could you explain in words how it's constructed, instead of just listing a bunch of terms?
You start from $3=1+2#$, then add $2#=2$ till you remain under $3#=6$ (that is $5$). Now you continue adding $3#$ unti you remain under $5#=30$. This way you obtain the elements $11,17,23,29$. At this point you have to continue adding $5#$, obtaining $59$ and so on ...
I checked: the sequence is not present in the OEIS database.
|
2025-03-21T14:48:29.810360
| 2020-02-08T23:05:13 |
352242
|
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|
Stack Exchange
|
Duality relation of Lorentz space $L^{p,1}$
want to prove the duality relation:
$$||f||_{L^{p,1}} =C_{p} \cdot \sup\{\int_X fg d\mu: \text{ for any } ||g||_{L^{p',\infty}}\le 1 \} $$
where $\frac{1}{p}+\frac{1}{p'}=1, p>1, \mu$ is $\sigma$-finite meaure on $X$, $C_{p}\in (0, \infty)$ and
$$||g||_{L^{p,q}}:=(p \cdot \int_{0}^{\infty} \lambda^{q-1}\mu(|f|>\lambda)^{\frac{q}{p}}d\lambda)^{\frac{1}{q}}, p>1, q \in (0,\infty].$$
There are several versions of proofs online, but they are all wrong:
https://www.math.ucla.edu/~tao/247a.1.06f/notes1.pdf or
https://www.math.ucla.edu/~killip/247a/Lorentz_space_notes.pdf
You may consult the book by Grafakos, Classical Fourier Analysis. A detailed study of Lorentz dual spaces is in Theorem 1.4.17, p.52
Thanks. This is a good reference!
|
2025-03-21T14:48:29.810449
| 2020-02-08T23:43:15 |
352243
|
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|
Stack Exchange
|
On Gromov's proof of the systolic inequality $\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$
In the page 10 of the paper "Filling Riemannian manifolds" by Gromov (ProjetEuclid link), the author proves the following inequality (1.2) relating the systole and the filling radius of manifolds.
$$\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$$
During the proof, he uses an argument something like this:
"$M$ is a closed $n$-dimensional manifold and $[M]$ is the fundamental class of it. In some ambient space $X$, the fundamental class $[M]$ is null-homologous. Therefore, one can find $n+1$-dimensional singular chain $c$ inside $X$ such that $\partial c=[M]$. Moreover, using a $\textbf{piecewise linear approximation}$ of $c$, one can construct a polyhedron $P$ such that $M\subset P \subset X$ and the fundamental class $[M]$ is null-homologous inside of this $P$."
So, somehow he is constructing a triangulation of realization of the singular chain $c$ (i.e., union of continuous images of $\Delta_{n+1}$) using "piecewise linear approximation". But I don't understand this step. How do we guarantee this realization of singular chain is triangulable? It is not necessarily manifold. What is precise meaning of his "piecewise linear approximation" of singular chain?
Edit: I changed the title of this question from "When a topological space (not necessarily manifold) has a triangulation?" to "A question on the Gromov's proof of $\operatorname{Sys}_1(M)\leq 6\operatorname{FillRad}(M)$". Even though the original question is interesting in its own right, but I want to focus on one specific topic: Understanding Gromov's argument.
Do you seek an answer to the precise question you ask, or an explanation of why Gromov's argument is correct? These likely have different answers.
@John Pardon: Mainly, explanation of Gromov's argument.
@KGEO Then you should probably change your title accordingly. Both are very interresting though.
@M. Dus Makes sense. I will edit my question to focus on one topic, soon.
About the question in the title (at this time), namely "When does a topological space (not necessarily manifold) have a triangulation?", it is an open question whether the (compact) space of closed subgroups of $\mathbf{R}^n$ has a triangulation for $n\ge 3$ (for $n\le 2$ it's true).
I think slight modification of the argument in Hatcher's algebraic topology book (pg 108-109) works.
Let $[M]$ be a $n$-singular cycle representing the fundamental cycle of $M$. Then $[M]=\delta_1+\dots+\delta_k$ where each $\delta_i:\Delta_n\rightarrow M\subset B_r(M,L^\infty(M))$ is a singular $n$-simplex. From the assumption, there is a singular $(n+1)$-chain $c$ such that $\partial c= [M]$. One can express $c=\sum_i\varepsilon_i\sigma_i$ with $\varepsilon_i=\pm 1$, allowing repetitions of singular $(n+1)$-simplices $\sigma_i$. Now, when we compute $\partial c$ as a sum of singular $n$-simplices with signs $\pm 1$, there may be canceling pairs consisting of two identical singular simplices with opposite sign. Choosing a maximal collection of such canceling pairs, construct an $(n+1)$-dimensional $\Delta$-complex $K$ from a disjoint union of $(n+1)$-simplices $\Delta_{n+1}^i$ one for each $\sigma_i$, by identifying the canceling pairs.
Then, we have a continuous map $\phi:K\rightarrow B_r(M,L^\infty(M))$ induced from $\sigma_i$'s. $K$ is $\Delta$-complex, not simplicial complex. $\phi$ is not homeomorphism, just continuous map. But I think this is enough. Let $P:=\mathrm{Im}(\phi)$ (then, $M\subset P\subset B_r(M,L^\infty(M))$). By subdividing the complex $K$ if necessary, one can make distance between $\phi(u)$ and $\phi(v)$ arbitrarily small (whenever $u,v$ are vertices). After that, just follow Gromov's argument.
The following may work.
$M$ is a smooth manifold so it can be triangulated smoothly. Now taking a fine enough (smooth) barycentric subdivision, there is an isotopy taking $M$'s embedding in $L^\infty$ to a linearly-embedded simplicial complex $M'$ which doesn't move any point too far (since $M$ is compact and bounds within an open $\epsilon$-neighborhood, it bounds in some $\epsilon'<\epsilon$-neighborhood; take $M'$ such that no point moves further than $(\epsilon-\epsilon')/2$ or so).
Denote the image of $c$ under the isotopy by $c'$. Those faces of $c'$ in $M'$ are already piecewise-linear, and the map can be approximated by a piecewise-linear one $c''$ from a fine enough barycentric subdivision of the domain of $c'$: send the vertices to their images under $c'$, and extend linearly. It is an approximation in the sense that it is the same up to homotopy, and within an $\epsilon$-neighborhood of $M'$. Since the set of linear simplices in $M' \cup c''$ has a finite-dimensional affine span, we may now work in a vector space $W$ of sufficiently high dimension, and we may assume $M'$ bounds within an $\epsilon$-neighborhood inside $W$.
Triangulate a sufficiently small neighborhood of $M'$ in $W$ in which $M'$ bounds, and call the result $P$. A sufficiently fine derived subdivision of $P$ refines the complex $M'$ (see for instance the first 15 pages or so of Glaser's book). Since singular and simplicial homology coincide, $M'$ is nullhomologous (simplicially) in those simplices of this complex contained in an $\epsilon$-neighborhood of $M'$. Now one can work with this $P$.
(Edit: closer to the original meaning than the paragraph above: a sufficiently fine derived subdivision of $P$ refines $M' \cup \text{im}(c'')$).
Since isotopy = ambient isotopy in high codimension, we may take $P$ "back" through the ambient isotopy to a smooth simplicial complex containing $M$.
|
2025-03-21T14:48:29.810827
| 2020-02-09T00:20:50 |
352245
|
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|
Stack Exchange
|
The formula for (and computation of) the inverse p-adic mellin transform
So, after scouring the entirety of the internet, I managed to find one (and, so far, only one) source that actually explains how to invert the $p$-adic mellin transform:
$$\mathscr{M}_{p}\left\{ f\right\} \left(s\right)\overset{\textrm{def}}{=}\frac{p}{p-1}\int_{\mathbb{Q}_{p}^{\times}}\left|\mathfrak{z}\right|_{p}^{s-1}f\left(\mathfrak{z}\right)d\mathfrak{z},\textrm{ }\forall s\in\mathbb{C}$$
where $d\mathfrak{z}$ is the haar probability measure on $\mathbb{Z}_{p}$, and where $s$ is a complex variable. The source in question are these notes from the University of Chicago, specifically, pages 72 and 73. However, being an analyst, the word "(un)ramified" gives me heart palpitations; I'll be honest, I don't know exactly how to interpret equations (4.15) and (4.16) from the notes (pages 72 & 73), nor their accompanying text. I know just enough to know that the integral I wrote above is what the writer meant in writing (4.14).
However, because of the maddening $t$ business in the notes—among other things—I cannot understand how to correctly write down the inversion formula, among other things. Before I ask my questions, let me just say:
i. I have no interest in integrating over anything other than complex-valued functions on $\mathbb{Z}_{p}$. For what I'm trying to learn, all the business about field extensions are needless complications in these notes that I'm trying to do away with as I explain the material to myself.
ii. I have no interest in Representation theory; I'm just an analyst whose work has led him into non-archimedean waters, and would like to know what the rules are for swimming in these circumstances.
Anyhow...
Is the correct way of writing (4.15):
$$\mathscr{M}_{p}^{-1}\left\{ F\right\} \left(\mathfrak{z}\right)=\frac{1}{2\pi i}\oint_{p^{-\sigma}\partial\mathbb{D}}\frac{F\left(s\right)}{\left|\mathfrak{z}\right|_{p}^{s}}ds$$ where $p^{-\sigma}\partial\mathbb{D}$ is the circle in $\mathbb{C}$ centered at $0$ of radius $p^{-\sigma}$, and where $\sigma$ is a positive real number.
Or is it: $$\mathscr{M}_{p}^{-1}\left\{ F\right\} \left(\mathfrak{z}\right)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{F\left(s\right)}{\left|\mathfrak{z}\right|_{p}^{s}}ds$$ where the contour is the line $\textrm{Re}\left(s\right)=\sigma$ in $\mathbb{C}$?
Or is it: $$\mathscr{M}_{p}^{-1}\left\{ F\right\} \left(\mathfrak{z}\right)=\frac{1}{2\pi i}\oint_{p^{-\sigma}\partial\mathbb{D}}\left|\mathfrak{z}\right|_{p}^{-s}F\left(-\frac{\ln s}{\ln p}\right)ds$$
Or is it something else, entirely?
Next, as a test-run, I tried to compute and then invert the transform of the constant $\mathbb{Z}_{p}$. Like in the notes, I computed: $$\mathscr{M}_{p}\left\{ \mathbf{1}_{\mathbb{Z}_{p}}\right\} \left(s\right)=\frac{1}{1-p^{-s}}$$ where $\mathbf{1}_{\mathbb{Z}_{p}}$ is the indicator function for $\mathbb{Z}_{p}$. This is the same as the notes, albeit they use $t=p^{-s}$ and write this as $\frac{1}{1-t}$.
However, when I try to use either of the above two attempts at interpreting the inversion formula (4.15), I end up with gobbledygook.
• The first formula I gave yields the constant function $$f\left(\mathfrak{z}\right)=\frac{1}{\ln p}$$
• The second formula yields (using the residue theorem): $$f\left(\mathfrak{z}\right)=\frac{1}{\ln p}\sum_{k\in\mathbb{Z}}\left|\mathfrak{z}\right|_{p}^{-\frac{2k\pi i}{\ln p}}=\frac{1}{\ln p}\sum_{k\in\mathbb{Z}}e^{2k\pi i\textrm{val}_{p}\left(\mathfrak{z}\right)}$$ which is always divergent.
• The third formula yields $f\left(\mathfrak{z}\right)=0$, because the integrand: $$\left|\mathfrak{z}\right|_{p}^{-s}F\left(-\frac{\ln s}{\ln p}\right)=\frac{\left|\mathfrak{z}\right|_{p}^{-s}}{1-p^{--\frac{\ln s}{\ln p}}}=\frac{\left|\mathfrak{z}\right|_{p}^{-s}}{1-s}$$ is holomorphic inside the unit disk.
None of these seem right to me, which makes me worry that none of the inversion formulae I've proposed are correct.
As such, I ask:
(1) What is the correct formula for the inversion of the $p$-adic mellin transform?
(2) What is the procedure for evaluating said integral? (Ex. Do I use the residue theorem, but ignore the existence of certain poles—if so, which ones?)
(3) More generally, given an $f:\mathbb{Z}_{p}\rightarrow\mathbb{C}$ so that the integral: $$F\left(s\right)=\int_{\mathbb{Z}_{p}\backslash\left\{ 0\right\} }\left(f\left(\mathfrak{z}\right)\right)^{s}d\mathfrak{z}$$ exists and has an analytic continuation to a meromorphic or entire function of $s\in\mathbb{C}$, how would I go about inverting it to re-obtain $f$? What would be the inversion formula, are there any special cares I should take in computing it (ignoring certain singularities when computing residues, etc.)? And to what extent can I re-obtain $f$ in this way?
To anyone who has read this far: thank you very much for your time!
I think that the correct answer is somewhere between your first and second formula: $$\mathscr M_p^{-1}{F}(\mathfrak z) = \frac1{2\pi i}\int_{\sigma-i\pi}^{\sigma+i\pi}\frac{F(s)}{\lvert\mathfrak z\rvert_p^s}\mathrm ds.$$ I'll check details and post later. However, in the meantime, while your obvious frustration is understandable, I think that it's not necessary to give such an apparently hostile pre- and post-lude to your question; perhaps it suffices to respond to such bad behaviour only after it happens here?
I've deleted the hostility. xD I had intended it more as a safe-guard against someone giving a patronizing "oh, just look it up yourself" type response. Anyhow, thanks for your time. :D
In those notes, taking $F = \mathbf Q_p$, the scary term "unramified character" of $\mathbf Q_p^\times$ means a continuous homomorphism $\chi \colon \mathbf Q_p^\times \rightarrow \mathbf C^\times$ that is trivial (equal to $1$) on the units $\mathbf Z_p^\times$. The simplest example of such a character is the $p$-adic absolute value: $x \mapsto |x|_p$. This is continuous on $\mathbf Q_p^\times$ and it is definitely trivial on $\mathbf Z_p^\times$ since those are exactly the $p$-adic number of $p$-adic absolute value 1. A complex power $x \mapsto |x|_p^s$ for $s \in \mathbf C$ is also an unramified character of $\mathbf Q_p^\times$, and he is saying all unramified characters of $\mathbf Q_p^\times$ look like this for some $s$. Why is that?
Every nonzero $p$-adic number has the form $p^nu$ for some $n \in \mathbf Z$ and $u \in \mathbf Z_p^\times$. For an unramified character $\chi$ of $\mathbf Q_p^\times$, we have $\chi(u) = 1$, so $\chi(p^nu) = \chi(p^n) = \chi(p)^n$. The number $\chi(p)$ is in $\mathbf C^\times$, so we can write $\chi(p)$ as $1/p^s$ for some $s \in \mathbf C$. (This $s$ is not unique, but is well-defined up to adding an integer multiple of $2\pi i/\log p$ on account of looking at the complex solutions $s$ to $p^s = 1$.) Then $\chi(p^nu) = \chi(p)^n = (1/p^s)^n = (1/p^n)^s = |p^nu|_p^s$, so $\chi(x) = |x|_p^s$ for all $x \in \mathbf Q_p^\times$: $\chi$ is the $s$-th power of the basic unramified character $x \mapsto |x|_p$, where
$s$ satisfies $\chi(p) = 1/p^s$. When a homomorphism $\mathbf Q_p^\times \rightarrow \mathbf C^\times$ is trivial on $\mathbf Z_p^\times$, it is continuous since it is locally constant (it is constant near 1 and a homomorphism) and is completely determined by its value at $p$.
The value of $\chi(p)$ can be arbitrary in $\mathbf C^\times$: for each $t \in \mathbf C^\times$ set $t = 1/p^s$ for some $s \in \mathbf C$ and define $\chi_t \colon \mathbf Q_p^\times \rightarrow \mathbf C^\times$ by the rule $\chi_t(p^nu) = t^n$ for $u \in \mathbf Z_p^\times$ and $n \in \mathbf Z$. This is a homomorphism, its value at $p$ is $t$, it is trivial on $\mathbf Z_p^\times$ ($\chi_t$ is "unramified"), and it is continuous since it is locally constant. Since $\chi_t(p^nu) = t^n = (1/p^s)^n = (1/p^n)^s = |p^nu|_p^s$, we have $\chi_t(x) = |x|_p^s$ for all $x \in \mathbf Q_p^\times$. This is why he says for each $t \in \mathbf C^\times$ there is a unique unramified character $\chi$ of $\mathbf Q_p^\times$ with $\chi(p) = t$: that $\chi$ is $\chi_t$.
The connected components of the group of characters $\Omega(\mathbf Q_p^\times)$ are entirely determined by how the characters look on $\mathbf Z_p^\times$: two characters of $\mathbf Q_p^\times$ are in the same connected component exactly when they are equal on $\mathbf Z_p^\times$, and by continuity a character $\mathbf Z_p^\times \rightarrow \mathbf C^\times$ is trivial on some neighborhood $1 + p^n\mathbf Z_p$ of 1 (a subgroup!) since $\mathbf C^\times$ has no subgroup in a neighborhood of 1 other than $\{1\}$. Therefore a character on $\mathbf Z_p^\times$ is a homomorphism to $\mathbf C^\times$ on some quotient group $\mathbf Z_p^\times/(1 + p^k\mathbf Z_p)\cong (\mathbf Z/p^k\mathbf Z)^\times$, which is finite. Going the other way, for each homomorphism $(\mathbf Z/p^k\mathbf Z)^\times \rightarrow \mathbf C^\times$ we can lift it to a character $\eta$ on $\mathbf Z_p^\times$ using the composite map
$$
\mathbf Z_p^\times \rightarrow \mathbf Z_p^\times/(1+p^k\mathbf Z_p) \cong (\mathbf Z/p^k\mathbf Z_p)^\times \rightarrow \mathbf C^\times
$$
(this is automatically continuous since it is a homomorphism and it is trivial on the neighborhood $1 + p^k\mathbf Z_p$ of 1) and then we can multiply this by an unramified character $\chi$ to get a character of $\mathbf Q_p^\times$: $p^nu \mapsto \chi(p)^n\eta(u)$. In other notation, since unramified characters of $\mathbf Q_p^\times$ are just complex powers $|\cdot|_p^s$, each character of $\mathbf Q_p^\times$ is $|\cdot|_p^s\eta$ where $s \in \mathbf C$ and $\eta$ is a character of $\mathbf Z_p^\times$: each connected component can be labeled by the common $\eta$ (restriction to $\mathbf Z_p^\times$) for all characters in that component. (The choice of $s$ for a character really is in $\mathbf C/(2\pi i/\log p)\mathbf Z$, a cylinder, which topologically is the same as $\mathbf C^\times$ using $s + (2\pi i/\log p)\mathbf Z \mapsto 1/p^s$.)
For $x \in \mathbf Q_p^\times$, written as $p^nu$ where $n \in \mathbf Z$ and $u \in \mathbf Z_p^\times$, write $u$ as $u_x$ to indicate its dependence on $x$. Then
for each character $\eta$ of some $(\mathbf Z/p^k\mathbf Z)^\times$ and $s \in \mathbf C$, we get a character $\chi$ of $\mathbf Q_p^\times$ by $\chi(x) = |x|_p^s\eta(u_x\bmod p^k)$. (Note that $\eta(p)$ makes no sense.) All characters of $\mathbf Q_p^\times$ look like this.
Prop. 4.6 is about continuous functions $\mathbf Q_p^\times \rightarrow \mathbf C$ with compact support. Note that your test-run example of the characteristic function of $\mathbf Z_p$, viewed as a function on $\mathbf Q_p^\times$ by taking $0$ out of its domain, does not have compact support in $\mathbf Q_p^\times$: the set $\mathbf Z_p - \{0\}$ is not compact in $\mathbf Q_p^\times$ just as $(0,1]$ and $[-1,1] - \{0\}$ are not compact in $\mathbf R^\times$. Therefore this is not a good example for a test-run for Prop. 4.6 (unlike Prop. 4.7).
For a better choice of test-run for Prop. 4.6, let $\xi_A$ be notation for the characteristic function of a set $A$ (1 if the variable is in $A$ and 0 otherwise). For $a \in \mathbf Q_p^\times$ and $n \in \mathbf Z$ chosen large enough so that $|a|_p > 1/p^n$, set $\phi = \xi_{a + p^n\mathbf Z_p}$: this is the characteristic function of the ball $a + p^n\mathbf Z_p$, which is a subset of $\mathbf Q_p^\times$ since we can't have $a + p^nx = 0$ for $x \in \mathbf Z_p$, as $|p^nx|_p \leq 1/p^n < |a|_p$. (If you don't like the characteristic function of a general ball in $\mathbf Q_p$ not containing $0$, consider the special case $a = 1$: $\xi_{1 + p^n\mathbf Z_p}$ for $n \geq 1$.)
Letting $|a|_p = 1/p^m$, so $m<n$, the Mellin transform $(M\phi)(\chi)$ of a character $\chi$ of $\mathbf Q_p^\times$ is the following integral
$$
(M\phi)(\chi) = \int_{\mathbf Q_p^\times} \xi_{a + p^n\mathbf Z_p}(x)\chi(x)d^\times x,
$$
where I write $d^\times x$ for the (multiplicative) Haar measure on $\mathbf Q_p^\times$ that you write as $p/(p-1) dx/|x|_p$. It is the Haar measure on $\mathbf Q_p^\times$ that gives the compact open subgroup $\mathbf Z_p^\times$ measure 1. After some calculation that I omit (tell me if you can work this out), we get
$$
(M\phi)(\chi) = \chi(a)\int_{1 + p^{n-m}\mathbf Z_p} \chi(y)d^\times y,
$$
which is $\chi(a)$ times the integral of the multiplicative character $\chi$ over the compact multiplicative group $1 + p^{n-m}\mathbf Z_p$. The value of this is determined by whether or not $\chi$ is trivial on $1 + p^{n-m}\mathbf Z_p$: if $\chi \not\equiv 1$ on $1 + p^{n-m}\mathbf Z_p$ then
$$
(M\phi)(\chi) = 0,
$$
and if $\chi \equiv 1$ on $1 + p^{n-m}\mathbf Z_p$ then
$$
(M\phi)(\chi) = \chi(a)\int_{1 + p^{n-m}\mathbf Z_p} d^\times y =
\frac{\chi(a)}{[\mathbf Z_p^\times:1 + p^{n-m}\mathbf Z_p]} =
\frac{\chi(a)}{p^{n-m-1}(p-1)}.
$$
There are only finitely many connected components containing the $\chi$ where $\chi \equiv 1$ on $1 + p^{n-m}\mathbf Z_p$ since such $\chi$ are determined on $\mathbf Z_p^\times$ (not on $\mathbf Q_p^\times$!) by their values on $\mathbf Z_p^\times/(1 + p^{n-m}\mathbf Z_p)$, which is a finite group (so it has only finitely many homomorphisms to $\mathbf C^\times$). Thus $M\phi$ is a "polynomial" in the sense defined above Prop. 4.6.
I have not yet addressed your question about the Mellin inversion formula. Consider this a partial answer so far. I will save what I have written and come back to this later.
Woo! Thanks so much! :D
Some questions: (1) does any of this simplify (even slightly) if we assume that the functions we are transforming are defined only on $\mathbb{Z}{p}$ (if so, how)? (2) How can we write a generic $\eta$ here "explicitly"? (Ex: like how, for a character $\xi$ on $\mathbb{Z}{p}$, the "explicit" form of $\xi$ is $\xi\left(\mathfrak{z}\right)=e^{-2\pi i\left{ t\mathfrak{z}\right} _{p}}$ for some $t\in\mathbb{Z}\left[\frac{1}{p}\right]/\mathbb{Z}$, treated here as a rational number in $\left[0,1\right)$). I can't know if I can do the integral or not until I see it written explicitly. xD
So far, I've got $\chi\left(x\right)=\left|x\right|_{p}^{s}\times?$ as the formula for the $\chi$ in the integral. Need to know what the ? is.
I am using $\xi_A$ in my post for the characteristic function of a set $A$, so please don't use $\xi$ in a comment to mean a character on $\mathbf Z_p$ (or $\mathbf Q_p$).
So far, what I've got for the integral is: $\int_{\mathbb{Q}{p}^{\times}}\left[\mathfrak{z}\overset{p^{n}}{\equiv}a\right]\left|\mathfrak{z}\right|{p}^{s-1}\eta\left(\mathfrak{z}\right)d\mathfrak{z}=\frac{1}{p^{ms}}\int_{\frac{a}{p^{m}}+p^{n-m}\mathbb{Z}{p}^{\times}}\left|\mathfrak{z}\right|{p}^{s-1}\eta\left(p^{m}\mathfrak{z}\right)d\mathfrak{z}$ where $\left|a\right|{p}=p^{-m}>p^{-n}$, and where the $\eta$ is the one you mentioned above. $\left[\mathfrak{z}\overset{p^{n}}{\equiv}a\right]$ is the iverson bracket for $a+p^{n}\mathbb{Z}{p}$ (1 if true, 0 if false).
I don't know how to simplify it further without an explicit formula for the $\eta$. I'm also unsure of how to get rid of the $\times$ in the $\mathbb{Z}{p}^{\times}$, or make the $\mathbb{Z}{p}$-unit $\frac{a}{p^{m}}$ into a 1.
Also, if the constant function on the integer ring of the field isn't a good candidate for (4.6), why does (4.7) then appear to immediately proceed to apply (4.14) to said function?
Please don't use the exact same notation for equation references and theorem/proposition references. Your "(4.14)" means equation 4.14, but your similar notation "(4.6)" and "(4.7)" mean Prop. 4.6 and Prop. 4.7, not equations 4.6 and 4.7.
The difference between Prop. 4.6 and Prop. 4.7 is that the integral in Prop. 4.6 holds for all $\sigma \in \mathbf R$ while the integral in Prop. 4.7 holds for $\sigma > 0$.
Let us continue this discussion in chat.
Edit:
The inversion of $F(s)$, in your notation is
$$f(x) = \frac{\mathrm{ln}(p)}{2 \pi i} \int_{ \{ s=ir \ : r \in (-\pi/\mathrm{ln}(p), \pi/\mathrm{ln}(p)] \}} F(s)|x|^{-s} ds = \frac{\mathrm{ln}(p)}{2 \pi i} \int_{ \{ s=ir \ : r \in (-\pi/\mathrm{ln}(p), \pi/\mathrm{ln}(p)] \}} F(s)p^{\mathrm{val}(x)s} ds$$
Here if $x=p^nu$ where $u \in \mathbb{Z}_p^{\times}$ is a unit $\mathrm{val}(x)=n$. You can evaluate this integral in the normal way, you don't ignore poles. It seems like the closest answer you got was the constant $\frac{1}{\mathrm{ln}(p)}$, but this is incomplete because $f(x)$ is not supported on $\mathbb{Q}_p^{\times}$ but rather on $\{ \mathrm{val}(x) \geq 0 \} $. You can see this because in the series expansion of the rational function $\frac{1}{1-p^{-s}} = 1 + p^{-s}+p^{-2s} + \cdots $ the terms consist of polynomials in the variable $p^{-s}$ and each integral $\int_{ \{ |s|=1 \} } p^{-ns}|x|^{-s}$ is non-zero only when the valuation of $x$ is $n$ for $n \geq 0$. Finally the $\mathrm{ln}(p)$ terms just comes to account for the fact that we are parametrizing the complex circle by $r \mapsto p^{-ir}$ so as to give complex circle measure $1$.
=================
It is helpful to state things first in terms of abstract harmonic analysis on locally compact abelian (=LCA) groups might make things clearer.
For any LCA group $A$ let $\widehat{A}$ be the group of it's unitary characters. If $f$ is a function on $A$ it's Fourier transform is $\chi \mapsto \widehat{f}(\chi) = \int_{A} f(x)\chi^{-1}(x)dx$ with $dx$ a fixed Haar measure on $A$. My Pontryagin duality is a unique measure $d\chi$ on the LCA group $\widehat{A}$ dual to $dx$ satisfying the Fourier inversion
$$
f(e) = \int_{\widehat{A}} \widehat{f}(\chi) d\chi
$$
$$
f(y) = \int_{\widehat{A}} \widehat{f}(\chi)\chi(y) d\chi
$$
$\mathbb{Q}_p^{\times}$ is your abelian group, under multiplication. It's Pontryagin dual is the space of unitary complex characters $ \widehat{\mathbb{Q}_p} =\{ \chi: \mathbb{Q}_p^{\times} \to \mathbb{T} \}$ with $\mathbb{T}$ the complex circle. For any smooth compactly supported function on $\mathbb{Q}_p^{\times}$ it's abstract Fourier transform is the Mellin transform $\widehat{f}(\chi) = \int_{\mathbb{Q}_p^{\times}} f(x) \chi(x) dx$
Because $\mathbb{Q}_p^{\times} = \mathbb{Z}_p^{\times} \times p^{\mathbb{Z}}$ it's unitary dual is $\widehat{\mathbb{Z}_p^{\times}} \times \widehat{\mathbb{Z}}$. The unramified characters are the characters which are trivial on $\mathbb{Z}_p^{\times}$, i.e. they are restrictions to $\{ 1 \} \times \widehat{\mathbb{Z}}$. The unitary characters of $\mathbb{Z}$ can be identified with unit norm complex numbers $\mathbb{T} = \{ |z|=1 \}$. The complex parameter refers to the exponent $s$ in the unitary character determined by $ | \cdot |^{s}:p \mapsto |p|_{p}^s=p^{-s}$ (or however you want to normalize it, but this is natural for number theory). To conform with Ngo's noration let us make a change of variables $t = p^{-s}$ and call this unitary character $\chi_t$. So we should think of the function "$\frac{1}{1-p^{-s}}$" more precisely as a function $L$ on the space of unitary characters:
$$\chi_t \mapsto L(\chi_t) = \frac{1}{1-\chi_t(p)}$$ if $\chi_t$ is an unramified chatacter
$$ \eta \mapsto L(\eta) = 0 $$ If $\eta$ is not unramified.
The dual measure $d\chi_{t}$ is always a Haar measure on the unitary dual, which in our case is $\mathbb{T}$, so is a multiple of $dt/t$. Explicitly it looks like $\frac{\mathrm{ln}(p)}{2\pi i} \frac{dt}{t}$ (because the contour is over a parameter between $\frac{-\pi}{\mathrm{ln}(p)} $ and $\frac{\pi}{\mathrm{ln}(p)}$).
We can write $L(\chi_t) = \chi_{t}(e) + \chi_{t}(p) + \chi_{t}(p^2) + \cdots = \chi_{t}(e) + l_p \chi_{t}(e) + l_{p^2} \chi_{t}(e) + \cdots$ where $l_p$ is translation by $p$. Strictly speaking this is not a polynomial function on the space of unitary characters but a rational one, but formally we can do everything for the terms in the sum and then take a limit.
Thus it is enough to Mellin-invert the function $\widehat{f_0}: \chi_t \mapsto \chi_{t}(e) = 1$ and deduce the rest of it by translating. Using the Fourier inversion formula and the fact that $\widehat{f_0}$ is supported on $\mathbb{Z}_p^{\times}$-invariant characters implies that $f_0$ is $\mathbb{Z}_p^{\times}$-invariant. Thus it is enough to know it's value on $\mathbb{Z}_p^{\times}$ and it's translates $p^n\mathbb{Z}_p^{\times}$.
$$f_0(e)=\mathrm{const} \int_{ |t|=1 } \chi_t(e) \frac{dt}{t} = \mathrm{const} \int_{ |t|=1 } 1 \frac{dt}{t}$$
Thus $f_0(\mathbb{Z}_p^{\times})=1$. Arguing by translation by either $p$ or $p^{-1}$ we obtain
$$f_0(p) = \mathrm{const} \int_{ |t|=1 } \chi_t(p)\frac{dt}{t} = \int_{ |t|=1 } dt=0$$ or
$$f(p^{-1}=\mathrm{const} \int_{ |t|=1 } \chi_t(p^{-1})\frac{dt}{t} = \int_{ |t|=1 } t^{-2}dt =0.$$
So $f_0 = 1_{ \mathbb{Z}_p^{\times}}$.
Arguing similarly with $\chi_{t} \mapsto l_p(\chi)(e)$ produces a function $f_{1}$ such that $l_p(f)(\mathbb{Z}_p^{\times}) =1$ and is invariant under and supported in $\mathbb{Z}_p^{\times}$. In other words it is the characteristic function of the coset $p \mathbb{Z}_p^{\times}$. And so on for the rest of the terms. Adding them all up yields $ \sum_n f_n = 1_{\mathbb{Z}_p - \{ 0 \}}$ as exepcted.
I am fine with pontryagin duality. The moment anyone breathes a word about group representations, I run screaming for the hills.
You completely lose me by paragraph 3. I don't know the formula for a "character of $\mathbb{Q}{p}^{\times}$ that is invariant under $\mathbb{Z}{p}^{\times}". I also don't understand your notation from that point onward, and cannot follow any of it.
I know the algorithm for computing $\int_{\mathbb{Z}_{p}}f\left(\mathfrak{z}\right)e^{-2\pi i\left{ t\mathfrak{z}\right} {p}}d\mathfrak{z}$ when $f$ is a schwartz-bruhat function on $\mathbb{Z}{p}$ and when $t\in\mathbb{Z}\left[\frac{1}{p}\right]/\mathbb{Z}$. I don't know how to do what you are describing. At this point, I just want to know the formula-rules for computing these p-adic mellin integral transforms and their inverses.
I tried to clarify my answer: the unramified characters are trivial on the unit group $\mathbb{Z}_p^{\times}$, and so are determined by their values at $p$. This introduces the complex variable. The inversion done by the contour integral, however you choose to parametrize it and keep in mind to use the Haar measure on the unit circle.
I don't understand your notation, you aren't parameterizing the integrals, you aren't parameterizing the characters, and you've said absolutely nothing about the heart of my question: giving me the correct formula for the inverse transform and explaining how (if at all) one can use the residue theorem to evaluate it. Although I appreciate you trying to explain the ideas behind the computations, I have no room for that in my head at the moment. Right now, I need formulae and algorithms to memorize. I can focus on understanding the reasoning behind them once I know how to do them.
Think of it like this: right now, I am like the student who asks "what is the derivative of sin(x)?" Your answer, to me, is like saying "Well, first you need to compute the difference quotient..." and giving a lengthy rambling explanation of how to derive the formula for the derivative of sin(x). Meanwhile, all I want to hear is "the derivative of sin(x) is cos(x)". If I know how the formulae work, I can eventually attain understanding of what they meaning and the reasoning behind them. But without knowing the purely formal rules for manipulating the symbols, I can't do anything.
Finally, just as a piece of friendly advice, when someone asks a question about being unable to understand the notation in a document, using the notation of the question-asker (or something close to it) maximizes the chance of your answer being understood and appreciated by them. :) Note that all of the formulae I gave use p-adic absolute values and a complex variable $s$. I specifically avoided using the $\chi$ notation because I find it confusing. It's difficult for me to understand what you've written when you use the same ambiguous, abbreviated notation as the notes.
So, the inverse transform is $$\mathscr{M}{p}^{-1}\left{ F\right} \left(\mathfrak{z}\right)=\frac{\ln p}{2\pi i}\int{-\frac{i\pi}{\ln p}}^{\frac{i\pi}{\ln p}}F\left(s\right)\left|\mathfrak{z}\right|{p}^{-s}ds$$ where the integral is taken over the vertical line segment from $s=-\frac{i\pi}{\ln p}$ to $s=\frac{i\pi}{\ln p}$? So, if I evaluate this integral directly, it will give me the function $f$ such that $$F\left(s\right)=\int{\mathbb{Z}{p}}\left|\mathfrak{z}\right|{p}^{s-1}f\left(\mathfrak{z}\right)d\mathfrak{z}$$ Is that correct?
Also, for what $\mathfrak{z}\in\mathbb{Z}_{p}$ does this inversion formula hold? For almost every $\mathfrak{z}$? For every $\mathfrak{z}$? And what will the inversion formula output at values of $\mathfrak{z}$ where $f$ has a discontinuity?
Edit: I meant $\mathbb{Z}{p}\backslash\left{ 0\right}$ in the integral in the previous comment. $f$ is supported in $\mathbb{Z}{p}$.
Yes that is the inverse transform. The inversion formula in this setting apriori holds for functions that are locally constant and compactly supported in $\mathbb{Q}_p^{\times}$, i.e. on sets of the form $-n \leq \mathrm{val}(x) \leq n$. Their Mellin transforms correspond to Laurent polynomials in $p^{-s}$. The fact that $F$ has a pole is reflected in the fact that it's inverse is the characteristic function of $\mathbb{Z}_p$ which is not compact. This is in some sense the content of the so called "unramified calculation" in Tate's thesis.
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2025-03-21T14:48:29.812646
| 2020-02-09T00:40:35 |
352246
|
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|
Stack Exchange
|
Singular value decomposition of random rectangular matrices
Let $A$ be a $m\times n$ real matrix, whose entries are independent, identically distributed random variables, following standard normal distributions (mean zero and unit variance).
What is the distribution of the singular values and singular (left and right) vectors of $A$?
For a symmetric square matrix, the concept of rotational invariance is helpful because (for ensembles where it holds) it allows us to focus on the eigenvalues since the eigenvectors are essentially uniformly distributed.
What is the analogue of "rotational invariance" in the context of a rectangular matrix like $A$? Is it helfpul here too?
Most papers/books I've read on the topic focus on square matrices so if anyone can point out relevant references I'd also appreciate that.
To be specific, let me assume $m\leq n$. The $m\times n$ matrix $A$ then has $n-m$ singular values equal to zero. The remaining $m$ singular values $s_i$, $i=1,2,\ldots m$ are the positive square roots of the eigenvalues $\sigma_i$ of the symmetric matrix product $AA^t$. The distribution $P(\sigma_1,\sigma_2,\ldots\sigma_m)$ of the $\sigma_i$'s is the Wishart distribution. For $m\gg 1$ the eigenvalue density is the Marchenko–Pastur distribution.
The eigenvectors of $AA^t$ are the left eigenvectors in the singular value decomposition. These are uniformly distributed with the Haar measure in $O(m)$. The right eigenvectors are $m$ rows of length $n$ of an $n\times n$ matrix that is uniformly distributed in $O(n)$, independently of the left eigenvectors.
Thanks! But how can you tell that the right singular vectors are independent from the left ones? Also, which of these statements rely upon the components of $A$ being Gaussian? I presume that for a non-Gaussian distribution, the left and right vectors become relevant (because the rotational invariance is broken in some sense).
your distribution $P(A)\propto\exp(-\tfrac{1}{2},{\rm tr},AA^t)$ is both left-invariant and right-invariant under multiplication by an arbitrary orthogonal matrix, $P(AO)=P(O'A)$, so the left- and right-eigenvectors can be chosen independently and uniformly; all of this is for a Gaussian distribution.
I see now, thanks!
What is the analogue of the Vandermonde determinant, if I want to change variables using a singular value decomposition? (Please see https://mathoverflow.net/questions/352309/jacobian-of-changing-of-variables-to-singular-value-decomposition)
That’s the measure of the Wishart ensemble (Vandermonde times product of eigenvalues to a power)
Any recommendations on what to read to gain an understanding on all this stuff?
Forrester is highly recommended: contains detailed derivations.
