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2025-03-21T14:48:30.411302
| 2020-04-25T14:21:05 |
358508
|
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|
Stack Exchange
|
Can one-sided derivatives always exist, but never match?
Is there a continuous function $f : \mathbb{R} \to \mathbb{R}$ which has left and right derivatives everywhere, but where those derivatives are unequal at every point?
No. Check out Denjoy-Saks-Young Theorem https://mathworld.wolfram.com/Denjoy-Saks-YoungTheorem.html
Thanks! Happy to accept if you want to make that an answer.
It seems that Denjoy-Saks-Young says that this is true even if $f$ is not assumed to be continuous (though in the end it has to be continuous a.e.)
In fact this cannot even happen on any uncountable set.
No, that cannot happen.
Let's use a Baire category argument. More precisesly: a pointwise limit of a sequence of continuous functions $\mathbb R \to \mathbb R$ is continuous everywhere except for a meager set [= set of first category]. ref.
Let $f : \mathbb R \to \mathbb R$ be continous. Assume the left-hand derivative $f^-(x)$ and the right-hand derivative $f^+(x)$ exist everywhere. Let
$$
f_n(x) = \frac{f(x+1/n)-f(x)}{1/n}
$$
Then each $f_n$ is continous and $f_n(x) \to f^+(x)$ everywhere. Therefore, $f^+$ is continuous everywhere except for a meager set. Similarly, $f^-$ is continuous everywhere except for a meager set. So there is a point $a$ such that $f^+$ and $f^-$ are both continuous at $a$.
By assumption, $f^-(a) \ne f^+(a)$. Replacing $f$ by $-f$, if necessary, we may assume WLOG that $f^-(a) > f^+(a)$. Adding a linear function to $f$, if necessary, we may assume WLOG that $f^-(a) > 0 > f^+(a)$. Because $f^+, f^-$ are continuous at $a$, there is $\delta > 0$ so that
$$
\forall u \in [a-\delta,a+\delta] \quad f^-(u) > 0\text{ and }
f^+(u) < 0 .
$$
Now, consider a point $u \in [a-\delta,a+\delta]$. Because $f^-(u) > 0$, there is $\alpha_u < u$ so that
$$
\forall x\in(\alpha_u,u),\quad \frac{f(u)-f(x)}{u-x} > 0, \text{ so } f(x) < f(u).
$$
Because $f^+(u) < 0$, there is $\beta_u > u$ so that
$$
\forall x\in(u,\beta_u),\quad \frac{f(x)-f(u)}{x-u} < 0, \text{ so } f(x) < f(u) .
$$
Thus, there is a neighborhood $(\alpha_u,\beta_u)$ of $u$ such that $f(x) < f(u)$ for all $x \in (\alpha_u,\beta_u)\setminus\{u\}$. So $f$ has a strict local maximum at $u$. But $f$ is continuous on $[a-\delta,a+\delta]$, and therefore achieves its minimum value at some point $u \in [a-\delta,a+\delta]$. A contradiction.
+!, but I think "adding a constant to $f$ if necessary" should be "adding a linear function to $f$ if necessary", because you want to shift the derivatives, not the function itself.
Looking at the argument, we get something stronger. If $f : \mathbb R \to \mathbb R$ is continuous and has left and right derivatives everywhere, then the left and right derivatives agree except on a meager set.
|
2025-03-21T14:48:30.411503
| 2020-04-25T14:22:12 |
358509
|
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|
Stack Exchange
|
What is meromorphic differentials like on Riemann Sphere?
There is a proposition that every meromorphic differential on Riemann Sphere (or $\mathbb{P}^1 = \mathbb{C} \cup \{ \infty \}$) can be written as $f dz$ where $f$ is a meromorphic function on $\mathbb{P}^1$, and that $f dz$ is meromorphic if and only if $f$ is holomorphic on $\mathbb{C}$ and $z^2 f(z)$ tends to a finite limit as $z \rightarrow \infty$.
This is Exercise 6.3. on Page 178 of Frances Kirwan's Complex algebraic curves. I know how to prove the former half of the proposition (every meromorphic differential can be written as $f dz$), but I cannot find a proof of the second half. In fact, I'm not sure whether the propostion is true, because I seem to find a meromorphic differential which is not holomorphic on $\mathbb{C}$ (e.g. $1/z$ $dz$).
Does anyone can answer my question?
Is it not a typo? Surely $fdz is holomorphic iff those conditions are satisfied.
I'm not sure whether there is a typo in the original book. Additionally, it seems that there is no non-zero holomorphic differentials on $\mathbb{P}^1$?
As you have observed there are meromorphic differentials on $\P^1$ which are not holomorphic on $\C$ (like $dz/z$). So it seems that there is a typo in the original. Also, you can probably prove that the statement is true after replacing meromorphic by holomorphic. Then, indeed, you can show that there are no non-zero functions satisfying that condition...
Maybe. Well anyway thanks for your help!
Oh, I got it. The exercise following this one is "To deduce that there are no holomorphic differentials on $\mathbb{P}^1$", and this is just a proof of it (I know another proof previously, so I didn't get what the writer means). Well, thank you very much!
The second half of the statment is not true, but if we replace "meromorphic" by holomorphic, it becomes a true proposition:
$f(z)dz$ is holomorphic if and only if $f(z)$ is holomorphic on $\mathbb{C}$ and $z^2f(z)$ tends to a finite limit as $z \rightarrow \infty$.
This is easy to prove because $$f(z_1)dz_1=-f(1/z_2)/z_2^2 dz_2$$ when changing local coordinate, and $$\lim\limits_{z_1\to\infty}f(z_1)z_1^2 = \lim\limits_{z_2\to 0}f(1/z_2)/z_2^2$$
In fact, there is no non-zero holomorphic function $f(z)$ that satisfies the condition above, so there is no non-zero holomorphic differential on $\mathbb{P}^1$.
|
2025-03-21T14:48:30.411667
| 2020-04-25T14:27:38 |
358510
|
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|
Stack Exchange
|
Good upper-bound for $\mathbb E_A[e^{-t\|A\|_2}]$, for $t\ge0$ and random m by n matrix with iid entries with law $N(0,1)$
Let $A$ be a random $m$-by-$n$ matrix with iid $N(0,1)$ entries, $m$ and $n$ large with $n/m \longrightarrow \alpha \in (0, 1)$ . Let $\|A\|_2$ be the largest singular value of $A$ (i.e the spectral norm of $A$) and let $t \ge 0$.
Question. What is a good upper bound for $\mathbb E_A[e^{-t\|A\|_2}]$ ?
I've attempted to make the title identical to the body. I've also added details about the size of shape of the matrix. Hope you're fine with the current version. Thanks.
The probability distribution of the largest singular value of $A$ (or the largest eigenvalue $\lambda_1$ of $AA^T$) is derived in Distribution of the largest eigenvalue for real Wishart and Gaussian random matrices and a simple approximation for the Tracy–Widom distribution. There are exact answers for finite $n,m$ and asymptotic forms for large $n,m$. From here finding the desired average of $e^{-t\lambda_1}$ is a matter of quadrature.
Great. Thanks for the reference. At the moment, I also came across this reference https://arxiv.org/pdf/1003.2990.pdf. A little twist: largest singular value of A is the square root of the largest eigenvalue of $AA^T$.
|
2025-03-21T14:48:30.411787
| 2020-04-25T15:07:46 |
358514
|
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"Andjela Todorovic",
"Nate Eldredge",
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|
Stack Exchange
|
If a sequence $X_n$ of RVs converges in probability to $X$, does the sequence $\mathbb{E}(X_n)$ also converge to $\mathbb{E}(X)$?
I couldn't find the answer in literature so any idea would be helpful.
This should actually have been posted to http://math.stackexchange.com instead as it is not research level. I've voted to migrate it there.
(I forgot which site I was on - I wouldn't have answered the question if I had noticed it was on MathOverflow.)
Not in general. The standard counterexample is to let $U \sim U(0,1)$, $X_n = n 1_{\{U \le 1/n\}}$, and $X=0$.
There are several basic theorems giving sufficient conditions for this to hold, e.g. monotone convergence, dominated convergence, uniform integrability. They can be found in any graduate-level probability textbook.
Thank you for the notice. What I am working on concretely are variables in form of e^tXiY, where t is a real number, and Xi, Yj are independent sequences of random variables having binomial distribution. How to figure out would the condition hold in this case?
Maybe look in the literature for results on sufficient conditions, as the answer suggests...
|
2025-03-21T14:48:30.411904
| 2020-04-25T17:59:37 |
358529
|
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|
Stack Exchange
|
On properties of an algebra as a bimodule
Let $A$ be a two-sided artinian ring.
Recall that a module $M$ is said to have dominant dimension at least $n$ in case the terms $I_i$ in the minimal injective coresolution of $M$ are projective for $i=0,1...,n-1$.
$A$ is said to have dominant dimension at least $n$ in case the regular module $A$ has dominant dimension at least $n$ (which is equivalent to the bimodule $A$ having dominant dimension at least $n$).
$M$ is said to be $n$-torsionfree in case $Ext_A^i(Tr(M),A)=0$ for $i=1,...,n$.
Note that being 1-torsionfree is the same as being torsionfree and being 2-torsionfree is the same as being reflexive.
Here $Tr(M)$ is the Auslander-Bridger transpose of $M$.
Consider the 3 conditions:
a) $A$ has dominant dimension at least $n$.
b) $A$ as a bimodule is $n$-torsionfree.
c) $A$ is an $n$-syzygy as a bimodule, that is $A \cong \Omega^n(U)$ for some other $A$-bimodule $U$.
Note that an equivalent condition to condition a) for $n=1,2$ was provided by Colby and Fuller in https://www.jstor.org/stable/2043764?seq=1 in terms of conditions on the double dual functor, which also gives a strong link to condition b).
Question: Is it true that the conditions a), b) and c) are equivalent? I can show that they are equivalent for $A$ being a finite dimensional algebra. But I had to use that $A$ has a duality and one deep result of Auslander-Reiten (and some other recent results on the dominant dimension). I wonder whether a more direct proof is possible for general artinian rings.
We always have that a) implies b) and that b) implies c). So the question is only about whether c) implies a).
|
2025-03-21T14:48:30.412034
| 2020-04-25T18:21:04 |
358531
|
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|
Stack Exchange
|
Universal property for derived category of coherent sheaves
Let $X$ be a scheme, and let $D^{*}(X)$ be the unbounded (resp. unbounded, resp. bounded below/above, etc) derived category of coherent sheaves on $X$.
The work of Robalo establishes a universal property for the motivic stable homotopy category of a scheme.
My question is simple: is it reasonable to think that the more classical derived categories may have similar universal properties?
Perhaps if you explain what the universal property of the motivic stable homotopy category is, it will make the question more clear.
|
2025-03-21T14:48:30.412106
| 2020-04-25T18:26:26 |
358532
|
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"authors": [
"Pierre PC",
"Wlod AA",
"https://mathoverflow.net/users/110389",
"https://mathoverflow.net/users/129074"
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|
Stack Exchange
|
Connected Hausdorff spaces with large collection of disjoint open sets
Is there for every infinite cardinal $\kappa$ a connected Hausdorff space $(X,\tau)$ with $|X| = \kappa$ and a collection ${\cal D}$ of mutually disjoint open sets with $|{\cal D}| = \kappa$?
Based on the previous answer of Wlod AA, such a space always exists for $\kappa$ infinite (it is clear that it cannot exist for $2\leq\kappa<\omega$).
Let $I$ be a countable Hausdorff connected space (see A connected countable Hausdorff space, R. H. Bing, 1952), and $x$ any point in $I$. Then the quotient $(I\times\kappa)/(\lbrace x\rbrace\times\kappa)$ is such a space ($\kappa$ is considered as a discrete space), with $\mathcal D$ the collection of open sets $(I\setminus\lbrace x\rbrace)\times\lbrace\alpha\rbrace$ for $\alpha<\kappa$.
For every $\ \kappa\ge|\Bbb R| \ $ the cone
$\ X=\text{cone(D)}\ $ over the discrete space $\ D\ $ of cardinality $\ \kappa\ $ is an example.
The required open sets are the cone lines minus the vertex.
When it comes to lower cardinalities then I'd look into the Urysohn countable connected Hausdorff space.
In particular, let's apply the Bing's example of Hausdorff connected countable space, $\ B\ $ (see https://www.ams.org/journals/proc/1953-004-03/S0002-9939-1953-0060806-9/S0002-9939-1953-0060806-9.pdf)
Any infinite family of pairwise disjoint euclidean-open subsets of $\ \Bbb Q\times\{0\}\ $ are open also in $\ B.\ $
Thus the OP's problem is solved for all cardinalities but the weird ones, between $\ |\Bbb Q|\ $ and $\ |\Bbb R|,\ $ that exist only when we don't accept the continuum hypothesis.
Can't you just take the “$B$-cone” over the discrete space $\kappa$ then? Just taking the quotient of $\kappa\times B$ by $\kappa\times\lbrace x\rbrace$ for $x$ any point in $B$.
@PierrePC, that's exactly what I've written in my answer above (for ≥|ℝ|).
I know, I got all the ingredients from your answer. But this would work for any cardinality if you replaced $[0,1]$ by the $B$ you describe in the construction of the cone. Or maybe I'm missing something?
@Pierre, yes, what you did is great!!! I should read with greater attention and a positive attitude. Please, post your solution since it's elegant and complete! #### (I didn't pay any attention to your B like Bing; to me, "B" is for Banach :) ).
No worries. :) But don't you want to just edit your answer? Everything is already here really. You can cite the comment if you're worried about copyright issues. ;)
Please, It is your job, your answer! BTW, I see that @bof got a similar idea.
|
2025-03-21T14:48:30.412414
| 2020-04-25T19:34:00 |
358538
|
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|
Stack Exchange
|
In which cases $E(e^{t S_n S_m})$ converges to $E(e^{t X Y})$
Suppose that $S_n$ and $S_m$ are two random binomial variables, which are independent and with the same distribution parameter $p$. I am wondering, in which cases $E(e^{t S_n S_m})$ converges to $E(e^{tXY})$ when $n, m$ tend to infinity, where $t$ is real and $X$ and $Y$ are normally distributed variables with parameters $N(np,\sqrt{npq})$ and $N(mp,\sqrt{mpq})$, respectively.
However, it is clear that $e^{tS_nS_m}$ converges in probability to $e^{tXY}$, according to the Moivre-Laplace theorem, but I am not sure that any of the sufficient conditions which can be found in literature are satisfied considering the expectation.
In fact, I am trying to prove $E(e^{tS_nS_m}) \sim E(e^{tXY})$ when $n, m$ tends to infinity.
Could you please clarify what you mean when you say ``$E(\text{something})$ converges in probability to $E(\text{something else})$''? These are numbers, not random variables.
@Ron P This was a mistake, I meant only converges. Thank you for the notice :)
The answer is "no", and the reason is that $E[e^{tXY}]=\infty$ when $nm$ is large enough. To see that, note that $X=\sqrt{Var[X]}N+E[X]$ and $Y=\sqrt{Var[Y]}M+E[Y]$, where $N$ and $M$ are independent standard normal random variables. If $nm$ is large enough so that $t\sqrt{Var[X]Var[Y]}\geq 1$, we have
$$
E[e^{tXY}]\geq E[\mathbf 1_{\{N\geq 0\}}1_{\{N\geq 0\}}e^{NM}]\geq \tfrac 1 4 E[e^{NM}].
$$
Direct computation shows
$$
E[e^{NM}]= \int\int\frac{1}{2\pi}e^{-\frac{(x-y)^2}{2}}\mathrm dx\mathrm dy=\int\frac{\mathrm dy}{\sqrt{2\pi}}=\infty.
$$
|
2025-03-21T14:48:30.412535
| 2020-04-25T20:20:39 |
358540
|
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|
Stack Exchange
|
Spectral Theorem for compact self-adjoint operators on real Hilbert spaces
Is the spectral theorem for self-adjoint compact operators on a Hilbert space also true if the Hilbert space is real (instead of complex)?
Wikipedia says this is true.
However, it seems to me that in none of the references of that article, the theory is established for real Hilbert spaces.
If it is true I would be grateful for a reference.
Actually, now I think of it, one probably does better to just follow the proof of the spectral theorem for the complex case, rather than appeal to it as a black box. If you look at how the theorem is usually proved : just like the finite-dimensional case one produces a non-zero eigenvalue (assuming your operator is non-zero) and uses self-adjointness to prove that the eigenvalue is real & that the perp of the eigenspace is invariant under your operator. Then you induct, using compactness to show that your sequence of eigenspaces (together with the kernel) eventually exhausts the whole space.
Now in the sketch above, the only place where I relied on having complex scalars was to produce a non-zero eigenvalue; everything else works for real inner products in exactly the same way as for complex inner products. So you are in business as soon as you can prove that a self-adjoint compact operator $T$ on a real Hilbert space always has a (real) eigenvalue. But there are (standard) arguments which show that at least one of $\Vert T\Vert$ or $-\Vert T\Vert$ will be an eigenvalue, and these don't rely on complex-variable techniques
Answered in greater generality here.
Yes, Raleigh-Ritz (sp?) method of proving existence of an eigenvalue/eigenvector is an essentially "real" method, anyway.
|
2025-03-21T14:48:30.412688
| 2020-04-25T20:31:43 |
358542
|
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|
Stack Exchange
|
Lower bound for mutual inner products of N random unit vectors in $\mathbb{R}^n$, N > n
I have $N$ independent random unit vectors $\{v_i\}$ in $\mathbb{R}^n$, where N > n. I need a concentration inequality of the form
$$\text{P}(|v_i \cdot v_j| > \epsilon \,\,\,\, \forall i, j = 1, \dots, N: i \neq j)\leq \psi(\epsilon)$$
where hopefully $\psi(\epsilon)$ is something small.
I think that I can use Johnson-Lindenstrauss to do this for isotropic vectors (e.g. by choosing orthogonal basis for $\mathbb{R}^N$ and projecting into $\mathbb{R}^n$ with a random subgaussian matrix).
Are there results of this form that hold when the $\{v_i\}$ are not distributed isotropically, for instance Gaussian with covariance $\Sigma$? For instance, when there is some weak correlation/dependence between the components of each of the $v$ --- maybe $|\Sigma_{ij}| \leq \alpha$ when $i\neq j$?
(Any seemingly related results in this area are much appreciated!)
what is a quantifier for $i,j$?
Thanks. They run from $1\dots n$ and for the concentration they're not equal. I edited the question.
The first sentence in your post is inconsistent with the rest. What happened to the "independent" ?
Would this help $P(\cap_{i\ne j}{|v_i^Tv_j| > \epsilon}) \le \min_{(i,j) \mid i \ne j} P(|v_i^Tv_j| \le \epsilon)$ to begin with ?
@dohmatob You are right, I will remove the word 'independent'.
Thank you for your second comment. I am not sure this is what I was looking for, but I will see whether I can make something from this observation.
I have a partial answer: Lemma 2.2 (rephrased below) from [1] gives a deterministic lower bound to the inner product by considering the gram matrix
Let $M \in \mathbb{R}^{p\times p}$ be a rank $d$, real, symmetric
matrix with $M_{ii} = 1$ $\forall i$ and $|M_{ij}| \le \epsilon$
$i\neq j$, then $$ \epsilon^2 \ge \frac{p - d}{d(p-1)}. $$
[1] Perturbed identity matrices have high rank: Proof and applications, 2009, Noga Alon
Interesting result to know. How do you intend to use it, given that for a full rank gram matrix (which is the precondition in the question) this would yield a trivial lower bound ($p-d = 0$)?
Very sorry. I made a typo in the question. I have $N$ vectors in $\mathbb{R}^n$ and want to concentrate on all of them, not just $n$ of them. If $N>n$ then this lemma means I can't make the inner products arbitrarily small (which is also obvious, thinking geometrically).
$$|v_i \cdot v_j| \ge \sqrt{\frac{N-n}{n(N-1)}} \ge \sqrt{1/n}$$
In OP's setting, a simple union bound argument gives that $\max_{i\neq j}|v_i\cdot v_j|=O(\sqrt{(\log N)/n})$ with probability $1-O(1/N)$, so this bound (known as the Welch bound) is optimal up to logarithmic factors.
@DustinG.Mixon. Thank you. I looked at the wikipedia page for welch bounds. Can you explain how I would turn them into a probabilistic statement please? (If so then you could give an answer so I could accept.)
|
2025-03-21T14:48:30.412928
| 2020-04-25T21:16:25 |
358543
|
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"AG learner",
"Phil Tosteson",
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|
Stack Exchange
|
A relative version of Ehresmann's theorem
Edited: Phil Tosteson suggested Thom's first isotopy lemma, but it does not seem to be in the direction that I'm trying to generalize. Let me reformulate my question again.
Let $N\subset M$ be a pair of the smooth manifolds and $\pi:M\to B$ a proper submersion to a smooth base. Assume moreover that the restriction $\pi_{|N}:N\to B$ is also proper and submersive, then both $\pi$ and $\pi_{|N}$ are locally trivial by applying Ehresmann's theorem separately, but I'm wondering if they can be simultaneously locally trivialized.
More precisely, I'm looking for a proof of the following argument:
Claim: For each $b\in B$, there is a neighborhood $U$ of $b$, and a fiber preserving diffeomorphism
$$\psi:\pi^{-1}(U)\xrightarrow{\cong} M_b\times U,$$
satisfying $$\psi(N\cap \pi^{-1}(U))=N_b\times U$$
where $M_b=\pi^{-1}(b)$ and $N_b=\pi_{|N}^{-1}(b)$.
The example that I have in mind is a smooth family of pairs (cubic surface, cubic curve). Explicitly take a general pencil $\mathbb P^1$ of hyperplane sections of a smooth cubic threefold $X$ over $\mathbb C$ with base locus a cubic curve $E$ (as the intersection of any two hyperplane section in the pencil). Project the incidence variety $I=\{(y,t)|y\in X\cap H_t\}\subset X\times \mathbb P^1$ to the second factor produces a family of cubic surfaces. Now $M\to B$ will be restriction of $I\to \mathbb P^1$ to the locus where cubic surfaces are smooth and $N:=E\times B$.
(For a smooth family of compact complex manifolds, one can even ask for transversely holomorphic trivialization, see page 2 in this notes, I would also like to know if it works for a pair.)
There is a general theorem that goes under the name of "Thom's first isotopy lemma." You can read about it in Mather's "Notes on topological stability."
@PhilTosteson Thanks for your comment, I checked Thom's first isotopy lemma in Mather's notes. However, I'm not sure how this lemma is related to my question since I already assumed both $\pi$ and $\pi_{|N}$ are proper and submersive, so Ehresmann's theorem applied to $M$ and $N$ separately tell us they are both locally trivial over $B$, but the argument that I am looking for is the existence of a (local) trivialization $\psi: \pi^{-1}(U)\xrightarrow{\cong} M_b\times U$ which satisfies $\psi (N\cap \pi^{-1}(U))=N_b\times U$.
Sorry, you need the "second isotopy lemma" to see your desired outcome as a direct consequence. However, I think that the proof of the first isotopy lemma involves proving a version of your statement.
@PhilTosteson Thank you! It makes more sense to me now. I hope I can dig an elementary proof out of it, since I am working on smooth submanifold case compared to the general stratified subsets considered in the first/second isotopy lemma.
The answer is positive. There are several proofs of Eheresmann's genuine lemma; I think that each of them can be straightforwardly generalized and gives your relative version. But you can also, alternatively, deduce the relative version from the absolute one together with a classical theorem of Cerf, as follows. After restricting $B$ to a small compact ball centered at a point $b$, by a first application of the genuine Ehresmann lemma to $(M,\pi)$, you can assume without loss of generality that $M=M_b\times B$ and that $\pi$ is the second projection. Then, after restricting $B$ to a smaller compact ball centered at $b$, by a second application of the genuine Ehresmann lemma to $(N,\pi\vert N)$, there is a local trivialization of $\pi\vert N$ over $B$, in other words a diffeomorphism
$$\psi_N:N\to N_b\times B$$
such that $\pi\circ\psi_N=\pi\vert N$. You can also see $\psi_N$ as a parametric family of smooth embeddings $$f_y:N_b\hookrightarrow M_b$$ parametrized by $y\in B$, namely $$(f_y(x),y)=\psi_N^{-1}(x,y)$$
It is classical (Cerf) that a parametric family of embeddings of $N_b$ in $M_b$ extends to a parametric family of self-diffeomorphisms (isotopies) of the ambiant manifold $M_b$: there is a smooth family $(F_y)$ of self-diffeomorphisms of $M_b$ such that $f_y=F_y\vert N_b$; and $F_b$ is the identity. Finally, the trivialization $\psi$ that you are looking for is just the inverse of $$(x,y)\mapsto(F_y(x),y)$$
($x\in M_b$, $y\in B$)
I think that one can prove the relative version of Ehresmann's lemma (and also versions for manifolds with corners, which is my interest) without any technology as follows:
Let us first recall the following simple proof of Ehresmann's lemma:
Let $U\subset B$ be a coordinate neighborhood $U\simeq \mathbb{R}^n$ centered at $b\in B$.
Consider the coordinate vector fields $\partial_1,\dotsc,\partial_n\in \mathfrak{X}(U)$, and pick their smooth lifts $X_1,\dotsc,X_n\in\mathfrak{X}(\pi^{-1}(U))$ under $\pi$, respectively.
This is possible if and only if $\pi$ is a submersion (more below).
The properness of $\pi$ implies that $X_i$ are complete, so they admit complete flows $\phi_i\colon \pi^{-1}(U)\times \mathbb{R} \to \pi^{-1}(U)$. Let $f\colon \pi^{-1}(b)\times U \to \pi^{-1}(U)$ be the map defined for all $\xi\in \pi^{-1}(b)$, $t=(t_1,\dotsc,t_n)\in U$ by
$$
f(\xi,t) \colon =(\phi_1^{t_1}\circ\dotsb\circ\phi_n^{t_n})(\xi).
$$
We have $\pi(f(\xi,t))=t$ because $X_i$ lift $\partial_i$, and $f$ is a diffeomorphism with inverse
$$
f^{-1}(x) = (t, (\phi_n^{-t_n}\circ\dotsb\circ\phi_1^{-t_1})(x)),
$$
where $t=f(x)$.
The relative version of Ehresmann's lemma---given a submanifold $N\subset M$ and assuming that both $\pi\colon M\to B$ and its restriction $\pi|_N\colon N\to B$ are submersions---can be obtained by choosing lifts $X_1,\dotsc,X_n$ of $\partial_1,\dotsc,\partial_n$ that are tangent to $N$ as follows:
Denote $s:=\dim(N)$, $m:=\dim(M)$. For each $x\in \pi^{-1}(U)\cap N$, let $V_x\subset M$ be a coordinate neighborhood $V_x\simeq \mathbb{R}^m$ of $x$ such that
$$
V_x \cap N = \{ (x_1,\dotsc,x_m)\in V_x \mid x_{s+1}=\dotsb =x_m = 0\}.
$$
The existence of such "slice charts" for a submanifold $N\subset M$ is a standard fact.
Because the differential $(D\pi)_x\colon T_x N \to T_{\pi(x)}B$ is surjective, we have $n\le s$, and we can pick $n$ coordinates, w.l.o.g. $x_1,\dotsc,x_n$, such that the restriction
$$
(D\pi)_x: \mathrm{span}\{\partial^x_1,\dotsc,\partial^x_n\} \to \mathrm{span}\{\partial_1,\dotsc,\partial_n\}
$$
is an isomorphism. This is a simple fact from linear-algebra, namely, that any generating system contains a basis.
By shrinking the neighborhood $V_x$ if necessary, we can assume that the above isomorphism holds for $(D\pi)_y$ for all $y\in V_x$.
For each $y\in V_x$ we can now define the vectors
$$
X_{x,i}(y) := ((D\pi)_y|_{\mathrm{span}\{\partial_1^x,\dotsc,\partial_n^x\}})^{-1}(\partial_i) \in T_y M.
$$
We obtain smooth vector fields $X_{x,i}\in \mathfrak{X}(V_x)$ that lift $\partial_i$ and are tangent to $N$.
We proceed similarly for $x\in \pi^{-1}(U)\backslash N$ to obtain neighborhoods $V_x$ and vector fields $X_{x,i}\in\mathfrak{X}(V_x)$ that lift $\partial_i$.
Pick a cover $(V_\alpha)\subset \{ V_x \mid x\in \pi^{-1}(U)\}$ of $\pi^{-1}(U)$ with a subordinate partition of unity $\lambda_\alpha\colon V_\alpha\to [0,1]$.
Define $X_i := \sum_\alpha \lambda_\alpha X_{\alpha,i}$.
Then $X_i\in \mathfrak{X}(\pi^{-1}(U))$ are smooth vector fields that lift $\partial_i$, so they can be used in 1 to construct the trivializing diffeomorphism $f$.
Because $X_i$ are moreover tangent to $N$, their flows through points in $N$ remain in $N$ for all times, so the diffeomorphism $f\colon \pi^{-1}(b)\times U \to \pi^{-1}(U)$ restricts to the diffeomorphism
$$
f|_N\colon \pi^{-1}(b)\cap N\times U\to \pi^{-1}(U)\cap N.
$$
This proves the relative Ehresmann lemma as stated in the question.
Analogously, one can prove Ehresmann's lemma in the case that $M$ is a manifold with corners.
Here one has to assume that the restriction of $\pi$ to any open codimension-$k$ boundary strata of $M$ is a submersion.
Given $x\in M$, one picks a corner chart on $M$ that intersects only those strata that contain $x$, and one chooses the coordinates $x_1,\dotsc,x_n$, which were used to define $X_{x,i}$, in the strata of highest codimension.
In this way, the vector fields $X_{x,i}$ are tangent to any strata containing $x$.
|
2025-03-21T14:48:30.413394
| 2020-04-25T21:49:50 |
358544
|
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|
Stack Exchange
|
Can conservativity depend on the universe?
Question 1: Let $F: C \to D$ be a conservative, $\kappa$-cocontinuous functor between small, $\kappa$-cocomplete categories. Is the induced functor $Ind_\kappa(F): Ind_\kappa(C) \to Ind_\kappa(D)$ also conservative?
Terminology: I think this is pretty self-explanatory, but to be clear:
$\kappa$ is a regular cardinal. Things are perhaps most familiar when $\kappa = \aleph_0$.
A $\kappa$-cocomplete category is a category with $\kappa$-small colimits, i.e. colimits indexed by categories with fewer than $\kappa$-many morphisms.
A $\kappa$-cocontinuous functor is a functor preserving $\kappa$-small colimits.
$Ind_\kappa(C)$ is obtained from $C$ be freely adjoining $\kappa$-filtered colimits, or by the formula $Ind_\kappa(C) = Fun_\kappa(C^{op},Set)$, where $Fun_\kappa(A,B)$ is the category of $\kappa$-continuous functors from $A$ to $B$.
The induced functor $Ind_\kappa(F)$ is defined by left Kan extension along the Yoneda embedding.
I'm a little worried that Question 1 is too much to ask for, so here's an even milder variant.
Let $\kappa$ be inaccessible.
Let $Pr^L(\kappa)$ be the category (really a $(2,1)$-category) of categories which are locally presentable with respect to the universe $V_\kappa$. That is, $Pr^L(\kappa)$ consists of categories of the form $Ind_\lambda^\kappa(C)$ where $C$ is a $\kappa$-small and $\lambda$-cocomplete category with $\lambda < \kappa$; here $Ind_\lambda^\kappa$ is the free cocompletion under $\kappa$-small, $\lambda$-filtered colimits. The morphisms are left adjoint functors.
Similarly, let $Pr^L$ be the $(2,1)$-category of locally presentable categories and left adjoint functors.
Then we have a functor $Ind_\kappa: Pr^L(\kappa) \to Pr^L$.
Question 2: Does the functor $Ind_\kappa: Pr^L(\kappa) \to Pr^L$ preserve conservative functors?
Question 2 is asking whether the property of a left adjoint between locally presentable categories being conservative depends on which universe we work in.
In the setting of either question, it's clear that if $Ind_\kappa(F)$ is conservative, then $F$ is conservative. So in either case, if the answer to the question is affirmative, then we have "$F$ conservative $\Leftrightarrow$ $Ind_\kappa(F)$ conservative", which would be reassuring.
Probably not the best one can do, and what follows might be a bit 'overkill', but it answer the question about dependency on universe, and it is a nice argument.
Also if you know how the proof of the left transfer results I will use below works, it might give some idea on how to prove more general case of the result.
Theorem: Let $\kappa$ be an uncountable regular cardinal. Let $F: \mathcal{C} \to \mathcal{D}$ be a strongly $\kappa$-accessible left adjoint functor between locally $\kappa$-presentable categories. Then $F$ is conservative if and only if the restriction of $F$ to the full subcategory of $\kappa$-presentable objects is conservative.
(Here strongly $\kappa$-accessible = sends $\kappa$-presentable objects of $\mathcal{C}$ to $\kappa$-presentable objects of $\mathcal{D}$)
Proof: The weak factorization system (isomorphisms, all maps) on $\mathcal{D}$ is ($\kappa$-)combinatorial: it is generated by the empty set of maps.
Hence the left transfered weak factorization system on $\mathcal{C}$ along $F$ exists and is also $\kappa$-combinatorial (I mean by this that is is cofibrantly generated by maps between $\kappa$-presentable objects).
By definition, the left class of this weak factorization system is the set of maps such that $F(i)$ is an isomorphism. But if the restriction of $F$ to $\kappa$-presentable objects is conservative it means that all generators are isomorphisms hence, the left class only contains isomorphisms, so $F$ is conservative.
Note: the finer version of the left transfer theorem that I'm using which specify the presentability rank can be found as Theorem B.8.(4) in this paper of mine, but at the end of day it mostly follows from an analysis of the proof of the existence of left transfer available in the litterature (for e.g. in Makkai and Rosicky Cellular categories.)
Here's a proof of a more general statement, which unfortunately still fails when $\kappa = \aleph_0$. The key tool is Fodor's lemma:
Proposition: Let $\kappa$ be an uncountable regular cardinal, and let $F: C \to D$ be a filtered-colimit-preserving functor between $\kappa$-accessible categories with filtered colimits, which preserves $\kappa$-presentable objects. Then $F$ is conservative iff its restriction to the $\kappa$-presentable objects is conservative.
Proof: Because each category has filtered colimits, it suffices by induction on presentability rank to show that if $(f_i: c_i \to c_i')_{i < \kappa}$ is a map of smooth $\kappa$-chains between $\kappa$-presentable objects with colimit $f: c \to c'$, and if $Ff: F(c) \to F(c')$ is an isomorphism, then $f: c \to c'$ is an isomorphism. (Here "smooth" $\kappa$-chain means that $c^{(')}_i = \varinjlim_{j < i} c^{(')}_j$ for $i < \kappa$ a limit ordinal. Because $F$ preserves filtered colimits, the image chains $(F(c_i^{(')}))_{i < \kappa}$ are also smooth.) Let $g: F(c') \to F(c)$ be inverse to $F(f)$, and write $\gamma_{ij}^{(')}: c_i^{(')} \to c_j^{(')}$ for the linking maps.
Define $\phi: \kappa \to \kappa$ by taking $\phi(j) = \sup S_j$ where $S_j$ is the set of $i< \kappa$ such that $g\gamma_{i\kappa}$ factors through $F(c'_j)$. If $\phi(j) < j$ on a stationary set, then by Fodor's lemma $\phi$ is constant on a stationary set; it follows that $\phi(j)$ is bounded for sufficiently large $j$, which is impossible because every $i$ lies in some $S_j$. Therefore we have $\phi(j) \geq j$ on a club set $S$ of $j$'s. By restricting to the limit ordinals in $S$ and using smoothness of our chains, we may assume that $g: F(c') \to F(c)$ is induced "levelwise": there is a map of chains $g_i: F(c_i') \to F(c_i)$ such that $g = \varinjlim_i g_i$.
A similar argument shows that we may assume that $g_i$ is a levelwise retract of $F(f_i)$, and similarly that $g_i$ is a levelwise section of $F(f_i)$. Thus $F(f_i)$ is invertible. Because $F$ is conservative on $\kappa$-presentable objects, $f_i$ is invertible, and so $f$ is invertible as well.
Note that we needed $\kappa$ to be uncountable and regular in order to apply Fodor's lemma. We also needed this to pass to the limit ordinals in $S$.
My suspicion is that a more general version of Fodor's lemma working on more general $\kappa$-directed posets ought to allow one to take $C,D,F$ to be arbitary $\kappa$-accessible categories and functors, but I think $\kappa$ will still need to be uncountable (and regular).
Actually the argument shows that we may assume that $g\gamma_i$ factors through some $g_i : F(c_i') \to F(c_i)$, but not that the $g_i$ cohere into a map of chains. With a bit more care, I think this can be shown though.
|
2025-03-21T14:48:30.413802
| 2020-04-25T21:58:15 |
358546
|
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"Dror Speiser",
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|
Stack Exchange
|
Subdividing a sequence such that sum is somewhat equally distributed
I have a sequence ( n, n-1, n-2,...,1). I need to find numbers in this sequence in this order that somewhat approximately divide it into M parts- within each M subgroup the sum is somewhat the same. Is there a general way to do it?
For example, 5,4,3,2,1 - for M=3, it could be 5; 4; 3,2,1 where the sums are somewhat close (5,4,6).
Or 8,7,6,5,4,3,2,1 - For M=3 maybe 8;7+6=13;(rest)15.
8;13;15 is good enough split(15;11;10 is also okay and arguably better than 8,13,15). I don't need it to be exactly equal but for a value of n, M is there a general way to approximately split it? In my case, n is very large( 1000s) and M is probably no greater than 8 at most.
This will be hard to answer, unless you tell us precisely what "somewhat the same", "somewhat close". "approximately split" mean. Arnold Ross used to ask, rhetorically, "What's an approximation to five?" and the correct answer was "any number but five".
So this is actually a programming thing where I have to split the jobs to processors. I don't know the value of n upfront and I am just trying to optimize the utilization of all processors. Somewhat same is me trying to make sure all the processors spend time and it doesn;t overload one processor while the other is sitting free. unfortunately that;s all i got.
