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2025-03-21T14:48:31.924712
| 2020-09-02T01:55:58 |
370644
|
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|
Stack Exchange
|
How to write down a generic genus $g$ curve in $\mathbb{P}^n$ as an intersection of hypersurfaces?
Let $g, n \geq 1$ be positive integers. How can we describe a generic genus $g$ curve inside $\mathbb{P}^n$ as an intersection of hypersurfaces?
For example when $g = 1$ and $n = 2,3,4$, we have the following descriptions:
$n = 2$: a generic genus 1 curve is given by a plane cubic;
$n = 3$: a generic genus 1 curve is given by the intersection of two quadrics;
$n = 4$: a generic genus 1 curve is given by the intersection of the five quadrics defined by $4 \times 4$ sub-Pfaffians of a $5 \times 5$ skew-symmetric of linear forms.
As one can see, the $(g,n) = (1,4)$ case becomes complicated, as a generic genus one curve can no longer be expressed as a complete intersection.
For $g = (d-1)(d-2)/2$, we always have a nice description of genus $g$ curves in $\mathbb{P}^2$: in particular, a smooth plane curve of degree $d$ will have genus $g$.
Are there any other pairs of $(g,n)$ which admit an explicit description of generic genus $g$ curves as the intersection (possibly not complete) of hypersurfaces?
Petri's theorem.
Depending how one interprets your question it has different answers. For $g\geq 2$ and $(g,n)=(3,2)$, $(4,3)$ and $(5,4)$, the generic curve is a complete intersection (of type (4),(2,3) and (2,2,2)), and those are the only cases where this happens. If you don't ask for a complete intersection, any projective variety is actually an intersection of quadrics in an appropriate embedding, but writing down these quadrics is usually intractable.
For $(g,n)=(g,g-1)$, i.e. the case of curves embedded by their complete canonical system for $g\geq 3$, one has a good understanding of the equations of a general $C$ for $g$ up to about $9$. The cases $g=3,4,5$ are classical and already described in the comments. The breakthrough was I believe Mukai's Curves and symmetric spaces I,II ($g=7,8,9$); there is a lot of followup work such as Ide-Mukai Canonical curves of genus 8, von Bothmer's Geometric Syzygies of Mukai Varieties and General Canonical Curves with Genus at most 8, etc. The main result (quoted from the last paper) is as follows.
Theorem (Mukai). Every general canonical curve of genus $7\leq g \leq 9$ is a general linear section of an embedded rational homogeneous variety $M_g$. General canonical curves of genus $6$ are cut out by a general quadric on a general linear section of a homogeneous variety $M_6$.
The equations of these homogeneous varieties can be written down reasonably explicitly using representation theory.
In a different direction, for $(g,n)=(1,m-1)$, one is talking about elliptic normal curves of degree $m$. Their equations are studied for example in Fisher's Pfaffian representations of elliptic normal curves, which generalises your examples for $g=1$.
For more general $(g,n)$ lots of things can happen. I recommend Eisenbud's A Mystery Variety in $P^3$ in the collection Computations in algebraic geometry with Macaulay 2.
|
2025-03-21T14:48:31.924917
| 2020-09-02T02:09:53 |
370645
|
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"AmorFati",
"Donu Arapura",
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|
Stack Exchange
|
What is the meaning of the monodromy theorem in Hodge theory?
Let $f : X^m \to Y^n$ be an algebraic fiber space (between projective manifolds) whose discriminant locus is denoted by $E$. Let $U$ be a polydisk in $\mathbb{C}^n$ (with coordinates $(y_1, ..., y_n)$) such that $U \backslash E \simeq (\Delta^{\ast})^{\ell} \times \Delta^{n-\ell}$, where $1 \leq \ell \leq n$. Let $x_0$ be a point in $U\backslash E$. Associated to $x_0$ is the monodromy operator $$T_k : H^{m-n}(f^{-1}(x_0), \mathbb{C}) \longrightarrow H^{m-n}(f^{-1}(x_0), \mathbb{C})$$ for a loop based at $x_0$ around the $k$th copy of $\Delta^{\ast}$, where $1 \leq k \leq \ell$.
The monodromy theorem states that $T_k$ is quasi-unipotent, i.e., there are positive integers $m_k$ and $d_k$ such that $$(T_k^{m_k} - I)^{d_k} =0.$$ Here, $m_k$ is the least common multiple of the multiplicities of the irreducible components over the generic point of $\{ y_k =0 \}$.
Question: What does the quasi-unipotence of the monodromy transformation tell us qualitatively?
While I don't know a great deal about Hodge or representation theory, a cursory examination of the relevant concepts suggests something to the effect of "the monodromy action on the fiber always fixes some $(m-n)$-cycle class."
It says the eigenvalues are roots of unity, which means, since the coefficients of the characteristic polynomial are integers, there are only finitely many possibilities, depending on the dimension, for the Jordan normal form of the local monodromy. Is that the kind of thing you're looking for? What do you mean by "qualitative"?
@WillSawin Thank you for your comment! Personally, "qualitatively" means "geometric picture".
I believe this is saying that you can choose an ordered basis of the cohomology so that monodromy on a basis element gives you back the basis element (multiplied by a root of unity from the multiplicity of the irr. component) plus only basis elements that come later in the list. Think of it as a generalization of Picard-Lefschetz theory (https://en.wikipedia.org/wiki/Picard%E2%80%93Lefschetz_theory for a very concrete calculation)
I'm not completely sure what you're after, but perhaps it's simply some insight. So let make a few remarks. For simplicity, assume $Y$ is a curve. (In general, you need to assume that $E$ has normal crossings, which you did implicitly.) Replace it by a small disk. So $f:X\to D$ becomes a $C^\infty$ fibre bundle when restricted to $D^*= D-\{0\}$, which can be replaced by the circle, since we can work up to homotopy. To construct a such a fibre bundle, you just take a manifold $F$ and a diffeomorphism $\phi:F\to F$, and the glue the ends of $F\times [0,1]$ using $\phi$. Now ask, given a fibre $F$ diffeomorphic to a projective manifold and a self diffeomorphism, when can it arise from a projective family? The monodromy theorem tells you that the answer is almost always no, even when $F$ is $2$-torus, because a necessary condition is that the eigenvalues of $\phi^*H^i(F)\to H^i(F)$ must be roots of unity for all $i$. I still find this statement pretty striking.
Does this help?
Thank you very much for your comment. This is exactly what I was after. Perhaps a (further vague) question: Do you think the main content of the monodromy theorem then is that it tells you how rigid projective families are (in comparison with fiber bundles)? Is this what you mean by your remark that you find it "pretty striking". P.S. I would give (+2) if MO permitted it, I really appreciate this.
It's interesting that, in your comment, you are pretty much keeping inside the smooth category. That is, Ehressman's theorem gives you the smooth fiber bundle structure that you mentioned, but we do not seem to be concerned about whether the fibers are biholomorphic. Of course, $f$ is a holomorphic fiber bundle iff the fibers are all biholomorphic. There is a question somewhere in this rambling, but I can't formulate it at present.
You're welcome. I wouldn't call it rigidity exactly, but it certainly says the topology of projective manifolds is very special.
|
2025-03-21T14:48:31.925207
| 2020-09-02T05:18:37 |
370652
|
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"Ville Salo",
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|
Stack Exchange
|
Can someone please help me understand the concept of twins?
I am unable to understand Lemma 2.3 of
Carmen Hernando, Mercè Mora, Ignacio M. Pelayo, Carlos Seara, David R. Wood, Extremal Graph Theory for Metric Dimension and Diameter, Electronic J. Combinatorics 17 (2010) #R30, doi:10.37236/302, arXiv:0705.0938.
It says that if $u, v$ are twins in a connected graph $G$, then $dist(u, x) = dist(v, x)$ for every vertex $x\in V(G)\setminus \{u, v\}$.
I tried like this: Suppose that $x\in V(G)\setminus \{u, v\}$. Let $P$ be the shortest path from $x$ to $u$. We need to show that $P$ is also the shortest path from $x$ to $v$.
How can we show that?
Can someone please elaborate the issue?
Lemma 2.3 is immediate from the definition given on the line immediately above (which is why its proof is omitted). Your idea does not work, as $P$ need not even be a path from $x$ to $v$ (under the definition of "path" used in the cited paper). You already got an answer, but for future reference (or if the given answer is not sufficient), the correct forum for this question is math.stackexchange.com.
You can use this: if $u\ne x$, $$dist(u,x)=1+\min\{dist(w,x):uw\in E\}.$$
|
2025-03-21T14:48:31.925317
| 2020-09-02T05:50:10 |
370654
|
{
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"sort": "votes",
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}
|
Stack Exchange
|
Interpretation of "27" lines for cubic surface with rational double points
It is well known that a smooth cubic surface has $27$ distinct lines. Explicitly, if we choose a planar representation, i.e., blowup $\mathbb P^2$ at $6$ general points $p_1,...,p_6$, the $27$ lines are (1) $E_i$, $1\le i\le 6$, the exceptional divisors, (2) $F_{ij}$, $1\le i<j\le 6$, the proper transform of lines joining $p_i$ and $p_j$, and (3) $Q_i$, $1\le i\le 6$, the proper transform of conics passing $5$ points except $p_i$.
When a cubic surface acquires with one node ($A_1$ singularity), it has $21$ lines. One can think this happens in a specialization as the $6$ points become to lie on a single conic, and the line $E_i$ and $Q_i$ coincide in the limit as a double line, for $i=1,...,6$, while the rest of the $15$ lines $F_{ij}$ stays simple. So $27$ is interpreted as $2\times 6+15$.
What happens in general? My understanding is that, since the number $27$ (or $2875$ for quintic threefolds) is calculated via the intersection theory, it should be interpreted as the length of the Hilbert scheme of lines, especially when the cubic surface is not too singular and the number of lines is still finite.
According to Dolgachev's book section 9.2.2, all cubic surfaces with at worst rational double point singularities have finitely many lines. (e.g., a cubic surface with an $A_2$ singularity has $15$ lines; a cubic surface with an $E_6$ singularity has only $1$ line.)
So my question is, is there work been done to describe the Hilbert scheme of lines for cubic surfaces with rational double point singularities, or is there a geometric interpretation of how the number $27$ are attributed to the multiplicities of geometric lines in those cubic surfaces?
I have a vague memory that Nikos Tziolas discussed this case as an example in a talk around 2001, based on work in his thesis. His thesis and papers "Multiplicities of smooth rational curves on singular local complete intersection Calabi-Yau threefolds" and "Infinitesimal extensions of P^1 and their Hilbert schemes" does take the point of view you are describing, but I cannot access the first one right now and the second one is mostly about threefolds (though a singular cubic surface shows up in Ex.1.1.(b)).
One nice example is discussed here: https://math.stackexchange.com/questions/1209963/lines-on-a-singular-cubic-surface
A further search also gives https://www.math.upenn.edu/~siegelch/Notes/cagii.pdf which discusses another example on p34
One last comment: this http://wwwf.imperial.ac.uk/~apal4/summer/programme.html#Kass suggests that Jesse Kass has also thought about this problem but again I cannot see a paper; their "An arithmetic count of lines on a smooth cubic surface" with Wickelgren is clearly related.
Dear @Balazs, thank you very much for helping me to find these useful references!
As mentioned in Dolgachev's book, Schläfli classified cubic surfaces according to their singularities. In A Memoir on Cubic Surface Cayley tabulates for each type of singular cubic surface the number of distinct lines and their multiplicity. The multiplicity of a line in the Hilbert scheme of lines depends on whether it passes through a singularity and the type of that singularity. I'll illustrate this in some examples.
(II) In the case you mention (one $\mathrm{A}_1$-singularity $p$) $15$ lines don't pass through $p$ (which have therefore multiplicity 1), and $6$ do (and each has multiplicity $2$).
(IV) If you consider a cubic surface with two $\mathrm{A}_1$-singularities $p$ and $q$, then $7$ lines miss both $p$ and $q$, $8$ lines pass through one of $p,q$, and exactly one line passes through both $p$ and $q$ (which has multiplicity $2\times 2=4$).
(III) If a cubic surface has just one $\mathrm{A}_2$-singularity $p$, then $9$ lines miss $p$, and the $6$ lines who pass through $p$ have multiplicity $3$.
(XXI)
As mentioned by Balazs in the comments, the case XXI of three $\mathrm{A}_2$-singularities is particularly nice. The singularities form the vertices of a triangle, whose edges are the three lines in the cubic surface, each of which has multiplicity $3\times 3=9$ in the Hilbert scheme. In this case it is particularly simple to write down the equations cutting out the Hilbert scheme as a subscheme of the Grassmannian $\mathrm{Gr}(2,4)$; one obtains that the Hilbert scheme of lines is the spectrum of three copies of $\mathbf{C}[x,y]/(x^3,y^3)$ (which confirms that each line has multiplicity $9$).
(Note the following consequence: as the universal Hilbert scheme of lines is flat over the locus of cubic surfaces which contain finitely many lines, and since the Hilbert polynomial is constant in flat families, this computation shows that if a cubic surface has finitely many lines, then the number of lines must be 27, counted with multiplicity of course.)
I guess you could wonder which finite $\mathbf{C}$-algebras occur as rings of functions of Hilbert schemes of lines of singular cubic surfaces; I don't think Cayley tabulated these.
I just want to give an argument for the flatness, since I don't think it was mentioned before: The cubic equation defines a section of the symmetric cube of the tautological bundle on the Grassmanian, and the Hilbert scheme of lines is its vanishing locus. Thus it is locally the intersection of $ \operatorname{dim} \operatorname{Sym}^3(\mathbb C^2)=4$ equations in a $4$-dimensional Grassmanian variety, hence is flat if its dimension is $0$.
It's nice to have the classical intepretation of multiplicity of a line as the type of singularities it passes. I'm not suprised the result dates back to 1869, but thanks for picking the reference! Besides, the flatness argument is also what I am looking for. Thanks!
|
2025-03-21T14:48:31.925678
| 2020-09-02T06:12:50 |
370655
|
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|
Stack Exchange
|
Definition of Q gorenstein variety
I have a question about the definition of Q-Gorenstein variety.
I saw a definition of Q-Gorenstein variety：for a normal variety $X$, it's Q-Gorenstein if the canonical divisor is Q Cartier. I wonder how to use this definition (about canonical divisor) to see the structure ring is Q-Gorenstein. Through some duality?
Maybe there's no such thing like $\mathbb Q$-Gorenstein ring? Basically, I want to know from canonical divisor is $\mathbb Q$ Cartier, what can we say about the structure ring R. Here I mean locally X=Spec(R).
The structure ring is $\Bbb{Q}$-Gorenstein, not Gorenstein.
@abx Thanks,edited it
What is your definition of $\Bbb{Q}$-Gorenstein? For me, it is the property that you quote.
@abx. On a normal variety, I want to ask if the canonical divisor is $Q$-Cartier, how to show that on on affine piece X=Spec(R), the ring R is $Q$-Gorenstein.
Again: what is, for you, the meaning of "the ring R is Q-Gorenstein"? What definition do you have in mind?
@abx, Sorry for the impreciseness. I think there's a standard definition of Gorenstein ring. Maybe no definition of Q-Gorenstein ring? Basically, I want to know from the Q-Cartier property of canonical divisor, what can we say about the structure ring?
I do not know of any other definition of $\Bbb{Q}$-Gorenstein. For Gorenstein it is equivalent to finite injective dimension, as you will find in any textbook in commutative algebra.
@abx Thanks! I know what's Gorenstein ring. As I said, I don't know how it's related to the Cartier property of canonical divisor.
See Hartshorne's Residues and duality, Ch. V, §9.
@abx Great, that's what I want!
|
2025-03-21T14:48:31.925801
| 2020-09-02T08:22:18 |
370657
|
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|
Stack Exchange
|
Field of algebraic functions
We assume $K$ as a field of characteristic zero. By a field of algebraic functions of one variable over $K$ we mean a field $R$ satisfying $R=K(x,y)$ with $x$ being transcendental over $K$, and $R$ is algebraic over $K(x)$.
My question is: Whether there exist a subfield $F \subset R$ and $a \in R$ such that $R=F(a)$ with $a$ being transcendental over $F$. (In the case of rational function fields, this result follows easily.)
Almost never. $R$ is the field of rational functions on a smooth projective curve $C$ over $K$, and your assertion implies that $C$ has genus $0$. As soon as $g(C)\geq 1$, this will not happen.
@abx: indeed, if $F$ is assumed to contain $K$. It is possible though that $R$ is non-rational over $K$ but rational over some other subfield, but of course also that doesn't happen very often.
@ArnoFehm Just to be clear, that cannot happen if $K$ is algebraically closed, correct?
@Kevin Casto: indeed, can't happen if $K$ is algebraically closed
|
2025-03-21T14:48:31.926111
| 2020-09-02T10:39:48 |
370661
|
{
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"Arno Fehm",
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|
Stack Exchange
|
Issue with "definition" of pseudo algebraically closed fields
I'm having an issue with a sentence in Chapter 11 of Fried & Jarden's Field Arithmetic. As a "motto" for pseudo algebraically closed (PAC) fields, they say they are fields $K$ such that "each nonempty variety defined over $K$ has a $K$-rational point". No mention of absolute irreducibility is made at this point.
My issue is: how does this differ to algebraically closed fields? In particular, considering one variable polynomials, demanding that each $f(x) \in K[x]$ has a zero in $K$ would surely satisfy the standard ACF axiomatization?
If I remember correctly, that is just a terminology issue. In that book, "variety defined over $K$" includes "absolutely irreducible". See section 10.2 where this is explained, although maybe not very clearly.
Yes, I think this is it exactly. Thank you.
In any case I've seen several times pseudo-X in model theory to mean "satisfying all sentences that hold all for all structures in X". In this respect the terminology is confusing.