Thanks for your help!
|
2025-03-21T14:48:29.812918
| 2020-02-09T02:16:23 |
352249
|
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|
Stack Exchange
|
Nontrivially fillable gaps in published proofs of major theorems
Prelude: In 1998, Robert Solovay wrote an email to John Nash to communicate an error that he detected in the proof of the Nash embedding theorem, as presented in Nash's well-known paper "The Imbedding Problem for Riemannian Manifolds" (Annals of Math, 1956), and to offer a nontrivial fix for the problem, as detailed in this erratum note prepared by John Nash. This topic is also discussed in this MO question.
Of course, any mathematician who has been around long enough knows of many published proofs with significant gaps, some provably irreparable, and some perhaps authored by himself or herself. What makes the above situation striking--and discomforting to many of us--is the combination of the following three factors:
(1) The theorem whose proof is found faulty is a major result that was published in 1950 or after, in a readily accessible source to experts in the field . (I chose the 1950 lower bound as a way of focusing on the somewhat recent past).
(2) The gap detected is filled with a nontrivial fix that is publicly available and consented to by experts in the field (so we are not talking about gaps easily filled, or about gaps alleged by pseudomathematicians, or about false publicly accepted theorems, as discussed in this MO question).
(3) There is an interlude of 30 years or more between the publication of the proof and the detection of the gap (I chose 30 years since it is approximately the age difference between consecutive generations, even though the interlude is 42 years in the case of the Nash embedding theorem).
Question to fellow mathematicians: what is the most dramatic instance you know of where all of the three above factors are present?
I caught an error in the proof of the main result of https://arxiv.org/abs/1309.4516 It's been a while, but I seem to recall that Prop 3.5 is the invalid culprit. I didn't say anything at first, because my PhD advisor told me it would be inappropriate to do so (???) and I deferred to his judgement. But after it passed peer review and was published, I finally pointed out the error to one of the authors. He agreed that the proof was invalid, but insisted the conclusion is true. It's not a huge deal I guess, but sorta disheartening to know that mathematicians won't admit their errors publicly.
Though you say you're not looking for "false publicly accepted theorems" as discussed in the MO question you link to, this answer there: https://mathoverflow.net/a/35864/25028 seems somewhat applicable.
Specifically the paper: https://doi.org/10.1112/S0024610705022416
"Dulac's theorem" on the finiteness of the number of limit cycles of a two-dimensional polynomial system is an extreme example of difficulty in "filling the gap", but it does not qualify since Dulac's "theorem" was published in the 1920s.
@BenW The advice given to you by your former advisor is very unfortunate and rather atypical of the attitude of most mathematicians I know of.
I chose 30 years since it is approximately the age difference between consecutive generations - I think mathematicians reproduce at a much faster rate, actually.
This doesn't meet your 30-year threshold but I think it's an interesting example nonetheless: The Fight to Fix Symplectic Geometry.
The question reminded me of the four color theorem, for which a couple early proofs had gaps that took 11 years to recognize, but then it took nearly 100 years after that to actually fill the gaps.
@DanielSchepler : Along similar lines, the gap in Dehn's lemma was detected in 1929, after "only" 19 years, but was not filled until 1957.
In 1970, I. N. Baker published a proof of a basic result in holomorphic dynamics:
a transcendental entire function cannot have more than one completely invariant domain.
A completely invariant domain is an open connected set $D$ such that $f(z)\in D$ if and only if $z\in D$.
Baker "proved" a more general statement that: there cannot be two disjoint domains whose preimages are connected.
The "proof" was a simple topological argument which occupied less than one page.
Since then this result has been used and generalized by extending his simple argument. In summer 2016 I was explaining Baker's argument to Julien Duval, he was somewhat slow in understanding and kept asking questions. Few weeks later he found a gap in the proof. It also took him some time to convince me that there is a gap indeed. Specialists were informed.
Half a year later an amazing counterexample has been constructed in https://arxiv.org/abs/1801.06359 by Lasse Rempe-Gillen and David Sixsmith. This paper contains the full account of the story.
This is a counterexample to Baker's more general statement only, not to the highlighted theorem itself, which is now an important open question.
(Was any gap filled?)
@Francois Ziegler: No. The counterexample I refer to shows that the gap cannot be filled. Some completely new idea is required. So this example is "NON (trivially-fillable)" rather than "(non-trivially) FILLABLE" :-)
So technically this doesn't answer Ali Enayat's question, I guess? It seems to be a requirement that the gap be filled, since Enayat explicitly excludes false publicly accepted theorems. (Mind you, I like your answer and I upvoted it; I'm just pointing out an apparent technicality.)
@AlexandreEremenko Thank you for your enlightening answer (which I upvoted). As Francois Ziegler and Timothy Chow have pointed out your answer does not meet criteria 3 (the gap is fillable), however, it is very strong on criteria (1) and (2) of my question.
I agree with all comments. But I will not remove my answer since it has many votes:-)
@AlexandreEremenko I am all for keeping your answer, since what you reported in your answer deserves wide dissemination.
This nice answer might fit better in https://mathoverflow.net/questions/35468/widely-accepted-mathematical-results-that-were-later-shown-to-be-wrong.
@Zach Teitler: Perhaps. But it also does not fit completely, because the main result (the highlighted statement) is not known to be wrong. It is an open question. Only the published proof is wrong.
In 2017 an erratum to the 1973 paper Isotopies of homeomorphisms of Riemann surfaces by Birman and Hilden appeared in Annals of Mathematics which satisfies your three criteria. That's a 43-year gap! The way Birman and Hilden tracked all papers citing theirs is admirable.
The error was found by Ghaswala, and a fix was provided by Ghaswala and Winarski in Lifting Homeomorphisms and Cyclic Branched Covers of Spheres, published the same year as the erratum.
Thank you very much for your enlightening answer, it nicely meets the three criteria stipulated in my question.
A lightweight counterpart: a false argument (ascribed to Hilbert) and a false statement by Cauer in 1910s: one cannot find a center of a circle [Hilbert] or two disjoint circles [Cauer] using a ruler only. A wrong argument can be found in most popular books [e.g. Courant/Robbins, or Rademacher/Toeplitz], the error was noted just a few years ago:
Arseniy Akopyan, Roman Fedorov, Two circles and only a straightedge, Proc. AMS 147 no. 1 (2019) pp. 91-102, doi:10.1090/proc/14240, arXiv:1709.02562.
That's an impressively long gap - over 100 years!
If a 25-year interlude will do, there is
R. F. Coleman has sent me his preprint ["Manin's proof of the Mordell conjecture'', Preprint, 1988; per bibl.] concerning my proof of Mordell's conjecture for function fields (see the paper cited in the heading). Coleman has discovered and corrected inaccuracies in my paper. Below I explain what changes should be made in the original paper in the language of that paper.
(If not, then maybe this.)
In 1980, Micali and Vazirani published An $O(\sqrt{|V|}\cdot |E|)$ algorithm for finding maximum matching in general graphs. I regard this as a major result in theoretical computer science. By Vazirani's own account, a complete proof of the running time claimed in the title was not provided until his 2012 arXiv preprint. That is a gap of 32 years.
However, one could object that the 1980 paper was technically just an "extended abstract" that did not claim to provide a full proof of correctness. In 1994, Vazirani published a paper claiming to give a proof (but which, as he himself acknowledged in his 2012 preprint, contained gaps and errors). So the gap is arguably "only" 18 years.
In 1990, Ravi Kannan wrote a paper giving an algorithm deciding $\forall \exists$ sentences of integer programs. As an intermediate claim, he "proved" the "Kannan Partition Theorem". Because his proof was unreadable to Eisenbrand and Shmonin who wanted to extend his result, they proved their own slightly weaker version of KPT. In 2017, Nguyen and Pak showed that if the KPT is true, then Short Presburger Arithmetic sentences can be decided in polynomial time, but a few months later, they showed that this is in fact hard and discovered the bug in the proof of KPT. The weaker version of Eisenbrand and Shmonin holds and is sufficient to prove Kannan's original final result.
So here the bug was fixed by Eisenbrand and Shmonin before it was discovered by Nguyen and Pak, and a (conditional) positive result was derived from it by the same authors who (a few months later) disproved it.
Since some answers include the results published before 1950, let me include the famous story of Hilbert's Problem on finiteness of the number of limit cycles of a polynomial vector field in the plane (Problem 16, second part). In 1923 Henri Dulac published a proof. The proof was accepted, it was published in a top journal and Dulac even obtained a prize for it. The huge gap was found in 1981, and fixed in 1992, independently by Yulii Ilyashenko and Jean Ecalle.
Smal correction: the top journal was not Acta but the Bulletin de la Société Mathématique de France.
@YangMills Yes "Sure les cycle limit" http://archive.numdam.org/article/BSMF_1923__51__45_1.pdf
In 1979, Dobkin and Snyder published an algorithm that purported to give the largest-area triangle inscribed in a convex n-gon in O(n) time. In 2017, Keikha, Löffler, Urhausen, and van der Hoog showed that this algorithm was in fact wrong. Two different authors were quickly and independently able to fill the gap and give a correct linear-time algorithm (I was one of them). It later turned out that the gap was already filled in a 1992 paper by Chandran and Mount, where they describe a linear-time algorithm to simultaneously construct the largest inscribed and smallest circumscribed triangles. Because the 1979 algorithm was not known to be wrong at the time, the fact that it gave an O(n) algorithm for the largest inscribed triangle was not a selling point of the 1992 paper and was not emphasized.
I think that the formula for the Littlewood-Richardson rule (how to expand a product of Schur functions into Schur functions) qualifies. It was first claimed to be proved 1934,
then an error was discovered and fixed in 1938.
The first complete proof was given in 1977, and nowadays there are many different short proofs.
I seem to remember somebody (possibly Gian-Carlo Rota) summarizing this situation by saying something like "Putting men on the moon was easier than proving the Littlewood-Richardson rule".
"Unfortunately the Littlewood–Richardson rule is much harder to prove than was at first suspected. The author was once told that the Littlewood–Richardson rule helped to get men on the moon but was not proved until after they got there."
--Gordon James (1987)
When I originally posted the question, my colleague Jim Schmerl and I had just discovered a major gap (as well as a fix for the gap) in the proof of a "classical" characterization (1975) by Barwise and Schlipf of recursively saturated models of PA (Peano arithmetic). This result of Barwise and Schlipf inaugurated the study of recursivey saturated models of PA, a topic that boasts a rich literature.
More specifically, the aforementioned Barwise-Schlipf theorem states:
Theorem. The following are equivalent for a nonstandard model $M$ of PA:
(1) $M$ is recursively saturated.
(2) There is $\mathfrak{X}$ such that $(M,\mathfrak{X})$ satisfies $\Delta^1_1$-Comprehension.
This recently published paper of Schmerl and me shows that the Barwise-Schlipf proof of $(2)\implies(1)$ has a serious gap. This problematic direction is established via an alternative argument in our paper using a coding method introduced by Kaufmann and Schmerl (1984).
For nonexperts: this recent note by John Baez on recursive saturation sings the praises of recursively saturated models of PA.
In this blog post Terry Tao discusses some corrections to a 2010 paper of his and Ben Green's "An arithmetic regularity lemma, an associated counting lemma, and applications". Daniel Altman found some problems with the arguments, and they can only be repaired by making additional assumptions. (This doesn't meet the 30 year gap though, although maybe it will take that long to repair the proof of the full conjecture of Gowers and Wolf which Tao and Green claimed to have resolved.)
EDIT: I think the errors may have now been fixed by Altman, see: A non-flag arithmetic regularity lemma and counting lemma.
In 1986, Partha Dasgupta and economic Nobel laureate Eric Maskin published a paper with an incorrect definition of symmetric games in it. The paper has over 1200 citations and I'm pretty sure it took 25 years until I pointed out on Wikipedia in 2011 that there's a problem (see the edits of this page). The problem is outlined in this with more work on the topic in this more recent draft (currently developing software though).
Could you clarify why "incorrect definition" counts as "gaps in published proofs"?
It has had over 1200 citations since, of which have citations themselves, that can easily lead to all sorts of gaps, and seems like a noteworthy gap in and of itself, though yes is not for a proof itself.
Not too long to wait to find the mistake, but the following might be interesting:
In 1905, Henri Lebesgue claimed to have proved that if B is a subset of $\mathbb{R}^2$ with the Borel property, then its projection onto a line is a Borel subset of the line.
This is false. In 1912 Souslin and Luzin defined an analytic set as the projection of a Borel set. All Borel sets are analytic, but, contrary to Lebesgue's claim, the converse is false.
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2025-03-21T14:48:29.814105
| 2020-02-09T03:22:51 |
352253
|
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Stack Exchange
|
Compact complex surface that admits a Kodaira fibration is Kahler
A Kodaira fibration is a compact complex surface X endowed with a holomorphic submersion onto a Riemann surface $\pi: X\to\Sigma$ which has connected fibers and is not isotrivial.
Is there an easy way to see why a compact complex surface that admits a Kodaira fibration is Kahler? I know for a complex compact surface is Kahler if and only if its first Betti number is even. I wonder it's possible to deduce that a compact complex surface that admits a Kodaira fibration has even Betti number?
What is a Kodaira fibration?
@abx I added a definition
It depends on what you call easy... A standard result in surface theory is that a surface with a line bundle $L$ such that $c_1(L)^2>0$ is projective. Consider a fiber $F$ of your fibration; its genus $g$ is $\geq 2$, hence by the genus formula $(K_X\cdot F)=2g-2>0$. Then $(K_X+nF)^2>0$ for $n\gg 0$, hence the result.
@abx Sorry why is $g>1$?
If $g=1$ the $j$-invariant of the fibers must be constant, hence $X$ is an elliptic fiber bundle. I assumed that you exclude this case — otherwise, as observed by Nick L, you can get non Kähler surfaces.
All Kodaira fibrations are projective algebraic. This is standard material, see the book by Barth, Hulek, Peters, Van de Ven, beginning of chapter V.14 (same argument as abx, supplemented by the fact that indeed g>1 by Theorem III.15.4)
@YangMills thank you, that helps a lot!
Let $f \colon S \longrightarrow B$ be a Kodaira fibration, and let $F$ be a general fibre. Then by [Kas68, Thm. 1.1] we have $g(B) \geq 2$ and $g(F) \geq 3$.
In particular, $S$ contains no rational or elliptic curves: in fact, such curves cannot neither dominate the base (because $g(B) \geq 2$) nor be contained in fibres (because the fibration is by assumption smooth).
So every Kodaira fibred surface $S$ is minimal and, by the superadditivity of the Kodaira dimension, it is of general type.
In particular, it is not only Kähler but actually algebraic (i.e., projective).
References.
[Kas68] A. Kas: On deformations of a certain type of irregular algebraic surface, American J. Math. 90, 789-804 (1968). ZBL0202.51702.
Can you explain the theorem 1.1 of Kas' paper, why the map $U\to T$ is holomorphic?
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2025-03-21T14:48:29.814308
| 2020-02-09T03:38:43 |
352255
|
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|
Stack Exchange
|
Consequences of existence of a certain function from $\omega_1$ to $\omega_1$
In his book [1], Paul Larson remarks (Remark 1.1.22) that in L there is a function $h:\omega_1\rightarrow\omega_1$ such that for any countable elementary submodel $X$ of $V_\gamma$ (where $\gamma$ is the first strong limit cardinal), we have the order-type of $X\cap Ord$ is strictly less than $h(X\cap\omega_1)$.
What is known about the relationship between the existence of such a function and large cardinal phenomena?
Larson remarks that the existence of such a function is consistent with many large cardinals, and later in the book sketches a result of Velickovic showing that no such function exists in the presence of a precipitous ideal on $\omega_1$. What more is known? Happy to also learn results concerning other related functions, or be pointed to associated references.
[1] Larson, Paul B., The stationary tower. Notes on a course by W. Hugh Woodin, University Lecture Series 32. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3604-8/pbk). x, 132 p. (2004). ZBL1072.03031.
I don't know if Boban's argument in Larson's book is the same as that in his paper below, but I think it is worth taking a look at least at the introduction of his paper since in his forcing construction he uses some approximations of height functions, which I think resemble things you are looking for.
www.logique.jussieu.fr/~boban/pdf/PFA_and_NS.pdf
Seems related to bounding by canonical functions (the example you pointed out in $L$ is a function that is, in a very strong way, NOT bounded by a canonical function). Some relevant sources might be: Deiser-Donder "Canonical functions..."; Schimmerling-Velickovic "Collapsing functions"; Jech-Shelah "A note on countable functions".
Yes, I think the canonical function angle is exactly the sort of thing I was looking for!
In the paper "The consistency strength of the perfect set property for universally Baire sets of reals", Ralf Schindler and I show that the negation of a very similar statement is equiconsistent with the existence of what we call a virtually Shelah cardinal.
We say that a cardinal $\kappa$ is virtually Shelah if for every function $f:\kappa \to \kappa$ there is an ordinal $\lambda > \kappa$, a transitive set $M$ with $V_\lambda \subset M$, and a generic elementary embedding $j : V_\lambda \to M$ with critical point $\kappa$ and $j(f)(\kappa) \le \lambda$.
(By "generic elementary embedding" we mean an elementary embedding between structures in $V$ that exists in some generic extension of $V$. By absoluteness it suffices to consider any generic extension in which the domain structure is countable.)
This is a fairly weak large cardinal hypothesis. If $0^\sharp$ exists, then every Silver indiscernible is a virtually Shelah cardinal in $L$. On the other hand, every virtually Shelah cardinal is ineffable and a limit of ineffable cardinals.
By Theorem 1.2 of our paper, the following statements are equiconsistent modulo ZFC:
There is a virtually Shelah cardinal.
Every universally Baire set of reals has the perfect set property.
For every function $f : \omega_1 \to \omega_1$ there is an ordinal $\lambda > \omega_1$ such that for a stationary set of $\sigma \in \mathcal{P}_{\omega_1}(\lambda)$ we have $\sigma \cap \omega_1 \in \omega_1$ and the order type of $\sigma$ is at least $f(\sigma \cap \omega_1)$.
More specifically, we showed:
Statement 1 implies that after forcing with $\text{Col}(\omega,\mathord{<}\kappa)$ where $\kappa$ is a virtually Shelah cardinal, statement 2 holds.
Statement 2 implies statement 3.
Statement 3 implies that $\omega_1^V$ is a virtually Shelah cardinal in $L$.
I would also be interested to hear what else is known about statement 3 or the related statement in your question. In the paper, we only considered statement 3 because it came up naturally as a convenient intermediary between statements 1 and 2.
In the terminology of the Schimmerling-Velickovic "Collapsing functions" paper mentioned in Sean's comment, statement 3 above becomes "there is no collapsing function for Ord".
I’d love to accept both answers, but have to choose one!
Not sure if this is what you're interested in, but let me take up the question of what large cardinal axioms are known to be consistent with the existence of $h$. The answer is all large cardinal axioms known to have a canonical inner model. Beyond that, the consistency question remains open: indeed, it remains open whether large cardinal axioms beyond the reach of inner model theory could outright imply the existence of a precipitous ideal on $\omega_1$, and thereby refute the existence of $h$.
It seems plausible, however, that $h$ in fact exists in the canonical inner models yet to be discovered. This is because one can construct a function that is almost exactly like $h$ (see the fourth-to-last paragraph below) using a general condensation principle due to Woodin called Strong Condensation, which likely holds in all canonical inner models. If $\kappa$ is a cardinal, Strong Condensation at $\kappa$ states that there is a surjective function $f : \kappa\to H(\kappa)$ of such that for all $M\prec (H(\kappa),f)$, letting $(H_M,f_M)$ be the transitive collapse of $(M,f)$, $f_M = {f}\restriction H_M$.
All the known canonical inner models satisfy Strong Condensation at their least strong limit cardinal. (Strong condensation cannot hold at any cardinals past the first Ramsey cardinal.) Moreover by a theorem of Woodin, under $\text{AD}^+ + V = L(P(\mathbb R))$, for a Turing cone of reals $x$, $\text{HOD}_x$ satisfies Strong Condensation at its least strong limit cardinal. This heuristically argues that Strong Condensation should hold at the least strong limit cardinal in canonical inner models satisfying arbitrarily strong large cardinal axioms, assuming that such models exist. The reason is that the pattern observed in inner model theory to date suggests that these models should locally resemble the $\text{HOD}$s of determinacy models.
Here is the approximation to the existence of $h$ one gets assuming Strong Condensation at the least strong limit cardinal $\gamma$: There is a set $a\subseteq \omega_1$ and a function $g:\omega_1\to\omega_1$ such that for any $N\prec V_\gamma$ with $a\in N$, $g(N\cap \omega_1) > \text{ot}(N\cap \gamma)$. The rest of this answer consists of a proof of this fact.
Fix $f : \gamma\to H(\gamma)$ witnessing Strong Condensation at $\gamma$. The first step of the proof is cosmetic. One uses a theorem of Woodin which states that $f$ is definable over $H(\gamma)$ from the parameter $f \restriction \omega_1$. Let $a\subseteq \omega_1$ code $f\restriction \omega_1$. (It is an easy exercise to show that $f\restriction \omega_1\in H(\omega_2)$.) Then every $N\prec V_\gamma$ with $a\in N$ has the property that $N\cap H(\gamma)\prec (H(\gamma),f)$.
For $\alpha < \gamma$, let $P_\alpha = f[\alpha]$. Note that the $P_\alpha$ are increasing with union $H(\gamma)$, and if $M\prec (H(\gamma),f)$ and $\text{ot}(M\cap \gamma) = \alpha$, then the transitive collapse of $M$ is equal to $P_\alpha$. The structures $P_\alpha$ will play the role of the $L_\alpha$ hierarchy in Larson's proof.
For every $\xi < \omega_1$, let $g(\xi)$ be the least ordinal $\alpha$ such that there is a surjection from $\omega$ to $\xi$ in $P_\alpha$. Suppose $N\prec V_\gamma$ and $a\in N$. We will show that $g(N\cap \omega_1) > \text{ot}(N\cap \gamma)$. Let $M = N\cap H(\gamma)$, so $M\prec (H(\gamma),<)$. Clearly it suffices to show that $g(M\cap \omega_1) > \text{ot}(M\cap \gamma)$. Let $H_M$ be the transitive collapse of $M$. Then $M\cap \omega_1 = \omega_1^{H_M}$ and letting $\beta = \text{ot}(M\cap \gamma)$, $H_M = P_\beta$. Assume towards a contradiction that $\beta \geq g(\omega_1^{H_M})$. By the definition of $g$, there is a surjection from $\omega$ to $\omega_1^{H_M}$ in $P_\beta$. But since $P_\beta = H_M$, this contradicts that $\omega_1^{H_M}$ is uncountable in $H_M$.
Thank you for this contribution!
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2025-03-21T14:48:29.814843
| 2020-02-09T04:07:45 |
352258
|
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|
Stack Exchange
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Measure, volume and cardinality on Minlos' book on statistical physics
The following content was based on Minlos' book on statistical physics. Let $\Lambda \subset \mathbb{R}^{d}$ be fixed (Minlos takes $d=3$ but I think the ideas follow without change to $d \ge 1$). We denote by $\Lambda^{N}$ the $N$-fold cartesian product of $\Lambda$ with itself and $(\Lambda^{N})'$ $N$-uples $(x_{1},...,x_{N})$ in $\Lambda$ with different entries, i. e. $x_{i}\neq x_{j}$ if $i\neq j$. Also $\Gamma_{\Lambda, N}:=\{\omega \subset \Lambda, \hspace{0.1cm} \mbox{card}(\omega) = N\}$, where $\mbox{card}(\omega)$ is the cardinality of the set $\omega$. Define $\Pi: (\Lambda^{N})' \to \Gamma_{\Lambda, N}$ by $(x_{1},...,x_{N}) \mapsto \{x_{1},...,x_{N}\}$. For every subset $A$ of $\Gamma_{\Lambda, N}$, Minlos set:
$$ \mu_{\Lambda}^{(N)}(A) := \frac{\mbox{Vol}(\Pi^{-1}(A))}{N!} $$
Then, he states that
$\mu_{\Lambda}(\Gamma_{\Lambda, N}) = \frac{|\Lambda|^{N}}{N!}$. The problem is that he doesn't seem to define $\mbox{Vol}$ or $|\cdot|$ anywhere and it is getting me a little confused. At first, I thought $\mbox{Vol}$ was just the Lebesgue measure on $\mathbb{R}^{dN}$. But it would be a little odd because if I take $A = \{x_{1},...,x_{N}\}$ it seems that $\mu_{\Lambda}^{(N)}(\{x_{1},...,x_{N}\}) = 0$. Besides, how come does the second statement about $\mu_{\Lambda}(\Gamma_{\Lambda, N})$ follow? If $|\Lambda|$ is the cardinality of $\Lambda$ (which I don't know for sure), does it follow from de definition of $\mu_{\Lambda}$?
Note that there are pedagogically better sources to read about such topics. For instance, these lecture notes by S. Adams. In the lattice case, I'd recommend our book.
If $\text{Vol}$ denotes the Lebesgue measure on $(\mathbb R^d)^N$, if $|\cdot|$ denotes the Lebesgue measure on $\mathbb R^d$, and if $\mu_\Lambda=\mu_\Lambda^{(N)}$, then indeed $\mu_\Lambda(\Gamma_{\Lambda,N})=\frac{|\Lambda|^N}{N!}$. This follows because (i) $\Pi^{-1}(\Gamma_{\Lambda,N})=(\Lambda^N)'$ and (ii) (say, by the Fubini--Tonelli theorem) $\text{Vol}((\Lambda^N)')=\text{Vol}(\Lambda^N)=|\Lambda|^N$.
As for $\mu_\Lambda^{(N)}(A)$ with $A=\{x_{1},...,x_{N}\}$, it is undefined. Indeed, for $\mu_\Lambda^{(N)}(A)$ be defined, $A$ must be a subset of $\Gamma_{\Lambda,N}$, whereas $\{x_{1},...,x_{N}\}$ is a subset of $\mathbb R^d$ but not of $\Gamma_{\Lambda,N}$. On the other hand, if $A$ is the singleton set
$\{(x_{1},...,x_{N})\}$ with pairwise distinct $x_j$'s in $\mathbb R^d$, then indeed $\mu_\Lambda^{(N)}(A)=0$, and there is nothing odd about that.
This is a very good answer! In summary, it looks like $|\cdot|$ is not the cardinality of the set, but it $|\cdot|$ and $\mbox{Vol}$ seem to be two Lebesgue measures on different spaces. It makes sense to me!
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2025-03-21T14:48:29.815029
| 2020-02-09T04:34:07 |
352260
|
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|
Stack Exchange
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Is the set of non-escaping points in a Julia set always totally disconnected?
I am looking for examples of transcendental entire functions $f:\mathbb C\to \mathbb C$ such that the set of non-escaping points in the Julia set of $f$ is not totally disconnected. I denote this set $J_r(f)=\{z\in J(f):f^n(z)\not\to\infty\}$ because in some interesting cases it is the same as the "radial Julia set'' of $f$.
For two types functions, including the exponential family of $\exp(z)-2$, I have recently shown that $J_r(f)$ is not only totally disconnected, but is zero-dimensional in the topological sense: my paper. The Julia sets of these functions have a relatively simple "Cantor bouquet'' structure which was used in the proofs.
Question. Can $J_r(f)$ contain non-degenerate or unbounded connected sets?
If $f$ has order $<1/2$ then there is a sequence $r_k\to\infty$ with the property that
$$\min_{|z|=r_k}|f(z)|>r_k.$$
Restricting $f$ on $\{ z:|z|<r_k\}$ we obtain a polynomial-like map in the sense of Douady and Hubbard. If $J_k$ is the Julia set of this map, then evidently $J_k\subset J(f)$, and the points of $J_k$ are not escaping. Now, if $f$ has an attracting cycle, then for large $k$, $J_k$ contains the boundary of the attraction
domain of this cycle, which is a continuum. Thus $J(f)$ contains a continuum consisting of non-escaping points.
Thank you for answering my question. Just to make sure I understand, could you provide a few concrete examples?
Examples of functions of order less than $1/2$? Plenty. Take an appropriate power series, for example. To arrange an attracting cycle, multply on $z^2$.
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2025-03-21T14:48:29.815186
| 2020-02-09T07:03:55 |
352267
|
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|
Stack Exchange
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On the proportion of simplicial $d$-polytopes on $n$-vertices
I have a question regarding estimates for the proportion of simplicial $d$-polytopes on $n$-vertices.
Let $c_s(n,d)$ denote the number of combinatorial types of simplicial $d$-polytopes on $n$ labelled vertices and let $c(n,d)$ denote the number of combinatorial types of (general) $d$-polytopes on $n$ labelled vertices. I am interested in estimates for the following limits:
For a fixed $d$, $\lim_{n\to\infty} \frac{c_s(n,d)}{c(n,d)}$, and
For a fixed $n=g(d)$, say $n=d+3$ for instance, $\lim_{d\to\infty} \frac{c_s(n,d)}{c(n,d)}$.
Is this known? Any help would be much appreciated.
Thank you very much in advance, and best regards,
Guillermo
I assume that $d$-polytope means full-dimensional $d$-dimensional convex polytope. Then $c_s(n,d) = c(n,d)=0$ if $d>n-1$, and so the second limit is not meaningful.
Good point M. Winter. I was thinking of n=g(d), say n=d+3. I have made that clear.
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2025-03-21T14:48:29.815278
| 2020-02-09T08:30:20 |
352271
|
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|
Stack Exchange
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Events of supremum and infimum of random variables
Let $Y_n$ be a sequence of real random variables, and let $c \in \mathbb{R}$ be some constant. I know that $\inf_n Y_n$ and $\sup_n Y_n$ are also random variables. I'm interested in trying to represent events of these sup and inf random variables as unions or intersections of events of $Y_n$. I've come up with the following relations intuitively, but I'm not a 100% sure that they are true:
$\{\sup_n Y_n > c\} = \bigcup_n \{Y_n > c\}$
$\{\inf_n Y_n > c\} = \bigcap_n \{Y_n > c\}$
and
$\{\sup_n Y_n < c\} = \bigcap_n \{Y_n < c\}$
$\{\inf_n Y_n < c\} = \bigcup_n \{Y_n < c\}$
Additionally, if these are true, in which cases can the $>$ or $<$ be relaxed into $\geq$ or $\leq$?
$\{\text{inf}_n Y_n > c \}= \bigcup_k \bigcap_n \{Y_n > c +\frac{1}{k} \}$. And similar for $\{\text{sup}_n Y_n < c\}$.
$\{\text{inf}_n Y_n \geq c \}=\bigcap_n \{Y_n \geq c \}$ is true, but $\{\text{sup}_n Y_n \geq c\} =\bigcap_k \bigcup_n \{Y_n \geq c -\frac{1}{k} \}$ and similar for $\{\text{inf}_n Y_n \leq c \}$.
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2025-03-21T14:48:29.815375
| 2020-02-09T10:45:18 |
352275
|
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|
Stack Exchange
|
Vertex connectivity of join of two graphs
Does there exist any results on the vertex connectivity of join of two graphs?
If $G_1$ and $G_2$ are two graphs what can we say about the vertex connectivity of $G_1\lor G_2$ where $G_1\lor G_2$ is the join of $G_1$ and $G_2$? The join of $G_1$ and $G_1$ is the graph obtained from the disjoint union of $G_1$ and $G_2$ by adding all edges between $V(G_1)$ and $V(G_2)$.
If $\kappa(G)$ denotes vertex connectivity of a graph $G$ how is $\kappa(G_1\lor G_2)$ related to $\kappa(G_1)$ and $\kappa(G_2)$?
How is the join defined?
@M.Winter; https://math.stackexchange.com/a/1769191/665065; check this
The vertex connectivity of $G_1 \vee G_2$ is $\min \{|V(G_1)|+\kappa(G_2), |V(G_2)|+\kappa(G_1)\}$. To see this, note that if $X \subseteq V(G_1 \vee G_2)$ is such that $V(G_1) \setminus X$ and $V(G_2) \setminus X$ are both non-empty, then $(G_1 \vee G_2) - X$ is connected. Thus, a smallest vertex separator $X$ of $G_1 \vee G_2$ must be of the form $V(G_1) \cup X_2$ or $V(G_2) \cup X_1$, where $X_i$ is a smallest vertex separator of $G_i$.
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2025-03-21T14:48:29.815466
| 2020-02-09T10:49:23 |
352276
|
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|
Stack Exchange
|
Functorial interpretation of formal completion
Let $S$ be a scheme, $X/S$ a "nice" scheme (I think "nice" = separated) and $e : S \to X$ a section.
Let $\hat{X}$ be the formal completion of $X$ along the section $e$.
(i.e., = $\lim_n S_n$, where $S_n$ is the "$n$-th-thickened of $S \to X$".)
Then does $\hat{X}$ represent (or pro-represent) a nice functor?
First, the case that $S$ is a field $k$:
Let $\mathscr{C}_k$ be a category of artin local $k$-algebras with the residue field $k$. (Maps are local $k$-homomorphisms.) and
$\hat{\mathscr{C}}_k$ be a category of complete local noetherian $k$-algebra with the residue field $k$.
And consider a functor
$$\mathscr{C}_k \to \mathscr{S}et \\
A \mapsto \{ f : \operatorname{Spec}A \to X, \text{ an $S$-morphism which takes the only closed point to $e$ } \}. $$
Then this is $\operatorname{Hom}_{\mathscr{C}_k}(\mathscr{O}_{X,e}, A) = \operatorname{Hom}_{\hat{\mathscr{C}}_k}( \hat{\mathscr{O}_{X,e}}, A) = \hat{X}(A).$
So this functor is pro-represented by $\hat{X}$.
And if $X$ is a group scheme, then this functor is $\operatorname{ker} (X(A) \to X(A/\mathfrak{m}_A))$, so inherits the group structure of $X$.
How about the general case?
Here is what I have tried:
Let $\mathscr{C}_S$ be the full subcategory of $\mathscr{S}ch/S$ whose objects are the spectrum of artin local ring.
Then by the same arguments, it seems that the functor
$$ \mathscr{C}_S^{\text{op}} \to \mathscr{S}et \\
T \mapsto \{ f : T \to X, \text{ an $S$-morphism whose set-theoritically image is contained in $e(S)$} \}$$
is equals to $\hat{X}(T)$.
However $\hat{X}$ is not in " $\hat{\mathscr{C}}_S$ ", so this functor is "bad" in order to study $\hat{X}$.
If $S$ is affine and if we take $\mathscr{C}_S$ to be the category of finite $S$-schemes of finite length, then it seems to $\hat{X} \in \hat{\mathscr{C}_S}$.
But I cannot show that $\hat{X}$ pro-represents that functor,
and this does not coincide with the case that $S$ is a field.
Thank you very much!
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2025-03-21T14:48:29.815763
| 2020-02-09T11:10:07 |
352279
|
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"url": "https://mathoverflow.net/questions/352279"
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|
Stack Exchange
|
Completely positive, unital maps acting on unitary operators [solved]
Call a completely positive, irreducible, unital map $E$ on the operators on a finite-dimensional Hilbert-space primitive if there is only a single eigenvalue with modulus 1 (all others have modulus <1) and furthermore the corresponding eigenspace is one-dimensional. This eigenvalue is then necessarily equal to 1 and the eigenspace spanned by the identity operator. Irreducible in this case means that the (unique) fixed point $\rho$ of the Hilbert-Schmidt dual $E^*$ is strictly positive.
(This last clarification was added after I thought to have found an example to my question below, which, however, was not irreducible. See below.)
Now consider such a map $E$ and suppose that there exists a unitary and hermitian operator $U$ (hence $U^2=1$) that is itself mapped to a unitary (and hermitian) operator by $E$.
If it is mapped to a multiple of itself, then $U=\pm 1$ by primitivity, so let's assume that it is mapped to a unitary that is not proportional to the identity.
My basic question is whether this is possible (i.e., an example of such a map $E$ would make me happy, as would a proof that $U$ would have to be $\pm 1$).
Note that by equality in Kadison-Schwarz, we have that $E[UX]=E[U]E[X]$ and $E[XU]=E[X]E[U]$ for all $X$. Since $U^2=1$, we further have $U=P-Q$, with $P$ and $Q$ orthogonal projectors such that $1=P+Q$. It's not very difficult to deduce that $E[P]^2=E[P]$ using equality in the Kadison-Schwarz inequality and similarly for $Q$. It follows that also $E[PX]=E[P]E[X]$ and similarly for $Q$ (and the opposite ordering). If one could show $E[P]\leq P$, then any $PYP$ would be mapped to $PE[PYP]P$. Since $E$ is primitive and hence irreducible, this would imply $P\in\{1,0\}$ and therefore $U=\pm 1$.
It is easy to construct examples of non-primitive, unital CP-maps where $E[P]\leq P$ is violated and yet $E[P]^2=E[P]$ for a non-trivial projector $P$. But for primitive ones I somehow did not succeed yet, although I would not be too surprised if there is a simple counter-example.
Any insight would be very much appreciated!
I asked this question previously on stack-exchange, but maybe I have more luck here. Maybe unsurprisingly, the problem appeared in an argument in quantum information theory, but has now become a problem on its own :-)
Update 2: The following provides an example that is not irreducible, but fulfills all other requirements.
Consider $\mathbb{C}^3$ as Hilbert-space with the Euclidean inner product $\langle \cdot,\cdot\rangle$.
Pick an orthonormal basis $e_i$ and choose $E$ as
$ E[A] = \begin{pmatrix}
\langle e_1,A e_1\rangle & 0&0\\
0& \langle e_1,A e_1 \rangle &0\\
0& 0& \frac{1}{2}(\langle e_2,A e_2\rangle + \langle e_3,A e_3\rangle)
\end{pmatrix}$
This is clearly unital and completely positive. A small calculation shows 1 is the unique eigenvalue with modulus 1 and the eigenspace is one-dimensional. However, it maps
$$\begin{pmatrix} 1 & 0 & 0\\
0 &-1 & 0\\
0 & 0 &-1
\end{pmatrix}
\mapsto
\begin{pmatrix} 1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{pmatrix}.$$
Note that the map is not irreducible, since the projector onto the space spanned by $e_1$ is an eigenvector of $E^*$.
Update 3: The above example can in fact be generalized to the irreducible case. To do this, follow the above map $E$ with conjugation by a unitary $V$ that maps the basis $\{e_i\}$ to a basis that is mutually unbiased with respect to the $\{e_i\}$:
$$\tilde E[A]=V E[A]V^*.$$
Then the map becomes irreducible, since the only fixed point of $\tilde E^*$ (up to normalization) is $\mathrm{diag}(2/3,1/6,1/6)$ in the basis $\{e_i\}$, but $\tilde E$ maps
$$\begin{pmatrix} 1 & 0 & 0\\
0 &-1 & 0\\
0 & 0 &-1
\end{pmatrix}
\mapsto
V \begin{pmatrix} 1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{pmatrix} V^*, $$
showing that indeed it is possible to find an example with $U\neq \pm 1$.
What is a primitive map?
An irreducible positive map with trivial peripheral spectrum, i.e., where 1 is the unique, non-degenerate eigenvalue with modulus 1. I realize this might not be standard terminology and have edited the post accordingly.
When you say non-degenerate eigenvalue, do you mean "the eigenspace is one-dimensional"? (Not complaining about your terminology, just wanted to make sure)
Yes, that's right.
sorry to keep nitpicking, but in the new version, do you mean the following: (a) 1 is an eigenvalue, and all other eigenvalues have modulus less than 1, and the eigenspace of the eigenvalue 1 is dimensional; or (b) there is only one eigenvalue with the properties "has modulus 1 and has 1-dimensional eigenspace", and that eigenvalue is 1 ?