For $M=4$ and n divisible by 8 such that $n+1$ is prime, it's an exercise to show that dividing $[1...n]$ into their quartic Jacobi symbols mod $p$ gives equal sums. Meaning, you can take the $(p-1)/4$-th power of each number mod $p$, and assign a processor for each of the four possible results. If $n$ doesn't satisfy the requirement, then find the largest prime less than $n$ that is $1\mod{8}$, apply the procedure, and you're left with $O(log(n)^2)$ terms of size $O(n)$ (assuming Cramer's hypothesis), which are negligible in comparison to the $O(n^2)$ work you have.
You can solve the problem exactly as a shortest path problem in a layered network. The nodes are $(i,k)$, where $i\in\{n,\dots,1\}$ and $k\in\{1,\dots,M\}$. The (directed) arcs are from $(i,k)$ to $(j,k+1)$ with $j<i$. The idea is that traversing arc $(i,k)\to (j,k+1)$ means that $\{i,i-1,\dots,j+1\}$ comprise a part. The cost of this arc is the (absolute, or squared, or any type of) deviation from the target $n(n+1)/(2M)$. Because the network is acyclic, you can solve the problem in $O(M n^2)$ time, that is, linear in the number of arcs.
Alternatively, a greedy heuristic could be to start a new part $p+1$ whenever including the next number would make the cumulative sum exceed $pn(n+1)/(2M)$. For $n=5$ and $M=3$, this yields $5;4;3+2+1$. For $n=8$ and $M=3$, this yields $8;7+6;5+4+3+2+1$.
The greedy heuristic makes sense to me - if I fixed M=4, is there a general formula for any n maybe?
|
2025-03-21T14:48:30.414024
| 2020-04-25T22:31:03 |
358547
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Gerry Myerson",
"Max Alekseyev",
"Pruthviraj",
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|
Stack Exchange
|
partial sum of a power number equal to a power number with same power
Can it be shown that there exist least one pair for every positive integer $m$ such that
$$a_1^m+a_2^m+\cdots+a_n^m=b^m$$
Where $a_i,b,m,n\in \mathbb{Z}_+$ and $a_i\ne a_j $ for $1\le i,j\le n$ and $n>1$
Example:
$$\begin{split}1+2+7&=10\\1^2+2^2+3^2+5^2+19^2 &=20^2\\ 3^3+4^3+5^3&= 6^3\\ 30^4+120^4+272^4+315^4&=353^4\\ 7^5+43^5+57^5+80^5+100^5&=107^5\\
8^6+12^6+30^6+78^6+102^6+138^6+165^6+246^6&=251^6 \end{split}$$
An example for $7$'th powers found by Mark Dodrill:
$$ 127^7 + 258^7 + 266^7 + 413^7 + 430^7 + 439^7 + 525^7 = 568^7$$
An example for $8$'th powers found by Scott Chase:
$$90^8 + 223^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 = 1409^8$$
$9$'th and $10$'th from Jaroslaw Wroblewski:
$$42^9 + 99^9 + 179^9 + 475^9 + 542^9 + 574^9 + 625^9 + 668^9 + 822^9 + 851^9 = 917^9$$
$$62^{10} +115^{10} +172^{10} +245^{10} +295^{10} +533^{10} +689^{10} +927^{10} +1011^{10} +1234^{10} +1603^{10} +1684^{10} = 1772^{10}$$
This question was posted in MSE(27/2/20) got a answer by Robert Israel but without getting proof hence posting in MO link to MSE post
Edit
Related problem
Claim $$\sum_{q=0}^{u}(n+qd)^m\ne a^m \ \ \ \ \forall n,u,d,a\in\mathbb{Z}_+, \ \ \ \ m\in \mathbb{Z}_{\ge 4} $$
In this post
I take it you are aware of Waring's Problem, https://en.wikipedia.org/wiki/Waring%27s_problem and the distinction here is that you insist on distinct summands. I suspect that if you dive into the Waring literature you'll find that's been taken care of.
@GerryMyerson thanks for your reference it's helpful to better visualise.
It's not so interesting when $n$ is unbounded. For bounded $n$, see desults at http://euler.free.fr/
|
2025-03-21T14:48:30.414259
| 2020-04-25T23:35:57 |
358552
|
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"ste"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358552"
}
|
Stack Exchange
|
Integral with 4 Bessel functions and an exponential
I would like to solve the following integral
$$
\int_0^\infty e^{-a k^2} J_{3/2}(b k) J_{3/2}(c k) J_{3/2}(f k) J_{1/2}(r k) k^{-3} dk,
$$
where $a,b,c,f,r > 0$, and $J_\nu(x)$ is the Bessel function of order $\nu$.
An equivalent (within proportionality) integral in terms of spherical Bessel functions is
$$
\int_0^\infty e^{-a k^2} j_1(b k) j_1(c k) j_1(f k) j_{0}(r k) k^{-1} dk,
$$
So far I haven't found the integral in any integration tables. Any guidance on how to solve it would be most appreciated!
Let's consider the second integral, which can be written in the following form:
$$
I(p, q, i, j, k, l; a, b, c, d)
:=
\int_0^\infty
dt\,
\exp(-p t^2)
t^q
j_i(a t) j_j(b t) j_k(c t) j_l(d t)
$$
where in your case, $i = j = k = 1$, $l = 0$, and $q = -1$.
These kinds of integrals (as well their generalization to a product of arbitrarily many spherical Bessel functions) are discussed in Fabrikant - Elementary exact evaluation of infinite integrals of the product of several spherical Bessel functions, power and exponential, where the main idea is to use the following identity:
\begin{align}
I(p, q, i, j, k, l; a, b, c, d)
&=
(-1)^{i+j+k+l}
a^i
b^j
c^k
d^l
\frac{\partial^i}{(a \partial a)^i}
\frac{\partial^j}{(b \partial b)^j}
\frac{\partial^k}{(c \partial c)^k}
\frac{\partial^l}{(d \partial d)^l}
\biggl[\\
&\int_0^\infty
dt
\exp(-p t^2)
\frac{
j_0(a t) j_0(b t) j_0(c t) j_0(d t)
}
{
t^{i + j + k + l - q}
}
\biggr].
\end{align}
The key point is to now expand the zeroth order spherical Bessel functions into trigonometric functions, and converting the products of the trigonometric functions into sums:
\begin{align}
\sin(ax)
\sin(bx)
\sin(cx)
\sin(dx)
=&
\frac{1}{8}
\biggl\{
\cos[(a + b + c + d)x]
+
\cos[(a + b - c - d)x]
+
\cos[(a - b + c - d)x]\\
&+
\cos[(a - b - c + d)x]
-
\cos[(-a + b + c + d)x]
-
\cos[(a - b + c + d)x]\\
&-
\cos[(a + b - c + d)x]
-
\cos[(a + b + c - d)x]
\biggr\}
\end{align}
followed by the use of the following integral, which is not considered in the reference above, but can be found in Gradshteyn and Ryzhik, 7th ed., formula 3.953.8:
$$
\mathcal{I}(p, s; n)
:=
\int_0^\infty dt\,
t^n
\exp(-p t^2)
\cos(s t)
=
\frac{1}{2}
p^{\frac{-(n + 1)}{2}} \,
e^{-s^2 / 4 p}
\Gamma \left(\frac{1}{2} + \frac{n}{2}\right) \,
_1F_1\left(-\frac{n}{2}; \frac{1}{2}; \frac{s^2}{4 p}\right).
$$
Note that the formal requirement is that $\operatorname{Re}(n) > -1$, but the above result can be understood as an analytic continuation for general values $p, s, n$.
Additionally, it can happen that one of the "angles" above is zero, in which case we have the integral:
$$
\mathcal{I}(p, 0; n)
:=
\int_0^\infty dt\,
t^n
\exp(-p t^2)
=
\frac{1}{2} p^{-\frac{n}{2}-\frac{1}{2}} \Gamma \left(\frac{n+1}{2}\right)
$$
with the same condition on $n$ as above.
The result in your specific case is then:
\begin{align}
I(p, -1, 1, 1, 1, 0; a, b, c, d)
&=
-
a
b
c
\frac{\partial}{(a \partial a)}
\frac{\partial}{(b \partial b)}
\frac{\partial}{(c \partial c)}
\int_0^\infty
dt
\exp(-p t^2)
\frac{
j_0(a t) j_0(b t) j_0(c t) j_0(d t)
}
{
t^4
}\\
&=
-
a
b
c
\frac{\partial}{(a \partial a)}
\frac{\partial}{(b \partial b)}
\frac{\partial}{(c \partial c)}
\int_0^\infty
dt
\exp(-p t^2)
\frac{
\sin(a t) \sin(b t) \sin(c t) \sin(d t)
}
{
a\, b\, c\, d\, t^8
}\\
&=
-
\frac{\partial}{\partial a}
\frac{\partial}{\partial b}
\frac{\partial}{\partial c}
\bigg[
\frac{1}{a\, b\, c\, d}
\int_0^\infty
dt
\exp(-p t^2)
\frac{
1
}
{
t^8
}
\frac{1}{8}
\bigg\{
\\
&\cos[(a + b + c + d)t]
+
\cos[(a + b - c - d)t]
+
\cos[(a - b + c - d)t]
\\
&+
\cos[(a - b - c + d)t]
-
\cos[(-a + b + c + d)t]
-
\cos[(a - b + c + d)t]
\\
&-
\cos[(a + b - c + d)t]
-
\cos[(a + b + c - d)t]
\bigg\}
\bigg]
\\
&=
-
\frac{1}{8}
\frac{\partial}{\partial a}
\frac{\partial}{\partial b}
\frac{\partial}{\partial c}
\biggl\{\frac{1}{a\, b\, c\, d}
\\
&\mathcal{I}(p, a + b + c + d; -8)
+
\mathcal{I}(p, a + b - c - d; -8)
+
\mathcal{I}(p, a - b + c - d; -8)
\\
&+
\mathcal{I}(p, a - b - c + d; -8)
-
\mathcal{I}(p, -a + b + c + d; -8)
-
\mathcal{I}(p, a - b + c + d; -8)
\\
&-
\mathcal{I}(p, a + b - c + d; -8)
-
\mathcal{I}(p, a + b + c - d; -8)
\biggr\}.
\end{align}
The explicit result is fairly cumbersome to fully write out; below is an example Mathematica code which can be used as a starting point to generate the full solution (when $a \pm b \pm c \pm d \neq 0$) and compare it with the numerical result:
numeric[p_, a_, b_, c_, d_] := NIntegrate[
Exp[-p t^2] SphericalBesselJ[1, a t] SphericalBesselJ[1,
b t] SphericalBesselJ[1, c t] SphericalBesselJ[0, d t]/t,
{t, 0, Infinity}
];
integral[p_, s_, n_] :=
1/2 p^(-(n + 1)/2) Exp[-s^2/(4 p)] Gamma[
1/2 + n/2] Hypergeometric1F1[-n/2, 1/2, s^2/(4 p)];
result = -1/8 Table[
Series[
expression,
{epsilon, 0, 0}
] // Normal // D[#/(a b c d), a, b, c] &,
{
expression,
{
integral[p, a + b + c + d, -8 + epsilon],
integral[p, a + b - c - d, -8 + epsilon],
integral[p, a - b + c - d, -8 + epsilon],
integral[p, a - b - c + d, -8 + epsilon],
-integral[p, -a + b + c + d, -8 + epsilon],
-integral[p, a - b + c + d, -8 + epsilon],
-integral[p, a + b - c + d, -8 + epsilon],
-integral[p, a + b + c - d, -8 + epsilon]
}
}
] // Total;
No idea if the solution which the code above generates can be simplified though.
I think the GR reference should be formula 3.952.8. Great find with Fabrikant!
|
2025-03-21T14:48:30.414533
| 2020-04-26T01:14:35 |
358554
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358554"
}
|
Stack Exchange
|
How to solve a problem from Frank Olver's book
I'm learning Frank Olver's book, called Asymptotics and Special Functions. There is an difficult exercise.
Problem. Suppose that $f,\frac{1}{f}$ possess the following asymptotic expansions :
$$f(z)\sim \sum_{s=0}^{\infty}f_s z^{-s},\frac{1}{f(z)}\sim \sum_{s=0}^{\infty}k_s z^{-s}$$
as $z\to \infty$ in the region F, contained in $\mathbb{C}$. That is to say
\begin{align*}
f(z)& =\sum_{s=0}^{n-1}f_s z^{-s}+\phi_n(z),\\
\frac{1}{f(z)}& =\sum_{s=0}^{n-1}k_s z^{-s}+\psi_n(z),
\end{align*}
where $\left|\phi_n(z)\right|\leqslant K_n\left|z\right|^{-n},\left|\psi_n(z)\right|\leqslant L_n\left|z\right|^{-n}$, and $K_n,L_n$ are the least constants satisfying last two inequalities. Show that
$$L_n\leqslant \frac{1}{m}\sum_{s=0}^{n-1}|k_s|K_{n-s},\ n\geqslant 1,$$
where $m=\inf_{z\in \mathbf{F}}|f(z)|$.
I have got the recurrence relation :
$$f_0 k_s=-\left(f_1 k_{s-1}+f_2 k_{s-2}+\cdots+f_s k_0\right)$$
for $s=1,2,\cdots$. For example,
$$k_0=\frac{1}{f_0},k_1=-\frac{f_1}{f_0^2},k_2=\frac{f_1^2-f_0 f_2}{f_0^3}.$$
But I absolutely have no idea to prove the origin problem.
|
2025-03-21T14:48:30.414649
| 2020-04-26T01:29:14 |
358555
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hans",
"Paata Ivanishvili",
"Todd Trimble",
"Ville Salo",
"YCor",
"dhy",
"domotorp",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/32660",
"https://mathoverflow.net/users/50901",
"https://mathoverflow.net/users/51424",
"https://mathoverflow.net/users/7402",
"https://mathoverflow.net/users/955",
"juan"
],
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"provenance": "stackexchange-dolma-0006.json.gz:628493",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358555"
}
|
Stack Exchange
|
For $x$ irrational, is $a_{n} =\sum_{k=1}^{n}(-1)^{⌊kx⌋}$ unbounded?
For $x$ irrational, define $a_{n} :=\sum_{k=1}^{n}(-1)^{⌊kx⌋}$. Can you prove that $\left\{a_n\right\}$ is unbounded?
I feel that it is not easy to treat every irrational $x$.
I have asked in S.E. and it seems that it is an advanced question. Thus I go for help here.
I must confess that it's not obvious to me that it's unbounded for any irrational $x$. Can someone set me straight?
I am a little bit curious where is this question coming from?
@ToddTrimble If the infimum of $q^2|x-\frac{p}{q}|$ over all rational approximations $\frac{p}{q}$ with $q$ odd is zero, then this sum is unbounded. (For each fixed integer $k$, consider $\displaystyle\sum_{i=m}^{m+kq-1}(-1)^{\lfloor ix\rfloor}$ for large $q$ and "generic" $m$.) Almost every $x$ (in the sense of Lebesgue measure) satisfies this - this follows from the Duffin-Schaeffer conjecture (now the Koukoulopoulos-Maynard theorem), but I'm sure that's massive overkill.
It's a very natural question from the point of view of ergodic theory. One has the dynamical system $u:t\mapsto t+x$ on the circle /2. In turn induces an operator on functions or measures by $(())=(())=(+)$. In ergodic theory it's natural to estimate the average $\mathcal{A}nT=\frac1n\sum{i=1}^nT^n$ in various ways. Here the question is just evaluating $(\mathcal{A}nT)g$ at $0$, for $g=\mathbb{1}{[0,1]}-\mathbb{1}_{[1,2]}$ (namely asking if it's (1/) or not).
In particular, if $T$ is considered as operator on $L^2([0,2])$, then estimating $|(\mathcal{A}_nT)g|_2$ should be an exercise using Fourier transform.
Here's a related reference: https://pdfs.semanticscholar.org/9cb4/96498be476290bfa737f2f4ba337c5b5d37d.pdf The case they mention as settled ($g(x)={2x}$ sounds at least as hard as the current case. So I believe the result is true and known.
It's indeed unbounded for every irrational $x$.
Let me identify points of $\mathbb{R}/\mathbb{Z}$ with their representatives on $[0,1)$, and order it by the usual order $<$ of $\mathbb{R}$ applied to the representatives.
Replace $x$ by $y = x/2$, and the question becomes whether for all $m$ there exists $k$ such that the orbit of $0$ in the irrational rotation on $\mathbb{R}/\mathbb{Z}$ by $y$ is in $[0,1/2)$ at least $m$ more times than in $[1/2,1)$, in the first $k$ time steps $y,2y,3y,...,ky$; or that this happens with $[0,1/2)$ and $[1/2,1)$ interchanged.
Since the irrational rotation by $2y$ is topologically transitive, we can find odd $k$ with $ky > 0$ arbitrarily small. For odd $k$ the set $Y_k = \{y,2y,...,ky\}$ has to intersect either $[0,1/2)$ or $[1/2, 1)$ more times than the other. Let's suppose the first case happens for infinitely many $k$. (The other case is symmetric, and one happens by the pigeonhole principle.)
Let now $P_m$ be the statement that there are $ky > 0$ arbitrarily small such that $[0,1/2)$ contains $m$ more elements of $Y_k$ than $[1/2,1)$ does. We have that $P_1$ holds. Observe that if $P_m$ holds and $k$ is as in the definition, then $[0, 1/2)$ also contains $m$ more elements of $Y_{k,a} = \{a+y,a+2y,...,a+ky\}$ than $[1/2,1)$ whenever $a$ is small enough, because $Y_k$ is disjoint from $\{0,1/2\}$.
Let now $m \geq 1$ be any integer such that $P_m$ holds. Let $\epsilon > 0$ be arbitrary and pick $k$ such that $0 < ky < \epsilon/2$ and $[0,1/2)$ contains at least $m$ more elements of $Y_k$ than $[1/2,1)$ does. Let $a$ be as in the previous paragraph.
Using $P_m$ again, take $0 < k'y < \min(\epsilon/2, a)$ such that $[0, 1/2)$ contains $m$ more elements of $Y_{k'} = \{y,2y,...,k'y\}$ than $[1/2,1)$ does. Then $[0,1/2)$ contains $2m$ more elements of $Y_{k'+k}$ than $[1/2,1)$. We have $0 < (k+k')y < \epsilon$, so $P_{2m}$ holds.
Thus $P_m$ holds for all $m$, and the claim follows.
Nice argument. I guess you can make things a little quantitative to show that $c_x=\limsup |a_n|/\log(n)>0$, and even $\inf_xc_x>0$ when $x$ ranges over irrationals. Also it should work without change for $\lfloor ix\rfloor$ replaced with $\lfloor ix+x'\rfloor$ (that the irrational arithmetic progression passes through 0 plays no role).
I had similar thoughts, but didn't have time to work more on it. (My main job ATM is changing diapers.)
Two stupid questions: 1) Why should $k'$ presumably distinct from $k$ in the last paragraph exist should $k$ exist? 2)Why does "$[0,1/2)$ contains $2m$ more elements of $Y_{k'+k}$ than $[1/2,1)$. We have $0 < (k+k')y < \epsilon$"? Is it because for any $iy, jy\in[0,\frac12)$, $(i+j)y\in[0,\frac12)$? That does not seem right.
Regarding 1) Maybe you assume as part of the mathematical induction there are infinitely many $k$'s such that $P_m$ holds and $k'$ denotes any other such natural number. But then how do you conclude that there is always such $k'$ in that set that satisfies $k'<\min(\frac\epsilon2,a)$. 3) Why do we need $ky<\frac\epsilon2$ for an arbitrarily given positive $\epsilon$ in the first place?
$k'$ can be the same as $k$ for all I care. It exists by $P_m$. In any case, the definition is $P_m \iff \forall \epsilon > 0: \exists k > 0: 0 < ky < \epsilon$, and this implies that there are infinitely many such $k$, by picking smaller $\epsilon$.
The fact $[0,1/2)$ contains $2m$ more elements of $Y_{k'+k}$ than $[1/2,1)$ is seen as follows: ${y, 2y, ..., k'y}$ hits $[0,1/2)$ $m$ more times than it does $[1/2,1)$. By the choice of $k'$, we're in $(0,a)$ now, so by the choice of $a$, the choice of of $k$ applies and the $k$ following steps hit $[0,1/2)$ $m$ more times than $[1/2,1)$.
We have $0 < (k+k')y < \epsilon$ because $0 < ky < \epsilon/2$, $0 < k'y < \epsilon/2$, and because $x<y \implies 2x<2y$ holds for small $x, y$. (Hmm, maybe I need $\epsilon$ to be small. Anyway it's meant to be arbitrarily small.)
"Why do we need $ky < \epsilon/2$ for arbitrarily given positive $\epsilon$?" We need to do this to establish $P_{2m}$, as $P_{2m}$ talks about infinitely many $\epsilon$s. Hopefully this clears some things up.
Erm, in my response to 1) above, I write a formula for $P_m$ to show the quantifiers. That formula is of course missing the statement concerning $Y_k$. I cannot edit it anymore it seems.
Oh, I understand the need for "arbitrary" now. Thank you. +1. Such a clever proof!
We have
Theorem. Let $\psi(x)$ and $\varphi(x)$ be positive increasing functions such that
$$\int_1^\infty \frac{dx}{\psi(x)}=+\infty,\qquad \int_1^\infty \frac{dx}{\varphi(x)}<+\infty.$$
Then for almost all $\alpha\in(0,1)$ we have
$$\Omega(\log N\cdot \psi(\log\log N))\le\sup_{n\le N}\sum_{j=1}^n(-1)^{\lfloor j\alpha\rfloor}=O(\log N\cdot \varphi(\log\log N)).$$
This is proved in my paper with Jan van de Lune On Some oscillating sums, Uniform Distribution Theory 3 (2008) 35--72.
In this paper it is contained an algorithm to compute the sums. We obtain for example
$$S_{\sqrt{2}}(10^{1000})=-10,\quad S_{\sqrt{2}}(10^{10000})=166,\quad S_{\pi}(10^{10000})=11726.$$
With $S_\alpha(N)=\sum_{j=1}^n(-1)^{\lfloor j\alpha\rfloor}$.
So this answers the question for almost all irrational number, but not for all, right?
@domotorp In the paper we show also that the sums are not bounded for any irrational.
@pisoir Thanks for the edits
Here is an argument essentially due to fedja I learned about thirteen years ago on artofproblemsolving.com.
Proposition: if $f$ is $2$-periodic Riemann integrable such that $\sup_{n \geq 1} \left|\sum_{k=1}^{n}f(kx)\right|<C<\infty$ for some irrational $x \in \mathbb{R} \setminus \mathbb{Q}$ then necessarily
$$
\sum_{m\neq 0}\, \left| \frac{\hat{f}(m)}{e^{i \pi m x}-1}\right|^{2}<2C^{2}. \quad (1)
$$.
Notice that the proposition solves the question. Indeed, $f(t)=(-1)^{\lfloor{t}\rfloor}$ is 2-periodic with Fourier coefficients
$\hat{f}(m) = \left| \frac{1}{2}\int_{0}^{2}f(t)e^{-i\pi m x}dx \right| \approx \frac{1}{m}$ for $m$ odd, and $\hat{f}(m)=0$ for even $m$. There are infinitely many odd numbers $m$ such that $\mathrm{dist}\left(mx, 2\mathbb{Z}\right) < \frac{C'}{m}$ (it does not follow directly from Dirichlet's rational approximation theorem, however, I think it is not difficult to adapt it here or use Minkowski's theorem on product of two linear forms). Therefore, in the left hand side of (1) there are infinitely many terms comparable to 1 so we get a contradiction.
Proof of the proposition:
Let me first assume that $f$ is continuous and then I will explain later what to do in the arbitrary case.
Let $S_{m} = \sum_{k=1}^{m-1}f(kx)$. Then the boundedness of $|S_{M+n}-S_{M}|$ implies
$$
|\sum_{k=0}^{n-1} f(Mx+kx)|<2C \quad \text{for all} \quad M>1 \quad (2)
$$
and all $n\geq 1$. By density of the sequence $\{Mx\, \mathrm{mod}\, (2)\}_{M>1}$ in $[0,2)$ we conclude $$
|\sum_{k=0}^{n-1} f(t+kx)|\leq 2C \quad \text{for all} \quad t \in [0,2). \quad (3)
$$
Let us convolve $f$ with Fejer Kernel so that the new function $F$ is now a trigonometric polynomial with almost the same Fourier coefficients, and clearly it also satisfies the inequality (3). After expanding $F$ into its Fourier series $F(s) = \sum_{m} \hat{F}(m) e^{i\pi m s}$ (finite sum), and using (3) for $F$ we obtain
$$
\left| \sum_{m} \hat{F}(m)e^{i \pi m t} \, \frac{e^{i \pi m n x} -1}{e^{i \pi m x} -1} \right| <2C
$$
In particular its $L^{2}$ norm is bounded, i.e.,
$$
\sum_{m\neq 0} \left|\hat{F}(m)\, \frac{e^{i \pi m n x} -1}{e^{i \pi m x} -1}\right|^{2} <4C^{2}\quad (4)
$$
Invoking the density of the sequence $\{n x\, \mathrm{mod}\, (2)\}_{n \geq 1}$ again we can replace $nx$ by an arbitrary $s \in [0,2)$, after that we integrate (4) with respect to $s$ over the interval $[0,2]$, and we use the identity $\int_{0}^{2} |e^{i \pi m s} -1|^{2}ds=4$, hence, we conclude (1) for $F$. Finally it remains to remove the convolution with Fejer kernel to conclude (1) for $f$ (we have nonnegative terms in the sum on the left hand side of (1), so we can just cut the sum and take the limit $\hat{F}(m) \to \hat{f}(m)$)
In general, if $f$ is not continuous, let $K_{N}$ be the Fejer kernel. Split $Mx=M_{1}x+M_{2}x$ in (2), multiply (2) by $K_{N}(M_{1}x)$ and take the average in $M_{1}$ and use the Riemann integral criteria for equidistribution of $\{M_{1} x\, \mathrm{mod}\, (2)\}_{M_{1}\geq 1}$ to conclude $\left| \sum_{k=0}^{n-1} F(M_{2}x + kx)\right|<2C$ and now due to continuity of $F$ and density of $\{M_{2} x\}$ we conclude (3) for $F$ and the rest of the argument proceeds in the same way. $\square$
As Ville Salo already wrote in his answer, your question can be phrased in terms of the difference between the number of elements of the sequence $y, 2y, \dots, ky$ which are contained in $[0,1/2]$, and $k$ times the length of $[0,1/2]$. Here $y = x/2$. In the language of discrepancy theory, you are asking whether the interval $[0,1/2]$ is a so-called bounded remainder set of the sequence $(n y~\text{mod}~1)_{n \geq 1}$. However, it is known that it is not: the only intervals with bounded remainder are those whose length is in $\mathbb{Z} + y \mathbb{Z}$, and since $y$ is irrational in your question the length of $[0,1/2]$ is not of such type. Bounded remainder sets were classified in this paper: H. Kesten, On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arith. 12(1966), 193–212.
|
2025-03-21T14:48:30.415583
| 2020-04-26T02:44:35 |
358559
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"RobPratt",
"dineshdileep",
"https://mathoverflow.net/users/141766",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358559"
}
|
Stack Exchange
|
Fractional values in linear programming
Consider the linear programming problem
\begin{align}
f^* = \max_{x}&~p^Tx~\\~&A^Tx\leq b~,~0\leq x_i\leq 1
\end{align}where $p$ is a $n\times 1 $ vector, $A$ is a $n\times c$ matrix and $b$ is a $c \times 1$vector. Here $x=[x_1,\dots,x_n]$ is the $n \times 1$ vector to be found. Thus, in addition to the box constraint $0\leq x_i\leq 1$, we have $c$ constraints. It is also known that $p,A,b$ are all element-wise positive.
I randomly generated $p,A,b$ such that each entry is i.i.d and is uniformly distributed in the interval $[0,1]$. Note that this meets the problem set-up. I observed that whenever the problem was feasible, the optimal solution $x^*=[x_1^*,\dots,x_n^*]$ had a interesting property. The number of fractional $x_i^*$ (i.e. $0<x_i^*<1$) were atmost $c$. Every other $x_i^*$ belonged to $\{0,1\}$ (accounting for rounding). Is there anything going on here?
Yes, this is a known consequence of the fact that there always exists an optimal solution that is basic (an extreme point of the feasible region). This property is the foundation of the simplex method, which moves from one basis to another in each iteration.
Can you please point me to a proof. FWIK, simplex method should work regardless of randomness in $p,A,b$ as long as it is feasible. Are you suggesting that, whenever it is feasible, solution is going to obey this property?
There is probably a proof in Chvatal's Linear Programming (1983). Yes, the property holds for problems that have an optimal solution (feasible and not unbounded).
|
2025-03-21T14:48:30.415721
| 2020-04-26T03:08:48 |
358562
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Blur",
"Gerry Myerson",
"Ville Salo",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/156981",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358562"
}
|
Stack Exchange
|
Expanding Hall's theorem
I'm trying to get a "feel" about Hall's theorem and try to expand it for one to many matching.
So my question is:
Given a bipartite graph, what would be a neccessary and sufficient condition for that it would be possible to match every vertex on one side, to two vertices on the other side, that would belong only to him.
Iv'e "cloned" the vertices on the "one side", and for each edge from v to u where v is on the "one side" and u is on the other side, Iv'e connected an additional edge between v_clone and u.
I'm trying to figure out what would be the condition(s)? when I return to my original graph.
And how can I prove it?
Thanks!
What's wrong with the answer you got on m.se before you posted here? https://math.stackexchange.com/questions/3644024/one-to-many-matching-using-halls-theorem
Let $G$ be a bipartite graph with bipartition $(L,R)$. A necessary and sufficient condition for each vertex on the left to be matched to two vertices on the right is $|N_G(X)| \geq 2|X|$ for all $X \subseteq L$, This can be proved by applying Hall's theorem to the auxiliary graph that you defined.
There are many related results. For example, $G$ contains a forest $F$ such that $\deg_F(v)=2$ for all $v \in L$ if and only if $|N_G(X)| > |X|$ for all non-empty $X \subseteq L$.
This is an old result of Lovász. There is also a $(2-\epsilon)$ version of Hall's theorem for the existence of VW-matchings (these are subgraphs where every connected component looks like a V or a W), due to Bennett, Bonacina, Galesi, Molloy, Wollan, and myself. Finally, see this paper of Roberts for a very general result about 'tree matchings', which implies both Lovász's theorem and the result on VW-matchings.
Iv'e been trying to prove that but I can't quite prove that (I get lost in translation from the auxiliary graph to the original one), can you please quickly mention the steps?
If I'm not mistaken you can find a detailed explanation of the first paragraph in the book Cellular Automata and Groups by Tullio Ceccherini-Silberstein and Michel Coornaert. (Presumably in many other places as well.)
|
2025-03-21T14:48:30.415885
| 2020-04-26T04:54:51 |
358566
|
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|
Stack Exchange
|
Kac-Moody Lie algebra as derivations of associative algebras
The set of derivations of an algebra $\Bbb A$ forms a Lie algebra. This is one aspect of why Lie algebras are interesting. When $\Bbb A$ is polynomial algebra in $n$ variable then $\text{Der } \Bbb A$ is well studied. For example, these Lie algebras are infinite-dimensional simple. These include Witt's algebra as a particular case. Witts algebras and Kac-Moody Lie algebras have some similarities I think.
So I was wondering, are there any algebra $\Bbb A$ for which $\text{Der }\Bbb A$ is a Kac-Moody Lie algebra? The question is vague but wants to explore this direction. Kindly share some thoughts or references. Thank you.
|
2025-03-21T14:48:30.415960
| 2020-04-26T05:28:02 |
358569
|
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|
Stack Exchange
|
Picard group of double cover branched along reducible divisor
Let $X$ be the double cover of $\mathbb{P}^2$ branched along a divisor which is union of two lines. Then what will be the $\text{Pic}(X)$ ? Is it torsion free ? If yes, then what is its generator ?
This double cover is a quadratic cone in $\mathbb{P}^3$, its Picard group is $\mathbb{Z}$, it is generated by the hyperplane class, equivalently by the pullback of a line class from $\mathbb{P}^2$.
Suppose it is branched along $l_1 \cup l_2$ and let $C_1, C_2$ are the inverse image of $l_1, l_2$ respectively. Then how to write $\mathcal{O}_X(C_1)$ in terms of pullback of the hyperplane class ?
What do you precisely mean by the inverse images of $l_i$? If the pullbacks, then these are just hyperplane classes.
By the inverse image i mean $\pi^{-1} (l_i)$ where $\pi : X \to \mathbb{P}^2$ is the covering map. I did not mean the pullback.
Being the inverse image of a closed subset $l_i$, $C_i$ is a closed subset (hence a divisor in $X$)
As @Sasha explained, $X$ is a quadratic cone. Your $C_i$'s are just 2 generatrices of this cone. They are Weil divisors, not Cartier.
If you take $\pi^{-1}(l_i)$ as a scheme then it gives you the hyperplane class; if you take it as a set, this is a Weil divisor (not Carier) whose double is the hyperplane class.
If we resolve the singularity of $X$, i.e., if we consider the blow up of $X$ at the intersection points $C_1 \cap C_2$, and consider the proper transform of $C_i$, then it would be a Cartier divisor. I guess this proper transform will be linearly equivalent to the exceptional divisor. Am I correct ?
No — the exceptional divisor does not move linearly. Your blown up surface fibers onto $\Bbb{P}^1$, the proper transforms are fibers, while the exceptional divisor is horizontal.
How we write down the line bundle associated to the divisor $\tilde{C_1}$ in terms of the generators which are the pullback of hyperplane bundle and associated to the exceptional divisor, where $\tilde{C_i}$ is the proper transform of $C_i$ ?
Is it the pullback of hyperplane bundle tensored with $\mathcal{O}_{\tilde{X}}(-E)$ ?
|
2025-03-21T14:48:30.416119
| 2020-04-26T05:52:47 |
358571
|
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|
Stack Exchange
|
A Whitehead theorem for 3-categories
Let $F:\mathscr{C}\rightarrow \mathscr{D}$ be a 3-functor between 3-categories. Are the following two properties known to be equivalent?
$F$ is a 3-equivalence, meaning that there is a 3-functor $G:\mathscr{D\rightarrow\mathscr{C}}$ and natural equivalences $GF\Rightarrow Id$, $Id\Rightarrow FG$.
$F$ is essentially surjective, and for every object $A,B$ of $\mathscr{C}$, $F$ induces a 2-equivalences of 2-categories $\mathscr{C}(A,B)\rightarrow \mathscr{D}(F(A),F(B))$.
Let me add that 1. might not be exactly the statement one wants. More precisely, maybe we need to specify the inverses of the natural equivalence?
Regarding your last comment, this might be the difference between an adjoint equivalence, and one where the various data are not chosen to be coherent, or even only postulated to exist. I would start with Nick Gurski's book Coherence in Three-Dimensional Category Theory, or other of his work. The paper Biequivalences in tricategories is not enough, but might give you hints: http://www.tac.mta.ca/tac/volumes/26/14/26-14abs.html
That said, I would guess the theorem is true, for the correct definitions of all the things you mention. The trick is more getting the definition completely written down.
There should be such a theorem for $n$-categories for any $n$ (with condition 1 exactly as you stated), in fact even for $(\infty,n)$-categories. There are several equivalent definitions of the latter, and I think this theorem is more or less straightforward using Barwick's complete $n$-fold Segal spaces (see section 14 in https://arxiv.org/pdf/1112.0040.pdf for the definition).
By "3-category", "3-functor", etc. do you mean weak ones or strict ones? Traditionally in "low-dimensional higher category theory" the words "2-category" and "3-category" mean the strict versions, with "bicategory" and "tricategory" for the weak ones.
|
2025-03-21T14:48:30.416396
| 2020-04-26T06:18:21 |
358573
|
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|
Stack Exchange
|
cohomology classes of complex submanifolds
I was wondering if there were restrictions in what the cohomology classes corresponding to complex submanifolds of a complex manifold could be.
For example, say $T^4$ is regarded as a complex manifold as $\mathbb{C}^2/\mathbb{Z}[i] \times \mathbb{Z}[i]$, we have $H^2(T^4) = \Lambda^2(\mathbb{Z}^4) =\mathbb{Z}^6$. If $C$ is a ($1$ complex dimensional) complex submanifold, and we let $[C] \in H^2(T^4)$ be the corresponding cohomology class, could $[C]$ be any cohomology class in $H^2(T^4)$, or are there some restrictions we could impose just by knowing that it is a complex submanifold?
Your question is ambiguous. Are you asking about properties of cohomology classes of complex subvarieties of $(\mathbb{C}/\mathbb{Z}[i])^2$, in which case the answer is well-known (Lefschetz theorem), or of classes of submanifolds which are complex subvarieties of $T^4$ for some complex structure of $T^4$? (also well-known, but might be more interesting in general).
https://en.wikipedia.org/wiki/Hodge_conjecture
This is related to the Steenrod realizability problem, though that's for the laxer question of representing classes by oriented manifolds.
@abx I just meant cohomology classes of complex subvarieties of $(\mathbb{C}/\mathbb{Z}[i])^2$. Thanks for letting me know about the condition that it has to be in $H^{1,1}(T^4,\mathbb{C}) \cap H^2(T^4,\mathbb{Z})$.
I'm sure I'm being very stupid, though, but is it really true that any element in the subgroup generated by complex subvarieties can itself be represented by a complex subvariety? I can see that if C_1 and C_2 are subvarieties, then [C_1] + [C_2] is represented by an actual subvariety, but is the same true of [C_1] - [C_2]?
No, of course. What I meant is that this is a much studied problem, with proven methods to decide. Look at "divisors and line bundles" in any algebraic geometry book.
|
2025-03-21T14:48:30.416543
| 2020-04-26T07:12:57 |
358574
|
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|
Stack Exchange
|
Monotonity of an integral
It seems that for $p>4/3$ the following function $$g(a)=\sec a\int_{-\pi}^\pi |\cos t|^p (1 - \sin(a - t))^{3p-2}\sin tdt $$ achieves supremum when $a\to \pi/2-0$ for $a\in[0,\pi/2)$. How to prove this.
$\sec\pi/2=\infty,$ therefore $g(\pi/2-0)=+\infty$ and finite at all other points, therefore $\pi/2$ is the unique maximum on $[0,\pi/2]$.