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2025-03-21T14:48:31.926216
| 2020-09-02T10:55:45 |
370662
|
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"Dattier",
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|
Stack Exchange
|
$\pi(x+200)-\pi(x)\leq 50$?
Is it true, that $\forall x \in \mathbb N, \pi(x+200)-\pi(x) \leq 50 $ ?
$$\pi(x)=\text{card}(\{n \in [0,x] \cap \mathbb N, n\text{ is prime}\})$$
Yes.
Up to $207$ there are $46$ primes. Hence, the inequality is true for $x \le 7$.
Let $$\pi_{210}(x) = \textrm{card}(\{n \in [0,x] \cap \mathbb{N}, \, \gcd(n,210)=1\}).$$
For $x>7$, $\pi(x+200)-\pi(x) \le \pi_{210}(x+200)-\pi_{210}(x)$. Since $\pi_{210}$ is $210$-periodic, it is enough to verify that $\pi_{210}(x+200)-\pi_{210}(x) \le 50$ for $x \le 210$, which can be done by hand or by computer. Here is SageMath code:
L = [int(gcd(i,210)==1) for i in range(420)]
max(sum([numpy.array(L[i:][:210]) for i in range(200)]))
and the last line outputs $47$. So the bound can be improved to this number.
The number $210=2 \times 3 \times 5 \times 7$ was chosen because $\prod_{p \mid 210}(1-1/p) < 50/200$.
We have $\pi(x+210)-\pi(x) \leq 48$
Indeed. By considering coprimality with $2310 = 2 \times 3 \times 5 \times 7 \times 11$, the same method produces $\pi(x+210)-\pi(x) \le 46$ for $x>11$; small $x$ can be checked by hand, and in fact there are cases of equality. The second Hardy-Littlewood Conjecture predicts $\pi(x+210)-\pi(x) \le \pi(210)=46$ also (for $x \ge 2$), but the general form of the conjecture is believed to be false.
for x=1, we have $\pi(211)-\pi(1)=47$
Yes, sorry. That's the only counterexample though.
|
2025-03-21T14:48:31.926337
| 2020-09-02T11:30:50 |
370665
|
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|
Stack Exchange
|
Residually finite Lie groups
My question is about the definition of residually finite Lie groups and their properties. I couldn't find any reference that contains exactly the meaning of residual finiteness of a given Lie group.
For a topological group you have a priori two notions: residually finite as discrete group (which also mean residually finite as topological group, allowing non-Hausdorff finite quotients), and residually finite as topological group, which more properly would mean residually (finite Hausdorff), and means that the intersection of open normal subgroups of finite index is trivial. This is a stronger condition. Actually, a non-discrete Lie group is never residually finite even as discrete group, as it contains an isomorphic copy of $\mathbf{Q}$, so there is not much to say.
Surely Lie groups are almost the opposite of residually finite groups? For instance the additive group of $\mathbb{R}$ has no finite quotients.
Thank you YCor for explaining. I was so confused about this notion of residually finite Lie groups.
|
2025-03-21T14:48:31.926429
| 2020-09-02T12:44:55 |
370674
|
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|
Stack Exchange
|
Spherical average of $\frac{1}{x}$
Let $X_1,...,X_n$ be points on $\mathbb S^1.$
We then define the expectation value $E(X)=\frac{1}{n}\sum_{i=1}^n X_i.$
Let $\frac{dS(X_1)}{2\pi}$ be the normalized surface measure of $\mathbb S^1,$ i.e. $X_i$ are uniformly distributed random variables on the circle.
I am curious to know:
How does
$$\int_{(\mathbb S^1)^n } \frac{1}{\vert E(X) \vert}\frac{dS(X_1)}{2\pi}...\frac{dS(X_n)}{2\pi}$$
scale with $n$?
A quick guess would be the Central limit theorem $\sim \sqrt{n}$
@RaphaelB4 would you mind elaborating a bit on this point?
See the distribution in Carlo.s answer : $e^{-R^2/n}$: the Gaussian with variance $n$.
The probability distribution $P(R)$ of $R=n|E(X)|$ was calculated by Kluyver (1906), it is given by
$$P(R)=\frac{1}{2\pi}\int_0^\infty [J_0(x)]^n J_0(rx)x\,dx.$$
For $n\gg 1$ one has a Rayleigh distribution (here is derivation including higher order corrections):
$$P(R)=\frac{2R}{n}e^{-R^2/n}.$$
The desired integral then becomes
$$I=\int_0^{\infty}\frac{n}{R}P(R)\,dR\rightarrow \sqrt{\pi n}$$
in the limit $n\rightarrow\infty$.
|
2025-03-21T14:48:31.926537
| 2020-09-02T13:28:51 |
370678
|
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|
Stack Exchange
|
Collapse successor of singular while preseving supercompactness
Suppose $\kappa$ is a supercompact cardinal. Is it possible to find a forcing which collapses $\kappa^{+\omega+1}$ to $\kappa^{+\omega}$ (all those $\kappa^{+n}$'s are preserved) while the supercompactness of $\kappa$ is preserved?
Suppose that there is a Woodin cardinal above $\kappa$, can you do some tricks to ensure that the stationary tower will preserve the supercompactness?
@AsafKaragila, I doubt it. Isn't it the case that such a forcing, let's say $P_\lambda$ always singularize some inaccessible $\theta$ which lives in between $\kappa$ and $\lambda$ to be of confinality $\omega$?
You can use the proper class version of the stationary tower. Do you want a set-sized forcing?
@MonroeEskew, will it make a big difference? If there are proper class of Woodin cardinal isn't it true by density argument that the class forcing factorize at some initial segment $P_{\lambda}$ for some Woodin cardinal $\lambda$?
@JiachenYuan Yes that should be true. But it’s possible that you need to add the full proper-class sized generic to restore supercompactness.
The point is that $\kappa$ is supercompact in the generic ultrapower by elementarity, but with the proper-class tower, the ultrapower is the generic extension.
@Monroe: Do you have a good reference for that?
@AsafKaragila Paul Larson’s book.
|
2025-03-21T14:48:31.926655
| 2020-09-02T13:33:31 |
370680
|
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|
Stack Exchange
|
$\mathbb{A}^1$-invariance of $K$-theory involving automorphisms
For a variety $X$ let's $K_0(X,\mathbb{G}_m^t)$ denote the Grothendieck group generated by vector bundles with $t$ commuting automorphisms on $X$. Subject to the relations coming from short exact sequence of vector bundles with $t$ commuting automorphisms (The morphisms of vector bundles has to be compatible with automorphisms) plus if one of the automorphisms is identity it is considered to be zero. Is it true that $K_0(X,\mathbb{G}_m^t)$ is $\mathbb{A}^1$-invariant for $X=\mathbb{A}^n$ where $n>t$?
(This question is inspired by putting together the Beilinson-Soulé conjecture plus another conjecture that claims that the motivic cohomology defined by Grayson filtration is unchanged if we switch from direct sum $K$-theory to ordinary $K$-theory)
|
2025-03-21T14:48:31.926747
| 2020-09-02T14:11:42 |
370682
|
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|
Stack Exchange
|
Fixed point scheme of finite group Cohen-Macaulay?
Let $X$ be a quasi-projective scheme over a field $k$.
Let $G$ be a finite group acting on $X$ whose order is invertible in $k$.
If $X$ is Cohen-Macaulay, can we conclude that the subscheme of fixed points $X^G$ is Cohen-Macaulay?
I went down a rabbit hole looking for a counter-example with $X = \mathbb{C}^n$, and I'll report my failure. According to https://imsc.uni-graz.at/baur/AGIT/Talks/Kraft_Ascona.pdf (slide 33, bullet point 3), every known algebraic action of a finite group on $\mathbb{C}^n$ is holomorphically equivalent to a linear action. So the fixed point locus is holomorphically isomorphic to $\mathbb{C}^k$, and hence smooth, and hence Cohen-Macaulay. I haven't traced the references in these slides, but this makes me suspect the problem is hard.
That's interesting! I'm not sure if I understand you're suggestion correctly, but $X^G$ is certainly smooth if $X$ is, see for example Proposition 3.5 of Edixhoven 'Neron models and tame ramification'.
Thanks for the reference! And, in the positive direction, I think I have a counter-example now.
@David E Speyer. Near each fixed point of the (full) group action, the action is linearizable. Smoothness of the fixed point scheme for an action of a tame, linearly reductive group on a smooth scheme is usually attributed to Iversen (who only considered the complex case).
Here is a simpler example than the one I left before, using the same strategy. Let
$$X = \{ x_1 x_3 = x_1 x_4 = x_1 x_5 = x_2 x_4 = x_2 x_5 = x_3 x_5 = 0 \} \subset \mathbb{C}^5.$$
This is the reduced union of four $2$-planes. Here is a projective picture, where $j$ represents the point where $x_j$ is the sole nonzero coordinate:
$$1 - 2 - 3 - 4 - 5.$$
The graph above is shellable, so this is Cohen-Macaulay.
Now, let $C_2$ act on $X$ by $(x_1, x_2, x_3, x_4, x_5) \mapsto (x_1, x_2, - x_3, x_4, x_5)$. Then the fixed locus of $C_2$ (even scheme-theoretically) is
$$Y = \{ x_1 x_4 = x_1 x_5 = x_2 x_4 = x_2 x_5 = x_3 = 0 \}.$$
This is the reduced union of two $2$-planes; we can visualize it as
$$1 - 2 \phantom{- 3 -} 4 - 5.$$
That is a standard example of a non-Cohen-Macaulay ring.
Thanks for the great answer! Small typo: $\mathbb{C}^2$ should be $\mathbb{C}^6$.
Fixed, thank you!
|
2025-03-21T14:48:31.926910
| 2020-09-02T14:29:01 |
370684
|
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|
Stack Exchange
|
Proof of Tian's constant
Basically Tian's invariant relies fundamentally on a lemma of Homander which says that $e^{- \phi}$ is integrable for $\phi$ plurisubharmonic on a ball of radius $1$ under some extra assumption. I am reading through a detailed proof of it on this note, p15, proof of theorem 3.10. I am confused as to why it is necessary to choose geodesic balls to cover the manifold, and it is not clear to me where the property of geodesic balls are used in the proof. Any clarification/help on this is much appreciated!
it's not important, you can cover $X$ by any (small enough) charts.
Thank you. I just want to make sure that I am not missing any important information here.
|
2025-03-21T14:48:31.926985
| 2020-09-02T15:02:17 |
370686
|
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|
Stack Exchange
|
An inequality of KL Divergence for two different distributions passing through a same channel
Let $X$ be a random variable which takes values in $\mathcal{X}$. Assume that we pass $X$ through two independent conditional pdf $p_{X_1|X}$ and $p_{X_2|X}$ and choose $X_1$ with probability $\lambda$ and $X_2$ with $\bar{\lambda}=1-\lambda$. Then, we have the following random variable $Z$:
\begin{align}
Z=\begin{cases}
X_1&\text{with prob.}~ \lambda\\
X_2&\text{with prob.}~ \bar{\lambda}.
\end{cases}
\end{align}
Therefore, the resulted probability density function of $Z$ is $\lambda p_{X_1}+\bar{\lambda}p_{X_2}$. Assume that $X_1,X_2$ also belong to $\mathcal{X}$. Now assume that both $X$ and $Z$ experience a conditional pdf $p_{Y|X}$ and produce random variables $Y_1$ and $Y_2$ respectively, i.e.,
\begin{align}
p_{Y_1}&=\sum_{x\in\mathcal{X}}p_{Y|X=x}p_X(x)\\
p_{Y_2}&=\sum_{x\in\mathcal{X}}p_{Y|Z=x}p_Z(z).
\end{align}
How one can find $\lambda$ such that the following inequality for KL divergrnce is satisfied
\begin{align}
\mathrm{D}(p_{Y|Z}||p_{Y_2})\leq \mathrm{D}(p_{Y|X}||p_{Y_1}).
\end{align}
Is it possible?
Can you restate your question in formal mathematical terms, without using such terms as "channels", "cross", and "produce"? Also, even though one could guess what you mean by $\bar\lambda$, can you still define this symbol? Also, by $X\in\mathcal X$ you apparently mean "$X$ takes values in $\mathcal X$", which is of course quite different from $X\in\mathcal X$.
Math_Y, your definitions are too lax. Nothing about X1, X2 enforces that Z resemble or preserves information about X in any way. This leaves the two sides of your desired inequality unrelated.
What's more, as you've written it, the LHS $D(P_{Y|Z}|P_{Y_2})$ is going to be a random variable depending on $Z$ and the RHS is a RV depending on $X$. I'm not sure that this is what you are really after.
I'm guessing that you're trying to formalize a notion similar to this: $${}$$ "$X_1$ and $X_2$ are like $X$, but with different types of noise. $Z$ is a mixing of $X_1$ and $X_2$. There is also some process $Y(\cdot)$ that acts on RVs that live in $\mathcal{X}$. Can I always mix $Z$ so that I learn less about $Y(Z)$ from $Z$ than I learn about $Y(X)$ from $X$?". $${}$$ The answer seems like yes, but what you've written isn't quite the right formalization yet. In particular you need to more closely specify the nature of $X_1$ and $X_2$ (and more cleanly specify the entire problem).
|
2025-03-21T14:48:31.927159
| 2020-09-02T15:22:20 |
370687
|
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|
Stack Exchange
|
Looking for a paper on transfinite diameter by David Cantor
I have been reading about transfinite diameter and its applications to number theory and have been hunting for the following paper for quite a while:
Cantor D.: On an extension of the definition of transfinite diameter and some
applications, J. reine Angew. Math., vol. 316 (1980), pp. 160-207.
Unfortunately, I am an undergraduate and have no access to MathSciNet or other resources which provide papers for free. The only links I got results from are ones accepting a payment which is completely outside my financial capacity. From some advice I got in a previous similar reference-request post, I tried an interlibrary loan but due to the pandemic and no nearby libraries, that hasn't been successful so far. I would really appreciate a link to the full text of the above reference. Thank you.
To anyone thinking of closing this question, please remember that not everyone has access to institutional support which makes literature access easy. I strongly feel that people who are eager to do research but lack such support should be helped and encouraged. Good answers like Benjamin Dickman's could be really helpful.
Thank you for your support.
There is a website called Sci Hub where you can get pretty much any article immediately as long as you know the DOI (which can usually be found very easily), however Sci Hub is possibly illegal so I definitely do not recommend that you use it to gain easy, no-hassle access to any paper which you need to read whenever you want.
Hopefully this works:
Cantor D.: On an extension of the definition of transfinite diameter and some applications
Since you said that you had "been hunting for the following paper for quite a while," here are the specific steps that I took to find this paper. Probably this will make it seem simple in retrospect, but I don't want to give off the impression that this required a complex knowledge of databases.
I googled the title of the paper
I clicked the link to EUDML (with which I wasn't previously familiar)
The sidebar says In Other Databases and I didn't find ZBMath helpful, so I used GDZ
I've never heard of GDZ before, but I saw the paper listed on p. 160 of that document and, indeed, the paper was there
For future reference, there is also an often used website called sci-hub [dot] ee that can pull papers for which you have the DOI. You can read more about Sci-Hub on wikipedia and decide whether that is a route that you are willing to take when searching out future materials.
The link works for me, thank you so much.
Hi, thank you so much for the tip! I tried EuDML too, I kept on getting an "unsafe" message from my browser. It could be that the site is safe (could you confirm this please?) and it is some additional layer of security that is not necessary here (however I didn't have the courage to click since I had lost data before ignoring an "unsafe" message for a previous site). I didn't find this paper on zbMath either, but thank you for the information on GDZ and sci-hub, I'll be sure to check them out.
@asrxiiviii I, too, got the unsafe message; I chose to bypass it manually but do not make that suggestion, in general. A question around whether an individual website is safe is probably better directed elsewhere (definitely not on topic at MO). I can confirm that I used the site but cannot confirm that it was, is, or will be secure.
|
2025-03-21T14:48:31.927446
| 2020-08-29T19:14:44 |
370411
|
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"Andreas Blass",
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|
Stack Exchange
|
Set-theoretic geology: controlled erosion?
I have to say that after the two last posts by Timothy Chow on Forcing I got so intrigued that I am trying to rethink the little I know about this formidable chapter of mathematics.
I have also to add that, although aware of the new field of set-theoretic geology, I am far from having a full grasp of it, so pre-emptive apologies to experts if I ask something that is either trivial or false.
Onto the point. Suppose I start from a transitive model of set theory $M$, and, rather than trying to expand it, I would rather do something opposite, namely the following: given an element $G$ of the model, try to "yank it out", to remove it surgically so that what remains is still a transitive model $M_0$.
In other words, try to establish $M= M_0[G]$.
Of course things are not so easy: I want to eliminate $G$ from $M$, but obviously I have to get rid of a lot of other sets in $M$ which are associated to $G$, for instance other sets which would imply its existence. Moreover, I have to choose judiciously whether or not $G$ is removable in such a way that after its removal (and of its "peers" ) the remaining set is still a model of $ZF$ of the same ordinal height.
I would call this operation selective erosion (if there is a canonical name for this operation please supply it) .
I understand that this may not be possible in some scenarios: for instance if $M$ is the minimal model, it is too "skinny" to allow for removals. But, unless intuition fails me, there should be plenty of "fat" models which should be liable to erosion.
MOTIVE
The way I look at this scenario is kind of the reciprocal of forcing: I would like to yank out some specific $G$ which codes some specific truths in $M$, for instance get rid of some map which collapses some cardinals.
QUESTION:
Are there methods that can be employed to do the surgery I sketched
? Notice that I do not ask whether a model is liable to erosion, rather whether some specific sets can be removed, and if so how.
NOTE: if I already know that $M$ is a forcing extension by $G$, then the problem is already trivially solved. Rather, suppose I only know that $M$ is a transitive model and someone comes along and gives me a $G$ in the model, and asks: is $G$ removable? I want to answer yes or no. Again, in some particular case the negative answer is obvious (example if G is an ordinal in $M$). But what about less trivial cases?