I meant option (a). I have rephrased the question and hope that it is clear now.
A simple example where $E[P] \not\leq P$ is the following: Let $E: \mathbb{C}^{2 \times 2} \to \mathbb{C}^{2 \times 2}$ be given by $E[A] = \frac{1}{2}\begin{pmatrix} \operatorname{tr}(A) & 0 \ 0 & \operatorname{tr}(A) \end{pmatrix}$. For $P = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}$ we then have $E[P] \not\le P$. Unfortunately, this does not answer your original question, though.
While this example doesn't answer my problem, it lead me to finding a simple example that does answer my problem, so thanks a lot! :-)
@HenrikWilming: That's funny: for the last 30 minutes or so I was quite convinced that such an example could not exist and that I only had to find a proof for this... :-) Anyway, you're welcome! (And I really like your counterexample!)
@JochenGlueck: And in fact you might still be right! I just realized that my example is not irreducible, as I was asking for (see updated question).
@HenrikWilming: Oops... I think I was buying to easily your former claim that primitivity plus the existence of a "strictly positive" fixed point implies irreducibility. This is, for instance, indeed true for non-negative contractive matrices that operate on $\mathbb{C}^n$ endowed with the $\ell^1$-norm; but it is not true on $C^*$-algebras (as your example shows).
@JochenGlueck: Yes, I think I was fooling myself similarly. But the newly added example solves this loophole.
|
2025-03-21T14:48:29.816168
| 2020-02-09T12:52:02 |
352282
|
{
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"Mark Wildon",
"https://mathoverflow.net/users/38068",
"https://mathoverflow.net/users/7709",
"spin"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352282"
}
|
Stack Exchange
|
Homomorphism induced by the second exterior power of a linear map
Consider the map from $M(n, \mathbb Z) \rightarrow M(\binom{n}{2}, \mathbb Z)$ taking a matrix A to its second compound, i.e, $\bigwedge^2 A$.
Restricting this map to the invertible matrices we get a homomorphism of groups from $\mathrm{GL}(n, \mathbb Z)$ to $\mathrm{GL}(\binom{n}{2}, \mathbb Z)$.
How can we determine if a given matrix $B \in \mathrm{GL}(\binom{n}{2}, \mathbb Z)$ is contained in the image of this map?
First, let us discuss the same question over an algebraically closed field (e.g., over $\overline{\mathbb{Q}}$). Let $V$ be a vector space of dimension $n$. The question is to understand the image of the homomorphism
$$
\lambda \colon \mathrm{GL}(V) \to \mathrm{GL}(\wedge^2V).
$$
Note that $\mathrm{GL}(\wedge^2V)$ acts naturally on the projective space $\mathbb{P}(\wedge^2V)$, which contains as a subvariety the Grassmannian
$$
\mathrm{Gr}(2,V) \subset \mathbb{P}(\wedge^2V).
$$
Clearly, it is preserved by the action of $\mathrm{GL}(V)$. The converse is also true for $n > 4$, i.e., if $g \in \mathrm{GL}(\wedge^2V)$ is such that
$$
g(\mathrm{Gr}(2,V)) \subset \mathrm{Gr}(2,V),
$$
then $g \in \mathrm{Im}(\lambda)$. This follows immediately from the isomorphism
$$
\mathrm{Aut}(\mathrm{Gr}(2,V)) \cong \mathrm{PGL}(V).
$$
Over $\mathbb{Z}$, I guess, the equality
$$
\det(\wedge^2g) = \det(g)^{n-1}
$$
gives an extra constraint; so besides preserving the Grassmannian one should impose the condition that the determinant is $(n-1)$-st power.
Why is the combination of the Plücker relations (for the Grassmannian) and the determinant condition sufficient?
To be invertible over $\mathbb{Z}$, the determinant should be $\pm 1$.
|
2025-03-21T14:48:29.816326
| 2020-02-09T14:12:04 |
352286
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"John Greenwood",
"Qfwfq",
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"url": "https://mathoverflow.net/questions/352286"
}
|
Stack Exchange
|
When is a virtual bundle an actual bundle?
As far as I understand, one can make a monoid from the space of vector bundles on a (compact) manifold $M$, with respect to the direct sum operation $\oplus$. In order to make this into a group, one needs to add 'inverses' and this leads to the consideration of formal differences $V-W$ of vector bundles. One can again compute Euler classes, Chern classes, etc... for these objects called 'virtual bundles'.
Now my question is:
When is such an object an actual vector bundle? Is there an easy criterion to check? For instance, if I have a complex virtual vector bundle and I know that its Chern classes are integral i.e. in $H^*(M,\mathbb{Z})$, rather than $H^*(M,\mathbb{Q})$, is that sufficient?
Disclaimer: I am only familiar with basic aspects of $K$-theory, so the above question might be completely trivial, but I couldn't find an answer.
Let $c$ denote the total Chern class. Then the Whitney formula is $c(E\oplus F)=c(E)c(F)$. So, to be consistent, I think you should have $c(E-F)=c(E)c(F)^{-1}=c(E)\cdot 1/(1+c_1(F)+c_2(F)+\ldots + c_r(F))=c(E)\sum_{k\geq 0}(-1)^k(\sum_{i=1}^r c_i(F))^k$ which seems integral.
@Qfwfq 's argument shows that it is necessary to have $c_i(E)=0$ for $i$ large enough. But it is clearly not sufficient because any bundle of the form $V-nC$, where $nC$ is the trivial $n$-dimensional complex bundle verifies this.
Over a compact space every map $f: X \to BO$ factors through a finite-dimensional Grassmannian. In particular every virtual v.b. is of the form $V - n$ for some vector bundle $V$, and $n$ is determined by the rank of the virtual vb.
And once you've used Mike Miller's observation, a necessary condition is that the last chern classes of are zero. If the rank and the dimension of the manifold interact well you can appeal to some obstruction theory, but I think it's a pretty hard problem in general, since it's the same as asking for $n$ linearly independent sections of $V$.
|
2025-03-21T14:48:29.816496
| 2020-02-09T16:16:54 |
352291
|
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"url": "https://mathoverflow.net/questions/352291"
}
|
Stack Exchange
|
Completion and extension by scalars
Let $R\subset S$ be commutative rings, $I\trianglelefteq R$ an ideal and $M$ be an $R$-module. Suppose that
1) $R$ is Noetherian and $I$-adically complete.
2) $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)
3) $S$ is a flat $R$-algebra.
4) $S$ is $I$-adically complete
5) $M/IM$ is free module over $R/I$,
Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?
I am only able to prove the above under the assumption that
6) $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$
in the following way:
Consider a resolution of $M$ in $R$-$\operatorname{Mod}$
$$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$
by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence
$$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$
using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences
$$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$
by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude
$\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel
of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.
Can one show this without assumption 6) at least for the case when $I$ is principal, perhaps involving the second half of condition 1) and condition 5) ?
This follows by adapting the proof of Tag 00MA, even without assumption 5. We also don't need $S$ to be an algebra; a complete $R$-module suffices. Finally, we never use that $R$ is $I$-adically complete!
Indeed, by assumption 2 there exists a short exact sequence $0 \to K \to F \to M \to 0$ with $F$ finite free. By assumption 3, the sequence
$$0 \to K_S \to F_S \to M_S \to 0$$
is exact, where $(-)_S = (-) \otimes_R S$. For each $n$, this gives a short exact sequence
$$0 \to K_S/(I^nF_S \cap K_S) \to F_S/I^nF_S \to M_S/I^nM_S \to 0.$$
Since all systems satisfy the Mittag-Leffler condition, the limit sequence
$$0 \to \lim_{\substack{\longleftarrow \\ n}} K_S/(I^nF_S \cap K_S) \to (F_S)^\wedge \to (M_S)^\wedge \to 0\label{Eq 1}\tag{1}$$
is exact. By the Artin–Rees lemma, there exists $c \geq 0$ such that $I^nK \subseteq I^nF \cap K \subseteq I^{n-c}K$ for all $n \geq c$. By flatness of $S$, for any $R$-module $N$ and any $n \in \mathbf Z_{\geq 0}$, the surjection
$$N_S \twoheadrightarrow N_S/I^nN_S \cong (N/I^nN)_S$$
identifies $I^nN_S$ with $(I^nN)_S$ as submodules of $N_S$. Similarly, the map
$$F_S \to F_S/I^nF_S \oplus F_S/K_S \cong (F/I^nF \oplus F/K)_S$$
identifies $I^nF_S \cap K_S$ with $(I^nF \cap K)_S$ as submodules of $F_S$. Thus, we conclude that
$$I^nK_S \subseteq I^nF_S \cap K_S \subseteq I^{n-c}K_S$$
for all $n \geq c$. In particular, (\ref{Eq 1}) reads as
$$0 \to (K_S)^\wedge \to (F_S)^\wedge \to (M_S)^\wedge \to 0.$$
We now get a commutative diagram with exact rows
$$\begin{array}{ccccccccc}0 & \to & K_S & \to & F_S & \to & M_S & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & (K_S)^\wedge & \to & (F_S)^\wedge & \to & (M_S)^\wedge & \to & 0.\! \end{array}$$
By assumption 4, the middle vertical arrow is an isomorphism. This immediately implies that the right vertical arrow is surjective, and applying the same reasoning to $K$ gives the same statement for the left vertical arrow. Then the five lemma shows that all vertical maps are isomorphisms. $\square$
|
2025-03-21T14:48:29.816721
| 2020-02-09T17:49:04 |
352294
|
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"authors": [
"ACR",
"Piyush Grover",
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"https://mathoverflow.net/users/30186",
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}
|
Stack Exchange
|
Earliest reference on the calculation of derivatives by Fourier transform
I was looking for an earliest reference or the name of the mathematician who showed calculating the derivatives is possible in the Fourier domain?
The Fourier transform of the derivative is (Wikipedia)
$$
\mathcal{F}(f')(\xi)=2\pi i\xi\cdot\mathcal{F}(f)(\xi).
$$
Thank you.
Has to be fourier since he invented this to solve a differential eqn
To compute $f'$ this way, you have to first compute a Fourier transform and then compute an inverse Fourier transform, which is quite a bit of work to do. Ignoring historical firstness, what references do you have for computing derivatives this way?
Wojowu, I am a chemist by training. Our instruments generate data as time vs. signal. In that case, this is the only way to determine the derivative using FT method as far as I know and one can include smoothing before inverse FT. Is there a better way to calculate derivatives of the data $using$ FT methods?
Despite this being complicated by Fourier’s use of sine and cosine transforms, I think @PiyushGrover is right: Théorie analytique de la chaleur, Art. 419:
$$
\begin{align}
\frac{d^{2i}}{dx^{2i}}fx
=\pm&\int d\alpha f\alpha\int dp\,p^{2i}\cos.(px-p\alpha)\\
\frac{d^{(2i+1)}}{dx^{2i+1}}fx
=\mp\frac1{2\pi}&\int d\alpha f\alpha\int dp\,p^{2i+1}\sin.(px-p\alpha)
\end{align}
$$
and Art. 422 “Expression générale de la fluxion d’ordre $i$”.
|
2025-03-21T14:48:29.816851
| 2020-02-09T17:51:22 |
352295
|
{
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"authors": [
"David Roberts",
"Maksym Voznyy",
"Noam D. Elkies",
"https://mathoverflow.net/users/14830",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/95511"
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"sort": "votes",
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|
Stack Exchange
|
Hard: One more generator needed for a Z/6 elliptic curve
We are searching for rank 8 elliptic curves with the torsion subgroup $\mathbb{Z}/6$ using newly discovered families similar to Kihara's as described in
A. Dujella, J.C. Peral, P. Tadić, Elliptic curves with torsion group $\mathbb{Z}/6\mathbb{Z}$, Glas. Mat. Ser. III 51 (2016), 321-333 doi:10.3336/gm.51.2.03, 1503.03667
and came across a curve
[1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]
Both Magma Calculator and mwrank return $7$ generators for this curve:
SetClassGroupBounds("GRH");
E:=EllipticCurve([1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]);
MordellWeilShaInformation(E);
Both Magma and mwrank return $8$ for the upper bound on rank:
E:=EllipticCurve([1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]);
TwoPowerIsogenyDescentRankBound(E);
8 [ 4, 4, 4, 4, 4 ]
[ 6, 6, 6, 6, 6 ]
mwrank -v0 -p200 -s
[1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]
Version compiled on Oct 29 2018 at 22:35:09 by GCC 7.3.0
using NTL bigints and NTL real and complex multiprecision floating point
Enter curve: [1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]
Curve [1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988] : selmer-rank = 9
upper bound on rank = 8
Considering parity, there should be one more generator on the curve.
Is there a way to find it?
We would greatly appreciate any hint leading to the discovery of the extra generator.
A bounty of $100$ has been offered for obtaining it.
Also, if you can compute an extra generator, your name will be published at the bottom of the page here: https://web.math.pmf.unizg.hr/~duje/tors/z6.html
Did you try to use FourDescent as in Jeremy Rouse's accepted answer for your previous question of this kind (https://mathoverflow.net/questions/337621)?
Yes. Jeremy Rouse himself tried it for us with no extra generator produced. That is why "Hard: " was put in the title.
A set of points that generate $E(\mathbb{Q})$ modulo torsion is given by
(1955516573881233507049678279 : -86467145649172260650105545143411861089140 : 1),
(49225691888888099223656060329/10201 :<PHONE_NUMBER>5993353685065159554645568700902610/1030301 : 1),
(61339810590192565389735634 : -440289331793622522908840423931186017125 : 1),
(301884243790342804873202050999/1681 :<PHONE_NUMBER>03197903219089875947912899634054060/68921 : 1),
(12495717670305680867142229 : -24031745881863415519418908823242701040 : 1),
(48812081421189741670987918753619270029/14228919471376 : -3895612939954697213016286372117889003488190324193605593985/53673248632044722624 : 1),
(5561842419887590167868100830494509281/162696869449 :<PHONE_NUMBER>012663087305509196041719017978015930195439090/65624921170340293 : 1),
(-24644413733187137559835573003063695698428162289232517969749039/810893447144357785058346728220801409 :<PHONE_NUMBER>0076383865716266151756242512110696731502256770076024073253839003102120576612459770/730206486187013450403786627354716551758061149557632577 : 1)
My guess is that you were missing the final one.
We can find the final point by applying $4$-descent to the $2$-covering $C_2$ given by
$$y^2 =<PHONE_NUMBER>862039958255625x^4 -<PHONE_NUMBER>4576368607091450x^3 +<PHONE_NUMBER>86468859778411799x^2 +<PHONE_NUMBER>10722380421312754x +
<PHONE_NUMBER>8020667812024521$$
of $E$, which is one of the 511 $2$-coverings returned by running the command TwoDescent(E) magma. Running the command FourDescent(C2) returns 256 $4$-coverings of $E$. One of them is the curve $C_4$ defined by the intersection of the quadrics
$$46500x^2 + 74693xy + 54170xz + 647076xw - 121026y^2 - 196538yz + 862965yw - 212375z^2 - 238791zw + 333744w^2$$
and
$$722768x^2 - 2936122xy + 3336517xz - 2782182xw - 2731148y^2 - 13360024yz - 950117yw - 4385375z^2 + 2688700zw - 199207w^2$$
in $\mathbb{P}^3(\mathbb{Q})$. Running PointsQI(C4,2^11) returns a single point $Q = (2834:53:2444:376)$. We can map $Q$ to a rational point on $E$ using the map given as the second return value of the command AssociatedEllipticCurve(C4:E:=E), and the point we get is the final point above.
The curve with all generators and credit to Zev Klagsbrun is published at the bottom of the page here: https://web.math.pmf.unizg.hr/~duje/tors/z6.html
Thanks @Martin for fixing a bug in my edit
|
2025-03-21T14:48:29.817079
| 2020-02-09T18:30:01 |
352296
|
{
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"Gerald Edgar",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352296"
}
|
Stack Exchange
|
Measure on a set and its value on $\emptyset$
After my first post here, I have one more doubt which is bothering me. It concerns Minlos's book Introduction to mathematical statistical physics again. To fix the notation, we have $\Lambda \subset \mathbb{R}^{d}$, and we set $C_{\Lambda}^{(N)}:=\{\omega \subset \Lambda, \lvert\omega\rvert = N\}$ where $\lvert\omega\rvert$ denotes the cardinality of $\omega$. Suppose $\mu_{\Lambda}^{(N)}$ is a measure on $\mathbb{P}(C_{\Lambda}^{(N)})$, the set of all subsets of $C_{\Lambda}^{(N)}$. As you can see here, Minlos introduces $C_{\Lambda}:=\bigcup_{N=0}^{\infty}C_{\Lambda}^{(N)}$ (with $C_{N}^{(0)} = \{\emptyset\}$) and his aim is to define a measure $\mu_{\Lambda}$ on $C_{\Lambda}$. But he states $\mu_{\Lambda}(\emptyset) = 1$. So, I have two questions concerning this definition. The first is: what does it mean to define a measure on a set? Should I assume the underlying $\sigma$-algebra is $\mathbb{P}(C_{\Lambda})$? Second, how can it be that $\mu_{\Lambda}(\emptyset) = 1$ if a measure has to satisfy $\mu_{\Lambda}(\emptyset) = 0$?
EDIT: I think I understood my second question. The point is that, in this case, $\emptyset$ is viewed as an element not a subset of the $C_{\Lambda}$.
Yeah, it seems like this is just slightly sloppy notation, and Minlos really means $\mu_{\Lambda}({\emptyset})$.
Thank you!! I'm new here!
Perhaps you copied something wrong ...$C_{\Lambda}^{(N)}:={\omega \subset \Lambda, \lvert\omega\rvert<+\infty}$ does not depend on $N$.
Yes. It should be $C_{\Lambda}^{(N)}:={\omega \subset \Lambda, |\omega| = N}$. I'll edit it.
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2025-03-21T14:48:29.817246
| 2020-02-09T18:33:11 |
352298
|
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|
Stack Exchange
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Could groups be used instead of sets as a foundation of mathematics?
Sets are the only fundamental objects in the theory $\sf ZFC$. But we can use $\sf ZFC$ as a foundation for all of mathematics by encoding the various other objects we care about in terms of sets. The idea is that every statement that mathematicians care about is equivalent to some question about sets. An example of such an encoding is Kuratowski's definition of ordered pair, $(a,b) = \{\{a\},\{a,b\}\}$, which can then be used to define the cartesian product, functions, and so on.
I'm wondering how arbitrary the choice was to use sets as a foundation. Of course there are alternative foundations that don't use sets, but as far as I know all these foundations are still based on things that are quite similar to sets (for example $\sf HoTT$ uses $\infty$-groupoids, but still contains sets as a special case of these).
My suspicion is that we could instead pick almost any kind of mathematical structure to use as a foundation instead of sets and that no matter what we chose it would be possible to encode all of mathematics in terms of statements about those structures. (Of course I will add the caveat that there has to be a proper class of whichever structure we choose, up to isomorphism. I'm thinking of things like groups, topological spaces, Lie algebras, etc. Any theory about a mere set of structures will be proved consistent by $\sf ZFC$ and hence be weaker than it.)
For concreteness I'll take groups as an example of a structure very different from sets. Can every mathematical statement be encoded as a statement about groups?
Since we accept that it is possible to encode every mathematical statement as a statement about sets, it would suffice to show that set theory can be encoded in terms of groups. I've attempted a formalization of this below, but I would also be interested in any other approaches to the question.
We'll define a theory of groups, and then ask if the theory of sets (and hence everything else) can be interpreted in it. Since groups have no obvious equivalent of $\sf ZFC$'s membership relation we'll instead work in terms of groups and their homomorphisms, defining a theory of the category of groups analogous to $\sf{ETCS+R}$ for sets. The Elementary Theory of the Category of Sets, with Replacement is a theory of sets and functions which is itself biinterpretable with $\sf ZFC$.
We'll define our theory of groups by means of an interpretation in $\sf{ETCS+R}$. It will use the same language as $\sf{ETCS+R}$, but we'll interpret the objects to be groups and the morphisms to be group homomorphisms. Say the theorems of our theory are precisely the statements in this language whose translations under this interpretation are provable in $\sf{ETCS+R}$. This theory is then recursively axiomatizable by Craig's Theorem. Naturally we'll call this new theory '$\sf{ETCG+R}$'.
The theory $\sf{ETCS+R}$ is biinterpretable with $\sf ZFC$, showing that any mathematics encodable in one is encodable in the other.
Question: Is $\sf{ETCG+R}$ biinterpretable with $\sf ZFC$? If not, is $\sf ZFC$ at least interpretable in $\sf{ETCG+R}$? If not, are they at least equiconsistent?
Is the category of Groups is not even a topos, which makes bi-interpretability very iffy (certainly very strange).
@cody I don't see what being a topos has to do with biinterpretability. For example the category of topological spaces isn't a topos, but one can interpret $\sf{ETCS+R}$ in $\sf{ETCTS+R}$, by looking at the full subcategory on the discrete spaces. One identifies the discrete spaces in the categorical language by saying that they are precisely the objects such that any morphism into them which is both monic and epic is an isomorphism.
What intuition would you be trying to capture by using groups as foundations? Also, I cannot imagine how you would axiomatise groups without first axiomatising their carriers. Groupoids and Homotopy Type Theory, well that would be a different matter.
@PaulTaylor The idea would be to demonstrate that the choice of sets as foundations was arbitrary (or at least motivated only be ease-of-use rather than necessity). Alternatively, if you want to show that the choice of sets wasn't arbitrary then you could prove that $\sf{ETCG+R}$ isn't biinterpretable with $\sf ZFC$.
As you can see from my other contributions to this site, I am not going to defend set theory. Indeed I said carriers not sets. But I challenge you to axiomatise (the category of) groups in whatever foundational system you like without first axiomatising their carriers.
I apologise for being unfriendly earlier. Apart from limits and colimits, the category of groups does not have any interesting categorical structure (so far as I am aware) that could stand in for function spaces or powersets. Abelian groups would be a better bet, since they form a symmetric monoidal closed category; maybe you could characterise the symmetric tensor algebra functor, which would be the start of something more interesting.
Could you confirm that you really mean to refer to bi-interpretability in the question, as opposed to mutual interpretability?
@JoelDavidHamkins When I say that $\sf{ETCS+R}$ is biinterpretable with $\sf ZFC$, I mean exactly that. (You have to get the language right for it to work out perfectly; morphisms have equality but objects don't.) In the questions at the end I ask for biinterpretability, then mutual interpretability, then equiconsistency, which I believe are in decreasing order of strength. Proving any of those would be interesting, but biinterpretability would be best. (Conversely disproving any of those would be interesting but disproving equiconsistency would be best.)
Two comments: 1) the category of groups does have some interesting categorical structures, that have been abstracted and studied a lot by people working on semi-abelian categories, proto-modular categories, etc... One of them is that you can give purely categorically a definition of the automorphism group of a group. Aut(G) is universal for "group action on G", and a "group action of H on G" can be defined as a split exact sequence connecting G and H.
Have you try to instead recover the category of sets as the full subcategory of free groups and morphisms preserving generators ?...
... In theory this should be possible as Set is comonadic over both groups and abelian groups... If one could characterize abstractly this comonad that would prove the result immediately. One can also try a more elementary approach: I think Free groups can be characterized as the projective objects, but this is not enough as that does not fix the set of generators, but maybe there is a structure definable in the language that can do that ?
Also: One can characterize sets as certain co-algebras in abelian group (as the free abelian group with their "diagonal" coalgebra structure). But I don't know if the tensor product of abelian group can be characterized abstractely.
@SimonHenry Free groups are examples of cogroups internal to $\mathsf{Grp}$, and the cogroup maps probably preserve the generators.
This follows from a more general theorem about the classification of comonoid objects in $\mathsf{Mon}$ via $E$-systems, see Bergman's "Invitation to general algebra", Section 9.6.
@MartinBrandenburg That sounds like it answers my question.
Oups. :-) $\mbox{}$
As for which categories of algebras are such that the free functor $Set \to Alg$ is comonadic: these are quite general. See this section in the nLab: https://ncatlab.org/nlab/show/comonadic+functor#comonadic_adjoints_to_monadic_functors
@ToddTrimble That's a nice result! Is there always a way to define the comonad within the categorical language?
Excellent question!
A fun puzzle is to ask how far you can get using cyclically ordered sets as your primitive. Can you construct unordered pairs? triples? 4-tuples? 5-tuples?
The answer is yes, in fact one has a lot better than bi-interpretability, as shown by the corollary at the end. It follows by mixing the comments by Martin Brandenburg and mine (and a few additional details I found on MO). The key observation is the following:
Theorem: The category of co-group objects in the category of groups is equivalent to the category of sets and partial map between then, or equivalently to the category of pointed sets..
(Thanks to Martin Brandenburg for pointing out the mistake in an earlier version)
(According to the nLab, this is due to Kan, from the paper "On monoids and their dual" Bol. Soc. Mat. Mexicana (2) 3 (1958), pp. 52-61, MR0111035)
Co-groups are easily defined in purely categorical terms (see Edit 2 below).
The equivalence of the theorem is given by free groups as follows: if $X$ is a set and $F_X$ is the free group on X then Hom$(F_X,H)=H^X$ is a group, functorially in H, hence $F_X$ has a cogroup object structure. As functions between sets induce re-indexing functions: $H^X \rightarrow H^Y$ that are indeed group morphisms, morphisms between sets indeed are cogroup morphisms.
Explicitly, $\mu:F_X \rightarrow F_X * F_X$ is the map that sends each generator $e_x$ to $e_x^L * e_x^R$, and $i$ is the map that sends each generators to its inverse.
An easy calculation shows that the generators and the units (Edit: that's where the mistake was) are the only elements such that $\mu(y)=y^L*y^R$ and hence that any cogroup morphism comes from a partial function between sets: each generator is sent either to a generator or to the unit.
And with a bit more work, as nicely explained on this other MO answer, one can check that any cogroup object is of this form.
As pointed out by Martin Brandenburg in his answer - once we have the category of sets and partial maps, (or equivalently the category of pointed sets) one can easily characterize the category of sets and map in purely categorical language: this is the (non-full) subcategory containing all objects and whose maps are the $f:X \to Y$ such that the square obtained by adding the maps $0 \to X$ and $0\to Y$ (where $0$ is the initial object) is a pullback square.
So let's call this the category of "cogroup objects and total maps between them" (where by total, I mean the map that satisfies this pullback condition).
Now, as all this is a theorem of $\sf{ETCS}$, it is a theorem of $\sf{ETCG}$ that all the axioms (and theorems) of $\sf{ETCS}$ are satisfied by the category of "cogroup objects and total maps between them" in any model of $\sf{ETCG}$, which gives you the desired bi-interpretability between $\sf{ETCS}$ and $\sf{ETCG}$. Adding supplementary axioms to $\sf{ETCS}$ (like R) does not change anything.
In fact, one has more than bi-interpretability: the two theories are equivalent in the sense that there is an equivalence between their models. But one has a lot better:
Corollary: Given $T$ a model of $\sf{ETCS}$, then $Grp(T)$ is a model of $\sf{ETCG}$. Given $A$ a model of $\sf{ETCG}$, then the category $CoGrp(A)^{total}$ of cogroups object and full map between them is a model of $\sf{ETCS}$. Moreover these two constructions are inverse to each other up to equivalence of categories.
Edit: this an answer to a question of Matt F. in the comment to give explicit example of how axioms and theorems of $\sf{ECTS}$ translate into $\sf{ECTG}$.
So in $\sf{ECTS}$ there is a theorem (maybe an axioms) that given a monomorphism $S \rightarrow T$ there exists an object $R$ such that $T \simeq S \coprod R$.
In $\sf{ECTG}$ this can be translated as: given $T$ a cogroup object and $S \rightarrow T$ a cogroup monomorphism* then there exists a co-group $R$ such that $T \simeq S * R$ as co-groups**.
*: It is also a theorem of $\sf{ECTG}$ that a map between cogroup is a monomorphism of cogroup if and only if the underlying map of objects is a monomorphisms. Indeed that is something you can prove for the category of groups in $\sf{ECTS}$ so it holds in $\sf{ECTG}$ by definition.
** : We can prove in $\sf{ECTG}$ (either directly because this actually holds in any category, or proving it for group in $\sf{ECTS}$) that the coproduct of two co-group objects has a canonical co-group structure which makes it the coproduct in the category of co-groups.
Edit 2: To clarify that the category of cogroup is defined purely in the categorical language:
The coproduct in group is the free product $G * G$ and is definable by its usual universal property.
A cogroup is then an object (here a group) equipped with a map $\mu: G \rightarrow G * G$ which is co-associative, that is $\mu \circ (\mu * Id_G) = \mu \circ (Id_G * \mu)$, and counital (the co-unit has to be the unique map $G \rightarrow 1$), that is $(Id_G,0) \circ \mu = Id_G$ and $(0,Id_G) \circ \mu = Id_G$, where $(f,g)$ denotes the map $G * G \rightarrow G$ which is $f$ on the first component and $g$ on the other component, as well as an inverse map $i:G \rightarrow G$ such that $(Id_G ,i ) \circ \mu = 0 $.
Morphisms of co-groups are the map $f:G \rightarrow H$ that are compatible with all these structures, so mostly such that $ (f * f) \circ \mu_H = \mu_G \circ f $.
If you have doubt related to the "choice" of the object $G * G$ (which is only defined up to unique isomorphisms) a way to lift them is to define "a co-group object" as a triple of object $G,G *G,G * G *G$ with appropriate map between them satisfying a bunch of confition (includings the universal property) and morphisms of co-group as triple of maps satisfying all the expected conditions. This gives an equivalent category.
This is really nice!
Just to make sure I understand correctly, are your objects here co-groups (ie free groups) or groups equipped with a co-group structure (ie free groups with a fixed basis)?
When I say "the category of cogroups" I mean the category of object endowed with a cogroup structure, and cogroup morphisms between them (exactly as when you talk about the category of group you mean groups and groups morphisms). So, (as the "theorem" says) free groups equiped with a basis.
@SimonHenry. Many thanks. Would this work for any category in which the forgetful functor to set has a left adjoint?
@HJRW : I don't really know, but that's a great question. As pointed out by Todd Trimble in the comment to the original theory, it is very common that given such a left adjoint, the adjunction is comonadic (essentially, as soon as the left adjoint is non-degenerated enough). But while comonadicity was a very natural way to say that "one can reconstruct the category of set from the category of groups", it was not completely clear that this was enough to solve the problem, as you also need to be able to construct the comonad using only the language of categories. This was the part where...
... the remark by Martin Brandenburg about this specific way of characterizing free groups was very helpful, but this is very specific to the category of group, and I don't see how to generalize this... Also the fact that the trick was to look at "cogroup object in the category of group" is probably an isolated coincidence (for example, every abelian group is automatically an abelian cogroup in the category of abelian group).
Correction, its not completely an isolated coincidence: it is always the case that given an aglebraic theory T, the free T-models are co-T-models in the category of T-models. But the fact that this induces an equivalence of categories between co-T-models in T-models and sets is not generally true as the example of abelian groups shows.
@SimonHenry: are you really "reconstructing the category of set from the category of groups"? Or are you reconstructing the category of sets from the category of cogroups inside the category of groups? I ask because the latter seems to me to be much more closely related to sets than the full category of groups. I don't know if this makes sense, but it seems to me that using cogroup morphisms (which preserve a distinguished subset) is really sneaking sets in through the back door.
But perhaps that's not the spirit in which the question is meant. I'm not a logician, and neither am I a category-theorist.
@HJRW : Cogroup morphisms are not defined as the map the preserves a distinguished subsets, they are defined as the map that are compatible to the co-group structure, i.e. the $f:G \rightarrow H$ such that $\mu_G \circ f = (f * f) \circ \mu_H$ (where $\mu_G$ and $\mu_H$ are the comultiplication maps). It is then a theorem that these are the same as the map preserving the generators. That is what I mean by "reconstructed from the category of groups": we only use concept elementarily formulated in the language of category theory. Otherwise that would not solve the question.
@MattF. : done in the edit.
"For all $T$ and $S$ two objects, $\mu_T:T \rightarrow TT$ a map satisfying ..., $\mu_S:S \rightarrow S * S$ satisfying..., and$f: S \rightarrow T$ satisfying ... ". Of course $T * T$ is not part of the language either, so if I want to be fully rigorous I should add "given an object $T * T$ endowed with two maps $T \rightrightarrows T T $ satisfying the universal property of a coproduct..." And the same for $S S$ and $T T T$ and $SS*S$ (because they are needed in the formulation of associativity of the cogroups operations")
@MattF. : what would be missing is what you say is a purely categorical definition of what it means for a map of free group to sends "generators to generators." (I'm not sure whether your characterization of free group is correct, but I'm assuming it is). And with this sort of definition it would be hard because your characterization of free groups do not fix the set of generators (while the one in terms of cogroups structures does).
@SimonHenry, thanks for the explanation. This is very interesting to me; I've thought a lot about bases for free groups, but never seen them re-interpreted in this way.
@MattF., your characterisation of free groups is not correct. Every group that surjects $\mathbb{Z}$ is "almost free" in your sense. (Group theorists call this property "indicable".) There are many "almost free" groups that aren't free and yet don't split as a direct product; a simple example is $\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}$. (And your claim about when maps of free groups take generators to generators is incorrect too, I'm afraid.) Bases of free groups are quite subtle! The situation isn't at all analogous to what happens in free abelian groups.
@SimonHenry I would suggest to edit your answer (see my answer).
I've removed the green tick until we get an answer with all the right pieces in it.
@MartinBrandenburg : thanks I have edited ! I had missed that the the group unit also satisfies the condition that $\Delta(x) =x_L * x_R$.
@OscarCunningham The solution proposed by Martin Brandenburg in his answer already solve the problem. But I have edited to include his argument and put everything together.
Let me point out that the equivalence mentioned in Simon Henry's answer is not correct. (Edit. It's now corrected.)
There is a unique map of sets $\{1\} \to \{1\}$, but there are two cogroup homomorphisms $F(\{1\}) \to F(\{1\})$, namely $x \mapsto x$ and $x \mapsto 1$. Even more trivially, there is no map $\{1\} \to \emptyset$, but there is (exactly) one cogroup homomorphism $F(\{1\}) \to F(\emptyset)$.
This is a good lesson that a category is more than just its objects.
The category of cogroups in $\mathbf{Grp}$ is equivalent to $\mathbf{Set}_*$. The equivalence maps a pointed set $(X,x_0)$ to the free group on $X$ modulo $\langle \langle x_0 \rangle \rangle$ (which is isomorphic to the free group on $X \setminus \{x_0\}$, but with this POV the functoriality is less clear) equipped with the usual cogroup structure. This is explained (in a much larger framework) in Bergman's An Invitation to General Algebra and Universal Constructions in Section 10.7. I also mention this in Example 6.1 of my paper on limit sketches.
So it remains to recover $\mathbf{Set}$ from $\mathbf{Set}_*$. This can be done:
$\mathbf{Set}$ is equivalent to the subcategory of $\mathbf{Set}_*$ which has the same objects but only those morphisms of pointed sets $f : X \to Y$ such that
$\begin{array}{ccc} \ast & \xrightarrow{~~} & \ast \\ \downarrow && \downarrow \\ X & \xrightarrow{~f~} & Y \end{array} $
is a pullback, where $\ast$ is the zero object of $\mathbf{Set}_*$.
Somehow related to the question is Shelah's paper Interpreting set theory in the endomorphism semi-group of afree algebra or in a category. There are further papers on this direction, see for example Set theory is interpretable in the automorphism group of an infinitely generated free group.
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2025-03-21T14:48:29.818941
| 2020-02-09T18:33:19 |
352299
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|
Stack Exchange
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Flatness of certain quotient rings
Let $p,q \in \mathbb{C}[x,y]$ be two polynomials such that $p_xp_yq_xq_y \neq 0$
(namely, each partial derivative is non-zero).
Assume that the following four conditions are satisfied:
(1) $\frac{\mathbb{C}[x][y]}{\langle p \rangle}$ is a flat $\mathbb{C}[x]$-module.
(2) $\frac{\mathbb{C}[y][x]}{\langle p \rangle}$ is a flat $\mathbb{C}[y]$-module.
(3) $\frac{\mathbb{C}[x][y]}{\langle q \rangle}$ is a flat $\mathbb{C}[x]$-module.
(4) $\frac{\mathbb{C}[y][x]}{\langle q \rangle}$ is a flat $\mathbb{C}[y]$-module.
Question: When $\frac{k[x,y]}{\langle p,q \rangle}$ is a flat $\mathbb{C}[x]$-module or a flat $\mathbb{C}[y]$-module?
Or slightly more generally, when $\langle p,q \rangle= \langle r \rangle$ for some $r \in\mathbb{C}[x,y]$?
"Answer": Perhaps the followig additional condition (5) would guarantee
flatness of $\frac{\mathbb{C}[x,y]}{\langle p,q \rangle}$ over $\mathbb{C}[x]$
or $\langle p,q \rangle= \langle r \rangle$:
(5) $\{p_x,q_x\}$ are linearly independent over $\mathbb{C}$ and $\{p_y,q_y\}$ are linearly independent over $\mathbb{C}$.
Remarks:
(i) If $(p,q)$ is an automorphic pair so $\mathbb{C}[p,q]=\mathbb{C}[x,y]$, then $x-\lambda, y-\mu \in \langle p,q \rangle$, hence $\langle x-\lambda, y-\mu \rangle \subseteq \langle p,q \rangle$. This impies that $\langle p,q \rangle$ is either non-proper or maximal which equals $\langle x-\lambda,y-\mu \rangle$. If it is maximal, then $\frac{\mathbb{C}[x,y]}{\langle p,q \rangle}=k$ which is not $\mathbb{C}[x]$-flat nor $\mathbb{C}[y]$-flat.
(ii) If $\frac{k[x,y]}{\langle p,q \rangle}$ is $\mathbb{C}[x]$-flat, then it can be shown that there exists $r \in k[x,y]$ such that $\langle p,q \rangle =\langle r \rangle$; indeed, apply Corollary 1.3 and Theorem 178(3).
(iii) In condition (5) both $\{p_x,q_x\}$ and $\{p_y,q_y\}$ should be $\mathbb{C}$-linearly independent, as can be seen in my second comment below.
An attempts to find a counterexample:
$p=x, q=y+x^2$: Condition (1) is not satisfied, since
$\frac{\mathbb{C}[x][y]}{\langle x \rangle}=\mathbb{C}[y]$ is not a flat $\mathbb{C}[x]$-module, by this criterion (the action of $\mathbb{C}[x]$ on $\mathbb{C}[y]$ is $x r=0$ for every $r \in \mathbb{C}[y]$ and scalars of $\mathbb{C}[x]$ act as usual multiplication in $\mathbb{C}[y]$. $x \otimes_{\mathbb{C}[x]} 1$ is in the kernel).
Any comments and hints are welcome! Thank you.
Try $p=x+y$, $q=x-y$.
@LaurentMoret-Bailly, thank you! Of course you are right, $\frac{\mathbb{C}[x,y]}{\langle x+y,x-y \rangle}=\frac{\mathbb{C}[x,y]}{\langle x,y \rangle}=k$ is not $\mathbb{C}[x]$-flat and not $\mathbb{C}[y]$-flat. Actually I had another condition in mind: (5) Each of the following Jacobian matrices has rank two over $\mathbb{C}$: $(p_x,q_x)$, $(p_y, q_y)$ (partial derivatives). Please, do you think that there is any chance that my question + condition (5) has a positive answer?