"monotonity"? I prefer; "monotony" or "monotonicity".
for $p=2$ the integral can be evaluated and $g(a)=\frac{1}{2} \pi (4-\cos 2 a)$ --- with a finite maximum at $a=\pi/2$.
I find the language confusing—the supremum on a set $[0, \pi/2)$ is achieved at a point $a = \pi/2$ outside the set?
Ask it at MSE which is a right forum for such type questions.
|
2025-03-21T14:48:30.416643
| 2020-04-26T08:05:58 |
358578
|
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|
Stack Exchange
|
What is the definition of a prorelation?
In the context of quasi-uniform spaces, what is a prorelation?
In the text I'm reading, they're defined as a down-directed upper set on relations X->Y.
Now, I'm fine with a down-directed up-set, but don't know of and couldn't find any preorder on a collection of relations from X to Y, let alone a partial order in order to define an up-set
The set of relations has a natural order : the inclusion. It's the one that is used in the first paragraph of the introduction, when they say $1_X\leq a , a\cdot a \leq a$ for preorders
Turns out, a prorelation is just a filter on the set of relations X to Y -_-
|
2025-03-21T14:48:30.416719
| 2020-04-26T09:34:16 |
358583
|
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|
Stack Exchange
|
On the smallest solution of a linear congruence
I have the following question. First, consider the following congruence for primes $p\geq 5$:
$24x\equiv -1\;(\mbox{mod}\;p)$.
The smallest $x$, that is, $1\leq x\leq p-1$ for which the above congruence is true is $\dfrac{p^2-1}{24}-p[p/24]$, where $[.]$ is the integral part. Is there any explicit form (similar to the one above) of smallest $x$ for a general congruence: $ax\equiv b\;(\mbox{mod}\;p)$ for primes $p$, $(p,ab)=1$ and fixed $a, b$?
Any insight or comment is highly appreciated! Thanks!
I'm not sure there's any explicit form for a solution, even if you don't insist on it being the smallest.
"Explicit form" is a subjective term. At any rate, one can calculate the smallest solution of $ax\equiv b\pmod{b}$ quickly, using the Euclidean algorithm.
Adding on to @GHfromMO 's comment, it then isn't hard to do the Euclidean algorithm for every remainder modulo $b$, and so find a general formula once given $p$. It may be worth looking into whether there is a simple-ish formula with $p$ as an input.
|
2025-03-21T14:48:30.416837
| 2020-04-26T10:20:34 |
358585
|
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|
Stack Exchange
|
Isomorphism for Ext spaces for finite dimensional algebras
Let $A$ be an Artin algebra with enveloping algebra $A^e$.
Then we have $Hom_{A^e}(X,A^e) \cong Hom_A(D(A) \otimes_A X,A)$ for a bimodule $X$. (see for example in the article "A theorem of Green on the dual of the transpose" by Auslander and Reiten in corollary 4.2.)
Question:
When do we have the Ext analogue:
$Ext_{A^e}^i(X,A^e) \cong Ext_A^i(D(A) \otimes_A X,A)$ for all $i \geq 1$?
This holds for example for $X=A$. Maybe there is a nice condition and a reference for such isomorphisms.
My guess is that it should be true for $X=A^{*}$ (the dual of the bimodule $A$) for $i=1,..,n-2$ in case $A$ is $n$-torsionfree for $n \geq 3$.
Question 2: Is this true?
This would prove the equivalence of conditions a) and b) for finite dimensional algebras in question 2 in On properties of an algebra as a bimodule
edit:
Here is an example found with the computer that shows that the formula might not hold in general:
Let $A$ be the Nakayama algebra with Kupisch series [ 2, 3, 2, 1 ] and $X=D(A)$.
Then QPA says that $Ext_{A^e}^1(D(A),A^e)$ has dimension 1 while $Ext_A^1(D(A) \otimes_A D(A) , A)$ should have dimension 0.
The code:
A:=NakayamaAlgebra([2,3,2,1],GF(3));D:=DualOfAlgebraAsModuleOverEnvelopingAlgebra(A);B:=EnvelopingAlgebra(A);RegB:=DirectSumOfQPAModules(IndecProjectiveModules(B));
t:=Size(ExtOverAlgebra(D,RegB)[2]);
CoRegA:=DirectSumOfQPAModules(IndecInjectiveModules(A));U:=NakayamaFunctorOfModule(CoRegA);RegA:=DirectSumOfQPAModules(IndecProjectiveModules(A));
tt:=Size(ExtOverAlgebra(U,RegA)[2]);
I used that $D(A) \otimes_A D(A)$ is isomorphic to the Nakayama functor applied to $D(A)$.
Another example is again $X=D(A)$ and A linear oriented of Dynkin type $A_n$.
Can you give an example of when it does not hold please?
First, I do not understand your first formula. What is $X$ there? You start with $D(A)\otimes_A Tr(X)$ and arrive at something that does not depend on $X$. Second, can you be more specific of how you derive that from the results of the article? Another little remark: in your question, $Ext^i(D(A)\otimes_A X,A)$ is the Ext of right modules (see Corollary 4.2 that you quote); this is not apparent from the way you write it.
@VladimirDotsenko Sorry, I think I made a mistake. $- \otimes_A Tr(A)$ does lift $\tau^{-1}(-)$ for an Artin algebra (which is the main result of the cited article) but must not take the value 0 when $X$ is injective in it seems (so $\tau^{-1}(X)$=0 when $X$ is injective but $X \otimes_A Tr(A)$ might be nonzero). So I guess I have no explicit counterexample and it seems reasonable that the formula should hold in general with the standard proof. Ill think about that. Sorry for the confusion.
@VladimirDotsenko I added another counterexample in an edit.
thanks. I didn't run the calculations myself, but I have no reason to question them. Presumably (since for Hom's everything works), the non-exactness of tensoring with $D(A)$ is the only fathomable issue, right? (And that IS an issue.)
The natural way for proving such an isomorphism would be to take a projective resolution of $X$ (over $A^e$) and using the Hom-isomorphism to translate to one-sided modules. But then, even if $\operatorname{Tor}^A_i(D(A),X) = (0)$ for $i>0$, then to get an isomorphism on Ext-groups one would need either that (i) $D(A)\otimes_A A^e$ is projective as a left $A$-module or (ii) that $\operatorname{Ext}^1_A(D(A)\otimes_A A^e,A) = 0$. Then (i) is equivalent to $A$ being selfinjective, while (ii) is equivalent to $\operatorname{Ext}^1_A(D(A), A) = 0$.
Thanks, I made my guess 2 more precise, so we do not need the iso for all $i$. Here an example where guess 2 should be true so that we have an iso with $X=A^{*}$ and $i=1$ with $Ext^1(D(A),A)$ nonzero: The Nakayama algebra with Kupisch series [ 2, 2, 3, 2, 2, 2, 1 ] (or any other algebra with dominant dimension at least three and $Ext^1(D(A),A)$ nonzero).
Question 2 has an easy answer in case I made no mistake:
Let $A$ be $n$-torsionfree as an $A$-bimodule for $n \geq 2$, then also $X=A^{*}$ in $n$-torsionfree.
Choose a minimal projective resolution $(P_i)$ of $X$, then we get an exact sequence
$0 \rightarrow X^{*} \rightarrow P_0^{*} \rightarrow P_1^{*} \rightarrow ... \rightarrow P_{n-2}^{*}$.
Now we can apply the Hom-isomorphisms to get the result for $i=1,..,n-2$.
I was probably just confused because I first thought it should hold for all $i \geq 1$.
|
2025-03-21T14:48:30.417125
| 2020-04-26T11:14:09 |
358587
|
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|
Stack Exchange
|
Distribution of degree in graphs: when is the friendship paradox the paradox it wants to be?
$\DeclareMathOperator\deg{deg}\DeclareMathOperator\ndeg{ndeg}\newcommand\abs[1]{\lvert#1\rvert}$The friendship paradox goes most people have fewer friends than their friends have on average. The original paper Feld - Why your friends have more friends than you do has a simple counter example: (Figure 5, p. 1474).
My question is how exceptional is this example?
Even in some random graph models where some answer is tractable, the analysis of the distribution of a neighbour's degree deals with expectation. That is, in some models it is explicitly shown that the average difference between the mean number of friends of friends and the number of friends is positive, which is much less likely to get press attention I think.
Given any graph $G=(V,E)$ and $v\in V$, let $N(v) = \{w\in V:\{v,w\} \in E\}$ and $\deg(v) = \abs{N(v)}$. This is the number of friends of $v$. Now consider $\ndeg(v)= \sum_{w \in N(v)}\deg(w)$, the total friends of friends of $v$ and the distribution of
$$
f(v) = \frac{\ndeg(v)}{\deg(v)} - \deg(v).
$$
For a friendship paradox,
$$
g = \abs{\{v:f(v)>0\}}-\abs{\{v:f(v)<0\}}>0.
$$
For the graph pictured above,
$$
\{f(v)\}_{v\in \{A,B,C,D,E,F\}}=\left\{ 1,1,-\frac13-\frac13,-\frac13,-\frac13 \right\}.
$$
Clearly $g = -2$, and no paradox. But, the mean of $f$ is $1/9>0$.
The mean of $f$ is shown to be positive in the popular configuration model. And this is usually said to be evidence/proof of the friendship paradox. (Tangential question: is $g>0$ in this model?)
Now the question(s):
Is the mean of $f$ positive for any graph?
What are some equivalent conditions stated in terms of graph properties to $g >0$?
I can answer question 1, at least. The mean of $f$ is always positive (unless every connected component of the graph $G$ is regular, in which case the mean of $f$ is zero). To see this, we rewrite the sum of the values of $f$ as
$$\sum_{v \in V} f(v) = \sum_{v \in V} \sum_{w\text{ s.t. }\{v,w\} \in E} \left(\frac{\deg(w)}{\deg(v)} - 1\right) = \sum_{\{v,w\} \in E} \left(\frac{\deg(w)}{\deg(v)} + \frac{\deg(v)}{\deg(w)} - 2\right),$$
and each summand is positive by AM-GM. More explicitly, we have
$$\sum_{v \in V} f(v) = \sum_{\{v,w\} \in E} \frac{(\deg(v)-\deg(w))^2}{\deg(v)\deg(w)} \ge 0.$$
|
2025-03-21T14:48:30.417292
| 2020-04-26T11:27:54 |
358588
|
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|
Stack Exchange
|
Alternate characterization of floquet multipliers: Floquet theory
Given an autonomous ode $\dot{x}=f(x)$ in $\mathbb{R}^n$ possessing a period-p time-periodic solution $\bar x(t)$, one can use the so-called variational equation about $\bar x$ to study its stability. This equation satisfies \begin{align} \dot{y}(t)=A(t)y \end{align}, where $A(t)=D_xf(\bar x(t))$.
Now the typical definition of Floquet multipliers $\lambda$ is that they are the e.values of the time-p flow map U(p) of the variational equation, i.e., of the matrix/operator U(p) where U(t) itself satisfies the variational equation, with $U(0)=I_{n\times n}$. The Floquet exponents $\mu$ are then defined as $\frac{1}{p}\log{\lambda}$.
An alternative definition can apparently be given by starting with the exponents instead. Here one defines $\mu$ as the eigenvalues of the operator $L[x](t)=(\frac{d}{dt}-A(t))x(t)$ on the space of p-periodic functions. And then the multipliers are defined analogously as $\lambda=e^{-\mu p}$.
I am looking for a proof that both the definitions are the same (modulo some signs if need be).
|
2025-03-21T14:48:30.417396
| 2020-04-26T12:31:06 |
358594
|
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|
Stack Exchange
|
From a constructive perspective, what are the ordinal numbers?
From a constructive and computational perspective, what are the ordinal numbers?
On the one hand, it seems you can represent ordinal numbers symbolically using something like Cantor Normal Form notation. Then the ordinals are discrete, symbolic entities, and things like "$\leq$" become decidable.
On the other hand, you can represent the ordinals recursively as monotonically nondecreasing sequences of smaller ordinals, and then define an undecidable $\leq$ relation: An ordinal $(\alpha_i)$ is greater than or equal to $(\beta_i)$ if for every element of $\beta_j$ there is an element $\alpha_i$ such that $\alpha_i \geq \beta_j$. I struggle to see how this could be useful though.
nLab also mentions the Plump Ordinals, but I'm not sure what they are.
For an application, let $\mathrm{On}$, denote the class of all ordinals. Let $\alpha : \mathrm{On} \to \mathbb{R}$ be a mapping from $\mathrm{On}$ to real numbers. For any finite set $S = \{i_1, \ldots, i_n\} \subseteq \mathrm{On}$ where $i_1 < \cdots < i_n$, define the quantity
$${\mathcal K}_S(\alpha) = \sqrt{\alpha_{i_1}^{2^1} +
\sqrt{\alpha_{i_2}^{2^2} + \cdots \sqrt{\alpha_{i_n}^{2^n}}}
}$$
Say that a real $\mathcal{K}(\alpha)$ is the limit of $S \mapsto \mathcal{K}_S(\alpha)$ when for every $\epsilon > 0$ there exists a finite $S \subseteq \mathrm{On}$ such that for all finite $T \subseteq \mathrm{On}$, if $S \subseteq T$ then $|\mathcal{K}_S(\alpha) - \mathcal{K}(\alpha)| < \epsilon$.
Examples:
if $\alpha_n = 2$ then $\mathcal{K}(\alpha) = \sqrt{2^{2^1} + \sqrt{2^{2^2} + \sqrt{2^{2^3} + \dotsb}}} = 2\phi$ where $\phi$ is the Golden ratio.
given $x \in \mathbb{R}$, take
$$\alpha_n = \begin{cases}
1 & \text{if $n \neq \omega$,}\\
x & \text{if $n = \omega$}
\end{cases}$$
Then $\mathcal{K}(\alpha)$ is the limit if the sequence
$x, \sqrt{1+x^2}, \sqrt{1+\sqrt{1+x^4}}, \sqrt{1+\sqrt{1+\sqrt{1+x^8}}}, \ldots$. This is a continued radical which is "transfinite".
Somewhat related: https://mathoverflow.net/questions/325876/ordinal-valued-sheaves-as-internal-ordinals
In a different world this would not be a research-level question.
@AndrejBauer, I'm sure it's clear to some, but not to me. What does your comment mean? That this should be common knowledge but isn't?
What does Cantor normal form have to do with decidability of $\leq$?
@VilleSalo If you represent an ordinal as a finite expression in CNF, then $\leq$ is clearly decidable. The downside is you can only express ordinals up to $\epsilon_0$ in this way
Ok. That works only for small ordinals though, usually CNF doesn't "terminate" if you try to open it up recursively.
Sorry I did not read your comment, you say that yourself.
@LSpice; Suppose someone asked about the ordinal numbers in classical mathematics, would the question stand a chance? This is not a research-level question, as any PhD student of contructive mathematics will simply know how to use Google to look things up.
@ogogmad: Could you be a bit more precise about your application, in particular, how is the limit supposed to work? Is it $n \to \infty$? In which case, in what sense is $\Omega$ finite? You fixed $\Omega$ at the beginning and it is not allowed to change, and neither are the finitely many ordinals $i_1, \ldots, i_n$, so there's nothing to compute a limit of. It would help if you were precise about how precisely you quantify things. And what does $[i = \omega]$ mean, what's an Inversion Bracket?
@AndrejBauer An Iverson Bracket (not an "inversion" bracket) is the notation $[p]$, which equals $1$ when $p$ is true, and equals $0$ when $p$ is false.
@AndrejBauer I've clarified that it's the limit of $\Omega \mapsto {\Large\kappa}_{i \in \Omega}(\alpha_i)$. I then immediately go on to explain what that means
@AndrejBauer I don't think your suggestion to use filters is enough
You'll never have to go beyond the ordinal $\omega_1$, right? Your notation is still quite unintelligible. I will edit your question and please reject my edit if it is completely off the mark, or fix it, or whatever.
@AndrejBauer Don't need to for my application
I edited the question so that there aren't any dangling and unspecified variables, unecessasry indices etc. Is my interpretation of what you wrote correct?
More or less. I'm about to make an edit. Do you mind if I ping you when I'm done to get your opinion of it?
Sure, go ahead. I edited my answer to explain that you're not looking for anything related to constructive ordinals. What you have is the limit of a net, as known in topology.
Let us continue this discussion in chat.
@AndrejBauer Addressing your first remark: I did Google the topic, but unlike you I don't have decades of experience in this area.
That remark wasn't a reflection on your asking. It was a reflection on the fact that such a question isn't shot down by the users of MO. This just shows that "research-level" is a very much a social category (which of course is to be expected).
The ordinals in constructive mathematics are not as well-behaved as in classical mathematics. For example, if they are linearly ordered the excluded middle holds. There are appropriate substitutes, such as well-founded orders and inductive types. If you told us what you need the constructive ordinals for, we might be able to tell you what to use instead.
There are several possible definitions of ordinals which are classically equivalent but are intuitionistically distinct. For further reading I recommend
the material available on Paul Taylor's web page summarizing his work on induction, recursion, replacement and the ordinals. Plump ordinals were defined in "Intuitionistic Sets and Ordinals", available on the web page.
Supplemental: If I understand your application correctly, then it has nothing to do with ordinal notations, ordinal representations, or intuitionistic ordinals. It is a case of the limit of a net from topology. Specifically, let $D$ be the set of all finite subsets of the index set $I$ of ordinals (from your application), ordered by $\subseteq$. Then $D$ is a directed set, and the number $\mathcal{K}(\alpha)$ is precisely the limit of the map $f : D \to \mathbb{R}$ defined by $f(S) = \mathcal{K}_S(\alpha)$.
OK, what about the representations I proposed in my question?
From what I understand about your description of an application, you are not looking for any constructive ordinals. Are you familiar with general notions of convergence, such as net convergence and filter convergence? That's where you should be looking, as far as I can tell.
The purpose of using the ordinal numbers in my example was to define a transfinite sequence, not to define any kind of limit. That said, now I can see that I don't need the well-ordering property of the ordinals, and can just as well use any totally ordered set
|
2025-03-21T14:48:30.417851
| 2020-04-26T13:43:29 |
358601
|
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|
Stack Exchange
|
Detecting weak equivalence on free loop space homology
Given $f:X \to Y$ a continuous map between two spaces (unpointed CW-complexes) such that $f$ induces an isomorphism in homology with integer coefficient, and $f$ induces an isomorphism on homology of the free loop spaces: $H_*(X^{S^1}) \to H_*(Y^{S^1})$ (also with integer coefficient). Does $f$ has to be a weak homotopy equivalence ?
If I further assume that $f$ induces an equivalence $H_*(X^{K}) \to H_*(Y^{K})$ for each finite CW-complex $K$, is $f$ a weak homotopy equivalence ?
Note that working in the unpointed settings makes a big difference here: It was showed by Arlin and Christensen that, contrary to the case of pointed connected space, for any small set of spaces $E$, there are maps that induces an isomorphism $\pi_0(X^K) \to \pi_0(Y^K)$ for all $K \in E$ without being weak equivalences. It is unclear to me if adding higher homology groups makes a big difference or not.
Edit: The case of groupoids $X=BG$ and $Y=BH$ is already an interesting example. $X^{S^1}$ is the (classyfing space of) the groupoid corresponding to the action of $G$ on itself by conjugation. Hence given $f:G \to H$ a morphism of group. Saying that it is a bijection on the $H_0$ of the loop space gives that $f$ induces a bijection between the set of conjugation classes of $G$ and $H$, and the fact that it is a bijection on the $H_1$ of the loop spaces gives that for all $g \in G$, $f$ induces an isomorphisms between the ablianization of the centralizer of $g$ and the abelianization of the centralizer of $h$.
That does not seem to be quite enough to conclude that $f$ is an isomorphisms, but this is already pretty restrictive. I havn't been able to formulate the fact that $f$ induce bijections on the higher $H_i$ in simple terms.
Now, if we start looking a $H_0(X^K)$ and $H_0(Y^K)$ for $K$ a finite complex, it seems that in this case all the interesting information is in the case where $K$ is a wedge of circle. Being a bijection on $H_0(X^K)$ and $H_0(Y^K)$ means that:
1) $f$ induces a bijection between $G^n/G$ and $H^n/H$ for all $n \in \mathbb{N}$, where $G$ and $H$ acts on $G^n$ and $H^n$ by the diagonal conjugation action.
2) For every finite family of elements $g_1,\dots,g_n$ in $G$, $f$ induce a bijection between the abelianization of the centralizer of $\{g_1,\dots,g_n\}$ and the abelianization of the centralizer of $\{f(g_1),\dots,f(g_n) \}$.
For your first question, what if $X=BG$ and $Y=pt$, with $G$ an acyclic group ?
That does not seems enough to me: $BG^{S^1}$ is the homotopy quotient of the action of $G$ on itself by conjugation. So just its $H_0$ and $H_1$ already remember the set of conjugation class of $G$ and the abelianization of the centralizer of each elements.
I believe these questions were studied and answered in the this 1984 paper:
HIROSHIMA MATH. J. 14 (1984), 359-369
On the set of free homotopy classes and Brown's construction
Takao MATUMOTO, Norihiko MINAMI and Masahiro SUGAWARA
They also have counterexamples that seem to be the same as in the recent preprint of Arlin and Christensen that you mention.
Take a look and see if they do what you want.
Just to clarify, the example in this paper is well known, and is what we generalized beyond finite complexes in our paper.
I might be missing something, but I don't understand how this paper answer the question ? They give a counter-exemple to the case where we look at free homotopy class [K, _ ] for K a finite CW-complex (which was already mentioned in the question), but I don't think it is a counter-example to the case where we also look at homology group, and it proves that it works when considering [K,X] when K is an arbitrary CW-complexes, whose special case relevant for the question is clear by Yoneda lemma ?
But their theorem 1 is definitely relevant: it shows that it is enough to show that $f$ is surjective on $\pi$ to conclude that it is an equivalence.
|
2025-03-21T14:48:30.418131
| 2020-04-26T15:28:31 |
358605
|
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|
Stack Exchange
|
Residually finite group with dense finite index subgroups
Let $G$ be a locally compact Hausdorff topological group whose underlying abstract group is residually finite. Let $H\subset G$ denote the intersection of all finite-index, closed subgroups. Is there an example of such a $G$ where $H$ is not trivial? Is there an example where $H=G$ (i.e., every finite-index subgroup is dense)?
My motivation for asking this question comes from the study of automorphism groups of connected, locally finite graphs. Therefore, the closer the example is to being of this type the better. E.g., an example which is an automorphism group of some structure on $\mathbb{N}$, say a hypergraph structure, would be extremely interesting; an example where $G$ is separable would be more useful than an example with cardinality larger than the continuum, and so on.
@Ycor thank you for your suggestion, I edited the question accordingly.
Consider Z with the indiscrete topology? What kind of example are you really looking for?
@Sempliner it is implicit that $G$ is Hausdorff. (There's a huge literature on locally compact groups, and it's a standing assumption.)
That's reasonable.
I think I can produce an example of such a $G$, actually abelian, but it's far from "simple".
@YCor Yes, I am assuming the group is Hausdorff, thank you for the remark.
@Sempliner, As for "what kind of example are you looking for?", I originally came to this question studying automorphism groups of connected, locally finite graphs, so in some sense that would be the best possible answer. Barring that, the closer the group is to that family the better; e.g., it would be more interesting it the example is separable. I would gladly accept any l.c. Hausdorff group as an answer, especially since it doesn't seem that easy to come up with examples.
Writing is there a "simple" example suggested that you had a complicated/elaborate example in mind, but it seems you're just asking for an example, right?
@YCor yes, now that you mention it maybe the question was poorly worded. I am looking for an example (I don't have any). I am going to edit the question.
I'm afraid I'm unable to produce an example so far. My idea was to make variants of those examples of LCA groups $G$ in which $pG$ is a dense proper subgroup (and $qG=G$ for all other primes), which have a clopen subgroup isomorphic to $\mathbf{Z}_p^\mathbf{N}$, for $\mathbf{Z}_p$ the $p$-adics, and maybe playing with two primes... so far it's not successful.
BTW I should mention that I'm aware of nontrivial locally compact groups $G$ in which there is no proper closed finite index subgroup, while the intersection of finite index subgroups is countable (hence a dense countable normal subgroup). But this does not answer the question. Also these examples are not compactly generated.
Yes it exists (at least the weaker version). Namely, here is a way to produce second-countable abelian, totally disconnected locally compact groups whose underlying abstract group is residually finite, but in which the intersection of finite index open subgroups is not trivial.
Fix $p$ prime and an infinite countable set $I$. Let $\mathbf{Z}_p$ be the $p$-adic group. Let $A$ be the divisible closure of $K=\mathbf{Z}_p^I$ in $\mathbf{Q}_p^I$ (i.e., those sequence with bounded denominator) and endow $A$ with the group topology making $K$ a compact open subgroup (with its usual product topology). The example will be a suitable subgroup of $A$ containing $K$.
Lemma: let $F$ be a finite subset of $A$ and $\bar{F}$ its image in $A/K$. Let $U$ be a nonempty open subset of $K$, disjoint from $\langle F\rangle+pK$, and $n\ge 1$. Then there exists $x\in A$ whose image $\bar{x}$ in $A/K$ has order $p^n$, such that $\langle \bar{x}\rangle\cap \bar{F}=\{0\}$, and such that $p^nx\in U$.
Proof: choose $z\in U-pK$ (this is possible: as $I$ is infinite $pK$ has empty interior in $K$) and set $x=p^{-n}z$. Then $\bar{x}$ has order $p^n$.
If $\langle \bar{x}\rangle\cap \bar{F}=\{0\}$ fails, then $p^\ell\bar{x}\in\bar{F}$ for some $\ell$ with $1\le\ell<n$. Hence we can write $p^\ell x=s+k$ with $s\in F$ and $K\in K$. Thus $z=p^{n-\ell}s+pk'$, with $k'=p^{n-\ell-1}k\in K$, contradiction.
Now let $(K_n)_{n\ge 0}$ be an enumeration of clopen subsets of $K$. Define by induction a sequence $(x_n)$: define $F_n=\{x_i:i<n\}$. If $K_n\cap (\langle F_n\rangle+K)$ is nonempty set $x_n=0$. Otherwise, choose $x=x_n$ as in the lemma (for $U=K_n$).
Let $B$ be the (open) subgroup of $A$ generated by $K$ and $\{x_n:n\ge 0\}$. I claim it satisfies the required properties.
(a) $B$ is residually finite as abstract group. Indeed, since it is a torsion-free abelian group, residually finite means it does not contain any isomorphic copy of $\mathbf{Q}$. If we had such a copy, it could not be contained in $K$ (which is residually finite), hence would have nontrivial projection on $B/K$.
By construction, the cyclic subgroups $\bar{x_n}$ generate their direct sum. So $B/K$ is also residually finite. We get a contradiction.
(b) $B$ is not residually (discrete finite) as topological group. Namely, let us check that ($\sharp$) $K=\bigcap_n \overline{n!B}$. This implies that in any discrete quotient of $B$, the image of $K$ belongs to every finite index open subgroup, and in particular, $K$ is contained in every finite index open subgroup of $B$.
Let us prove the assertion ($\sharp$). Since $B$ is a $\mathbf{Z}_p$-submodule of $A$, $B$ is $p$-divisible for every prime $\neq p$. So we only have to prove that $K=\bigcap_n \overline{p^nB}$. Indeed, let $U$ be some nonempty open subset of $K$, and let us check that $p^nB\cap U$ is not empty.
Suppose by contradiction the contrary: $p^n B\cap U$ is empty. So for every $m\ge n$, we have $p^n(K+\langle F_m\rangle)\cap U$ empty, and hence $p^m(K+\langle F_m\rangle)\cap U$ is empty as well. For some $m\ge n$, we have $K_m\subset U$. Since $p^m(K+\langle F_m\rangle)\cap U$ is empty, by construction $x_m$ is nonzero and $p^mx_m\in K_m$, so $p^mB\cap U$ is non-empty, and in turn $p^nB\cap U$ is non-empty, contradiction.
Notes:
a) if $G$ is a locally compact group and not totally disconnected, then it has a 1-parameter subgroup, hence its underlying abstract group is not residually finite. So examples will be totally disconnected.
b) I'm not sure whether the above example can be modified to satisfy the stronger conclusion ($\neq \{0\}$ and no proper open subgroup of finite index)
c) I don't know how to find a compactly generated example. It certainly can't be abelian: indeed a compactly generated LCA-group either maps onto $\mathbf{Z}$ with open kernel, or is profinite. In both case, unless $\{0\}$, it has a proper open subgroup of finite index.
|
2025-03-21T14:48:30.418675
| 2020-04-26T15:40:04 |
358606
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Stack Exchange
|
On which regions can Green's theorem not be applied?
In elementary calculus texts, Green's theorem is proved for regions enclosed by piecewise smooth, simple closed curves (and by extension, finite unions of such regions), including regions that are not simply connected.
Can Green's theorem be further generalized? In particular, are there regions on which Green's theorem definitely does not hold?
I'll concede that the phrasing of the question is a little naive, and from context I would guess that the author is not a mathematical researcher, but nevertheless a question like "how far can you push Green's theorem using state of the art geometric measure theory" probably is a good MO question. Then again it might require a little more care to formulate that question well.
Thank you both for the advice.
I have attempted to rewrite the question and have nominated it for reopening.
@YCor: The latest title edit is ungrammatical.
@R..GitHubSTOPHELPINGICE do you confirm that "On which regions can GT not be applied" is the correct wording? if so I'll edit.
@YCor: Yes, that's correct. This is a place where the grammatical structure of the clause imposes conditions on the placement of both the "can" and the "not" that prevents them from being joined or contracted.
@R..GitHubSTOPHELPINGICE your last comment is a bit of an overstatement — “On which regions can’t Green’s theorem be applied?” is perfectly grammatical (to my native BrE-speaker ear). But I agree the original phrasing with “can … not” better, emphasising the “not” more clearly.
@PeterLeFanuLumsdaine: Perhaps. AmE here and it at least sounds non-idiomatic and sufficiently awkward I'd have to stop to think about it.
I think this is an interesting and sort of deep question, so I'm going to answer it in part with the hope that my answer attracts even better answers.
I'll start with my first thought: surely there's no hope of formulating Green's theorem for an unbounded region, say the region $y > 0$. But then I thought about it for a moment, and observed that if you consider a smooth vector field $F(v)$ on the plane such that $F(v) \to 0$ rapidly as $v \to \infty$ then we can extend $F$ to the sphere by stereographic projection; this sends $y > 0$ to a hemisphere and the boundary curve $y = 0$ to the bounding great circle, and you can apply Stokes' theorem to this situation. Unwinding the calculations, this would give you a version of "Green's theorem" even for unbounded regions, albeit one that applies only to a certain class of vector field.
Then I thought about regions whose boundary is pathological, like the interior of the Koch snowflake. Here the boundary has infinite length, so surely there is no real hope of even defining the "boundary side" of Green's theorem. But then I noted that the Koch snowflake - like many pathological plane curves - has a very nice polygonal approximation, and it didn't sound insane that the boundary side could be defined as a limit of integrals over these approximations (again, maybe not for all vector fields). Sure enough, this has been worked out, and there is indeed a version of Green's theorem for fractal boundaries:
Jenny Harrison and Alec Norton, The Gauss-Green theorem for fractal boundaries, Duke Math. J. 67 Number 3 (1992) pp. 575-588. doi:10.1215/S0012-7094-92-06724-X, author pdf.
There are other crazy things to try, like removing a non-measurable set from the plane or something. But Green's theorem (and its parent, the fundamental theorem of calculus) is based on a very resilient idea, something like "when you sum differences, things cancel". So in the spirit of the principle, "The fastest way to find something is to assert that it doesn't exist on the internet", I'll make a bold conjecture: Green's theorem can be generalized to any subset of the plane.
Thank you. Your response has been very helpful.
To extend the first idea: the entire plane? Or the plane except a single point?
"The best way to get the right answer on the Internet is not to ask a question; it's to post the wrong answer." — Ward Cunningham's law.
"I'll start with my first thought: surely there's no hope of formulating Green's theorem for an unbounded region" Don't tell people solving external acoustic problems that. The entire foundation of their boundary element numerical methods is Green's functions on unbounded regions of $\mathbb{R}^3$.
It can be generalized even beyond subsets. A subset of the plane may be viewed as a function $f : \mathbb{R}^2 \to {0,1}$, but we can also generalize the range to be $\mathbb{R}$, to allow fractional points to be in that subset. We can then define the integral over the "generalized subset" $\int_f g$ = $\int f g$. In this case Greene's theorem becomes the same as integration by parts. Now we can generalize $f$ further from a function to a Schwartz distribution, and perhaps beyond.
As mentioned elsewhere on this site, Sauvigny's book Partial Differential Equations provides a proof of Green's theorem (or the more general Stokes's theorem) for oriented open sets in manifolds, as long as the boundary has capacity zero, and the differential form that you integrate has compact support. (Sauvigny just assumes that the open set is bounded in Euclidean space, but the same proof works without that hypothesis, as long as the differential form is compactly supported, $C^1$ in the interior and $C^0$ up to the boundary). The precise definition of capacity is complicated, so you would need to read the book to get that, but it includes reasonable things like corners and cone points. He also indicates the problems that arise with capacity nonzero, and (if I remember correctly) there are always problems. Of course, there is a problem with making sense of the integration if you allow objects that are too wildly behaved.
There is a fun reverse definition that is used for so called "currents" in geometric measure theory, objects for which then in Green's theorem always ends up trivially being true. But then using the resulting theory, one can then show that Green's theorem is always true in a more proper sense, whenever the two integrals in it are defined, even if only in a very weak measure theoretic way.
Let $\Omega \subset \mathbb{R}^2$ be a $\mathcal{H}^2$-measureable set ($\mathcal{H}^2$ is the 2-dimensional Hausdorff measure).¹ Then we can define the corresponding linear operator
$$
\begin{array}{rccl}
[\Omega] : & C_c^\infty(\mathbb{R}^2) &\to &\mathbb{R}\\
& f & \mapsto & \displaystyle\int_\Omega f dx
\end{array}
$$
This is what is called a $2$-current, i.e. an element of the topological dual $\mathcal{D}_2 := C_c^\infty(\mathbb{R}^2)'$. The theory of currents is quite similar to those of distributions, but a bit more geometrical. In particular, for any $T \in \mathcal{D}_2$ we define its boundary by
$$
\begin{array}{rcl}
\partial T: & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) & \to & \mathbb{R}\\
& F &\mapsto & T(\operatorname{curl}F)
\end{array}
$$
The resulting object is what is called a $1$-current.² As one would expect, those correspond to integral along suitably generalized curves. Using this definition, Green's theorem is always automatically true, since $\partial[\Omega](F) = [\Omega](\operatorname{curl} F)$ by definition.
One can of course then reformulate the question to make it interesting again. Let $\Gamma$ be a $\mathcal{H}^1$-measurable set of locally finite measure and $\tau: \Gamma \to \mathbb{R}^2$ some $\mathcal{H}^1$-measurable "unit tangent" orienting that set. Then we can define similarly
$$
\begin{array}{rccl}
[\Gamma] : & C_c^\infty(\mathbb{R}^2;\mathbb{R}^2) &\to &\mathbb{R}\\
& F &\mapsto& \displaystyle\int_\Gamma F \cdot \tau d \mathcal{H}^1
\end{array}
$$
which is likely the weakest way to give some sense to the other integral in Green's theorem. Now the question is, for which $\Omega$ does this commute, i.e. $\partial [\Omega] = [\partial \Omega]$? Here the topological boundary $\partial \Omega$ is easy enough to define, but it turns out that the key-question here is what is the tangential vector $\tau$? The resulting notion is that of rectifiability. Roughly the condition for $\tau$ to be "the" tangential vector of $\Gamma$ at $x$ is that for any double-cone in direction $\tau$, most of $B_\epsilon(x)\cap \Gamma$ lies in that cone (the details are technical). If such a $\tau$ exists $\mathcal{H}^1$-almost everywhere, then $\Gamma$ is called rectifiable.
Now there are some further minor details involving the orientation of $\tau$ and possible multiplicity, but fundamentally it turns out that whenever the topological boundary $\partial \Omega$ is rectifiable, then there is a matching $\tau$ such that $\partial$ and $[\,]$ commute, i.e. Green's theorem holds.
The proper citation for all of this is Federer's "Geometric measure theory", but as it is one of those books, I'd recommend picking up Morgan's "Geometric measure theory: A beginner's guide" first.
¹Asking if Green's theorem holds for non-measurable sets should probably only be done by Zen Buddhists.
²As you can see, the number refers to the "dimensionality" of the object. To be more precise one should actually use differential forms of the dimension 2 in place of $f$ (resp. dimension 1 in place of $F$). This, and using the exterior derivative instead of $\operatorname{curl}$, also give the right generalisation to higher dimensions.
|
2025-03-21T14:48:30.419360
| 2020-04-26T18:30:44 |
358616
|
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|
Stack Exchange
|
A naive question about the prime number theorem
Let $\psi(x)=\sum_{n\leq x} \Lambda(n)$, where $\Lambda(n)$ is the von Mangoldt function.
Then as Chebyshev showed, the following equality holds
$$\sum_{n\leq x} \psi(x/n)=x\log(x)-x+O(\log(x)).$$
My question is, how far can one go towards proving the prime number theorem by only
using the above estimate and the fact that $\psi$ is increasing? Alternatively, is there a well-known example of an increasing function $\psi(x)$ for which the above equality holds, but $\lim_{x\to \infty}\frac{\psi(x)}{x}\neq 1?$ Thank you very much.
David Speyer has described interesting sets of natural numbers for which the (analogue of the) prime number theorem fails but that share some of the density properties of the set of primes. You might try to concoct a suitable $\psi$ function out of his "fake primes."
Theorem 1 in the following paper of Ingham shows that the stated estimate, together with $\psi$ being positive and nondecreasing, is 'enough' to deduce that $\psi(x)/x \to 1$:
A. E. Ingham: Some Tauberian theorems connected with the prime number theorem. J. London
Math. Soc. 20, 171–180 (1945). Full text (paywalled)
In fact, this remains true even if one only has the weaker error term $o(x)$ in place of $O(\log{x})$. Of course, one has to be careful about what `enough' means here; Ingham's proof still uses the nonvanishing of $\zeta(s)$ on $\Re(s)=1$, which is the key input in the usual proofs of the prime number theorem.