ADDENDUM: After the comments of Asaf, and especially after the great first answer by Jonas, time to take stock: The first thing that comes to my mind is that there are at least TWO candidate strategies to tackle this problem (and perhaps neither of them is the good one). You can call them BOTTOM-UP, which is the one I have sketched very loosely in my "debate" with Asaf, and the one which I would call TOP-DOWN which is the one advocated by Jonas.
Let us briefly recap them:
BOTTOM UP. Start from a minimal model $W_0$ such that $A\notin W$ (for instance the constructibles in $M$) , and look at the set of extensions $W$ of the bottom $W_0$ such that $W[A] \neq M$, ordered by inclusion, then try to take the colimit of this ordered set (in other words, you hope that the sup of all of them is a model and does not contain A, but adding A you get M) . Is such a beast exists you found your A-eroded M
TOP DOWN See Jonas's answer (I would call it the "take the limit " method).
Notice that both could be considered a form of selective geology:
1 is like growing the "earth", from some core, till a layer where A is present is reached.
2 is actually more in line with erosion, getting rid of as much as you can, as so aptly Jonas said.
So, either going from non A-grounds and looking for their union, or from A-grounds and looking for their intersection
PROBLEM: Both methods rely on looking at a certain ordered set of models of ZF in the "universe' M, and on certain lattice operations which can be performed. I have absolutely no clue whether such operations (taking sups or infs) are admitted in all cases (my gut feeling is no).
The story continues....
Your question and the terminology "selective erosion" suggest that you want the submodel $M_0$ to be in some sense close to $M$. Otherwise, you could simply say that $G$ is yankable iff it's not in the minimal model, and the result of yanking is the minimal model. But in what sense should $M_0$ be close to $M$? The same ordinals? Then you can yank any $G$ that isn't in $L^{(M)}$. Maybe $M$ should be a forcing-extension of $M_0$? More specifically, a forcing extension by re-adjoining $G$? Or maybe $M$ should satisfy some form of the covering lemma over $M_0$? Or maybe $\dots$?
If you weren't here already to help me either tuning my often sloppy questions, or "yank them out" altogether, I should pay you to do so. So, let me be a bit more precise: YES, the idea is to get M_0 as close as possible to M (hence the name, I could have as well called judicious surgery) Secondly, I do not just want to "peel off" anything, rather suppose taht someone (say the devil) comes up to you and say: Andreas, you have this wonderful M. Let me pick a set in it, say G. Now, you have to tell me if this set is yankable or not, meaning that I can yank it out,
take out also a minimal set of sets in M which are related to G, in such as way that after that I end up with a transitive submodel of M of the same height and moreover if I throw in G back, close with respect to definable operations, again BANG! I get M. Now, it is clear to me that tons of Gs are not yankable, right? But some possibly are. So, again, think of this as putting a MINUS sign to forcing: M+ G = M[G], now the equatin would be : M -G is a M_0 such that M_0[G] = M. Kind of an inverse operation
Post Scriptum I am sure you already see where I am trying to go. The inverse of forcing, or "selective erosion"< assuming of course that is feasible and not trivial, would play the reversal role of forcing: rather than judiciously "fattening" a model a bit, and add new facts unknown to the ground model (but "true" outside) here you would like to have a method for removing facts from a model. Some kind of inner model construction to obliterate some facts from your initial model
Possibly related: https://math.stackexchange.com/q/291088/622
@AsafKaragila Todah Rabbah! Yes, it is related, especially your answer there. Now, problem is, we do not know if there is an inner model inside, although in fact we do (the constructibel minimal model) . Now, M_minimal is definitely inside M, but perhaps it is too "thin". But we may be able to salvage your idea: how about expanding M_minimal by forcing extensions all the while remaining inside M? If we are lucky, at some point we may meet G, in which case we win. But I suspect there are situations in which there is an ascending chain inside M, and none contains G...
elaborating this idea a bit further: M_minimal , M1, M2,..... etc all forcing extensions of the previous one. What one would like to see is whether our G is reachable by this chain.Even if it is, though, there is an issue: the extension which contains G may not reach the "crust" of M...
emendation to what I have said: actually, if G appears at some point of the expansion, it is certainly removable, but unless it is so to speak in the outermost boundaries of the expansion, it will not qualify to describe M as M_0[G] for some M_0 submodel of M. It is, so to speak, "buried inside some layer of M".
In general, there's no reason to expect a minimal model. In $L[0^#]$, for example, I don't believe there is a maximal model without $0^#$ itself. Or even adding a Cohen real, there's no reason to expect a maximal model where "only the Cohen real is missing", because you will therefore still want the Cohen real restricted to various subsets of $\omega$.
You might find the notion of grounds and mantle relevant here. It’s a nontrivial result that the statement “V is a set-generic extension of some inner model” is actually first-order expressible, and moreover, the intersection of all such inner models, the mantle, is also first-order definable. Usuba proved that the intersection of set-many grounds is itself a ground.
Asaf, I apologize for my ignorance, but are you saying that if I start from $M =L[0^#]$ L truncated to M is not a model of ZF?
@MonroeEskew as I mentioned in the question, I am a newbie in Set Theoretical Geology, but I do know a few fact, like the ones you mentioned (not their proof!) by reading a nice slides intro. Definitely relevant. I hope some expert will chime in (included you and Asaf) to tell me what can be done and what is unknown. Meanwhile, thanks to you both for the useful comments
@Mirco: I don't really understand your question. My point is that there is no maximal model without $0^#$. You said that you want to avoid removing "too many sets". Going all the way down to $L$ is going to remove a lot of sets that you could have kept.
@Asaf, sorry I was not clear. What I meant is not to erode till I get to L, rather, I want to have a ground INSIDE M to start my campaign, and if this ground exists, grow it (add forcing layers) in the hope that at some point I stumble upon my selected G. Notice that assuming that there is a ground, there are many ascending chains of forcing extensions, and so I will have to be able to see if G gets intercepted in any of them. Makes sense?
Well, then let's go to the case of $L[c]$ where $c$ is a Cohen real.
If by that you mean M = L[c] then obviously the case G= c is trivial: Just define M_0 = L (relativized to M) and we are done. If G is not c, then it is not so trivial, at least to me. Intuitively, the simple minded approach would work is G has , so to speak, the same definability strength of c.: c is in L[G] and conversely.
But you removed too much. You can keep the subset of the Cohen real with the even coordinates. You can even save a large number of subsets, provided you're willing to forego the axiom of choice.
I see your point now. Well done. But, rather than deter me, this makes me even more confident that if you want to erode without taking out "too much", the best strategy would be to start from some basic ground, grow it in all possible ways as long as it does stay inside M, and just try to detect where and when G finally pops up.
So how would you erode $0^#$ from $L[0^#]$?
Answer: I would not erode it :) Rather, assuming there is a CORE (not a mantle, something much more solid, such a L itself restricted to M ) I would begin growing the core in all possible forcing directions (think of a directed lattice of extensions). Now, if I understand you (remember you are the set theorist, not I) , zero sharp is in none of the extensions in the lattice. What about then taking the minimal submodel of M which contains ALL the forcing extensions?
Two cases: either this fat submodel is M itself, in which case zero sharp is NOT erodable, or it is a proper submodel of M. In that case, I call it M_0 and add zero sharp. PS I would bet that the envelope of the lattice is M itself
Your bottom up really can't work. Suppose that $x$ is a generic real over $L[0^#]$, make it even minimal, just for kicks. Now you want to remove $x$ from $L[0^#,x]$. By minimality we even know what the result must be, $L[0^#]$. But if you start with $L$, no matter what you do, you will never, ever, ever, reach $L[0^#]$ by adding more and more generics.
Asaf, to begin with, even if your counterexample worked, it would simply imply that in SOME cases my bottom up method does not work, not that it "can't work". Basic logi. Secondly, if you have an answer to this question which goes beyond what Jonas has already put forward, I strongly invite you to post it. MO has already complained that our back and forth goes beyond the standards of acceptable comments, and suggested continuing the conversation in chat. If you wish to do it, bring the laser saber... :)
now, your counterexample: let us say that we work in $M =L[0^#, x]$ without knowing that it is a forcing extension of $M =L[0^#]$ and the devil chooses x. I consider the set of all submodels of M which are generic extensions of L and which omit x, ordered by inclusion. This set is not empty (trivial). Assume I take the UNION of all of them. The union is exactly your friend, namely $M =L[0^#]$.
emendation: the set of all extensions of L which do not contain A AND such that their are NOT A-grounds (in the sense specified by The answer)
Wait, are you claiming that $0^#$ lies in some set generic extension of $L$?
You like cheap shots, don't you? No I do not claim that. What I claim is this: assume you start with $M =L[0^#, x]$. You want to erode x from M, presumably reducing it to $M =L[0^#]$ . Now, let us call $ $M_0 =L[0^#]$. Now I want somehow reach M_0 from below. I believe that , one can express M_0 as a union of forcing extension , NOT as any one of them . PS This is my last reply to you as a comment: either you post an answer, I promise that if it is well written I will give you my vote, or light sabers in private chat
What a fantastic question, and thanks to Asaf and Mirco for the great discussion in comments! I love the idea of “removing” a given set from a model of ZFC, to obtain a smaller model of ZFC - some kind of inner model method analogous to the outer model method of forcing. This may not be a complete answer, but I think that geology does offer a useful framework for attacking this question, at least when the “erosion” is strictly due to forcing (the more general question, when is a set removable at all while leaving behind a model of the same height, can be answered I think by looking to see whether the set in question is in $L$).
(Recall from set-theoretic geology: an inner model $W$ is a ground of our universe V if it is a transitive proper class satisfying ZFC, such there exists $G \in V$ which is generic over $W$ and $W[G]=V$. The foundational theorem of geology says the grounds of $V$ form a uniformly first-order definable collection of inner models in $V$).
Given a candidate set $A\in V$, we can ask whether $A$ is forcing-erodable by asking “Is there a ground $W$ that omits $A$”? Any such ground $W$ is a candidate for the model obtained by removing $A$ from $V$.
How do we identify a single, canonical inner model by removing $A$? In contrast to forcing, in which we want add as little as possible to $V$ in order to obtain $V[G]$, here we are doing the inverse - I argue that we want to remove the absolute maximum possible from $V$, while still retaining the property that everything we remove can be added back by adding $A$ itself.
For example, given a Cohen extension $V[c]$, we can eliminate $c$ by going to an inner model $V[c^\prime]$ that contains only the real $c^\prime$ that lies on the even digits of $c$... but this is unsatisfying, because although we removed $c$ it feels as though we only removed half of the information contained in $c$. To “erode $c$”, we want to go all the way down to the inner model $V$.
Geology gives us an approach. For a set $A \in V$, call a ground $W$ of $V$ an $A$-ground if:
$A\notin W$ (we are eroding $A$)
$W[A] =V$ (we are not going ‘too far’ - everything we remove can be added back by adding $A$)
Is there a minimal such $A$-ground? I am not certain of the answer, but the natural candidate is the intersection of all $A$-grounds (let’s call this the $A$-mantle).
Questions: If $M_A$ is the $A$-mantle, then
is $M_A$ an $A$-ground ? If so, this is the right candidate for “eroding $A$ from $V$”.
If $M_A$ is not an $A$-ground, then is $M_A$ a model of ZFC? If that is the case, then does $M_A[A]=V$?
These are analogous to the questions in geology “Is the Mantle a model of ZFC” and “Is the mantle necessarily a ground”.
Jonas, I have no time to read your answer now (but I will!). You already got my like for 1) taking the first stab 2) because I "smelled " your answer and it is definitely going where I would like it to go. Detailed comments forthcoming . Meanwhile I can tell you the REAL reason motivating my question (Siths never tell you their real agenda, but I suspect that Andreas had already understood, his is simply too good to fool ). My question is there because i want to really understand forcing, and by that I mean really really really. So I am trying first to understand its opposite :)
Removing AC from the equation is so much more interesting. Especially in the case of a Cohen real, where adding the one set will invariably add a lot of grounds (and over $L$ and other "small models" a proper class of them).
Thanks, @MircoA.Mannucci - this is something like my interest in this question as well. The introduction to your question above pointed me in the direction of Timothy Chow's two recent questions about forcing - thank you, they were great reads and your answers were as well.
@AsafKaragila - I must admit, questions about forcing without AC seem like a terrible quagmire to me - but that just says I need to work more with models without choice! Did I read your comment correctly - adding a Cohen real can add class-many grounds?
Will there generally be a proper class of A-grounds and is it ok to intersect them all in that case?
https://arxiv.org/abs/2006.04514, Yes, adding a Cohen real to L has a proper class of grounds. However their intersection is L. More interestingly, their union is a ZF model which is not a ground of the Cohen extension.
@jonasreitz my answer to your answer is in the ADDENDUM. Hopefully there will be more "digging", just to remain on geology...
@jonasreitz when u have a chance, take a look at my new "geological" question. Perhaps you have something to contribute there too. All the best, Mirco
@MircoA.Mannucci - just responded - great question!
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2025-03-21T14:48:31.928747
| 2020-08-29T19:21:48 |
370412
|
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Stack Exchange
|
PA, provably total function, Gödel's T
If I am not mistaken, the provably total functions of $PA$ are exactly those definable in terms of the functionals of higher type $T$, as introduced by Gödel in his 1958 Dialectica paper.
In particular, for theories $\Sigma_n-IA$, the fragments of $PA$ where induction is restricted to $\Sigma_n$ formulae, and for the classes of functions definable in terms of $T_n$, the fragments of $T$ limited to recursors $R_\sigma$, for $\sigma$ of rank at most $n$ (where $\operatorname{rank}(0)=0$, $\operatorname{rank}(\sigma(\tau))= \operatorname{max}(\operatorname{rank}(\sigma)+1,\operatorname{rank}(\tau))$, the provable total function of $\Sigma_n-IA$ are characterized as those definable in terms of the functionals of higher type $T_n$.
If it is so, I am looking for a detailed, direct proof of it (both directions). In particular, I do kind of remember that to prove that any function definable in term of $T_n$ is provably total in $\Sigma_n-IA$, the way to go is to formalize Tait's computability predicate. I am interested to double check that that is the case, and –if it is– it would be great to look at a reference where all the details has been spelled out.
('direct' means without any detour to ordinal recursion, or Grzegorczyk functions.)
Using notation of simply-typed $\lambda$-calculus, does $\sigma(\tau)$ mean $\sigma \to \tau$ or $\tau \to \sigma$? The definition of rank looks strange. The ranks of $(\mathbb{N} \to \mathbb{N}) \to \mathbb{N}$ and $\mathbb{N} \to (\mathbb{N} \to \mathbb{N})$ should not be the same.
I would say $\sigma \to \tau$. Yes, those two ranks should be different. Maybe rank($\sigma(\tau)$)= max(rank($\sigma$)+1,rank($\tau$))?
Yes, that would be the usual definition of rank.
Good, thanks. I edit the question.
In Shoenfield's mathematical logic, pp.224--227, you will find a reasonably detailed and direct proof of the general case (provably recursive function of $PA$ iff designated by some constant of $T$). The converse implication is proved by formalizing the reducibility predicate. It may be possible, by a careful analysis of the proof, to extract a direct proof of the result you are looking for.
Yes, thanks. I did not check Shoenfield's, my fault. If I am not misreading him, for the PA to T direction he goes thru the Kreisel's no-counterexample interpretation; for the T to PA direction, he claims that the relevant propositions about Tait's reducibility predicate can be proved within PA (of course, for each single type $\sigma$). I understand that it is moderately easy to fill in all the details but I still hope that somebody has already done all the work (for fragments too).
|
2025-03-21T14:48:31.929210
| 2020-08-29T21:47:34 |
370417
|
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|
Stack Exchange
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Is Axler's Linear Algebra done right insufficient for an introduction of the subject?
Is Axler's Linear algebra done right insufficient for an introduction of the subject?
I have recognized the similar question, but I have a somewhat different situation. My familiarity with linear algebra is pretty much limited to basic matrix computation . I am currently in Multivariate Calculus and am looking for an introduction to linear algebra to prepare for future classes. The reason I mention Axler's is because it is the only book I have on hand, and I worry it is too abstract for an initial treatment. If any of you could clarify whether starting with this book is the right choice I would be delighted. Additionally, if you would recommend a different book for an introduction, what would you say?
Axler is intended as a second course in Linear Algebra, and my experience is that it works much better as a second course than a first one.
Thank you! Are there any texts you would recommend for first course? Some texts I have pondered those by the authors of: Lay, Strang, Hoffman, Friedberg, Halmos.
I don't have a good sense of which of those would be best. I tried teaching an honors linear algebra class out of Axler and thought it was a mistake, but haven't had a chance to try a different book yet.
I have no strong feelings on this, but wouldn't this question be better on MathStackExchange?
Axler is not good for first course, especially for a self-study. Go with Strang.
As a first course for math/CS majors I prefer Lay who gives a good mix of theory and numerics. I hear that engineers like Strang. Personally, I find Strang handwavy and needlessly subtle/brief, especially in later chapters. Keep in mind that Strang writes for MIT students.
When I was a first year student we used Fraleigh-Beauregard. It was a pretty good first introduction, as I recall. https://www.amazon.com/Algebra-Paperback-Fraleigh-Raymond-Beauregard/dp/B00CDXJH7Y
Check https://elblogdecontar.files.wordpress.com/2017/01/john-b-fraleigh-raymond-a-beauregard-linear-algebra-addison-wesley-1995.pdf
(Pending moving this question to MathStackExchange, perhaps...)
Ironically, it may be best to learn/read "linear algebra done wrong" first, not only because it's what nearly everyone sees (so you'll have an idea what other people know/think), but because "doing it wrong" is in many ways very natural, even if considerably suboptimal in the long run. As with many mathematical things.