The following is not a counterexample to my edited question, namely, the one with additional condition (5): $f_1=x+y^2, f_2=y+x+y^2$, $R=k[x]$, $n=1$. $\frac{k[x][y]}{\langle x+y^2,y+x+y^2 \rangle}= \frac{k[x,y]}{\langle x,y \rangle}=k$.
The matrix $(p_y, q_y)=(2y, 1+2y)$ has rank two, but the matrix $(p_x, q_x)=(1,1)$ has rank one.
Think about the geometry: let $X = V(p)$ and $Y = V(q)$ in $\mathbf A^2$. The nonvanishing of partial derivatives of $p$ and $q$ is equivalent to the statement that $X$ and $Y$ project surjectively onto each coordinate axis.
Since $\mathbf C[x]$ and $\mathbf C[y]$ are principal ideal domains, flatness is equivalent to torsion-freeness. Thus, a closed subscheme $Z \subseteq \mathbf A^2$ is flat over both coordinate axes if and only if every associated prime of $Z$ maps to the generic points in both coordinate $\mathbf A^1$s. In other words, every irreducible component of $Z$ is a curve, $Z$ has no emdedded points, and $Z$ does not contain a horizontal or vertical line. Then primary decomposition, Krull's principal ideal theorem, and factoriality of $\mathbf C[x,y]$ show that $Z = V(r)$ is principal. The condition that $Z$ contains no horizontal or vertical lines means that no irreducible factor of $r$ is of the form $x - a$ or $y - b$ for $a, b \in \mathbf C$.
Now $V(p,q)$ is the intersection $V(p) \cap V(q)$. By the discussion above, it is flat over both coordinate axes if and only if $(p,q) = (r)$ for a polynomial $r$ that has no irreducible factor of the form $x-a$ or $y-b$. But $r$ is a product of common factors of $p$ and $q$, so since $p$ and $q$ have no factors $x-a$ and $y-b$, the same goes for $r$. In this case $V(r)$ is the union of the components that $V(p)$ and $V(q)$ have in common.
Let $p = rp'$ and $q = rq'$ where $p'$ and $q'$ have no factors in common (but they could contain further powers of factors of $r$). Then $V(p) \cap V(q)$ is flat over both coordinate $\mathbf A^1$s if and only if $(p,q) = (r')$ for some $r'$. We claim that this is only possible if $(r') = (r)$ and $(p',q') = (1)$, i.e. $V(p') \cap V(q') = \varnothing$.
Indeed, suppose $V(p',q') \neq \varnothing$. Since $p'$ and $q'$ have no factors in common, $V(p',q')$ is a finite union of closed points. Then $r \in \mathbf C[x,y]/(p,q)$ is killed by $(p',q')$, hence all primes in $V(p',q')$ occur as associated primes for $(p,q)$, which is impossible when $(p,q)$ is supposed to be principal (see discussion above). Thus we conclude that $(p',q') = (1)$ and hence $(p,q) = (r)(p',q') = (r)$.
Conclusion. If $V(p)$ and $V(q)$ are flat over both coordinate axes, then the same holds for $V(p,q)$ if and only if $(p',q') = (1)$ where $p = rp'$ and $q = rq'$ where $p'$ and $q'$ have no factors in common. Geometrically, this means that $V(p') \cap V(q') = \varnothing$.
Example. If $f$ and $g$ are irreducible polynomials not of the form $x-a$ or $y-b$, then $p = f^2g$ and $q = fg^2$ satisfy the hypotheses if and only if $(f,g) = (1)$.
For example, in Moret-Bailly's example $f = x + y$, $g = x - y$, this is not satisfied, because $(x+y,x-y) = (x,y)$ is the origin. We see that also $\mathbf C[x,y]/(f^2g,fg^2)$ is not flat over $\mathbf C[x]$, because the element $fg$ is killed by $x$.
A case where it does work is $f = x+y$ and $g = x + y + 1$.
Thank you very much for your interesting answer!
|
2025-03-21T14:48:29.819321
| 2020-02-09T21:18:02 |
352304
|
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|
Stack Exchange
|
Integral involving associated Laguerre polynomial and Bessel function
In a quantum mechanics problem I encountered the following integral
\begin{equation*}
\int_0^\infty t^{\nu+1}J_\nu(\beta t)L_{\mu-\nu}^{2\nu}(t)e^{-t/2}dt\,,
\end{equation*}
where $L$ denotes the associated Laguerre polynomial, $J$ a Bessel function of the first kind, $\beta\geq0$ a real parameter, and $\mu\geq\nu\geq0$ positive integers. I've tried to find this integral in tables such as Gradshteyn and Ryzhik
Table of Integrals, Series, and Products but haven't found anything. The closest match seems to be a paper by A.D.Alhaidari (App. Math. Letters 2007, ScienceDirect link) however the exponent in the first term is different. Does anyone have any ideas?
Let $n=\mu - \nu$ be an integer, as specified in the problem. Then I'll indicate how to prove that
$$I:=\int_0^\infty e^{-a\,t} t^{\nu+1} J_\nu(b\,t) L_n^{2v}(t) \,dt =\frac{(2b)^\nu \Gamma(\nu+1/2)}{\sqrt{\pi}}\sum_{k=0}^n \frac{(-1)^k}{k!}\binom{n+2\nu}{n-k} $$
$$
(a^2+b^2)^{-(\nu+k+3/2)}a^{k+1}{}_2F_1(-k/2, -k/2-1/2,1+\nu,-(b/a)^2)(2\nu+1)_{k+1}$$
The ${}_2F_1$ is the Gauss hypergeometric and it is a polynomial, because either $-k/2$ or $-k/2-1/2$ is a negative integer or zero. The $(\cdot)_n$ is a Pochhammer symbol.
The first step is to use the expansion for the Laguerre polynomials,
$$ L_n^{2v}(t) = \sum_{k=0}^n \frac{(-t)^k}{k!} \binom{n+2\nu}{n-k} $$
The parameter $a$ is not the specific $1/2$ of the problem, because we will differentiate with respect to it:
$$I=\sum_{k=0}^n \frac{(-1)^k}{k!} \binom{n+2\nu}{n-k} (-1)^{k+1}\frac{d^{k+1}}{da^{k+1}} \int_0^\infty e^{-a\,t} t^{\nu} J_\nu(b\,t) \,dt
$$
This is done because Grashteyn 6.623.1 says
$$\int_0^\infty e^{-a\,t} t^{\nu} J_\nu(b\,t) \,dt = \frac{(2b)^\nu \Gamma(\nu+1/2)} {\sqrt{\pi}}(a^2+b^2)^{-(\nu+1/2)}$$
To complete the proof we need to show that
$$\frac{d^n}{da^n}(a^2+b^2)^{-(\nu+1/2)}=$$
$$= (a^2+b^2)^{-(\nu+n+1/2)}
(-a)^n{}_2F_1(-n/2, -n/2+1/2,1+\nu,-(b/a)^2)(2\nu+1)_n$$
I did this by starting with the Rodrigues formula for the Gegenbauer polynomials
$$\frac{d^n}{da^n}(1-t^2)^{-(\nu+1/2)}=(1-t^2)^{-(\nu+n+1/2)}C_n^{-\nu-n}(t)
\frac{{(1/2-\nu-n)}_n\,2^n}{\binom{2(\nu+n)}{n}}$$
Now, scale $t \to ia/b,$ use a hypergeometric representation for the Gengenbaur polys, use several gamma function ID's, and simplify.
Thank you very much!
|
2025-03-21T14:48:29.819512
| 2020-02-09T21:35:47 |
352305
|
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|
Stack Exchange
|
Can filtered colimits be computed in the homotopy category?
For $\mathcal{S}$ the $(\infty,1)$-category of spaces its homotopy category $h\mathcal{S}$ does not have pushouts or pullbacks. Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in $\mathcal{S}$.
Generally, filtered colimits are much better behaved than general ones. For example filtered colimits in the (Quillen) model category $\mathit{sSet}$ compute homotopy colimits in $\mathcal{S}$.
In light of this I was wondering the following:
Does $h\mathcal{S}$ have filtered colimits and, if so, does the functor $\mathcal{S} \to h\mathcal{S}$ preserve them?
More generally one could ask the same for any presentable $(\infty,1)$-category $\mathcal{C}$; I particularly care about the case of the derived category $\mathcal{D}(R)$ of a ring $R$.
No, for a diagram $X: I \to \mathcal{S} \to h\mathcal{S}$ the colimit in $h\mathcal{S}$ would satisfy $[\mathrm{colim} X(i),Y] \cong \lim [X(i),Y]$ where brackets denote morphisms in $h\mathcal{S}$. Informally, the set of maps in $h\mathcal{S}$ from the (homotopy) colimit in $\mathcal{S}$ should involve also higher derived limits.
For a concrete example, you could take $I = (\mathbb{N},<)$, $X(i) \simeq S^1$ for all $i$, $X(i) \to X(i+1)$ a map of degree 2. Then $\mathrm{colim}(X(i)) \simeq K(\mathbb{Z}[1/2],1)$ is not the colimit in $h\mathcal{S}$ since it has distinct morphisms to $Y = K(\mathbb{Z},2)$ which become equal upon restricting to each $X(i)$.
Similar examples exist in $D(\mathbb{Z})$.
Thanks! I suppose the colimit of the diagram you're suggesting is $K(\mathbb{Z}[\frac{1}{2}],1)$ and you're using that $H^2(\mathbb{Z}[\frac{1}{2}]; \mathbb{Z})$ is non-trivial?
As has already been said, the homotopy category does not admit filtered colimits in general, but it’s much worse than that. Even colimits in an $\infty$-category which don’t give rise to colimits in the homotopy category sometimes do give rise to weak colimits. (A weak colimit cocone gives the existence, but not the uniqueness, of the factorizations a colimit cocone gives.) This is the case, for instance, with sequential colimits, as well as those along any free category, at least in spaces.
So one might ask whether at least every filtered colimit in $\mathcal S$ gives rise to a weak filtered colimit in $h\mathcal S$. Alas, this is still not true. In our paper kindly referenced by Tim, Christensen and I give an $\aleph_1$-indexed sequence of spaces whose homotopy colimit is not a weak colimit in the homotopy category, namely the sequence mapping a countable ordinal $\alpha$ to the wedge of $\alpha$ circles. The homotopy colimit is a wedge of $\aleph_1$ circles, and the problem is that a map out of that just requires too much coherence to be constructed out of a cocone over countable wedges in $h\mathcal S$. So there is not much hope for filtered colimits in $h\mathcal S$. I expect the same counterexample would work, though I have no idea how to make the argument, in higher homotopy categories $h_n\mathcal S$.
Regarding “minimal” or “distinguished” weak colimits, the general idea is that you want some weak colimits which are distinguished up to at least non-unique isomorphism, as occurs for cones in triangulated categories. Since “homotopy colimits” of sequences in triangulated categories with countable coproduct a are constructed out of those coproduct together with cones, they are also distinguished in this sense.
It is possible to get at the idea of minimal weak colimit of at least a filtered diagram in a category which may not be triangulated, but which has some set of objects detecting isomorphisms, by asking that $Hom(S,\mathrm{wcolim} D_i)\cong \mathrm{colim} Hom(S, D_i)$ for every $S$ in your isomorphism-detecting set. Such weak colimits are then indeed determined up to isomorphism, and they’re also nice because they see the objects $S$ as compact.
However, this is not to say that such distinguished weak colimits are common! Our diagram from above actually admits no weak colimit which views even $S^1$ as compact in this way. (Though note that some weak colimit always exists-homotopy pushouts give weak pushouts, coproducts exist, and then the usual construction applies.)
If your category actually has a set of compact generators in a model, as for $D(R)$, then a distinguished weak colimit must come from a homotopy colimit. Franke gives an argument, cited in our paper, that on these grounds distinguished weak colimits of uncountable chains should essentially never exist in $D(R)$. The problem is that there’s a spectral sequence converging to homs out of a homotopy colimit indexed by $J$ whose $E_2$ page involves the derived functors $R^n\mathrm{lim}^J$ for all $n$. These derived functors were shown by Osofsky to be non vanishing up through $n$ when $J=\aleph_n$, and the homotopy colimit is a weak colimit only if the spectral sequence collapses, so this probably shouldn’t happen. However Franke doesn’t give an argument that there couldn’t in principle be enough unlikely differentials to produce the collapse.
Christensen and I tried for a while to work with the analogous spectral sequence for spaces, but it seemed to require a proficiency with calculating higher derived limits unsupported by the literature-Osofsky gives a special example, and for all I can tell nobody else has ever calculated a derived limit over $\aleph_n$. So our approach turns out to be entirely different and doesn’t immediately apply to the stable case. Thus I think it’s unknown, though highly doubtful, whether $D(R)$ admits minimal filtered colimits in general.
Ironically, I was wondering something similar earlier this week (the irony is that I was sitting next to the OP while doing my wondering). Here's another reason why this can't be the case. In general, if $C\subseteq D$ is a full subcategory and every object in $D$ is a colimit of a diagram in $C$, then it follows that $C$ is a strong generator for $D$ -- i.e. the hom functors $\{Hom(c,-): D \to Set\}_{c \in C}$ are jointly faithful and conservative. If $C$ is essentially small, then one can take a product (or coproduct) of these functors to obtain a faithful, conservative functor $D \to Set$.
At the $\infty$-categorical level, every space $X$ is the filtered colimit $X = \varinjlim_i X_i$ of finite complexes $X_i$. If this colimit were also a colimit in the homotopy category, then $X$ would be a colimit of finite complexes in the homotopy category. Since the homotopy category of finite complexes is essentially small, we would get a faithful, conservative functor $hS \to Set$ by the above.
But as Freyd famously showed, there does not exist a faithful functor $hS \to Set$! I learned recently that Carlson and Christensen have shown there also does not exist a set of spaces such that the corresponding functor $hS \to Set$ is conservative!
We can conclude something stronger from this: there does not exist a regular cardinal $\lambda$ such that $\lambda$-filtered colimits in $S$ can be computed in the homotopy category. More generally, let $A$ be a $\lambda$-accessible $\infty$-category, and suppose that $A \to hA$ preserves $\lambda$-filtered colimits. Then, since $A \to hA$ is obtained by applying the filtered-colimit-preserving functor $\pi_0: Spaces \to Sets$ to the homsets, one sees that any $\lambda$-presentable object in $A$ is also $\lambda$-presentable in $hA$. Moreover, every object is a $\lambda$-filtered colimit of such objects. It follows that the functor $hA \to Ind_\lambda(hA_\lambda)$ is fully faithful (here $A_\lambda \subseteq A$ is the full subcategory of $\lambda$-presentable objects); by Rosicky's theorem (see below), this functor is also essentially surjective. So $hA = Ind_\lambda(hA_\lambda)$ is also $\lambda$-accessible! (As an alternative to Rosicky's theorem, if we assume that $hA$ has $\lambda$-filtered colimits, then it since every object is a $\lambda$-filtered colimit of $\lambda$-presentable objects, it follows directly that $hA$ is accessible) That is, if you have an accessible $\infty$-category and you can compute sufficiently-filtered colimits in the homotopy category, then the homotopy category is also accessible. Beyond the existence of classic counterexamples like Spaces, my sense is that this situation is comparatively rare in practice (except when $A$ was a 1-category to start with). In particular, I think it's rare for the homotopy derived category of a ring to be accessible. I don't know about the existence of filtered colimits in these homotopy categories though, but I suspect they typically don't exist.
A notable exception would be the derived category of field $D(k)$. In this case, the homotopy category is the category of graded $k$-vector spaces, which is accessible (indeed, it's locally presentable -- though not all colimits are preserved by the forgetful functor) and the forgetful functor can be identified with homology, which does preserve filtered colimits.
I haven't thought about this much -- it's possible that there are a lot of rings $R$ for which the homotopy category of $D(R)$ is accessible...
I think there's a lot more to be said about understanding (highly) filtered colimits by looking at the homotopy category.
One commonly-used trick is that (co)products can be computed in the homotopy category, so if you have a triangulated category with countable coproducts, then you can compute a sequential colimit $\varinjlim_n X_n$ as the cofiber (using the triangulated structure) of the map $\oplus_n X_n \to \oplus_n X_n$ given by the difference of the identity and the coproduct of the maps $X_n \to X_{n+1}$. This allows you to compute countable filtered colimits using only the homotopy category + triangulated structure, by passing to a cofinal chain. But I don't think this generalizes to larger filtered colimits.
There are Brown representability-type results which look at the functor $ho(Ind_\lambda(C)) \to Ind_\lambda(ho(C))$. Rosicky showed that this functor is always essentially surjective (perhaps after passing to a higher $\lambda$?). In some cases, Adams apparently showed that this functor is also full, and I think there's more work on this in the triangulated categories literature. But maybe this is not so close to the original question.
Another tangentially related point -- not every idempotent in the homotopy category splits! (This is also a theorem of Freyd). The splitting of an idempotent is an example of a colimit of a diagram which is $\lambda$-filtered for every $\lambda$. This is only tangentially related, because if you know that your idempotent arises from a genuine homotopy-coherent idempotent (i.e. an idempotent in the $\infty$-category of spaces) then it does split; in particular the splitting in the homotopy category exists and agrees with the $\infty$-categorical splitting.
I think in the triangulated categories literature, people deal with this issue by various "weak colimit" and "minimal weak colimit" constructions, but I don't know exactly how these work. Perhaps an expert can comment.
Maybe it's implicit in things I've said above, but with these sorts of questions there can be a big difference between $\infty$-categories like $Spaces$ and $\infty$-categories like $D(R)$. For example, there does exist a conservative functor $D(R) \to Set$ (even already there is a conservative functor $h(pointed spaces) \to Set$ by Whitehead's theorem) -- although I think still not a faithful functor.
Fun fact, in case you intended the implication in your last point that $h\mathcal S$ admits no conservative functor whatsoever to Set: actually every locally small category admits such a functor! I didn’t believe that when I heard it, but it’s yet another coup de Freyd. Of course there is still a big gap between $h\mathcal S$ and $D(R)$ in admitting such a functor which is a coproduct of representables.
@KevinCarlson Whoa -- that's surprising! When you say "every locally small $C$ admits a functor $C \to Set$", do we have to assume that $C$ has no more isomorphism classes of objects than the size of the universe?
Nope! As, er, tcamps recalls in the comments here: https://math.stackexchange.com/a/2128004/31228... the functor is roughly “the set of split subobjects with a disjoint basepoint.” In particular any object lacking nontrivial split subobjects will be mapped to a two-point set, so the functor is not injective on isomorphism classes even when applied to Set itself. If there are a large number of such objects, then the functor will have a large fiber on isomorphism classes while still being conservative.
Haha! I can't believe I forgot such a striking fact!
|
2025-03-21T14:48:29.820438
| 2020-02-09T21:50:06 |
352306
|
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|
Stack Exchange
|
Representation theorem for matrices (reference request)
Motivation. If $A \in \mathbb{C}^{n \times n}$ is self-adjoint (or, more generally, normal), then we all know that
$$
A = \sum_{k=1}^n \lambda_k \, h_k \otimes h_k,
$$
where $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $A$ (counting multiplicities), $(h_1,\dots,h_n)$ is a corresponding orthonormal basis of eigenvectors, and $h_k \otimes h_k$ denotes the matrix given by $(h_k \otimes h_k)x = \langle h_k, x\rangle h_k$ for each $x \in \mathbb{C}^n$ (here I used the "physical" convention that the inner product is linear in the second component).
A representation result. The following result for general (i.e. also non-normal) matrices is - very loosely - reminiscent of the above quoted spectral theorem:
Let $S$ denote the (Euclidean) unit sphere in $\mathbb{C}^n$ and let $\lambda$ denote the surface measure on $S$ (more precisely, we identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$, consider the surface measure on the unit sphere there and pull it back to $S$). Now, normalize $\lambda$ such that $\lambda(S) = n$.
Theorem. For every matrix $A \in \mathbb{C}^{n\times n}$ we have
$$
A = (n+1) \int_{S} \langle h, Ah \rangle \; h \otimes h \; d \lambda(h) - \operatorname{tr}(A) \, I;
$$
here, $\operatorname{tr}(A)$ denotes the trace of $A$ and $I \in \mathbb{C}^{n\times n}$ denotes the identity matrix.
One can prove the above theorem by using the answers to this MathOverflow question.
The question (a reference request). I have no idea whether the above representation theorem is of any use - but given its very symmetric and rather simple nature, it is natural to suspect that the theorem should already be somewhere in the literature.
So my question is: Do you know any reference where the above representation theorem is stated and proved?
@FrancoisZiegler: Well, it's precisely the computation of those two scalars that took me a page or two. But admittedly, "lengthy" might be a bit of an exagaration. I think I'll better remove this word from the post.
If I'm not mistaken then it's also possible to use the coarea formula to rewrite the integral over $S^{2n-1}$ into a double integral over the unit disk and $S^{2n-3}$. An induction over $n$ then should do the trick. Alternatively: Pick a basis and do everything in coordinates. But this might be "lengthy" again.
@JohannesHahn: Thank you for your comment! I agree that one can probably show the formula by concretely computing some integrals. The major motivation of my question though was not to look for an alternative proof (probably the original wording of the question was somewhat misleading, and suggested that I was looking for a simpler or less "lengthy" proof). The main reason for my reference request was that I was wondering in which contexts this formula might occur in the literature.
@FrancoisZiegler: Thank you very much for the references! (I particularly like the first one.) If you post the reference as an answer, I'll certainly accept it.
My comments converted to an answer:
1st comment. I know no reference but the proof need not be [as per OP] lengthy — one just computes the two scalars by which the map $\mathscr I:$
$$
\textstyle A\mapsto\mathscr I(A)=\int_S h\langle h,Ah\rangle\langle h,\cdot\rangle d\lambda(h)
\tag1
$$
acts on the irreducible components of $\mathfrak{gl}(n,\mathbf C)=\mathfrak{sl}(n,\mathbf C)\oplus\mathbf C I$ (Schur’s lemma).
(In detail: write $r$ and $s$ for the scalars in question. Taking the trace of $sI=\mathscr I(I)$ in (1) gives $s=1$. It follows that we have
$
\mathscr I(A)= r\left(A-\tfrac{\operatorname{tr}A}{n}I\right)+\tfrac{\operatorname{tr}A}{n}I
$
and hence
$$
\operatorname{tr}(\mathscr I(A)B)=r\operatorname{tr}(AB)+\tfrac{1-r}n\operatorname{tr}(A)\operatorname{tr}(B)\rlap{\qquad\quad\forall A, B.}
\tag2
$$
Writing this out for $(A,B)=(E_{12},E_{21})$, resp. $(E_{11},E_{22})$ where $E_{ij}=e_i\langle e_j,\cdot\rangle$, one obtains that $\int_S|\langle e_1,h\rangle|^2|\langle e_2,h\rangle|^2d\lambda(h)$ equals both $r$ and $\frac{1-r}n$. Therefore $r=\frac1{n+1}$, and with that (2) becomes (3) below. QED)
2nd comment. For a reference: your desired formula is equivalent to
$$
\mathrm{tr}(AB)+\mathrm{tr}(A)\mathrm{tr}(B)=(n+1)\int_S\langle h,Ah\rangle\langle h,Bh\rangle\,d\lambda(h)
\tag3
$$
which is e.g. (3.8) of Gibbons (1992), taking differences of notation into account. (The integrand descends to $P^{n-1}(\mathbf C)$ where he uses the measure of total volume $\pi^{n-1}/(n-1)!$) I think it should also follow from Archimedes-Duistermaat-Heckman (1982, Prop. 3.2).
More context: Generally I think you’ll find many similar formulas in the (somewhat repetitive) literature on “coherent states” or “quantum mechanics as classical mechanics on $P\mathscr H$”. Also compare Schur’s proof of his orthogonality relations (1924, p.199 or Bröcker-tom Dieck 4.5i), specialized to the adjoint representation.
This is an easy consequence of the $k=2$ case of the complex version of the Isserlis-Wick theorem for moments of Gaussian measures, i.e., the identity
$$
\int_{\mathbb{C}^n} z_{i_1}\cdots z_{i_k}\ {\bar{z}}_{j_1}\cdots {\bar{z}}_{j_k}
\ e^{-|z|^2}
\prod_{a=1}^{n}\frac{d(\Re z_a) d(\Im z_a)}{\pi}\
=\ \sum_{\sigma\in\mathfrak{S}_k} \delta_{i_1 j_{\sigma(1)}}\cdots \delta_{i_k j_{\sigma(k)}}\ .
$$
Going to spherical coordinates produces the integral $\int_{S}\cdots d\lambda(h)$, while the sum over the permutation $\sigma$ gives the other two terms of the wanted identity.
Indeed, it is good to see this really as an integral over $\mathbb{C}\mathbb{P}^{n-1}$
for the Fubini-Study metric, but appealing to Duistermaat-Heckman is not necessary (as known to François).
That’s essentially Gibbons’ proof. Mine (“3rd comment”) uses neither Duistermaat-Heckman (mentioned for context) nor really any integral formula.
Well, yes and no. For $V=\mathbb{C}^n$, finding that the space of $U(n)$ invariants in $V\otimes V\otimes V^{\vee}\otimes V^{\vee}$ is spanned by two elements can be done by averaging over the group. See my two answers to https://mathoverflow.net/questions/255492/how-to-constructively-combinatorially-prove-schur-weyl-duality?noredirect=1&lq=1 So I would say an integral formula is hiding behind the representation theoretic proof. Once that is known, finding the scalars by contracting with suitable "test functions" is not hard.
Don’t get me wrong, I like this too (and, someone should write “Elementary mathematics from an Atiyah-Singer standpoint”). But I felt like I managed with nothing more than $d\lambda$’s invariance + Schur’s lemma, so now I’m curious: at which precise step do you see me smuggling in any computed integral at all?
A proof of course must start somewhere. Yours starts from some basic results in representation theory, whereas mine begins from a more rudimentary starting point, namely, multivariate calculus. The way I see it, the hidden use of integrals is in the used results from representation theory, not in the follow up explicitly written in your answer. Look at I how I prove Schur-Weyl duality in the link in the comment above and you will see what I mean.
Food for thought, thanks. [Later:] I think I see now. A computation is hidden, not in the representation theory, but in the OP’s phrase “Now, normalize such that () = .” (This cannot be done without computing ().) What we can prove with absolutely no integral evaluated is: for any ()-invariant on ,$$A+\mathrm{tr}(A)I=\frac{n(n+1)}{\mu(S)}\int_S h\langle h,Ah\rangle\langle h,\cdot\rangle,d\mu(h).$$
Here is a down-to-earth proof. Remark that for complex matrices, proving $A=B$ is equivalent to proving $\langle x,Ax\rangle=\langle x,Bx\rangle$ ; the superiority of the complex numbers over the real ones !
We must prove that
$$\langle x,Ax\rangle=(n+1)\int_S|\langle h,x\rangle|^2\langle h,Ah\rangle d\lambda(h)-({\rm Tr}\,A)|x|^2.$$
By rotational invariance, it is enough to prove this for $x=\vec e_1$, that is
$$a_{11}=(n+1)\int_S|h_1|^2\langle h,Ah\rangle d\lambda(h)-{\rm Tr}\,A.$$
This amounts to verifying the following identities, all of which being classical:
$$\int_S|h_1|^4d\lambda(h)=\frac2{n+1},\quad\int_S|h_1|^2|h_2|^2d\lambda(h)=\frac1{n+1}$$
and
$$\int_S|h_1|^2h_j\bar h_kd\lambda(h)=0,\quad j\ne k.$$
Thank you for your answer! If I could, I would give you a second upvote for "the superiority of the complex numbers over the real ones!"
@JochenGlueck. It deserves to be more known. This property is equivalent to saying that the numerical radius is a norm over ${\bf M}_n({\mathbb C})$. Even if a matrix has real entries, it is crucial to compute its numerical radius from complex vectors, not only real ones !
|
2025-03-21T14:48:29.821054
| 2020-02-09T22:36:28 |
352307
|
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|
Stack Exchange
|
When homology isomorphism implies homotopy isomorphism
Let's suppose that
$f:X\rightarrow X$ is a continuous map such that
$H_{\ast}(f): H_{\ast}(X)\rightarrow H_{\ast}(X)$ is a homology isomorphism (with integral coefficients)
$X$ is a finite connected CW-complex.
$\pi_{1}(f): \pi_{1}(X)\rightarrow \pi_{1}(X)$ is an isomorphism of fundamental groups.
$\pi_{1}(X)$ is a finitely presented group.
$\pi_{n}(f)=0$ for $n>1$.
the homotopy colimit $$hocolim(X\rightarrow_{f} X\rightarrow_{f} X\dots)$$ is homotopy equivalent to a finite CW-complex.
Does it imply that $f$ has to be a weak homotopy equivalence ?
My guess is that the answer should be no but I don't have a counterexample.
Are you aware of Whitehead’s theorem? And, what role does the limit object play or is ought to play?
Here's a counterexample.
Set $X'=S^1\vee S^2$.
Consider the following map $F':S^2\vee S^2\vee S^2\rightarrow X'$: It maps the first $S^2$ summand to the $S^2$ summand of $X'$ via a map that represents $2\in\pi_{2}S^2$; it maps the second summand once around the $S^1$ factor of $X'$, and maps the third $S^2$ summand to the $S^2$ summand in $X'$ by a map that represents $-1\in\pi_{2}S^2$.
Let $F$ be the composition of $F'$ with the map $S^2\rightarrow S^2\vee S^2\vee S^2$ which collapses 2 different latitudinal circles.
Form $X$ by attaching a 3-cell to $X'$ by the map $F$. Note that $\pi_{1}X=\pi_{1}S^1$, and the inclusion $S^1\hookrightarrow X$ is a homology isomorphism.
The map $f:X\rightarrow X$ which collapses $X$ to its $S^1$ summand satisfies all the requirements. In this case the hocolim in requirement 6 is $\simeq S^1$.
|
2025-03-21T14:48:29.821190
| 2020-02-09T23:01:50 |
352309
|
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|
Stack Exchange
|
Jacobian of changing of variables to singular value decomposition
It is well known that changing variables from a symmetric matrix to its eigenvalue decomposition involves a Jacobian which is just the Vandermonde determinant of the eigenvalues.
Now suppose I have a rectangular matrix and I want to change variables to its singular value decomposition. What is the Jacobian of this transformation?
For an $m\times n$ real matrix $A=U\Sigma V^t$ with $m\leq n$, diagonal matrix of singular values $\Sigma={\rm diag}\,(\sqrt\sigma_1,\sqrt\sigma_2,\ldots\sqrt\sigma_m)$, orthonormal left and right eigenvector matrices $UU^t=VV^t=\mathbb{1}$, the Jacobian $J$ in the measure $dA=JdUdV\prod_{i=1}^m\sigma_i$ follows from the Wishart distribution,
$$J=\prod_{i<j}|\sigma_i-\sigma_j|\prod_k\sigma_k^{(n-m)/2}.$$
Is there anything like $\frac{1}{2^mm!}$ in $J$ to account for the fact that by permuting $\sigma_i$ and columns of $U,V$ simultaneously can lead to the same $A$? Also, by changing the signs of each column of $U,V$ simultaneously can also lead to the same $A$? Basically this change of variable is not unique unless we restrict the order of $\sigma_i$ and the sign of the first column of $U$ and $V$ to be positive.
|
2025-03-21T14:48:29.821300
| 2020-02-09T23:29:46 |
352310
|
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|
Stack Exchange
|
On some integral involving the Liouville sum
Define $L(x) = \sum_{n\leq x} \lambda(n)$, where $\lambda$ denotes the Liouville function. If $c$ is the supremum of the real parts of the zeros of the Riemann zeta function and $\lfloor x\rfloor$ denotes the integer part of $x$, is it true that
$$\Bigl\lvert\int_{1}^{x} L(y)\lfloor x/y\rfloor\frac{\mathrm{d}y}{y} \Bigr\rvert > x^{c-\epsilon}$$ for arbitrarily large $x$ and any $\epsilon>0$?
$\int_{1}^{x} L(y)\lfloor x/y\rfloor\frac{\mathrm{d}y}{y} = \int_1^x \frac{\lfloor \sqrt{t}\rfloor}{2t}dt$
It does not seem to be true. If you try $x=10^4$ and $\epsilon=10^{-5}$, either directly in your inequality, ,or in the inequality obtained from the comment above, Mathematica will tell you this is not true. You have to be careful with your inequality because the integral fails to converge adequately.
|
2025-03-21T14:48:29.821382
| 2020-02-09T23:34:34 |
352311
|
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|
Stack Exchange
|
Embedding random variables in infinite-dimensional spaces
Let $H$ be a reproducing kernel Hilbert space of functions $f:E\to F$ with kernel $k$. A point in $E$ may be embedded into $H$ via the canonical embedding $x\mapsto k(x,\cdot)$. Similarly, a random variable that takes values in $E$ defines a canonical $H$-valued random variable via the mapping $X\mapsto k(X, \cdot)$. Let $X_H := k(X, \cdot)$. Then $X_H$ "encodes" much of the information about $X$ via its mean and covariance operator. In particular, $X_H$ has properties that make it "nice" to deal with, e.g. it leads to nice calculations with means and covariances, and connects abstract covariance operators to the covariance of $X$. For example, it is not hard to show that if $C_X$ is the covariance operator of $X_H$, then
$$
\text{cov}(f(X)) = \langle f, C_Xf\rangle.
$$
My question: Is there a similar construction that holds for arbitrary Hilbert spaces, or say even Banach spaces? Another way to think of this: Is there a way to embed $X$ into an RKHS $H$ that doesn't explicitly involve the kernel $k$ (but still leads to "nice" properties)?
(I am aware that one could define many different types of embeddings ad hoc, but my question is geared more towards whether or not there is a canonical way to do this that leads to something similar to $k(X,\cdot)$ and if so, what type of results are out there.)
I don't think I understand the construction you mention in the first paragraph. Can you give precise definitions and statements, or a reference?
Hopefully fixed now. I shouldn't have referred to $X_H$ as an embedding into $H$, strictly speaking it embeds $X$ into the space of RVs over $H$.
|
2025-03-21T14:48:29.821532
| 2020-02-10T00:03:53 |
352312
|
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|
Stack Exchange
|
How to check whether a given matrix is in the image of a representation?
Let $G$ be a compact simple Lie group, and let $\rho$ be a (faithful, unitary) irreducible representation thereof of $\mathbb K$-dimension $n$, where $\mathbb K=\mathbb C/\mathbb R/\mathbb H$ if $R$ is real/complex/pseudo-real, respectively. It then follows that there is a subgroup of $SU(n)/SO(n)/Sp(n)$, respectively, isomorphic to $G$. One can think of $\rho$ as a map from $G$ to this subgroup.
How can I check whether a given matrix $M\in SU(n)/SO(n)/Sp(n)$ is in the image of $\rho$? In other words, given one such matrix $M$, how can I decide whether there exists some $g\in G$ such that $\rho(g)=M$?
For the sake of concreteness, say $G=G_2$ is the first exceptional simple group, and let $\rho$ be the representation with highest weight $2\omega_2$ (which is real and $27$-dimensional). This means that for any $g\in G_2$, $\rho(g)$ is a $27$-dimensional orthogonal matrix. If I take some arbitrary $27$-dimensional orthogonal matrix $M$, how can I check whether it can be written as $M=\rho(g)$ for some $g\in G_2$?
Note: I am particularly interested in the case where $M$ is diagonal, but I'd be interested in hearing about the general case as well. In the diagonal case, where everything is abelian, and one can essentially focus on a Cartan subalgebra, I assume one can be quite explicit about the image of $\rho$. In the general case, I wouldn't be surprised if one has to work harder.
FWIW: I posted this on math.SE (cf. https://math.stackexchange.com/q/3531073/289977). After a week with no activity whatsoever I decided to cross-post here. Hopefully the question is not as easy as to be off-topic here, but if so please let me know and I'll delete. Cheers!
The image is Zariski closed, and probably after conjugation can be chosen defined over a number field ($\mathbf{Q}$ itself?). The issue is then to determine an explicit family of polynomials whose zero set equals $\rho(G)$. Possibly some computational commutative algebra machinery does this in a somewhat complicated way, and maybe there are clever ways to do so too.
The group G2, in its compact real form, has explicit Euler angles, worked out a few years ago by some physicists. You can immediately see using trigonometry if a matrix belongs.
The paper I was thinking of: https://arxiv.org/abs/hep-th/0503106
@BenMcKay: Actually, checking the Euler angles would only show you that the matrix is conjugate in $\mathrm{SO}(27)$ to an element of $\rho(\mathrm{G}_2)$, wouldn't it? That's not enough to show that it actually lies in $\rho(\mathrm{G}_2)$ because $\rho(\mathrm{G}_2)$ is not a normal subgroup of $\mathrm{SO}(27)$.
Is $G_2$ self-normalizing in $SO(27)$? Because it would be easy to test if $g G_2 g^{-1} = G_2$ -- it's enough to check the corresponding equality of Lie algebras, which is just a Linear Algebra computation.
@DavidESpeyer very nice!. Its 27-dim rep is absolutely irreducible, so has a centralizer reduced to scalars, hence has trivial centralizer in $\mathrm{SO}(27)$. Moreover $G_2$ (in its compact form or any other) has a trivial Out (and trivial center), so its normalizer in any overgroup is the direct product with its centralizer. Hence $G_2$ equals its own normalizer. (This also works with the split form, in $\mathrm{SL}_{27}(\mathbf{R})$).
Suppose that $G$ is a compact Lie group (or, more generally, an algebraic group over some field $F$), ${\mathfrak g}$ is its Lie algebra. You are given a linear representation $\rho: G\to GL(n, F)$. From this, compute the representation of the Lie algebra $\rho': {\mathfrak g}\to End(F^n)$. In fact, the representation $\rho$ is frequently given in terms of the highest weight of $\rho'$. Now, you check section 4.5 "Computing defining polynomials of an algebraic group in terms of its Lie algebra" in
W. de Graaf, "Computing with Linear Algebraic Groups", CRC, 2017.
He describes an algorithm for computing defining polynomial equations $p_i, i=1,...,N$, for $\rho(G)$ in terms of $\rho'({\mathfrak g})$ (more precisely, in terms of a basis for this subalgebra in $End(F^n)$). He even has software (check his webpage https://www.science.unitn.it/~degraaf/) for practical computations of this type.
Lastly, to verify if the given matrix $A\in GL(n, F)$ belongs to $\rho(G)$, evaluate the polynomials $p_i$ on $A$.
I can answer the diagonal case, at least, where the answer is fairly easy. This assumes that you have everything set up 'nicely', which I'll define. I'm going to do the case of an algebraic group, as that's what I have used in the past, but a real Lie group should be similar.
Assume that you have a maximal torus $T$ of your simple algebraic group $G$ and a representation $\rho$. Assume that the elements of $T$ are sent to diagonal matrices under $\rho$. (This is what I mean by a nice setup.)
Notice that $\rho(g)$ is diagonal if and only if $g\in T$, so it suffices to check that. If $t_1(a_1)\ldots t_n(a_n)$ is an expression for a general element of $T$, then we have $g=t_1(a_1)\ldots t_n(a_n)$. Using $\rho(t_i(a_i))$ one obtains a system of equations, one for each diagonal entry of the equation $\rho(g)=\rho(t_1(a_1)\ldots t_n(a_n))$. Because there are more than $n$ equations (in general) one obtains an overspecified system of equations in the $a_i$, which one may then solve via elimination (or fail to find a solution for).