Thank you very much!
|
2025-03-21T14:48:30.419501
| 2020-04-26T21:16:11 |
358620
|
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|
Stack Exchange
|
Does nefness carry over through flips?
Suppose $\pi: X \dashrightarrow Y$ is a birational map which is an isomorphism in codimension 1 (such as a flip). Also suppose both $X$ and $Y$ have reasonable (say log terminal) singularities. We know that divisors in $X$ and $Y$ correspond bijectively to each other. If $D$ is a nef divisor in $X$, is it true that $\overline{\pi(D)}$ is also nef in $Y$?
No, $\overline{\pi(D)}$ will typically have a non-empty base locus where it is negative - try to see what happens in the Atiyah flop..
More precisely, for a klt flip, it carries through iff $D\cdot \Sigma =0$ where $\Sigma$ is the flipping curve (else $D\cdot \Sigma >0$ and so $\overline{\pi(D)} \cdot \Sigma ^+<0$ where $\Sigma ^+$ is a flipped curve....if $D\cdot \Sigma=0$, use the base point free thm). This fact is important when running the mmp with scaling.
|
2025-03-21T14:48:30.419601
| 2020-04-26T22:39:48 |
358626
|
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|
Stack Exchange
|
Prove the supremum of Mittag-Leffler function
I find two interesting limits :
\begin{align*}
\frac{1}{2}& =\lim_{s\to 1^-}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{\Gamma(1+ns)}{\Gamma(1+n)}\\
& =\lim_{s\to 1^+}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{\Gamma(1+n)}{\Gamma(1+ns)}.
\end{align*}
But I have no idea about the second. The process that I investigated it produces a problem :
$$\underset{x\geqslant 0}{\sup}\left|\sum_{n=0}^{\infty}\frac{\left(-x\right)^n}{\Gamma(1+\alpha n)}\right|=\begin{cases}
1& 0<\alpha\leqslant 2\\
+\infty& \alpha>2
\end{cases}.$$
But I have no way to prove it. Thanks for any help.
@LSpice Thanks.
Your original limits can be generalized: for $0<\alpha<\beta$, if we let $$G(\alpha,\beta):=\sum_{n=0}^\infty(-1)^n\frac{\Gamma(1+n\alpha)}{\Gamma(1+n\beta)},$$ then $\lim\limits_{\alpha\to\beta^-}G(\alpha,\beta)$ and $\lim\limits_{\beta\to\alpha^+}G(\alpha,\beta)$ are both equal to $1/2$ (see the last section).
For the "produced" problem, we have the following integral representation: $$F_\alpha(x):=\sum_{n=0}^\infty\frac{(-x^\alpha)^n}{\Gamma(1+n\alpha)}=\frac{1}{2\pi\mathrm{i}}\int_\lambda\frac{e^{xz}\,dz}{z(1+z^{-\alpha})},$$ where the contour $\lambda$ encircles the negative real axis and the poles of the integrand (we assume that the principal value of $z^{-\alpha}$ is taken here). This is shown using the geometric series for $(1+z^{-\alpha})^{-1}$, with $\lambda$ deformed if needed for $\Sigma\leftrightarrow\smallint$, and Hankel's contour integral for $1/\Gamma$.
Using residues, we can shrink $\lambda$ to encircle the negative real axis closely (or even evaluate $F_\alpha$ completely if $\alpha$ is an integer). For $\alpha>2$, the residues bring exponentials of the form $e^{cx}$ with $c>0$, which imply the unboundedness of $F_\alpha$. For $0<\alpha<1$, the integrand has no poles, and we obtain (in the limit of "closely") $$F_\alpha(x)=I_\alpha(x):=\frac{\sin\alpha\pi}{\alpha\pi}\int_0^\infty\frac{\exp(-xt^{1/\alpha})\,dt}{1+2t\cos\alpha\pi+t^2},$$ positive and decreasing in $x$. For $1<\alpha<2$, there are poles at $z=\exp(\pm\mathrm{i}\pi/\alpha)$, so this time $$F_\alpha(x)=I_\alpha(x)+\frac{2}{\alpha}\exp\left(x\cos\frac\pi\alpha\right)\cos\left(x\sin\frac\pi\alpha\right),$$ which makes a proof of $|F_\alpha(x)|\leqslant 1$ tedious this way (but still possible I believe).
The same approach applies to our "generalized limits". We have $$G(\alpha,\beta)=\frac{1}{2\pi\mathrm{i}}\int_0^\infty e^{-t}\int_\lambda\frac{\exp(t^{\alpha/\beta}z)}{z(1+z^{-\beta})}\,dz\,dt$$ and, if we deform $\lambda$ to fit in $\Re z<1$ strictly, then both $\alpha\to\beta^-$ and $\beta\to\alpha^+$ can be taken under the integrals, and moreover, the integrations can be interchanged, resulting in $$\frac{1}{2\pi\mathrm{i}}\int_\lambda\frac{dz}{z(1-z)(1+z^{-\alpha})}=-\operatorname*{Res}_{z=1}\ldots=\frac12$$ (by completing the part of $\lambda$ inside $|z|=R$ with the larger arc, and taking $R\to\infty$).
I'm still looking for a simple way to show $|F_\alpha(x)|\leqslant1$ for $1<\alpha<2$. The above shows that $F_\alpha$ is a difference of two functions, an exponentially decaying oscillating one, and a decreasing tending-to-zero one. Perhaps an alternative integral representation would do, or even the original contour integral could be modified suitably.
|
2025-03-21T14:48:30.419816
| 2020-04-26T23:17:02 |
358627
|
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|
Stack Exchange
|
What do Sylow 2-subgroups look like for Schur covering groups of finite simple groups?
What do Sylow 2-subgroups look like for Schur covering groups of finite simple groups?
Are there any references in which we can find the stucture of Sylow 2-subgroups of Schur covering groups of finite simple groups?
How is this different from your question https://mathoverflow.net/questions/357300/in-which-books-we-can-find-structure-information-for-finite-simple-groups-and-th ?
For the sporadic groups, I'd look in Aschbacher's SPORADIC GROUPS, although there are easier books for, say, the Mathieu groups. For the classical groups that aren't orthogonal, then you're just looking at matrix groups which are pretty easy to deal with -- Taylor's CLASSICAL GROUPS is a good option. For the orthogonal groups, you need to understand spin groups. For these and for the exceptionals I'd probably start with Gorenstein-Lyons-Solomon Vol 3 although I don't have it in front of me so can't double-check. Finally double-covers of alternating groups are covered in many places.
Thank you very much!
|
2025-03-21T14:48:30.419926
| 2020-04-27T02:28:03 |
358629
|
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|
Stack Exchange
|
Acting with a finite number of rotations on a set of positive measure can you fill almost the whole circle?
Let $E\subset S^1$ have positive Lebesgue measure. Do there exist finitely many rotations
$r_1, r_2, \dots ,r_n$ such that $r_1E\cup r_2E\cup \dots\cup r_nE$ has measure $2\pi$? Or is there a counterexample?
The answer is no. Take a compact set $E$ with positive measure buth empty interior and assume that $K=r_1 E\cup r_2 E \cdots \cup r_n E$ has measure $2\pi$. Then $K$ would be dense in $S^1$, hence equal to $S^1$, since it is closed. But this is impossible, since $K$ has empty interior, too.
Do you have an exmple of a compact set with positive mesure but empty interior?
You can take a Cantor set of positive measure
Of course I wanted to delete my comment but to late, you answered already. Also, your answer to the main question is very nice!
Nice answer! What can be said if we relax the condition to: for every $\varepsilon>0$, there exist $n$ rotations such that $r_1E\cup\ldots r_nE$ has measure greater than $2\pi-\varepsilon$?
@Capublanca Thanks. In that case it is true, by a limit argument, since using a countable dense set of rotations one can cover almost all the circle (see https://mathoverflow.net/questions/358506/acting-with-all-rational-rotations-on-a-subset-of-the-circle-having-positive-mea).
|
2025-03-21T14:48:30.420051
| 2020-04-27T02:54:42 |
358630
|
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"Praphulla Koushik",
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|
Stack Exchange
|
Why not consider categorical quotient in stacks?
Let's say we have a $k$ group scheme $G$ acting on a $k$ scheme $X$, we can consider its quotient in the category of stacks, the usual definition of it would be the quotient stack $[X/G]$ defined by
$[X/G](S)$ is the groupoid of $G$ bundles over $S$ together an equivariant maps to $X$.
My Question 1 is why don't we take the categorical quotient?
I think the categorical quotient exists because we can take the categorical quotient in the categories of 2-functors from Schemes to Groupoids, and then sheafify (stackify).
I understand that the stack quotients enjoy some very good properties, e.g. it makes every action look like a free action and it the theory of sheaves downstair is equivalent to the theory of sheaves.
but why people (or do they) don't study categorical quotient as well?
My Question 2 is how do stack quotients differentiate from categorical quotients?
Stack quotient is equal to the categorical quotient in the case free action, and I have the feeling that the two notions agree only when the action is free, is it true?
Thank you very much!
I am not sure if I understand your question correctly.. You say "Schemes to Groupoids, and then sheafify (stackify)"... I thought this is what happening.. Isn't it so? See page 29 for definition of $\widetilde{y}(\mathcal{G})$ for a Lie groupoid $\mathcal{G}$ and page number 69 how sheafification/stackification comes into picture when defining the stack $B\mathcal{G}$... This is about stacks coming from Lie groupoids.. I am sure the same can be done for stacks coming through schemes...
See https://mathoverflow.net/a/339953/118688 It looks like it holds for schemes as well
answer in https://mathoverflow.net/questions/319038/motivation-for-definition-of-quotient-stack may be of useful..
@PraphullaKoushik, I am sorry, I should be more clear. I always think about stacks coming from schemes. What I mean is thinking a stack as a (pseduo) functor from the categories of schemes to the categories of groupoids. We can take categorical quotient in this larger category, and the stackification of it is the categorical quotient in the category of stacks. My question is that why people don't care about categorical quotient stack even they always exist.
It sounds like you are asking for a map from $X$ to a stack that is $G$-invariant and universal with respect to $G$-invariant maps to stacks. The quotient map $X \to [X/G]$ satisfies this property, so stack quotients and categorical quotients in stacks are the same thing.
I do not think I still understand it... Lets wait for some one more experienced to comment...
@S.Carnahan I think the categorical quotient should be universal with respect to $G$ invariant maps FROM $X$. A simple example to see the two concepts are different is the case of the trivial action acting on point, the stack quotient is $BG$, and categorical quotient is just the point.
You seem to be working with some sort of 1-category of stacks. I'm unclear on how exactly it is defined, but regardless that is not the right thing to consider. What you want to consider is the (2,1)-category of stacks, and there $BG$ is the categorical quotient. In general any notion that does not identify two equivalent stacks is unlikely to have geometric meaning.
|
2025-03-21T14:48:30.420282
| 2020-04-27T03:05:31 |
358632
|
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|
Stack Exchange
|
Coherent sheaf with big enough support is non-zero in K-theory
Let $X$ be a noetherian, separated, integral scheme of dimension $d < \infty$ and $\mathcal F \in \mbox{Coh}(X)$ coherent sheaf on $X$ with support $\operatorname{Supp} \mathcal F = X$. Is it true that $0 \neq [\mathcal F] \in {K'_0}(X) := K_0(\mbox{Coh}(X))$? How one can provide a simple counterexample?
The image of $\mathcal F$ in the $K'_0$ of the generic point is non-zero (equal to the rank).
Thank you! But how this consistent with the nontriviality of Serre's positivity conjecture? If I am not wrong, the conjecture asked to prove that for any f.g. $R$-modules $M,N$, over local ring $(R,m)$, with appropriate supports, the number $\chi(M,N) = [M]\cdot[N] = [M \otimes N] \in {K'_0}^{{m}}(\operatorname{Spec} R) \cong \mathbb{Z}$ is positive? Coherent sheaf associated to $M \otimes N$ is naturally non-zero coherent sheaf on point ${m}$, and by your argument, it must be positive in K-theory, so, where is the problem?
The equality $[M].[N] = [M\otimes N]$ is not true unless you assume strong extra conditions.
@HailongDao I see, I misconceived a little how the general theory work. Thank you!
|
2025-03-21T14:48:30.420500
| 2020-04-27T04:31:31 |
358633
|
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|
Stack Exchange
|
Expectation of multiplied random variables given their individual expectations
Suppose that I have two non-negative real valued random variables $x, y \in Z_+$ that always satisfy $$x+y \leq 1.$$ Also suppose that $E[x] = 1/2$ and $E[y] = 1/4$. What is the maximum possible value of $E[xy]$? Can it be larger than $1/8$?
More generally, is there a systematic way of analyzing $E[xy]$ for say other values of $E[x]$ and $E[y]$? (You can assume the assumptions $x, y \in Z_+$ and $x+y\leq 1$ continue to hold.)
If $x$ is uniform $[0,1]$ and $y=x/2$, then $E(xy)=1/6\gt 1/8$ I believe.
@BrendanMcKay But that wouldn't satisfy $x+y \leq 1$ since e.g. for $x=1$ you get $y=0.5$ and then $x+y = 1.5$.
Right, I forgot that condition, sorry.
|
2025-03-21T14:48:30.420863
| 2020-04-27T04:48:43 |
358634
|
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|
Stack Exchange
|
Differentiability of eigenvalues of positive-definite symmetric matrices
Let $A\in M(n,\mathbb{R})$ be an invertible matrix. Consider the (real) eigenvalues $\lambda_1,\cdots,\lambda_n$, in increasing order, of the positive-definite symmetric matrix $A^t A$. We shall denote the eigenvalues as $\lambda_i(A)$.
Question What can be said about the differentiability of the functions $\lambda_i:GL(n,\mathbb{R}) \to \mathbb{R}$?
[We may assume that the domain is $GL^+(n,\mathbb{R})$ for differentiability/smoothness.]
Any reference for this or relevant results would be appreciated.
After writing my answer below, I noticed that similar questions have been asked on MO: link, link.
In the open subset of $M_n(\mathbb{R})$ where the $\lambda_i$ are distinct, they are $C^{\infty}$ functions: this follows from the implicit function theorem.
On the other hand, when some eigenvalue has multiplicity $>1$ you don't get more than continuity. For example if $A=\begin{pmatrix} 0 & 1\\ 1 & t
\end{pmatrix}$ the largest $\lambda_i$ is $\dfrac{1}{2}\left(t^2+2 +|t|\sqrt{t^2+4}\right)$, which is not differentiable (as a function of $t$) at $t=0$.
Note that the question is about the eigenvalues of $A^\dagger A$ not about the eigenvalues of $A$ itself. But basically the same example works for that case too.
I am dicussing the eigenvalues of $A^{\dagger}A$, not of $A$ (I am using the notation of the OP). The eigenvalues of $A$ are distinct for $t=0$.
In the simple $A$ given as example, it appears that if we disregard $\lambda_i$ and use a new numeration $\mu_1,\mu_2$ of the eigenvalues that is wrt. the sign of the square root of the discriminant, not wrt. the relative size of the roots, then you can have that both $\mu_1$ and $\mu_2$ are smooth functions of $t$. Possibly see this Wolfram Alpha illustration. Under what conditions is it possible to pick $\mu_i$ with $i=1\ldots n$ "smart" so they are smooth?
Sorry, I should have read the answers by Denis Serre and Marc Nardmann before writing this comment. They address it.
The keyword is the Cartan decomposition in the theory of symmetric spaces.
In short, when an eigenvalue is simple (its multiplicity is $1$) it is locally an analytic function. But when the eigenspace is degenerate (the multiplicity is greater than $1$), the eigenvalue function is not differentiable. The problem is essentially one of choosing branches: if you try to deform the identity matrix, there is no consistent way to say which of the resulting distinct eigenvalues after deformation is the eigenvalue that you should have kept track of.
Let $K = \mathrm{O}(n)$, and let $A$ be the group of diagonal matrices with positive entries. You then have $G=KAK$ and if $g=k_1 a k_2$ then the eigenvalues of $g^\dagger g$ are exactly the squares of the eigenvalues of $a$. The problem is that the decomposition is not unique: you can conjugate $a$ by a permutation matrix, and there will be problems when $a$ is fixed by a permutation matrix.
If I follow what you are saying, then does it follow that $a$ (in the decomposition of $g$) is unique up to conjugation by permutation matrices? In that case, although each $\lambda_i$ may not be well-defined, any smooth function $f$ of $\lambda_i$'s which is invariant under permutations will be smooth.
Yes. A smooth symmetric function will be smooth. The best way to think about it for analytic functions is as follows: the elementary symmetric polynomials are polyomials in the traces of powers of the matrix (that's the Newton identities), and the Taylor expansion of an analytic symmetric function is a sum of elementary symmetric polyomials.
As mentionned by other answers, simple eigenvalues are $C^\infty$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of linear operator.
The first one is that each $\lambda_j$ is a Lipschitz function. This statement is still valid if you replace ${\bf Sym}_n({\mathbb R})$ by a subspace $E\subset{\bf M}_n({\mathbb R})$ with the property that the eigenvalues are always real.
The second one is that if $t\mapsto A(t)$ is a smooth curve in ${\bf Sym}_n({\mathbb R})$, then there is a labelling of the eigenvalues $t\in{\cal V}\rightarrow(\mu_1(t),\ldots,\mu_n(t))$ such that each $\mu_j$ is smooth. Mind that this labelling does not respect the order between eigenvalues when the multiplicities vary. Mind also that this becomes false if we replace a curve by a surface.
A smooth eigenvector labelling (where "smooth" means $C^\infty$, not just $C^1$) does not exist in the generality you claim (and, as far as I can tell, this is not claimed in Kato's book); see the references in my answer below.
@MarcNardmann. Thank you for the precision. I am stuck at home, without access to our library.
Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).
If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2(t) = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).
(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)
Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:
If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
Let $A$ be $C^\infty$. Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.
Very clear answer (+1). Do all the bullet points still hold for hermitian matrices?
@lcv: Yes. I just cited a few theorems that can be stated succinctly. Many of these results hold in greater generality, for instance for normal matrices or even unbounded normal operators on Hilbert spaces. The reference in Ben McKay's answer contains much more information.
Your last bullet is false. Your condition about $\lambda_i,\lambda_j$ is used to avoid $\exp(-1/t^2)$-type functions. Then there is a numbering of the eigenvalues $(λi)i≤n$ s.t. the associated functions are $C^{\infty}$. However, the natural ordering of the eigenvalues is not necessarily met; consider, for example, $B(t)=diag(t+2,2t+2)$ when $t$ goes through $0$. See my post in https://math.stackexchange.com/questions/3601351/gradient-of-a-mapsto-sigma-i-a/3630189#3630189
I do not understand your objection; I rather suspect that we mean the same thing. The last bullet point is correct. There might be a misunderstanding: Of course I do not claim that the ordered tuple $(\lambda_1,\dots,\lambda_n)$ is $C^\infty$. I claim that one can choose another tuple of eigenvalue functions which is $C^\infty$. I mention the ordered tuple only to state the sufficient condition under which this other choice is possible.
OK I agree with your comment.
For the best known positive results under mild hypotheses, you might want to look at
Armin Rainer, Perturbation theory for normal operators, Trans. A.M.S., Volume 365, Number 10, October 2013, Pages 5545–5577
link
In this paper by Xuwen Zhu it is shown that, after resolution by radial blow-ups, the eigenvalues can be made to be smooth:
https://arxiv.org/abs/1504.07581.
|
2025-03-21T14:48:30.421484
| 2020-04-27T05:15:50 |
358636
|
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|
Stack Exchange
|
Anomaly in QFT physics v.s. determinant line bundle
In a quantum field theory (QFT) lecture, a math-physics professor explains the anomaly in physics, say the non-invariance of the partition function of an anomalous theory under background field transformation, can be regarded as a section of a complex line bundle over the space of background field.
In math terminology, he explained that the so-called anomaly in physics is the determinant line bundle in math.
How precise is the analogy:
anomaly in QFT physics v.s. determinant line bundle
can you provide a few examples between the twos?
The answer by Severin Bunk is pretty good but I think more answers could be useful too, hopefully someone could go further on the parallel between physics and maths terminology or share another perspective or give a illustrative example, etc.
Almost-duplicate question on physics:SE: https://physics.stackexchange.com/q/254984/50583
I’m addition to the complete works of Dan Freed mentioned below, you might like Moore and Nelson’s [The Aetiology of Sigma Model Anomalies] (https://repository.upenn.edu/cgi/viewcontent.cgi?article=1536&=&context=physics_papers&=&sei-redir=1)
To be a little more specific on Dan’s stuff, start with Determinants, Torsion and Strings (https://projecteuclid.org/download/pdf_1/euclid.cmp/1104116145) and On Determinant Line Bundles (https://projecteuclid.org/download/pdf_1/euclid.cmp/1104116145)
@AaronBergman: Thanks for pointing out Moore's paper -- I had not seen it before, but it looks great!
The partition function should assign to each possible field configuration $\Phi$ (or field history) in your quantum field theory a number $Z(\Phi)$. That is, it should be a function on the collection of fields configurations, and from that function you can derive lots of quantities in the field theory.
It can happen, however, that in order to come up with, or write down, the number $Z(\Phi)$, you need to make some auxiliary choices. Often these choices work only for certain fields, rather than for all fields at the same time. For instance, in a gauge theory with fermions you might need to choose a real number $\lambda$ which is not in the spectrum $\sigma(D_A)$ of the Dirac operator coupled to the gauge field $A$. In general, such a choice cannot be made for all gauge potentials $A$ simultaneously, and hence exists "only locally" on the collection of fields. Different local choices of auxiliary information will lead to different values of what you compute as $Z(\Phi)$, and it usually turns out that the transformation law between the values of $Z(\Phi)$ for different auxiliary choices is that of a section of a line bundle on the collection of fields.
What people normally want in gauge theory is for the partition function to be well-defined on the space $\mathcal{A}/\mathcal{G}$ of gauge potentials modulo gauge transformations. While any line bundle on $\mathcal{A}$ is trivialisable (since $\mathcal{A}$ is an affine space), this is not true on the quotient $\mathcal{A}/\mathcal{G}$. Let us say that our partition function can be understood as a section $Z$ of a line bundle $L \to \mathcal{A}/\mathcal{G}$. Then, any trivialisation of $L$ allows us to translate $Z$ into a function on $\mathcal{A}/\mathcal{G}$, and hence into an actual partition function. The QFT anomaly can hence be described as the obstruction to the existence of a trivialisation of $L$ -- this is a class in $H^2(\mathcal{A}/\mathcal{G};\mathbb{Z})$. Often this class can be computed, like in the case of the Dirac anomaly.
Some nice mathematical references, in my opinion, are https://arxiv.org/abs/hep-th/9907189, https://arxiv.org/abs/math-ph/0603031v1, and for a more conceptual perspective, https://arxiv.org/pdf/1212.1692.pdf.
Journal versions for the first and third preprint are also available for free: "Anomalies in string theory with D-branes" by Freed and Witten (Asian J. Math. 1999) and "Relative Quantum Field Theory" by Freed and Teleman (CMP 2014)
Thanks voted up, so in one sentence, how is the anomaly in QFT in physics = determinant line bundle? Or is that a class in $H^2(\mathcal{A}/\mathcal{G};\mathbb{Z})$?
also do you know is this to say bordism group in Dai Freed theorem?
Yes, you can interpret the QFT anomaly, or at least the chiral anomaly, as the Chern class in $H^2(\mathcal{A}/\mathcal{G};\mathbb{Z})$. There are more anomalies which have different descriptions, for instance they do not have to live on a space $\mathcal{A}/\mathcal{G}$ of gauge equivalence classes.
|
2025-03-21T14:48:30.421806
| 2020-04-27T07:02:28 |
358647
|
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|
Stack Exchange
|
Can every quadrangle whose corners are in little squares be more square-like when we allow corners to move in the little squares
Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$.
Where $AB$ is the distance betwin $A$ and $B$.
Let $A$, $B$, $C$, $D$ be four points of the plane $P$, respectively in $S_A$, $S_B$, $S_C$, $S_D$ that are areas delimited by squares. We also suppose that no point is the corner of its square—but it can be on the egde.
Suppose that $f((A,B,C,D))>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A$, $B'\in S_B$, $C'\in S_C$, $D'\in S_D$ such that $f((A',B',C',D'))<f((A,B,C,D))$?
I found the quantification a little ambiguous as stated it, so I changed it, I think while preserving your meaning. Please feel free to undo it if I failed (although I encourage you to keep the spell-checking). Also, do $S_A$, $S_B$, $S_C$, $S_D$ have to be disjoint (or at least different)?
Thank you very much LSpice, it is much better this way, with the definition of $f$ at the beginning. Maybe there is still ambiguities, indeed, i ask if it is true for any $S_A,S_B,...$ (typically as small as we want) .
What does $AB$ mean, when $A$ and $B$ are points in a plane?
This is a stadard notation in France for the distance betwin $A$ abd $B$ , but must admit i never met it in any graduate text, I'm going to edit
I'm curious about the close votes. This looks to me like a question about elementary subject matter, but it's not clear that it's an elementary question—at least, I don't see how to solve it. If it's trivial, then I think it might be a good idea for a close-voter to say so.
|
2025-03-21T14:48:30.421963
| 2020-04-27T07:16:22 |
358649
|
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|
Stack Exchange
|
Graph metric approximating Euclidean metric
I've been reading Wolfram's recent articles about graph/mesh/grid structures as an analogy for physical space, and it seems to me that there will be a problem getting the notion of distance to work out, since the natural metric on eg a square grid or triangle grid is L1, even in the limit of a very dense grid becoming a patch of the plane.
Is there any simple rule for generating a mesh-like graph that, with the length of all edges considered to the be the same, in the limit of very fine scale, closely approximates Euclidean distance in the plane?
Looking around, I was able to find this paper (https://projecteuclid.org/download/pdf_1/euclid.cmp/1104286245) which shows that if you allow the length of each edge in the graph to match its actual length in the plane, there is a solution with a simple production rule, but is it possible with all graph edges considered to be the same length?
I really have no expertise to answer this, so I just state my intuition. I would say an appropriate infinite random graph model in which neighborhoods grow polynomially (say, with power $n-1$) feels like a good way to model $n$-dimensional Euclidean space, since there is no preference of any direction over another.
I think actually just solved this on square grids using a kind of family of supergraphs! I've been trying to find out if it's already been done and landed here in my search. I hope to publish the results this December 2020. PLEASE Email me, tterlep at purdue d0t edu, I'd really appreciate some feedback as this result came as somewhat of a surprise from working on simple Go Board problems!
@BenCrowell Why do you say that?
Some useful references are: Pach, J., Pollack, R. and Spencer, J., 1990. Graph distance and Euclidean distance on the grid. In Topics in combinatorics and graph theory (pp. 555-559); Burago, D. and Ivanov, S., 2015. Uniform approximation of metrics by graphs. Proc. Amer. Math. Soc. 143 (2015), 1241-1256; Benjamini, I., 2013. Euclidean vs. graph metric. In Erdős Centennial (pp. 35-57).
Pick some small $\epsilon>0$, take the dilated integer lattice $\epsilon^2\cdot\mathbb{Z}^2$ to be our vertex set, and draw an edge between two vertices if their Euclidean distance is between $\epsilon-2\epsilon^2$ and $\epsilon+2\epsilon^2$. Let $d(u,v)$ denote the distance between $u$ and $v$ in this graph. Given the scaling, we might expect $\|u-v\|_2\approx \epsilon d(u,v)$.
We claim that every $u,v\in \epsilon^2\cdot\mathbb{Z}^2$ satisfies
$$\epsilon d(u,v)-2\epsilon \leq \|u-v\|_2\leq (1+2\epsilon)\cdot \epsilon d(u,v).$$
The right-hand inequality follows from the triangle inequality. For the left-hand inequality, let $k$ denote the smallest number of steps of length $\epsilon$ it takes to traverse from $u$ to $v$ in $\mathbb{R}^2$. For example, if $\|u-v\|_2\leq 2\epsilon$, then $k=2$. In general, $k\leq\|u-v\|_2/\epsilon+2$. Now take $u_0,\ldots,u_k\in\mathbb{R}^2$ such that $u_0=u$, $u_k=v$ and $\|u_{i+1}-u_i\|_2=\epsilon$ for each $i$. Then rounding each $u_i$ to the nearest point in $\epsilon^2\cdot\mathbb{Z}^2$ gives a path in our graph of length $k$. The claim follows.
@BenCrowell: "in the limit of very fine scale" seems precise enough, at least to discourage discussion of distances less than $\epsilon$. And yet, I'm not sure the taxicab metric fares well apples-to-apples. As in this answer, suppose you connect two points in the plane if they are taxicab distance $\epsilon$ apart. If you only consider the same grid of points $\epsilon^2 \mathbb{Z}^2$ as in this answer, then your "chessboard-neighbor points" aren't even connected! (The graph is bipartite.) Allowing all points of the plane as potential intermediate stops, you also get "2000% error".
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2025-03-21T14:48:30.422227
| 2020-04-27T08:14:09 |
358651
|
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|
Stack Exchange
|
MOOC's courses on Functional analysis
Several years ago, Coursera offered a MOOC (Massive Online Open Course) on Functional Analysis:
An Introduction to Functional Analysis. Taught by John Cagnol.
Ecole Centrale Paris.
The duration was 8 weeks:
1 Topology; continuity and convergence of a sequence in a topological space.
2 Metric and normed spaces; completeness.
3 Banach spaces; linear continuous functions; weak topology.
4 Hilbert spaces; The Riesz representation theorem.
5 The Lax-Milgram Lemma.
6 Properties of the Lp spaces.
7 Distributions and Sobolev Spaces.
8 Application: simulating a membrane.
Now it is possible to find the videos in the following youtube link:
https://www.youtube.com/playlist?list=PLSpInro6Ys2IHve6oN9h005zmwfLnOSp1
However, the possibilities of doing short exams, asking questions, etc. are no longer available.
Is there any comparable MOOC on Functional analysis currently active?
MIT provides homework and solutions https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/pages/assignments/
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2025-03-21T14:48:30.422323
| 2020-04-27T08:14:50 |
358652
|
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|
Stack Exchange
|
A map similar to a foot map on a cap of a strictly convex surface
Consider a strictly convex surface $\Sigma$ in $\mathbb{R}^3$ homeomorphic to a sphere. When $p$ is a point not in a convex hull of $\Sigma$, then $\Sigma'$ is the boundary of convex hull of $p$ and $\Sigma $. Define $$C=\overline{ \Sigma' -\Sigma },\ D= \overline{\Sigma-\Sigma'}$$
When $|\ |_X$ is an intrinsic metric on the set, then let $x\in
\partial C$. When $\gamma :[0,l]\rightarrow
D$ is a unit speed shortest path in $D$ s.t.
$$\gamma (0)=x,\ \gamma'(0)=\frac{ p-x}{|p-x|}$$ where $|\ |$ is a
Euclidean metric in $\mathbb{R}^3$, then we let $l\in [0,|p-x|]$ to
be the largest.
For any $y\in \partial C$, then $$|\gamma (l)-y|_{D}\leq |p-y|$$ This is
true ?
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2025-03-21T14:48:30.422412
| 2020-04-27T09:23:22 |
358653
|
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|
Stack Exchange
|
How much time does a function spend above or below its average value around a point?
Given a locally integrable function $f: \mathbb R \to \mathbb R$, define $
K: \mathbb R \times \mathbb R+ \to \mathbb R$ by
$$
K(x, r) :=
\begin{cases}
1, & \text{if }f(x) > \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\
-1, & \text{if }f(x) < \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\
0,& \text{if }f(x) = \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x
\end{cases}
$$
Define $U: \mathbb R \to [-1, 1]$ and $L: \mathbb R \to [-1, 1]$ by
$$
\begin{split}
U(x) &:= \limsup_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t\\
L(x) &:= \liminf_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t
\end{split}
$$
Intuitively, $U$ and $L$ represent the weighted proportion of time a function spends above or below its average value in an infinitesimal neighbourhood of a point.
i) Is it true that for any locally integrable function, $U = L$ a.e?
ii) Is it true that $U = 1, 0, or -1$ a.e.?
For i) the Brownian motion gives a counter example.
We choose $f(t)=W_t$ with $W_t$ a Brownian on $\mathbb{R}$. Because one can calculte $U(0)$ and $L(0)$ from the Brownian motion restricted to any neighbourhood of $0$, we can apply Blumenthal’s 0-1 law: There exists $c_1,c_2\in \mathbb{R}$ such that $\mathbb{P}(U(0)=c_1)=1$ and $\mathbb{P}(L(0)=c_2)=1$. By symetry we have $c_1 =-c_2$. Moreover for any $0<a<1$ there exists $\epsilon$ such that for all $r>0$
$$\mathbb{P}\Big(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \geq a\Big)>\epsilon$$ Therefore $c_1\geq a$ and we conclude that $c_1=1$ and $c_2=-1$.
Finally by translation invariance of the Brownian motion we have for all $t\in \mathbb{R}$ $$\mathbb{P}(U(t)=1)=\mathbb{P}(L(t)=-1)=1$$
How do you show that for each $a\in(0,1)$ there is some $\epsilon[>0?]$ such that for all $r>0$
$P(\frac{1}{2r}\int_{B_r(0)}K(0,t)dt \ge a)>\epsilon$? Also, even if your have $P(U(t)=1)=P(L(t)=-1)=1$ for all $t$, how do you deduce from this that $P(U(t)=1\ \forall t)=1$?
@losif Pinelis: The Brownian motion is scale invariant ($f^{(r)}(t):=\frac{1}{\sqrt{r}}W_{rt}$ is also a brownian motion). So $K(0,t)$ have the same law as $K(0,r^{-1}t)$. So $\mathbb{P}(\frac{1}{2r}\int K(0,t)dt\geq a)$ is independant of $r$.
@losif Pinelis : The second point you mention is not true for all t but almost all $t$. Calling $\Omega$ the probability space of the brownian motion then $\int_{-L}^L\int_{\Omega} 1_{U(t)=1}dt d\mu=2L=\int_{\Omega} \int_{-L}^L1_{U(t)=1}dt d\mu$. So with probability 1, for almost all $t\in [-L,L]$ $U(t)=1$
All right, thank you.
By Hartogs‘ theorem it suffices to show that it is separately analytic in $s$ and $t$, presumably a tad easier.
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2025-03-21T14:48:30.422595
| 2020-04-27T09:52:42 |
358655
|
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|
Stack Exchange
|
Topological Hochschild homology using equivariant orthogonal spectra
In the Hesselholt-Madsen paper "On the K-theory of finite algebras over Witt vectors of perfect fields", the authors develop some results concerning the Topological Hochschild homology (THH) of functors with smash products.
They use the language of equivariant spectra in the sense of Lewis-May-Steinberger.
My question is if there is a more modern reformulation of this work using equivariant orthogonal spectra (or other (one-categorical) variants).
I am quite certain that using the results from Mandell, May, Schwede and Shipley, one can translate the results from Hesselholt-Madsen to orthogonal spectra. My Question then is if this has already been done. At this moment, I am not looking for an infinity categorical treatment.
To be more precise, I would like to see the construction of THH and topological cyclic homology using orthogonal spectra.
Preferably I would like to see the construction of the Tate spectral sequence as well, where for a finite cyclic group $C$ and $T$ an equivariant $C$ spectrum, the Tate spectral sequence is
$$E^2_{r,s}=\hat{H}^{-r}(C,\pi_sT)\Rightarrow \pi_{r+s}\hat{\mathbb{H}}(C,T),$$
where the $E^2$ page is the ordinary Tate cohomology of $C$ and $E^{\infty}$-page is the homotopy groups of the Tate spectrum.
This has a partial positive answer.
The theory of cyclotomic spectra using orthogonal spectra is developed in 'The homotopy theory of cyclotomic spectra' by Blumberg and Mandell. See https://arxiv.org/abs/1303.1694.
Moreover, Nikolaus and Scholze have an exposition of the Böckstedt construction of THH using orthogonal spectra in their paper 'On topological cyclic homology'.
See https://arxiv.org/abs/1707.01799.
|
2025-03-21T14:48:30.422847
| 2020-04-27T10:09:36 |
358656
|
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|
Stack Exchange
|
Freeness of completed homology over universal deformation ring
In Theorem 7.4 of the paper "patching and the $p$-adic Langlands program for $\mathrm{GL}2(\mathbf{Q}_p)$" (arXiv link), it is proved that in the minimally ramified case, the completed homology of given level is the completed tensor of the dual family of Banach representation (denoted as $\tilde{P}$) with the Galois representation. It seems the result should be valid as long as the local deformation rings outside $p$ are smooth. (Is it right?) It is remarked that the method does not apply beyond the locally formally smooth case.
My question is: in which generality can one get such result, by putting additional restrictions on the deformation problems outside $p$? For example can one (under reasonable assumptions) look at one irreducible component of the universal deformation space and prove such free rank one result of the corresponding quotient space of completed homology with appropriate level?
|
2025-03-21T14:48:30.422935
| 2020-04-27T10:14:31 |
358657
|
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"DSM",
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|
Stack Exchange
|
Hypothesis: An injection from polygons into $SO(2) \times S_n$
I have stumbled upon a possible representation of polygons by a concise description of their behaviour under rotation. I would like to know more about it and, obviously, if it is actually a bijection. I will be somewhat informal to epxlain things more intuitively.
Consider the following: We call $\mathcal{P}_n$ the set of polygons in the plane with n points that do not self-intersect.
Furthermore, we want the vetrices of the polygons to have pairwise different x-components.
This means no two vertices can lie above eachother.
We can represent a $P \in \mathcal{P}$ as a tuple:
\begin{equation}
P = ((x_1,y_1),...,(x_n,y_n)),
\end{equation}
where connecting the points in order yields the boundary of the polygon.
Now we define $\sigma(P) \in S_n$ as the permutation that is the "ascending ordering of $(x_1, ..., x_n)$". (the permutation that returns the ordered list).
If we let $\tau \in SO(2)$ be a rotation, then $\tau(P) \in \mathcal{P}$ will always be the case except for a finite number of rotations $\tau_1, ..., \tau_k$. This is the case because the polygon has finitely many points, so two poimts vertically align only for finitely many rotations.
For one of these $1 \leq j \leq k$, we can't define $\sigma(\tau_j(P))$ because the rotated polygon 's vertices' x-values are not pairwise different.