I myself have not kept up with introductory linear algebra texts, but, ... although years ago I was impatient with Strang's book because it was insufficiently modern/abstract, with some hindsight I tend to think that it is understandable by the students in the class... while my notion of "correct" linear algebra (notation, terminology, etc.) is not. The "right" version does not make sense until later, after having gone through the transitional stage of "doing things wrong".
I appreciate your response greatly. I have lightly glossed over LADW in the past, and my only worry with it is, if it is viable as an introduction. Namely, does it provide suffient exercises and applications that are integral for understanding, or does it go straight into theory without computation. Pardon my potential naivety, as I am still not acquainted with the subject matter.
I don't have a copy of it to look at, and never really did look at its exercises and such... so, while I am confident that it aims to be pedagogically substantial, it may have some grudges or similar that skew the exercises. It is true that... with hindsight, as in many things... the exercises can be construed as reasonable, but my last impression from some years ago is that beginners would have trouble. Strang's book, the one I mainly remember, is much more accessible to beginners... even though in the long run I'd absolutely NOT take up that book's literal attitude about linear algebra.
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2025-03-21T14:48:31.929513
| 2020-08-29T21:52:10 |
370418
|
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|
Stack Exchange
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Public games for solving open problems
Many people in the world like playing games. Also, maybe we can design special interesting games for some open problems in mathematics (or physics).
In quantum mechanics, there are some games (for example, Quantum Moves; which it is about quantum control and hybrid intelligence About the Game) that if you play a game and can finish it, you solved an open problem. For example (simplified explanation), there is a bucket of water and a lot of obstacles (hills and pits). If you can move the bucket to the end of the road and there are specific amount of water on the bucket, you solved an open problem in quantum mechanics (you solved an equation in quantum mechanics). The game is so funny and everyone can play it without any knowledge about the quantum mechanics Quantum Moves.
My question is:
Are there similar games for solving open problems in mathematics (or in physics)?
Jigsaw puzzles and the Riemann Hypothesis, https://www.dr-mikes-math-games-for-kids.com/blog/2014/05/jigsaw-puzzles-and-the-riemann-hypothesis/
Not directly related, but some time ago I gamified a problem I had solved by computer to see if it'd be fun to solve by hand. It was NOT http://villesalo.com/kuluma/e57game.html
Do you mean games that have actually been advertised to the public as something fun to play, or are you talking about games in an abstract sense? Because there are some open problems that are about games.
@PyRulez I mean the games that are fun to play.
I believe the start of citizen-science games for scientific research
was FoldIt,
asking users to find a low potential energy configurations of protein molecules.
Some publishable results were achieved.1
Wikipedia has a list of citizen-science projects here.
Only one is marked as "Mathematics":
SZTAKI Desktop Grid,
closed two years ago.
There are several labeled Physics, including
Quantum Moves
as mentioned by the OP.
Astronomy projects are prevalent, e.g., Planet Hunters.
1Cooper S, Khatib F, Treuille A, Barbero J, Lee J, Beenen M, et al. (August 2010). "Predicting protein structures with a multiplayer online game." Nature. 466 (7307): 756–60.
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2025-03-21T14:48:31.929722
| 2020-08-29T23:29:47 |
370423
|
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|
Stack Exchange
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What are conditions to satisfied by rational prime p so that every prime lying above p is a prime of order 1 and generates class group?
I was reading a paper on Euclidean ideals by H Graves and M. Ram Murthy. I have a problem in understanding one of the claims.
setup
Let $K$ be a number field and $H(K)$ is its Hilbert class field. Suppose $H(K)/\Bbb Q$ is abelian and $f(K)$ is the conductor of K which is defined to be the smallest even number $n$ such that $K \subset \Bbb Q(\zeta_n)$. Suppose that the class group $\mathrm{Cl}_K$ of $K$ is cyclic and let $C\in\mathrm{Cl}_K$ be an ideal class generating the ideal class group.
On page 2 they claim that there exists some integer $a$, $0<a<f(K),(a, f(K)) = 1,$ such that if $p \equiv a\pmod{f(K)},$ then $\mathfrak{p}$ is of degree $1$ and $[\mathfrak{p}]=[C]$. Here $\mathfrak{p}$ is a prime in $K$ lying above $p$.
I tried to establish a relation between the class $p \pmod{f(K)}$ and the ideal class of the primes above $p$ by tracking down the map given there, but I was not successful. Any help in this regard is very appreciable.
Please define the objects in your question clearly. The link you provided is behind a paywall, so as written the question is unclear and is not self-contained.
https://www.ams.org/journals/proc/2013-141-09/S0002-9939-2013-11602-3/S0002-9939-2013-11602-3.pdf
First of all, as they observe, the assumption that $H(K)/\mathbb{Q}$ is abelian ensures that $H(K)\subseteq\mathbb{Q}(\zeta_{f(K)})$ and not only $K\subseteq\mathbb{Q}(\zeta_{f(K)})$. Also, they work under the assumption that the class group of $K$ is cyclic, and I have added this to your question.
Now, Artin reciprocity gives an isomorphism between the ideal class group $\mathrm{Cl}_K$ of $K$ and the Galois group $\operatorname{Gal}(H(K)/K)$: this is one of the main results of Global Class Field Theory. In particular, there is an isomorphism (let me call it $\varphi$) between $\mathrm{Cl}_K$ and a subgroup of $\operatorname{Gal}(\mathbb{Q}(\zeta_{f(K)})/\mathbb{Q})$, as a consequence of the inclusion $H(K)\subseteq\mathbb{Q}(\zeta_{f(K)})$.
On the other hand, the theory of cyclotomic fields yields an isomorphism
$$
\tau\colon (\mathbb{Z}/f(K))^\times\cong \operatorname{Gal}(\mathbb{Q}(\zeta_{f(K)})/\mathbb{Q}).
$$
But more is true: the isomorphism $\tau$ associates to every prime $p\nmid f(K)$ (hence, whose class $[p]$ lies in $(\mathbb{Z}/f(K))^\times$) the element $\mathrm{Frob}(p,\mathbb{Q}(\zeta_{f(K)}/\mathbb{Q})$.
Now we have this class $C$: through $\varphi$ it corresponds to a unique element $\varphi(C)=\gamma\in\operatorname{Gal}(H(K)/K)$. The Chebotarev Density theorem says that there exists a positive density of primes $\mathfrak{q}$ in $K$ such that their Frobenius element $\operatorname{Frob}(\mathfrak{q},H(K)/K)$ equals $\gamma$. Since primes of degree $>1$ have density $0$, the above set of primes contains infinitely many primes of degree $1$, lest its density would be $0$. So we can chose one prime $\mathfrak{p}$ in that set, and it will
Have degree $1$ in $K/\mathbb{Q}$;
Have Frobenius element $\mathrm{Frob}(\mathfrak{p},H(K)/K)$ equal to $\gamma=\varphi(C)$;
Setting $p=\mathrm{Norm}^K_\mathbb{Q}(\mathfrak{p})$, this number will be prime (by 1.) and its Frobenius element $\tau(p)=\mathrm{Frob}(p,H(K)/\mathbb{Q})$ will restrict to $\mathrm{Frob}(\mathfrak{p},H(K)/K)=\varphi(C)$ and will depend only on $p\pmod{f(K)}$: letting $a$ be the integer satisfying $a\equiv p\pmod{f(K)}$ and $0<a<f(K)$ you get your element.
How can we say primes of degree >1 have density 0?
Look at the answer in https://math.stackexchange.com/a/1597230/277479
|
2025-03-21T14:48:31.930015
| 2020-08-29T23:48:46 |
370424
|
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|
Stack Exchange
|
Periods of Coxeter transformation associated to root posets
$\DeclareMathOperator\Co{Co}$Let $P$ be the root poset associated to a simple Lie algebra.
Let $L=L(P)$ denote the distributive lattice of order ideals of $P$ and let $\Co_L$ denote the Coxeter matrix of $L(P)$ which is defined as $-C^{-1} C^T$ when $C$ is the matrix with entries $c_{i,j}=1$ when $i \leq j$ and $c_{i,j}=0$ else for $i,j \in L$.
Question: For which types is the matrix $\Co_L$ periodic? That is when do we have $\Co_L^t=\operatorname{id}$ for some $t \geq 1$? What is the period (the smallest such $t$) in case this is true for a given type?
The quesiton has a positive answer for all simple Lie algebras of rank at most 6. So maybe it is always periodic?
For $A_n$ it is $2n+4$ for $n=2,3,4,5,6$ but for $A_1$ it is 3. Is it $2n+4$ in general for $n \geq 2$?
For $D_4$ it is 14 and for $D_5$ it is 18 and for $D_6$ it is 22. Is it given by $4n-2$?
For $B_2$ it is 10 and for $B_3$ it is 7 and for $B_4$ it is 18 and for $B_5$ it is 11 and for $B_6$ it is 26. Is it $8n+2$ for $B_{2n}$? Is it $4n+3$ for $B_{2n+1}$?
For $G_2$ it is 14 and for $F_4$ it is 26 and for $E_6$ it is 26.
edit: I saw that in the introduction of https://arxiv.org/pdf/1710.10632.pdf there is a conjecture of Chapoton stated that says the the incidence algebras of those distributive lattices are fractionally Calabi-Yau. The truth of this conjecture would imply a positive answer to the question of this thread.
Is the period always $2*(h+1)$ where $h$ is the Coxeter number?
@SamHopkins For $B_3$ it is 7.
It seems that the period could be $h+1$ or $2(h+1)$.
It seems the first nonnegative entry in the row of $-C^{-1}C^T$ indexed by $i \in L$ is $\mathrm{Row}(i)$, the rowmotion of the order ideal (a.k.a., the Panyushev map). Since root posets are known to have good behavior of rowmotion (see https://arxiv.org/abs/0711.3353 and https://arxiv.org/abs/1101.1277), this may be key to your observation (although note that the order of rowmotion is $2h$ or $h$, slightly different than what you see here). Rowmotion appeared in another related Q of yours: https://mathoverflow.net/questions/343203/permanent-of-the-coxeter-matrix-of-a-distributive-lattice
@SamHopkins I added more terms and some guesses for the types. For type $A_n$ and $D_n$ it seems to behave regular, while for type $B_n$ one might have to split into even and odd cases.
Let $M:=-C^{-1}C^T$. Let $D$ denote the permutation matrix of the action of rowmotion. Thus $MD$ is lower-triangular by my previous comment (which is actually proved in the linked to MO question). It seems that we may always have that $(MD)^2$ is the identity. I feel like this must be related to your problem somehow.
You can see Sage code here: https://cocalc.com/share/0efb266aca1eef099315e87e1750671ee5737323/root_poset_coxeter_transformation.sagews?viewer=share
@SamHopkins That is a nice observation. I will try to see where it might come from in case this identity is true.
Another comment: the $h+1$/$2(h+1)$ order also appears in the paper of Yildirim you link to. She shows $\tau^{h+1}=\pm 1$, in the context of minuscule posets (rather than root posets).
@SamHopkins Isnt this conjecture 5.5. in the article?
You're right. It's proved there in most cases, but not quite all.
|
2025-03-21T14:48:31.930262
| 2020-08-29T23:49:30 |
370425
|
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|
Stack Exchange
|
Wightman reconstruction theorem-details of the proof
First of all forgive me if this question is not well suited for this forum: it is motivated by physics however after all my concerns are mathematical so I hope it would be appropriate to post it here. I'm interested in understanding the proof of Wightman reconstruction theorem: the standard literature where this proof can be found is the book by Streater and Wightman ,,PCT, Spin and Statistics and all that''. However I found several problems in this proof and I would be grateful if someone can clarify those issues.
First problem which I have is the equality $(f,g)=\overline{(g,f)}$ (see page 119). Authors claim that this equality follows from hermicity condition namely that $W_n(f_1 \otimes f_2 \otimes ... \otimes f_n)=\overline{W_n(f_n \otimes f_{n-1} \otimes ... \otimes f_1)}$ for simple tensors. However in the definition of this scalar product we sum terms of the form $W_n(\overline{f_k} \otimes g_{n-k})$ and I don't see why are we allowed to swap the first $k$ ,,variables'' with the other $n-k$ (we are allowed only to reverse all variables). Should we assume something more about our Wightman distributions?
The second problem which I have is somewhat similar:
On page 120 it is stated that $(\varphi(\overline{h})f,g)=(f,\varphi(h)g)$ i.e. the formal adjoint of $\varphi(h)$ is $\varphi(\overline{h})$. However I would rather say that the formal adjoint multiplies by $\overline{h}$ from the right as opposed to $\varphi(h)$ which multiplies from the left by $h$. But after all-do we need to check this equality in order to construct quantum field theory? There is no requirment for the form of $\varphi(h)^*$ in the Wightman axioms (or am I wrong? Maybe it has something to do with the fact whether we want to construct neutral or charged field etc?)
And the last one
On page 125 it is claimed that the fact that the support of the (joint) spectral measure for $\{U(a,I)\}_{a \in \mathbb{R}^4}$ lies in/on the future light cone is an immediate consequence of Axiom b) (spectral conditions) on page 117. However it is not clear for me how one can reverse the argument given on page 109 where it is proven that once we start with the quantum field obeying Wightman axioms then this spectral condition is satisfied. Roughly speaking on page 109 they show that $4$-dimensional Fourier transform of $W$ is zero (in other words that Fourier transform applied to one variable is zero) which implies that the $4n$-dimensional Fourier transform is $0$-I don't see how to reverse this argument.
I will be grateful if someone could shed some light upon my issues.
|
2025-03-21T14:48:31.930589
| 2020-08-30T00:55:14 |
370427
|
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|
Stack Exchange
|
When does Matiyasevich's theorem "kick in"?
Hilbert's 10th problem was famously resolved by the Matiyasevich–Robinson–Davis–Putnam theorem: the theorem implies that there is no algorithm which decides whether a given polynomial equation with integer coefficients has an integer solution.
However, if one restricts to various subsets of polynomials, then Hilbert's 10th admits a positive solution: for example, if one only looks at polynomials in one variable then such an algorithm does indeed exist: given $f(x) = a_d x^d + \cdots x_0, x_i \in \mathbb{Z}$ for $i = 0, \cdots, d$, this equation admits an integer solution if and only if $f(k) = 0$ for some divisor $k$ of $a_0$, so the algorithm reduces to factoring $a_0$ and then checking $f(k)$ for each $k | a_0$.
I believe Baker's theorem on linear forms in logs imply that an algorithm also exists for polynomials in two variables, but such an algorithm is highly impractical, due to the astronomical bounds Baker's method produces for the heights of possible solutions.
Hilbert's original question really only concerns hypersurfaces in $\mathbb{A}^n$ for $n \geq 1$. The previous two paragraphs show that for $n = 1,2$ the conclusion of Hilbert is actually true, while the first paragraph shows that the conclusion is false in general. My question is, for each $n \in \mathbb{N}$ what is the largest class of hypersurfaces in $\mathbb{A}^n(\mathbb{Z})$ for which the conclusion of Hilbert's 10th problem is true? Indeed, do we expect the conclusion of the MRDP theorem to hold as soon as $n \geq 3$?
The wikipage states that we know that it kicks in for $9$ natural variables, or 11 integer variables. Are you asking if these quantities have been lowered?
This is basically an open problem. For instance it should be solvable for rational points on curves (i.e. homogeneous polynomials in three variables) but we don't know.
See also this: https://mathoverflow.net/q/51987/6085
Levent Alpoge's thesis might be interesting to look at.
There is still no known algorithm for deciding whether a given integer is the sum of three cubes.
|
2025-03-21T14:48:31.930778
| 2020-08-30T03:23:49 |
370430
|
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|
Stack Exchange
|
If $\mathcal C$ is a $\kappa$-accessible $\infty$-category, then is $Mor \mathcal C$ $\kappa$-accessible?
If $\mathcal C$ is a $\kappa$-accessible 1-category, then the category of morphisms $Mor \mathcal C$ is a $\kappa$-accessible 1-category, with the $\kappa$-presentable objects being those morphisms whose domains and codomains are each $\kappa$-presentable.
In the context of $\infty$-categories, the best result I know of is HTT Proposition <IP_ADDRESS>, which shows that if $\mathcal C$ is a $\kappa$-accessible $\infty$-category and $\kappa \ll \tau$ (meaning that $\lambda < \tau \Rightarrow \kappa^\lambda < \tau$ and $\kappa < \tau$), then $Mor \mathcal C$ is $\tau$-accessible.
Lurie's proof, via HTT Lemma <IP_ADDRESS> (note that this lemma's proof has been revised since the printed edition), seems to really use the full strength of the assumption $\kappa \ll \tau$. Can this be improved to $\kappa = \tau$? Or at least to the "sharply below" relation $\kappa \triangleleft \tau$ familiar from the theory of accessible 1-categories?
This boils down to asking: if $\mathcal C$ is $\kappa$-accessible, then is every morphism of $\mathcal C$ a levelwise $\kappa$-filtered colimit of morphisms between $\kappa$-presentable objects?
In the case of 1-categories, a follow-your-nose argument works: you just take colimiting diagrams for the domains and codomains and factor the original map through stages of the colimit. I suspect that the same must be true in $\infty$-categories, with the same argument in principle working. But the question seems to be much more subtle $\infty$-categorically.
By the way -- Lurie systematically uses the relation $\kappa \ll \tau$ in place of the relation $\kappa \triangleleft \tau$ familiar from accessible 1-categories. I don't know whether this is the only place this matters, but I think it does here (in Lemma <IP_ADDRESS>'s construction of the $K(\alpha)$'s)
$Mor(\mathcal C)$ is also $\kappa$-accessible by HTT Prop. <IP_ADDRESS>.
@MarcHoyois Thanks -- just a few pages from where I was looking, too!
Marc Hoyois answered in the comments: the answer is affirmative, by HTT <IP_ADDRESS>.
|
2025-03-21T14:48:31.930949
| 2020-08-30T06:41:42 |
370434
|
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|
Stack Exchange
|
Is $\limsup_{x\to\infty}\big(\sum\limits_{d|3^x-1}{1/d}\big)/\big(\sum\limits_{p<x}1/p\big)<\infty$?