Of course, if you don't have things set up in this way, then I think in general it becomes incredibly complicated.
Edit: For the single concrete example you give of the $27$-dimensional representation for $G_2$, there is a better way, but this won't be available in general. That representation gives an embedding of $G_2$ into $E_6$ (for algebraic groups only! See comment below), and so stabilizes the $E_6$ trilinear form. Over positive characteristic, so I guess over characteristic $0$ as well, $G_2$ stabilizes exactly a $2$-space of symmetric trilinear forms.
An explicit basis for this 2-space can be constructed by solving a bunch of linear equations, if you can generate a few elements of your group. Note that $G_2$ must be the exact stabilizer of this $2$-space of symmetric trilinear forms, because it is a maximal subgroup of $E_6$, which stabilizes a unique $3$-form.
What do you mean by "a basis for $T$"? Have you left the algebraic-group setting and gone back to the compact-group setting, and chosen a minimal set of closed-subgroup generators? (Or maybe you mean a basis for $X_*(T)$, and then $t_1^{a_1}\dotsc t_n^{a_n}$ means $t_1(a_1)\dotsc t_n(a_n)$?)
Ah yes, I mean $t_1(a_1)$, sorry. I'm actually thinking in algebraic groups in positive characteristic, where you can always put $\rho(g)$ into a finite group, and then use a basis for the finite abelian torus. I forgot to translate through to algebraic groups in characteristic $0$ first. I've edited it to reflect this, and think I haven't missed any other instances.
@DavidCraven: Actually, the compact form of $\mathrm{G}_2$ represented in $\mathbb{R}^{27}$ as $\rho(\mathrm{G}_2)$ is maximal in $\mathrm{SO}(27)$ (by Dynkin), so if you already know that the matrix is orthogonal, it suffices to check whether it preserves one of the two $\mathrm{G}_2$-invariant cubic forms on $\mathbb{R}^{27}$. If you don't know that it's orthogonal, then, yes, it might just lie the noncompact $\mathrm{E}_6\subset\mathrm{SL}(27,\mathbb{R})$ that contains $\rho(\mathrm{G}_2)$.
I was thinking about an arbitrary matrix, yes. But I am not an expert on Lie groups, as is increasingly clear from this discussion, and was just thinking about algebraic groups.
|
2025-03-21T14:48:29.822140
| 2020-02-10T00:53:03 |
352315
|
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|
Stack Exchange
|
program to compute hurwitz numbers
Is there a computer program available to compute Hurwitz numbers easily? In fact I only care about counting covers $C\to\mathbb{P}^1$ branched over $0,1,\infty$, and am even willing to restrict to the case $C=\mathbb{P}^1$.
In combinatorial language, I would like to input integers $d\ge1,g\ge0$ and three partitions $\lambda_1,\lambda_2,\lambda_3$ of $d$ such that the total number of parts of the $\lambda_i$ is $d-2g+2$, and compute, up to simultaneous conjugation, the (weighted) number of permutations $\sigma_1,\sigma_2,\sigma_3\in S_d$, where $\sigma_i$ has cycle type $\lambda_i$, for which the $\sigma_i$ generate a transitive subgroup of $S_d$ and $\sigma_1\sigma_2\sigma_3=1$.
|
2025-03-21T14:48:29.822218
| 2020-02-10T02:12:40 |
352320
|
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"url": "https://mathoverflow.net/questions/352320"
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|
Stack Exchange
|
grobner basis of an ideal dependent on some parameter
Suppose $I = \langle f_1, ... , f_l \rangle$ is an ideal generated by polynomials $f \in k[x_1,\dots,x_n]$, where $k$ is a field of rational functions in some parameters $s_1,\dots,s_m$.
What are the techniques to study the dependence of the grobner basis of $I$ (with respect to some order) on the parameters $s$? Is it possible to find "regions" in the parameter space where the basis is doesn't change?
The relevant notion is that of a "Groebner system," closely related to comprehensive Groebner bases introduced by Weispfenning. The regions of the parameter space are constructible sets in general. One generally expects these calculations to be immense, though I believe implementations do exist.
|
2025-03-21T14:48:29.822294
| 2020-02-10T02:50:10 |
352321
|
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|
Stack Exchange
|
A question on the Dieudonné property
Recall that a Banach space $X$ is said to have the Dieudonné property if for every Banach space $Y$, an operator $T:X\rightarrow Y$ that transforms weakly Cauchy sequences into weakly convergent sequences is weakly compact. We denote by $H(X)$ the subset of $X^{**}$ formed by all the $\sigma(X^{**},X^{*})$-limits of weakly Cauchy sequences in $X$. A. Grothendieck (Canad J. Math, 1953) characterized the Dieudonné property as follows:
Let $X$ be a separable Banach space. The following statements are equivalent:
(1) $X$ has the Dieudonné property;
(2) Every $\sigma(X^{*},H(X))$-convergent sequence in $X^{*}$ is weakly convergent.
Question: Is there a self-contained or simple proof of this result?
|
2025-03-21T14:48:29.822374
| 2020-02-10T03:29:29 |
352325
|
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|
Stack Exchange
|
3-balls with the same boundary in $S^4$ differ up to diffeomorphism
I am looking at this recent paper by Budney and Gabai and I am confused by a certain sentence in it. Theorem 4.7 states that if $\Delta_1$ and $\Delta_2$ are two 3-balls smoothly embedded in $S^4$ that are identical near there boundary, then there is a diffeomorphism $\Phi: S^4 \to S^4$ that is the identity on the neighborhood of the boundary of these two balls where they agree and $\Phi(\Delta_1) = \Delta_2$.
The proof of this theorem is one sentence long, and this is the sentence where I am seeking some clarification. The sentence cites two facts:
(1) Regular neighborhoods are unique
(2) $\operatorname{Diff}_0(S^2)$ is connected
and two papers:
(1) J. Cerf, Topologie de certains espaces de plongements, Bull. Soc. Math. France, 89 (1961), 227–380.
(2) R. Palais, Extending diffeomorphisms, Proc. Amer. Math. Soc. 11 (1960), 274–277.
I am not sure exactly what parts of these two references are being used. I believe for the Palais paper, maybe the authors are using Corollary 2 to say that there is some diffeomorphism of $S^4$ taking $\Delta_1$ to $\Delta_2$. I am not sure what is in the Cerf paper (it is written in french and over 100 pages so that has kept me away). However, maybe this is the standard reference for the fact that every diffeomorphism of $S^3$ extends over $B^4$?
I'm also not exactly sure what the statement "regular neighborhoods are unique" means. I suppose it means that any two regular neighborhoods for a smooth submanifold differ up to homotopy rel the submanifold.
I would love it if someone could tell me how to fit these pieces together and understand the proof.
I think you have your question answered. The issue your question perhaps did not emphasize is that while $\Delta_1$ and $\Delta_2$ are isotopic, they are generally not isotopic through embeddings that preserve the boundary, or slightly weaker, isotopic through embeddings that are linear inclusions on the boundary.
The cited (early) work by Cerf proves that, given a submanifold Y in a manifold X, the obvious map Diff(X)->Emb(Y,X) is a locally trivial fibration.
I guess that Budney and Gabai mean the following. By Palais, all embeddings D^3->S^4 are isotopic. Hence, for i=0, 1, the complement C_i of a small
open tubular neighborhood U_i of Delta_i in S^4 is diffeomorphic with the compact 4-ball B^4. One has two disjoint embeddings phi_i, psi_i of B^3 in the boundary S^3 of B^4=C_i, one orientation-preserving, the other orientation-reversing, representing the two sides of Delta_i. It remains to extend the diffeomorphism between C_0 and C_1 through U_0
and U_1. This amounts to find a diffeomorphism f:B^4->B^4
such that f o phi_0=phi_1 and f o psi_0=psi_1. The papers by Palais and Cerf precisely give this. The connexity of Diff_+(S^2), and the unicity of the tubular neighborhood up to isotopy, serve to arrange that
the extension goes well on a small neighborhood of the 2-sphere bounding Delta_0 and Delta_1.
I think Cerf proves the map is a Serre fibration. Palais went the extra step to prove it is locally-trivial. But that's broadly the interpretation we meant.
I insist that Cerf proved that it is a locally trivial fibration. See Jean CERF,
"Topologie de certains espaces de plongements",
Bulletin de la S. M. F., tome 89 (1961), p. 227-380, Corollaire 2 p. 294.
Thanks. It appears Palais and Cerf both proved this result in short succession, with the Palais article appearing the year before (1960). Perhaps I should be crediting Cerf as well.
Palais writes in the introduction of "Extending diffeomorphisms": "The author understands that the theorem has been proved independently by J. Cerf in his thesis (not yet published)." Also, note that this paper by Palais only proves that the isotopies act transitively on subballs. The local triviality theorem is actually proved in another paper by Palais in the same year:
"Local triviality of the restriction map for embeddings",
Comment. Math. Helv. 34 (1960), 305–312.
|
2025-03-21T14:48:29.822680
| 2020-02-10T04:12:48 |
352331
|
{
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|
Stack Exchange
|
Request for an exact formula related to a partition in number theory
The Frobenius equation is the Diophantine equation $$
a_1 x_1+\dots+a_n x_n=b,$$
where the $a_j$ are positive integers, $b$ is an integer, and a solution $$(x_1, \dots, x_n)$$
must consist of non-negative integers, i.e.
$$
x_j \in \mathbb{N}
$$
as Natural numbers. For negative $b$, there are no solutions.
My question: Is there any known formula that counts the number of solutions, by giving $a_1, \dots, a_n$, $b$, and $n$? Let us call this function as $F (a_1, \dots, a_n; b, n)$, what is known for this:
$$F (a_1, \dots, a_n; b, n)=?$$
For all the $a_j=1$, we can simplify the above Frobenius equation to:
$$
x_1+\dots+x_n=b, \tag{1}$$
where $b \in \mathbb{Z}^+$ is a positive integer.
Here is another simpler question: Is there a general formula for Eq.(1) counting all the possible solutions $$(x_1, \dots, x_n)$$
for given the positive integer $n \in \mathbb{Z}^+$ and $b \in \mathbb{Z}^+$? This should be related to the Partition, but I am not sure the exact forms are known? Say, can we find the total number of soultions as a function $f(n,b)$, and what is
$$
f(n,b)=?
$$
It seems the answer is known:
$$
f(n,b)= \binom{b+n-1}{n-1}.
$$
p.s. Sorry if this question is too simple for number theorists. But please provide me answer and Refs if you already know the answer. Many thanks!
https://www.researchgate.net/publication/27187071_On_the_number_of_solutions_of_the_Diophantine_equation_of_Frobenius_-_General_case
For second questions - see https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
See also a technical question for Mathematica: https://mathematica.stackexchange.com/q/214865/11223
A related paper is https://doi.org/10.37236/1592.
As usual, it depends on what you call a "formula". There are several approaches used to study the general equation: numerical semigroups, Fourier analysis, partial fraction decomposition, generating funtions, counting lattice point in polytopes, multiple zeta functions, etc.
For $n=2$ (and $a_1$, $a_2$ relatively prime) there is a simple formula, namely
$$F(a_1,a_2;b,2)=\frac{b}{a_1a_2}
-\left\{\frac{{a_2}^{-1}b}{a_1}\right\}
-\left\{\frac{{a_1}^{-1}b}{a_2}\right\}+1,$$
where ${a_1}^{-1}$ is an inverse modulo $a_2$, ${a_2}^{-1}$ is an inverse module $a_1$,
and $\{x\}$ denotes the fractional part of $x$. There is no known similar formula for $n\ge3$ (and there is probably no hope in finding a simple one). In the simplest case $a_i=1$ for all $i$, your formula is correct.
There is a beautiful book that discusses this topic in Chapter 1: "Computing the continuous discretely" by Beck and Robins. Also, as a reference, you may look at any book about "Numerical semigroups".
(There are several people in MO that are far better qualified at answering this, and I hope they will see your question and answer it.)
thanks +1, this is helpful
$F(a_1,\dots,a_n;b)$ equals the coefficient of $z^b$ in the generating function
$$f(z):=\frac{1}{1-z^{a_1}}\frac{1}{1-z^{a_2}}\cdots \frac{1}{1-z^{a_n}}.$$
For a fixed choice of $a_1,\dots,a_n$, explicit formula for $F(a_1,\dots,a_n;b)$ as a function of $b$ can obtained via partial fraction decomposition.
For the second question, see Stars and bars.
thanks +1 for the short formula and refs.
You can take generating function $$f(z):=\frac{1}{1-z^{a_1}}\frac{1}{1-z^{a_2}}\cdots \frac{1}{1-z^{a_n}}$$ as in Max Alekseyev's answer and calculate $F (a_1, \dots, a_n; b, n)$ as $$
\frac{1}{2 \pi i} \int_{|s|=\rho} f(s) \frac{d s}{s^{b+1}} \quad (0<\rho<1).
$$
It gives the answer
$$
F (a_1, \dots, a_n; b)=\frac{b^{n-1}}{(n-1) ! a_{1} \ldots a_{n}}+\sum_{k=0}^{n-2} c_{k} b^{k}.
$$
It is a classical applications of contour integration taken from the book "Residues and their applications" by A.O. Gelfond (1966, pp. 98-99, Russian). If $(a_j,a_k)=1$ ($j\ne k$) then all poles (excepting $s=1$) are simple and formula can be simplified:
$$
F (a_1, \dots, a_n; b)=\frac{(-1)^{n-1}}{(n-1) !} \frac{d^{n-1}}{d s^{n-1}}\left[s^{-b-1} \prod_{k=1}^{n} \frac{1-s}{1-s^{a_{k}}}\right]_{s=1}+R
$$
where $|R|<C$ for some constant $C$.
What are the $c_k$ in your formula for $F(a_1,\dots,a_n;b)$?
@BlaCa They are some coefficients. I can't say anything about them.
Some number theory terminology for your second question is the number of integer compositions of $b$ with $n$ parts, where the parts are required to be positive integers. There are $\binom{b-1}{n-1}$ of these: think of having $b$ 1s in a row and, among the $b-1$ spaces between them, placing $n-1$ plus signs. Combine the adjacent 1s and separate parts by +, e.g., $11+1+1+11 \sim 2+1+1+2$ is one of the $\binom{5}{3}=10$ 4-part compositions of 6.
By the way, integer partitions are equivalent to solutions where the order of the summands does not matter. Equivalently, if the summands are placed in a specified order, typically nonincreasing. For example, $2+1+1+2$ and $1+2+1+2$, etc., would all correspond to $2+2+1+1$. There are generally fewer partitions than compositions and there is not such a simple formula for the number of them.
Your formula, $\binom{b+n-1}{n-1}$, is for the number of compositions with nonnegative integer parts. The "stars & bars" argument in combinatorics verifies the formula: Any arrangement of $b$ 1s and $n-1$ plus signs gives one of these compositions, e.g., $1111+11++ \sim 4+2+0+0$ is one of the $\binom{9}{3} = 84$ 4-part weak compositions of 6.
As @EFinat-S wrote, there's a nice formula for two variables, but for arbitrary linear Diophantine equations, there is nothing like that. The end of Beck & Robbins chapter 1 touches on this (e.g., the "chicken McNugget problem") and goes on to more advanced approaches---it's a great book.
thanks +1, this is good.
|
2025-03-21T14:48:29.823069
| 2020-02-10T04:57:42 |
352333
|
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"url": "https://mathoverflow.net/questions/352333"
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|
Stack Exchange
|
What is the left adjoint to base change of schemes?
Restriction of Scalars and Functoriality of Presheaves.
Let $\phi\colon R\longrightarrow S$ be a morphism of rings. There is associated to $\phi$ a natural functor from $\mathrm{Alg}_S$ to $\mathrm{Alg}_R$, called the restriction of scalars functor:
$$f\colon\mathrm{Alg}_S\longrightarrow\mathrm{Alg}_R.$$
In detail, this is the functor taking an $S$-algebra $S\rightarrow A$ to the $R$-algebra $R\rightarrow S\rightarrow A$, which we denote $A_R$.
As remarked in this nLab page (and developed in detail in SGA IV, Exposé I, Section 5), there exists an induced adjoint triple of functors between the corresponding presheaf categories:
where $f^*\colon\mathrm{PSh}(\mathrm{Alg}_R)\longrightarrow\mathrm{PSh}(\mathrm{Alg}_S)$ is given by precomposition with $f$.
Base Change of Schemes.
Consider the restriction $f^*|_{\mathrm{Aff}_R}$ of $f^*$ to the full subcategory $\mathrm{Aff}_R$ of $\mathrm{PSh}(\mathrm{Alg}_R)$ spanned by the representable presheaves on $\mathrm{Alg}_R$, i.e. by affine $R$-schemes.
This functor takes an $R$-scheme $h_A$ to the presheaf
$$f^*h_A\colon\mathrm{Alg}_S\longrightarrow\mathrm{Alg}_R\longrightarrow\mathrm{Sets}$$
defined by
$$B\mapsto\mathrm{Hom}_{\mathrm{Alg}_R}(B_R,A)\cong\mathrm{Hom}_{\mathrm{Alg_S}}(B,A\otimes_RS),$$
where the isomorphism comes from the adjunction between restriction and extension of scalars.
That is, $f^*h_A=h_A\times_R h_S$ and the functor $f^*$ is therefore base change of schemes.
Adjoints to Base Change.
As R. van Dobben de Bruyn points in the comments, the right adjoint $f_*$ of $f^*$ is called Weil restriction. While it can fail to be a scheme in general, it is representable by schemes under nice conditions. One may then ask about the left adjoint $f_!$:
Question 1. Is the left adjoint $f_!$ of $f^*$ representable by schemes? Moreover, if it isn't, are there conditions we can require of an $S$-scheme $X$ guaranteeing the presheaf $f_!X$ to be a scheme?
Question 2. What about the non-affine case?
(I gather from the Wikipedia page on Weil restriction that these have been studied in the very general case of schemes over ringed topoi. What reference/s is Wikipedia alluding to?)
Could someone please embed the linked image into this question? (MO unfortunately prevents me from doing it myself)
For $f_*$ this is studied under the name Weil restriction. It always exists for a finite flat ring extension, but in general is not representable (I don't recall the most general statement right now). This is also related to the more general construction of the Hom scheme $\mathbf{Hom}_S(X,Y)$, which is representable for a pair of proper flat morphisms $X, Y \to S$, but not in general.
@R.vanDobbendeBruyn Thanks! I have rewritten the question, both correcting some mistakes I made and making modifications which take into account your comment.
Isn't $f_!$ just given by postcomposing the structure morphism with $f$?
@DenisNardin You're right. Combined with R. van Dobben de Bruyn's comment, this settles the question. Thanks! (And sorry for taking so long to reply!)
|
2025-03-21T14:48:29.823290
| 2020-02-10T07:48:30 |
352341
|
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"Dominic van der Zypen",
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|
Stack Exchange
|
Thinning directed sets ${\frak P}$ of partitions of $\omega$ with no ${\frak P}$-discrete subsets
This question branches from Taras Banakh's recent question on a cardinal characteristic connected to families of partitions that are directed in the ordering of partition refinement.
A partition $\mathcal P$ of $\omega$ is said to be finitary if there is $k\in \omega$ such that for every $P\in {\cal P}$ we have $|P|\leq k$.
A family $\mathfrak P$ of partitions of $\omega$ is called directed if for any two partitions $\mathcal A,\mathcal B\in\mathfrak P$ there exists a partition $\mathcal C\in\mathfrak P$ such that each set $S\in\mathcal A\cup\mathcal B$ is contained in some set $C\in\mathcal C$.
Let $\mathfrak P$ is a family of partitions of $\omega$. An infinite subset $D\subset\omega$ is called $\mathfrak P$-discrete if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset D$ such that for any $P\in\mathcal P$ the intersection $P\cap (D\setminus F)$ contains at most one point.
Let ${\frak P}$ be a directed family of finitary partitions admitting no ${\frak P}$-discrete set. Is there ${\frak C}\subseteq {\frak P}$ with the following properties?
${\frak C}$ admits no ${\frak C}$-discrete subset, and
for all ${\cal C}_1, {\cal C}_2\in {\frak C}$ we have that either ${\cal C}_1$ refines ${\cal C}_2$, or vice versa. (If ${\cal P}, {\cal Q}$ are partitions of $\omega$, we say that ${\cal P}$ refines ${\cal Q}$ if every member of ${\cal P}$ is contained in some (i.e. exactly one) member of ${\cal Q}$.)
The answer to this question is negative.
Given a linearly ordered family $\mathfrak C$ of finitary partitions of $\omega$, write $\mathfrak C=\bigcup_{n=1}^\infty\mathfrak C_n$ where $\mathfrak C_n=\{\mathcal P\in\mathfrak C:\sup_{P\in\mathcal P}|P|\le n\}$.
For a partition $\mathcal P$ of $\omega$ and a point $x\in\omega$, let $\mathcal P(x)$ be the unique set in the partition $\mathcal P_n$ that contains $x$.
Taking it account that for every $n\in\mathbb N$ the family $\mathfrak C_n$ is linearly ordered, we conclude that for every $x\in \omega$ the family $\{\mathcal P(x):\mathcal P\in\mathfrak C_n\}$ is linearly ordered and consists of sets of cardinality $\le n$.
Consequently, the union $$\mathcal P_n(x):=\bigcup_{\mathcal P\in\mathfrak C_n}\mathcal P(x)$$ is a set of cardinality $\le n$. It can be shown that $\mathcal P_n:=\{\mathcal P_n(x):x\in\omega\}$ is a partition of $\omega$ such that every partition $\mathcal P\in\mathfrak C_n$ refines $\mathcal P_n$.
Since the family $\mathfrak P=\{\mathcal P_n\}_{n\in\mathbb N}$ is countable, it possesses a countable $\mathfrak P$-discrete set, which remains $\mathfrak C$-discrete.
In light of this solution let us ask another
Question. Is any linearly ordered family of finitary partitions of $\omega$ at most countable?
Thanks for the beautiful answer - and if your question in turn has a negative answer, we might see yet another cardinal characteristic :-) . If an answer to your question is not found over the next few days, it could be worthwhile that you ask this as a separate MO question
|
2025-03-21T14:48:29.823509
| 2020-02-10T07:58:52 |
352343
|
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|
Stack Exchange
|
Where can I find a table of the exponents of the sporadic groups?
Is there a table showing Sporadic Groups and their exponents, and, perhaps, other basic properties.
In particular, I'm interested in what the exponent of the Monster Group is. (Obviously the order is well publicised, but not the exponent, as far as I can tell.)
Thanks!
This information can be calculated easily from the printed character tables in the ATLAS of Finite Groups (which include orders of elements in conjugacy classes) or, perhaps more conveniently, using the same information online via GAP or Magma. From there you can just load the character table from the library and calcualte the lcm of the orders of the elements.
I had a quick look at the character table of the Monster in the Atlas, and its exponent appears to be $<IP_ADDRESS>.<IP_ADDRESS>.<IP_ADDRESS>.59.71$.
Or, using the version of the ATLAS tables in Gap's character table library, Exponent(CharacterTable("F1")); (returns<PHONE_NUMBER>878376600800) for the exponent of the Monster.
Oh dear, I missed out the factor 41 in my previous comment! Well I said it was a quick look!
Hi chaps, thanks, I really appreciate your answers :)
For the Monster the answer is<PHONE_NUMBER>878376600800, which is amazing.
Does anyone know of an existing table online showing the exponents of all the sporadic groups?
From the comments,
This information can be calculated easily from the printed character tables in the ATLAS of Finite Groups (which include orders of elements in conjugacy classes) or, perhaps more conveniently, using the same information online via GAP or Magma. From there you can just load the character table from the library and calculate the lcm of the orders of the elements.
I had a quick look at the character table of the Monster in the Atlas, and its exponent appears to be <IP_ADDRESS>.<IP_ADDRESS>.<IP_ADDRESS>.59.71.41
- Derek Holt
Or, using the version of the ATLAS tables in Gap's character table library, Exponent(CharacterTable("F1")); (returns<PHONE_NUMBER>878376600800) for the exponent of the Monster.
– Gro-Tsen
I couldn't find an online table of exponents for sporadic groups, so I used GAP to produce one:
$$
\begin{align*}
\mathbf{Group}&&\mathbf{Exponent}&&\mathbf{Factorization}\\
M_{11}&&1320&&2^3\cdot3\cdot5\cdot11\\
M_{12}&&1320&&2^3\cdot3\cdot5\cdot11\\
J_1&&43890&&2\cdot3\cdot5\cdot7\cdot11\cdot19\\
M_{22}&&9240&&2^3\cdot3\cdot5\cdot7\cdot11\\
J_2&&840&&2^3\cdot3\cdot5\cdot7\\
M_{23}&&212520&&2^3\cdot3\cdot5\cdot7\cdot11\cdot23\\
HS&&9240&&2^3\cdot3\cdot5\cdot7\cdot11\\
J_3&&116280&&2^3\cdot3^2\cdot5\cdot17\cdot19\\
M_{24}&&212520&&2^3\cdot3\cdot5\cdot7\cdot11\cdot23\\
McL&&27720&&2^3\cdot3^2\cdot5\cdot7\cdot11\\
He&&14280&&2^3\cdot3\cdot5\cdot7\cdot17\\
Ru&&633360&&2^4\cdot3\cdot5\cdot7\cdot13\cdot29\\
Suz&&360360&&2^3\cdot3^2\cdot5\cdot7\cdot11\cdot13\\
O'N&&10884720&&2^4\cdot3\cdot5\cdot7\cdot11\cdot19\cdot31\\
Co_3&&637560&&2^3\cdot3^2\cdot5\cdot7\cdot11\cdot23\\
Co_2&&1275120&&2^4\cdot3^2\cdot5\cdot7\cdot11\cdot23\\
Fi_{22}&&720720&&2^4\cdot3^2\cdot5\cdot7\cdot11\cdot13\\
HN&&2633400&&2^3\cdot3^2\cdot5^2\cdot7\cdot11\cdot19\\
Ly&&10651271400&&2^3\cdot3^2\cdot5^2\cdot7\cdot11\cdot31\cdot37\cdot67\\
Th&&57886920&&2^3\cdot3^3\cdot5\cdot7\cdot13\cdot19\cdot31\\
Fi_{23}&&845404560&&2^4\cdot3^3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot23\\
Co_1&&16576560&&2^4\cdot3^2\cdot5\cdot7\cdot11\cdot13\cdot23\\
J_4&&607938537360&&2^4\cdot3\cdot5\cdot7\cdot11\cdot23\cdot29\cdot31\cdot37\cdot43\\
F_{3+}&&24516732240&&2^4\cdot3^3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot23\cdot29\\
B&&234033344344800&&2^5\cdot3^3\cdot5^2\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\cdot31\cdot47\\
M&&1165654792878376600800&&2^5\cdot3^3\cdot5^2\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\\
&&&&\cdot29\cdot31\cdot41\cdot47\cdot59\cdot71\\
\end{align*}
$$
It would be more intuitive if you also paste the prime decomposition of these numbers.
@JeremyRickard Thanks for adding the factoring and the nice layout. It's much better than what I posted. You should have posted it as your own answer and gotten the points.
It wasn’t me, @StevenStadnicki deserves the credit. All I did was correct the table header from “Order” to “Exponent”.
@StevenStadnicki Sorry for not paying more attention to the edit history. Thank you for your improvements! Your edit comment says that I had a typo. What was it?
You didn't have a typo — my first edit did (an incorrect factorization), and I made a second edit to fix my mistake but they may have been folded into one. And you're very welcome!
|
2025-03-21T14:48:29.823763
| 2020-02-10T08:18:04 |
352344
|
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|
Stack Exchange
|
Can we have cyclic generalized positive comprehension?
In positive set theory, the axiom scheme of generalized positive comprehension in $GPK^+_\infty$ [of Olivier Esser] is stated in a manner as to forbid the symbol of the asserted set to occur in the defining formula.
To quote from the above source:
Positive Comprehension: For any positive formula $\phi$ in which $A$ does not appear, $(\exists A.(\forall x.x \in A \leftrightarrow
\phi)).$
Where a positive formula is defined as:
Definition: Let the class of (bounded) positive formulas be the smallest class containing the formula $x \neq x$ (useful because
uniformly false), all atomic formulas, and closed under conjunction,
disjunction, bounded universal quantification $(\forall x \in A.\phi)$
and existential quantification $(∃x ∈ A.φ)$.
Question 1: Is there a clear inconsistency involved with permitting symbol $A$ to occur in $\phi$, in the statement of positive comprehension?
Question 2: Suppose that there is no clear inconsistency, then would that allowance results in increment in consistency strength over the original system?
Yes, there is a clear inconsistency, take the formula:
$\exists A.\ (\forall x.\ (x\in A \leftrightarrow A=\emptyset))$.
Clearly, such a set $A$ cannot exist: if $A\neq \emptyset$, then any $x\in A$ would give a contradiction. If $A=\emptyset$, then this formula would imply that any $x$ satisfies $x\in A$, but then $A\neq \emptyset$.
|
2025-03-21T14:48:29.824268
| 2020-02-10T09:08:37 |
352348
|
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|
Stack Exchange
|
Given functions $A(x), B(x)$ find $f(x)$ s.t. $A\big(f(x)\big)=f\big(B(x)\big)$
Currently, I am facing this problem:
Given two real functions $A( \vec x )$ and $B( \vec x ):\Bbb R^N\to \Bbb R$, I want to find a third real, monotonic function $f(x):\Bbb R\to\Bbb R$ such that:
$$A\big(f( \vec x )\big)=f\big(B( \vec x )\big)$$
where the simplified the notation writing $f( \vec x )$ means
$$
f(x_1,x_2,\ldots,x_n) = \big(f(x_1), f(x_2),\ldots, f(x_n)\big).
$$
I am interested in either having a formula/method for finding $f$, or even just having a proof that $f$ exists (or doesn't) under some specific conditions. Eventually, I am interested also in the solution in the case $N=1$.
Also, does this type of problem have a specific name?
Thank you very much!
Fixed! I hope now it's fun enough
you might also want to consider changing the name of your profile; the adjectives you use to characterise Schrödinger are inappropriate, IMO.
Ok, I also changed the nick to a more appropriate one. I guess the next answer I'll receive is not gonna be on the question I asked, but on the way, I use the comma. Maybe I should ask this same question on a grammar forum...
It seems too easy: if $A(x_1,x_2)=1+x_1$ and $B(x_1,x_2)=x_1+x_2$ then you are looking for $f$ such that $1+f(x_1)=f(x_1+x_2)$ which cannot be, for example if $x_2=0$. Maybe you missed some conditions? (Or did I?)
Thanks for answering, Yaakov!
Regarding the assumptions, I am looking to see under which conditions f exists and when it doesn't. In your example, we saw that f doesn't exist, however, there are cases in which it does. E.g.
$$ A( \vec x ) = \prod x_i$$
$$ B( \vec x ) = \sum x_i$$
In this case the solution is:
$$ f( x ) = e^x $$
So, I am looking to see if there are certain conditions under which f exists (and eventually how to find it).
I'll be happy also just to know if there is a class of similar problems and their name.
Are your functions supposed to be continuous?
More formally one can write the equation as $A\circ\vec{f}=f\circ B$, where $\vec{f}=f^{\times n}$.
Also, does this type of problem have a specific name?
Given two maps $A$ and $B$, trying to find an $f$ such that $A \circ f =f \circ B$ is a problem of semi-conjugacy.
Alexandre, I was checking if I could add some extra assumptions which are not too stringent.
Assuming they all are continuous is acceptable. Regarding f, we can also assume it to be derivable.
Existence of such $f$ is a very strong condition on $A$ and $B$ which is called semi-conjugacy. Of course, for generic $A$ and $B$ function $f$ does not exist. Suppose for simplicity that $N=1$. Then it is clear that the image of any fixed point of $B$ under $f$ is a fixed point of $A$. Furthermore, when $N=1$, your equation implies that
$$A^n\circ f=f\circ B^n$$
where $A^n$ means the $n$-th iterate. This implies that periodic points of $B$ are mapped to periodic points of $A$ of the same period. So we have some very complicated relation between $A$ and $B$.
Alexandre, thank you very much!
I think I need to study a little more the theory of topological conjugacy, as there are some points that I don't fully understand.
By adding slightly more stringent conditions (i.e. A() and B() continuous and f derivable) is it possible to find a solution to the problem? Otherwise, what are the typical hypotheses to solve it? (if they exist)
Eventually, could you suggest me a book or article that could help me in understanding better how to solve the problem?
Again, thanks a lot!
|
2025-03-21T14:48:29.824517
| 2020-02-10T12:30:13 |
352357
|
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|
Stack Exchange
|
Non-singular rings which are Rickart
A ring $R$ is said to be a right Rickart ring if the right annihilator of any element in $R$ is of the form $eR$ for some idempotent $e \in R$.
It turns out that a ring $R$ is right Rickart iff every principal right ideal of $R$ is projective as a right $R$-module. (Prop. 7.48 in Lam - "Lectures on Modules and Rings".)
Recall that a ring $R$ is a right non-singular ring if the left annihilator of each essential right ideal of $R$ is zero.
Question:
For a ring $R$, being right Rickart implies being right non-singular (loc.cit., p.262).
The converse (right non-singular $\Rightarrow$ right
Rickart), however, does not hold in general. What would be (a candidate for) the mildest possible
assumption that one can impose on $R$ and get that the converse holds?
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2025-03-21T14:48:29.824628
| 2020-02-10T13:29:37 |
352362
|
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|
Stack Exchange
|
What is a coordinate less definition of differentiable manifolds
![enter image description here]
From Clifford algebra to geometric calculus by d. Hestenes
https://en.wikipedia.org/wiki/Universal_geometric_algebra
The attempt above is to have the base manifold like its tangent bundle be a "vector manifold". This seems very practical since you don't need a function angebra, coordinates or large expressions to define tangent vectors since simply by subtracting two points you get a meaningful value because your points already are vectors.
I'm curious why exactly and how this would work
The way you get geometry of your manifold is by looking how the psudoscalar slides along it. In this particular case both on wiki and in the book string emphasis is made on intrinsic nature of vector manifolds and to me it doesn't seem that this is justified.
To justify the intrinsic nature of vector manifolds means, as far as I can see, to demonstrate that every manifold, in the sense of the usual definition in differential geometry, arises as a vector manifold, i.e. admits a smooth embedding into Euclidean space, i.e. the Whitney embedding theorem.
The question you ask in the title is different from the question you ask in the body above. Which is the real question: do you want a definition of manifold free of coordinates, or an understanding of the intrinsic nature of vector manifolds?
I don't think he does that. I think I might have a pretty good idea now what he means. But would still like it to be a it more precise.
What Hestenes calls a vector manifold is more commonly known as affine space. https://en.wikipedia.org/wiki/Affine_space
@ben mckey if vector manifolds are not intrinsic than I don't consider them a coordinate free alternative to atlases and charts formulation of manifolds.
@DeaneYang: Hestenes's definition of a vector manifold is: an embedded submanifold of a Clifford algebra, strangely. It is not merely an affine space. Hestenes has many followers who apparently feel that this definition is the best way forward for differential geometry, and that more abstract notions of pseudo-Riemannian metric are unnecessarily complicated.
@BenMcKay: in regarding to the first comment, maybe we want Nash instead? A lot of the "power" of the vector manifold stuff seems to be using the Clifford structure and this requires an ambient pseudo-Riemannian structure, so probably the correct embedding theorem to think about is the isometric embedding. (Related: https://mathoverflow.net/a/127735/3948 )
@WillieWong: yes, you are right: Nash.
@Ezio: incidentally, the idea of studying the geometry of an embedded submanifold through sliding the pseudo-scalar is quite classical. In the case of Riemannian hypersurfaces of Euclidean spaces, this is essentially studying the geometry through the Gauss map, which, in the surface case, dates back to (unsurprisingly) Gauss in the 1820s. The theory has been extended to treat arbitrary codimensional embeddings into pseudo-Euclidean spaces by taking the target of the Gauss map as the appropriate pseudo-Grassmanian.
@BenMcKay, I was responding to image posted above.
The book of Hestenes does not give a definition of vector manifold. It says that a vector manifold is a set of vectors in a Clifford algebra, with some additional properties. It defines those properties in terms of a notion of interior, and a notion of boundary, neither of which are defined. So it relies on intuitive notions of interior and boundary. It defines tangent space at a point (of any set of vectors), as the set of velocities of curves through those vectors. Presumably the curves are required to be smooth enough to have velocity vectors. It states that at interior points, the tangent space is a vector space. But that can't be enough to define a manifold in the usual sense, because it doesn't say what an interior point is. A set of vectors much worse than a submanifold can have points at which all velocities of differential curves in the set are zero. So the property of having a vector space as tangent space does not decide for us how to define the notion of interior point. In the end, the authors are relying on prior experience with manifolds, especially surfaces, as is often the case for authors working close to classical physics, engineering or statistics.
So to answer the questions: there is no coordinate free definition of manifold, because every definition we currently have relies on some charts, or on being a submanifold of some other previously defined manifold, ending us up with Euclidean space. There is no way to fully justify Hestenes's definition, because he doesn't really have one, but we can say that Nash's embedding theorem proves the existence of isometric embeddings of Riemannian manifolds, and subsequent authors have generalized to pseudo-Riemannian manifolds.
You just said in a comment some people say this is the alternative way to do differential-geometry, using vector manifolds instead of pseudo riemann metrics. If hestenes doesn't have a precise definition of what he means who has?
@Ezio: I am only really guessing that Hestenes's intension is to study manifolds embedded into Clifford algebras (subject to a nondegeneracy hypothesis which he explains in that book). You might ask Hestenes for a precise definition.
Yes that's what's he's doing, but I don't see how that's different than an embedding into an Euclidean space. Embedding is still embedding. When you talk about standard charts and coordinates manifolds there is no embedding whatsoever. So a priori his approach can't be an alternative to the standard way no matter how clumsy he says the standard coordinate approach is.
There seems to be quite a bit of confusion regarding what Hestenes means exactly by "geometric algebra". The subject matter of "geometric algebra" can be defined and understood precisely in the standard language of spin geometry as that part of the theory of Clifford bundles which relies on their isomorphic presentation as Kahler-Atiyah bundles, a presentation which follows from the classical Chevalley-Riesz isomorphism.
This definition of "geometric algebra" makes perfect sense in the ordinary language of intrinsic differential geometry and does not require an embedding into an ambient space -- in particular, the entire subject can be defined without using the notion of "vector manifold". In this formulation, "geometric algebra" is merely that part of spin geometry which relies on systematic application of the Chevalley-Riesz isomorphism in order to translate a certain class of problems in spin geometry into equivalent problems for differential forms and polyvector fields --- an approach which can be very useful in certain situations.
See also the answers to the following questions:
What's "geometric algebra"?
Clifford algebras as deformations of exterior algebras
I don't see how this answers the problem. The main point hestenes tried to make is that exactly what you seem to be talking about is just a redundancy ridden formalism that he tries to simplify in what he calls geometric algebra. Also geometric algebra is coordinate less, so it doesn't use coordinates. Thaugh his calculations are coordinate less, they seem to unequivocally imply an imbedding. I don't see how this is not the case, that is the question
@Ezio I disagree that intrinsic differential geometry would be redundant. Since the 19th century, it has been the general opinion of mainstream mathematicians that extrinsic formulations based on embedding a manifold into an ambient space should be avoided if one is interested in the intrinsic content of differential geometry. The formulation of Hestenes appears to be coordinate independent only because it relies on an embedding of a manifold into a Clifford algebra, but such embeddings are highly non-unique. That's why his followers appear to be disconnected from the mathematics mainstream.