However, for a small $\epsilon > 0$ we know that
\begin{equation}
(\tau_j + \epsilon)(P) \in \mathcal{P} \text{ and } (\tau_j - \epsilon)(P) \in \mathcal{P}.
\end{equation}
This means that we can define a $\sigma_j \in S_n$ as the permutation
\begin{equation}
\sigma_j := \sigma((\tau_j + \epsilon)(P)) \; \sigma((\tau_j - \epsilon)(P))^{-1}.
\end{equation}
Basically, $\sigma_j$ is the permutation that describes how $\sigma(\tau(P))$ changes at $\tau_j$.
Now our representation of $P$ is:
\begin{equation}
\mathcal{R}(P) = \{(\tau_1, \sigma_1), ..., (\tau_k, \sigma_k)\} \subset SO(2) \times S_n.
\end{equation}
Now my question is: How much of $P$ can we recover from $\mathcal{R}(P)$? Is the representation unique up to congruence? In other words: Is
\begin{equation}
\mathcal{R}: \mathcal{P} / \sim \quad \longrightarrow SO(2) \times S_n
\end{equation}
an injection for some $\sim$?
I think this might be true in some way because if you know that $(\tau, \sigma) \in \mathcal{R}(P)$, then you know that after rotating by exactly $\tau$ you can study $\sigma$ and know that all the points whose position in the permutation changed must have been on a straight line.
This might be sufficient to recover enough "straight-line-relations" between the points to construct the polygon (up to translation, rotation and scale, of course).
The use would be the following: When we have a transformation of polygons $f: \mathcal{P} \longrightarrow \mathcal{P}$, it might be the case that we can do this:
\begin{equation}
\mathcal{f}: \mathcal{R}(\mathcal{P}) \overset{\text{recover}}{\longrightarrow} \mathcal{P} \overset{f}{\longrightarrow} \mathcal{P} \overset{\mathcal{R}}{\longrightarrow} \mathcal{R}(\mathcal{P}).
\end{equation}
This can then be studied algebraically.
If, instead of polygons, you just consider a discrete set of points, then it sounds close to Goodman & Pollack's circular sequence of permutations. A source is the Handbook of Discrete & Computational Geometry, Chapter 5: PDF Chap.5.
Thanks a lot! It seems like Theorem 5.2.2 is surprisingly close to being an answer to my question.
Yes. The language is quite different, but the underlying concepts appear to be analogous.
For simplicity, restrict to polys with no parallel sides. If I am correct, the $\sigma_k$'s (or transpositions) give the order in which line segments at angle $\theta_k$'s (with x-axis) join together to form a polygon. Let the lengths of line segments be $r_k$'s. WLOG, the first line segment starts at origin along the x-axis. This leads to the two linear equations (over $r_k\geq 0,~\forall k$) - net displacement along x and y axis must be zero. This under-determined system of linear equations will likely have multiple solutions, which might not all be congruent to each other. Hope this helps.
The title is misleading because you produce from each polygon a subset of $SO(2)\times S_n$.
|
2025-03-21T14:48:30.423229
| 2020-04-27T10:32:18 |
358659
|
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"Adam P. Goucher",
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|
Stack Exchange
|
Ramsey theory in infinite-dimensional projective spaces
Let $\mathbb{F}_q$ be a finite field and $k$ be a positive integer. If we colour each point of the infinite-dimensional projective space $\mathbb{F}_q \mathbb{P}^{\infty}$ with one of $k$ colours, can we necessarily find an infinite-dimensional monochromatic projective subspace?
There are a couple of observations that hint that the answer is 'yes':
Observation 1: The statement for $\mathbb{F}_2$ is true, and is equivalent to Hindman's theorem. The reason for asking this question is that it would provide an interesting generalisation of Hindman's theorem.
Observation 2: If we weaken the problem to finding an $n$-dimensional monochromatic projective subspace (where $n$ is finite), the statement is also true (and follows, for example, from the 'Vector Space Ramsey Theorem' in the paper Ramsey's Theorem for Spaces by Joel H. Spencer).
"infinite-dimensional" means (infinite countable)-dimensional? it was proved some time ago that there are several infinite cardinals :)
Yes: for concreteness, define $\mathbb{F}_q \mathbb{P}^{\infty}$ to be the union of the sequence $\mathbb{F}_q \mathbb{P}^1 \subseteq \mathbb{F}_q \mathbb{P}^2 \subseteq \mathbb{F}_q \mathbb{P}^3 \subseteq \cdots$.
After further investigation, it appears that disappointingly the answer is 'no', and a proof appears in Lemma 2.4 of Partition Theorems for Subspaces of Vector Spaces (Cates and Hindman, 1975).
|
2025-03-21T14:48:30.423354
| 2020-04-27T10:40:42 |
358660
|
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"JP McCarthy",
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"https://mathoverflow.net/users/35482"
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"url": "https://mathoverflow.net/questions/358660"
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|
Stack Exchange
|
Showing a product on a character space is continuous
Quoting from Timmermann's An invitation to quantum groups and duality:
Prop. 5.1.3 Let $A$ be a commutative algebra of functions on a compact
quantum group. Then there exists a compact group $G$ and an
isomorphism $A\cong C(G)$ of $\mathrm{C}^*$-algebras.
Proof: By the Gelfand theorem, there exists a compact space $G$, a continuous
map $m:G\times G\rightarrow G$...
I might reduce this to:
Let $A$ be a commutative unital $\mathrm{C}^*$-algebra with a $\mathrm{C}^*$-morphism $\Delta:A\rightarrow A\otimes A$. By the Gelfand theorem, where $\Phi(A)$ is the Hausdorff weak*-compact character space, the map $m:\Phi(A)\times \Phi(A)\rightarrow \Phi(A)$, $(\chi,\varphi)\mapsto (\chi\otimes\varphi)\Delta$ is continuous...
My difficulties are more appropriate to MSE, although the content knowledge might be more appropriate to here.
My natural inclination is to take sequences $(\chi_n)\rightarrow \chi \in \Phi(A)$ and $(\varphi_n)\rightarrow\varphi\in\Phi(A)$ and to look at $(\chi_n,\varphi_n)$, and show:
$$m(\chi_n,\varphi_n)\rightarrow m(\chi,\varphi),$$
and using the dense subalgebra generated by the matrix elements of unitary irreducible (co)representations $(\rho_{ij}^\alpha)_{\alpha\in\text{Irr}(A)}$ shows:
$$(\chi_n\otimes\varphi_n)\Delta(\rho_{ij}^\alpha)\rightarrow (\chi\otimes\varphi)\Delta(\rho_{ij}^\alpha),$$
but I don't think that $\Phi(A)$ is in general a sequential space. I am happy that it is when $A$ is separable, but I understand that the opening proposition is believed to be true in the non-separable case also.
Woronowicz originally assumed that the algebra of functions on a compact quantum group was separable in order to deduce the existence of a Haar state. Van Daele removed this assumption.
How can we show that the map $\Phi(A)\times \Phi(A)\rightarrow \Phi(A)$ is continuous when $A$ is unital, commutative, non-separable?
Yeah, this isn't research level. I think there are people on MSE who would have no trouble answering this. The problem has nothing to do with groups: it's the general fact that any $*$-homomorphism from $C(X)$ to $C(Y)$ (where for you $X = G$ and $Y = G\times G$) arises as composition with a continuous function from $Y$ to $X$. Separability isn't needed. Just use nets instead of sequences.
@NikWeaver that answers the question perfectly, thank you. Should I flag the question for closure, or should I delete the question, or should you answer and I accept? Or should I answer community wiki? There is no need at this point to ask the question at MSE now that it is answered.
I'd say just leave the question here as is. If it bothers people enough to want to close, they can vote to do so. Just probably try MSE first next time.
As per Nik Weaver's comment, this is a simple consequence of the fact that for compact Hausdorff spaces $X$, $Y$, for every unital *-homomorphism $\pi:C(X)\rightarrow C(Y)$, there exists a continuous function $\phi:Y\rightarrow X$ such that, for $f\in C(X)$:
$$\pi(f)=f\circ \phi.$$
A reference for this fact is given by Nik's text, Mathematical Quantisation.
|
2025-03-21T14:48:30.423572
| 2020-04-27T11:20:02 |
358663
|
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"ABIM",
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|
Stack Exchange
|
Use of this space of very rapidly decreasing continuous functions
Let $C_n$ denote the subspace of continuous function on $[0,\infty)$ supported on $[n,n+1]$. Denote the $\ell^p$-direct sum Banach space
$$
V_p
:=
\left\{
f \in C([0,\infty)):\,
\sum_{n=1}^{\infty} \left(
\sup_{x \in [n,n+1] } |f(x)|
\right)^p <\infty
\right\}.
$$
In the case where $p=\infty$ we regain the space of function vanishing at infinity but what about $p<\infty$? Are these spaces studied? If so what are their use?
The notation in your first line is extremely confusing. $C([n,n+1])$ should mean, and can only mean, the continuous functions on $[n,n+1]$. You then seem to switch to functions defined on $[-n,n]$
I think it would be much clearer to avoid the notation on the LHS and just define your Banach space $V_p$ to be the space on the RHS. Then $V_p$ embeds isometrically inside the $\ell_p$-sum of copies of $C[0,1]$, as a complemented subspace
What's the other summand then and is this object useful?
If I understand the definitions correctly, $V_p$ is the kernel of the map $Q:\left(\bigoplus_{n=0}^\infty\right){\ell_p} C[0,1] \to \ell_p$ defined by $Q((f_n){n=0}^\infty) = (f_{n}(0)-f_{n-1}(1))_{n=1}^\infty$. In other words, you're just taking an equalizer of a bunch of a functionals, expressing the idea that a continuous bounded function on $[0,\infty)$ is obtained by gluing together a sequence of functions on $[0,1]$ which "match up" at the endpoints
It is clear that $Q$ has an isometric right inverse, and so the kernel of $Q$ is complemented. As to whether this object is useful: there are, in some sense, uncountably many Banach spaces that one could define, so the question is whether you have something you want to do with this space, either in terms of theorems or in using the space as a natural home for particular functions arising in e.g. probability or ODE
Thanks Yemon. This is basically what I needed from an answer. If you'd like to post exactly this it I'd accept it gladly.
|
2025-03-21T14:48:30.423736
| 2020-04-27T11:35:37 |
358664
|
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"Brendan McKay",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358664"
}
|
Stack Exchange
|
4-cycles vs eigenvalue information on quasi-random graphs
My (philosophical) question arises from reading the wonderful paper of Chung-Graham-Wilson where the authors introduces the notion of quasi-random graphs.
The main purpose of the paper is to show that many properties of a graph are in fact equivalent. I focus in this question in two properties, namely (I will use the notation of the paper):
P3 property gives information concering the second largest eigenvalue of the adjacency matrix of the graph G
where $N_G(C_t)$ is the number of copies of the cycle on t vertices. The paper shows that, among several other properties, G satisfies $P_2(4)$ (See that t=4 here) if and only if G satisfies $P_3$.
In order to prove that $P_2(4)$ implies $P_3$ one can use the following easy property: the number of 4-cycles in G can be obtained by estimating
$$\sum_{i=1}^{n}\lambda_i^4,$$
which is, up to lower order terms, the number of 4-cycles. Hence, having upper bounds from this sum would give upper bounds on the second largest element in the sum.
In the paper the proof that $P_3$ implies $P_2(4)$ is not direct, and uses at least 5 intermediate implications. So my question is the following:
Is there a direct way to bound the number of 4-cycles directly and just using spectral information on the graph (namely, about the second largest eigenvalue, or even its multiplicity)?
It seems to me that something could be said here, even more because we have a direct and easy relation between the number of 4-cycles and the spectrum of the matrix.
$\sum_i \lambda_i^4$ is not the number of 4-cycles. It is the number of closed walks of length 4. For example, it is positive for a tree, which has no 4-cycles.
completely true: this sum is equal to 4-cycles+o(n^4), and hence, having information on a bound for this gives information on the eigenvalues...the converse I do not know how to do it directly (if possible)
|
2025-03-21T14:48:30.423882
| 2020-04-27T12:53:21 |
358665
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358665"
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|
Stack Exchange
|
Categorification of "Every domain embeds into a field"?
In the category of commutative rings, every domain embeds into a field. Is this true in the category of presentably symmetric monoidal stable $\infty$-categories? Here's what I mean by that.
Let $PrSt^{L,\otimes}$ be the $\infty$-category of symmetric monoidal closed presentable stable $\infty$-categories and left adjoint strong symmetric monoidal functors.
Say that $F: \mathcal C \to \mathcal D \in PrSt^{L,\otimes}$ is an embedding if $F$ is conservative.
Say that $F: \mathcal C \to \mathcal D \in PrSt^{L,\otimes}$ is a localization if its right adjoint is fully faithful. A localization is trivial if it is the identity or if it is the map to the terminal object $0 \in PrSt^{L,\otimes}$.
Note that (just as for commutative rings) we have an orthogonal factorization system on $PrSt^{L,\otimes}$ given by the localizations and the embeddings.
Say that $0 \neq \mathcal C \in PrSt^{L,\otimes}$ is a field if there are no nontrivial localizations of $\mathcal C$.
Say that $0 \neq \mathcal C \in PrSt^{L,\otimes}$ is a domain if $X \otimes Y = 0 \Rightarrow X = 0 \text{ or } Y = 0$ for $X,Y \in \mathcal C$.
So now the question makes sense:
Question: In $PrSt^{L,\otimes}$, does every domain embed into a field?
Notes:
It's not hard to show that the converse holds -- if $\mathcal C \to \mathcal D$ is an embedding and $\mathcal D$ is a field, then $\mathcal C$ is a domain.
One can characterize the fields $\mathcal C \in PrSt^{L,\otimes}$ by the following property: for every nonzero $X \in \mathcal C$, there exist objects $Y_i$ and a colimit diagram $I = \varinjlim_i (Y_i \otimes X)$ where $I$ is the monoidal unit of $\mathcal C$.
There doesn't seem to be a "field of fractions" construction which automatically turns a domain into a field.
(At least under Vopenka's Principle) Every nontrivial localization $\mathcal C \to L\mathcal C \in PrSt^{L,\otimes}$ admits a further localization $LC \to \mathcal D$ which is maximal, i.e. such that $\mathcal D$ is a field. This is analogous to every ideal in a commutative ring being contained in a maximal one.
It's possible (under large cardinal hypotheses) that some sort of ultraproduct of all maximal localizations of a domain $\mathcal C$ might be a field into which $\mathcal C$ embeds diagonally, but I'm not sure.
|
2025-03-21T14:48:30.424036
| 2020-04-27T12:55:05 |
358666
|
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"url": "https://mathoverflow.net/questions/358666"
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|
Stack Exchange
|
Comparison of inductive limit topology with very-rapidly decaying non-convex $L^{!/2}$-space topology
This is related to these posts and here.
Let $L^1([n,n+1])$ denote the subspace of $L^p$-functions on $[0,\infty)$ essentially supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space
$$
\bigoplus_{n \in \mathbb{N}}^{\ell^1} L^{1/2}([0,n])
:=
\left\{
f \in L^{1}([0,\infty)):\,
\sum_{n=1}^{\infty} \left(
2^n\int_{x \in [n,n+1] } \sqrt{|f(x)|} dx
\right) <\infty
\right\}.
$$
Alternatively, equip $L^1_{\mathrm{comp}}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system
$$
\left\{f \in L^1_{\mathrm{loc}}: \,
\operatorname{esssupp}(f)\subseteq [0,n]
\right\}
\to
\left\{f \in L^1_{\mathrm{loc}}: \,
\operatorname{esssupp}(f)\subseteq [0,m]
\right\}\qquad n\leq m;\, n,m \in \mathbb{N}
,
$$
in the category of topological vector spaces. Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{\mathrm{comp}}$ on their intersection?
It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{\mathrm{comp}}$.
Just as a general point, your space is just equal to $L^{1/2}([0,\infty))$ when you change measure to $\nu$, defined by the Radon-Nikodym derivative $\frac{d\nu}{dm}(x) = \sum_{n=1}^{\infty} \frac1{2^n} I_{[n,n+1]}(x)$ (here $m$ is the Lebesgue measure).
|
2025-03-21T14:48:30.424138
| 2020-04-27T13:26:31 |
358669
|
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"authors": [
"LSpice",
"Leyli Jafari",
"Nick Gill",
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|
Stack Exchange
|
Is $A_5$ the only finite simple group with only 4 distinct sizes of orbits under the action of the automorphism group?
Given a finite group $G$, let $\eta(G)$ denote the number of distinct
sizes of orbits on $G$ under the action of ${\rm Aut}(G)$.
It happens that there are infinitely many non-abelian finite simple groups
$G$ such that $\eta(G) = 5$. For example, this holds for the groups
${\rm PSL}(2,p)$ for prime numbers $p \geq 7$.
In contrast, is it true that $G = {\rm A}_5$ is the only non-abelian finite
simple group such that $\eta(G) = 4$? --
GAP computations suggest this may be true.
Remark: I know that among the alternating groups, the sporadic groups and
the simple groups of Lie type except for those of type $^2{\rm A}$, ${\rm A}$, ${\rm B}$,
${\rm C}$, ${\rm D}$, $^2{\rm D}$ and $^3{\rm D}_4$, there are no further
groups $G$ with $\eta(G) \leq 4$.
Update (April 30, 2020): Meanwhile this question has 3 deleted answers.
I guess for a group of Lie type of rank $r$ over $\mathbb{F}_q$, if you take $r$ large enough and $q$ large enough, then $\mu(G)>4$ -- there will be regular semisimple elements with centralizers that are different maximal tori... And a bit of argument with Zsigmondy primes will ensure that the outer automorphisms don't yield any "accidental" coincidences in orbit size. Doing this more carefully, and lobbing some unipotent elements into the mix, might get you down to something like $r\leq 6$ and $q\leq 11$, say. Just a guess...
By the way your remark suggests you know this to be true for the unitary groups. Is this correct, or have you omitted them in error?
@ Nick Gill Indeed what one can see from some properties of such groups, is that if we take r large enough and q large enough, then μ(G)>4.... As to the Unitary groups, no I haven't dealt with these groups... Thank you for reminding me..I edited the question.
I would have thought you have a reasonable chance of pushing this approach to get a full proof (that the answer is YES). For instance, I looked at Kleidman's paper on ${^3!D_4}(q)$ where, amongst other things, he lists the maximal tori. There are seven classes up to isomorphism, so there should be no problem in excluding this case. There is also a handy paper by Buturlakin and Grechkoseeva that describes the cyclic structure of maximal tori for the classical groups. As soon as you are past rank 3, you have lots and lots of these. You'd need to be careful with small $q$ but still...
What is added by commenting on deleted answers?
Let me see if I can write down a proof that the answer is YES for most of the $A_n$-groups. The method should work for the rest of the $A_n$-groups and, indeed, all of the other groups you mention. Specifically, I'll prove
Proposition: If $G={\rm PSL}_n(q)$ with $n\geq 5$ and $q\geq 2$, then $\mu(G)>4$.
Proof: In what follows, I'll write $q=p^f$ where $p$ is a prime, $f$ a positive integer. Then $|{\rm Aut}(G)|/ |{\rm PGL}_n(q)|=2f$. Recall that a primitive prime divisor of $p^{df}-1$ is a prime that divides $p^{df}-1$ but not $p^k-1$ for any integer $k<df$. Zsigmondy's theorem asserts that such a prime always exists unless $df=2$ or $(p,df)=(2,6)$.
An easy argument shows that if $r$ is a primitive prime divisor of $p^{df}-1$, then $r>df$. It's written down as Lemma 2.7 of a paper of mine with Azad and Britnell. This is important because it means that primitive prime divisors of $p^{df}-1$ do not divide $2f$ provided $df>1$.
So now our job is to find five elements, $g_1,\dots, g_5\in G$, with different orbit sizes under ${\rm Aut}(G)$. Write $o(g_i)$ for the the orbit size of $g_i$. In what follows we note down what makes each $o(g_i)$ definitely different to the others.
Let $g_1=1$. Then $o(g_1)=1$.
Let $g_2$ be central in a Sylow $p$-subgroup of $G$. Then $o(g_i)$ is not divisible by $p$.
Let $g_3$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $\frac{q^n-1}{q-1}$. Then $o(g_3)$ is not divisible by a ppd of $q^n-1$.
Let $g_4$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $q^{n-1}-1$. Then $o(g_4)$ is not divisible by a ppd of $q^{n-1}-1$.
Let $g_5$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ isomorphic to $(q^{n-2}-1)\times(q-1)$. Then $o(g_5)$ is not divisible by a ppd of $q^{n-2}-1$.
Some remarks:
the existence of tori of this size can be seen directly, but is written down explicitly in the paper of Buturlakin and Grechkoseeva.
The fact that a ppd of $q^{n-2}-1$ does not divide $q^n-1$ follows from the fact that $n\geq 5$ and the fact that $gcd(q^{n-2}-1, q^n-1)$ divides $q^2-1$. Similarly for the other pairs of ppds.
We are implicitly using the lower bound on ppd's used above -- so that the action of $|{\rm Out}(G)|$ doesn't mess things up.
In theory one should check what happens when $(q,n)\in\{(2,6), (2,7), (2,8)\}$ -- here Zsigmondy's theorem fails for one of the tori mentioned above and a ppd does not exist. But I've excluded $q=2$ so this doesn't arise (see next comment).
Finally the fact that there really are elements that are centralized by these maximal tori is a fairly straightforward eigenvalue argument. The only problem arises when $q=2$ and we have the torus $(q^{n-2}-1)\times (q-1)$, hence I've excluded $q=2$ from the statement of the theorem. QED
Extra remarks
Dealing with $q=2$ should be easy. For instance if $n$ is odd, then you can substitute that torus of size $(q^{n-2}-1)\times (q-1)$ with one of size $(q^{n-2}-1)\times (q+1)$ (maybe this will work for $n$ even too but I haven't checked).
Likewise $n=2,3,4$ can be done by hand (I guess).
The whole argument carries over to the unitary groups just by changing a few signs in the size of the tori. The other classical groups will require that you choose tori of different sizes.
Thanks a lot for your interesting partial answer.
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2025-03-21T14:48:30.424517
| 2020-04-27T15:21:12 |
358675
|
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|
Stack Exchange
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"Simple" proof of irreducible characters of finite groups being non-zero
A search brought up this, with reference to a book by I. M. Isaacs. However, the proof in the book leverages on a lot of field theory knowledge. I am wondering, is there a simpler proof (or a proof requiring only materials covered in a standard representation theory course) of characters being non-zero, if I only care about about irreducible characters of a group $G$ over a field $F$ where $char(F)$ divides the order of $G$?
This is answered in https://mathoverflow.net/questions/252855/can-an-irreducible-representation-have-a-zero-character but requires some knowledge.
Yes I mentioned that in my question. =]
Sorry. I hadn't clicked your link. See also https://math.stackexchange.com/questions/819466/the-division-algebras-arising-in-the-wedderburn-decomposition-of-a-finite-group
Here is a solution, which may or may not be simple, based on a suggestion in Glasby’s comment to my question https://math.stackexchange.com/questions/819466/the-division-algebras-arising-in-the-wedderburn-decomposition-of-a-finite-group
First of all, since $G$ is finite, assuming $F$ has characteristic $p>0$, your irreducible representation is defined over a finite extension of the prime field so you can assume without loss of generality that $F$ is a finite field. (Edit. See Lemma 6 of https://archives.maths.anu.edu.au/people/Kovacs/K095.pdf for a proof). By Weddderburn theory and Wedderburn’s little theorem that any finite division ring is a field, we have that the image of $FG$ under you irreducible representation is isomorphic to $M_k(L)$ where $L/F$ is a finite extension and your simple module is isomorphic to $L^k$ with the natural action of $M_k(L)$ on it. If $A(g)$ is the $k\times k$ matrix over $L$ corresponding to $g\in G$, then the value of your character $\chi$ on $g$ is $tr_{L/F}(tr(A))$ where $tr_{L/F}(a)$ is the trace of left multiplication by $a$ on $L$ as a linear operator over $F$.
Thus the value of $\chi$ on any element of $FG$ is obtained by taking $tr_{L/F}$ of the trace of the corresponding $k\times k$ matrix over $L$. By considering the rank $1$ matrix $aE_{11}$ with $a\in L$, we see that if $tr_{L/F}$ is not identically zero, then $\chi$ does not vanish on $FG$ and hence $G$. But this is easy since $tr_{L/F}(a)$ is the sum of its Galois orbit. In fact, since $L/F$ is separable and Galois, as finite fields are perfect and all their finite extensions are cyclic, standard field theory tells you the trace form is non-degenerate. Alternatively, the normal basis theorem tells you that there is an element $a\in L$ whose Galois conjugates from an $F$-basis for $L$ and hence cannot sum to $0$.
This is of course simpler in characteristic $0$ because $M_k(L)$ is replaced by $M_k(D)$ for a division algebra $D$ but $tr_{D/F}(1)=\dim D$ doesn’t vanish.
Is it obvious why the "irreducible representation is defined over a finite extension of the prime field"?
It's not obvious but is proved in Lemma 6 of https://archives.maths.anu.edu.au/people/Kovacs/K095.pdf using wedderburn theory.
Thanks for the reference!
It seems to be a generalization of Brauer's argument on splitting fields in characteristic p.
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2025-03-21T14:48:30.424857
| 2020-04-27T15:41:28 |
358678
|
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|
Stack Exchange
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Entire even functions of order 1 have infinitely many zeros?
Let $f$ be an entire even function of order 1 such that $f(0)\neq 0$. Does $f$ have infinitely many zeros?
If $f(z)$ has just finitely many zeros, there would be a polynomial $P(z)$ such that $f(z) / P(z)$ is entire and non-vanishing, hence equal to $\exp(g(z))$ for some $g(z)$ entire. For $f(z)$ to have order $1$, the only possibility is that $g(z)$ be a linear function. But then $f(z)$ cannot be even. The contradiction shows that your $f(z)$ has indeed infinitely many zeros (and that can be made quantitative: $\sum 1 / |\rho| = +\infty$ is divergent over these zeros taken with multiplicities).
Alternatively, $g(z)=f(\sqrt{z})$ is entire and has order $1/2$, so has infinitely many zeros if non-constant.
Yes. Any even entire function $f$ can be written as $f(z)=g(\sqrt{z})$, where $g$
is entire. To prove this just consider the power series. If $f$ is or order $1$, then
$g$ is of order $1/2$ (also very easy). And every function of non-integral order
has infinitely many zeros. This follows from the Hadamard factorization theorem.
In fact this is a very special case of the Hadamard theorem: if $g$ has finitely
many zeros and finite order then $g(z)=e^{P(z)}Q(z)$, where $P$ and $Q$ are polynomials.
In fact this shows that any even function with finitely many zeros must have even or infinite order, which generalizes your statement.
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2025-03-21T14:48:30.424974
| 2020-04-27T16:34:31 |
358686
|
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Stack Exchange
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Definition of 1st degree obstruction class
Recently I go through obstruction class illustrated by Milnor.
He defined $\mathfrak{o}_i$by an element in $H^i(M; \pi_{i-1}(V_{n-i+1}(F))$, which is cohomology with local coefficients.
But the 0th homotopy group has no group structure, and the definition for $\mathfrak{o}_1$ doesn’t work. So does 1st degree obstruction class exist and if exists, how do you define it?
When $i=1$, the Stiefel manifold is $V_n({\mathbb R})$ which is the bundle of $n$-frames in an $n$-dimensional vector space. As such it is homeomorphic to $GL(n,{\mathbb R})$ and hence $\pi_0$ has a group structure. It has two components and it is easy to see that ${\mathfrak o}_1$ measures orientability in the way that you might expect.
Probably this question is more suited to MSE than Mathoverflow.
Then $\pi_0 (V_n(R^n)) = Z/2$, right?
That's right; there are two components, distinguished by the sign of the determinant.
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2025-03-21T14:48:30.425101
| 2020-04-27T17:24:04 |
358691
|
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Tarski's truth theorem — semantic or syntactic?
I was reading the sketch of the proof of Tarski's theorem in Jech's "Set Theory", which appears as Theorem 12.7, thinking that it would be an interesting result to really understand. As stated in the book, it is essentially a syntactic result (after fixing a Gödel numbering). However, after reading other proofs of Tarski's result, and really delving into the sketched proof, I believe that there is a serious error in Jech's proof, and now I'm not sure the result holds at the syntactic level.
Here is the problem as I see it. In the second sentence of the proof the formulas are enumerated as $$\varphi_0,\varphi_1,\varphi_2,\ldots.$$ Now, this is an enumeration outside ZFC, so the subscripts are metamathematical numbers. But in the next formula, which reads, $$x\in \omega \land \neg T(\#(\varphi_x(x))),$$ the subscript $x$ on $\varphi$ is being treated as a formal natural number---an element of $\omega$.
If we have a model of set theory, where $\omega$ matches the metamathematical natural numbers, maybe we could make this formula work. My question is whether or not we can somehow avoid making such a strong assumption. If not, what's the easiest way to assert such a matching (say, without forcing an interpretation of all of ZFC, just of the natural number part)?
Let the enumeration be recursive. Implement it inside the theory.
@MonroeEskew In fact we just need the enumeration to be definable.
@MonroeEskew Your comment doesn't seem to address the two questions I asked.
Yes it does. Don't take the enumeration to be "outside." Let $M$ be your model of a sufficiently strong theory. We are proving that there is no formula $T(x)$ such that $M \models T(x)$ iff $x$ is a natural number of the model which codes a real sentence $\sigma$ such that $M \models \sigma$. The reason is that $M$ can implement the recursive numeration of all formulas, which it may interpret as extending beyond the real integers, but which agrees on the real ones, and by Godel's argument we can construct a real sentence $\sigma$ such that $M \models \neg T(#\sigma)$ iff $\sigma$.
@MonroeEskew The fact that you start with "Let $M$ be a model..." suggests that somehow were are not communicating here. Let me try again. Jech lists two formulas that define his truth condition. The first is $\forall x, (T(x)\rightarrow x\in \omega)$. This is a formula in the FOL in the signature $(\in)$. Similarly, however you define a Godel numbering $#$ of the sentences in this FOL (whether recursively, or not), for each sentence $\sigma$, the statement $\sigma\leftrightarrow T(#\sigma)$ is also a formula in this language. Thus, it is valid to ask whether or not...
...a predicate $T$ can be defined, satisfying these conditions, without reference to any model. That is my first question. Your comment may address my second question, if the answer to my first question is "no".
This is a theorem about models. It is about undefinability of "truth" which means satisfaction. We are showing that for any given model $M$ of ZFC (although this holds in more generality) is no formula $\phi(v)$ in the language of set theory such that $M \models \phi(x)$ iff $x$ is a standard natural number coding a sentence that is true in $M$. If you like, we can restrict our attention to one preferred model, say "V". I don't know how to phrase this result as a purely syntactic / provability result-- it's about satisfaction.
@MonroeEskew "If you like, we can restrict our attention to one preferred model, say "V"" FWIW I think (per my answer) that this is what Jech's doing - working implicitly in NBG and proving "$Th(V)$ is not definable in $V$."
@MonroeEskew No, Jech doesn't define his truth predicate in terms of satisfaction. That's what motivated my question in the first place. He defines it via a (metacountable) schema of statements holding. For all I know, it is perfectly possible for there there be a definable predicate, satisfying his list of conditions, compatible with the axioms of ZFC.
If I recall correctly, Jech is using as his metatheory the class theory $\mathsf{NBG}$. In this context, "true" is a proxy for "true in the (class-sized) structure $V$."
Specifically, the (more) formal version of the natural-language Theorem $12.7$ is the following:
$Th(V)$ is not definable in $V$.
The definition of $Th(V)$ is taking place on the class level: it's a set of natural numbers defined by quantifying over classes. The same is true for the property "definable in $V$." So even though it looks like Jech is using a weirdly un-referring notion of "truth," it is in fact just the usual notion of truth with respect to a specific structure - that structure being $V$, and that whole facet of the argument being (annoyingly, perhaps) kept implicit. Note that this makes the whole "correctness-about-$\omega$" issue moot: Theorem $12.7$ is about a structure which by definition has the right $\omega$.
An in-my-opinion more satisfying version of the result, which makes correctness-about-$\omega$ nontrivial, is the following:
$T$ proves that for all $\mathcal{M}\models\mathsf{ZFC}$, $Th(\mathcal{M})$ is not the standard part of a definable subset of $\mathcal{M}$.
Here $T$ is a very weak theory indeed: $\mathsf{ACA_0^+}$ suffices (really the only need for strength being the requirement that the theory of a structure is actually a thing that makes sense in the first place - see e.g. here). Note that this version of the result does not apply only to models which are correct about $\omega$.
EDIT: And as Monroe Eskew pointed out below, if we drop models entirely we can go even lower. We can prove over a very weak base theory (e.g. $I\Sigma_1$ is already overkill) the following:
If $\mathsf{ZFC}$ is consistent, then there is no formula $\varphi$ such that for all sentences $\psi$ $\mathsf{ZFC}$ proves $\varphi(\#\psi)\leftrightarrow\psi$.
But I don't care what ZFC proves about its internal formulas. I want, from a metatheoretical point of view, to know what it takes to show that there is not definable truth predicate for ZFC. My metatheory doesn't necessary assume ZFC, per se.
@PaceNielsen What is a truth predicate for ZFC? Truth predicates only make sense for structures.
The natural-language version of Tarski's theorem is "$Th(\mathbb{N};+,\cdot)$ is not definable in $(\mathbb{N};+,\cdot)$." To even express that, our metatheory needs to be able to talk about $(\mathbb{N};+,\cdot)$ as a structure appropriately.
Regarding truth predicates requiring structure---that's what my question was asking about! Jech defines a truth condition without any reference to a structure. (See his condition (12.17) on page 162.)
@PaceNielsen No, there is a reference to a structure there although it's implicit: the book (if I recall correctly) uses $\mathsf{NBG}$ as the metatheory throughout. So $\mathsf{NBG}$ is proving "The set of sentences true in the structure $V$ is not a definable element of $V$." The structure in question is $V$ itself, which is a directly-handleable object in $\mathsf{NBG}$.
We can "de-classify" this along the usual lines of course: if we want to use $\mathsf{ZFC}$ as our metatheoy, Theorem $12.7$ is then understood as a natural-language version of a $\mathsf{ZFC}$-proof of "If $\mathcal{M}\models \mathsf{ZFC}$ then $Th(\mathcal{M})$ is not (the standard part of) a definable subset of $\mathcal{M}$."
Note that if we take the $\mathsf{NBG}$ approach - which I think is what Jech is implicitly doing - then we are (in the metatheory $\mathsf{NBG}$) proving a result about the actual universe of all sets ($V$), so the issue of "correct natural numbers" doesn't come up. The "de-classed" version of the result in my previous comment is applicable more broadly.
Noah, rather than making guesses, perhaps you should find a copy of the book and look at the relevant sections. In case you don't have a copy, here are the conditions: (i) $\forall x, (T(x)\rightarrow x\in \omega)$ and (ii) for every sentence $\sigma$, then $\sigma\leftrightarrow T(#\sigma)$. As you can see, there is no reference to modeling here. These conditions are independent of the choice of model.
@PaceNielsen I'm looking at the book right now. The only thing I'm guessing at is the book's specific choice of metatheory, which I'd have to search for. But regardless of whether he uses $\mathsf{NBG}$ or $\mathsf{MK}$ or similar, the point is the same: there is a specific model he's (implicitly) talking about there, namely the true $V$, and the truth definition is with respect to that (proper class sized) structure. That is, for Jech "true" is a proxy for "true in $V$."
@PaceNielsen I've edited my answer to hopefully better address your question; let me know if I'm still not getting your point.
No, it is still not right. His definition of $T$ (the purported truth predicate) is purely syntactical (up to choice of Godel numbering). He doesn't say $M\models \sigma$ if and only if $M\models T(#\sigma)$. He says that for each sentence $\sigma$, the formula $\sigma\leftrightarrow T(#\sigma)$ is valid (i.e., true in all worlds).
@PaceNielsen No, he doesn't: he says "if $\sigma$ is a sentence [then] $T(\ulcorner\sigma\urcorner)$ holds if and only if $\sigma$ holds." "Holds" here is implicitly with respect to $V$. Where are you getting the contrary?
The wording in my version of the book is exactly the following: "if $\sigma$ is a sentence, then $\sigma\leftrightarrow T(#\sigma)$." This is a schema of assertions. For each statement $\sigma$ (in the FOL language in the signature $(\in)$, in the theory given by the axioms of ZFC and the usual logical axioms), it is being asserted that $\sigma\leftrightarrow T(#\sigma)$ holds.
@PaceNielsen No, it's a single assertion and it's referring to what's true in $V$. (What does "holds" mean otherwise? It can't possibly refer to holding across all models of ZFC, since that is definable.)
The formal version of that definition is: "$T$ is the Godel number of a truth predicate iff ... (ii) if $\sigma$ is a sentence, then $V\models\sigma$ iff $V\models T(#\sigma)$." This indeed uses satisfaction with respect to a specific structure, namely $V$.
@PaceNielsen You can also take from this argument the syntactic result that there is no formula $\phi(v)$ such that for every sentence $\sigma$, ZFC proves $\phi(#\sigma) \leftrightarrow \sigma$. Because given $\phi(v)$, you can construct a particular $\sigma$ that is provably equivalent to $\neg\phi(#\sigma)$. This is just Goedel's fixed point lemma.
@MonroeEskew And that syntactic result - or more accurately, that result with the hypothesis that ZFC is consistent - is itself provable in a very weak theory (way below $I\Sigma_1$ certainly).
@MonroeEskew Are you asserting the following metamathematical statement? For every formula with one free variable $\varphi(v)$ in the signature $(\in)$, and every Godel numbering $#$ of sentences in that language, there exists a sentence $\sigma$ such that ZFC proves $\sigma\leftrightarrow \neg\varphi(#\sigma)$. If so, where is the easiest place to find a complete (easy) proof of this?
And to clarify, by a Godel numbering, I mean an injective map from the true formulas of that signature (not the internal formulas) to the elements of $\omega$ (inside the formal system).
@PaceNielsen Yes, that's just Godel's diagonal lemma (and it's provable in $I\Sigma_1$ for example).
@PaceNielsen- See for example the end of the first chapter of Kunen’s book.
Great, I think this is what I was looking for.