Can anyone provide any suggestion if the following is true for natural number $x$?
$$\limsup_{x\to\infty}\dfrac{\sum\limits_{d|3^x-1}{\frac{1}{d}}}{\sum\limits_{p<x}\dfrac{1}{p}}<\infty$$ where $p$ runs over primes.
P.S I know that
the bound ${\sum\limits_{d|n}{\frac{1}{d}}}<e^{\gamma}\log\log n +O(1)$is true so maybe it could help.
(Using Sage, checking for $x<180$) The given fraction seems to always be in $[3/4,3]$, with no obvious tendency. For $x$ prime, it seems to decrease and is $<0.8$ for $x\ge 149$, while for (apparently) most other values it's $\ge 1$, with apparent jumps (e.g., $x=160$ yields $1.58...$).
BTW liminf should be zero: for $x$ prime, one would expect the sum in the numerator to be $1+(1/2)+o(1)$ (maybe easy), so the fraction would be $\sim 3/2\log\log x\to 0$.
Your upper bound already shows a $\preceq \log x / \log\log x$ bound. So a slow partial convergence to infinity would not be much visible from such a small range as above.
The inequality in your "P.S." is Robin's criterion for the Riemann Hypothesis. So we don't know that it is true, we only expect (hope, conjecture etc.) that it is true.
@GH from MO I did an error.I guess now with a big Oh it can be proven.
@Superguy: The inequality in your updated "P.S." is known indeed.
The OP's first display is true. Indeed, Erdős (1971) proved that
$$\sum_{d\mid 2^x-1}\frac{1}{d}\ll\log\log x\ll\sum_{p<x}\frac{1}{p},$$
and he noted that the proof works equally well for the divisors of $a^x-1$.
You seem to write $f\ll g$ for $f=O(g)$. Is this standard? I always understood it as $f=o(g)$.
@YCor the Vinogradov notation $f(x) \ll g(x)$ is indeed the same as the Landau big O notation $f(x) = O(g(x))$ in analytic number theory.
One remark for the OP. On certain well-believed conjectures, the exact value of the limsup can be determined. The answer involves $\gamma$, $\log 3$, and an integral expression with Dickman's rho function. See this arXiv paper: https://arxiv.org/abs/2006.02373 (Theorem 1.2 and section 6)
|
2025-03-21T14:48:31.931127
| 2020-08-30T08:23:46 |
370435
|
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|
Stack Exchange
|
How to use inverse Fourier transform to get the original signal
Suppose I have a signal $u(x)$ where $x\in[0,2\pi]$ and the signal has following form in Fourier space,
$\hat{u}(\kappa)=\begin{cases}
\frac{1+i\tan{(\varphi_u(\kappa))}}{\sqrt{1+\tan^2{(\varphi_u(\kappa))}}}\kappa^{\alpha/2},\ \ 0<\kappa\le\kappa_c \\
0+i0,\ \ \ else
\end{cases}$
where $\varphi_u(\kappa)$ is a random number for each $\kappa$, and $\alpha$ is a constant.
How can I get $u(x)$ using inverse Fourier transform?
Hi and welcome to the MathOverflow. I suspect that the sole knowledge that $\varphi_u$ is a random number is not of great help: do you know something about its probability distribution as a random variable?
@DanieleTampieri $\varphi_u(\kappa)$ is a random variable which is uniformly distributed in $[-\pi, \pi]$ for each $\kappa$.
The expectation value $E[u](x)$ can be obtained by inverse Fourier transform of the expectation value of $\hat{u}(k)$, since the Fourier transform is a linear operation; If I interpret the statement in the OP that the phase $\phi$ is a "random number" as saying that $\phi$ is uniformly distributed, then the integral
$$\frac{1}{\pi}\int_0^\pi \frac{1+i\tan\phi}{\sqrt{1+\tan^2\phi}}\,d\phi=\frac{2}{\pi}$$
gives $E[u](x)$ as the inverse Fourier transform of $(2/\pi)\kappa^{\alpha/2}$, which is an incomplete Gamma function.
I should note that the statement in the OP that both $u(x)$ and $\hat{u}(k)$ have compact support is inconsistent.
|
2025-03-21T14:48:31.931263
| 2020-08-30T08:55:13 |
370437
|
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|
Stack Exchange
|
Pushforward of measure under surjective map
Let $X, Y, Z$ be measurable spaces with measures $\mu_X, \mu_Y, \mu_Z$ respectively. Let $\pi_Y : Y \times Z \rightarrow Y$ be the projection on $Y$ and $\pi_Z : Y \times Z \rightarrow Z$ the projection on $Z$. Let $\psi : X \rightarrow Y \times Z$ be a surjective map such that $(\pi_Y \circ \psi)_{*} \mu_X = \mu_Y$ and $(\pi_Z \circ \psi)_{*} \mu_X = \mu_Z$. Can we deduce that $\psi_{*}\mu_X = \mu_Y \otimes \mu_Z$ (this is a notation for the product measure of $\mu_Y$ and $\mu_Z$)?
If yes how? If not, what other conditions are required?
In case the answer to the above question is no, the following is a more precise account of my problem: Let $G$ be a Lie group and $H_i, K_i \leq G$ Lie subgroups for $i = 1,2,3$. Let $X_i = K_i \setminus G /H_i$ be the double coset space given by quotienting from the right by $H_i$ and from the left by $K_i$,i.e. an element of $X_i$ is a double coset of the form $K_igH_i$. Endow $X_i$ with the unique $G$ invariant measure and denote this measure by $\mu_{X_i}$. I have a map $\psi: X_1 \rightarrow X_2 \times X_3$ which satisfies the properties above. Namely, $\psi$ is surjective and $(\pi_{X_i} \circ \psi)_{*}\mu_{X_1} = \mu_{X_i}$ for $i =2,3$. I would like to show that $\psi_{*}\mu_{X_1} = \mu_{X_2} \otimes \mu_{X_3}$. Thank you in advance for your answer.
For the identity $\psi$ on $Y\times Z$, you ask whether a measure $\mu$ on the product with given marginals $\mu_Y=(\pi_Y)*\mu$ and $\mu_Z=(\pi_Z)*\mu$ is necessarily the product measure which, of course, is wrong.
The answer to the first part is no. The point is that surjectivity is not as strong a condition in this context as one might wish for.
Let $X=Y=Z=[0,1]$ with the Borel $\sigma$-algebra and $\mu_X=\mu_Y=\mu_Z$ be the uniform distribution. Consider the non-surjective function $x\mapsto (x,x)$. Its push-forward is clearly not $\mu_Y\otimes\mu_Z$, it is the uniform distribution on the diagonal. But you can take an uncountable subset $N$ of $X=[0,1]$ of measure zero (the Cantor ternary set will do). Then $N$ will actually be measurably isomorphic to the set $\{(x,x)\mid x\in N\}\cup\{(x,y)\mid x,y\in [0,1], x\neq y\}$ by Kuratowski's Borel isomorphism and the fact that all uncountable Borel sets in Polish spaces have the cardinality of the continuum. Let $\phi$ be such a measurable isomorphism. Now define $\psi:X\to Y\times Z$ by $\psi(x)=(x,x)$ for $x\notin N$ and $\psi(x)=\phi(x)$ for $x\in N$. The pushforward of $\mu_X$ under the surjection $\phi$ is the uniform distribution on the diagonal.
|
2025-03-21T14:48:31.931457
| 2020-08-30T11:16:53 |
370442
|
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|
Stack Exchange
|
Finding local algebra and relations lottery
This can be seen as an attempt for a mini Polymath project on homological properties of (local) finite dimensional algebras. You only need to know what a finite dimensional algebra is and have GAP to contribute. You can win nice prizes (see the end of the question).
A local quiver algebra is just an algebra of the form $A=K\langle x_1,...,x_p\rangle/I$ where there are at least $p \geq 2$ variables (we avoid trivial cases) and $K\langle x_1,...,x_p\rangle$ is the non-commutative polynomial ring in $p$ variables and $I$ is an ideal with $J^r \subseteq I \subseteq J^2$ where $J=\langle x_1,...,x_p\rangle$ and some $r \geq 2$.
Let $D(A)=\operatorname{Hom}_K(A,K)$ and define $e_i:=\dim_K(\operatorname{Ext}_A^i(D(A),A))$ for $i \geq 1$. $A$ is called Frobenius in case $A \cong D(A)$ or equivalently $A$ has a simple socle. We are only interested in non-Frobenius algebras in the following and assume $A$ is non-Frobenius.
You can enter such an algebra and calculate $e_i$ for $i=1,..,n$ as follows in the GAP-package QPA (a specific example, that should be easy to modify to enter your favorite local algebra):
n:=3;Q := Quiver(1, [[1,1,"a"],[1,1,"b"]]);KQ := PathAlgebra(GF(3), Q);AssignGeneratorVariables(KQ);rels := [a^2, ab+b^2-aba, ab^2, bab, ab+b^2+b^2a, b^3];A := KQ/rels;IsSelfinjectiveAlgebra(A);P:=IndecProjectiveModules(A)[1];;I:=IndecInjectiveModules(A)[1];;W:=[Size(ExtOverAlgebra(I,P)[2])];;for i in [1..n-1] do Append(W,[Size(ExtOverAlgebra(NthSyzygy(I,i),P)[2])]);od;W;IsSelfinjectiveAlgebra(A);
(you need to write a * between the variables in the relations to enter it in QPA, which is sadly not shown in the Mathoverflow output using the code )
The output at the end is $[ 0, 4, 4 ]=[e_1,e_2,e_3]$ and a "false" (meaning that your algebra is not Frobenius). The general output will be $[e_1,...,e_n]$ where $n$ is the first specified parameter in the input in case your algebra is indeed finite dimensional (in case it is not finite dimensional you should get an error and in case it is selfinjective you will see a "true" at the end in which case your example is not of the needed form). You can specify the number of variables in the input Q := Quiver(1, [[1,1,"a"],[1,1,"b"]]), here we have 2 variables a and b. And you can specify the relations in the input rels := [a^2, ab+b^2-aba, ab^2, bab, ab+b^2+b^2a, b^3]; . Thus the algebra here is $K\langle a,b\rangle/\langle a^2, ab+b^2-aba, ab^2, bab, ab+b^2+b^2a, b^3\rangle$.
You can easily specify the parameter $n$ (taking $n=3$ will be enough for some first experiments) and define your own algebra.
Here are the questions/prizes:
(Main prize): Find such an algebra $A$ with $e_i=0$ for all $i \geq 1$. You then disproved the second Tachikawa conjecture, the Nakayama conjecture and the finitistic dimension conjecture (and several other conjectures), which are the most important homological conjectures for finite dimensional algebras.
A discussion of the finitistic dimensions can be found in https://link.springer.com/article/10.1007/BF02100610 where also another related conjecture was disproven.
(Second prize): Find an infinite class of algebras $A_t$ with $e_i=0$ for $i=1,...,a_t$ ($a_t$ depending on $A$), where $a_t$ gets arbitrary large. You then disproved the Yamagata conjecture.
For Yamagata's conjecture, see for example https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/rigidity-dimension-of-algebras/6A8979068A79714D91394B746C42BA31 .
Find a commutative such $A$ with $e_1=0$. No such example is known, so you answer an open question.
Find an $A$ with $e_1=e_2=0$. No such example is known, so you answer an open question.
Find such an algebra $A$ having a d-cluster tilting module for some $d \geq 2$ (for this you need $e_1=e_2=...=e_{d-1}=0$). Prize (by me): 50 Euros for the first new (not isomorphic to $K\langle a,b\rangle/\langle a^2, ab+b^2-aba, ab^2, bab, ab+b^2+b^2a, b^3\rangle$) such algebra.
(note that it is also not known whether there is a local Frobenius algebra having a $d$-cluster tilting module for $d \geq 2$. Not even a non-projective module $M$ with $Ext_A^1(M,M)=0$ is known for local Frobenius algebras $A$)
This is the only non-elementary question. There is no fixed method to find cluster-tilting modules but when you find a nice algebra with $e_1$=0 and complexity one it might be a good candidate. Here complexity one can for example be checked by the property of the simple module having bounded projective resolution. To do this enter this code after you entered your algebra:
S:=SimpleModules(A)[1];G:=[];for i in [1..10] do Append(G,[Dimension(NthSyzygy(S,i))]);od;G;
In case this can be done quickly by the computer and the numbers are small, it is likely you found an algebra with complexity one (extremely rare for such algebras. I think for commutative (non-Frobenius) algebras no such example might be known). The output for the algebra as above is [ 5, 7, 5, 7, 5, 7, 5, 7, 5, 7 ]. For a general local algebra the output will be a quickly increasing sequence tending to infinity.
The only example known to mankind for a d-cluster tilting module for such algebras $A$ is the algebra $K\langle a,b\rangle/\langle a^2, ab+b^2-aba, ab^2, bab, ab+b^2+b^2a, b^3\rangle$ entered above, which appeared in Does this algebra have finite global dimension ? (Human vs computer) . It was found by Jan Geuenich by searching through some (random?) relations and I found a 2-precluster tilting module and Oyvind Solberg indeed proved that it was even cluster tilting in the cited thread. The relations are very weird, but it would be strange if there is really just one local algebra with (higher) cluster tilting modules. Noone knows how to construct another example, so some creativity and experiments might help to find more such examples! The algebra has just vector space dimension 6, so maybe there are other algebras out there of small dimension that can be found using some weird relations.
Note that the relations of this example can also be written in the form $rels := [a^2, (a+b)^2-(b^2+b)a]$ to give an isomorphic algebra.
|
2025-03-21T14:48:31.931822
| 2020-08-30T11:51:56 |
370443
|
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|
Stack Exchange
|
Connections in the setting of algebraic geometry
My level is at the beginning of a second year master. I'm interested in the project of translating some features of differential geometry to algebraic geometry. I'd like to know if there is an equivalent of the notion of connection in algebraic geometry. I've heard about Grothendieck connections, but I did not know if it is the only possible way to abstract the notion of connecton, nor I found any accessible material on the web. Can someone help me?
Some references here https://ncatlab.org/nlab/show/Grothendieck+connection might be useful.
Give this page a try.
If you can read french, I recommend Equations Différentielles à Points Singuliers Réguliers by Pierre Deligne, Springer Lecture Notes 163. The notion of connection in algebraic geometry is very clearly explained.
I think if you can read french the best source is Deligne book but there are several interesting articles by katz you can read for example:On the differentiation of de rham cohomology classes with respect to parameters, Algebraic solutions of differential equations (p-curvature and the hodge filtration),On the differential equations satisfied by period matrices.
there is also a beautiful article by Coleman where he introduces basic properties of connections and uses them to prove an interesting conjecture in number theory: Manin’s proof of the mordell conjecture over function fields.
Hopefully, the beautiful article mentioned above is this one
You should also have a look into Atiyah's 'Complex analytic connections in fibre bundles'. Even though he considers complex manifolds you can gain a lot insights from reading this paper.
One possible way to define connections is to use the language of sheaves and differential forms; see e.g. Chap. III, Definition 1.5, on p. 70 of [1]. Since these notions are equally well at one's disposal within the category of schemes, the notion of a connection on vector bundles can be defined for Algebraic Geometry along entirely analogous lines. For this, see e.g. this paper of Brian Osserman.
[1] Wells, R.O.,
Differential Analysis on Complex Manifolds
(Graduate Texts in Mathematics 65), Third Edition. Springer 2008.
|
2025-03-21T14:48:31.932022
| 2020-08-30T12:27:37 |
370445
|
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|
Stack Exchange
|
Non-zero, bounded, continuous, differentiable at the origin, compactly supported functions with everywhere non-negative Fourier transforms
Do there exist functions $F(x) \! : \, \mathbb R \to \mathbb R$ which are non-zero and bounded:
$$ \mathrm {Range} (F) = [l, u] \, , \quad \mathrm {where} \quad l, u \in \mathbb R \land u > l \, ; \tag 1 $$
continuous; differentiable at the origin; and compactly supported:
$$ \mathrm {supp} (F) = (a, b) \, , \quad \mathrm {where} \quad a, b \in \mathbb R \, ; \tag 2 $$
such that the Fourier transform, $\tilde F(t) \! : \, \mathbb R \to \mathbb C \, , \,$ defined as
$$ \tilde F(t) = \int_{-\infty}^\infty \! e^{i t x} F(x) \, \mathrm d x \, ; \tag 3 $$
exists and is everywhere real and non-negative:
$$ \mathrm {Range} \! \left ( \tilde F(t) \right ) \subseteq \mathbb R_{{\ge}0} \, ? \tag 4 \label {Condition} $$
I believe one can easily show that $\tilde F(t)$ must be bounded and non-zero:
$$ \mathrm {Range} \! \left ( \tilde F(t) \right ) = [0, c] \, , \quad \mathrm {where} \quad c \in \mathbb R_{{>}0} \, ; \tag 5 $$
and converge to zero:$~~\tilde F(t \to \infty) \to 0^+ \, .$
In order to have a real Fourier transform:
$$ \mathrm {Range} \! \left ( \tilde F(t) \right ) \subseteq \mathbb R \, , \tag 6 $$
$F(x)$ must be even:
$$ \forall x \in \mathbb R \! : \, F(x) = F(-x) \, , \tag 7 $$
which implies $b > 0 \, , \, $ $a = -b \, , \, $ $F' \! (0) = 0 \, , \, $ and that $\tilde F(t)$ is also even:
$$ \forall t \in \mathbb R \! : \, \tilde F(t) = \tilde F(-t) \, . \tag 8 $$
So, without loss of generality, we can ask the same question of the cosine transform, $ \tilde F^c \! (t) \! : \, \mathbb R_{{\ge}0} \to \mathbb R \, , \, $ defined as
$$ \tilde F^c \! (t) = \int_0^b \! \cos{(t x)} \, F(x) \, \mathrm d x \, ; \tag 9 $$
namely,
$$ \mathrm {Range} \! \left ( \tilde F^c \! (t) \right ) \subseteq \mathbb R_{{\ge}0} \, ? \tag {10} $$
Furthermore, $\tilde F^c \! (t)$ should obey the same conditions as $\tilde F(t)$ laid out in the previous paragraph.