This is exactly what I'm saying. He explicitly states his formulation is intrinsic yet I can't see how. At no poit did I say intrinsic geometry is redundant if anything, sometimes it's the other way around. I said that the mainstream intrinsic formulation has a lot of redundancy. Hestenes managed to remove this redundancy of structure in which he failed to retain the intrinsic formulation despite claiming otherwise.
@Enzio I think that Hestenes's formulation is not intrinsic and that his claims to the contrary do not satisfy basic standards of mathematical clarity and rigor. Can you clarify what you mean when you state that the standard intrinsic formulation of differential geometry is redundant ?
So I would like a simple, coordinate free manifold geometry theory without embedding. Hestenes has the coordinate free but not intrinsic , and mainstream formulation has a lot of redundancy and coordinates though it's intrinsic.
Well the redundancy part is quite subtle. For example in GA stokes theorem implies both divergence and rotation theorems, and at the same time applies to vector valued differential forms. So this is how redundancy is cut down.
@Ezio I don't think that Hestenes has produced a coordinate free formulation of the notion of differentiable manifold without using an embedding and I doubt that what you ask for is possible. Hestenes makes claims to that effect, but his claims are vague, somewhat circular and in my opinion incorrect. Unfortunately, he is so vague with his definitions that his claims are not proper mathematics so they cannot be disproven rigorously since he is so ambiguous about what he means. This is why I prefer the mainstream definition of "geometric algebra" as the theory of Kahler-Atiyah bundles.
@Ezio This type of Stokes theorem (and every other rigorous result that Hestenes discusses) can also be formulated in the Kahler-Atiyah approach and, in my opinion, is much clearer when understood in that approach.
Iv never heard of that and don't think it's exactly mainstream either. The main attraction to GA is that it's coordinate free. But I guess it fails to provide an intrinsic differentual geometry
This discussion is getting too long, but let me mention that Kahler-Atiyah algebras and bundles are classical and known to experts in spin geometry. See the questions on Math Overflow that I linked above and references therein.
|
2025-03-21T14:48:29.825290
| 2020-02-10T13:48:26 |
352365
|
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|
Stack Exchange
|
Group actions and "transfinite dynamics"
$\DeclareMathOperator\Sym{Sym}$I have a question about what I shall name here "transfinite dynamics" because it involves iterating a topological dynamical system $G \curvearrowright X$ through ordinals. I do not know whether this concept already has a name. If so, I would appreciate being directed to some references.
The intuition behind this construction is that the phase space $X$ will be the "board" of some game and $G$ will be the set of moves that the player is allowed to execute. I wish to understand what happens if the player is allowed to execute transfinitely many moves as long as the board "stabilizes" at limit stages.
To be more precise, let $G$ act on a Hausdorff space $X$. Extend $X$ to $\overline{X}$ by adding a point $\ast \notin X$ as an isolated point and extend the action of $G$ to $\overline{X}$ trivially. Let $\lambda$ be an ordinal and $\mathbf{g}=(g_{\alpha})_{\alpha < \lambda}$ be a sequence of elements of $G$. For each $x \in X$, define $(x_{\alpha})_{\alpha \leq \lambda}$ by transfinite recursion as follows.
$x_0=x$,
$x_{\alpha+1}=g_{\alpha}\cdot x_{\alpha}$ for all ordinals $\alpha<\lambda$, and
$x_{\theta}=\begin{cases} \ast & \text{ if } \lim_{\alpha \rightarrow \theta} x_{\alpha} \text{ does not exist}\\ \lim_{\alpha \rightarrow \theta} x_{\alpha} & \text{ otherwise }\end{cases}$
for every limit ordinal $\theta \leq \lambda$. Let us say that a sequence $(g_{\alpha})_{\alpha < \lambda}$ is valid if $x_{\lambda} \neq \ast$ for all $x \in X$. In other words, valid sequences are those along which every point can be transfinitely iterated.
For each valid sequence $\mathbf{g}=(g_{\alpha})_{\alpha < \lambda}$, consider the map $K_{\mathbf{g}}: X \rightarrow X$ given by $K_{\mathbf{g}}(x)=x_{\lambda}$.
Here are my questions.
Under what (not very restrictive) conditions can we guarantee that the set of
$K=\{K_{\mathbf{g}}: \mathbf{g} \text{ is valid of length} <\omega_1\}$ forms a subgroup of $\Sym(X)$?
Can we see this set as some kind of "closure" of the image of
$G$ in $\Sym(X)$?
Here are some basic facts that I was able to show
The maps $K_{\mathbf{g}}$ are not necessarily bijections. For example, consider the shift action of $\mathbb{Z}$ on the set $X$ of bi-infinite 0-1 sequences that are constant after some index. Then all sequences are mapped to constant sequences under the maps corresponding to the valid sequence $(1,1,1\dots)$ of length $\omega$.
If $G$ acts on a compact metric space $X$ by isometries, then each $K_{\mathbf{g}}$ is an isometry. Preservation of distances is trivial and surjectivity can be shown with some effort using transfinite induction and the compactness of the space. However, I cannot show that $K_{\mathbf{g}}^{-1}$ is of the form $K_{\mathbf{h}}$ and so, I cannot guarantee that these maps form a subgroup.
For an example of a non-trivial case where the maps $K_{\mathbf{g}}$ give us more bijections of $X$ than those induced by $G$, consider the action of $\mathbb{Q}$ on $S^1$ by rotations. In this case, the maps $K_{\mathbf{g}}$ induced by valid sequences of length $\omega$ gives all the real rotations because for any real $r$, we can form a sequence $(q_0,q_1,\dots)$ of rationals with $\sum_{i=0}^{\infty} q_i=r$. On the other hand, no other valid sequence gives us a bijection other than these real rotations.
Iterating such a system along ordinals seems very natural to me, however, I was not able to find much on this. As I said, if anything along these lines were investigated before, I would appreciate being directed to the correct references.
Although distinct, let me point out a slightly related well-studied concept: let $G$ act on a compact space $X$ (usually by homeomorphisms). Let $X^X$ be endowed with the product compact topology. The Ellis semigroup of the action is the closure of (the image of) $G$ in $X^X$. For instance, one can characterize when it's indeed a group of homeomorphisms. In opposite cases such as convergence actions, it consists of $G$ union functions that are constant outside a singleton.
You're idea reminds me of this paper by Küster where she relates the fixed space of the Koopman operator on a topological dynamical system to transfinite properties of the orbits of the underlying dynamic.
Here is a partial answer extracted from a Twitter user's answer that works for actions with uniformly bounded finite orbits, which indeed solves a special case that initially motivated this question.
For every sequence $\mathbf{g}=(g_{\alpha})_{\alpha < \lambda}$ and every $i \in \mathbb{N}$, set $\mathbf{g}^i$ to be the sequence of length $\lambda \cdot i$ given by $g^i_{\lambda j + \alpha}=g_{\alpha}$ for all $\alpha < \lambda$ and $j < i$. In other words, $\mathbf{g}^i$ is the concatenation of $i$-many $\mathbf{g}$'s back to back. Observe that $\mathbf{g}^i$ is valid whenever $\mathbf{g}$ is valid. Moreover, one can show by induction that for every valid sequence $\mathbf{g}$ and every $i \in \mathbb{N}$, we have $K_{\mathbf{g}}^i=K_{\mathbf{g}^i}$.
Lemma: Let $\mathbf{h}$ be a valid sequence such that $K_{\mathbf{h}}: X \rightarrow X$ is a bijection. If there exists $k \in \mathbb{N}$ such that $|\text{Orb}(x)| \leq k$ for all $x \in X$, then the map $K_{\mathbf{h}^{k!-1}}$ is the inverse of $K_{\mathbf{h}}$.
Proof: Assume that $k \in \mathbb{N}$ is a uniform bound for orbit sizes. It follows from the finiteness of orbits that the trajectory of a point under $\mathbf{h}$ must be eventually constant and hence, we must have $K_{\mathbf{h}}(x) \in \text{Orb}(x)$ for every $x \in X$. Consequently, $K_{\mathbf{h}} \upharpoonright \text{Orb}(x) \in \text{Sym}(\text{Orb}(x))$ for every $x \in X$. By Lagrange's theorem, $\left(K_{\mathbf{h}} \upharpoonright \text{Orb}(x) \right)^{k!} = \mathbf{id}_{\text{Orb}(x)}$ for every $x \in X$ and hence $K_{\mathbf{h}}^{k!} = \mathbf{id}_X$. By the previous lemma, $K^{k!-1}_{\mathbf{h}}=K_{\mathbf{h}^{k!-1}}$ is the inverse of $K_{\mathbf{h}}$.
|
2025-03-21T14:48:29.825669
| 2020-02-10T14:15:02 |
352366
|
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|
Stack Exchange
|
Discrete version of Helmholtz decomposition
In The curl of graphs and networks (Gustafson and Haray, 1984) it is claimed to be shown that any digraph $G$ can be decomposed as the sum of three graphs $U_1 + U_2 + U_3$, where $U_1$ is divergence-free, $U_2$ is curl-free and $U_3$ is both curl and divergence free.
However I am having a hard time understanding the proposed definitions for curl and divergence, and the decomposition.
It would be immensely helpful if somebody could help me decipher the paper by providing an example of a digraph decomposition.
For example, how could one decompose the following weighted digraph?
isn't this fully solenoidal (divergence-free) ?
@CarloBeenakker is it? How I was undertstanding it (and this may betray how little I do understand), in a solenoidal graph we would have that each node satifies that the inflow equals the outflow
|
2025-03-21T14:48:29.825803
| 2020-02-10T15:08:53 |
352368
|
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|
Stack Exchange
|
What's "geometric algebra"?
Sometimes one bumps into the name "geometric algebra" (henceforth "GA"), in the sense of this Wikipedia article. Other names appear in that context such as "vector manifold", "pseudoscalar", and "space-time".
A very superficial look at that Wikipedia article, or books on the topic, gives me the idea that it is essentially about Clifford algebras and related calculus. One impression that I got (but I could be wrong) is that there is a relatively small group of authors (I don't know if mathematicians or physicists or both) that have produced work in "GA" and that this group is probably disjoint from the set of mathematicians who wrote about the algebraic foundations of Clifford algebra theory or about Clifford algebras featuring in contexts such as the Atiyah-Singer index theorem or Clifford analysis (the study of Dirac-type operators). Also, I'm not aware if any reference to the field "GA", as such, appears outside works specifically designated as "GA" and written by people in that group. It's also not clear to me if there is an intersection (or even subset relation) between "GA" and the above mentioned areas of mathematics and the extent of such intersection.
1. Is there anything in the field "geometric algebra" that is distinct from usual Clifford algebra theory and/or Clifford analysis, or is it just a different name for the same set of mathematics? Or maybe does it provide a slightly different viewpoint on the same mathematics (like, e.g., probability theory having a completely different viewpoint from measure theory despite being formally measure theory)? If so, what are the advantages of this viewpoint?
2. Are there mathematical applications of "geometric algebra" outside the field itself? There seem to be applications to physics: are these applications mathematically rigorous?
Are you aware of Emil Artin's book of that title? It mostly has to do with the axiomatic investigation of projective geometries and (among other things) their coordinatization derived from the postulated symmetries. It's possible that what you are asking is inspired or developed from this text, but I don't know enough to give a proper answer. Gerhard "Starting A Change..." Paseman, 2020.02.10.
@GP: no, I was aware of that book, thank you. But I think probably the "GA" in my OP isn't related to axiomatic projective geometries.
It is essentially the same as Clifford algebras, quaternions, etc. equipped with non-standard notation. Usually the context for this is indeed applications to physics where it is argued that their notation is nicer
@GP Typo: I meant to write "wasn't aware of the book, thank you"
I have been wondering this question a lot lately, so I'm glad you asked it. I know the theory of Clifford algebras, and I don't understand the miraculous qualities attributed to them.
I'd just like to add that ncatlab has a brief page on this issue: https://ncatlab.org/nlab/show/geometric+algebra. Essentially it is an issue of presentation, trying to avoid both explicit representation by matrices and also quotients of tensor algebras, though I should also note that Hestenes in his public comments has been very adamant that somehow this distinction is very important in terms of new applications (see https://www.physicsforums.com/insights/interview-mathematician-david-hestenes/).
In my answer, I suggest it's a name for an arbitrary Clifford algebra + a geometric interpretation of it. I don't know whether this can be captured formally in a way that covers the full breadth of what gets called "geometric algebra". It's a bit like how a group can be equipped with an action which makes it more than just an abstract group.
I include all possible geometric interpretations. Maybe therefore Geometric Algebras are Clifford algebras equipped with additional structure. Different structures on the same CA give rise to different GAs.
The label "geometic algebra" was William Cifford's name for the algebra he discovered (invented). This bit of history is recounted by David Hestenes, who has perhaps been most responsible for promoting this lovely mathematics for physics, and has many books and papers on this subject. There is also a Cambridge lot, including Chris Doran and Anthony Lasenby, who have a physics book, and a gauge theory of gravity. These men are perhaps the most influential authors, but I would also want to include Perti Lounesto, and William Baylis, in the geometric algebra crowd. I have personally found these authors particularly accessible and a pleasurable learning experience.
What seems to me the most unifying element of those who use the term "geometric algebra" for "Clifford algebra" is the emphasis on a real exposition of the subject. To my mind, since complex numbers are in fact themselves a real geometric algebra, there is no small bit of confusion generated by insisting on the mathematical development of the subject over a complex field. But this is exactly what Emile Cartan did in defining spinors. With spinors deeply embedded in physical theories of elementary particles, a Clifford algebra defined over a complex field has a strong tradition, especially among mathematically inclined theorists. Most algebraists (Chevalley, Cartan, Atiyah) would consider complex numbers the truest form of 'number'.
Before closing on the subject, a serious researcher should include a third subject: "noncommutative algebra". In many respects, this is a similar development but with a pedigree more after Hermann Grassmann than William Clifford. Alain Connes, has developed this subject recently with physical applications.
In conclusion, I offer the advice that 'geometric algebra' is the most accessible and intuitive approach offering a physically descriptive mathematics. It is just multivariate linear algebra from a practical vantage point. I think it should be taught in high school. The 'Clifford algebra' and 'noncommutative algebra' approaches are more abstract and mathematically rigorous, but greatly expand the available literature when used as search terms.
Elie Cartan, not Emile.
I briefly met David Hestenes once, and he said something like that Clifford algebras are the language in which all of physics should be expressed.
The essence of "geometric algebra" (better known as Kähler-Atiyah
algebra) is the classical Chevalley-Riesz isomorphism, which presents
the Clifford algebra of a quadratic space $(V,h)$ as a deformation
quantization of the exterior algebra of $V$. The systematic use of this
presentation allows for the automatic translation of various
computations in spin geometry into computations with differential
forms, since it identifies the Clifford bundle of a pseudo-Riemannian
manifold with its Kähler-Atiyah bundle, which is a certain deformation
quantization of the exterior bundle. This affords a particularly
'rigid' approach to certain problems in spin geometry, which is quite
useful for example in supergravity and string theory, where one often deals with various kinds of generalized Killing spinors. So this relates to a specific isomorphic presentation of Clifford algebras and Clifford bundles which is useful for certain problems. As such, it is not merely the basic theory of Clifford algebras and Clifford bundles (since it concerns certain specific realizations of such through polyvectors and forms), nor is it something radically different.
This point of view is pursued in quite some detail in the preprint:
Cortes, Lazaroiu and Shahbazi: "Spinors of real type as polyforms and
the generalized Killing equation", arXiv:1911.08658 [math.DG], https://arxiv.org/abs/1911.08658,
where the Chevalley-Riesz isomorphism is discussed in Section 3.2. Other references on the subject are:
Calin-Iuliu Lazaroiu, Elena-Mirela Babalic, Ioana-Alexandra Coman,
"Geometric algebra techniques in flux compactifications", Adv. High
Energy Phys. 2016, 7292534,
https://www.hindawi.com/journals/ahep/2016/7292534/, https://arxiv.org/abs/1212.6766
Calin-Iuliu Lazaroiu, Elena-Mirela Babalic, "Geometric algebra
techniques in flux compactifications (II)", JHEP06(2013)054,
https://link.springer.com/article/10.1007%2FJHEP06%282013%29054, https://arxiv.org/abs/1212.6918
C. I. Lazaroiu, E. M. Babalic, I. A. Coman, "The geometric algebra of
Fierz identities in arbitrary dimensions and signatures",
JHEP09(2013)156,
https://link.springer.com/article/10.1007/JHEP09(2013)156, https://arxiv.org/abs/1304.4403
A nontrivial application of this approach (the application is rigorous in that they prove some hard theorems) can be found in the papers:
Elena Mirela Babalic, Calin Iuliu Lazaroiu, "Foliated eight-manifolds
for M-theory compactification", JHEP01(2015)140,
https://link.springer.com/article/10.1007%2FJHEP01%282015%29140, https://arxiv.org/abs/1411.3148
Elena Mirela Babalic, Calin Iuliu Lazaroiu, "Singular foliations for
M-theory compactification", JHEP 03 (2015) 116,
https://link.springer.com/article/10.1007/JHEP03(2015)116, https://arxiv.org/abs/1411.3497
whose main results are summarized in these proceedings:
https://arxiv.org/abs/1503.00373
https://arxiv.org/abs/1503.00273
Lazaroiu and collaborators are string theorists, mathematical
physicists and mathematicians, so they wouldn't be interested in merely talking about Clifford algebras by another name (they do make plentiful use of Clifford algebras and bundles in their papers). Here are some examples of the kind of work which they do in spin geometry:
C. Lazaroiu, C. S. Shahbazi, "Complex Lipschitz structures and bundles of complex Clifford modules", Differential Geometry and its Applications, Vol. 61, Dec. 2018, pp. 147-169, https://www.sciencedirect.com/science/article/abs/pii/S0926224518302018?via%3Dihub
C. Lazaroiu, C. S. Shahbazi, "Real spinor bundles and real Lipschitz structures", Asian Journal of Mathematics, Vol. 23, No. 5 (2019), pp. 749-836, https://www.intlpress.com/site/pub/pages/journals/items/ajm/content/vols/0023/0005/a003/
Vicente Cortés, C. I. Lazaroiu, C. S. Shahbazi, "N=1 Geometric Supergravity and chiral triples on Riemann surfaces", Communications in Mathematical Physics volume 375, pp. 429–478(2020), https://link.springer.com/article/10.1007%2Fs00220-019-03476-7
Ok so this, like in the above answer by AlexArvanitakis, is another context in which Clifford/geometric algebras appear, of which I wasn't aware. Thank you both for your answers.
@Qfwfq: That's the same context.
@AlexArvanitakis: I was referring to the context in my question (which is the same as that in this question https://mathoverflow.net/questions/352362/what-is-a-coordinate-less-definition-of-differentiable-manifolds) vs that of your answer (which, by the way, is surely the same as the context of amathematician's answer). I haven't read through anything at all, but my impression was that the two answers (present so far) deal with string theory and supergravity, while the work of Hestenes et al. only involves general relativity and classical mechanics. Was my impression inaccurate?
@Qfwfq The applications in the papers of Lazaroiu et al are to string theory and supergravity, though the wider approach which they develop works for many other problems in spin geometry and spinorial global analysis. "Geometric algebra" is a topic in the theory of Clifford bundles, which is itself a subject in ordinary spin geometry. As shown in the papers I mentioned, it is a rich subject which provides computationally and conceptually useful perspectives on certain problems but it is not a separate or new theory in mathematics. Lazaroiu et al are very clear on that point.
@amathematician: fair enough
@Qfwfq and amathematician: agreed with both of you, sorry for the confusion...
This answer is my opinion. I don't have references with me at the moment. Feel free to suggest some by making edits.
Geometric algebra is a school of thought towards linear algebra, geometry, and applications thereof, consisting of the following pedagogical idea:
An exposition using generators and relations, which is easier for relative mathematical novices to understand than the usual account of Clifford algebra.
That is combined with various contrasting geometric interpretations of Clifford algebra chosen whenever convenient:
The elements of a Clifford algebra can be thought of as rotations of some space. The space is often the familiar Euclidean space in $n$ dimensions, or a more exotic pseudo-Euclidean space given by a bilinear form. The latter possibility allows us to use GA to study the geometry of relativistic spacetime, which is given by the bilinear form $x^2 + y^2 + z^2 - (ct)^2$.
As representing the isometry group in various non-Euclidean geometries, some of which are not pseudo-Euclidean spaces. These geometries include famous examples like hyperbolic and elliptic geometry, which are all Cayley-Klein geometries. These transformation groups includes more than just the rotations and reflections that fix the origin of a vector space, but can also include non-origin-preserving transformations like translations that can be useful in things like computer graphics.
As a foundation for synthetic geometry a la Euclid. There are resemblances to this in Conformal Geometric Algebra where the elements of a Clifford algebra are sometimes interpreted as familiar geometric objects like circles, line and points. A product between two lines can mean the intersection of the two lines, which is a point. Notice that the elements of a Clifford algebra are no longer thought of as being transformations necessarily, but also as subsets of the plane.
As an extension of exterior algebra, where instead of interpreting elements of a Clifford algebra as (say) circles, we interpret them as scalars, vectors, bivectors, trivectors, and so forth. This might be helpful when working with differential forms, but I'm not sure.
As a slick way of expressing some exotic geometries like Laguerre's geometry. This enables you to do things like study these exotic geometries by visualising them on a computer, as is illustrated by the Wikipedia article I linked to. This is helpful because these geometries are fairly unintuitive, but tools like computer animations can make up for that.
Stated more briefly (and with an eye towards a more formal treatment), GA $\approx$ Clifford algebra equipped with a geometric interpretation which can be any of the above. Much of the difficulty of the theory is giving a complete account of each geometric point of view. These geometric interpretations contrast with each other, as I've tried to explain above.
Apart from any one else's hijacking of the phrase "Geometric Algebra", my own professional sense of this (from early 1970s) was more based in the sense of it in E. Artin's "Geometric Algebra" notes.
Further, "geometric algebra" would have been the way that many people then (and now) understand the features of many so-called "classical groups": orthogonal groups, unitary groups, symplectic groups, and somewhat more exotic groups defined using quaternions and so on... all this "over $\mathbb R$" or "over $\mathbb C$")
For number-theoretic purposes, the p-adic analogues of archimedean theorems about quadratic or unitary forms are important. And the Hasse-Minkowski theorem about local-to-global in that regard.
To understand the "adelic" and/or "global vs. local" aspects of the isometry groups of such structures, it is indeed very useful to present them in terms of "geometric algebra".
Also, in a somewhat different direction, on $\mathbb R^n$, or on a symmetric space $G/K$, a "Dirac operator", conceptually a kind of square root of the Laplacian/Casimir operator, is beautifully (and possibly not-otherwise) defined, very naturally, as a linear differential operator having coefficients in a Clifford algebra related to the ambient structure. It is exactly the right thing.
I gather that physicists' analogues of this have been used for decades, without much formalization.
I've had some PhD students do interesting work on "automorphic forms" versions of "Dirac operators", and this exactly requires expression in terms of various Clifford algebras.
For that matter, D. Vogan and others conjectured, a few decades ago, some structural facts about "Dirac cohomology", in this setting.
So, sure, Clifford, Dirac, and other physicists certainly showed the utility of such ideas... and, by this year (and for a few decades) mathematicians have appreciated the physical content...
I am unsure how appropriate this answer is. I hope to provide some explanation for the recent interest in geometric algebra within computer graphics.
My understanding of this viewpoint comes from the article "Geometric Algebra" by Eric Chisolm (https://arxiv.org/pdf/1205.5935.pdf), and the article "Course notes Geometric Algebra for Computer Graphics, SIGGRAPH 2019" (https://arxiv.org/pdf/2002.04509.pdf).
The story as I understand it, as a computer scientist, is that one starts with a real vector space $V$ of dimension $n$, and from this one constructs a much larger vector space $\mathbf{G}(V)$ which is freely generated by the union of all exterior powers $\bigcup_{i=0}^n \bigwedge^i V$. This allows us to talk about oriented subspaces, as well as "linear combinations" of subspaces of different dimensions.
This space has an enormous amount of structure. It's naturally a vector space. It is graded by the exterior product, is equipped with an involution (hodge duality), and forms an algebra with the geometric product. These structures interact very well, and allow us to model geometric transformations such as rotations and reflections with the same algebra, instead of what is usually done wthin computer graphics, where one uses matrices to represent scaling and quaternions to represent rotation.
As the next step, it's possible to projectivize the entire construction above. This enriches all of the above structures with projective duality, and also enables us to represent translations on the original space as linear transforms on the projective space. This unifies all three of scaling, rotation, and translation which makes it quite attractive for describing transformations in computer graphics.
The first part of your answer is otherwise known as Clifford algebra.
Sometimes, the Hodge dual is not an involution. Thanks for the links anyway.
That's interesting, could you drop a reference to a case where Hodge duality isn't involutive? (In my mind, it wouldn't be a duality if it weren't involutive...)
I think the space freely generated by the union should be replaced by a more mundane direct sum
This is too long for a comment.
I found the article
Lazaroiu, Calin Iuliu; Babalic, Elena Mirela; Coman, Ioana Alexandra, Geometric algebra techniques in flux compactifications, Adv. High Energy Phys. 2016, Article ID 7292534, 42 p. (2016). ZBL1366.83098.
on the classification of Killing (s)pinors using geometric algebra with applications to $\mathcal N=1$ M-theory compactifications to 3D.
Their perspective on geometric algebra is explained in section 3. The central object seems to be what they call the Kaehler-Atiyah algebra over some (pseudo)Riemannian manifold $M$ which as far as I can tell will reduce to geometric algebra as in Wikipedia when $M$ is Minkowski space. They also sketch how the KA algebra is obtained by a quantization procedure.
EDIT: In light of the sociological comments in the third paragraph in the question I should point out that I don't think the authors of that paper would say they are in the group of people who identify as "GA" (Lazaroiu, whose work I've read before, is a string theorist)
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2025-03-21T14:48:29.827193
| 2020-02-10T15:15:17 |
352369
|
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|
Stack Exchange
|
Bounding and domination numbers for relation $\leq$ modulo $\omega$-nullsets
We say that $A\subseteq \omega$ is a nullset if $$\lim\sup_{n\to \infty} \frac{|A\cap n|}{n+1} = 0.$$
Let $\omega^\omega$ denote the set of functions $f:\omega\to\omega$. We define a pre-ordering relation $\leq^0$ on $\omega^\omega$ by saying that $f\leq^0 g$ if $f(x) \leq g(x)$ for all $x\in\omega\setminus N$ where $N\subseteq \omega$ is a nullset.
Similarly to the bounding number and the dominating number respectively, we define
${\frak b}^0 = \min\{|B|: B\subseteq \omega^\omega \land \forall f\in \omega^\omega\; \exists b\in B(b\not\leq^0 f)\}$, and
${\frak d}^0 = \min\{|D|: D\subseteq \omega^\omega \land \forall f\in \omega^\omega\; \exists d\in D(f\leq^0 d)\}$.
Do we have ${\frak b}^0={\frak b}$? And what about ${\frak d}^0={\frak d}$?
The answer seems to be positive according to this paper: Barnabás Farkas, Lajos Soukup: The zero density ideal, cardinal invariants and related forcing problems.1
Theorem 2.3. If $\mathcal I$ is a rare ideal on $\mathbb N$ then $\mathfrak b = \mathfrak b_{\mathcal I}$ and $\mathfrak d = \mathfrak d_{\mathcal I}$.
Just before this theorem the authors mention that the ideal $\mathcal Z_0$ of the sets with zero density is a rare ideal.
A similar result is shown for analytic P-ideals in Corollary 5.5 of More on cardinal invariants of analytic P-ideals by the same two authors (arXiv, eudml). Again, this class of ideals includes $\mathcal Z_0$.
1I wasn't able to find whether the paper was published somewhere, but a preprint is available here (Wayback Machine). The same paper was also mentioned in this answer: Are these two quotients of $\omega^\omega$ isomorphic?
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2025-03-21T14:48:29.827340
| 2020-02-10T15:26:06 |
352370
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|
Stack Exchange
|
hypergraph product that preserve expansion properties
I am looking for a hypergraphs product of hypergraph H1,H2 that preserves some expansion properties of H1,H2.
The expansion property I am looking at is HD-random walk.
The product I am looking for is similar to normal hypergraph product, that is similar to a cartesian product, but not very much.
I work in the specific case where H2 is 4-complete hypergraph and H1 is a 3-uniform hypergraph that expands.
I am working on proving that the normal product is OK, but I don't want to reinvent the wheel.
Thanks.
What is "HD-random walk"? I can't find this anywhere.
High dimensional random walk. For example, see here: https://arxiv.org/abs/1604.02947. Basically, in the limited case of 3-uniform, it is a random walk on the graph of 2-edges where $e' \sim e$ if $e' \cup e \in T(H)$. $T(H)$ are the triplets/ 3-edges.
I just say that I found such a candidate construction and proved its expansion properties in the simplest case. To be declared in a future paper.
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2025-03-21T14:48:29.827445
| 2020-02-10T15:56:20 |
352373
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Stack Exchange
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Sum of products of irreducible characters of the symmetric group over a subgroup
When trying to build a dual formulation for lattice gauge theories using Weingarten integration I am getting sums of the kind
$$I^{m, n}_{\mu, \nu} (\sigma, \tau) = \sum_{\pi \in S_n} \chi_\mu (\pi \sigma) \chi_\nu(\pi \tau), $$
where $\sigma$ and $\tau$ are some fixed permutations in $S_m$ ($m \geq n$), $\mu$ and $\nu$ are some irreducible representations of $S_m$ defined by the partitions of $m$, and $S_n$ is a subgroup of $S_m$ leaving the last $m - n$ elements in place.
Now when $m = n$ the sum is trivial using the orthogonality rule for irreducible representations. Also when $\sigma$ and $\tau$ are in $S_n$, I can just use the Littlewood-Richardson rule to expand restrictions of $\mu$ and $\nu$ to $S_n$ into irreducible representations of $S_n$ and get the result. But can something be done in the general case?
I guess the answer cannot be written in terms of $S_m$ characters of $\sigma$ and $\tau$ since fixing the $S_n$ subgroup breaks the permutation symmetry -- in particular
$$ I^{m, n}_{\mu, \nu} (\rho^{-1} \sigma \rho, \rho^{-1} \tau \rho) = \Sigma^{m, n}_{\mu, \nu} (\sigma, \tau) $$
does not necessary hold for every $\rho \in S_m$, (just for $\rho$ in $S_n$). Still maybe there is some way to obtain the sum?
One possibility I am thinking of is taking the representation matrices for $\pi$, $\sigma$ and $\tau$ as $\Pi_\mu$, $\Sigma_\mu$, $T_\mu$ and rewrite
$$I^{m, n}_{\mu, \nu} (\sigma, \tau) = \sum_{\pi \in S_n} {\Pi_\mu}_{i j} {\Sigma_\mu}_{j i} {\Pi_\nu}_{k l} {T_\nu}_{l k}, $$
where $i$, $j$ enumerate Young tables of shape $\mu$ and $k$, $l$ enumerate Young tables of shape $\nu$.
After that if I could somehow express ${\Pi_\mu}_{i j}$ in terms of the matrix elements of irreducible representations of $S_n$, I would be able to use the orthogonality property for the matrix elements and get an answer in terms of matrix elements of $\sigma$ and $\tau$. But this would require some analogue of Littlewood-Richardson rule for matrix elements. (There should be a basis in which $\Pi_\mu$ is a block diagonal matrix, built from irreducible representation matrices for $S_n$, but I am not sure how to define such a basis).
Being a physicist, it is hard for me to understand if this is a simple problem, or an unsolvable one, so any advice is appreciated.
In the case when the restrictions of the two characters to $S_n$ don't have common irreducible constituents the answer is $0$, it follows from general group theory and is contained in the paper http://ftp.math.uni-rostock.de/pub/preprint/2005/pre05_10.ps
Thank you very much. This can be used to simplify the result i had (removing extra terms in the summation). Also intersting is the "only if" part -- so for all the terms that are left the result must be nonzero. Still, it would be very useful to know any kind of simplification for nonzero terms (though maybe it does not exist).
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2025-03-21T14:48:29.827670
| 2020-02-10T16:23:46 |
352377
|
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|
Stack Exchange
|
Identifications between different phase spaces
I've discovered Adam's lecture notes on statistical mechanics after posting my first question about Minlo's discussion on continuous Gibbs measures. Adam's lecture notes are really good, but there is some little difference between his and Minlo's book that I'd like to clarify.
On page 27, Adams introduces two sets, as follows (with $\Lambda \subset \mathbb{R}^{d})$:
$$\Gamma_{\Lambda, N} :=\{\omega \subset \Lambda \times \mathbb{R}^{d}, \omega =\{(q,p_{q}), q \in \hat{\omega}\}, |\hat{\omega}|=N\} \quad (1)$$
$$ \Gamma_{\Lambda} := \{\omega \subset \Lambda \times \mathbb{R}^{d}, \omega = \{(q,p_{q}), q \in \hat{\omega}\}, |\hat{\omega}|<+\infty\} \quad (2) $$
In addition, for each $\Delta \subset \Lambda$ Borel-measurable, define a counting variable $N_{\Delta}$ to be the function $N_{\Delta}:\Gamma_{\Lambda} \to \mathbb{R}$ by $N_{\Delta}(\omega) = |\omega \cap \Delta|$. The $\sigma$-algebra on $\Gamma_{\Lambda}$ generated by the family of counting variables $\{N_{\Delta}\}$ is denoted by $\mathcal{B}_{\Lambda}^{\infty}$.
Then, the notes procceeds by introducing the following definition.
Definition: Let $\Lambda \subset \mathbb{R}^{d}$, $\beta > 0$ and $\mu \in \mathbb{R}$. Define the phase space $\Gamma_{\Lambda}:=\bigcup_{N=0}^{\infty}\Gamma_{\Lambda, N}$, where $\Gamma_{\Lambda, N} := (\Lambda\times \mathbb{R}^{d})^{2N}$ is the phase space of exactly $N$ particles and equip it with $\mathcal{B}_{\Lambda}^{\infty}$. The probability mesure $\gamma_{\Lambda, \beta}$ on $(\Gamma_{\Lambda},\mathcal{B}_{\Lambda}^{\infty})$ such that the restrictions $\gamma_{\Lambda, \beta}|_{\Gamma_{\Lambda, N}}$ have densities:
$$\rho_{\Lambda,\beta}^{(N)}(x) = Z_{\Lambda}(\beta, \mu)^{-1}e^{-\beta H_{\Lambda}^{(N)}(x)-\mu N}$$
where $H_{N}^{(N)}$ is the Hamiltonian for $N$ particles in $\Lambda$ is called grand canonical ensemble in $\Lambda$.
Well, Adams seems to be using two different notions of phase space at the same time. In the above definition, $\Gamma_{\Lambda} = \bigcup_{N=0}^{\infty}\Gamma_{\Lambda,N}$ but the $\sigma$-algebra $\mathcal{B}_{\Lambda}^{\infty}$ only makes sense in $\Gamma_{\Lambda}$ given by (2). Thus, it seems to me that he's using some identification between $\Gamma_{\Lambda,N}$ as given by (1) and $(\Lambda \times \mathbb{R}^{d})^{2N}$. But what is this idetification? Note that nothing prevents us to take an element in $(\Lambda \times \mathbb{R}^{d})^{2N}$ which has equal entries and this would led to a single point in $\Gamma_{\Lambda,N}$ as given by (1). What m I getting wrong here?
$\newcommand\La{\Lambda} \newcommand\Ga{\Gamma} \newcommand\om{\omega} \newcommand\R{\mathbb R}$
To me, all this looks quite terrible. In formula (4.17) on page 28 of the linked lecture notes (corresponding to your formula (1)), Adams "defines" the "phase space for exactly $N$ particles in box $\La\subset\R^d$" as
$$\Ga_{\La,N} :=\{\om\subset\La\times \R^d\colon \om=\{(q,p_q), q\in\hat\om\}, |\hat\om|=N\},$$
"where $\hat\om$, the set of positions occupied by the particles, is a locally finite
subset of $\La$, and $p_q$ is the momentum of the particle at positions $q$." Here, unfortunately, Adams mixes a mathematical definition with its informal, physical interpretation. In particular, $p_q$ is only "defined" as a "momentum". Also, "locally finite" does not seem to be defined anywhere in the notes, and why one needs here "locally" is at best unclear, because we have $|\hat\om|=N$, which already implies that $\hat\om$ is just finite (assuming that the number $N$ of "particles" is a natural number). You can see how many problems just this one little "definition" has.
With this only physically "defined" $p_q$, my best guess is that $\Ga_{\La,N}$ was meant to be the set of all functions of the form $\om\colon\hat\om\to\R^d$, where $\hat\om$ is any subset $\La$ with $|\hat\om|=N$. Any such function is (or, if you prefer, can be identified with) a set of the form $\{(q_1,p_1),\dots,(q_N,p_N)\}$, where $q_1,\dots,q_N$ are pairwise distinct points in $\La$ and $p_1,\dots,p_N$ are arbitrary points in $\R^d$. So, $\Ga_{\La,N}$ is the image of the subset of the set $(\La\times\R^d)^N$ consisting of all $N$-tuples $((q_1,p_1),\dots,(q_N,p_N))\in(\La\times\R^d)^N$ with pairwise distinct $q_1,\dots,q_N$ under the map that maps the $N$-tuples $((q_1,p_1),\dots,(q_N,p_N))$ to the corresponding sets $\{(q_1,p_1),\dots,(q_N,p_N)\}$. So, $\Ga_{\La,N}$ may be viewed as a set of dimension $2Nd$.
On the other other hand, in Definition 4.6 on page 28 of the same lecture notes, $\Ga_{\La,N}$ is defined as the plain product set $(\La\times\R^d)^{2N}$ of dimension $2d\times2N=4Nd$, which is twice the dimension of $\Ga_{\La,N}$ according to the previous definition. So, there seems to be no way to reconcile these two different definitions of $\Ga_{\La,N}$, and then to reconcile the corresponding two different definitions of $\Ga_\La$. And indeed, the definition of the $\sigma$-algebra $\mathcal B_\La^\infty$ is applicable only to the first definition of $\Ga_\La$.
By the way, I strongly suggest that you avoid using commas in place of colons in definitions of sets; e.g., avoid writing $A:=\{x\in X,x>0,x<1,x^2>1/2\}$ instead of $A:=\{x\in X\colon x>0,x<1,x^2>1/2\}$. The colon here stands for "such as" and hence plays a role quite different from that of the commas (which stand for "and"), and that should be reflected in the notation.
First of all, thank you so much for the answer and also for the suggestion! Let me ask you something. You said that in Definition 4.6, $\Gamma_{\Lambda, N}$ is assumed to be the plain product $(\Lambda\times \mathbb{R}^{d})^{N}$ which has twice the dimension of your first definition of $\Gamma_{\Lambda, N}$. Actually, he writes $(\Lambda \times \mathbb{R}^{d})^{2N}$ instead of $(\Lambda\times \mathbb{R}^{d})^{N}$ (did you mistyped it?). This seems to have dimension $4Nd$. But I thought the problem was just a misprint, and it should be $(\Lambda\times\mathbb{R}^{d})^{N}$ instead. Could it be?