@PaceNielsen I've added the diagonal lemma to my answer.
Perfect! I've accepted the answer
@NoahSchweber There is one additional point. I believe Godel's lemma requires that the numbering is given by a nice enough function. What happens if you allow an arbitrary numbering?
@PaceNielsen Well, for an arbitrary numbering there's not much that can be said. For example, map all the true sentences onto the evens and the false sentences onto the odds. So some constraint is needed to get anything useful. The key properties we need for these sorts of results are representability, or invariant definability, properties: for example, we want there to be a formula $\psi$ such that for every standard $i,j,k$ we have $T\vdash\psi(i,j,k)$ iff $i=#(#^{-1}(j)\wedge#^{-1}(k))$ and $T\vdash\neg\psi(i,j,k)$ otherwise. (cont'd)
So intuitively "conjunction is the standard part of a definable relation in all models of $T$." This is all nuked by the theorem that (for $T$ appropriate) the (total) computable functions are exactly the functions which are representable in $T$ in this sense; although it's overkill for that specific purpose, this result is generally proved in modern expositions of Godel's incompleteness theorem.
Noah, I don't think Jech is working in NBG (at least, he does not hint this at any point.) @PaceNielsen I think the issue here depends on whether you see this theorem as a single theorem or a theorem scheme. If you treat this as a theorem scheme for each $T(x)$, then he explicitly produces a sentence $\sigma$, for which we can prove that $\sigma \leftrightarrow \neg T(# \sigma)$. If you want to see this a single theorem that quantifies over all formulas $T(x)$, then you need a satisfaction relation for proper classes which would require you to work in a class theory such as NBG.
@PaceNielsen: If you are bugged by this issue, it is also present in the definition of "truth definition". Is this a definition in ZFC, in which case all formulas in that definition are indeed the formal objects (sets) that represent those formulas, or is this a "metadefinition" that says that a formula $T(x)$ that I write down is a truth definition if I can prove the item (i) and every instance of item (ii)?
@Burak I'm no longer bugged by any issues. My confusion arose from my misunderstanding of how a formula was being used in Jech's proof. While $#\varphi_x(x)$ is not technically a formula in the language of set theory, it is representable by such a formula (when $#$ is recursive, etc...).
Tarski's theorem, as given in Undecidable theories, page 46, allows arbitrary numbering and is completely syntactic. I think this abstract version given by Tarski himself is the most clear. Let me summarize it here with some inessential variations.
Let $T$ be a consistent first-order theory (any consistent first-order theory). If $\varphi\mapsto \ulcorner\varphi\urcorner$ is a naming of formulas (any assignment of closed terms to formulas), then either the diagonalization function (the function $\varphi\mapsto \varphi(\ulcorner\varphi\urcorner)$) is not representable (under that naming), or the set of theorems is not representable (under the given naming), or both are not representable.
In the case of ZF, assuming it to be consistent, we know that if we choose a recursive naming, we can represent the diagonalization function but not the set of theorems. Also, we can easily choose a (nonrecursive) naming which allows us to represent the set of theorems, but, then, the diagonalization will not be representable.
The proof is quite simple. If the diagonalization is representable, the fixed-point lemma can be proved quite simply. Assume that $V$ is a formula representing the set of theorems. Apply the fixed point lemma to get $\varphi$, a sentence satisfying
$T\vdash\varphi\leftrightarrow \neg V(\ulcorner\varphi\urcorner)$.
If $T\vdash\varphi$, then, since $V$ represents the theorems, $T\vdash V(\ulcorner\varphi\urcorner)$, and $T$ is inconsistent. If $T\nvdash \varphi$, then, since $V$ represents the theorems, $T\vdash\neg V(\ulcorner\varphi\urcorner)$, and $T\vdash \varphi$ by the choice of $\varphi$. Therefore, $T\vdash \varphi$ and it is inconsistent by the previous argument.
EDIT
Motivated by the question in the comment, I will prove the fixed point lemma I have used above:
We are assuming that $T$ is a first-order theory and that the diagonalization is represented in $T$ under the arbitrary naming $\varphi\mapsto\ulcorner\varphi\urcorner$. It means that there is a formula $D(x,y)$ such that
$T\vdash\forall y(D(\ulcorner\phi\urcorner, y)\leftrightarrow y=\ulcorner\phi(\ulcorner\phi\urcorner)\urcorner)$.
Now, let $W(y)$ be an arbitrary formula. Let $\phi(x)$ be the formula $\exists y(D(x,y)\wedge W(y))$ and let $\varphi$ be the sentence $\phi(\ulcorner\phi\urcorner)$, the diagonalization of $\phi$. This sentence is a fixed point for $W(y)$.
Indeed, $\varphi$ is $\exists y(D(\ulcorner\phi\urcorner,y)\wedge W(y))$, which, from the hypothesis on the representation of the diagonalization, is equivalent to $\exists y(y=\ulcorner\varphi\urcorner\wedge W(y))$. The last sentence is logically equivalent to $W(\ulcorner\varphi\urcorner)$, and we are done.
Therefore, Tarski's result applies to arbitrary first-order theories and to arbitrary namings. The moral is that no matter what first-order theory and naming of formulas you choose, the representation of at least one of two simple metatheoretical notions
(diagonalization and theoremhood) within the object theory will always fail.
I like this trichotomy. Doesn't the fixed point lemma require more than just $T$ being a first-order theory? Also, this result is talking about validity rather than truth; am I right in saying that truth is representable iff validity is representable?
I have proved the fixed point lemma. Validity is more general than truth. Truth is validity when the theory is complete, that is, the theory of a model.
The Diagonalization Lemma (= the fixed point lemma) requires something like the Robinson Arithmetic Q. The diagonalization function must be (strongly) representable.
In the language of set theory, the very weak "baby set theory" i.e. "the adjunctive set theory" will do.
In arithmetic, you can use numbers ("Gödel numbers") to code formulas, and numerals in the very language to name them. In set theory, you can similarly use finite sets, e.g. von Neumann ordinals, to code formulas, and simple set theoretical expressions to name them.
This is done directly e.g. in Melvin Fitting: Incompleteness in the Land of Sets (Studies in Logic).
But because we know that arithmetic can be interpreted in set theory, and in that sense arithmetic can be done inside set theory, we can also just assume that all the Gödelian techniques can be transported to set theory.
|
2025-03-21T14:48:30.426336
| 2020-04-27T17:41:54 |
358692
|
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"Denis Nardin",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Is this a stack?
A continuous map $f:X\to Y$ and a vector bundle $E\to X$ seem to give rise to a presheaf of groupoids on $Y$ along the following lines. For an open $U\subseteq Y$, each section of $f$ over $U$ (i. e. a map $s:U\to X$ with $fs=$ inclusion) produces a vector bundle $s^*E$ on $U$ and for another section $t:U\to X$, a homotopy class $H$ of homotopies between $s$ and $t$ that induce trivial homotopy between $fs$ and $ft$ gives rise to a bundle isomorphism between $s^*E$ and $t^*E$.
Question 0: can this be finalized to indeed yield a presheaf of groupoids?
Question 1 (in case the answer to the previous one is "yes"): is this a stack? Are there some natural restrictions on $f$ and $E$ (like $f$ being some kind of fibration, etc.) which ensure that one gets a stack? Or maybe some modifications of the construction? For example, instead of $fs = $ inclusion one might add to the structure a homotopy between $fs$ and inclusion, and some coherence requirement; or, instead of demanding that $H$ induces a trivial homotopy, one might make part of the structure a choice of a homotopy between $fH:fs\sim ft$ and the trivial homotopy $fs = $ inclusion $ = ft$.
Question 2: is this an instance of some known construction?
Maybe I should add that I am actually interested in a very particular case, when $f$ is a fibration of smooth manifolds, with fibres all smooth, and $E$ is the tangent-along-the fibres bundle.
What are $X$ and $Y$? Spaces? Schemes? If spaces this is naturally a presheaf of ∞-groupoids, not of groupoids. Of course you can truncate, but you'll probably lose descent properties
@DenisNardin Spaces, as good as you want - say, finite CW-complexes.
Can you please clarify which category are you associating to an open subset $U$ of $Y$?.. sorry if it is already stated clearly but I am not able to see it.. Is it the category of vector bundles over $U$ obtained as pullbacks of sections $s:U\rightarrow X$?
@PraphullaKoushik You see it right - I don't know, this is part of the question 0 :D But at least whatever it can be it certainly can only contain vector bundles of the same rank as $E$
@მამუკაჯიბლაძე same rank? what is the first reason for imposing that condition?
@PraphullaKoushik We only have $E$ at hand, any of its pullbacks anywhere wil have the same rank, no way to get any other, right?
@მამუკაჯიბლაძე that was a stupid comment :D May be I was half sleeping :D You are absolutely correct...
|
2025-03-21T14:48:30.426808
| 2020-04-27T17:54:41 |
358694
|
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|
Stack Exchange
|
Recurrence relation for number of reduced words of longest element in $S_n$
Is there any recurrence relation known for the number of reduced words of the longest element in $S_n$ (not commutation classes)?
Edit: Sorry for unaccepting the answer, but I realized that I really would like to have a recurrence relation in $n$, so I would like to express the number of reduced words of the longest element in $S_n$ in terms of the numbers reduced words of the longest elements in $S_k$ for $k < n$.
Let $(W,S)$ be a Coxeter system. Let $R(w)$ be the number of reduced words for an element, let $D(w)\subseteq S$ be the set of right descents of $w$. Then
$$R(w)=\sum_{s\in D(w)}R(ws)$$
with $R(1)=1$.
This recurrence relation is not how you want to compute the number of reduced words for the longest element of $S_n$, though, because there is a closed formula for this, and using the recurrence relation would take forever.
If you want to build $R(w_0(n))$ from $R(w_0(n-1))$, note that the closed formula for these numbers comes from the hook length formula for standard tableaux on the shape $(n,n-1,\ldots,2,1)$. The product of the hook lengths of every box except for those in the first row is
$$\frac{\binom{n}2!}{R(w_0(n-1))}$$
The remaining hook lengths are $2n-1$, $2n-3$, $\ldots$, $3$, $1$. So
$$\frac{\binom{n+1}2!}{R(w_0(n))}=(2n-1)!!\frac{\binom n2 !}{R(w_0(n-1))}$$
Thx! I actually forgot to write that I mean recurrence relation in $n$ but now that there is a closed formula, all the better
@BipolarMinds The formula is well known, for example it's in Combinatorics of Coxeter groups by Bjorner and Brenti, though I'm not sure of the original source.
I found it in a paper of Richard Stanley
Well yes, fair enough :) My problem with this is that the quotient is not always a natural number, so it doesn't count anything. The recurrence relation should come from a bijection on sets, as for example $S_n= S_{n-1}\times [n]$ gives $n! =n(n-1)!$. In some sense, it should be a higher analogue of this identity, but I don't expect it to be easy. Sorry for my imprecise formulation of the question.
@BipolarMinds That's fair, but are you sure it's not always a natural number?
Yes, I actually did the same thing at first :)
@BipolarMinds: it sounds like you might be interested in higher Bruhat orders in the sense of Manin--Schechtman/Ziegler (https://doi.org/10.1016/0040-9383(93)90019-R). In that paper Ziegler says that, beyond the case covered by Stanley's work, little is known in terms of enumeration. But of course that paper is 30 years old so perhaps more is known now.
@BipolarMinds Not sure if it will be useful, but I combinatorially described a recurrence for $\frac{\binom{n+1}2!}{|R(w_0(n))|}$.
@SamHopkins thanks for the reference, this exactly the direction I'm aiming for
|
2025-03-21T14:48:30.427023
| 2020-04-27T18:43:50 |
358698
|
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|
Stack Exchange
|
Is this integral transform related to the Laplace transform?
The Laplace transform of a function $f(t)$, defined for all real numbers $t \geq 0$, is the function $F(s)$, defined by
$${\displaystyle F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt}.$$
Let $\varphi: {\mathbb R}_+ \to {\mathbb R}_+$ be an injective continuous increasing function (not necessarily linear). Is the function:
$${\displaystyle F_{\varphi}(s)=\int _{0}^{\infty }f(t)e^{- \varphi(st)}\,dt},$$
related to the Laplace Transform or another known integral transform?
After some calculations, I arrived to an integral of that kind, so I was wondering if some theoretical properties of that kind of functions are known already.
A generalization along these lines, but with the single function $\phi(st)$ replaced by the product of two invertible functions $\Phi(s)E(t)$, has been studied in The generalized Laplace transform and fractional differential equations of distributed order. The Mellin transform is a notable example of this type, with $\Phi(s)E(t)=(1-s)\ln t$.
Many thanks. I came across that paper, it looks interesting. I am starting to fear the generalisation I am considering has not been studied.
|
2025-03-21T14:48:30.427236
| 2020-04-27T18:51:36 |
358699
|
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|
Stack Exchange
|
Dependence rank: what is the size of the largest subcollection of random variables which is statistically independent?
Let $X_1,\ldots,X_p$ be random variables on the same space. Define their dependence rank, denoted $rank(X_1,\ldots,X_p)$ as the largest nonnegative integer $k$ such that there is a subcollection of $k$ out of the $p$ random variables which are mutually independent (i.e theire joint distribution funciton factors as a product of their individual distribution funcitons). Of course, every singleton $\{X_i\}$ forms an independent collection. Thus $1 \le rank(X_1,\ldots,X_p) \le p$, and the upper bound is attained if $X_1,\ldots,X_p$ are independent to begin with.
Question. Is the above notion of rank defined above studied in the literature ? Are there any interesting properties of quantity ? Are the any known proxies for this concept ?
Yes, indeed covarance matrices don't capture nonlinear dependence patterns. Removed my comment above cov matrix.
Recall that for $k\ge 3$, pairwise independence is weaker than full ("mutual") independence. What is your motivation for using "pairwise" in your definition?
It was a problem with terminology. Removed the "pairwise" adjective.
|
2025-03-21T14:48:30.427351
| 2020-04-27T23:50:06 |
358709
|
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"Ian Agol",
"Ken Baker",
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|
Stack Exchange
|
Homology torsion in the double branched cover of a tangle?
Let $T$ be a locally unknotted $2$-tangle in $B^3$ and $\Sigma(T)$ be its double branched cover.
Can $H_1(\Sigma(T))$ have a non-trivial torsion? (Obviously, not for rational tangles.)
I feel like there should be plenty of alternating tangles with this property, such as pretzel or algebraic tangles.
Studying this torsion has a nice application due to David Krebes (An obstruction to embedding 4-tangles in links, J. Knot Theory Ramif.,8 (1999), 321-352.) You can use it to show that a given 2-strand tangle doesn't embed as a subtangle of the unknot.
An example (similar to Ken's and related to Ian's comment): take the pretzel tangle P(-3,-3,3) drawn below; its double branched cover has homology $Z \oplus Z/3 \oplus Z/3$. This is from my paper (Embedding tangles in links, J. Knot Theory Ramif.,9, No. 4 (2000), 523-530).
$(T_3)^* + (T_3)^* + (T_{-3})^*$">
Yes. For example, start with the exterior of the Whitehead link and do p/q filling on one component where p is odd. This manifold has p torsion and is the double branched cover of a two strand tangle in a ball.
You can see this by taking the quotient of the Whitehead link by the strong inversion and doing the appropriate rational tangle replacement on one component. The exterior of the other component (in the quotient) is then a two-strand tangle in the ball.
Note that if p were even, then you'd also have a single closed component in the tangle in addition to the two strands.
To clarify the construction: The point is that you're starting with a manifold that you know is the double branched cover of a 2-strand tangle (here, the solid torus) and doing p/q surgery on a strongly invertible null-homologous knot (here, a genus one knot in the solid torus). The surgery dual curve will represent torsion of order p. The quotient is a rational tangle replacement on a neighborhood of an arc with ends on the original tangle. Hence you'll get another 2-strand tangle, or one with an extra closed component depending on the parity of p.
I wouldn't be surprised if you already had a picture of this somewhere on your blog.
|
2025-03-21T14:48:30.427537
| 2020-04-28T01:02:18 |
358715
|
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|
Stack Exchange
|
A tiling of $\mathbb{Z}^2$ from M. Barlow's paper
In M. Barlow's paper: arxiv.org/pdf/math/0302004.pdf, P17- (2.7) formula.
Let $k\geq 10$, and consider a tiling of $\mathbb{Z}^2$ by disjoint squares
$$T(x):=\{y\in \mathbb{Z}^2: x_i\leq y_i< x_i+k, i=1, 2\}$$
with side $k-1$.
Let $\widehat{Q}$ be a macroscopic square of side $m$, and associate with $\widehat{Q}$ the microscopic square $$Q=\cup\{T(x), x\in \widehat{Q}\}$$
How to understand $T(x)$ and $Q$? I am confused about why $T(x)$ is a disjoint tiling of $\mathbb{Z}^2$ and what is $Q$?
You have made a few copying errors ($T$ for $T^+$, for instance).
For the tiling, my guess is that the claim is not that $\{ T(x) \, : \, x \in \mathbb{Z}^d \}$ is a tiling (that would be false), but rather one just takes enough $x$'s to get a tiling, for instance $\{ T(x) \, : \, x \in k\mathbb{Z}^d \}$.
For the $Q$ vs. $\hat{Q}$ (Barlow uses $\tilde{Q}$, and I'm not sure why you changed this), note from Definition 2.1 that $T^+(x) = \mathbb{Z}^d \cap A_1$ where $A_1$ is the cube with the same center as $T(x)$ but side length $3k/2$. The centers of the cubes $T^+(x)$, for $x \in \tilde{Q}$, then form a cube inside $\tilde{Q}$ with $2m/3k$ cube centers per edge. In particular, this is smaller than $m$, which is why this is called microscopic. My guess is that this has something to do with the overall renormalization argument.
If this doesn't clear things up, I recommend emailing Barlow about it. I'm definitely not an expert in this area.
|
2025-03-21T14:48:30.427668
| 2020-04-28T04:59:28 |
358722
|
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|
Stack Exchange
|
Integral of an expression including a fraction having modified Bessel functions of the first kind on both numerator and denominator
I am looking for an analytic result of the following integral
$$\iint_0^\infty {{\rm{d}}x{\rm{d}}y{x^2}{y^2}\exp \left\{ { - {\alpha _1}{x^2} - {\alpha _2}{y^2}} \right\}\frac{{{\rm{I}}_1^2\left[ {\beta xy} \right]}}{{{{\rm{I}}_0}\left[ {\beta xy} \right]}}}$$
where ${\rm{I}}_0(\cdot), {\rm{I}}_1(\cdot)$ are the zero- and first-order modified Bessel functions of the first kind, respectively, and $\alpha_1, \alpha_2, \beta$ are positive constants.
I have referred to a book named "Tables of some indefinite integrals of Bessel functions of integer order" but did not find a solution.
It will be super appreciated if there are some ideas.
@student A revealing idea, thanks!
|
2025-03-21T14:48:30.427750
| 2020-04-28T05:24:34 |
358724
|
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"url": "https://mathoverflow.net/questions/358724"
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|
Stack Exchange
|
What are the generator for $K_1(C(\mathbb{T})\otimes\mathbb{K})$?
I've been working computing several K-groups associated to some $C^*$-algebras involved with my master's thesis, however I've just got stucked finding some generators for $K_1(C(\mathbb{T})\otimes\mathbb{K})$.
Just to elaborate my question let me explain the analogous problem associated to $K_0(C(\mathbb{T})\otimes\mathbb{K})$:
It is of course true that
$$K_0(C(\mathbb{T})\otimes\mathbb{K})\cong K_0(C(\mathbb{T}))\cong \mathbb{Z}.$$
Also, it is a well known fact that $[1]$ is a generator of $K_0(C(\mathbb{T}))$, so one would be tempted to find the generator of $K_0(C(\mathbb{T})\otimes\mathbb{K})$ using this information. Luckily, we have the next result:
The map from a $C^*$-algebra $A$ into $A\otimes\mathbb{K}$ given by $a\mapsto a\otimes e_{11}$, where $e_{11}$ is a rank one projection, induces an isomorphism between $K_0(C(\mathbb{T})$ and $K_0(C(\mathbb{T})\otimes\mathbb{K})$.
Finally, joining all the pieces it follows that a generator for $K_0(C(\mathbb{T})\otimes\mathbb{K}$ is $[1\otimes e_{11}]$.
Since the $K_1$-functor is also stable and finding generators for $K_1(C(\mathbb{T}))$ is not too hard, one would be tempted to do the same trick as above, however I have not found any analogous result for the $K_1$-groups. The main problem (I think) is that the proof (at least the one that I know) of the fact that the $K_1$-functor is stable is non-constructive, in the sense that the existence of the isomorphism follows from the continuity under direct limits of the $K_1$-functor.
With all this said, a generalized question is: Is there any known isomorphism between $K_1(A)$ and $K_1(A\otimes\mathbb{K})$? or more precisely (to my goals): Is there any way to find the generators of $K_1(A\otimes\mathbb{K})$ knowing the generators of $K_1(A)$?
Edited, thanks!
The morphism $f : A \to A \otimes \mathbb K$ which maps $a$ to $a \otimes e_{11}$ induce isomorphism $K_1(f) : K_1(A) \to K_1(A \otimes \mathbb K)$ on $K_1$ groups as well. In fact, we can construct very explicit inverse $KK_0(A \otimes \mathbb K, A)$ cocycle. Indeed, suppose $\mathbb K$ acts on separable Hilbert space $\mathcal H$, then $A \otimes \mathcal H$ is naturally Fredholm $A \otimes \mathbb K, A$ bimodule, which is naturally cocycle inside $KK_0(A \otimes \mathbb K, A)$. It's not so hard to see that Kasparov product with the morphism above is an identity.
Thank you so much! Do you know if I can find this information in some book or something? (Already accepted answer though)
E.g. Wegge-Olsen "K-theory and C*-algebras" Lemma 7.1.8 (it's about the isomorphism $K_1(A) \to K_1(M_n(A))$, but after passing to inductive limit nothing change, by proposition 7.1.7)
|
2025-03-21T14:48:30.427945
| 2020-04-28T05:38:34 |
358725
|
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|
Stack Exchange
|
Unknotting algorithm in higher dimensions?
Suppose we are given a 2-knot (say by a movie). Is there an algorithm to tell if it is unknotted ? I suppose that it could matter if I say "topologically" or "smoothly" here since those could be different - I am interested in results in either direction.
Is there an algorithm to tell if the fundamental group of the complement is $\mathbb{Z}$?
While I am mainly interested in the 4-dimensional case, I imagine the above problems are hard (although I can't find a reference) - maybe something is known in higher dimensions? Here there is an algebraic characterization (due to Kervaire) of the knot complements that can occur (namely finitely-presentable groups, generated by a single conjugacy class, has cyclic first homology, and 0 second homology), so maybe there is a result that says that in this class of groups there is no algorithm to recognize $\mathbb{Z}$?
The version of the problem in dimensions higher than 4 is undecidable, by work of Nabutovsky and Weinberger:
https://link.springer.com/content/pdf/10.1007/BF02566428.pdf
This is related to the undecidability of the triviality problem for group presentations.
The 4-dimensional version is thought to be undecidable, but this is not known. The relevant question in geometric group theory is: is it decidable whether a "balanced" group presentation (one that has the same number of generators and relations) presents the trivial group? Or $\mathbb{Z}$? For some limited recent progress on this problem, see this paper by Lishak and Nabutovsky:
https://arxiv.org/abs/1510.02773
This is all in the smooth or PL case. For topological knots, it's not clear to me how you would represent them computationally in a finite way. (Perhaps there are even uncountably many types of them? Someone who is better with topological topology should chime in here.)
|
2025-03-21T14:48:30.428091
| 2020-04-28T05:54:12 |
358726
|
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|
Stack Exchange
|
Proofs by Schubert calculus and combinatorics
Do you know some examples proved by two different methods: 1. Schubert calculus, 2. combinatorial method.
Unimodality of partitions inside a box of size $m\times n$ (i.e. partitions with number of parts at most $m$ and size of the largest part at most $n$) follows from applying Hard Lefschetz Theorem to the Grassmann variety $G_{m+n,n}$ (see the paper 'Combinatorial Applications of the Hard Lefschetz Theorem' by Stanley). A combinatorial proof of this was given by Kathy O'Hara (reference: Unimodality of Gaussian coefficients: a constructive proof, J. of Comb. Theory, Ser. A, 53 (1990) 29-52.
Thank you so much!
|
2025-03-21T14:48:30.428170
| 2020-04-28T06:05:26 |
358727
|
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"Daniele Tampieri",
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|
Stack Exchange
|
Analysis of solutions to a nonlinear ODE
Consider the following ODEs:
$\phi^2=\phi''\sqrt{1-\phi'^2}$, or $\phi^2=-\phi''\sqrt{1-\phi'^2}$.
Is there any theory (e.g. comparison theorems) which analyzes solutions of the above ODEs? I am only interested in nonnegative solutions, i.e. $\phi\geq 0$. Actually one can write down a solution $\phi(t)=\cos t,\ -\frac{\pi}{2}\leq t\leq \frac{\pi}{2}$ to the second equation. I want to know whether one can find explicit solutions to the first equations, as well as other solutions to the second one (of course excluding those coming from time translations of $\cos t$). If one can not find explicit formula for the solution, I am also interested in asymptotic properties/existence time of the solution. In particular, I want to know if there exist solutions defined on the entire real line.
Strictly speaking, the equation(s) you are analyzing seems more an algebraic differential equation (ADE) than an ODE: $$ \phi^4=\phi^{\prime\prime 2}(1-\phi^{\prime2}). $$ The properties of the solutions of this kind of equations depend heavily on the properties of the algebraic function involved. You'll probably find more useful information by looking on google for ADEs instead of ODEs
One thing that's not clear from your question: are you considering each equation individually? Or are you asking, like @DanieleTampieri suggested, for a $\phi$ that satisfies at least one of the two equations? If $\phi$ is allowed to follow exactly one (but not the other) equation, then by convexity there cannot be any global in time solutions.
@WillieWong Yes, I am considering each equation individually.
Edited on May 2, 2020: The OP pointed out that I had not addressed a special case (namely $C=1$ below), so I am amending my answer to address this and reorganizing so that the $C=1$ case gets addressed naturally when it comes up. — RLB
We can assume that $\phi$ is not constant, since the only constant solution is $\phi\equiv0$.
Thus, assume that the solution has $\phi'\not=0$ on some interval $I$. Multiply the equation $\phi^2 = \pm\phi''\sqrt{1-\phi'^2}$ by $\phi'$, then integrate both sides to get that there is a constant $C$ such that
$$
\phi^3 = C^3 \mp (1-\phi'^2)^{3/2}.
$$
The case $C=0$ corresponds to $\phi^2 = 1-\phi'^2$, which, since $\phi'$ is assumed nonzero on $I$ implies that $\phi = \cos(t-t_0)$ for some constant $t_0$. We can thus set $C=0$ aside and assume that $C\not=0$.
We have $C^3{-}1\le \phi^3\le C^3{+}1$, so $C\ge -1$, otherwise there cannot be any nonnegative solutions. If $C = -1$, then $\phi\le0$ and, since we are only interested in non-negative solutions, the only solution in this case is $\phi\equiv0$, so we can assume henceforth that $C>-1$.
Both equations (with either sign) can be studied as special cases of the polynomial differential equation
$$
(\phi^3-C^3)^2 - (1 - \phi'^2)^3 = 0,
$$
so that is what we will do. Set $\phi^3-C^3 = u^3$ where $|u|\le 1$
and note that this implies $(\phi')^2 = 1-u^2$. We then have
$$
\pm 1 = \frac{u^2u'}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}},
$$
so
$$
t_1-t_0 = \pm\int_{u(t_0)}^{u(t_1)}
\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}.
$$
For $C\not=0,1$ (remember that $C>-1$), the integral
$$
\int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}
= \int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C+u)^{2/3}(C^2-Cu+u^2)^{2/3}}
$$
converges. However, when $C=1$, the denominator contains a factor of $(1+u)^{1/2+2/3} = (1+u)^{7/6}$, so the integral diverges at $u=-1$.
In the case $C=1$, we can write down a parametrization of the graph $\bigl(t,\phi(t)\bigr)$ in the form $\bigl(t(v),\phi(t(v))\bigr)$ where
$|v|<\sqrt2$ with
$$
\phi(t(v)) = \bigl(1+(1-v^2)^3\bigr)^{1/3}
\quad\text{and}\quad
t(v) = \int_0^v \frac{2(1-s^2)^2\,ds}{(2-s^2)^{7/6}(1-s^2+s^4)^{2/3}}.
$$
Note that, while $t$ is a strictly increasing function of $v$ and $|t|\to\infty$ as $|v|\to\sqrt2$, $t'(\pm1) = 0$, and $\phi$ is not
a smooth function of $t$ where $v = \pm 1$. (It is, however, continuously once differentiable there, see below.)
Meanwhile, when $C\not=1$, a solution $\phi$ exists for all time and is periodic, as $\phi$ oscillates between $(C^3{-}1)^{1/3}$ and $(C^3{+}1)^{1/3}$. The period of $\phi$ is
$$
p(C) = 2\int_{(C^3{-}1)^{1/3}}^{(C^3{+}1)^{1/3}}\frac{d\xi}{\sqrt{1-(\xi^3-C^3)^{2/3}}}
= \int_{-1}^1\frac{2u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}.
$$
Of course, $p(0) = 2\pi$ and $p(C)$ has a series expansion $2\pi\,C^{-2}+\tfrac{175}{576}\pi\,C^{-8}+\cdots$ when $C>1$.
Near a minimum at $t=t_0$, the solution has a series expansion of the form
$$
\phi(t) = (C^3{-}1)^{1/3}\left(1+\tfrac12(C^3{-}1)^{1/3}(t-t_0)^2+\tfrac1{24}(C^3+1)(C^3{-}1)^{2/3}(t-t_0)^4 + \cdots\right)
$$
Near a maximum at $t=t_1$ the solution has a series expansion of the form
$$
\phi(t) = (C^3{+}1)^{1/3}\left(1-\tfrac12(C^3{+}1)^{1/3}(t-t_1)^2-\tfrac1{24}(C^3-1)(C^3{+}1)^{2/3}(t-t_1)^4 + \cdots\right)
$$
Note that, when $C\not=0$, $\phi$ is not $C^2$ when it attains the intermediate value $C$. In fact, if $\phi(t_2) = C$
and $\phi'(t_2) = 1$, then $\phi$ has a series expansion in powers of $(t{-}t_2)^{1/3}$:
$$
\phi(t) = C +(t{-}t_2) - \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3}
- \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots.
$$
Meanwhile, if $\phi(t_3) = C$
and $\phi'(t_3) = -1$, then $\phi$ has a series expansion in powers of $(t{-}t_3)^{1/3}$:
$$
\phi(t) = C -(t{-}t_2) + \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3}
+ \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots.
$$
Finally, note that $\phi''<0$ when $C<\phi<(C^3+1)^{1/3}$ while $\phi''>0$ when $(C^3-1)^{1/3}<\phi<C$.
Hi, Robert: Thanks for your answer! However I'd like to add something: it seems when C=1, the minimum $\phi(t_0)=0$, and the integral expressing $t-t_0$ diverges (the order of integrand is $\xi^{-3/2}$). Does that mean that $t_0=-\infty$ when $C=1$? When $C\neq 1$, the integral converges as the main term becomes $(\xi-(C^3-1)^{1/3})^{-1/2}$.
@YuhangLiu: You are right, of course. I didn't look at the $C=1$ case, and I should have. I'll fix that.
|
2025-03-21T14:48:30.428550
| 2020-04-28T07:57:56 |
358733
|
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|
Stack Exchange
|
How to turn a shuffled deck of card into bits
Suppose I fairly shuffle a deck of 52 playing cards, and I want to generate some bits. You could look at each pair of cards, see which is higher or lower, and output either a 0 or 1. That's 26 independent (you can convince yourself) and fair bits, guaranteed. If you were a computer, you could encode the permutation as a number between 0 and 2^225.58, and use standard methods to get 220+ expected bits out--but I'm not a human, so I can't do that.
What's a good way to generate lots of independent, fair bits using a single shuffled deck of cards as a human? Points for both guaranteed bits and expected bits.
Sorry for the kind of "wiki" nature of this question.
asked and answered at https://crypto.stackexchange.com/q/70386
asked but not answered in 'crypto'--out of three answers, only one makes an attempt to generate randomness, and it doesn't output bits (nor is it really hand-computable in practice)
Staying with your pairs, you can get more bits that are pretty independent by using characteristics of playing cards.
Your Bit 1: 0 increasing order, 1 decreasing order
Bits 2 & 3: Suit of lower card (say 00 club, 01 diamond, 10 heart, 11 spade)
Bit 4: 0 if the lower card is even (2, 4, 6, 8, 10, Q), 1 if odd (A, 3, 5, 7, 9, J, K)*
Notes: Towards the end, knowing suit and parity data for the lower cards of previous pairs could compromise independence, so stop somewhere before 104 bits. *Concerning Bit 4, there are more odds, but the lower card is a king only when the two cards are both kings--since you probably won't be using all pairs anyway, skip such a pair.
Depending on the "memory" of the human user, you could go from the 2 permutations of pairs (increasing or decreasing) to the 6 permutations applied to triples or the 24 permutations applied to quadruples along with more complicated patterns for suits and ranks. These are all somewhat imprecise on independence, but in a hopefully small enough way that could be quantified.
|
2025-03-21T14:48:30.428708
| 2020-04-28T09:06:44 |
358735
|
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|
Stack Exchange
|
Different definitions of local complete intersection
I cannot figure out (nor find a reference) whether the following statement is true or false.
Let $X$ be an excellent projective variety enjoying the property of being a local complete intersection (each local ring is the quotient of a regular local ring by an ideal generated by a regular sequence). Then there exists an immersion $\iota : X \rightarrow Y$ such that the ambient space $Y$ is regular and $X$ is a local complete intersection inside $Y$.
By saying that $X$ is a local complete intersection inside $Y$, I mean to say that there exists a locally free sheaf $\mathscr{E}$ of rank $r$ on Y together with a global section $s \in H^0(Y, \mathscr{E})$ such that the associated Koszul complex
$$ 0 \rightarrow \wedge^r \mathscr{E}^{\vee} \rightarrow ... \rightarrow \wedge^2 \mathscr{E}^{\vee} \rightarrow \wedge^1 \mathscr{E}^{\vee} \xrightarrow{s} \mathscr{O}_Y \rightarrow 0 $$
is quasi-isomorphic to $\iota_* \mathscr{O}_X$ via the natural map.
Intuitively, this condition requires $s$ to be a regular section of $\mathscr{E}$ cutting out $X$ inside $Y$ of codimension $r$.
https://mathoverflow.net/questions/27197/local-complete-intersections-which-are-not-complete-intersections
I don't think the question you link is relevant to this one. My understanding is that the counterexample provided in the link is not a global complete intersection in affine space. However, in this question, I am free to restrict to a smaller affine neighborhood.
I'm sorry for the misunderstanding. I little more relevant: https://arxiv.org/abs/1109.4921
This is very interesting. I will have a read. Thank you!
I wasn't aware of this local obstruction. I will change my question accordingly. Thank you for pointing it out.
What do you mean, "excellent"? The local rings of a projective variety are always excellent.
Even if it defined over a ring?
"variety" usually means "scheme of finite type over a field", or some further strengthening thereof ((geometrically) integral, separated, ...). However if $X$ is projective (more generally, locally of finite type) over an excellent base then it's also excellent.
|
2025-03-21T14:48:30.428868
| 2020-04-28T09:20:39 |
358737
|
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|
Stack Exchange
|
What is the distribution of the norm of the multivariate $X \sim \mathcal{N}(\mu, \Sigma) \in \mathbb{R}^d?$
Let $X \sim \mathcal{N}(\mu, \Sigma) \in \mathbb{R}^d$ follow a multivariate normal distribution. Then what's the distribution (PDF, CDF etc.) of $X?$ When $\mu = 0, \Sigma = I_d,$ we know that $||X||\sim \chi(d),$ the chi distribution, but I don't see why that'd directly tell me anything about the distribution of $||X||$ in general cases of $\mu, \Sigma?$
If needed, you can assume that $\mu=0$ or $\Sigma$ is diagonal (not identity) or a combination of both.
For the moment, I'm more particularly interested in $\mathbb{E}||X||, var[||X||]$. References very much appreciated as well!
https://en.wikipedia.org/wiki/Generalized_chi-squared_distribution me thinks
@BrendanMcKay Thank you, but I think it gives us the distribution of $||X||^2,$ but not of $||X||,$ so it seems we need to use PDF transformation for $W:=\sqrt{V}$ to obtain the PDF for $||X||...$
|
2025-03-21T14:48:30.428981
| 2020-04-28T10:14:33 |
358740
|
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|
Stack Exchange
|
inequivalent norms
I am thinking about the following question:
Let $X$ be a Banach space, say separable, e.g., $l_p$ or $c_0$.
When can I say that there exist inequivalent complete norms on $X$?
Just to clarify what the question is: We are given some vector space $X$ and the question is whether there are two norms which make $X$ into a Banach space (i.e., a complete normed space) and these two norms are not equivalent?
if the space is finite, then I think all norms are equivalent
@MartinSleziak Yes. I am particularly interested in the case of $l_p$
By "a complete norm on $X$" you mean a norm $|\cdot|'$ that is continuous on $(X,\cdot)$ (where $|\cdot|$ is the original norm) and such that $(X,|\cdot|')$ is complete?
@YCor dosent a banach space inply the space is complete with respect to the norm
@YCor Actually, I just meant that both $(X, |\cdot|')$ and $(X, |\cdot|)$ are Banach spaces
So you're not assuming that $|\cdot|'$ is continuous? Then it's quite easy to produce plenty of inequivalent norms.
@YCor Yes. I think if it is continuous, then by open mapping theorem, they have to be equivalent. So, please, show me, for example, are there any inequivalent norms for $l_p$?
Abstractly the $l_p$ ($p \neq \infty$) and $c_0$ both $\mathbb C$-vector spaces with basis of size $2^{\aleph_0}$, so there is some linear (not bounded) isomorphism $A : l_p \to c_0$ of $\mathbb C$-vector spaces, for $x \in l_p$ define $||x||{c_0} := ||Ax||{c_0}$, then $(l_p, || \cdot ||{c_0})$ isomorphic to $(c_0, || \cdot ||{c_0})$ as Banach space . In fact, you can do the same trick with any two infinitely-dimensional separable Banach spaces.