I understand that condition$~\eqref {Condition}$ is equivalent to requiring that $F(x)$ be a positive-definite function. Also, I am under the impression that this paper shows that if $F(x)$ is “convex”,
$$ \forall x > 0 \! : \, F'' \! (x) > 0 \, , \tag {11} $$
then it is positive-definite. I am doubtful, however, that such a convex $F(x)$ can satisfy the requirements laid out in the first paragraph. The Paley–Wiener theorem also seems potentially relevant. I have thusfar neither been able to use these results to construct an $F(x)$ satisfying those requirements nor to prove their non-existence.
Two functions which come close are
$$ F(x) = (|x| - 1)^2 \, \mathbf 1_{[-1, 1]} (x) \, , \tag {12} $$
and
$$ F(x) = -\ln{|x|} \, \mathbf 1_{[-1, 1]} (x) \, , \tag {13} $$
where $\mathbf 1_S (x)$ is the indicator function.
Both are non-differentiable at $x = 0 \, , \,$ and the latter is unbounded:$~~F(x \to 0) \to \infty \, .$
I am also interested in the generalization of this question to $D > 1$-dimensional isotropic Fourier transforms,
$$ t^{1 - D/2} \! \int_0^b \! J_{D/2 - 1} (t x) \, F(x) \, x^{D/2} \, \mathrm d x \, , \tag {14} $$
where $J_\alpha$ is a Bessel function.
Thanks!
As many as you want: just take any smooth even real-valued compactly supported function and convolve with itself.
@fedja That was remarkably fast. Neat trick. I get why it works. I also confirmed it numerically using the standard bump function. Thanks!
Additionally, I see that by adding two incommensurately scaled copies of the convolution, the Fourier transform can be made everywhere strictly positive.
Credit to @fedja who answered my question in a comment ten minutes after I posted it. Thanks!
As many as you want: just take any smooth even real-valued compactly supported function and convolve with itself.
This works because convolving the function with itself squares the Fourier transform. We get minima of zero wherever the original Fourier transform changed signs.
By adding the self-convolution of two functions with no shared Fourier roots, we can in fact strengthen the original requirement to an everywhere strictly positive Fourier transform, instead of just non-negative.
I confirmed this numerically using the standard bump function
$$ e^{\frac 1 {r^2 - 1}} \, \mathbf 1_{(-1, 1)} $$
and a version horizontally stretched by $\sqrt 2$.
|
2025-03-21T14:48:31.932300
| 2020-08-30T13:37:49 |
370447
|
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"Carlo Beenakker",
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|
Stack Exchange
|
I'm looking for a proof of Polya-Bertrandias Theorem
I'm looking for a proof of Polya-Bertrandias rationality criterion in english (not the one from Amice).
for reference, the proof in French is on page 175 and following of this book --- I would just pass the text through Google translate, it's mostly formulas and little text.
|
2025-03-21T14:48:31.932371
| 2020-08-30T15:02:57 |
370452
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Trace ideal of a projective module
In his 1969 paper "On projective modules of finite rank", Wolmer Vasconcelos writes
Let $M$ be a projective $R$-module... The trace of $M$ is defined to be the image of the map $M \otimes_R \operatorname{Hom}_R(M, R) \to R$, $m \otimes f \to f(m)$; it is denoted by $\tau_R(M)$. If $M \oplus N = F$ (free), it is clear that $\tau_R(M)$ is the ideal of $R$ generated by the coordinates of all elements in $M$, for any basis chosen in $F$. It follows that for any homomorphism $R \to S$, $\tau_S(M \otimes_R S) = \tau_R(M) S$.
A similar claim appears in his 1973 paper "Finiteness in projective ideals":
We recall the notion of trace of a projective module $E$ over the commutative ring $A$. It is simply the ideal $J(E) = J = \Sigma f(E)$ where $f$ runs over $\operatorname{Hom}_A(E, A)$. Equivalently, $J$ is the ideal generated by the “coordinates” of all the elements of $E$ whenever a decomposition $E \oplus G = F$ (free) is given. Under the second interpretation, it follows that if $h \colon A \to B$ is a ring homomorphism, then $J(E \otimes_A B) = h(J(E)) B$.
The first claim is easy to verify (albeit with a change of free module) as follows. If $F = M \oplus N$ has a basis $\{ v_i \}$, we can consider $F' = F \oplus R$, where $u$ is a generator for $R$. For any $f \colon F \to R$, $F'$ has a basis composed of $u$ and all $w_i := v_i - f(v_i)u$. With respect to this basis, $v_i = w_i + f(v_i) u$, hence the $u$-coordinate of $v_i$ is $f(v_i)$. So every homomorphism $M \to R$ is the restriction of a coordinate function on $F'$.
What is not clear to me is the reason for the second claim that $\tau_S(M \otimes_R S) = \tau_R(M) S$. The inclusion $\tau_R(M) S \subset \tau_S(M \otimes_R S)$ is obvious, so let me focus on the other one.
One can choose a decomposition $M \oplus N = F$ (free), so that $M \otimes_R S \oplus N \otimes_R S = F \otimes_R S =: F_S$, which is free over $S$. Up to adding a $S$ summand, one can also assume that every $f \colon M \otimes_R S \to S$ is the restriction of some coordinate function on $F_S$. But the coordinates on $F_S$ depend on the choice of a basis. If the basis is obtained from an $R$-basis of $F$, the claim is clear. But $F_S$ could have many choices of $S$-bases which are not derived from $R$.
I think I am missing something quite trivial, but I cannot see it right now, so I though I'd rather ask here.
To find the ideal generated by the coordinates of $M \otimes_R S$, it suffices to take the coordinates of any generating set. For example, the elements of the form $m\otimes 1$ for $m \in M$.
Yes, but these coordinates are not obtained with respect to an R basis.
You can take the same basis for $F \otimes_R S$ as you used for $F$. The point is that in the first statement, the choice of generating set for $M$ or the choice of basis for $F$ does not matter, because they both give the same intrinsically defined ideal. (It looks like you're trying to modify the proof of the first statement instead of just applying the statement.)
The ideal is obtained by considering all choices of bases. Some of these will be defined over R, but some will not
Oh now I'm confused: what does "any basis chosen in $F$" mean? I read it as "pick one", but it looks like you're reading it as "use all of them".
Yes it is the ideal generated by the coordinates of all elements for all choices of bases
Maybe the canonical isomorphism $\operatorname{Hom}_S(M\otimes_RS,S)\cong\operatorname{Hom}_R(M,S)$ can help?
The confusion is linguistic, as identified in the comments.
Lemma. Let $M$ be a projective $R$-module, and suppose $M \oplus N \cong F$ is free on a basis $\mathcal B$. For $b \in \mathcal B$, write $\varepsilon_b \colon F \to R$ for the 'dual' element taking $b$ to $1$ and all other basis elements to $0$. Then $\tau(M)$ is the ideal generated by $\varepsilon_b(m)$ for $b \in \mathcal B$ and $m \in M$.
(By abuse of notation, we write $\varepsilon_b(m)$ for what should properly be denoted $\varepsilon_b(m,0)$.)
Proof. Since $\varepsilon_b|_M$ is a homomorphism $M \to R$, we clearly have $\varepsilon_b(m) \in \tau(M)$ for all $b \in \mathcal B$ and all $m \in M$. We have to show that they generate. In the definition of $\tau(M)$, we may replace $\operatorname{Hom}(M,R)$ by $R^{\mathcal B} = \operatorname{Hom}(F,R) \twoheadrightarrow \operatorname{Hom}(M,R)$. Elements can be written as $f = (f_b)_{b \in \mathcal B}$, where $f_b = f(b)$ are constants. Now the idea is that $f(m)$ only depends on the coordinates of $f$ where $m$ is supported:
Let $f = (f_b)_{b \in \mathcal B} \in R^{\mathcal B}$ and $m = \sum_{b \in \mathcal B'} a_b b \in M$ for some finite subset $\mathcal B' \subseteq \mathcal B$. Write $f_{\mathcal B'}$ for the function whose $\mathcal B'$-coordinates agree with $f$ and whose other coordinates vanish. Then
$$f(m) = \sum_{b \in \mathcal B'} f(a_b b) = \sum_{b \in \mathcal B'} f_b \cdot a_b = \sum_{b \in \mathcal B'} f_b \cdot \varepsilon_b(m),$$
so $f(m)$ is expressed as a combination of the $\varepsilon_b(m)$. $\square$
Corollary. Let $M$ be a projective $R$-module, and let $R \to S$ be a ring homomorphism. Then
$$\tau\left(M \underset R\otimes S\right) = \tau(M)S.$$
Proof. Write $M \oplus N \cong F$ for some $R$-module $N$ and a free $R$-module $F$. Then
$$\left(M \underset R\otimes S\right) \oplus \left(N \underset R\otimes S\right) \cong F \underset R\otimes S.$$
If $F$ has basis $\mathcal B$, then the elements $b \otimes 1$ form a basis of $F \otimes_R S$. Moreover, $M \otimes_R S$ is generated by elements of the form $m \otimes 1$. Therefore, $\tau(M \otimes_R S)$ is exactly the ideal generated by $\varepsilon_{b \otimes 1}(m \otimes 1)$, which is $\tau(M)S$. $\square$
Thank you. In the meantime I also realized that if $m$ is any element of $M$, the coordinates of $m \otimes 1$ with respect to any basis of $F_S$ are S-linear combinations of the coordinates with respect to any fixed R- basis, which is an equivalent way of seeing this
|
2025-03-21T14:48:31.932904
| 2020-08-30T15:53:29 |
370456
|
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|
Stack Exchange
|
In a locally presentable category, is every object (a retract of) the colimit of a chain of smaller objects?
Let $\mathcal C$ be an accessible category. For $C \in \mathcal C$, define the presentability rank $rk(C)$ of $C$ to be the minimal regular $\kappa$ such that $C$ is $\kappa$-presentable. Following Lieberman, Rosicky, and Vasey, say that $C$ is filtrable if it is the colimit of a chain $C = \varinjlim_{\alpha < \lambda} C_\alpha$ of objects $C_\alpha$ of lower presentability rank $rk(C_\alpha) < rk(C)$, and almost filtrable if it is a retract of the colimit $D$ of such a chain such that $rk(D) = rk(C)$.
Question:
Let $\mathcal C$ be an accessible category. Under what conditions can we say that every object $C \in \mathcal C$ of sufficiently large presentability rank is almost filtrable? Does it suffice to assume that $\mathcal C$ is locally presentable?
(Of course, if "chain" is replaced with "highly-filtered colimit", then no conditions are necessary.)
In the above-linked preprint are given various conditions for filtrability dependent on $rk(C)$, but they are not really focused on the locally presentable case. In this case,
$rk(C)$ is always a successor (unless it's $\aleph_0$ or perhaps if it's smaller than the accessibility rank of $\mathcal C$);
there's a basic argument which shows that if $rk(C)$ is the successor of a regular cardinal, then $C$ is almost filtrable (and the last Remark in the above-linked paper asserts that the retract can be removed with the fat small object argument).
But I'm not sure how to say anything when $rk(C)$ is the successor of a singular cardinal.
Motivation:
It's important to me to be able to handle all $C \in \mathcal C$ of sufficiently large presentability rank, because this opens up the possibility of a new sort of inductive argument in the theory of locally presentable categories: induction on presentability rank using decomposition by chains. This sort of induction should be particularly well-suited to applications related to the small object argument, which interacts well with chains but not with general highy-filtered colimits.
I should also ask that the chain be smooth, i.e. that the canonical morphism $\varinjlim_{\beta<\alpha} C_\beta \to C_\alpha$ be an isomorphism for all $\alpha < \lambda$, because the small object argument only interacts well with smooth chains. This is indeed the case in the construction discussed in the answers, with just an iota of extra care (see below). And to clarify from the "Motivation": the small object argument does produce an accessible functor and so in some sense interacts well with highly-filtered colimits; it's just a weaker sense than when it comes to smooth chains.
The last Remark in my joint paper gives a positive answer to the Question.
Oh I see. I missed that it said "well $\lambda^+$-filtrable".
To spell it out for myself, Remark 8.10 talks about eliminating retracts, but the point is that Corollary 8.9(2) implies a positive answer to the Question. To flesh out the proof a bit, it is deduced from Theorem 8.8(1) with $\mu = \aleph_0$, since $\aleph_0 \triangleleft \lambda$ for all $\lambda$ (even singular $\lambda$ under the definitions in the paper). If $\mathcal C$ is locally $\theta$-presentable, then it is $(\aleph_0, \theta)$-accessible. So we get that $\mathcal C$ is almost $\lambda^+$-filtrable for all $\lambda \geq \theta$.
Since this result is derived by Lieberman, Rosicky, and Vasey as a corollary of some more sophisticated constructions with more sophisticated goals, I think it might be worth "compiling out" the proof here. It turns out to be not so bad. Let $\mathcal C$ be a locally $\lambda$-presentable category, and recall the following fact:
For any $C \in \mathcal C$, if $\mathrm{rk}(C) > \lambda$, then $\mathrm{rk}(C) = \kappa^+$ is a successor. (Proof hint: $\kappa$ is the smallest cardinal such that $C$ is a retract of a $\kappa$-sized colimit of $\lambda$-presentable objects.)
Theorem [Lieberman, Rosicky, and Vasey]
Let $\mathcal C$ be a locally $\lambda$-presentable category and let $C \in \mathcal C$ with $\mathrm{rk}(C) = \kappa^+ > \lambda$. Then $C$ is almost filtrable.
Proof: Write $C = \varinjlim_{i \in I} C_i$ as a colimit of $\lambda$-presentable objects. Then $C$ is a retract of the colimit of a $\kappa^+$-small subdiagram, so we may assume without loss of generality that the diagram $I$ is of cardinality $\kappa$. We may write $I = \cup_{\alpha < \mathrm{cf}(\kappa)} I_\alpha$ as the union of a $\mathrm{cf}(\kappa)$-sized increasing chain of subdiagrams of cardinality $|I_\alpha| <\kappa$. Setting $C_\alpha = \varinjlim_{i \in I_\alpha} C_i$, we have $C = \varinjlim_{\alpha < \mathrm{cf}(\kappa)} C_\alpha$, yielding the desired filtration.
Can this be simplified further by using Remark 1.30 in Adamek-Rosicky, that if $\mu\ge\lambda$ then every $\mu$-presentable object is a $\mu$-small colimit of $\lambda$-presentable objects? It seems that one could then express that $\mu$-small diagram as a union of subdiagrams to obtain a suitable chain. In particular, this would seem to eliminate the retract. (Although, the elimination of the retract in AR 1.30 depends on a highly technical argument in Makkai-Pare that I've never really understood.)
It just occurred to me that in order to be really useful, one should also ask that the chain be smooth, i.e. that the canonical morphism $\varinjlim_{\beta<\alpha} C_\beta \to C_\alpha$ is an isomorphism for all limit ordinals $\alpha < \kappa^+$. This is indeed the case in the construction discussed here, so long as we take $I_\alpha = \cup_{\beta < \alpha} I_\beta$ for limit ordinals $\alpha < \kappa^+$ (as we may).
|
2025-03-21T14:48:31.933245
| 2020-08-30T16:11:40 |
370458
|
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|
Stack Exchange
|
Borel conjecture and arbitrary surface
Before starting my question I want to write something that I already know.
Borel Conjecture: Any homotopy equivalence between two closed
aspherical manifolds is homotopic to a homeomorphism.
Now, my interest is in when our manifolds are surfaces. Here, I assume surface means connected and second countable, may or may not be orientable, may or may not have boundary, may or may not have punctures. In particular, I also consider two types of surfaces, finite-type or infinite-type, i.e. fundamental group is finitely generated or not.
Now, it is well known that any homotopy-equivalence between two closed surfaces is homotopic to a diffeomorphism. If the surface has boundary then one has to also require the homotopy-equivalence restricts to a homotopy-equivalence of the boundaries. A reference is Zieschang, Vogt and Coldeway, Surfaces, and planar discontinuous groups.
From here my question starts.
(1) Is any proper homotopy equivalence of surfaces is properly homotopic to a homeomorphism?
(2) Is there any sufficient condition on a homotopy equivalence (not necessarily proper) between two surfaces, so that it becomes homotopic to a homomorphism? If there is no such for the general case, is it possible to give such a sufficient condition in case our homotopy equivalence is in between two finite-type surfaces?
I am looking for references. Thanks in advance.
In the finite-type case, this is well-known. In the infinite-type case this is probably true since proper h.e. induces a bijection of the invariants of surfaces and the result is probably hidden in Richards' classification paper.
@MoisheKohan could you tell me a reference for the finite-type case?
See e.g. Theorem 8.8 in "Primer on MCG" by Farb and Margalit (although, they leave out some details). Let me know if you have trouble completing their proof. They state the result algebraically, but properness of the h.e. is what ensures that conjugacy classes of punctures map to conjugacy classes of punctures.
|
2025-03-21T14:48:31.933399
| 2020-08-30T16:27:56 |
370461
|
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"url": "https://mathoverflow.net/questions/370461"
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|
Stack Exchange
|
Is Sobolev limit of bijective maps surjective?
This is a cross-post.
Let $\Omega_1,\Omega_2 \subseteq \mathbb R^2$ be open, connected, bounded, with non-empty $C^1$ boundaries. Let $f_n:\bar\Omega_1 \to \bar\Omega_2$ be $C^1$ be bijective maps with $\det(df_n)>0$, and suppose that $f_n$ converges to a $C^1$ function $f: \bar\Omega_1 \to \bar\Omega_2$ strongly in $W^{1,2}$.