If $(\Lambda\times\mathbb{R}^{d})^{2N}$ was replaced by $(\Lambda\times\mathbb{R}^{d})^{N}$, it seems to me that the dimension would be consistent with your first definition.
@MathMath : Yes, I mistyped that (now corrected), and it is possible that Adams mistyped that too (in the opposite direction). Even then, there would remain the confusion between an $N$-tuple and the corresponding set. Overall, the entire presentation in the lecture notes seems to be pretty hopelessly confused.
Thank you! If found this other post where you answered too (curiously, couple days ago) https://mathoverflow.net/questions/352126/grand-canonical-gibbs-measure-for-continuous-systems
This could help me out. If we forget Adam's first definitions of $\Gamma_{\Lambda, N}$ and $\Gamma_{\Lambda}$ and suppose $\Gamma_{\Lambda, N}$ is given by $(\Lambda\times\mathbb{R}^{d})^{N}$, we could equip it with the Borel $\sigma$-algebra. But what would it be the $\sigma$-algebra associated to $\Gamma_{\Lambda}$ in this case? Is there a natural $\sigma$-algebra for such an union of sets?
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2025-03-21T14:48:29.828092
| 2020-02-10T16:25:46 |
352378
|
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|
Stack Exchange
|
Characterization of geodesic balls
In $\mathbb R^{n\geq3}$, spheres can be characterized by single layer potentials having constant eigenfunction. More precisely we have the following:
Theorem (H. Shahgholian) Let $\Omega\subset \mathbb R^{n\geq3}$ be a bounded smooth domain and for some $c>0$
$$\int_{\partial\Omega}\frac{1}{\|x-y\|^{n-2}}ds_x=\frac{c}{\|y\|^{n-2}},\,\,\forall y\not\in\bar{\Omega}$$
Then $\partial\Omega$ is a sphere centered at the origin. (here $ds$ represents the surface area).
Are there generalization of this statement for Riemannian manifolds and geodesic balls?
Have you tried working this out for spherically symmetric Riemannian manifolds with $\Omega$ some geodesic ball about the center? What are the results?
@WillieWong That seems like a great point to start. Currently, I am trying to see if the proof in Shagholian's can be adapted to $\mathbb S^n$.
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2025-03-21T14:48:29.828194
| 2020-02-10T16:59:25 |
352380
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Stack Exchange
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A numerical calculation for an integral
I am interested in the numerical calculation of
$$
F(\eta)=\frac{2}{π}\int_0^{+\infty}\sin(t\eta-t^3)\frac{dt}{t}\quad\text{for $\eta\ge 0$}.
$$
I believe that the function $F$ is bounded, but I do not know its upper bound. It is easy to get $F(0)=-1/3$ and various other values for $\eta \in [0,40]$ but getting above the value $40$ for $\eta$ seems to raise computational difficulties for Mathematica. It is also interesting to see that $F$ takes values strictly above 1 with the seemingly largest value $1.5$ for $\eta =3$.
As a motivation, one may note that with $g(t)= e^{it^3}\text{pv}\frac{1}{πt}$, the operator of convolution with $g$ is bounded on $L^2(\mathbb R)$ whenever its Fourier transform is bounded. We have here $F=\hat g$, up to some normalization, and a precise bound in $L^\infty$ for $F$ would provide the $\mathcal B(L^2)$ operator-norm of that convolution.
https://en.wikipedia.org/wiki/Airy_function http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/21/01/01/
@Nemo Thank you very much for that precise piece of information.
Nemo's representation of $F(\eta)$ in terms of a hypergeometric function can be evaluated without difficulty for large $\eta$:
$$F(\eta)=\frac{\sqrt{3} {\eta}^2 \Gamma \left(\frac{2}{3}\right) \; _1F_2\left(\frac{2}{3};\frac{4}{3},\frac{5}{3};-\frac{{\eta}^3}{27}\right)}{6\pi }-\frac{12 {\eta} \; _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{{\eta}^3}{27}\right)}{6\Gamma \left(-\frac{1}{3}\right)}-\frac{1}{3}$$
$$F(\eta)\rightarrow 1+\sqrt[4]{3} \sqrt{2/\pi}\frac{1}{\eta^{3/4}} \left[\sin \left(\frac{2 \eta^{3/2}}{3 \sqrt{3}}\right)-\cos \left(\frac{2 \eta^{3/2}}{3 \sqrt{3}}\right)\right],\;\;\eta\gg 1.$$
see plot, blue is $F(\eta)$, gold is the large-$\eta$ approximation (nearly indistinguishable for $\eta>5$). So $F(\eta)$ oscillates around 1, with an amplitude that decays as $\eta^{-3/4}$. The maximum 1.5487 is reached at $\eta= 3.37213$.
Great! Thank you very much.
In your asymptotic expression, it seems that the term $\sqrt{2π}$ should be replaced by $\sqrt{2/π}$.
Sorry to bother you again with that matter. I would be very grateful if you could double-check your constant $\sqrt{2\pi}$, which I believe should be replaced by $\sqrt{2/\pi}$. Thanks.
you are right, it was my typing error, apologies for the confusion.
Thanks a lot for caring to check this; of course I renew my thanks for your very helpful and thorough answer to my question.
|
2025-03-21T14:48:29.828493
| 2020-02-10T17:27:39 |
352381
|
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|
Stack Exchange
|
When is the model structure on functors correct, i.e. when does localization commute with taking functor categories?
Let $C$ be a small category and $M$ a model category. Then there are various "global" model structures (projective, injective, Reedy) on the category $Fun(C,M)$ of functors from $C$ to $M$, all with the same (levelwise) weak equivalences. The whole point of having such a model structure is that it should present the $\infty$-category $Fun(C,\tilde M)$, where $\tilde M$ is the $\infty$-category presented by $M$. But I'm not sure when this is actually the case.
Of course, by "the $\infty$-category presented by $M$", I mean $\tilde M = M[W^{-1}]$ is $M$ localized at the weak equivalences in the $\infty$-categorical sense, and similarly "the $\infty$-category presented by $Fun(C,M)$" is the $\infty$-categorical localization $Fun(C,M)[Fun(C,W)^{-1}]$.
Questions:
If $C$ is a small category and $M$ is a model category, then under what conditions do the standard model structures on $Fun(C,M)$ present the $\infty$-category $Fun(C,\tilde M)$, where $\tilde M = M[W^{-1}]$ is the $\infty$-category presented by $M$?
More generally, if $C$ and $M$ are relative categories, then under what conditions does the mapping relative category $\widetilde{Fun(C,M)} = Fun(\tilde C, \tilde M)$ where $\tilde{(-)}$ denotes taking the associated quasicategory?
In a more model-independent direction, when does localization of $\infty$-categories commute with taking functor categories? That is, when does $Fun(C,M[W^{-1}]) = Fun(C,M)[Fun(C,W)^{-1}]$ where $C,M$ are $\infty$-categories and $W \subseteq M$ is a subcategory?
Here's a possible way of setting up the problem. You have a functor from the localization of the functor category to the category of functors into the localization. You want to know when it's essentially surjective and fully faithful. The former is like a `rectification' question while the latter can be tackled using the end formula for natural transformations together with Lemma I.3.4 of Nikolaus-Scholze. I think this setup makes it clearer why the model category structures help give the equivalence, and I think you ought to get it for all the favorite model structures on functors.
I remember there are general results in Cisinski's work for the special case in which $M$ is a Brown category of Fibrant objects. I think it can be found in his book on higher categories, in the part on localization and categories with fibrations.
There is an article by Lenz, which formulates a positive answer nicely in the language of derivators: https://arxiv.org/abs/1712.07845
If your model structures are assumed to have small limits or colimits, the answer to the question of the title is: always. For any model category $M$ and any small category $C$, inverting levelwise weak equivalences in $Fun(C,M)$ is equivalent to considering the $\infty$-category of functors from $C$ to $M[W^{-1}]$. This is a special case of Theorem 7.9.8 (and Remark 7.9.7) in my book on higher categories. It is even possible to take for $M$ a model $\infty$-category in the sense of Mazel-Gee. In fact, Theorem 7.5.8 gives sufficient conditions on $M$ which are much more general: essentially, the mere existence of a class of well behaved fibrations is good enough (this includes Brown's categories of fibrant objects, but, more generally, a version where we do not assume all objects to be fibrant; in particular, all the variations on semi-model structures are OK) if we assume further properties: we mainly need this extra structure to exist on functor categories $Fun(C,M)$ as well (which is automatic in practice, as explained in Example 7.9.6 and Remark 7.9.7 of loc. cit.). If we restrict ourselves to those $C$ whose nerve is a finite simplicial set (e.g. finite partially ordered sets), this kind of properties is true in a much greater level of generality; see Theorem 7.6.17.
Observe that, if $M$ is good enough in the sense that $\widetilde{Fun(C,M)}\cong Fun(C,\tilde M)$ for any small category $C$, then, it is automatic that, for any subcategory $S\subset C$, the localization of the full subcategory of $Fun(C,M)$ whose objects are those functors sending the maps of $S$ to weak equivalences will autmatically be a model of the $\infty$-category of functors from $C[S^{-1}]$ to $\tilde M$.
All of this is obviously model independent.
|
2025-03-21T14:48:29.828796
| 2020-02-10T17:59:39 |
352382
|
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|
Stack Exchange
|
On the difference $\operatorname{Li}(\theta(x))-\pi(x)$
In G. Robin's paper, more precisely in Lemme12, how does he use formula (39) to prove formula (36)?
[1] Robin, Guy, "Estimation de la fonction de Tchebychef θ sur le k -ième nombre premier et grandes valeurs de la fonction ω(n) nombre de diviseurs premiers de n , Acta Arithmetica 42, 367-389 (1983). MR0736719, ZBL0475.10034.
The paper is available at
https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/42
Lemme 12 states,
Nous avons les majorations suivantes: Pour $x\ge2$
$${\pi(x)\log\theta(x)\over\theta(x)}\le1+{1\over\log\theta(x)}+{2,89726\over\log^2\theta(x)}\tag{36}$$ avec egalite pour $x=p_{442}$,
$${\pi(x)\log\theta(x)\over\theta(x)}\le1+{1\over\log x}+{2,96690\over\log^2 x}\tag{37}$$ avec egalite pour $x=p_{224}-0$.
Formula (39) is:
$$\int_2^x{\theta(t)\over t\log^2t}\,dt\le{x\over\log^2x}\left(1+{2,843\over\log x}\right)$$
(Okay, it seems the OP has tried to ask this question twice before; the first time it was closed with one reason being that more details are needed. I believe that OP is trying in good faith to provide details; the problem is in the deleting and then reasking in another post. Technically, this is a site violation.) To the OP: in future, please do not delete and reask; instead, edit the original question to improve it.
Why don't you start from a strong form of the PNT, say $|\theta(t)-t|\le A t/\log^2 t,|\pi(t)-\sum_{n\le t} 1/\log n|\le B t/\log^3 t$, you'll obtain that 36 is true for $t\ge T$ so you can check numerically for $t\le T$.
It doesn't seem clear to me either. If all the $\log\theta(x)$ terms were $\log x$, I can see Robin's method working; but at the very least, how $\log\theta(x)$ is dealt with is inadequately explained.
My problem is how to use them?
|
2025-03-21T14:48:29.828956
| 2020-02-10T18:38:55 |
352386
|
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|
Stack Exchange
|
Does there exist a compactly supported smooth function on $\mathbb{R}$ with all derivatives bounded by 1?
The question is the same as in the title. More precisely, does there exist a nonzero smooth mapping $f : \mathbb{R} \to \mathbb{R}$ such that for all $k \in \mathbb{N}$, the $k$-th derivative $f^{(k)}$ satisfies $\sup_{t \in \mathbb{R}} \lvert f^{(k)}(t)\rvert \le 1$ (with $f^{(0)}$ being understood as $f$ itself), and $f$ itself is compactly supported?
I suspects the answer is affirmative, but I can not find a proof at the moment. Here is some elementary observations:
If we drop the requirement of compact support, then functions like $\sin(x)$ and $\cos(x)$, as well as suitable linear combinations of them, all satisfy this property. Perhaps one can construct examples out of these functions.
Perhaps one can mollify piecewise linear functions satisfying the bound condition for all derivatives except on a finite number of points.
If all derivatives are bounded by 1, the Taylor series converges. This makes the function analytic, which is inconsistent with compact support.
@MichaelRenardy Convergence of the Taylor series by itself is insufficient; the series must converge to the function for it to be analytic. For example, $f(x) = e^{-\frac{1}{x^2}}$ has a Taylor series that converges in a neighborhood around each base point, but it is not analytic.
Yes, you are right.
However, if all derivatives are bounded by 1, you can use the remainder estimate of Taylor's theorem to show that the Taylor series actually represents the function.
Just expanding the comment of @MichaelRenardy. Such a function does not exists. Indeed, by Taylor's theorem with the Lagrange form of the remainder, we get for every $a\in\mathbb{R}$ and $x\in\mathbb{R}$ that $$f(x)=f(a)+f'(a)(x-a)+\dots+\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(k+1)}(y)}{(k+1)!}(x-a)^{k+1}$$
for some $y$ between $a$ and $x$.
Now choose $a$ outside the support of $f$ so that all derivatives vanish. Then the bound on the $(k+1)$-st derivative implies that $|f(x)|\leq (x-a)^{k+1}/(k+1)!$ and it remains to let $k$ tend to infinity.
|
2025-03-21T14:48:29.829117
| 2020-02-10T19:12:52 |
352389
|
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|
Stack Exchange
|
Infinite simple group of exponent a power of 2
Let G be a simple group of exponent 2^n for n>1. Is G necessarily finite? If not, what is an example of an infinite simple group of exponent a power of 2?
No (for $n$ large enough). For $n$ large enough it's known that the free Burnside group $B(2,2^n)$ is infinite, and also known that it's virtually aperiodic. (I call a group aperiodic if its only finite quotient is ${1}$). This finite index aperiodic subgroup being finitely generated and nontrivial, it has a simple quotient, necessarily infinite.
For every large enough $n$, there exist infinite simple groups of exponent dividing $n$.
(I call a group aperiodic if it has no nontrivial finite quotient.) Indeed, by the solution to the restricted Burnside problem the Burnside group $B(2,n)$ is known to be virtually aperiodic; let $B(2,n)^\circ$ be its (unique) minimal finite index subgroup, it is thus finitely generated. For $n$ large enough, $B(2,n)$ is known to be infinite. Hence $B(2,n)^\circ$ is infinite too, hence has a simple quotient, also aperiodic and hence infinite. $\Box$
The above proof produces groups of exponents dividing $n$, probably with some more efforts we can get exponent exactly $n$ but I think it's unimportant. I don't know if one can obtain quasi-finite groups in this context, when the exponent is $2^m$ (quasi-finite means infinite with all proper subgroups finite); these are close variants of Tarski monsters. Also probably one can produce infinitely generated countable examples, and uncountable too, but this requires other/additional arguments. Nevertheless:
$\forall n$, there exists no infinite locally finite simple group of exponent $n$. (Nor even in which every element has order dividing some power of $n$.)
For $n=2^k$ (or more generally $p^aq^b$ with $p,q$ prime, these would be locally solvable and indeed Malcev proved that there is no infinite, locally solvable simple group. In general, this is essentially due to Hartley (1995) (Springerlink behind paywall). Namely he proved that for every finite subset $F$ in a simple locally finite group $G$, there exists a finite subgroup $H$ containing $F$, and a normal subgroup $N$ of $H$ such that $H/N$ is simple and $F$ projects injectively into $H/N$. Since (by classification) finite simple groups of given exponent have bounded order, this implies the result. (Hartley also quotes Meierfrankelfeld for the same result.) The statement is explicit in a 2005 paper by Cutolo-Smith-Wiegold (ScienceDirect), Lemma 4.
To finite group theorists "$X$ is involved in $Y$", usually means$Y$ has subgroups $A,B$ with $B \lhd A$ such that $A/B \cong X$" (probably what you mean by "$X$ is a subquotient of $Y$"). . Hartley wasn't exclusively a finite group theorist, but I knew him, and I think he would have used "involved" in that sense.
Thank you for your reply. So for sure I can find an infinite simple groups with exponent a power of 2, as the simple factor of the Burnside group B(2,2^m) for m large enough, right? Do you have any good reference for this? Thanks again.
@marcos you missed the point: a priori these simple quotients of $B(2,2^m)$ might be finite. I pass to the aperiodic finite index subgroup and then take a simple quotient and then I'm sure that the quotient is infinite. The cost is that the number of generators is not $2$. Probably it's true that $B(2,2^m)$ has infinite simple quotients too, but would require a further argument.
I see, so I need to consider the aperiodic finite index subgroup to conclude. Is there any reference where I can find all of it? Thanks.
|
2025-03-21T14:48:29.829345
| 2020-02-10T19:22:04 |
352390
|
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|
Stack Exchange
|
Bending the hemisphere
Let $S^+$ be the upper hemisphere of the standard sphere in $\mathbb R^3$ and $b$ -- the boundary of $S^+$ (the equator).
Let $b'$ be a small isometric deformation of $b$ (a nearby curve of the same length). Is there a smooth (or of regularity higher than $C^2$) isometric immersion of $S^+$ which bounds $b'$?
I'm also interested in the infinitesimal version, i.e. if $X$ is a vector field along $b$ that does not stretch $b$, can $X$ be extended to a vector field along $S^+$ that is the derivative of a family of isometric immersions of $S^+$?
In other words: can an open set in the space of configurations of $b$ coincide with the subspace of configurations of $b$ induced from isometric immersions of $S^+$?
The answer to the infinitesimal version is 'no', which makes it very unlikely that the answer to the isometric deformation version is 'yes'. Here is how one can see this:
One can parametrize the upper hemisphere by the unit disk $x^2+y^2\le 1$ conformally by the well-known formula
$$
F(x,y) = \left(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{1-x^2-y^2}{1+x^2+y^2}\right).
$$
Let $V(x,y)$ be a vector field along the image of $F$, which can be expressed uniquely in the form
$$
V(x,y) = a(x,y)\,F_x(x,y) + b(x,y)\,F_y(x,y) + c(x,y)\,F(x,y)
$$
for some functions $a$, $b$, and $c$ on the unit disk. Then the condition that $V$ determine an infinitesimal isometric deformation, i.e., $\mathrm{d}F \cdot \mathrm{d}V = 0$, is easily seen to be the system of $3$ equations
$$
a_x-b_y = 0,\qquad a_y+b_x = 0,\qquad c = \frac{2(xa+yb)}{1+x^2+y^2} - a_x\,.
$$
Thus, $a+ib$ must be a holomorphic function of $z = x+iy$, and $c$ is determined in explicitly in terms of $a$ and $b$. In particular,
$$
a + ib = \sum_{n=0}^{\infty} c_n\,z^n
$$
for some complex coefficients $c_n$, $n\ge 0$.
Now, restrict everything to the boundary of the disk, set
$$
E_0(\theta) = F(\cos\theta,\sin\theta),\qquad
E_1(\theta) = F_x(\cos\theta,\sin\theta),\qquad
E_2(\theta) = F_y(\cos\theta,\sin\theta),
$$
so that $E_0$, $E_1$, and $E_2$ are an orthonormal frame field along the boundary curve (i.e., the equator). Let $W(\theta)$ be a vector field along the boundary. It can be written uniquely in the form
$$
W(\theta) = f_0(\theta)\,E_0(\theta)+f_1(\theta)\,E_1(\theta)+f_2(\theta)\,E_2(\theta)
$$
for some $2\pi$-periodic functions of $\theta$. The condition that $W$ furnish an infinitesimal isometric deformation of the boundary curve, i.e., $\mathrm{d}E_0\cdot\mathrm{d}W = 0$, is easily seen to be the differential equation
$$
f_0(\theta) = \frac{\mathrm{d}}{\mathrm{d}\theta}\bigl(\sin\theta\,f_1(\theta)-\cos\theta\,f_2(\theta)\bigr)
$$
Notice that $f_1$ and $f_2$ can be arbitrary $2\pi$-periodic functions of $\theta$. However, if $W$ is to be the boundary value of an isometric deformation vector field $V$ as above, we will have to have
$$
f_1 + i f_2 = a(\cos\theta,\sin\theta) + ib(\cos\theta,\sin\theta)
= \sum_{n=0}^{\infty} c_n\,\mathrm{e}^{in\theta},
$$
i.e., the 'negative' Fourier coefficients of $f_1+if_2$ must all vanish.
Thus, the generic infinitesimal isometric deformation of the boundary cannot be tracked by an infinitesimal isometric deformation of the hemisphere.
Does you description of infinitesimal deformations work for an arbitrary embedding of a disk to $R^3$ [that is, infin. deformations are parametrized by holomorphic functions]? There is an example of an infinitesimally rigid disk, "collander surface" of E. Rembs [take a convex surface with a strictly convex closed plane curve on it such that the surface touches the plane of the curve along the curve. suppose that the curvature is strictly positive on one of the caps. then it is infinit. rigid]. I don't see which part of your computation breaks down in this case.
@DmitryK: No, the description I gave above very much uses the specific parametrization to simplify the conditions for an infinitesimal deformation. I don't see how to generalize this to an arbitary surface.
|
2025-03-21T14:48:29.829597
| 2020-02-10T19:46:12 |
352393
|
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|
Stack Exchange
|
Book on manifolds from a sheaf-theoretic/locally ringed space PoV
I'm looking for an introductory (or rather, non-advanced) book on manifolds as locally ringed spaces, i.e., from the algebraic geometric viewpoint. Most introductory texts only introduce manifolds from the differentiable viewpoint; I wonder if a text introducing differential manifolds from an algebraic viewpoint exists.
Locally ringed spaces are not algebraic objects, they are algebraic-geometric objects.
@MartinBrandenburg That's what I meant, perhaps I was imprecise. I fixed my wording
Locally ring spaces are neither algebraic nor algebraic-geometric objects (they have nothing to do specifically with algebraic geometry); they are sheaf-theoretic objects.
There is a serious mismatch between the title and the question asked in the body of the post. The author should clarify whether he is looking specifically for a treatment using locally ringed spaces, as opposed to treatments that largely parallel those found in algebraic geometry books. Nestruev's book is a prime example of the latter, but not the former.
@DmitriPavlov Thanks for pointing this out. Since I'm looking for an introductory text, apparently this is a field which I know little about. Yes, I mean the former, not the algebraic geometry of complex manifolds covered in, e.g., Griffiths and Harris.
@xuq01: I am not sure why you brought up complex manifolds. In fact, complex manifolds need sheaves and locally ringed spaces much more than smooth manifolds, since in the holomorphic setting you do not have partitions of unity, which means you cannot get rid of sheaves and locally ringed spaces, unlike the case of smooth manifolds.
@DmitriPavlov Thanks for pointing this out (again, I have little idea how locally ringed spaces actually work). I brought this up simply because complex manifolds are commonly covered in algebraic geometry texts.
I'm not very familiar with this book (in particular, I don't know how introductory or not it is), but I think
Torsten Wedhorn, Manifolds, Sheaves, and Cohomology. Springer Studium Mathematik—Master. Springer Spektrum, Wiesbaden, 2016. xvi+354 pp. ISBN: 978-3-658-10632-4; 978-3-658-10633-1
would fit your description.
Ramanan's "Global Calculus" is a very nice introductory text which defines manifolds by their sheaves of differentiable functions. I don't know if this is what you're looking for: I don't see it as very algebraic, and tools from analysis have an important role.
Jet Nestruev (a collective author, I think) does this in "Smooth Manifolds and Observables".
In that book manifolds are defined by charts in the usual way and regarded as formal dual of certain commutative $\mathbb{R}$-algebras. But is a ringed space POV also present in that book?
This is the ringed space POV, no? (Although now that you mention it I don't think they prove that the $\mathbb R$-algebras they consider are obtained from sections of the structure sheaf of such-and-such ringed space.)
@Qfwfq For real smooth manifolds the ringed space POV is redundant, in that it suffices to deal with global sections. Bump functions and other special properties make everything affine, so to speak.
@Michael Bächtold: yes, I'm aware of that, I was just considering the OP
In Introduction to differential geometry (see the review) by R.Sikorski the author introduces the concept of (what is now called) Sikorski space. Sikorski spaces are "affine, reduced differential spaces" and hence they can be approached algebraically by looking at their coordinate rings. Differentiable manifolds are important examples Sikorski spaces. Unfortunately the book was not translated to english. Luckily there are some publications (in english and perhaps in french) by Sikorski in which he explains this very natural concept. One of them is Differential modules.
A book $C^{\infty}$ Differentiable Spaces by Navarro González and Sancho de Salas develop theory of differentiable spaces by first constructing real spectra for smooth algebras and then glue them in order to obtain general spaces. This is analogical to the development of algebraic geometry (scheme theory) by Grothendieck and his school. The book might be a bit advanced.
|
2025-03-21T14:48:29.830268
| 2020-02-10T20:58:02 |
352399
|
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"Iosif Pinelis",
"Nate Eldredge",
"Robert Furber",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/36886",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/61785"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352399"
}
|
Stack Exchange
|
Explicit examples of (probability) measures on $\prod \mathbb{R}$
Let $\prod_{n \in \mathbb{N}} \mathbb{R}$ be equipped with the Tikhonov product of the Euclidean topologies on $\mathbb{R}$ and let $B$ the corresponding Borel $\sigma$-algebra. What is are some concrete examples of:
Locally-positive Borel probability measures on $(\prod_{n \in \mathbb{N}} \mathbb{R},B)$,
Locally-positive $\sigma$-finite (but not finite) Borel measures on$(\prod_{n \in \mathbb{N}} \mathbb{R},B)$?
How does the situation change when the product is indexed over $\mathbb{R}$?
There is also this construction:
https://eudml.org/doc/281217
The situation changes dramatically when the product is indexed over $\mathbb{R}$, as the product space is no longer Polish. There are tons of interesting probability measures on the countable product of $\mathbb{R}$ (indeed, the study of these is practically the entire field of probability theory), and basically none on the uncountable product.
Wait a second - your bullet points have a product over $n \in \mathbb{R}$. Is that a typo? Are you mainly interested in countable or uncountable products, or both?
Here's a construction working for both cases (countable and continuum product): the product is then separable (though non-metrizable for an uncountable product), so there a dense sequence $(x_n)$, then the measures $\sum\delta_{x_n}$ and $\sum 2^{-n}\delta_{x_n}$ are positive on nonempty open subsets, are $\sigma$-finite and finite respectively. (Of course they have atoms, but this is not excluded.)
@NateEldredge This was a typo, I was initially interested in products over $\mathbb{R}$ both $\mathbb{N}$ but had simplified the question rapidly after to $\mathbb{N}$.
$\newcommand\R{\mathbb R}$ Let $T$ be any countable nonempty set. Let $B$ be the Borel $\sigma$-algebra over $\R^T$ generated by the Tikhonov product topology on $\R^T$. Take any $t_0\in T$. For each natural $k$ and each $t\in T$, let
$$\nu_{k,t}:=
\begin{cases}
N(0,1)&\text{ if } t\ne t_0,\\
N(k,1)&\text{ if } t=t_0.
\end{cases}
$$
By Kolmogorov's measure extension theorem , for each natural $k$ there is a product probability measure
$$\mu_k:=\bigotimes_{t\in T}\nu_{k,t} $$
on $B$, which will obviously be locally positive.
Moreover,
$$\mu:=\sum_{k=1}^\infty\mu_k$$
will be a locally-positive $\sigma$-finite (but not finite) measure on $B$.
The same constructions will work for uncountable sets $T$ if, instead of the Borel $\sigma$-algebra $B$, we will take the $\sigma$-algebra generated by the standard base of the Tikhonov product topology.
Indeed, the Kolmogorov extension theorem is the main tool for producing probability measures on this space, and is probably the first thing the OP ought to study if not already familiar with it.
But AFAIK, if $T$ is uncountable then Kolmogorov does not give an extension to the Borel $\sigma$-algebra, only to the product $\sigma$-algebra, which is much smaller. I do not know of any way to produce nontrivial measures on the Borel $\sigma$-algebra of an uncountable product of $\mathbb{R}$, except using things like measurable cardinals.
@NateEldredge : You are right. I have now corrected this.
@NateEldredge What is a measurable cadinal? (ie: a nice reference to them where I can read up?)
@NateEldredge How can measurable cardinals help produce measures on $\prod_\mathbf{R}\mathbf{R}$ (which is much smaller than any measurable cardinal)? On the other hand this space is separable so it's easy to produce fully supported Borel probabilty measures on this space (e.g. atomic, but non-atomic too, by tensoring with a fully supported proba on $\mathbf{R}$.
@YCor I believe Nate Eldredge is alluding to the product measure extension axiom, or PMEA. This allows us to extend the measures described by Iosif Pinelis not only to the Borel $\sigma$-algebra of $\mathbb{R}^{\mathbb{R}}$, but all the way to the power set $\sigma$-algebra. PMEA implies there exists a real-valued measurable cardinal $\kappa < \mathfrak{c}$, and the consistency of PMEA is implied by the existence of a supercompact cardinal.
@YCor: I really shouldn't have said that because I don't actually know what I'm talking about. All I mean is that in general I don't know of "interesting" ways to produce measures on such large spaces, certainly not any that give us such a rich class of measures as for countable products, and that the only examples I know of "large" spaces with really nontrivial measures are things like measurable cardinals.
@MrMMS: Usually a good resource is http://cantorsattic.info, but it seems to be down at present. Otherwise there's Wikipedia. But as mentioned I am really not a good person to ask about such things.
|
2025-03-21T14:48:29.830600
| 2020-02-11T04:25:04 |
352420
|
{
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],
"authors": [
"Anton Petrunin",
"D.-C. Cisinski",
"Gerrit Begher",
"Jiří Rosický",
"Martin Brandenburg",
"Paul Taylor",
"Pietro Majer",
"Wlod AA",
"YCor",
"Yemon Choi",
"https://mathoverflow.net/users/1017",
"https://mathoverflow.net/users/110389",
"https://mathoverflow.net/users/1261",
"https://mathoverflow.net/users/14094",
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"https://mathoverflow.net/users/2733",
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"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/73388",
"https://mathoverflow.net/users/763"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352420"
}
|
Stack Exchange
|
Category of metric spaces
Is there a standard/good reference text that does category of metric spaces?
Say, it seems that by looking at this category one can recover everything about particular metric space up to scaling --- is it written somewhere, or does it follow from something standard?
I had never seen "metric maps" but they're well often called "1-Lipschitz maps".
The obvious choice of morphisms between metric space is continuous functions, but then you have the category of metrisable topological spaces. Elsewhere, morphisms are supposed to preserve structure, which is what 1-Lipschitz maps do.
Bill Lawvere has a way to look at this as a particular case of enriched categories. See http://www.tac.mta.ca/tac/reprints/articles/1/tr1abs.html and the nLab entry on the subject.
@Denis-CharlesCisinski I don't think the Lawvere paper, seminal though it might be (waves to the Edinburgh magnitude crew) serves well as a reference for the category that the OP is asking for. I assume (but Anton Petrunin can correct me if I am wrong) that what is desired is a reference which will state the properties of this category (existence and nature of finite limits/colimits) and its relation with other categories via various adjoint functors
Anton: perhaps you could add some detail or examples of what you're looking for? as in, what kinds of definitions/theorems/facts are you hoping for such a text to contain?
In fact there are several classes of maps that define categories with metric spaces as objects: Bounded, Uniformly Continuous, Hölder, Lipschitz, 1-Lipschitz, Isometries,... and their local versions, e.g. Locally Lipschitz. Plus, those classes that are already defined between topological spaces (Continuous, Proper,...)
I would not call it a standard reference but we needed this category in https://arxiv.org/pdf/1608.05524.pdf.
@YemonChoi, I need some basics, a treatment of injective metric spaces for example.
@YemonChoi Regarding Lawveres paper: I always wanted to know more about the relationship between the $L^p$ business from functional analysis and the remark by Lawevere on $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$.
@GerritBegher Yes, but what does that have to do with Anton Petrunin's question?
Can I just point out to those who are, almost by reflex, reaching for the Lawvere paper, that Anton is concerned with the category whose objects are metric spaces, not metric spaces as categories, nor with the right "categorification" of the notion of metric spaces. I am not familiar with Isbell's papers but surely they would be more relevant to this particular question?
It is also worth pointing out that AFAICT Lawvere deals with V-enriched categories for V a suitable category, obtaining a notion of "metric space" which does not satisfy the symmetry axiom. It may be the case that Lawvere's category is from a certain POV the correct one to study, but it does mean that properties of the category Anton is asking about will have to be translated from properties of the category of "Lawvere metric spaces" by various symmetrization arguments
@YCor, ' I had never seen "metric maps" ' ==== I was using the term "metric maps" (and metric category) in my papers over half a century ago.
The question needs more focus.
@MattF. "up to rescaling". For $\lambda>0$ one has the obvious automorphism $(X,d)\mapsto (X,\lambda d)$ on objects, $f\mapsto f$ on maps.
@MattF. No --- I need a reference (it is easy to prove, and likely it is done somewhere).
@MartinBrandenburg, It seems that there is no such reference --- so the question has no answer, but I do not see a way to make it more focused.
|
2025-03-21T14:48:29.830982
| 2020-02-11T04:57:08 |
352421
|
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"url": "https://mathoverflow.net/questions/352421"
}
|
Stack Exchange
|
Second order differentiability of convex functions
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a convex function. Then $f$ is locally Lipschitz and hence differentiable a.e. (Rademacher). Let $E\subset\mathbb{R}^n$ be the set of points where $f$ is differentiable.
In fact, the second order distributional derivatives of $f$ are Radon measures (a simple consequence of the Riesz representation theorem). Let $D^2f$ be the absolutely continuous part of the distributional second order derivative.
Theorem 1. The classical Aleskandrov theorem states that for almost all $x\in\mathbb{R}^n$,
$$
\lim_{y\to x}
\frac{|f(y)-f(x)-Df(x)(y-x)-\frac{1}{2}(y-x)^TD^2f(x)(y-x)|}{|y-x|^2}=0.
$$
This is Theorem 6.9 in [EG]. The argument used there is purely analytic and is based on a careful analysis of weak derivatives.
In fact, using a very different and more geometric argument (see [AA] (7.3) and (7.4)) one can prove that in addition to the above second order differentiability:
Theorem 2. For almost all $x\in E$ we have
$$
(*)\ \ \ \ \ \ \ \ \ \ \ \ \ \
\lim_{E\ni y\to x}
\frac{|Df(y)-Df(x)-D^2f(x)(y-x)|}{|y-x|}=0.
$$
The proof given in [AA] is limited due to its geometric nature to the case of monotone operators (derivative of a convex function is an example of a monotone operator) while the proof given in [EG] seems to be more flexible.
Question.
Is it possible to modify the proof given in [EG] so that it would also include the result listed in $(*)$?
For a related question, see Aleksandrov's proof of the second order differentiability of convex functions.
[AA] L. Ambrosio, G. Alberti, A geometric approach to monotone function in $\mathbb{R}^n$. Math Z. 230(1999), 259-316.
[EG] L. C. Evans, R. F. Gariepy, Measure Theory and Fine Properties of Functions. CRC Press.
The following argument is used in my recent paper with D. Azagra and A. Cappello (work in progress).
The answer is yes. I asked about whether the proof of Theorem 1 can be modified so to cover Theorem 2. In fact the following stronger version of Theorem 2 is a direct cosequence of Theorem 1.
Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, and it is twice differentiable at $x$, then
$$
\lim_{y\to x}\sup_{\sigma_y\in\partial f(y)}\frac{|\sigma_y-Df(x)-D^2f(x)(y-x)|}{|y-x|}=0.
$$
We can always assume that $x=0$ by placing the origin at $x$.
If $f$ is twice differentiable at $0$ as in Theorem 1, then we have
$$
f(x)=f(0)+Df(0)x+\frac{1}{2}x^TD^2f(0)x+R(x)=
f(0)+Df(0)x+\langle Ax,x\rangle +R(x),
$$
where $A=\frac{1}{2}D^2f(0)$ and $R(x)=0(|x|^2)$.
Note that
$$
a(r):=\sup_{0<|x|\leq 2r}\frac{|R(x)|}{|x|^2}\to 0
\qquad
\text{as $r\to 0^+$.}
$$
Moreover,
$$
|R(x)|\leq a\Big(\frac{|x|}{2}\Big)\, |x|^2\leq a(|x|)|x|^2.
$$
It suffices to prove:
Proposition. If a convex function $f:\mathbb{R}^n\to\mathbb{R}$ is twice differentiable at $0$ as in Theorem 1, then
$$
\lim_{x\to 0}\frac{\sigma_x-Df(0)-D^2f(0)x}{|x|}=0,
$$
whenever $\sigma_x\in\partial f(x)$.
Proof. For points $x,y\neq 0$, we have
$$
f(x)=f(0)+Df(0)x+\langle Ax,x\rangle +R(x),
\quad
f(y)=f(0)+Df(0)y+\langle Ay,y\rangle +R(y).
$$
Since $f(x)+\langle \sigma_x,y-x\rangle\leq f(y)$, we have
$$
\langle\sigma_x,y-x\rangle
\leq
f(y)-f(x)=
Df(0)(y-x)+\langle A(x+y),y-x\rangle+R(y)-R(x).
$$
We used here the fact that $A$ is symmetric and hence $\langle Ax,y\rangle=\langle Ay,x\rangle$.
Let
$$
y=x+w,
\quad
\text{where}
\quad
w=\sqrt{a(|x|)}\,|x|z,\ |z|=1.
$$
Then
$$
\langle\sigma_x,w\rangle\leq Df(0)w+
\langle A(2x+w),w\rangle +R(y)-R(x),
$$
$$
\langle\sigma_x-Df(0)-2Ax,w\rangle\leq
\langle Aw,w\rangle +R(y)-R(x).
$$
If $|x|$ is sufficiently small, then $a(|x|)\leq 1$ and hence $|w|\leq |x|$, so $|y|\leq 2|x|$. Therefore,
$$
|R(y)|\leq a\Big(\frac{|y|}{2}\Big)\, |y|^2\leq 4a(|x|)|x|^2,
\qquad
|R(y)-R(x)|\leq 5a(|x|)|x|^2.
$$
Taking the supremum over all $z$ with $|z|=1$ we get
$$
|\sigma_x-Df(0)-2Ax|\sqrt{a(|x|)}|x|\leq |A|a(|x|)|x|^2+5a(|x|)|x|^2,
$$
and hence
$$
\frac{|\sigma_x-Df(0)-2Ax|}{|x|}\leq (|A|+5)\sqrt{a(|x|)}\to 0
\quad
\text{as $x\to 0$.}
$$
Since $2A=D^2f(0)$, the result follows.
|
2025-03-21T14:48:29.831243
| 2020-02-11T07:20:43 |
352427
|
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"Thomas Yang",
"Wojowu",
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}
|
Stack Exchange
|
Explicit construction of Kakeya sets using Perron tree
I have found many excellent notes online that illustrate how to construct a Kakeya needle set (with measure $<\varepsilon$.) Yet none of them gives full argument about the construction of a Kakeya set (with measure zero). The closest one is given in Page 6 of
https://web.stanford.edu/~yuvalwig/math/teaching/KakeyaNotes.pdf,
which unfortunately leaves the detail of arguing for the existence of a unit line segment in $\cap_{i=1}^\infty U_i$. It says this is proved subtly using a compactness argument.
What is this argument?