If $X$ is a separable Banach space then it has cardinal $c$, and so does its automorphism group (group of bicontinuous linear automorphisms). But its automorphism group as vector space has cardinal $2^c$. So you get $2^c$ pairwise non-equivalent norms. I think the question is better suited for MathSE.
@kp9r4d I am sorry. I can not really see the connection between your answer and my question
@YCor Well. Could you supply a concrete example?
@YCor Alright. Anyway, thank you very much for helping
No, this proof doesn't provide any concrete example. It might be true that "every norm on $\ell^2$ is continuous" is consistent with ZF+DC.
@YCor Indeed, your statement is exactly what I am looking for, and I can not find any references for it.
Answered here: https://math.stackexchange.com/a/2339334/17929
First, let me mention the result of Mackey: if $X$ is an infinite-dimensional Banach space, then its Hamel dimension (= dimension as abstract vector space, over $\mathbf{R}$ or $\mathbf{C}$) is $\ge c=2^{\aleph_0}$, with equality if $X$ is separable.
In the separable case, $X$ has cardinal $\le c$ so the upper bound is clear. The lower bound is trivial under CH (continuum hypothesis) using Baire. In general (i.e., not assuming CH), here's a proof of $\ge$ due to Lacey (mentioned at MathSE here and here). Using Hahn-Banach, there exist sequences $(x_n)$ and $(\ell_n)$ in $X$ and $X^*$ (the topological dual) such that $\ell_m(x_n)=\delta_{m,n}$ for all $m,n$ (Kronecker's $\delta$). It follows that for every $n$, $x_n\notin\overline{\mathrm{span}\{x_m:m\neq n\}}$. Now use the existence of a family $(J_t)_t$ of cardinal $c$ of infinite subsets of $\mathbf{N}$ with pairwise finite intersection. [To obtain this, take the countable set as the set of vertices of an infinite regular rooted binary tree, and the subsets as the set of infinite geodesic rays starting from the root.] Then the family $(y_t)$ defined by $y_t=\sum_{n\in J_t}2^{-n}x_n$ is linearly independent by an easy argument.
(Of course in $\ell^p$ or $c_0$ one can produce these elements without using Hahn-Banach, namely $x_n$ can be chosen as Dirac at $n$.)
Next, the assertion is that for a vector space $V$ of infinite dimension $\alpha$ over a field $K$ of cardinal $\beta$ with $\beta\le 2^\alpha$, the cardinal of $\mathrm{GL}(V)$ is $\beta^\alpha$, and in particular equals $2^\alpha$ if $\beta\le 2^\alpha$ (which is automatic if the field has cardinal $\beta\le c$).
For the latter fact, just use that $\beta^\alpha\le (2^\alpha)^\alpha=2^{\alpha\times\alpha}=2^\alpha$.
For the upper bound, do it for $\mathrm{End}(V)$, which is in bijection with the set of maps $\alpha\to V$, i.e., has cardinal $\beta^\alpha$.
For the lower bound, splitting a basis, write $V=W\oplus W$ with $W$ isomorphic to $V$. Then every $u\in\mathrm{End}(W)$ induces $f_u:(x,y)\mapsto (x+u(y),y)$, with $f\in\mathrm{GL}(V)$, these provide $\beta^\alpha$ automorphisms. (Alternatively we already get $2^\alpha$ automorphisms by permuting basis elements— say gather elements by pairs and for any subset of this set of $\alpha$ pairs transpose them and leave identity elsewhere.)
Combining, we obtain that for $X$ infinite-dimensional separable Banach space, the automorphism group $G_X$ of $X$ as abstract vector space (over $\mathbf{R}$ or $\mathbf{C}$) has cardinal $2^c$.
If $H_X\subset G_X$ is the subgroup of (bi)continuous automorphisms then $H_X$ has cardinal $\le c$ and hence $G_X/H_X$ has cardinal $2^c$, so the $g\cdot (\|\cdot\|)$, when $g$ ranges over $G_X/H_X$, yield $2^c$ pairwise non-equivalent norms on $X$.
One can elaborate using cardinal invariants about the non-separable case, but for free we get $2^c$ non-equivalent norms in $X$ without assumption: choose an infinite-dimensional separable subspace $Y$, and define $G$ as the group of abstract linear automorphisms of $X$ that preserve $Y$, and $H$ its subgroup of bicontinuous automorphisms: then looking at the action on $Y$ shows that $H$ has index $\ge 2^c$ in $G$, and again the $g\cdot (\|\cdot\|)$ for $g$ ranging over $G/H$ are pairwise non-equivalent norms on $X$ (they're non-equivalent since they're non-equivalent in restriction to $Y$).
Note: proving that there's a non-equivalent norm is easier than all this (no need to use the $2^{\aleph_0}$ fact). Namely, if $X$ is an infinite-dim normed space, find a basis $(v_i)_{i\in I}$ contained in the 1-sphere. For every family $(t_i)$ of nonzero scalars, $v_i\mapsto t_iv_i$ defines a linear automorphism. Choosing $(t_i)$ unbounded yields a non-continuous automorphism. Choosing different families easily yields a group of $c$ such automorphisms, each unbounded except identity, and hence this yields $c$ pairwise non-equivalent norms (but $2^c$ is hopeless if the dimension is countable).
Am I confused? The Hamel dimension of any infinite dimensional separable Banach space is $2^{\aleph_0}$, so they are all the same as vector spaces. So the underlying vector space of one of them can be the underlying vector space of any of them.
Actually, I see that kp9r4d already said this in the comments.
I actually provided a reference and a proof in my answer.
@YCor you're right, I should have read your answer before writing mine.
|
2025-03-21T14:48:30.429534
| 2020-04-28T10:18:19 |
358741
|
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|
Stack Exchange
|
Transposition Cayley graphs are planar
Consider the Cayley graph $G$ with vertex set the elements of the symmetric group $S_n$ and generating set the set of minimal transposition generators of the group $S_n$, that is the set $S=\{(12),(13),\ldots, (1n)\}$. Then, will the graph be always planar?
The graph is easily seen to be bipartite with one part the set of odd permutations and the other part the set of even permutations. Upto the group $S_3$ it is trivial to see why $G$ is planar. The first nontrivial part is when the group is $S_4$. The degree is $3$ in this case, but I don't see the complete regular bipartite graph $K_{3,3}$ explicitly. But can $G$ have a $K_{3,3}$ minor? I don't think so. Also, what if I replace the set $S$ with the set of all transpositions? Any hints? thanks beforehand.
You already have an answer regarding the first part of your question, but this uses the fact that with your given generating set, the Cayley graph is $(n-1)$-regular. What if we pick the generating set $\{ (12), (12\cdots n)\}$?
So: there is a full characterisation, due to Maschke (1896), of which finite groups admit planar Cayley graphs with respect to some generating set. Such groups are called planar.
The finite group $A$ is planar if and only if $A = B_1 \times B_2$ with $B_1 = 1$ or $\mathbb{Z}_2$, and $B_2 \in \{ \mathbb{Z}_n, D_n, S_4, A_4, A_5 \mid n \in \mathbb{N} \}$.
A proof can be found e.g. here, and the original article is the (aptly) named [Maschke, H.
The Representation of Finite Groups, Especially of the Rotation Groups of the Regular Bodies of Three-and Four-Dimensional Space, by Cayley's Color Diagrams.
Amer. J. Math. 18 (1896), no. 2, 156–194.]. If you'd like to read it, a stable link is here.
In particular, $S_n$ for $n \geq 5$ admits no generating set such that its Cayley graph is planar, answering the second part of your question.
thanks! just a small miss. The graphs have to connected right. otherwise we could have lot of counterexamples
@vidyarthi A Cayley graph is always connected, yes. What counterexamples do you have in mind?
why? just take $S_5$ with generating set $S={(12)}$. I mean, if the generating set of the graph is not a generating set of the group, then we have a disconnected graph
@vidyarthi If your set $S$ does not generate the group, then the graph is not a Cayley graph. This is part of the definition of a Cayley graph.
oh! I am used to thinking that the set of generator of the graph need not generate the group. Anyway, thanks for pointing out that
but still, consider the graph for $S_n$ with respect to ${(12),(123\ldots n)}$. Now, since degree is $2$, is it not a cycle and hence, planar?
ok, got it! the generating set should be symmetric. So the inverse of $(12\ldots n)$ will be $(1\ n\ n-1\ldots 2)$, whence the degree is $3$, right?
@vidyarthi The assumption that the generating set should be symmetric is optional (though always assumed in geometric group theory). If it needs to be symmetric, then yes, we need to include $(123\dots n)^{-1})$ as well in $S$, just as you say.
Hello after a long time! I just read the mentioned reference now. As I see, the paper is quite old and uses archaic terms. Anyways, I do not get the specific charecterisation of which groups admit a Cayley diagram which can be embedded in the sphere from the paper explicitly. Could you mention which theorem specifically mentions this? Is it the ones concerning the regular color groups?
@vidyarthi Yes, Mashke calls "planar groups" as "regular groups". A colour group diagram is a Cayley graph. The relevant theorem is summarised at the very beginning of §5. The Cayley graphs are drawn out at the very end of the paper (Figs I-X). Note that a graph is planar if and only if it can be embedded on the surface of a sphere.
Isn't this list exactly the finite subgroups of $O(3)$? It's tempting to try to come up with a uniform proof...
Your graph $G$ will not be planar for all $n \geq 5$. To see this, we use the well-known fact that every bipartite planar graph $H$ satisfies $|E(H)| \leq 2|V(H)|-4$, and hence has average degree less than $4$. Since your Cayley graph $G$ is $(n-1)$-regular and bipartite, it is not planar for all $n \geq 5$.
thanks, can it be said to $k-$ planar, for some $k$?
No, if $k$ is fixed, $k$-planar graphs also have bounded average degree, $O(\sqrt{k})$ if I recall correctly.
|
2025-03-21T14:48:30.429855
| 2020-04-28T11:32:18 |
358745
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brendan McKay",
"J.J. Green",
"LSpice",
"https://mathoverflow.net/users/157117",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/5734",
"https://mathoverflow.net/users/9025",
"leo_bouts"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358745"
}
|
Stack Exchange
|
Find occurrences of certain matrix inside a matrix
This problem occurred from my need to find all graphs with a certain topology inside a bigger one. I don't need the subgraphs but the graphs that have the exact topology I am searching.
We know for a fact that we can represent graph networks with a Laplacian matrix. If I wanted to check the entire graph it would take $n^2 \times n$ for the diagonal and $2^n$ for checking the row and column of each element.
Is there a better way to do this? Either by using matrices or graph theory?
For example below i want to find all the upper left matrices inside a big matrix and also check that the rest of their columns and rows are zero.
$$\begin{pmatrix}
2 & -1 & 0 & 0 & \cdots & 0 \\
-1 & 3 & -2 & 0 & \cdots & 0 \\
0 & -2 & 3 & -1 & \cdots & 0 \\
0 & 0 & 0 & 0 & \cdots & n \\
\vdots & \vdots & \vdots & \vdots & \ddots & n \\
0 & 0 & 0 & 0 & \cdots & n
\end{pmatrix}$$
Your description is inadequate. What exactly are you looking for?
@BrendanMcKay added an example
So given graphs $G$ and $H$, you're looking to see if you can find all subgraphs of $G$ which are isomorphic to $H$? Finding one is NP-complete I'm afraid.
I tried to turn your picture https://i.sstatic.net/kvGUD.png into TeX, which we usually encourage for searchability. However, I couldn't figure out what was supposed to be happening at the bottom of the picture. Please feel free to edit further if I got it wrong.
|
2025-03-21T14:48:30.429994
| 2020-04-28T12:24:45 |
358749
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Tony419",
"Willie Wong",
"https://mathoverflow.net/users/157356",
"https://mathoverflow.net/users/35593",
"https://mathoverflow.net/users/3948",
"https://mathoverflow.net/users/86198",
"u184",
"user35593"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358749"
}
|
Stack Exchange
|
A second order nonlinear ODE
In my research (in differential geometry) I recently came across the following nonlinear second order ode:
$$\frac{f''(x)}{f'(x)}-\frac{2}{x}+\frac{f'(x)+1}{2f(x)-x-1}+\frac{f'(x)-1}{2f(x)+x}=0$$
It actually arose from the symmetry reduction of some pde. I know from an analysis of the equation that there exists a 1-parameter family of solutions. Moreover I also know two explicit solutions;
$$f(x)=x+\frac{1}{2}$$
$$f(x)=\frac{1}{4}+\frac{1}{4}(1+3x)\sqrt{(1+2x)}$$
The existence of these 2 solutions, expressible in terms of elementary functions, makes me wonder if one can in fact find more (if not all) explicit solutions to this ode. Note that both these solutions are well defined at $x=0$, although the ode itself is singular at that point! It is not too hard to show that any solution well-defined at $x=0$ requires $f(0)=\frac{1}{2}$ and $f'(0)=1$.
As far as I am aware there are no standard tricks for these type of fully nonlinear odes. I have been trying to simplify the ode by various substitutions but without any success.
I was hoping that someone might be able to spot a clever transformation, or even argue that it is impossible to find any other explicit solutions. I would also be interested to know of any references where such a class of ode that might have been studied.
This also leads me to ask if there is any general theory known about when can a solution to an ode (say of second order) be expressed in terms of elementary functions, or is it just a case-by-case study? Thanks!
It appears that $f(x) = -x$ is another solution. Note that this solution is also "globally regular" but does not satisfy your condition that $f(0) = 1/2$ and $f'(0) = 1$.
Another solution is $$ f(x) = \frac14 - \frac14(1+3x) \sqrt{1+2x} $$ Note: my two solutions are obtained from yours by the symmetry transformation $$ f(x) \mapsto 1/2 - f$$
@WillieWong Thanks for pointing out that $\mathbb{Z}_2$ symmetry
If you use the ansatz $f(x)=a_0+a_1x+a_2 x^2 \dots$, put $a_0=1/2$, $a_1=1$ and multiply everything out you get that $a_2$ can be chosen arbitrary whereas then $a_3=-2a_2/5$, $a_4=a_3a_2+3/5$. You can define a formal power series by iteratively solving for the next coefficient. However I am not sure if that power series converges for every $a_2$. The solution with the square root seems to be the case $a_2=5/8$. I am not sure if the other power series can be simplified for other $a_2$'s.
it should be $a_4=a_3a_2+3/2a_3$ above.
If we multiply everthing out we get $4xf^2f''-8f^2f'+8xff'^2-2x^2ff''-xf'^2-(x^3+x^2)f''+(4x^2+3x)f'=0$
Did you check in Kamke E. -Differentialgleichungen Lösungsmethoden und Lösungen? In Chapter XVI (in my 1977 edition) there are solutions to some second order non-linear ODEs.
This ODE has some very interesting properties. If one clears fractions and writes it out as
$$
x(x+2y)(x-2y+1)\,y'' = (4x^2-8y^2+3x+4y)\,y' + x(4y-1)\,(y')^2,
\tag1
$$
one recognizes this as the equation for geodesics of a projective connection in the complement $D$ (which has $7$ components) of the three lines $x = 0$, $x+2y=0$, and $x-2y+1=0$ in the $xy$-plane. Moreover, because the right hand side of (1) has no terms of degree $0$ or $3$ in $y'$, it follows that the lines $x=x_0$ and $y=y_0$ are geodesics of this projective connection in $D$ and should be regarded as 'solutions' of the equation. This is probably most evident if one writes the equation in parametric form for a curve $\bigl(x(t),y(t)\bigr)$, in which case, the equation becomes
$$
x(x{+}2y)(x{-}2y{+}1)\,\bigl(\dot y\,\ddot x-\dot x\,\ddot y\bigr) + x(4y{-}1)\,\dot x\,\dot y^2 + (4x^2{-}8y^2{+}3x{+}4y)\,\dot x^2\,\dot y =0
\tag2
$$
Contrary to what the OP claims, there is a $2$-parameter family of solutions that are regular at $x=0$. If one looks for a formal power series solution in the form
$$
y(x) = a_0 + a_1\,x + a_2\,x^2 + a_3\, x^3 + \cdots,\tag3
$$
then one finds, by examining the three lowest terms in the equation, that one must have either
$$
(i)\ \ a_1 = a_2 = 0,\qquad (ii)\ \ a_0 = 0,\ a_1 = -1,
\quad\text{or}\quad (iii)\ \ a_0=\tfrac12,\ a_1 = 1.
$$
Since $y(x)$ is a solution if and only if $\tfrac12 - y(x)$ is a solution, the second and third cases are essentially the same, so I will treat only the first two cases henceforth.
In the first case, one finds that there is a formal power series solution of the form
$$
y(x) = \tfrac14(1{+}a) + \frac{(a^2{-}1)b}{12} x^3 -\frac{b}{4}\,x^4 - \frac{b}{5}\,x^5 - \frac{a(a^2{-}1)b^2}{72}\,x^6 + \cdots + p_k(a,b)\,x^k + \cdots,\tag4
$$
where $p_k(a,b)$ is a (unique) polynomial in constants $a$ and $b$. Moreover, this series has a positive radius of convergence for each $(a,b)$. [Proofs that these and similar series listed below have positive radii of convergence can be based on techniques in the book Singular Nonlinear Partial Differential Equations by R. Gérard and H. Tahara.] Note that the symmetry $y(x)\mapsto \tfrac12 - y(x)$ corresponds to the symmetry $(a,b)\mapsto (-a,-b)$.
In the second case (and, similarly, via the symmetry $y\mapsto \tfrac12 - y$, the third case), one finds that there is a formal power series solution
$$
y(x) = - x + \frac{b}{2}\,x^2 - \frac{b}{5}\,x^3 + \frac{b(b{+}3)}{10}\,x^4
- \frac{4b(13b{+}25)}{175}\,x^5 + \cdots + q_k(b)\,x^k + \cdots,
\tag5
$$
where $q_k(b)$ is a (unique) polynomial in $b$ of degree at most $\tfrac12 k$. This series has a positive radius of convergence for every $b$. The value $b=0$ gives the solution $y(x) = -x$ and the value $b=-5/4$ gives the solution $y(x) = \tfrac14 - \tfrac14(1+3x)(1+2x)^{1/2}$. Note that (4) with $a=-1$ and (5) give two distinct $1$-parameter families of solutions passing through the point $(x,y)=(0,0)$, where the two singular lines $x=0$ and $x+2y=0$ meet.
As for analytic solutions meeting the singular line $x+2y=0$, there exist two distinct $2$-parameter families of series solutions: The first is given in parametric form by
a formal power series
$$
\begin{aligned}
x(t) &= a + a(2a{+}1)\,t\,,\\
y(t) &= -\frac{a}{2} + a(2a{+}1)b\,t^2\left(1 + \frac{2(5a{-}4b{+}2)}{3}\,t +\cdots + p_k(a,b)\,t^k + \cdots \right),
\end{aligned}
\tag6
$$
where $p_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. The second $2$-parameter family can be written in the form
$$
\begin{aligned}
x(t) &= a + a(2a{+}1)\,t^3\,,\\
y(t) &= -\frac{a}{2} + (2a{+}1)\,t^2\left(b + a\,t + \frac{2b^2}{5}\,t^2 +\cdots + q_k(a,b)\,t^k + \cdots \right),
\end{aligned}
\tag7
$$
where $q_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. Interestingly, these solutions with $b\not=0$ have a cusp singularity at $t=0$, while, when $b=0$, only the terms involving $t^{3k}$ remain, so that $x(t)$ and $y(t)$ are analytic functions of $t^3$.
Note, however, that these two series solutions degenerate at the special values $a=0$ and $a=-\tfrac12$. The value $a = 0$ corresponds to the point $(x,y)=(0,0)$, where the singular lines $x=0$ and $x+2y=0$ cross, while the value $a=-\tfrac12$ corresponds to the point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross.
Finally, through the singular point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross, there are two convergent series solutions with one parameter: The first is
$$
\begin{aligned}
x(t) &= -\frac{1}{2} + t\,,\\
y(t) &= +\frac{1}{4} + b\,t^3 -3b\,t^4 + \cdots + f_k(b)\,t^k + \cdots ,
\end{aligned}
\tag8
$$
where $f_k(b) = -f_k(-b)$ is a polynomial in $b$. The second series is
$$
\begin{aligned}
x(t) &= -\frac{1}{2} + b\,t^2 + \frac{b^2(5b{+}32)}{16}\,t^4 + \cdots + g_k(b)\,t^{2k} + \cdots\,,\\
y(t) &= +\frac{1}{4} + t\, ,
\end{aligned}
\tag9
$$
where $g_k(b)$ is a polynomial in $b$. Note that the value $b=2$ in this latter series corresponds to the known solution(s) represented by solving $(y-\tfrac14)^2-(1+3x)^2(1+2x) = 0$ for $y$ as a function of $x$.
It is very interesting that through every point of the $xy$-plane, there passes at least one $1$-parameter family of solution curves, and, along two of the singular lines, there can be two distinct $1$-parameter families of solution curves.
One more interesting feature should be noted: Because the equation defines a projective structure on $D$, each solution curve in $D$ comes equipped with a canonical projective structure, i.e., a 'developing map' to $\mathbb{RP}^1$ that is unique up to linear fractional transformation and provides a local parametrization of the curve. This developing map extends analytically across the points where such a curve crosses one of the three singular lines, but the developing map is no longer a local diffeomorphism at such places; its differential vanishes to second or third order at such places.
Not a solution just the octave code related to the comment above
function nonlinear_ode
%number of new coefficients (starting at a_3)
n=4;
%initial guess
x0=rand(1,n);
%values of a_2 that are considered
as=-5:5
x=zeros(length(as),n);
for i=1:length(as)
x(i,:)=fsolve(@(x) fun(x,as(i)),x0);
end
for i=1:n
polyfit(as,x(:,i)',n)
end
end
function y = fun(x,a)
f=[1/2 1 a x];
df=f(2:end).*(1:length(f)-1);
d2f=df(2:end).*(1:length(df)-1);
%nominators
n{1}=d2f;
n{2}=-2;
n{3}=df;
n{3}(1)=n{3}(1)+1;
n{4}=df;
n{4}(1)=n{4}(1)-1;
%denominators
d{1}=df;
d{2}=[0 1];
d{3}=2*f;
d{3}(1:2)=d{3}(1:2)-[1 1];
d{4}=2*f;
d{4}(2)=d{4}(2)+1;
for k=1:4
yk=n{k};
for j=1:4
if j~=k
yk=conv(yk,d{j});
end
end
if k==1
y=yk;
else
y=y+yk;
end
end
%restrict to first terms
y=y(1:length(f));
end
Output
ans =
-1.1592e-18 -1.0962e-17 6.4646e-17 -4.0000e-01 -3.0447e-16
ans =
6.2764e-17 2.5432e-17 -4.0000e-01 6.0000e-01 3.3331e-15
ans =
-1.2474e-09 1.8926e-09 1.1886e+00 -1.1429e+00 -4.3242e-08
ans =
-2.8979e-09 5.0286e-01 -3.3600e+00 2.4286e+00 -1.2213e-07
|
2025-03-21T14:48:30.430538
| 2020-04-28T12:35:21 |
358750
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Praphulla Koushik",
"Robert Bryant",
"https://mathoverflow.net/users/118688",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358750"
}
|
Stack Exchange
|
Realization of a $\mathfrak{g}$-valued $2$-form as a curvature form
Consider the Lie group $S^1$. Recall that the associated Lie algebra is $\mathbb{R}$.
Let $M$ be a manifold. Consider the second de-Rham cohomology group $H^2(M,\mathbb{R})$.
Let $\Omega\in H^2(M,\mathbb{R})$ be an integral cohomology class; that is, it is image of an element of $H^2(M,\mathbb{Z})$ under the natural map $H^2(M,\mathbb{Z})\rightarrow H^2(M,\mathbb{R})$. Then, we know that we can construct a principal $S^1$-bundle $P\rightarrow M$ over the manifold $M$, a connection $1$-form $\omega$ on the manifold $P$ such that, the associated curvature form (which is a $2$-form on $P$, but can be projected uniquely to a $2$-form on $M$) is precisely the $2$-form $\Omega$ that we have started with.
Question : How far can we go relaxing the condition that the structure group is abelian?
Question : Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. Let $M$ be a manifold and $\Omega$ be a $\mathfrak{g}$-valued $2$-form on $M$. Under what conditions, can we find a principal $G$ bundle over the manifold $M$, and a connection on $P(M,G)$ whose curvature is $\Omega$?
The case of integral cohomology class is called quantization problem. So, I am calling this quantization problem for non abelian groups.. Please correct me if I misunderstood the name or anything in the post above..
As a start, you might want to have a look at this MO question: https://mathoverflow.net/questions/73439/when-is-a-given-matrix-of-two-forms-a-curvature-form/73451#73451
@RobertBryant Sir, Yes, I saw that now. I did not follow all calculations but I got the idea :) Now, I am thinking of what conditions I should add to expect a positive answer...
|
2025-03-21T14:48:30.430683
| 2020-04-28T13:14:39 |
358754
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Igor Belegradek",
"Kafka91",
"Kevin Casto",
"Moishe Kohan",
"https://mathoverflow.net/users/147200",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/39654",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358754"
}
|
Stack Exchange
|
Difference between the diffeomorphism classification of a manifold $M$ and the set of equivalences of homotopy smoothings $hS(M)$
In Lopez de Medrano "Involutions on manifolds", a homotopy smoothing of a Poincaré space $X$ is a homotopy equivalence $f:M^n\rightarrow X$, where $M^n$ is a smooth $n$-dim. manifold (everything is oriented and orientation-preserving). Two homotopy smoothings $f_i:M_i^n\rightarrow X$, $i=0,1$, are equivalent if there exists a diffeomorphism $\phi:M^n_0\rightarrow M^n_1$ such that $f_1\circ \phi\simeq f_0$. Denote by $hS(X)$ the set of equivalence classes of homotopy smoothings on $X$.
Now suppose $X$ is already a smooth, oriented $n$-dim. manifold. Under which conditions does $hS(X)$ actually correspond to a diffeomorphism classification, i.e. when is the homotopy condition $f_1\circ \phi\simeq f_0$ always satisfied?
Do you know a space $X$ where $hS(X)$ does not correspond to the diffeomorphism classification?
You are effectively asking for a smooth manifold $M$ such that there exists a self-homotopy-equivalence $M\to M$ which is not homotopic to a diffeomorphism. If one allows for noncompact $M$, then such examples exist already among surfaces (say, the triply-punctured 2-sphere). If you want compact examples, see my answer here.
One approach to the diffeomorphism classification of closed manifolds in a given homotopy type is to compute the action of the monoid of homotopy self-equivalences on the structure set. You can find a discussion of these matters in https://arxiv.org/abs/0912.4874. For example in section 10 you can find a homotopy self-equivalence of $S^7\times CP^3$ with nontrivial invariant. In Remark 5.4 we discuss a self-homotopy equivalence with trivial normal invariant that is not homotopic to a diffeomorphism.
It's probably worth mentioning Mostow rigidity (hyperbolic manifolds are smoothly rigid, and in fact hot-equiv implies isometric, for dim $\ge 3$) and the Borel conjecture (that aspherical manifolds are topologically rigid).
Assuming that $X$ is a smooth manifold, your question can be reformulated as: Under which conditions every self-homotopy-equivalence $X\to X$ is homotopic to a diffeomorphism?
I will say that $X$ satisfying this property is smoothly rigid. I will say that $X$ is rigid if every self-homotopy equivalence is homotopic to a homeomorphism.
Here are some positive and negative answers:
In dimensions 2 and 3 there is no difference between smooth rigidity and rigidity since TOP=DIFF in these dimensions.
If $X$ is 2-dimensional and closed (compact and has empty boundary) then $X$ is rigid. Ditto for the case when $X$ is 2-dimensional and has abelian fundamental group. However, if $X$ is noncompact, oriented connected, has nonebelian fundamental group (say, $X$ is the triply punctured sphere) and is different from the once punctured torus, then $X$ is not rigid.
For this reason, I will restrict to closed manifolds.
There are non-rigid 3-manifolds, say, lens spaces. David Gabai gives $L(8,1)$ as an example:
Gabai, David, On the geometric and topological rigidity of hyperbolic 3-manifolds, Bull. Am. Math. Soc., New Ser. 31, No. 2, 228-232 (1994).
Nevertheless, closed aspherical 3-manifolds are known to be rigid. (This is due to many people, starting with Waldhausen and concluding with Perelman.)
Starting from dimension 4, there are examples of homeomorphisms which are not homotopic to diffeomorphisms. See for instance here for some examples of 4-manifolds as well as among exotic 7-dimensional spheres.
Thanks for your answer! I am actually looking for conditions on $X$ in dimension greater than 4 such that $hS(X)$ corresponds to the different diffeomorphism types. For example, how could I determine if $hS(\mathbb{R}P^n)$ gives the diffeomorphism classification?
@Kafka91: The only way to interpret your question "such that ℎ() corresponds to the different diffeomorphism types" is that you are asking for (sufficient or necessary?) conditions on a (smooth manifold) $X$ such that $hS(X)$ is (naturally) bijective to the set of diffeomorphism classes of manifolds homotopy-equivalent to $X$. As I said in my answer, this happens if and only if every self-homotopy-equivalence $X\to X$ is homotopic to a diffeomorphism. As for the specific question about projective spaces, I do not know, but you should update your question.
@Kafka91: the group of homotopy self-equivalences of real projective spaces is computed in Corollary 6 of "Coverings of fibrations" Becker and Gottlieb,
Compositio Mathematica, Volume 26 (1973) no. 2, p. 119-128, http://www.numdam.org/item/CM_1973__26_2_119_0/. It is trivial in even dimensions and $\mathbb Z_2$ in odd dimensions $>1$.
@IgorBelegradek Thank you very much!
|
2025-03-21T14:48:30.430967
| 2020-04-28T13:16:08 |
358755
|
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"authors": [
"Asvin",
"Ben Smith",
"abx",
"https://mathoverflow.net/users/156215",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/58001"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358755"
}
|
Stack Exchange
|
A map on Jacobians coming from a correspondence explicitly
From this question, we know that every map of the form $J(C) \to J(C)$ for a curve $C$ and it's jacobian $J(C)$ comes from a correspondence between $C\times C$ and in fact we can take this correspondence as a divisor on $C\times C$.
Now, let us take a map $f: C\to D$ of curves and consider the induced map $\pi: J(C) \to J(D) \to J(C)$ which is a projection onto a factor. What correspondence does it correspond to, or more precisely, what is the divisor $X \subset C\times C$ with maps $\alpha,\beta: X \to C$ such that $\pi = \beta_*\alpha^*$?
$X=$ the inverse image of the diagonal in $D\times D$ by the map $(f,f):C\times C\rightarrow D\times D$.
Thanks, I was thinking about that too.
Note that $X$ is not unique: you can add fibres of the projections and principal divisors on $C\times C$ without changing the induced $\pi$. But @abx's $X$ is a good choice.
|
2025-03-21T14:48:30.431193
| 2020-04-28T13:17:00 |
358756
|
{
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"authors": [
"Dave L Renfro",
"Piotr Hajlasz",
"YCor",
"https://mathoverflow.net/users/100231",
"https://mathoverflow.net/users/121665",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/15780",
"https://mathoverflow.net/users/75761",
"vidyarthi",
"wlad"
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"url": "https://mathoverflow.net/questions/358756"
}
|
Stack Exchange
|
Do infinitely nested radicals have any applications?
There is a simple necessary and sufficient condition for a continued radical of the form $\sqrt{a_1 + \sqrt{a_2 + \dotsc}}$ to converge (where all terms $a_1, a_2$ etc. are nonnegative). Namely, that the sequence $n \mapsto a_n^{2^{-n}}$ should be bounded. This is known as Herschfeld's Convergence Theorem (though it was discovered independently of Herschfeld by Paul Wiernsberger thirty years before).
Lately, I've developed a constructive proof of this theorem, which you can find here.
It leads me to wonder: Is there any actual use for continued square roots, apart from studying them for their own sake? Googling doesn't show up anything.
may be, it has some use in galois theory, typically kummer theory, or serves as an example for the applications of galois theory
It might be useful to look through the 98 page (unpublished?) manuscript A chronology of continued square roots and other continued compositions, through the year 2016 by Dixon J. Jones.
Here's a paper (behind paywall, sorry: JSTOR still earns money on 45 year old papers) on the case of complex iterated radicals.
If it was discovered 30 years before Herschfeld, would you mention by who?
The first application goes back to Archimedes. Let me explain how.
$$
\underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n}\to 2.
$$
The question is how fast. It turns out that the rate of convergence can be caught quite precisely from the formula:
$$
2^{n+2}\cdot\sqrt{2-{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}}\to 2\pi.
$$
This formula has an easy geometric interpretation: the expression on the left hand side equals the circumference of the regular $2^{n+2}$-gon inscribed in the unit circle. In fact, Archimedes used this approach to find the approximate value of $\pi$. The method was mastered later by Ludolph van Ceulen who published in 1596 the first 20-decimals of $\pi$.
The method from my paper gives that $2 - \underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n} \leq 2^{2^{-n}}(2^{2^{-n}} - 1)$, but this is quite pessimistic
@ogogmad I think you should provide coordinates for your paper in the statement of the question. Otherwise nobody knows what you are talking about.
I'm not sure if it's clear, but the general idea is described in the section Overview and Strategy...
The following is not an answer.
In a comment to @PiotrHajlasz's answer, I say that a method from my paper shows that $2-\underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n} \leq 2^{2^{-n}}(2^{2^{-n}} - 1)$. You can find a description of this in the subsection Overview and Strategy of my proof. But in case it's not clear, I describe how to do this below:
The general idea is that given $U \geq L$ and a non-negative sequence $(a_i)_i$, we have that $$\begin{align}
&\sqrt{a_1 + \sqrt{a_2 + \dotsb \sqrt{a_{n-1} + \sqrt{U}}}} - \sqrt{a_1 + \sqrt{a_2 + \dotsb \sqrt{a_{n-1} + \sqrt{L}}}} \leq \sqrt{0 + \sqrt{0 + \dotsc\sqrt{0 + \sqrt U}}} - \sqrt{0 + \sqrt{0 + \dotsc\sqrt{0 + \sqrt L}}} = U^{1/2^n} - L^{1/2^n}.
\end{align}$$
In other words, we can upper bound the difference between an upper bound and lower bound by driving the terms $a_i$ down to $0$.
In the case of $\sqrt{2 + \sqrt{2 + \dotsb}}$, we have an upper bound in the form of $\sqrt{2 + \sqrt{2 + \dotsb \sqrt{2 + \sqrt{\color{red} 4}}}}=2$, and a lower bound in the form of $\sqrt{2 + \sqrt{2 + \dotsb \sqrt{2 + \sqrt{\color{red}
2}}}}$. The difference between upper and lower bound can thus be increased to $4^{1/2^n} - 2^{1/2^n} = 2^{1/2^n} (2^{1/2^n} - 1)$, which clearly goes to zero.
|
2025-03-21T14:48:30.431456
| 2020-04-28T13:21:19 |
358757
|
{
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"0xbadf00d",
"M. Dus",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358757"
}
|
Stack Exchange
|
A subadditive maximal ergodic theorem
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\tau:\Omega\to\Omega$ be a measurable map on $(\Omega,\mathcal A)$ with $\operatorname P\circ\:\tau^{-1}=\operatorname P$, $Y_n:\Omega\to[-\infty,\infty)$ be $\mathcal E$-measurable for $n\in\mathbb N$ with $\operatorname E\left[Y_1^+\right]<\infty$ and $$Y_{m+n}\le Y_m+Y_n\circ\tau^m\;\;\;\text{for all }m,n\in\mathbb N\tag1$$ and $$M_n:=\max(Y_1,\ldots,Y_n)\;\;\;\text{for }n\in\mathbb N.$$
It's easy to show the following extension of the maximal ergodic theorem: $$\operatorname E[Y_1;M_n\ge0]\ge0\;\;\;\text{for all }n\in\mathbb N.\tag2$$
The ordinary maximal ergodic theorem is given by the special case, where $$Y_n=\sum_{i=0}^{n-1}X\circ\tau^i\;\;\;\text{for all }n\in\mathbb N$$ for some integrable real-valued random variable on $(\Omega,\mathcal A,\operatorname P)$. In that special case, it can be deduced from $(2)$ that $$\operatorname P\left[\sup_{n\in\mathbb N}\left|\frac{Y_n}n\right|\ge c\right]\le\frac1c\operatorname E[|Y_1|]\;\;\;\text{for all }c>0\tag3.$$
Can we extend this result to the general case?
After a bit of searching, I found the answer to my question in comment. So it is true that (2) holds in general and it is proved in literature that (3) holds for the special case where $Y_n=\sum_k X\circ \tau ^k$ indeed, except that there is a mistake in your formulation : you have to remove the absolute values (and this is very important). This is actually written without the absolute values on the wikipedia page you're refering too. Also, if you edit the question, you should add the tag ergodic theory. I answer below your question (positively) with the correct formulation.
@M.Dus I'm sorry for my late response. What I mean is that $(2)$ and $(3)$ (with the absolute value) are known to be true when $(Y_n)$ is additive. It's easy to see (by more or less precisely the same proof as in the additive case) that $(2)$ remains to hold true when $(Y_n)$ is subadditive, but it's not clear to me how $(3)$ generalizes and how this generalization can be proven.
Could you provide a proof for (3) with the absolute values then ? I'm currently writing a proof as an answer for (3) without them in the general case.
see, for example, Lemma 2.6 (2) on p. 34 in https://wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Andreas_Eberle/MarkovProcesses1920/MarkovProcesses1920.pdf. (Note that you need to take $Y_n=nA_nF$.)
Right thank you ! I have a (simple) proof for the statement without the absolute values, which seems to be the important result in literature, but I don't know about yours. Do you still want me to write it up ?
Yes, sure. Maybe we can built up on that.
The answer is no in general, but yes if the sequence $Y_n$ is non-negative.