Question: Must $f$ be surjective?
Note that $f$ is surjective if and only if $|f^{-1}(y)| \le 1$ a.e. on $\Omega_2$:
By the area formula
$$
\int_{\Omega_1} \det df_n = \int_{\Omega_2} |f_n^{-1}(y)|=\text{Vol}(\Omega_2),
$$
so
$$
\int_{\Omega_2} |f^{-1}(y)|= \int_{\Omega_1} \det df =\lim_{n \to \infty} \int_{\Omega_1} \det df_n=\text{Vol}(\Omega_2).
$$
This implies the claim.
|
2025-03-21T14:48:31.933485
| 2020-08-30T17:07:19 |
370464
|
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"LSpice",
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|
Stack Exchange
|
Family of $p$-groups closed under products and subgroups: closed under quotients?
Let $p$ be a prime, and let $\mathcal{U}$ be a family of finite $p$-groups such that
Any group isomorphic to a group in $\mathcal{U}$ is also in $\mathcal{U}$
Any product of groups in $\mathcal{U}$ is also in $\mathcal{U}$
Any subgroup of a group in $\mathcal{U}$ is also in $\mathcal{U}$.
Is it automatically true that any quotient of a group in $\mathcal{U}$ also lies in $\mathcal{U}$? This seems unlikely but I could not think of a counterexample. (I tried various things involving dihedral groups and generalised quaternion groups, but did not go much beyond that.) Here is an initial result:
Lemma: if $A$ is abelian and is a quotient of a group $G\in\mathcal{U}$, then $A\in\mathcal{U}$.
Proof: we can write $A$ as a product of cyclic groups $C$. As $\mathcal{U}$ is closed under products, it will suffice to prove that $C\in\mathcal{U}$. Choose an element $g\in G$ that maps to a generator of $C$. Then $g$ generates a cyclic subgroup $C'\leq G$ whose order must be a multiple of $|C|$. It follows that $C'$ contains a subgroup $C''$ isomorphic to $C$. As $\mathcal{U}$ is closed under subgroups and isomorphisms, it follows that $C\in\mathcal{U}$. ☐
Maybe specify what you mean by product: direct products? or more general ones?
I just mean Cartesian product of groups. I'm not sure what other kind of products could be relevant here.
Semidirect products, etc. I guess it's less interesting then, but this was only a guess. You probably mean direct products. Cartesian is a set-wise definition (the underlying set of a semidirect product is the Cartesian product of sets), and "direct" precisely means that the law is not twisted.
No. Fix an odd prime $p$. Let $H_p$ be a non-abelian group of order $p^3$ and exponent $p$ (this is unique to isomorphism).
Let $\mathcal{C}_p$ be the class of $p$-groups not containing any subgroup isomorphic to $H_p$. Then $\mathcal{C}_p$ is stable under taking subgroups (obvious) and direct products (easy because $H_p$ has a unique minimal normal subgroup).
Then it is enough to show that $H_p$ is quotient of some group in $\mathcal{C}_p$. Let $G_p$ be the free group on 2 generators in the variety of groups satisfying that $x^{p^2}=[y,z^p]=[[x,[y,z]]=1$ for all $x,y,z$, so $H_p$ is a quotient of $G_p$. It is enough to show that $G$ has no subgroup isomorphic to $H_p$. Indeed, $G_p$ has order $p^5$, and it is easy to check ($*$) that all elements in $G_p$ of order $p$ commute; hence it does not contain any copy of $H_p$.
($*$) I see it by writing $G_p$ as a Lie algebra over $\mathbf{Z}$ using Malcev-Lazard correspondence, namely the free Lie algebra over $\mathbf{Z}$ satisfying the laws $p^2x=p[y,z]=[x,[y,z]]=0$ for all $x,y,z$. It's quotient of the Heisenberg Lie algebra over $\mathbf{Z}/p^2\mathbf{Z}$ (which is the free one in the variety of laws $p^2x=[x,[y,z]]=0$), which has order $p^6$, by the subgroup of order $p$ in its center, hence has order $p^5$ ("basis" $(u,v,w)$ with $p^2u=p^2v=pw=0$, $[u,v]=w$), and precisely the set of elements killed by $p$ is an $p$-elementary abelian subgroup of order $p^3$, with "basis" $(pu,pv,w)$.
This is more or less the same as YCor's answer but perhaps more elementary. Let $p$ be an odd prime, and let $\mathcal{U}$ be the class of $p$-groups in which all elements of order $p$ commute. This is clearly closed under isomorphisms, products and subgroups. Now let $G$ be the group of matrices over $\mathbb{Z}/p^2$ of the form
$$ g = \left[\begin{array}{ccc} 1&u&v \\ 0&1&w \\ 0&0&1 \end{array}\right], $$
and let $\overline{G}$ be the corresponding group over $\mathbb{Z}/p$. In $G$ we find that $g^p=1$ iff $u,v,w\in p.\mathbb{Z}/p^2$, and it follows easily that $G\in\mathcal{U}$. In $\overline{G}$ we find that all elements satisfy $g^p=1$, and so $G\not\in\mathcal{U}$. There is an evident surjective homomorphism $G\to\overline{G}$, so $\mathcal{U}$ is not closed under quotients.
Indeed, in my answer I could have taken this $G$ (of order $p^2$) instead of its quotient $G_p$ (of order $p^5$). Both don't contain any copy of their quotient $\bar{G}\simeq H_p$.
(As no one won't have realised, of course @YCor meant $G$ has order $p^6$, not $p^2$.)
@LSpice weird typo! thanks for noticing.
|
2025-03-21T14:48:31.933756
| 2020-08-30T17:31:14 |
370465
|
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|
Stack Exchange
|
Finite étale morphism from a scheme to an algebraic space
Let $f : X \to Y$ be a finite, surjective étale morphism of algebraic spaces (say, of finite type over some noetherian scheme). Assume that $X$ is a scheme. Does this imply that $Y$ is a scheme? Is $Y$ the quotient of $X$ by the relation $R = \{(x,y) \in X \times X \ : \ f(x) = f(y) \} \subseteq X \times X$?
The answer is no. A version of Hironaka’s examples of a proper nonprojective variety (see e.g. the appendix in Hartshorne’s Algebraic Geometry) has a fixed point free involution whose orbit does not admit an affine neighborhood. It follows that the quotient is not a scheme.
Is it known whether projectivity of X ensures that Y is a scheme?
Yes, projectivity is sufficient: the graph of the morphine is a flat family of closed subschemes of $X$ parameter used by $Y$. This determines a closed immersion of $Y$ into the Hilbert scheme of $X$.
"morphine" --> "morphism", "parameter used" --> "parameterized" (autocorrect error)
|
2025-03-21T14:48:31.933861
| 2020-08-30T18:18:39 |
370467
|
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|
Stack Exchange
|
A meet-semilattice with top element that is not a lattice?
I am reading Francis Borceux’s “Handbook of Categorical Algebra I” and on page 135 it says
In particular a finite version of 4.2.5 does not hold: a finitely complete and well-powered category certainly admits finite intersections of subobjects (see 4.2.3), but not in general finite unions of subobjects. Finite unions have been constructed in 4.2.5 using possibly infinite intersections. For a counterexample, just consider a ∧-semi-lattice with a top element which is not a lattice.
I get the argument for not being able to construct a union through finite intersection. But I have trouble proving it with the mentioned counterexample. Specifically, I can’t picture a “ ∧-semi-lattice with a top element which is not a lattice” in my mind. Why can’t the top element be the union for a finite set of elements if there is no other “better” element? Does the construction of this counterexample have anything to do with (in)finiteness?
It is well-known that a finite meet-semi-lattice with a maximum element is a lattice. The reason is that we can define $a \vee b := \wedge \{c\colon \textrm{$c$ is an upper bound for $a,b$}\}$, where this set is non-empty (since we have a maximum) and finite (since the poset is finite), and finite meets exist by supposition that we have a meet-semi-lattice.
But this is not true for infinite posets. Let $P := (\{(a,b)\colon 0\leq a,b \leq 1\}\setminus \{(1,1)\}) \cup \{(a,a)\colon 1 < a \leq 2\}$, with the usual partial order $(a_1,b_1)\leq (a_2,b_2)$ iff $a_1 \leq a_2$ and $b_1 \leq b_2$. Then $P$ is a meet-semi-lattice (with $(a_1,b_1)\wedge (a_2,b_2)=(\mathrm{min}(a_1,a_2),\mathrm{min}(b_1,b_2)$) and it has a maximum element $(2,2)$. But $(1,0)$ and $(0,1)$ lack a join.
Or the subset of $\mathcal P(\mathbb N)$ consisting of $\emptyset$, ${1}$, ${2}$, and the sets ${1,2}\cup{n,n+1,n+2,\dots}$ for $n\ge3$.
|
2025-03-21T14:48:31.934018
| 2020-08-30T18:58:00 |
370471
|
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|
Stack Exchange
|
Are there any identities for alternating binomial sums of the form $\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} $?
In equations (20) - (25) of Mathworld's article on binomial sums, identities are given for sums of the form $$\sum_{k=0}^{n} k^{p}{n \choose k}, $$ with $p \in \mathbb{Z}_{\geq 0}$. I wonder whether identities also exist for the alternating counterparts: $$\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} .$$ Furthermore, I'm interested in results for the same sum that is “cut off”, i.e. when the summands go from $k=0$ to some $D<n$.
A rewrite of formula (10) on MathWorld (replacing the summation index $k-i\mapsto i$) gives the desired formula:
$$\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} =(-1)^n n! S_2(p,n),$$
where $S_2(p,n)$ is the Stirling number of the second kind (the number of ways of partitioning a set of $p$ elements into $n$ non-empty subsets).
It is remarkable that the alternating sum equals zero for $p<n$.
It's not so remarkable that the alternating sum is zero for $p<n$. The sum is the $n$th difference at 0 of a polynomial of degree $p$. For any polynomial of degree $p$, the $p$th difference is constant and the $n$th difference for $n>p$ is 0.
I think letting the LHS sum run to infinity allows it to be defined for fractional $n$ and still maintain the same behavior on the integers. An optimist might hope that this suggests a definition of an analytic continuation of the Stirling numbers
For the cutoff version:
We can get a subtraction-free formula for the cutoff version, which should be sufficient to get asymptotics, by the same idea that gives a simple bijective proof of the identity that Carlo Beenakker mentioned. That is:
$k^p$ counts maps from a $p$-element set $[p]$ to a $k$-element set
Thus $\binom{n}{k} k^p$ counts pairs of a $k$-element subset $S$ of an $n$-elements set $[n]$ with a map from $[p]$ to $S$. In other words, it counts maps $f$ from $[p]$ to $[n]$ together with a $k$-element subset $S$ of $[n]$ containing the image of $f$.
So $\sum_{k=0}^d (-1)^k \binom{n}{k} k^p$ is the sum over maps $f: [p] \to n$ of the sum over subsets $S$ of $[n]$, containing the image of $f$, of size at most $k$, of $(-1)^{|S|}$. We may assume the image of $f$ has size $\leq d < n $ and thus that there is some element $e$ not in the image of $f$. We can cancel each subset with $e\notin S$ with the $S \cup \{e\}$, as these have opposite signs. The only subsets that fail to cancel are those that have size exactly $d$ and do not contain $e$, of which there are $\binom{n - | \operatorname{Im}(f) | -1}{ d - |\operatorname{Im}(f)| } $.
With $S_2(p,j)$ again the Stirling numbers of the second kind, the number of maps from $[p]$ to $[n]$ with image of size $j$ is $ \frac{n!}{ (n-j)!} S_2(p,j) $, so the sum is
$$ (-1)^d \sum_{j=0}^d S_2(p,j) \frac{n!}{(n-j)!} \binom{ n-j-1}{d-j} $$
$$= (-1)^d \frac{n!}{ (n-1-d)!} \sum_{j=0}^d S_2(p,j) \frac{1}{(n-j)} \frac{1}{(d-j)!} $$
(If $d=n$ then all subsets cancel and so only the terms with $| \operatorname{Im} f| =n$ remain, so we just obtain the count of surjections from $[p]$ to $[n]$, as in Carlo Beenakker's answer.)
Alternately, a formula-based proof:
we have $$ k^p = \sum_{j=0}^k S_2( p,j) \frac{k!}{ (k-j)!} $$ ( a standard identity.) so
$$\sum_{k=0}^d (-1)^k k^p {n \choose k} = \sum_{j=0}^d \sum_{k=j}^d (-1)^k S_2( p,j) \frac{k!}{(k-j)!} {n \choose k} $$
and $$\frac{k!}{(k-j)!}{n\choose k} = \frac{k! n!}{ (k-j)! k! (n-k)! } = \frac{n!}{ (k-j)! (n-k)!} = \frac{n!}{(n-j)!} \binom{n-j}{k-j} $$
so
$$ \sum_{k=0}^d (-1)^k k^p {n \choose k} = \sum_{j=0}^d \sum_{k=j}^d (-1)^k S_2( p,j) \frac{n!}{(n-j)!} \binom{n-j}{k-j}$$
$$ = \sum_{j=0}^d (-1)^d S_2( p,j) \frac{n!}{(n-j)!} \binom{n-j-1}{d-j} = (-1)^d \frac{n!}{ (n-1-d)!} \sum_{j=0}^d S_2(p,j) \frac{1}{(n-j)} \frac{1}{(d-j)!} $$
Thank you! It is at times like these that I wish I could accept two answers.
Up to the factor $(-1)^n$, the uncut sum is
$$s_{p,n}:=\sum_{k=0}^n(-1)^{n-k}\, k^p\,\binom nk.$$
As noted in the comment by Richard Stanley,
$$s_{p,n}=(\Delta^n f_p)(0),$$
where $f_p(x):=x^p$ and $(\Delta f)(x):=f(x+1)-f(x)$. Here and in what follows, $x$ denotes any real number.
It is easy to check by induction on $n$ that for any smooth enough function $f$ we have
$$(\Delta^n f)(x)=Ef^{(n)}(x+S_n),$$
where $f^{(n)}$ is the $n$th derivative of $f$, $S_n:=U_1+\cdots+U_n$, and $U_1,\dots,U_n$ are independent random variables uniformly distributed on the interval $[0,1]$. So,
$$s_{p,n}=n!\binom pn ES_n^{p-n} \tag{1}$$
for $p=0,1,\dots$ and $n=0,1,\dots$. In particular, it follows that $s_{p,n}=0$ for $n=p+1,p+2,\dots$, as noted in the answer by Carlo Beenakker.
In fact, (1) holds for all real $p\ge n$ (and $n=0,1,\dots$), and then, obviously,
$$0<s_{p,n}\le n!\binom pn n^{p-n}. \tag{2}$$
If $p-n\ge1$, then, in view of Jensen's inequality, the lower bound $0$ on $s_{p,n}$ in (2) can be greatly improved, to $$b_{p,n}:=n!\binom pn \Big(\frac n2\Big)^{p-n}.$$
Moreover, by the law of large numbers, $S_n/n\to1/2$ in probability (say). Also, $0\le S_n/n\le1$. So, by dominated convergence, from (1) we immediately get the following asymptotics: if $n\to\infty$ and $p-n\to a$ for some real $a>0$, then
$$s_{p,n}\sim b_{p,n}.$$
You can find plenty of documentation on Gould's site. Maybe it could be useful. The link is https://math.wvu.edu/~hgould/
Interesting files are Vol.1.PDF to Vol. 8.PDF.
|
2025-03-21T14:48:31.934358
| 2020-08-30T20:02:05 |
370473
|
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|
Stack Exchange
|
Rigorous stability analysis of infinite dimensional ODEs : How to bound the tails?
My question is about linear stability analysis of dynamical systems obtained by discretizing linear(ized) partial differential equations. Consider,
$\dot{x}=Ax$, where $x$ is the infinite dimensional vector of coefficients, and $A$ a linear operator in an countably finite basis of a nice function space (using, say Fourier series on a circle, or Hermite on $\mathbb{R}$). Assume that the PDE system is of the type that the full spectrum of $A$ determines the stability.
Usually, physics/engineering papers impose an arbitrary cut-off on number of modes considered, and give analytical/numerical results on the spectrum of this truncation of $A$ to prove/disprove stability.
My question is: How does one make such arguments rigorous ? Are there some standard tricks that can enable one to take into account the neglected modes in a rigorous manner ?
Specifically, I am looking for examples/papers where stability was deduced based on considering the operator acting on a subset of modes, and estimating the effect of "tails" (or neglected modes).
What you say is clear is not clear at all. Indeed, it is in general false. Stability of infinite dimensional systems is in general not determined by the spectrum. Even when it is, finite dimensional approximation of the spectrum is a topic for entire books.
@MichaelRenardy I have edited to restrict to only those systems where spectrum does determine stability. Note that I am not asking for computation of the full spectrum, but rather concluding stability using a combination of analysis of the truncated system and some estimate of the neglected modes.
|
2025-03-21T14:48:31.934613
| 2020-08-30T22:20:46 |
370480
|
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|
Stack Exchange
|
Why are root data a natural candidate for classifying connected reductive groups?
For the purpose of this question, you may assume that we are working over the complex numbers.
Given a connected reductive group $G$, one can choose a maximal torus $T$, and then let $T$ act on the Lie algebra $\mathfrak{g}$ of $G$. One can use this action to define the root datum, which in turn is invariant of the choice of $T$, and use it to classify connected reductive groups.
The action of $T$ on $\mathfrak{g}$ is nice in that it has more information than just the action of $T$ on the Lie algebra $\mathfrak{t}$ of $T$, and yet is simple enough so as to decompose into one dimensional weight spaces. But that is a far cry from saying that this is a natural action to consider when trying to classify connected reductive groups!