Fix some orientation of unit line segments. Let $S_n$ be the set of possible centers of intervals contained in $U_n$ with that orientation. Then $S_n$ is a decreasing sequence of closed sets which has a nonempty intersection by compactness.
I believe your question should be asking about the construction of a Besicovitch set, not a Kakeya set. I think this terminological hiccup caused some confusion in the answer below (which I don't think answers the question you intended to ask).
@Wojowu Thanks! I thought Besicovitch sets and Kakeya sets mean the same thing?
@Wojowu Following your idea, I have posted a proof in my answer. Thanks!
Let me answer my own question, following @Wojowu's idea and Page 6 of https://web.stanford.edu/~yuvalwig/math/teaching/KakeyaNotes.pdf,
after the lemma.
We use $|A|$ to denote the 2 dimensional Lebesgue measure of a set. Fix an equilateral triangle $T$, and a bounded open set $U_1 \supseteq T$ with $|U_1|=|\overline U_1| < 2 |T|$. By chopping up and translating $T$, we form a new set $S_1$ of area $<1/2 $, and by the lemma we can ensure that $S_1 \subseteq U_1$. Now pick a new open set $U_2$ such that $S_1\subseteq U_2\subseteq U_1$ and that $|U_2|=|\overline U_2|< 2 |S_1|$. Since $S_1$ is a union of triangles, we can apply the previous lemma to each such triangle and get a new set $S_2$ of area $<1/4$, such that $S_2 \subseteq U_2$. Again, we pick a new open set $U_3$ with $S_2\subseteq U_3\subseteq U_2$ and $|U_3|=|\overline U_3| < 2|S_2|$, and iterate this. Notice that when we do this, we get a sequence of sets $S_i$ with $|S_i| < 2^{-i}$, and a nested sequence of compact sets $\overline U_1 \supseteq \overline U_2 \supseteq \overline U_3 \supseteq \cdots$, with
$|U_i|=|\overline U_i| < 2 |S_{i-1}| < 2^{-i}$.
Therefore, if we set $B =\cap_{i=1}^\infty \overline U_i$, then $B$ will automatically have measure zero. We now prove that $B$ will contain a unit line segment in every direction. In particular, $B$ is nonempty.
Let $e$ be a direction. Let $C_i$ be the collection of all possible centres $c$ such that the unit line segments $\{c+te:t\in [-1/2,1/2]\}\subseteq \overline U_i$. Since $\overline U_i$ is compact, $C_i$ is also compact. Also, $C_i$'s are nested since $\overline U_i$'s are nested. By construction, each $C_i$ is nonempty since $\overline U_i\supseteq S_i$ and $S_i$ contains a unit line segment in $e$. Hence, $\cap_i C_i$ is nonempty. This means that their is a unit line segment in $e$ which is contained in $\cap_i \overline U_i=B$.
|
2025-03-21T14:48:29.831453
| 2020-02-11T08:26:12 |
352429
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"ABIM",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/36886"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352429"
}
|
Stack Exchange
|
Disjoint union of measures
This is a sort of follow-up question to this old post I came across.
Setup:
Let $\{X_n\}_{n \in \mathbb{N}}$ be a collection of Hausdorff topological spaces and let $\{\Sigma_n\}_{n \in \mathbb{N}}$ be their respective Borel $\sigma$-algebras. Let $\left\{\mu_n\right\}_{n \in \mathbb{N}}$ be a family of finite Borel measures on $X_n$, respectively.
Question:
Can one define the disjoint union of $\left\{(X_n,\Sigma_n,\mu_n)\right\}_{n \in \mathbb{N}}$?
What I have So far:
Then one can define the disjoint union $\sigma$-algebra on $\bigsqcup_{n \in \mathbb{N}} X_n$ as the subset $\Sigma$ of $\bigsqcup_{n \in \mathbb{N}} \Sigma_n$ consisting of all sets $S$ satisfying
$$
S \cap X_n \in \Sigma_n \,(\forall n \in \mathbb{N}).
$$
Therefore, one may define the coproduct measurable space $(X,\Sigma)$ to be $\left(\bigsqcup_{n\in \mathbb{N}} X_n, \Sigma\right)$.
Note: (I'm curious, does this correspond to the Borel $\sigma$-algebra on the disjoint union somehow?)
Does the following:
$$
\mu\triangleq \sum_{n =1}^{\infty} \mu_n,
$$
work?
I can't find a problem as $\Sigma$ restricted to (the copy of) $\Sigma_n$ (within $\Sigma$) is contained in $\Sigma_n$. So (informally written) $\mu\rvert_{\Sigma_n}=\mu_n$ but is defined on a (potentially) smaller $\sigma$-algebra than $\Sigma_n$ ….
Is this correct?
You're not introducing $\mu_n$.
@YCor I added in the required information. Thanks for pointing that out, of course it's crucial to the problem.
As written your definition doesn't quite make sense because $\bigsqcup_{n \in \mathbb{N}} \Sigma_n$ doesn't contain nearly enough sets (e.g., $X \mathrel{:=} \bigsqcup_n X_n$ itself is not in it). It seems better to me just to set
$\Sigma = \left\{ \bigsqcup_n A_n : A_n \in \Sigma_n \right\}.$ Equivalently, this is the set of all $A \subset X$ for which $A \cap X_n \in \Sigma_n$ for every $n$. You can easily check that this is a $\sigma$-algebra and that it equals the Borel $\sigma$-algebra of the disjoint union topology on $X$.
This definition of $\mu$ works fine, and in fact your restriction statement is true on the nose: for every $A \subset X$ with $A \subset X_n$, we have $A \in \Sigma$ iff $A \in \Sigma_n$, and in this case we have $\mu(A) = \mu_n(A)$. The $\sigma$-algebra is not smaller.
Note that everything here would also work just fine for uncountable disjoint unions, though of course the resulting measure $\mu$ need not be $\sigma$-finite.
The first issue: In order to talk about a coproduct / disjoint union we need a category for our objects to live in.
A natural candidate seems to be the following category (result of a Grothendieck construction):
Objects are triples $X=(S_X, \Sigma_X, \mu_X)$ where $S_X$ is a set, $X_\Sigma$ is a $\sigma$-algebra and $\mu_X$ is a measure.
Morphisms $f:X\to Y$ are maps $f:S_X\to S_Y$ that are measurable and satisfy $f_*\mu_X\leq \mu_Y$ (edit) up to extensional equality on measures (i.e. $f_*\mu=g_*\mu$ for all $\Sigma_X$-measures).
Now let's see if there is a coproduct for a family $X_i$ of such objects:
Underlying set: We take the direct sum $\coprod_i S_{X_i}$.
$\sigma$-algebra: We take the largest $\sigma$-algebra s.t. all the inclusions $S_{X_i}\to\coprod_i S_{X_i}$ are measurable (this should exist, if I am not mistaken, since the set of such $\sigma$-algebras is stable under ascending chains).
The measure: We take the smallest measure s.t. all the inclusions are morphisms in our category. This measure exists; explicitly: $$\mu_\amalg(E):=\inf_i \mu_{X_i}(E).$$ To show that this is a measure, I think the following should be true: We have $\sup_j(\inf_i a_{i, j})=\inf_i(\sup_j a_{i, j})$ given the $a_{i, *}$ are countable, ascending sequences.
By construction, all the inclusions $e_i:S_{X_i}\to\coprod_i S_{X_i}$ are morphisms in our category.
Now let $f_i:X_i\to Y$ be morphisms in our category. Then the induced map $f:S_{\coprod_i X_i}\to S_Y$ is a morphism in our category:
It is measurable since we have $e_i^*f^*\Sigma_Y=f_i^*\Sigma_Y\subseteq\Sigma_{X_i}$ and so $f^*\Sigma_Y\subseteq \Sigma_\amalg$ by construction.
It is also compatible with the measures since the pushforward of measures respects the order.
The uniqueness of this morphism is inherited from the category $\mathcal{Set}$.
Now, I am not sure if this is fully the answer you want since it is not specifically about Hausdorff spaces with finite Borel measures. But the question could then be tackled further by comparing the convergences of the Hausdorff topologies, Borel $\sigma$-algebras and respective measures.
A starting point could be to investigate if the subcategory $\mathcal{Haus}$ we get when restricting to Hausdorff spaces is reflective or coreflective. This could be used to show a lax or oplax distributivity in the following sense: The identity map is most likely a morphism $\coprod_{\mathcal{Haus}}X_i\to \coprod X_i$ (or the other way around?) given the coproduct in $\mathcal{Haus}$ exists.
In "D.H. Fremlin's "Measure Theory", vol. 2, 214L", Fremlin defines the disjoint union of measure spaces. He calls it "direct sum" of measure spaces. Here I quote from the book:
"214L Direct sums Let $\langle(X_i, \Sigma_i, \mu_i)\rangle_{i \in I}$ be any indexed family of measure spaces. Set $X =\bigcup_{i \in I}(X_i \times \{i\})$; for $E \subseteq X$, $i \in I$ set $E_i=\{x : (x,i) \in E\}$. Write
$$ \Sigma= \{E:E \subseteq X, E_i \in \Sigma_i \text{ for every } i \in I\}, $$
$$ \mu E = \textstyle {\sum_{i \in I}} \mu_i E_i \text{ for every } E \in \Sigma. $$
Then it is easy to check that $(X, \Sigma,\mu)$ is a measure space; I will call it the direct sum of the family $\langle(X_i, \Sigma_i, \mu_i)\rangle_{i \in I}$.
Note that if $(X, \Sigma,\mu)$ is any strictly localizable measure space, with decomposition $\langle X_i\rangle_{i \in I}$, then we have a natural
isomorphism between $(X, \Sigma,\mu)$ and the direct sum $(X', \Sigma',\mu')=\bigoplus_{i \in I}(X_i, \Sigma_{X_i},\mu_{X_i})$ of the subspace measures, if we match $(x,i) \in X$ with $x \in X'$ for every $i \in I$ and $x \in X_i$."
|
2025-03-21T14:48:29.831835
| 2020-02-11T08:26:26 |
352430
|
{
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"Alexandre Eremenko",
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"sharpe"
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|
Stack Exchange
|
Inverse of the Schwartz-Christoffel map and the continuity
I have a question on the Schwartz-Christoffel formula.
The map is a conformal map from the unit disk to polygons. Let me give you a specific example. In fact,
\begin{align*}
\phi(z)=\int_{0}^z (1-u^n)^{-2/n}\,du
\end{align*}
maps $\mathbb{D}$ onto the interior of a regular polygon with $n$ sides.
We can know the modulus of continuity of general conformal maps. The following result is well known.
Let $\mathbb{D} \subset \mathbb{C}$ be the unit disk centered at the
origin. A conformal map $f$ defined on $\mathbb{D}$ is $\alpha$-Hölder
continuous ($\alpha \in (0,1]$) if and only if there exists $L \in
(0,\infty)$ such that \begin{align*} |f'(z)| \le
L(1-\|z\|^2)^{\alpha-1},\quad z \in \mathbb{D}, \end{align*} where we
denote by $\|\cdot\|$ the Euclidean metric on $\mathbb{C}$.
According to this result, $\phi$ defined above is $(1-2/n)$-Hölder continuous. That is, there exists $C>0$ such that
\begin{equation}
\| \phi(x)-\phi(y)\| \le C\|x-y\|^{1-2/n}
\end{equation}
for any $x,y \in \mathbb{D}$.
However, I do not know the Hölder continuity of $\phi^{-1}$. I think that the index should be bigger than $1-2/n$. Can we show this? What is the specific and optimal index?
The inverse map satisfies $|\phi^{-1}(z)-\phi^{-1}(a)|\leq c|z-a|^{n/(n-2)}$
at any vertex $a$ (reciprocal to the exponent of the direct map). Since this exponent is $>1$ the inverse is just Lipschitz
everywhere. You do not need any general theorems for this. At the vertex the inverse map behaves like a power. At all other points it is smooth (analytic).
Thank you for your comment. If the inverse map $\phi^{-1}$ is $n/(n-2)$--Hölder continuous, it should be a constant function. Am I misunderstanding something?
Do you mean that $\phi^{-1}$ is a Lipschitz continuous function with respect to $|\cdot|$?
@Sharpe: yes, I edited.
Thank you for the clarification. I understood. If we consider a conformal map of the unit disk to a convex domain, the inverse map is likely to be Lipschitz continuous.
@sharpe: Yes, this generalization seems to be correct.
|
2025-03-21T14:48:29.832047
| 2020-02-11T10:34:53 |
352436
|
{
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"H1ghfiv3",
"Moishe Kohan",
"Stefan Witzel",
"YCor",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352436"
}
|
Stack Exchange
|
Finite models for torsion-free lattices
Let $G$ be a real, connected, semisimple Lie group and $\Gamma < G$ a torsion-free lattice. Then does there exist a finite $CW$-model for $B\Gamma$?
I know this to be true in many instances (e.g. when $\Gamma$ is uniform, when $G=\mathrm{SO}_0(n,1)$ or when $\mathrm{rank}_{\mathbb R}(G) \geq 2$ and $\Gamma$ is irreducible). In these instances, there always exist certain "canonical" finite CW-models for $B\Gamma$.
However, I am unaware of the situation for general such lattices, hence the question.
I guess many (all?) cases were done in the 70's by Borel-Serre (at least when $G$ has finite center): there's a model that's a compact manifold (with boundary). I'd thought it was known in general, but I'm not sure.
I am pretty certain that Borel-Serre constructed their finite models only for torsion-free, arithmetic lattices.
Let $\Gamma$ be such a lattice, and assume $G$ has trivial center (if $G$ has finite center this is no restriction). Then there is a unique product decomposition $G=G_1\times \dots G_n$ such that $\Gamma$ meets $G_i$ in an irreducible lattice. Passing to the overgroup of finite index generated by projections, we can reduce to the irreducible case. Then rank $\ge 2$ is fine (probably by arithmeticity + Borel-Serre), remains rank 1. This case is certainly OK too (removing a disjoint family of horoballs and modding out directly yields a manifold with boundary).
There is a Ph.D. thesis, I don't remember now exactly by whom, whose main result is the existence of compactifications in the rank $1$ case (basically generalizing the situation for hyperbolic lattices in the way that you described). The Lie groups I am interested in are all the real points of an algebraic group, so they all have finite center anyways.
I think your argument shows only that each $\Gamma$ virtually has a finite CW-model, i.e that there exists a finite-index subgroup $\Gamma' < \Gamma$ possessing a finite CW-model.
"I think your argument shows only that each $\Gamma$ virtually..." No, I was careful on that point: I project the lattices on factors, so I embed $\Gamma$ into a larger torsion-free lattice which is a direct product (I said "overgroup").
I think for Lie groups you should say "Raghunathan" where you said "Borel-Serre".
@StefanWitzel thanks for correcting. Indeed, I had in mind later papers of Borel and Serre. The statement is very precisely stated in Raghunathan's 1968 paper you're quoting.
I am afraid that I still don't understand the argument. It is clear to me that $G_i$ can be chosen such that each $\Gamma_i =G_i \cap \Gamma$ is irreducible. Moreover, the group $\Gamma' = \Gamma_1 \times \dots \times \Gamma_n$ has a finite CW-model, since each factor is irreducible torsion-free and therefore has one. It is also clear to me that $\Gamma$ is commensurable to $\Gamma'$, which is why $\Gamma$ virtually has a finite model as well. It is also clear that both $\Gamma$ and $\Gamma'$ embed as subgroups in the big group $\Gamma''$ generated by the projections of $\Gamma$.
However, neither do I understand why $\Gamma$ must have finite index in $\Gamma''$ nor why $\Gamma''$ has a finite CW-model, which is what I understand your argument to be.
@H1ghfiv3 This is slightly off-topic but you could avoid the technicalities by contenting yourself with a finitely dominated model for $B\Gamma$: Since $\Gamma$ has finite geometric dimension and a model for $B\Gamma$ of finite type, it has a finitely dominated one (Geoghegan, Prop 7.2.13).
@H1ghfiv3 I'm not sure I see your problem with YCor's argument. $\Gamma''$ is a discrete overgroup of $\Gamma'$, which is a lattice, right? The index is the quotient of the covolumes, so on particular it is finite. So $\Gamma''$ spits into factors to which you can apply the "references", so they have finite models. A finite direct product of groups with finite models has a finite model. Finally $\Gamma$ is a finite index subgroup of $\Gamma''$ so it has a finite model as well.
I suppose $\Gamma''$ might have torsion, so replace "finite model" by "proper, cocompact action on a contractible CW-complex". Since $\Gamma$ itself is torsion-free, you get the same conclusion.
How do you conclude that $\Gamma''$ splits into irreducible factors ?
@StefanWitzel indeed I claimed that the projection $\Gamma_i$ of $G_i$ is torsion-free, but this is not true. To give an example, choose two irreducible lattices $\Lambda_i$, $i=1,2$, with nontrivial homomorphisms $\pi_i:\Lambda_i\to C_2$, with $\Lambda_1$, $\mathrm{Ker}(\pi_2)$ torsion-free, but $\Lambda_2$ not torsion-free. It's easy to find such examples, e.g., as finite index subgroups of $\mathrm{PSL}_k(\mathbf{Z})$ for any $k\ge 2$. Choose $\Gamma={(x,y)\in\Lambda_1\times\Lambda_2:\pi_1(x)=\pi_2(y)}$. Then it's torsion-free, but $\Gamma_2=\Lambda_2$ is not torsion-free.
Because the torsion-free setting is unpleasantly unstable, I think the right statement to prove would be: Let $G$ be a semisimple Lie group with finite center, let $K$ be a maximal compact subgroup and $X=G/K$ (for every $G$-invariant Riemannian metric this is a symmetric space of non-compact type). Let $\Gamma$ be a lattice in $G$. Then there exists a $\Gamma$-invariant $\dim(X)$-dimensional submanifold "with corners" of $X$ on which $\Gamma$ acts cocompactly.
@H1ghfiv3: my understanding was that $\Gamma''$ is the product of projections of $\Gamma$ to the $G_i$s. So it splits by definition.
@YCor: Yes, or using Raghunathan's approach one can use Morse-theory to produce a CW-complex along the way. In any case I think proper cocompact actions are what one wants and if the group happens to be torsion-free then it has a finite model. Thanks for the explicit example!
I apologise for being so inquisitive, but why is the projection again an irreducible lattice ? It is clear to me that the intersection with each factor is one, but the intersection and the projection need not agree, as already visible in YCor's example.
Incidentally, it is an open problem that if G is a torsion free group whose finite index subgroup H admits a finite K(H,1) then there exists a finite K(G,1).
@MoisheKohan Yes that was also mentioned by the OP elsewhere in a comment. But of course you wouldn't be expected to read all the comments on this post :).
In fact, more is true and you do not need separate arguments for rank 1 and higher rank.
The following is Theorem 13.1(i) in the book of Ballmann, Gromov and Schroeder "Manifolds of nonpositive curvature":
Suppose that $(M,g)$ is a complete real-analytic Riemannian manifold of nonpositive curvature and finite volume. Then $M$ is tame: It is diffeomorphic to the interior of a compact manifold with boundary $M'$.
Moreover, the proof shows that $M'$ can be realized as a submanifold (with boundary) of $M$.
Applying this to the locally-symmetric space $(M,g)=X/\Gamma$, where $\Gamma$ is a torsion-free lattice in the isometry group of a nonpositively curved symmetric space $X$, after triangulating $M'$, we obtain a finite model for $\Gamma$.
Just one question: Do they also show that locally symmetric spaces of non-positive curvature are analytic ?
@H1ghfiv3 I think this will come from the fact that your groups are real points of algebraic groups.
@MoisheKohan Neat! You don't happen to know of any generalization to other lattices in locally compact CAT(0)-groups (like algebraic groups over local fields)? I'm asking because in certain cases there is an answer similar to the one I sketched, but it uses heavy machinery (reduction theory) and it is an open problem to understand how much of that one could do purely geometrically.
@H1ghfiv3 There are many arguments, for instance, use an equivariant isometric embedding in the symmetric space of $SL(n, R)$, which would have to be real analytic. Or use the fact that the metric on the symmetric space is obtained by applying action of it's isometry group (which is real analytic) to a K- invariant metric on the tangent soace at the point corresponding to K in G/K.
@StefanWitzel That I do not know, Gromov's argument is differential-geometric.
Do you know if there's a statement covering both results? Namely in the same setting but applying to possibly non-torsion-free lattices?
@YCor No, I do not. One can go through Gromov's argument and check if it can be modified to deal with orbifolds. My guess is that it works. Gelander, I think, would be the person to ask since he improved Gromov's proof.
I'll try to summarize YCor's comments into an answer (using big guns): Let $G$ be the real points of an algebraic group (a restriction by the OP in the comments) and assume $\Gamma$ irreducible.
Then Raghunathan shows that the answer is "yes" if $\Gamma$ is arithmetic. Margulis (Discrete subgroups of semisimple Lie groups) says that $\Gamma$ will be arithmetic if $G$ has rank at least $2$ (this is the sum over the ranks of the almost-factors). This leaves the case where $G$ is a single rank-1 factor. In that case $\Gamma\backslash G/K$ is a finite volume hyperbolic manifold from which one can cut off the cusps, see Theorem 12.7.2 in Ratcliffe, "Foundations of hyperbolic manifolds".
Edit: I'll also try to summarize the discussion about the reduction to the irreducible case. Suppose $G = G_1 \times \ldots \times G_k$ and the image of $\Gamma$ under the projection to $G_i$ is an irreducible lattice $\Gamma_i$. Let $\Gamma'' = \Gamma_1 \times \ldots \times \Gamma_k$. Now the above discussion shows that each $\Gamma_i$ acts properly and cocompactly on a contractible CW-complex $X_i$. Hence $\Gamma''$ acts properly and cocompactly on $X = X_1 \times \ldots \times X_k$. This action restricts to a proper and cocompact action of $\Gamma$. If $\Gamma$ happens to be torsion-free, the action is free and $\Gamma \backslash X$ is a finite model for $B\Gamma$.
I probably should have clarified that the irreducible case was clear to me, precisley because of the arguments that you employ. What I currently fail to understand is YCor's reasoning for general (reducible) lattices. Thanks anyways!
Your original question was a "reference request", I think these are the references. The reduction to the irreducible case may be technical but I don't think anything profound happens (no new "references").
The reduction on the irreducible case that YCor employed might be correct, I am not yet convinced (but that doesn't mean anything, I am not an expert in this area and all of what I've been asking might be very obvious to experts on Lie groups). However, the question on whether a general torsion-free discrete group $\Gamma$ has a finite CW-model for $B\Gamma$ whenever this virtually is the case is highly non-trivial, and as far as I know still open. A statement conforming this would indeed be of some profoundness (at least to me)
Oh yes, I didn't mean to claim anything in that direction. What you can get on abstract grounds is finite domination as I mentioned above. I didn't want to think about the details of YCor's reduction, but I probably should.
@H1ghfiv3 My attempt was to embed $\Gamma$ as subgroup of finite index in a product, so I didn't try to get finiteness of $B\Gamma$ from a finite index subgroup, but from a finite index overgroup. The problem with my approach is that the overgroup can have torsion even if the small group doesn't.
So here is the thing that I haven't quite sorted out: what if $\Gamma < G_1 \times G_2$ has discrete image in $G_1$ but dense image in $G_2$? This can happen for instance by choosing the wrong decomposition of $G_1 \times G_2 = \mathbb{R}^2$. I would like to say that it can only happen by choosing a wrong decomposition.
It is not clear to me why you can suppose, without loss of much generality, that each projection is an irreducible lattice (it is clear for the intersection).
@H1ghfiv3 Suppose $\Gamma < G_1 \times G_2$ with $\textrm{pr}_1(\Gamma)$ discrete (because $\Gamma$ is reducible). Then $\textrm{pr}_1(\Gamma)$ is a lattice and $\Gamma \cap G_2$ is a lattice and also $\overline{\textrm{pr}_2(\Gamma)}$ has finite covolume and $\Gamma \cap G_1$ is discrete, agreed? Now I think showing that $\Gamma \cap G_1$ has finite covolume is "equivalent" to showing that $\overline{\textrm{pr}_2(\Gamma)}$ is discrete. So if it is clear to you that $\Gamma'$ is a lattice, it should also be clear that $\Gamma''$ is a lattice.
Yes, I should have phrased my question more precisely: Why is it clear to you that $pr_i(\Gamma)$ is irreducible as a lattice ?
@H1ghfiv3 Well, in general of course it's not, apply induction.
Right! Now I get it. You proceed by induction.
|
2025-03-21T14:48:29.832930
| 2020-02-11T12:14:46 |
352440
|
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"Carlo Beenakker",
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}
|
Stack Exchange
|
Not unique eigenvalues in singular value decomposition
I have the following problem: I have a matrix $M\in \mathbb{R}^{3\times 3}$ and I consider two SVD's $U_1DV_1^T$ and $U_2DV_2^T$ of $M$ with $D = \mathrm{diag}(\lambda_1,\lambda_1,\lambda_2)$. Therefore, the first and second column vector of $U_1$ are rotated with a rotation matrix $R$ $$U_2=[Ru_1^1,Ru_1^2,u_1^3]$$ because we have a two dimensional eigenspace. What is the relationship between $V_1$ and $V_2$? Is there also (the same) rotation matrix which rotates the first two rows or columns?
Many greetings and thanks in advance.
I thought the SVD is unique (up to phase factors) if the $\lambda$'s are distinct?
The first two eigenvalues are equal, so the $\lambda$'s are not distinct.
shouldn't you be rotating the rows of $U_1$, rather the columns? since you are inserting the rotation matrix beween $U_1$ and $D$, so it acts on the second index of $U_1$; and then the transpose of $R$ acts on the columns of $V_1^T$, and therefore on the rows of $V_1$.
|
2025-03-21T14:48:29.833040
| 2020-02-11T12:30:42 |
352441
|
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"Dabed",
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"url": "https://mathoverflow.net/questions/352441"
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|
Stack Exchange
|
Why does the Lax pair formalism look so similar to the Hamiltonian equations, and what is the significance of this?
If we have a Lax pair for a system, which we'll call operators $L$ and $B$, then the system
\begin{align*}L\psi&=\lambda\psi\\
\psi_t&=B\psi\end{align*}
has as its integrability condition the equation
$$L_t=[B,L].$$
It then follows that the equation defined by the above is completely integrable, and what have you, because you can prove the existence of infinitely many first integrals in involution.
But this also makes the "equation of motion" for $L$, if we can call it that, look very similar to the dynamics of a quantity in a Hamiltonian system, where $B$ takes the role of a Hamiltonian $H$; just compare this with the equation of motion for any quantity $F$ with respect to time $t$:
$$\dot{F}=\{F,H\}+\frac{\partial F}{\partial t}$$
which when $F$ does not depend explicitly on $t$ is just the equation
$$\dot{F}=\{F,H\}.$$
Then this is suggestively similar to the behaviour of the Lax pair, with the Poisson bracket in the role of the commutator, and the operator $B$ in the role of the Hamiltonian, up to sign change.
Is this just a mundane mathematical coincidence, or is there some form of significance to the Lax operator formalism that makes it appear so similar to a Hamiltonian?
This is briefly mentioned on wikipedia so it doesn't seems to be a mundane coincidence, it seems the Heisenberg equation is basically the same as the integrability condition and the derivation of both looks equal too 1 2
|
2025-03-21T14:48:29.833177
| 2020-02-11T12:34:13 |
352442
|
{
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"Greg Martin",
"Mateusz Kwaśnicki",
"Sylvain JULIEN",
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|
Stack Exchange
|
Is the imaginary part of $t\mapsto\zeta(1/2+it)$ close to the derivative of its real part?
Plotting $t\mapsto\zeta(1/2+it)$ on Wolfram alpha, it seems that the maxima of its real part are close to the zeros of its imaginary part, while the maxima of the latter seem close to the inflection points of the former. Can this be made precise? For example, is there a canonical notion of distance between those two functions that attains only small values?
This can be rephrased as follows: the Hilbert transform of the real part is close to the derivative. Or, in yet another form: the Fourier transform of $t \mapsto \Re \zeta(1/2+i t)$ is concentrated around ${-1, 0, 1}$. This indeed is the case, and can be quantified in some sense; see, for example, this answer.
Well, I should have rather written "this indeed may be the case, and possibly could be quantified". I do not claim I can quantify this at that moment.
(An extended comment.) The derivative of the real part does not really seem to be close to the imaginary part, as seen in the following picture generated by Mathematica:
Code: Plot[{Re[I Zeta'[1/2 + I t]], Im[Zeta[1/2 + I t]]}, {t, 0, 80}]
The corresponding zeroes of the two functions indeed seem to be reasonably close to each other. This is no surprise, however: $\zeta(\tfrac{1}{2} + i t)$ essentially circles around (mostly in the right half-plane). In each "circle" the distance between a maximum or minimum of the real part and the corresponding zero of an imaginary part is roughly as large as the distance froth the "center" to the real axis. And most "circles" seem to be centered near the real axis.
Thank you for your answer. Denoting respectively the derivative of the real part and the imaginary part by $Zr'$ and $Zi$, could one expect to get a result along the lines of $\int_{x_{0}}^{x}(Zr'(t)-Zi(t))dt\ll_{\varepsilon}x^{\varepsilon}$ for some $x_{0}>1$ and all $\varepsilon>0$?
I doubt this is any more regular than the integral of $\Im z(t)$, but honestly I know next to nothing about the Riemann zeta function.
Do you think my question in my previous comment should be asked separately?
@SylvainJULIEN: Again, I do not know, you may ask someone more knowledgeable.
There certainly does seem to be some sort of correlation: consider ListPlot@Table[{Re[I Zeta'[1./2 + I t]], Im[Zeta[1./2 + I t]]}, {t, 0, 800}] and ListPlot@Table[{Re[I Zeta'[1./2 + I t]], Im[Zeta[1./2 + I t]]}, {t, 700, 800, 0.1}].
|
2025-03-21T14:48:29.833380
| 2020-02-11T12:35:42 |
352443
|
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|
Stack Exchange
|
How to prove monotonicity of such function?
Let $0<a \le 1, \alpha<0$ and $\beta>0$. How to prove that the function:
$$f(x)=\frac{(\Gamma(a)-\Gamma(a,\alpha \ln(\beta x))) (\alpha\ln(x))^a}{(\alpha\ln(\beta x))^a (\Gamma(a)-\Gamma(a,\alpha \ln(x)))},$$
is decreasing for $\beta <1$ and increasing for $\beta>1$.
By drawing the graph for some values with mathematica we can expect that the result is true. Also the sign of derivative is more delicate.
Maybe one would have a smart idea to do it.
$\newcommand{\Ga}{\Gamma}$The desired result actually holds for all real $a>0$ and, moreover, $f$ is monotonic on the entire interval $(0,\infty)$.
Indeed, let $c:=-\alpha>0$ and $b:=\beta>0$. Note that for any real $u\ne0$ we have
\begin{equation}
u^{-a}(\Ga(a)-\Ga(a,u))=u^{-a}\int_0^u t^{a-1}e^{-t}\,dt
=\int_0^1 s^{a-1}e^{-us}\,ds.
\end{equation}
So, we may let $0^{-a}(\Ga(a)-\Ga(a,0)):=\int_0^1 s^{a-1}e^{-0s}\,ds=1/a$, by continuity.
So, we may write
\begin{equation}
f(x)=J_b(x)/J_1(x),
\end{equation}
where
\begin{equation}
J_b(x):=\int_0^1 s^{a-1}x^{cs}b^{cs}\,ds.
\end{equation}
Hence, $f'(x)$ is equal in sign to
\begin{align*}
& \tfrac2c\,J'_b(x)J_1(x)-\tfrac2c\,J'_1(x)J_b(x) \\
& =\int_0^1 s^a x^{cs-1}b^{cs}\,ds\, \int_0^1 t^{a-1}x^{ct}\,dt \\
& +\int_0^1 t^a x^{ct-1}b^{ct}\,dt\, \int_0^1 s^{a-1}x^{cs}\,ds \\
&-\int_0^1 s^a x^{cs-1}\,ds\, \int_0^1 t^{a-1}x^{ct}b^{ct}\,dt \\
&-\int_0^1 t^a x^{ct-1}\,dt\, \int_0^1 s^{a-1}x^{cs}b^{cs}\,ds \\
&=\int_0^1\int_0^1 ds\,dt\,x^{cs+ct-1}(st)^{a-1}(sb^{cs}+tb^{ct}-sb^{ct}-tb^{cs}) \\
&=\int_0^1\int_0^1 ds\,dt\,x^{cs+ct-1}(st)^{a-1}(s-t)(b^{cs}-b^{ct}).
\end{align*}
Now the result follows, because $(s-t)(b^{cs}-b^{ct})$ is equal in sign to $b-1$ for all distinct real $s$ and $t$.
Thank you so much!
|
2025-03-21T14:48:29.833496
| 2020-02-11T13:15:43 |
352447
|
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|
Stack Exchange
|
Pages missing in Gromov: Rigid transformations groups
I cannot find a complete version of the 1988 paper Rigid transformations groups by Gromov. The only version I can find is the one on the authors homepage. However this version appears to be missing pages 2 and 3 (it starts on page 65 and the next page is 68, among other things the definition of $r$-rigidity is missing).
Is a complete version of the paper available online?
The paper with Giuseppina d'Ambra has the complete definitions, and more details.
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2025-03-21T14:48:29.833582
| 2020-02-11T13:47:06 |
352448
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"url": "https://mathoverflow.net/questions/352448"
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Stack Exchange
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Are there (enough) injectives in condensed abelian groups?
The question is very simple : does $Cond(\mathbf{Ab})$, the category of condensed abelian groups (as defined in Scholze's Lectures in Condensed Mathematics), have enough injectives ?
Does it, in fact, have any nontrivial injective ?
Recall that $Cond(\mathbf{Ab})$ is defined to be the colimit over strong limit cardinals $\kappa$ of $Sh(*_{\kappa-proet}, \mathbf{Ab})$, where $*_{\kappa-proet}$ is the site of $\kappa$-small profinite sets (it's easier to just use extremally disconnected spaces, and it is in fact equivalent)
Each of these sheaf-categories has enough injectives, but it's not clear that the colimit does, because a priori, the left Kan extension functor (along the inclusion of $\kappa$-small extremally disconnected spaces to $\kappa'$-small ones) $Sh(*_{\kappa-proet}, \mathbf{Ab}) \to Sh(*_{\kappa'-proet}, \mathbf{Ab})$ ($\kappa<\kappa'$) has no reason to preserve injectives, and any injective comes from one of these categories.
A lot of the time one can do without actual injectives (for instance to define $R\hom$, use projectives; or you can also a lot of time use the injectives of one of the sheaf-categories to get what you need), but I suspect that they might be useful at some point; and the question seems relevant regardless
One could argue that we don't care about set-theoretic complications but it seems to me that this is one situation where they're actually non-stupid complications (that can't be solved by just saying "fix a universe"), but maybe someone can explain why they don't matter here ?
EDIT : in this comment, Scholze seems to claim that there are not enough injectives : he says "A few things that exist in pyknotic condensed sets but not in condensed abelian groups (e.g., injective pyknotic abelian groups)". So the question could become : what would be a proof of that ?
Made the tag. Suggest that if you know other questions that qualify you edit the tag in.
@DavidRoberts : thanks !
But just to emphasize, pyknotic abelian groups do have enough injectives (because they are the category of sheaves of abelian groups on a site).
@TimCampion : I don't know a lot about those, but it seems from what I've heard that they just form the abelian group objects of a certain Grothendieck topos, in which case it's obvious, right ? (my intuition for the lack of injectives in condensed abelian groups is essentially that if you're coming from $\kappa$-condensed people, there's no way you're going to stay injective in the colimit, since the left Kan extension functors have no reason to preserve those )
@Tim Campion: Pyknotic abelian groups are just $\kappa$-condensed abelian groups ($Sh(\ast_{\kappa-proet},\mathrm{Ab})$) for some choice of (strongly inaccessible) cardinal $\kappa$, so arguably Maxime Ranzi stated this right away in this question.
Indeed, there are no nonzero injective condensed abelian groups.
Let $I$ be an injective condensed abelian group. We can find some surjection
$$ \bigoplus_{j\in J} \mathbb Z[S_j]\to I$$
for some index set $J$ and some profinite sets $S_j$, where $\mathbb Z[S_j]$ is the free condensed abelian group on $S_j$ -- this is true for any condensed abelian group. But now we can find an injection
$$\bigoplus_{j\in J} \mathbb Z[S_j]\hookrightarrow K$$
into some compact abelian group $K$, for example a product of copies of $\mathbb Z_p$ for any chosen prime $p$. Indeed, it suffices to do this for any summand individually (embedding into a product in the end), and each factor embeds into a product of copies of $\mathbb Z$ (by choosing many maps $S\to \mathbb Z$), thus into a product of copies of $\mathbb Z_p$.
We remark that it is in this step that we need to work in the condensed setting: In the pyknotic setting, $J$ can be larger than the relevant cutoff cardinal for the profinite sets, so $K$ would not be in the site of $\kappa$-small compact Hausdorff spaces.
By injectivity of $I$, we get a surjection $K\to I$. In particular, the underlying condensed set of $I$ is quasicompact.
Now assume that $I$ is $\kappa$-condensed for some $\kappa$, and pick a set $A$ of cardinality bigger than $\kappa$, and consider the injection
$$\bigoplus_A I\hookrightarrow \prod_A I.$$
The sum map $\bigoplus_A I\to I$ extends to $\prod_A I\to I$ by injectivity of $I$.
I claim that the map $\prod_A I\to I$ necessarily factors over a map $\prod_{A'} I\to I$ for some subset $A'\subset A$ where the cardinality of $A'$ is less than $\kappa$. To check this, we use the surjection $K\to I$; then it is enough to prove that the map $\prod_A K\to I$ factors over $\prod_{A'} K\to I$ for some such $A'$. But this follows from $I$ being $\kappa$-condensed and $\prod_A K$ being profinite.
Thus, the sum map $\bigoplus_A I\to I$ factors over $\prod_{A'} I\to I$ for some $A'\subset A$. But then restricting the sum map along the inclusion $I\to \bigoplus_A I$ given by some $a\in A\setminus A'$ gives both the identity and the zero map, finally showing that $I=0$.
I hope I didn't screw something up.
Thanks for answering ! The result (and the proof) seem to indicate that my intuition that set-theoretic issues aren't silly here isn't completely off. There's just a point in your argument that I'm not entirely sure about : I'm not entirely sure how you embed $\mathbb Z[S]$ into a product of $\mathbb Z$'s - I don't understand how you guarantee that you get an injection from sufficiently many maps $S\to \mathbb Z$ (surely you can get an injection $S\to \prod_A\mathbb Z$, but I'm not sure how you get to $\mathbb Z[S]$). One possible way would be
to use the map to the solidification (which is itself a product of $\mathbb Z$'s), but I'm not sure that map is injective (my scribblings don't get me anywhere, I seem to need to rely on $\mathbb Z^{presheaf}[S]$ being a seprated presheaf, which isn't clear to me). Could you explain that ?
See Proposition 2.1 in http://www.math.uni-bonn.de/people/scholze/Analytic.pdf
Ah ok, I had missed that one, great ! Thanks a lot !
If I understand Peter Scholze's answer correctly, a key feature of the category of condensed abelian groups is that from any projective object P there is a monomorphism to a compact projective object Q. The same argument applies to other condensed abelian categories such as the category of condensed vector spaces over a given finite field.
How does this address the question?
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