First, let us focus on the case where $Y_n$ is non-positive. Then, $\sup \frac{1}{n}|Y_n|=-\inf \frac{1}{n}Y_n$. If you assume moreover that all the $Y_n$ are $L^1$, then by Kingman's subadditive theorem, $\frac{1}{n}Y_n$ converges to $Y=\inf \frac{1}{n}Y_n$ almost surely. Notice then that $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=\mathbb{P}\left (Y\leq -c\right )$. Whenever $Y$ has positive probability of taking the value $-\infty$, you cannot bound $\mathbb{P}\left (Y\leq -c\right )$ by something converging to 0 as $c$ goes to infinity.
Here is a concrete counter-example. Let $Y_n$ be the constant function $Y_n=-n^2$. Then $Y_n$ is subadditive and satisifes all your assumptions. You have $\sup \frac{1}{n}|Y_n|=+\infty$ and so for any $c$, $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=1$, so you don't have $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|)=\frac{1}{c}$.
However, small remark : the answer is yes for non-positive $Y_n$ if you have the property that $\mathbb{E}(\frac{1}{n}|Y_n|)\leq \mathbb{E}(|Y_1|)$. Indeed, using Markov inequality, you get $\mathbb{P}\left (\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(\frac{1}{n}|Y_n|)\leq \frac{1}{c}\mathbb{E}(|Y_1|)$ and this is true for all $n$, so this is true for the almost sure limit, using dominated convergence.
About the non-negative case now. Inequality (3) is usually stated without the absolute values in literature :
$$\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|).$$
This statement is true in general and so in particular, the answer to your question is yes whenever $Y_n$ is non-negative.
Indeed, consider a subbaditive sequence $Y_n$, that is satisfying your condition $Y_{n+m}\leq Y_m+Y_n\circ \tau^m$. Let $Z_n=\sup_{k=1,...,n}\frac{1}{k}Y_k$. Also let $\widetilde{Y}_n=\sum_{j=0}^{n-1}Y_1\circ \tau^j$ and finally, let $\widetilde{Z}_n=\sup_{k=1,...,n}\frac{1}{k}\widetilde{Y}_k$. As you claim, the result is true for the sequence $\widetilde{Y}_n$.
Note that since $Z_n$ in non-decreasing, you have $\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )=\lim_n\mathbb{P}(Z_n\geq c)$ so we just need to prove that $\mathbb{P}(Z_n\geq c)\leq \frac{1}{c}\mathbb{E}(Y_1)$.
Now for fixed $n$, for every $x$, there exists $1\leq k(x)\leq n$ such that $Z_n=\frac{1}{k(x)}Y_{k(x)}$. Because of subadditivity, you have $Y_{k(x)}\leq \sum_{j=0}^{k(x)-1}Y_1\circ \tau^j(x)=\widetilde{Y}_{k(x)}(x)$.
So $\frac{1}{k(x)}Y_{k(x)}\leq \frac{1}{k(x)}\widetilde{Y}_{k(x)}(x)\leq \widetilde{Z}_n(x)$. This proves that for any $x$, $Z_n(x)\leq \widetilde{Z}_n(x)$ so $\mathbb{P}(Z_n\geq c)\leq \mathbb{P}(\widetilde{Z}_n\geq c)$. Using that $\widetilde{Z}_n$ also is non-decreasing, you get $\mathbb{P}(\widetilde{Z}_n\geq c)\leq \mathbb{P}\left (\sup\frac{1}{n} \widetilde{Y}_n\geq c\right )$ and so you can use the result for $\widetilde{Y}_n$.
@0xbadf00d I added an answer for the non-positive case : unfortunately the answer is no. However, I also added some comments and the right assumption might be that $\mathbb{E}(\frac{1}{n}|Y_n|)\leq \mathbb{E}(|Y_1|)$.
@0xbadf00d Oh but I realized that for non-postive $Y_n$, then $|Y_n|\geq \sum_{k=0}^{n-1}Y_1\circ \tau ^k$. So if $\mathbb{E}(\frac{1}{n} |Y_n|)\leq \mathbb{E}(|Y_1|)$, then we have $Y_n=\sum_{k=0}^{n-1}Y_1\circ \tau^k$ almost surely. In my opnion, this leads to the following question : Assume that $Y_n$ is non-positive and integrable. Let $Y$ be the almost sure limit of $Y_n$ and $Y'$ the almost sure limit of $\sum_{k=0}^{n-1}Y_1\circ \tau^k$. Do we have $Y=Y'$ almost surely ? BTW, sorry for the many messages, but I find this question very interesting !
I'm a bit out of time recently, but regarding your counterexample: Please note that it is only a counterexample for $c>1$, but usually one is interested in infinitesimal small $c>0$. So, it wouldn't be a restriction to assume $c\in(0,1)$.
@0xbadf00d No worries ! You're right, but also note that you can change $Y_n$ to be $-\frac{c}{2}n^2$. In this situation, you still have that $\inf \frac{|Y_n|}{n}$ is $+\infty$ so $\mathbb{P}(\sup \frac{1}{n}|Y_n|\geq c)=1$. On the other hand, $\frac{1}{c}\mathbb{E}(|Y_1|)=1/2$. This is not a counter-example for any $c$, but for any $c$, you get a counter-example. Now the question becomes "does (3) holds for any small enough $c$ ?" Also, in my last comment, I obviously meant "assume that $Y_n$ is non-positive and integrable AND THAT (3) HOLDS, then do we have $Y=Y'$ a.s. ?
|
2025-03-21T14:48:30.431875
| 2020-04-28T13:37:57 |
358758
|
{
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"authors": [
"Amir Sagiv",
"Guilherme",
"https://mathoverflow.net/users/156344",
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|
Stack Exchange
|
Snoidal wave solutions of the $\phi^4$ model
I want to prove the existence of snoidal wave solutions of the $\phi^4$ model, given by
$$u_{tt}-u_{xx}=u-u^3,\; (x,t) \in \mathbb{R}\times \mathbb{R}.$$
So, we are looking for solutions in the form $u(x, t) = \varphi_c(x−ct),
$ with $c \in \mathbb{R}$. Substituting into the $\phi^4$ model and integrating once, we obtain
$$\varphi'^2=\frac{1}{w}\left(\frac{\varphi^4}{2}-\varphi^2+2a\right),$$
where $\varphi:= \varphi_c$,$w:=1-c^2>0$ and $a \in \mathbb{R}$ is a constant of integration. Therefore,
$$\varphi'^2=\frac{1}{w}\frac{1}{2}\left(\varphi^2-B_1^2)(\varphi^2-B_2^2\right),$$
where $B_1^2=1+\sqrt{1-4a}, B_2^2=1-\sqrt{1-4a}$, with $a<\frac{1}{4}$. Hence,
$$\frac{d\varphi}{ds}=\frac{1}{\sqrt{2w}}\sqrt{\left(\varphi^2-B_1^2)(\varphi^2-B_2^2\right)} \Rightarrow\frac{d\varphi}{\sqrt{\left(\varphi^2-B_1^2)(\varphi^2-B_2^2\right)}}= \frac{1}{\sqrt{2w}} \; ds.$$
I thought about integrating this last equality and using, somehow, the formula $ 216.00 $ of $ [1] $ in order to obtain the snoidal solution sought, but I don't know how to proceed and if I can use this formula. Moreover, what I did is correct?
$[1]$ P.F.Byrd, M.D.Friedman. Hand Book of Elliptical Integrals for Engineers and Scientis. Springer-Verlag-New York-Heidelberg-Berlin, 1971.
Just to comment that a very good modern exposition to this equation may be found here: https://www.springer.com/gp/book/9783030118389, Keverkidis & Cuevas-Maraver, A Dynamical Perspective on the $\phi ^4$ Model, Springer 2019
@AmirSagiv Thanks!
|
2025-03-21T14:48:30.432001
| 2020-04-28T13:43:05 |
358759
|
{
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"authors": [
"Ofir Gorodetsky",
"Wojowu",
"https://mathoverflow.net/users/156584",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/31469",
"user156584"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358759"
}
|
Stack Exchange
|
On the asymptotics of the Chebyshev psi function
Denote by $\psi(x)$ the Chebyshev psi function over prime powers. Assuming the RH, it can be shown via the Riemann explicit formula that
$$\psi(x)-x \ll √x |\sum_{|\gamma| < T} \frac{1}{\rho} | + O(1)$$, where the sum is over the complex zeros $\rho$ of the Riemann zeta function $\zeta(s)$ with imaginary part $|\gamma| < T$. Let $N(T)$ denote the number of such zeros. Since $N(T)\ll T\log T$, it follows from the above that
$$\psi(x) - x \ll √x (\log T)^2.$$.
Montgomery-Vaughan, in their famous text "Multiplicative Number Theory" on p.419, then take $T=x$ and deduce that
$$\psi(x) - x \ll √x (\log x)^2.$$
My question is, what motivates this choice of $T$ ? Is there anything special about it ?
The error term in the explicit formula is not $O(1)$ but rather $O(\frac{x \log^2(xT)}{T} + \log x)$ (in fact, something slightly better can be said - see p. 400 in Montgomery-Vaughan). The sum $\sqrt{x} (\log T)^2 + \frac{x \log^2(xT)}{T}$ will always be of order at least $\sqrt{x} (\log x)^2$, and $T=x$ gives this order (also other choices, such as $T=x^{\alpha}$ for $\alpha>1/2$). So $T=x$ is special but not that special.
@Ofir, maybe am missing something, but doesn't the $O(1)$ term come from the explicit formula itself ? Recall that $\psi(x) = x - F(x, T) - \log 2\pi - (1/2)\log(1-x^{-2})$, where $F(x, T)$ denotes the sum over the complex zeros ?
The explicit formula in your last comment is false. The RHS only converges to $\psi(x)$ as you let $T\to\infty$.
|
2025-03-21T14:48:30.432467
| 2020-04-28T13:51:53 |
358761
|
{
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"Jiří Rosický",
"Simon Henry",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358761"
}
|
Stack Exchange
|
Can a locally presentable category have a proper class of accessible localizations?
Question: What is an example of a locally presentable category $\mathcal C$ such that there exists a proper class of accessible localizations $(\mathcal C \to \mathcal D_i)_{i < ORD}$?
In other words, $(D_i)_{i < ORD}$ should be a proper class of full reflective subcategories of $\mathcal C$ which are accessibly embedded in $\mathcal C$.
I'm also interested in the infinity-categorical setting, though I suspect it doesn't make much difference.
A limit closure of a set of objects of a locally presentable category $\mathcal K$ is reflective. In this way one gets an increasing chain of reflective full subcategories of $\mathcal K$. If this chain stops $\mathcal K$ has a cogenerator. Since a category of groups does not have a cogenerator, it has a proper class of reflective full subcategories. These subcategories are accessibly embedded under Vopěnka's principle. I do not expect that Vopěnka's principle is needed but, at this moment, I do not see how to avoid it.
Don't we also get easy example assuming the negation of Vopenka ? if A is a non-accessible localization of C, then we can consider the class of localization of C at any set S of morphisms inverted by the reflection $C \to A$. If these only forms a set, as $A$ identifies with the localization at their unions, it would make $A$ a localization at a set of arrows hence an accessible localization.
So, there is an example in any model of ZFC. Can one do it without mentioning VP at all?
|
2025-03-21T14:48:30.432586
| 2020-04-28T14:18:18 |
358767
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358767"
}
|
Stack Exchange
|
Dimension bound on invertible bimodules for blocks
A co-author and I have recently proved the following result:
Theorem.
Let $B$ be a block of a finite group defined over some algebraically closed field $k$ of characteristic $p>0$. If $M$ is a $B$-$B$-bimodule inducing a Morita auto-equivalence of $B$ then $\dim_k(M)\leq \dim_k(B)$.
In fact it holds for any finite dimensional $k$-algebra whose Cartan matrix is symmetric and positive definite. My question is whether this is already known/written down? If the block is defined over a suitable complete, discrete valuation ring, then the result is a bit easier as one can use ordinary characters but is it already known over $k$?
Any help would be much appreciated.
|
2025-03-21T14:48:30.432661
| 2020-04-28T14:23:12 |
358768
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358768"
}
|
Stack Exchange
|
Dixmier trace, Wodzicki residue and topological index
There are a well-known facts about Dixmier trace and Wodzicki residue. Let $P$ be an elliptic pseudodifferential operator of degree $−n$ on a compact Riemannian manifold $(M,g)$, than its Dixmier trace is related to the Wodzicki residue by the formula:
$$
\operatorname{Tr}^{+} P=\frac{1}{n(2 \pi)^{n}} \operatorname{Res} P
$$
Also, for an arbitrary smooth function $a \in C^{\infty}(M)$ the following equality
$$
\int_{M} a(x)\left|\nu_{g}\right|=\frac{n(2 \pi)^{n}}{\Omega_{n}} \operatorname{Tr}^{+}\left(a \Delta_{g}^{-n / 2}\right)
$$
holds. Is it possible to associate Wodzicki residue and topological index of the operator $a \Delta_{g}^{-n / 2}$? For example, can a something like $\operatorname{Tr}^{+}\left(a \Delta_{g}^{-n / 2}\right) \in H^*_{b, dR}(M)$ take place? Where $H^*_{b, dR}(M)$ can be understood as bounded John Roe's cohomology classes.
|
2025-03-21T14:48:30.432737
| 2020-04-28T14:55:04 |
358769
|
{
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"Donu Arapura",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/64302",
"user2520938"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358769"
}
|
Stack Exchange
|
What does it mean for a form to have a meromorphic extension?
In the article "a remark on a non-vanishing theorem of P. Deligne and G. D. Mostow." by Esnault and Viehweg, they consider the following setup. $X$ is an $n$-dimensional smooth complex manifold, and $D\subset X$ is a normal crossings divisor. We denote $U=X\setminus D$. We have a form $\omega\in H^0(U,\Omega_U^n\otimes L)$, for some rank 1 local system $L$ on $U$. The condition they need is then that $\omega$ has a meromorphic extension to $X$. They don't define what this means though?
My guess is that it means that if we take a general point $p\in D_i$, a component of $D$, and coordinates $x_1,\dots,x_n$ in which $D_i=\{x_1=0\}$, that we can then write
$$\omega = \frac{f}{x_1^m}dx_1\wedge\dots\wedge dx_n \otimes e,$$
where $e$ is a multivalued section of $L$ around $p$, $f$ is holomorphic, and $m$ is some integer. Is this indeed the meaning?
That sounds right.
@DonuArapura thanks. Does that mean that this condition is automatic if all objects involved are analytifications of algebraic ones?
@DonuArapura e.g. $\omega=f^q\psi$ with $q$ a rational number, and $\psi,f$ algebraic
The point is avoid sections with essential singularities. These could arise even if the bundles come form algebraic ones.
@DonuArapura If you write this as an answer I can accept it and the question can be marked as answered.
|
2025-03-21T14:48:30.432860
| 2020-04-28T14:58:11 |
358770
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hanul Jeon",
"Holo",
"Johannes Schürz",
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"https://mathoverflow.net/users/134910",
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|
Stack Exchange
|
Preserve unbounded sets between different cofinality
Working in ZFC, let $\kappa,λ$ be cardinals with $\kappa>λ$, and assume that $\kappa$ is regular.
We say that a function $F:\kappa^n→λ$, for some finite $n$, is preserving unbound, if for all $a⊆\kappa$ unbounded, we have that the closure of $F[a^n]$ under $F$ is unbounded in $λ$.
Is there always preserving unbound function?
For $\mbox{cof}(λ)=ω$ this is easy: take $n=1$ and map every element from $\kappa\setminusλ$ into some some element of $λ$, and map an element from $λ$ to somewhere greater using some fixed cofinal sequence of $λ$, i.e. fix some cofinal $(λ_i\mid i\in\omega)$, and for $k$ be the minimal $k$ such that $x∈λ_k$, map $x$ to some element of $\lambda_{k+1}\setminusλ_k$.
The problem arise when $\mbox{cof}(λ)>ω$. In this case we cannot "climb" a cofinal.
In this case for $n=1$ there are clearly no preserving unbound function. Indeed if $F:\kappa→λ$ be any function, then there exists some $x∈λ$ such that $F^{-1}(x)$ is unbounded in $\kappa$.
A special case is that $λ$ is regular, in this case preserving unbound becomes:
A function $F:\kappa^n→λ$, for some finite $n$, such that if for all $a⊆\kappa$ unbounded, we have $F[a^n]$ is unbounded in $λ$.
Indeed if $F[a^n]$ is bound, then $|F[a^n]|<λ$, then $|\bigcup\{F[a^n],F[F[a^n]],...\}|<λ$, so the closure is also bounded.
The I am most interested in the special cases where $λ$ is indeed regular and that $\kappa=λ^+$, in particular $\kappa=ω_{k+1},λ=ω_k$ for finite $k$
If $\kappa$ is measurable and $\lambda$ regular such a function cannot exist. Assume the contrary. W.l.o.g. let $n=2$. Then we can define the function $G({\alpha,\beta}):=\max(F((\alpha,\beta)), F((\beta,\alpha)))$. But there exists a measure 1 set $a$ such that $G\restriction [a]^2$ is constant. Therefore, $F$ cannot be preserving unbounded.
@JohannesSchürz doesn't similar thing works for $\kappa$ weakly compact? Regardless on $λ$
@HanulJeon but we are looking at the closure of the image, not just the image itself
"fix some cofinal $(λ_i\mid i\in\omega)$, and for $k$ be the minimal $k$ such that $x∈λ_k$, map $x$ to some element of $\lambda_{k+1}\setminusλ_k$" If $\kappa=ω_{ω+1}$ and $λ=ω_ω$, send all elements in $\kappa\setminusλ$ to some $x$. If $x\inω$, then $F(x)∈ω_1$, if $x∈ω_1$ then $F(x)∈ω_2$ etc. Then the closure of every subset of $λ$ under $F$ is cofinal
@HanulJeon $F(\xi)$ is a fixed value, but the closure of $F[\kappa\setminus\lambda]$ under $F$ will be $\bigcup{F(\xi),F(F(\xi)),...}$, which will be unbounded(where $\xi$ is some arbitrary value $>\lambda$)
I get it, I confuse the topological closure and the $F$-closure under the given set.
In the case that $\kappa=\lambda^+$, I claim that $n=2$ suffices to produce such a function, as follows. Fix, for each ordinal $\beta<\kappa$ a one-to-one function $f_\beta:\beta\to\lambda$. Then define $F:\kappa^2\to\lambda$ by setting $F(\alpha,\beta)=f_\beta(\alpha)$ if $\alpha<\beta$. (It doesn't matter how you define $F(\alpha,\beta)$ for $\alpha\geq\beta$.) Now consider any unbounded $A\subseteq\kappa$; I want to show that $F(A^2)$ is unbounded in $\lambda$, for which it suffices to show that $F(A^2)$ has cardinality $\lambda$. Since $A$ is unbounded in $\kappa$, it has a $\lambda$-th element $\beta$. As $\alpha$ ranges over the $\lambda$ members of $A$ that are $<\beta$, $F(\alpha,\beta)=f_\beta(\alpha)$ takes $\lambda$ distinct values, because $f_\beta$ is one-to-one. And all of these $\lambda$ values are in $F(A^2)$ because $\beta$ and all the $\alpha$'s under consideration are in $A$.
I'm reasonably sure a similar argument (iterating this idea) will show that $n=q+1$ suffices when $\kappa$ is the $q$-fold successor of $\lambda$. Unfortunately, I don't have time right now to write down the argument.
|
2025-03-21T14:48:30.433104
| 2020-04-28T15:02:26 |
358772
|
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"Tim Campion",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358772"
}
|
Stack Exchange
|
Need to know if a certain full subcategory of Top is cartesian closed
Consider the full subcategory of Top consisting of all spaces $X$ such that a subset $A$ of $X$ is closed if and only if $A \cap K$ is closed in $K$ for all subspaces $K$ of $X$ which are countably compact, that is every open covering has a countable subcovering. It would be convenient for a proof I'm trying to write if this were known to be cartesian closed; is it known whether this is the case?
I doubt it -- the usual way to show cartesian closure is to use that the category is generated under colimits by exponentiable spaces, which won't be the case here.
In fact -- doesn't the usual counterexample to cartesian closure of $Top$ just use the quotient of $\mathbb R$ by some countable subset? If I understand correctly, this construction should be perform-able in your category.
|
2025-03-21T14:48:30.433185
| 2020-04-28T15:14:14 |
358773
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358773"
}
|
Stack Exchange
|
Stability and capacity, error in the book of Adams-Hedberg?
I am struggling to understand the proof from the book of Adams and Hedberg, "Function spaces and potential theory". It seems to me that there is a serious flaw, and moreover, the statement is incorrect. However, I am not an expert in the field of potential theory, and I want to ask somebody to verify those thoughts of mine.
The theorem I am interested in is Theorem 11.5.10 on p.335. To recall its statement, I recall that a set $E \subset \mathbb{R}^n$ is called $(1, 2)$-stable if the following identity between Sobolev spaces hold
$$
W_0^{1,2}\big(\operatorname{int}(E)\big) = W_0^{1,2}(E),
$$
where the set $W_0^{1,2}\big(\operatorname{int}(E)\big)$ is defined as the closure of the smooth functions of compact support in $\operatorname{int}(E)$, and $W_0^{1,2}(E)$ is defined as the set of functions $f \in W_0^{1,2}(\mathbb{R}^n)$ with vanishing trace on the complement of $E$. Remark that the trace here is defined by using quasi-continuous representatives.
The statement Adams and Hedberg are making in Theorem 11.5.10 a),d) states that the $(1,2)$-stability of a Borel set $E$ is equivalent to the divergence of the sum
$$
\sum_{n = 0}^{\infty} 2^{n - 2} N_{1,2}\big(B(x, 2^{-n}) \setminus\operatorname{int}(E) ; B(x, 2^{-n+1})\big),
$$
for (1,2)-quasi every $x \in \partial E$. The quantity $N_{1,2}\big(B(x, 2^{-n}) \setminus \operatorname{int}(E) ; B(x, 2^{-n+1})\big)$ is defined as some sort of modified capacity, which, according to Proposition 11.5.9, is bigger that the classical capacity $C_{1,2}\big(B(x, 2^{-n}) \setminus \operatorname{int}(E)\big)$, defined for a given closed set $D$ by
$$
C_{1,2}(D) = \inf\left\{\| f\|_{1,2} : f \in W^{1,2}_{0}(\mathbb{R}^n), f \geq 1 \text{ on } D\right\}.
$$
I think that the correct statement would be that the $(1,2)$-stability of a Borel set $E$ is equivalent to the divergence of the sum
$$
\sum_{n = 0}^{\infty} 2^{n - 2} N_{1,2}\big(B(x, 2^{-n}) \setminus E ; B(x, 2^{-n+1})\big),
$$
for (1,2)-quasi every $x \in \partial E$, which is significantly weaker. The proof, presented by Adams and Hedberg, as it seems to me, shows exactly that, see the beginning of the p. 324. I think the counterexample for the statement of Adams and Hedberg would be the compact set $K$ from Theorem 11.5.5.
In the notes for this chapter, Adams and Hedberg say that this result is due to Netrusov, but his paper was never published...
|
2025-03-21T14:48:30.433456
| 2020-04-28T16:00:19 |
358777
|
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|
Stack Exchange
|
What are the Newton groupoids from Drinfeld's paper on the Grinberg-Kazhdan theorem?
The paper the Grinberg-Kazhdan formal arc theorem and the Newton groupoids by Drinfeld seems to contain many interesting things which are beyond me. For now, I am trying to get some intuition for the Newton groupoids. From the introduction:
The proof is based on the ideas that go back to the 17th century (namely, the implicit function theorem or equivalently, Newton's method for finding roots) and the 19th century (the Weierstrass division theorem).
The goal of the rest of the article is to clarify the geometric ideas behind the proof from §2. In particular, we introduce Newton groupoids. These are certain groupoids in the category of schemes, which are related to Newton's method for finding roots. They are associated to any generically étale morphism from a locally complete intersection to a smooth variety (both schemes are assumed separated).
The actual setting for the definition of Newton groupoids seems to me rather involved, so instead of writing down, I risk assuming many people read what Drinfeld writes.
Question. What is the geometric intuition behind Newton groupoids, and how are they related to the implicit function theorem/Newton's method of finding roots?
|
2025-03-21T14:48:30.433565
| 2020-04-28T16:26:34 |
358779
|
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|
Stack Exchange
|
Decide whether there are "linear" relations between quadrics
Let $k$ be an algebraically closed field of characteristic $0$. For a homogeneous ideal $I=(q_1,\dots, q_k)\subset k[x_0,\dots,x_n]$ generated by quadrics, is there a method to decide whether the generators of $I$ admit relations "of degree $1$" i.e. if there are linear polynomials $L_i\in k[x_0,\dots,x_n]_1$ such that $\sum_iL_ig_i=0$?
Can one compute (in explicit cases) the dimension of such $k$-uples of linear polynomial?
These relations are called "linear syzygies". A lot are known about them, and interesting questions are abound too. You can compute them with Macaulay 2 and read the number off the Betti table.
In this particular case, there is a convenient formula using Hilbert function. The number of linear syzygies is $(n+1)r - \dim_k I_3$, where $r$ is the number of generators. The reason is that the space of cubics contains $(n+1)r$ cubics from the generators, but each linear relation translates to a $k$-linear relation among these cubics and make the dimension of $I_3$ goes down $1$.
For instance, if $I=(x^2,y^2,x^2)$, the Hilbert series of $R/I$ is $1+3t+3t^2+1$. It follows that $\dim_k I_3=10-1=9$, so there are $3\times 3-9=0$ linear syzygies. If $I= (x^2,y^2,x^2,xy+yz+zx)$, then the Hilbert series of $R/I$ is $1+3t+2t^2$, so $\dim I_3=10$ and there are $4\times 3-10=2$ linear syzygies.
Thank very much for your detailed answer.
|
2025-03-21T14:48:30.433692
| 2020-04-28T16:27:26 |
358780
|
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"PHL",
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|
Stack Exchange
|
Torsion group with finitely many elements of order 2 but infinitely many elements of order 4
Does there exists a group $G$ satisfying all the following conditions?
$G$ is finitely generated,
$G$ is of bounded torsion (has finite exponent),
$G$ has finitely many elements of order $2$,
$G$ has infinitely many elements of order $2^l$ for some $l$ (for example $l=2$).
By 1. and 2. $G$ is a quotient of a free Burnside group.
Near-examples of such groups violating only Condition 2. are generalized dicyclic group $Dic(A,y)$, where $A$ is a finitely generated abelian group.
References on somehow related problems are welcome.
Note that for an infinite group $G$, 4 is implied by 2 (so 4 is just "$G$ is infinite"). Also no infinite locally finite group satisfies (2 and 3) because it would have an infinite abelian subgroup with the same property and this can be discarded. I don't know either if $2$ is replaced by an odd prime, which may make the question easier or harder, I don't know.
BTW: the assumption implies that the FC-center is non-trivial. Using the restricted Burnside, we can suppose that $G$ has no nontrivial finite quotient. Hence the FC-center is central. Also, because there's no locally finite example, the locally finite radical is finite, hence equals the FC-center, and hence equals the center. Then, modding out by a subgroup of index 2 in center yields an example in which the locally finite radical has order $2$. So we can reduce to this problem.
@YCor In order for the implication (G infinite) 2=>4 to be true, the condition that the exponent of G is even is necessary. On the other hand, I have trouble to see why it is sufficient.
Let $G$ have finite exponent, and finitely many elements of each 2-power order. So it has finitely many elements of 2-power order. Let $W$ be the subgroup generated by these elements. Since these elements of 2-power order have finite conjugacy classes, $W\subset FC(G)$, the FC-center of $G$, which is locally finite, and in particular $W$ is finite. This proves that (2 and not-4) implies (not-3). So (2 and 3) implies (4).
Here is an example with infinitely many $4$-torsion elements and only one $2$-torsion element. (The smallest number of generators I can do is $4$, and the exponent is pretty big and not a power of $2$.)
Example. Pick $n$ odd and $d \geq 2$ such that $B(d,n)$ is infinite. Let $p$ be an odd prime. Set
$$A = \mathbf Z/2 \oplus \bigoplus_{g \in B(d,n)} (\mathbf Z/p)e_g,$$
with distinguished element $y = (1,0,\ldots)$ or order $2$. Then the generalised dicyclic group $\operatorname{Dic}(A,y) = A \amalg Ax$ satisfies all criteria except 1:
$\operatorname{Dic}(A,y)$ is not finitely generated because it has an index $2$ subgroup $A$ that is not finitely generated.
Every element of $\operatorname{Dic}(A,y)$ is killed by $4p$.
The only element of order $2$ is $y \in A$.
Every element in $Ax$ has order $4$, and $Ax$ is infinite by assumption.
Finally, the group $B(d,n)$ acts on $A$ fixing $y$ by $(g,e_h) \mapsto e_{gh}$. Thus this extends to an action on $\operatorname{Dic}(A,y)$, and we take $G$ to be the semidirect product
$$G = \operatorname{Dic}(A,y) \rtimes B(d,n).$$
Then all criteria are satisfied:
If $x_1,\ldots,x_d$ are the standard generators of $B(d,n)$, then $x, e_1, x_1,\ldots,x_d$ generate $G$. Indeed, they generate the quotient group $B(d,n)$, hence we get all elements $ge_1g^{-1} = e_g$, hence we get everything.
Since $B(d,n)$ has exponent $n$ and $\operatorname{Dic}(A,y)$ has exponent $4p$, we conclude that $G$ has exponent dividing $4pn$.
and 4. Since $B(d,n)$ has odd exponent, all $2$-power torsion happens in $\operatorname{Dic}(A,y)$.
So $G$ is an example. $\square$
Remark. I have no idea if there are also examples of $2$-power exponent.
Oh nice. I tried something similar and this failed because I stuck trying with 2-power exponent locally finite groups for which this is hopeless (whence my initial comment).
Here's an elaboration on R. van Dobben de Bruyn's answer.
Let $A$ be an abelian group. Define the group $\mathrm{Di}(A)$ as follows: (a) consider the direct product $A'=A\times C_2$, denoting $y$ the nontrivial element of $C_2$. (b) perform the semidirect product $A''=C_4\ltimes A$, with $\pm$-action. Denote by $z$ the element of order 2 of the acting $C_4$: it remains central in $A''$, and so does $y$. (c) Obtain $\mathrm{Di}(A)$ (generalized dicyclic group on $A$) by modding out $A''$ by the central subgroup of order 2 $\langle z^{-1}y\rangle$. Still denote by $y$ the image of $y$ in $\mathrm{Di}(A)$, it it central of order $2$.
Note that $\mathrm{Di}(A)/\langle y\rangle$ is the dihedral product $C_2\ltimes_\pm A$.
In $A''$, denoting by $t$ a generator of $C_4=\{1,t,z,t^{-1}\}$ and writing $A$ additively, we have $(t^{\pm},a)^2=(z,0)$, $(z,a)^2=(1,a)^2=(1,2a)$ for $a\in A'$. In particular, $\eta=(z,y)$ (which is killed in $\mathrm{Di}(A)$ is not a square. Hence the elements of order $\le 2$ in $\mathrm{Di}(A)$ are the images of elements of order $2$ in $A''$, which are the elements of the form $(1,a),(z,2a)$ with $2a=0$. In particular, if $A$ has no element of order $2$, these elements are $(1,0),(1,y),(z,0),(z,y)$, which in $\mathrm{Di}(A)$ is reduced to $\{1,y\}$. That is, if $A$ has no element of order $2$ then the only element of order $2$ in $\mathrm{Di}(A)$ is $y$.
Also this shows that all elements $(t^\pm,a)$ map to elements of order $4$ in $\mathrm{Di}(A)$, which therefore has infinitely many elements of order $4$ if $A$ is an arbitrary infinite abelian group.
Now the construction $A\mapsto\mathrm{Di}(A)$ is clearly functorial under group isomorphisms. Hence, if $\Gamma$ acts on $A$ by automorphisms, then it naturally acts on $\mathrm{Di}(A)$ by automorphisms: in steps: (a) extend in the trivial way the action to $A'=A\times C_2$, then (b) extend in the trivial way to $C_4\ltimes A'$ (acting trivially on $C_4$): this works because the $C_4$-action, by $\pm$, commutes with the $\Gamma$-action on $A'$; finally this action fixes $\eta=(z,y)$ and hence passes to the quotient to an action on $\mathrm{Di}(A)$, defining a semidirect product $\mathrm{Di}\rtimes\Gamma$.
From what's preceding, we immediately get: if $\Gamma$ has no element of order $2$, then the only element of order $2$ in $\mathrm{Di}(A)\rtimes\Gamma$ is $y$; if $\Gamma$ is infinite then it contains $\mathrm{Di}(A)$ hence has infinitely many elements of order $4$.
Finally we can choose $A=C_n^{(\Gamma)}=\bigoplus_{\gamma\in\Gamma}C_n$ for some odd $n>1$, and choose $\Gamma$ of finite odd exponent $q>1$. Here $n$ and $q$ are unrelated, can be chosen equal or not (they could be chosen even for the construction but then this will produce infinitely elements of order $2$). Then the resulting group $$G=\mathrm{Di}(C_n^{(\Gamma)})\rtimes \Gamma$$ works: it has a single element of order $2$, infinitely of order $4$, and has exponent dividing $nq$. Actually modding out by $\langle y\rangle$, the resulting group admits, as subgroup of order $2$, the standard wreath product $C_n\wr \Gamma$ (in particular, all elements of order $4$ in $G$ lie in the nontrivial coset of the unique subgroup of index $2$).
(In general— arbitrary abelian group $A$, arbitrary $\Gamma$-action on $A$, we always get this subquotient killing the center $\langle y\rangle$ and passing to a subgroup of index $2$, which yields the semidirect product $A\rtimes\Gamma$. For instance, if we want $G$ to have Kazhdan's Property T, the permutational action is hopeless, but possibly some choice of $\Gamma$-module works, namely we need $A\rtimes\Gamma$ to have Kazhdan's Property T.)
|
2025-03-21T14:48:30.434141
| 2020-04-28T16:31:19 |
358781
|
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|
Stack Exchange
|
On existence of conjecture relating prime zeta function:
There is an article on Wikipedia about prime zeta function (PZF):
https://en.m.wikipedia.org/wiki/Prime_zeta_function
In that article , there is table of fairly accurate values of PZF for different $s$ .
We all know that
$\zeta(2n)=π^n\mathbb{Q}$.
So my question is :
Are there conjectured values of PZF such that they are the combination of well known transcendental numbers like $\pi$ and $e$ (like $\zeta(2n)$ above) and are very close to values given in the article ?
since these numerical PZF values are on OESIS, I presume if one of these had a closed form expression it would be mentioned there, don't you think so?
That would be OEIS, e.g., https://oeis.org/A085548 for digits of prime zeta function at 2.
|
2025-03-21T14:48:30.434231
| 2020-04-28T16:51:47 |
358783
|
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"Phil Tosteson",
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|
Stack Exchange
|
Visualising the moduli space of stable disks with interior marked point and 4 marked point on the boundary
Is there any nice description/picture of the moduli space of stable disks with 1 interior marked point and 4 marked points on the boundary?
I'm expecting it to be a 3-dimensional polytope, because if we fix the interior point and, say the green marked point, I have 3-degree of freedom for moving the remaining marked point.
Is it possible to know, say, the number of edges each faces should have? the number of vertices and so on?
I was wondering if there are some references where this space is already studied, before trying to reinvent the wheel.
If there are any trick for finding all the limiting configurations I'm very interested in knowing them as well.
Thanks in advance.
A generalization of this moduli space was used by Costello, in the paper https://arxiv.org/abs/math/0601130. In his terminology, your space is $D_{0,1,4,1} = \overline{\mathcal N}_{0,1,4,1}$.
The combinatorics of the boundary strata are described by isomorphism classes of genus $0$ stable ribbon graphs with $4$ labelled external edges, and one labelled internal vertex. And Lemma 2.2.4 describes each stratum as an orbicell.
Thanks for pointing out that paper! In figure 1 the author draw a picture of an element of $D_{0,2,4,2}$ (i.e. annuli with 4 marked points on the boundary and 1 interior) as three disk glued together at three nodes. Do you happen to know why? moreover that configuration is not a tree, and in a Mumford compactification of the moduli space we only have "tree"- like bubbling phenomena
I can't say much beyond the definitions in the paper. The figure is a degenerate configuration where the annuli is "squeezed". Only $\overline N_{0,2,4,2}$ will contain actual annuli. In your case-- the case of $D_{0,1,4,1}$-- only trees will appear.
|
2025-03-21T14:48:30.434387
| 2020-04-28T17:01:39 |
358784
|
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"url": "https://mathoverflow.net/questions/358784"
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|
Stack Exchange
|
Is the acyclic chromatic number bounded in terms of the book thickness?
ISGCI says that the chromatic number of a graph is upper bounded in terms of the book thickness.
https://www.graphclasses.org/classes/par_32.html
This can be improved by saying that the book thickness bounds the degeneracy.
A further improvement would be that the book thickness bounds the acyclic chromatic number.
Is it true? Is there a reference?
I believe that "book thickness bounds the acyclic chromatic number" was
established in this paper:
Dujmovic, Vida, Attila Pór, and David R. Wood. "Track layouts of graphs." Discrete Mathematics and Theoretical Computer Science 6, no. 2 (2004).
arXiv abs.
In the Abstract they say,
"As corollaries we
prove that acyclic chromatic number is bounded by both queue-number and stack-number."
And then later (Section 5), they say,
"Note that stack-number is also called page-number and book-thickness."
Not sure if this is relevant to your quesiton, but
the acyclic chromatic number is not bounded by geometric thickness
$\overline{\theta}(G)$.
The literature can be confusing with four names for the same quantity: book-thickness, page-number, stack-number, and fixed outer-thickness. I would be interested to learn how this proliferation occurred, as book-thickness is ~50 yrs old.
Hi Joe. Page-number is descriptive, but does not connect with related topics. Stack-number is used because in some sense it is dual to queue-number. Book-thickness is used to highlight the similarity with thickness (minimum k such that G is the union of k planar graphs). Fixed outer-thickness also make sense, since book-thickness is the minimum k such that G has an edge partition into k outerplanar graphs with a fixed ordering of the vertices.
... my personal favourite is "convex thickness", since it is really a straight-line drawing with the vertices in convex position.
An indirect answer to the original quesiton is that graphs of bounded book thickness have bounded expansion (in the sense of Nesetril and Ossona de Mendez). So they have bounded generalised colouring numbers. So they have bounded acylic chromatic number.
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