Is there a deep reason that root data, or more generally the action of a maximal torus on the Lie algebra of $G$, is a natural thing to consider? Does it correspond to some cohomological invariant? Does it arise naturally? Or is this entire theory a fluke?
The proof does not seem to bring much insight into this story, in that it boils down to a series of reductions, which reinforces for me the suspicion that root data are not in and of themselves natural, but rather that this was a guess for a way to classify connected reductive groups that just happened to work out...
I think you're right, that the success of this scheme in a certain domain was happy/lucky... but, after all, much of mathematics seems to me finding happy coincidences... structures which were not a-priori obvious, but which, in fact, govern certain domains of phenomena.
Are you happy with the theory of weights itself, and the question is more about why this particular action?
I might need convincing on the theory of weights itself... I did not delve deep into representation theory, and learned it for the sole purpose of understanding the classification of connected reductive groups. What type of motivation are you thinking of?
One of the most basic intuitions you can get about weights is: not every matrix in $\mathrm{GL}_n(\mathbb{C})$ is diagonalizable, of course, but the subset of diagonalizable matrices is dense, so one has the feeling that understanding how the diagonal matrices behave in a given representation should tell you a lot about the whole representation; and then the weights are really just the eigenvalues for these diagonal matrices.
From a Tannakian perspective, a reductive group is determined by its representations. If you want to understand a $G$ representation $V$, it makes sense to restrict to a large subgroup whose representation theory you understand (i.e. $T/\mathfrak t$) whose rep theory is governed by weights. After decomposing $V$ into weights, you might wonder how the action other elements of $\mathfrak g$ is related to this weight decomp. The Jacobi identity tells you that if you decompose the adjoint representation into roots, then these roots permute the weight decomp nicely.
I would say that connected reductive groups over an algebraically closed field are classified by based root data rather than by root data. Indeed, to a reductive group $G$ one associates its canonical based root datum ${\rm BRD}(G)$. Moreover, to any normal homomorphism $\varphi\colon G\to G'$ one associated the induced morphism of the based root data $\varphi_*\colon{\rm BRD}(G)\to{\rm BRD}(G')$. A homomorphism $\varphi$ is called normal if its image is normal in $G'$.
Terry Gannon has suggested that the deepest “moonshine” in mathematics is not the moonshine of the monster sporadic group, or the modern moonshines of K3 surfaces, but rather the overwhelming prevalence of root data, and particularly ADE root data. This was perhaps first observed in the classification of Lie groups, but it seems to be much primitive: it seems that there is some class of “groups” over the absolute base F_1, which really are “the same” as root data, and the complex reductive groups are merely their C-points.
I don't believe in F_1, but that does sound interesting. Where can I read about that?
I can't give you a very deep reason for why root data appear in this context (because, let's face it, root systems spring out everywhere), but there are some very elementary reasons to why the action in question is very natural with regard to the classification.
Let me start with the following two considerations:
When one tries to distinguish between the two objects, one usually looks for some simple properties which differ between them. For example, to show that two abstract groups are not isomorphic, one starts by comparing their orders, and procede by comparing the number of elements of a given order in each of them or which subgroups are there and how they fit together.
In a semisimple Lie algebra $L$ there is a Jordan decomposition, which tells that every elements $x$ is the sum of an $\operatorname{ad}$-semisimple element $x_s$ and an $\operatorname{ad}$-nilpotent part $x_n$. And there is a subalgebra consisting of semisimple elements (otherwise $L$ itself is nilpotent by Engel theorem). Such subalgebras are called toric, and it turns out they are always abelian. Thus when considered in their adjoint repesentation, the elements of a toric subalgebra form a commuting family of semisimple endomorphisms of $L$, hence are simultaneously diagonalizable, which is equivalent to $L$ decomposing into the direct sum of its weight subspaces, which gives rise to the root system.
So combining these two considerations, to distinguish (and ultimately classify) semisimple Lie algebras we essentially take the simplest type of elements of $L$ (the semisimple ones) and look at how we can fit them together in $L$ (so that they from a subalgebra, and a maximal such).
This looks somewhat abstract, but really just mimics what can be easily seen in the examples, namely, in the classical semisimple Lie algebras. The standard constructions in their minimal representations are equipped with some very simple bases (for example, what comes first to mind for $\mathfrak{sl}_n$?), and there is a very natural maximal toric subalgebra $H$, namely, the diagonal matrices. The non-zero-weight subspaces are the spans of individual off-diagonal basis elements, and the root system captures their configuration.
Now getting back to algebraic groups, they can be roughly classified by the isomorphism classes of the corresponding Lie algebras, but additional information is needed to account for the center. The center sits inside the torus, so to incorporate this missing data into the rough classification one translates the adjoint action of the corresponding toric subalgebra to the adjoint action of the torus.
|
2025-03-21T14:48:31.935339
| 2020-08-30T22:21:27 |
370481
|
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|
Stack Exchange
|
Hadamard factorization of L-functions
I have already asked this question here in a different form, but really need an answer.
Let $L(s)$ be a "standard" $L$-function, say with Euler product, functional equation, etc...
(Selberg class if you like), of order 1, and let $\Lambda(s)$ be the completed $L$-function
with gamma factors. We thus have $\Lambda(k-s)=\omega\Lambda^*(s)$, where $\Lambda^*$ is
the "dual" Lambda function (example: if $L(s)$ corresponds to a Dirichlet character $\chi$,
$\Lambda^*$ corresponds to its conjugate), and $\omega$ root number of modulus 1.
Assume for instance that there are no poles. Since $\Lambda$ has order $1$ it has a Hadamard
product $$\Lambda(s)=ae^{bs}\prod_{\rho}(1-s/\rho)\;,$$
where the product is over the zeros of $\Lambda$ and understood as the limit as $T\to\infty$
of the product for $|\rho|<T$ (on purpose I do not use the more standard $(1-s/\rho)e^{s/\rho}$).
My question is this: do we always have $b=0$ ? This is trivial if $\Lambda^*=\Lambda$
(self-dual), otherwise the only thing I can prove is that $b$ is purely imaginary.
I have experimented numerically with some non self-dual $L$ functions attached to Dirichlet
characters, and it seems to be true.
Remarks: 1) I may have a proof using the "explicit formula" of Weil, but I am not sure of its
validity, and it seems too complicated. 2) I have a vague memory of Harold Stark mentioning this
result 50 years ago.
At least I'd agree that it is not trivial to prove what you want...
Is it clear that $\sum_{|\rho|<T}1/\rho$ has a limit as $T\to\infty$? I only know how to prove this in the self-dual case.
@Aurel: I think this is standard, but otherwise include it in the conjectural statement.
If I'm not mistaken, your equality should be invariant under the map $s\mapsto 1-\bar{s}$. Does it help?
@Sylvain: yes, this proves that $b$ is purely imaginary, but nothing more as far as I can see.
I think the vanishing of $b$ would make the Hadamard product quite analogous to the one of the sine function used by Euler to solve the Basel problem, but for an aperiodic sequence of (real) zeros. So it may require the truth of GRH viewing those real zeros as the ones of the analogue of the Riemann Xi function. So an unconditional proof thereof seems unlikely.
@Sylvain: I would be happy to assume GRH if necessary.
I made a few back-of-enveloppe computations, and under GRH one has $\Lambda(s)=0\Longrightarrow s=1-\bar{s}$ hence $e^{bs}=e^{b(1-\bar{s})}$ which seems to imply, writing $s=x+iy$, that $2bx=b$ hence $b=0$ for $x\neq 1/2$ but it is to be taken with a grain of salt.
I believe you are correct and $b$ is zero, although I find it inexplicable why this is not better known (certainly I didn't know it before). Let's stick to a primitive Dirichlet character $\mod q$, but what follows should be applicable in general. If we take logarithmic derivatives, then
$$
\frac{\Lambda^{\prime}}{\Lambda}(s) = b + \sum_{\rho} \frac{1}{s-\rho},
$$
with the understanding that the zeros $\rho=\beta+i\gamma$ are counted with $|\gamma|\le T$, and then $T\to \infty$. Let's evaluate the above at $s=R$ for a large real number $R$, and focus just on the imaginary parts.
Now
$$
\text{Im} \Big( \frac{\Lambda^{\prime}}{\Lambda}(R)\Big)
$$
tends exponentially to $0$ as $R\to \infty$. So let's look at the imaginary part on the right hand side, which is
$$
\text{Im} (b) + \lim_{T\to \infty} \sum_{|\gamma|\le T} \frac{\gamma}{(R-\beta)^2 + \gamma^2}.
$$
Note that
$$
\sum_{|\gamma|\le T} \frac{\gamma}{(R-\beta)^2+\gamma^2}
= \sum_{|\gamma|\le T}\Big( \frac{\gamma}{R^2+\gamma^2} + O\Big( \frac{R|\gamma|}{(R^2+\gamma^2)^2}\Big)\Big). \tag{1}
$$
To handle the error term, split into the terms $|\gamma|\le R$ and $|\gamma|>R$, obtaining that the error term is
$$
\ll \sum_{|\gamma|\le R} \frac{1}{R^2} + \sum_{R<|\gamma|} \frac{R}{|\gamma|^3} \ll \frac{\log qR}{R},
$$
upon recalling that there are $\ll \log q(|t|+1)$ zeros in an interval of length $1$ (we will recall this more precisely next).
Now the main term in (1) can be handled by partial summation. For $t>0$, put $N^+(t)$ to be the number of zeros of $\Lambda$ with imaginary part between $0$ and $t$, and $N^{-}(t)$ to be the number of zeros with imaginary part between $-t$ and $0$. Then both $N^+$ and $N^-$ satisfy by the argument principle the well known asymptotic formula (for $t\ge 1$)
$$
N^+(t), N^{-} (t) = \frac{t}{2\pi} \log \frac{qt}{2\pi e} +O(\log (q(t+1))).
$$
Thus for all $t>0$
$$
|N^+(t) - N^-(t)| = O(\log (q(2+t))).
$$
Now by partial summation
\begin{align*}
\sum_{|\gamma|\le T} \frac{\gamma}{R^2+\gamma^2} &= \int_0^{T} \frac{t}{R^2+t^2} dN^+(t) - \int_0^T \frac{t}{R^2+t^2} dN^-(t) \\
&= \frac{T}{R^2+T^2} (N^+(T)-N^-(T)) - \int_0^T (N^+(t)-N^-(t)) \Big( \frac{t}{R^2+t^2}\Big)^{\prime} dt \\
&= O\Big(\frac{T\log qT}{R^2+T^2} \Big) + O\Big(\int_0^T (\log q(t+2)) \Big(\frac{1} {R^2+t^2} + \frac{2t^2}{(R^2+t^2)^2} \Big)dt \Big)\\
&= O\Big( \frac{\log qR}{R}\Big),
\end{align*}
upon letting $T\to \infty$.
We conclude that the quantity in (1) is $O((\log qR)/R$, and so tends to $0$ as $R\to \infty$.
Another derivation appears in Corollary 10.18 of Montgomery-Vaughan; see also the discussion at the end of Section 10.3, where the proof is attributed to Vorhauer. Related to this is this question and the long discussion in the comments: https://mathoverflow.net/questions/343248/b-chi-l1-chi-l1-chi-dotsc
@PeterHumphries: I recall that other MO question and the discussion, but I don't quite see that Corollary 10.18 is the same as this question. Here one seems to say that $B(\chi)$ is simply $-\sum_\rho 1/\rho$ with the zeros counted as $|\rho|\le T$ (not just that the real parts match, which is obvious). It's this part that I found surprising and don't recall seeing before.
I think you're right; I may have been overzealous. Is there a way to see this more directly via Corollary 10.18 though?
@PeterHumphries: I didn't see how (hence the write-up).
@Lucia: would you allow me to include a sketch of your proof in a book that I am writing (if you want your real name to be given, send me a private e-mail) ?
@HenriCohen: Of course you are welcome to use the proof in your book. Did you try asking Stark if he recalled this? I can't believe it is not known somewhere.
I believe that a slightly different argument due to Ihara-Murty-Shimura ("On the log derivatives of Dirichlet L-functions at $s=1$") implies $b=0$ for (finite order) Hecke L-functions. It's Theorem 2 of the paper (subsecs 2.3, 2.4, pp. 4-5), and a key part of the proof is Weil's explicit formula.
@Alufat: Thanks for that reference!
@Alufat: so my own proof may be correct after all (but too long to post). Thanks for the reference.
Dear @Lucia, I am writing a paper where I reproduce your argument for (extended) Selberg L-functions in an appendix. It is the exact same argument up to very minor complications (which vanish once one spells the argument out) and it will be on arxiv soon. Should I just cite this m.o. question or would you prefer something else? Thanks!
@Alufat: Please just cite this MO question. Glad it was useful.
|
2025-03-21T14:48:31.935811
| 2020-08-30T22:28:30 |
370482
|
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|
Stack Exchange
|
Polynomials for the alternating group $A_n$
It is my understanding that the polynomial $f_n(x)=x^n-1$ has the Galois Group $(\mathbb{Z}/n\mathbb{Z})^*$, group of units of order $\phi(n)$. In some sense, these are the "simplest" polynomials with that Galois Group. Is there a formula for the polynomial, say $g_n(x)$, whose Galois group is $A_n$? And, I mean $g_n(x)$ in the same sense as $f_n(x)$: that is, the "simplest" polynomials with that Galois Group of $A_n$.
a related question here, seems like a part of the inverse galois problem
The Galois group of $x^n-1$ (over $\mathbb{Q}$) is $(\mathbb{Z}/n \mathbb{Z})^{\times}$, the group of mulitiplicative units modulo $n$, not $C_n$.
@DavidESpeyer: I will update the question. Could you please provide a reference/proof/sketch-of-the-proof for that fact? Thank you.
See https://kconrad.math.uconn.edu/blurbs/galoistheory/cyclotomic.pdf , for example.
Note that there are distinct $n,m$ with $(Z/nZ)^\times$ and $(Z/mZ)^\times$ isomorphic (for instance, $m,n=1,2$, or $m,n=3,4$), so it's not a unique "simplest" polynomial.
In some sense the simplest polynomial with Galois group $S_n$ is the polynomial $\sum_{i=0}^n (-1)^i \sigma_ix^i$ over $\mathbf C(\sigma_1,\ldots,\sigma_n) = \mathbf C(x_1,\ldots,x_n)^{S_n}$, where $\sigma_i$ is the $i^{\text{th}}$ elementary symmetric polynomial. Replacing $S_n$-invariants with $A_n$-invariants gives an analogous polynomial for $A_n$ (possibly over a field that is not pure transcendental ― I'm not sure).
The question was asked recently at https://math.stackexchange.com/questions/3808650/the-cyclotomic-polynomial-analogue-for-a-n. Generally speaking, when posting the same question on both sites this should be mentioned to avoid duplicate answers/work.
@KConrad: the other question is more specific. I posted a generalized version here in the hopes that I may be pointed to some research on the topic. I will include a reference if I post again in a situation similar to this one.
@YCor: thank you for pointing that out. Non-uniqueness is not an issue, just need one, or more, family of functions parametrized by $n$ with $A_n$ as the Galois Group.
Schur proved that the truncated exponential polynomials $1 + x + x^2/2! + \cdots + x^n/n!$ are irreducible over $\mathbf Q$ for all $n \geq 1$ and their splitting field over $\mathbf Q$ has Galois group $A_n$ when $4 \mid n$. When $n$ is not divisible by $4$, the Galois group is $S_n$. The cases of $n$ being or not being divisible by $4$ corresponds to the discriminant of the polynomial being or not being a square, which is the minimial kind of information needed to know if a Galois group of an $n$th degree irreducible polynomial is or is not contained in $A_n$.
@KConrad: any available references to that work?
Just google "schur exponential galois" and you'll find a math.stackexchange page that answers your question.
@KConrad that works: http://people.math.gatech.edu/~mbaker/pdf/Coleman_GaloisNewton.pdf contains the proof in a very approachable way.
Hermez and Salinier, Rational trinomials with the alternating group as Galois group, Journal of Number Theory, Volume 90, Issue 1, September 2001, Pages 113-129 has the abstract,
For any integer $n\ge7$, we show how to explicitly build an infinite number of rational trinomals of degree $n$ whose Galois group over $\bf Q$ is isomorphic to $A_n$.
|
2025-03-21T14:48:31.936059
| 2020-08-30T22:30:24 |
370483
|
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|
Stack Exchange
|
Product decompositions and maps from product of initial object with itself
For any (commutative, unital) ring $A$, maps $\mathbb{Z}^2\to A$ correspond to idempotent elements of $A$. Also, idempotent elements of $A$ correspond to ways of expressing $A$ as a product of two rings, by mapping an idempotent $e$ to the decomposition $(e)\times (1-e)$, where we treat $(e)$ as a ring with identity $e$. This gives a bijection between $\text{Hom}(\mathbb Z^2,A)$ and the set of product representations of $A$.
More generally, in any category with finite products and initial object $I$, we can define a product decomposition of an object $A$ as an isomorphism class of tuples $(A,B_1,B_2,p_1,p_2)$ with $p_i:A\to B_i$ which form product diagrams. Then for any object $A$, there is a map $F$ from the "set" of product decompositions to $\text{Hom}(I^2,A)$ as follows: given a representative $(A,B_1,B_2,p_1,p_2)$, we have $I\times I \xrightarrow{\pi_i}I\to B_i$ which for $i=1,2$ gives two maps $I\times I\to B_i$, and thus a map $I\times I\to B_1\times B_2=A$.
As noted above, for any object of the category of rings, the $F$ is a bijection. (The set of product decompositions doesn't have a functor structure, so it doesn't seem like we could upgrade this to a natural bijection.) In other categories this is not the case: for example, in the category of sets, the initial object is $\emptyset$ and $\emptyset\times \emptyset=\emptyset$, so $\text{Hom}(I^2,A)$ is always a singleton. However, in $\text{Set}^{\text{op}}$, $F$ is again a bijection, because the coproduct of the final object with itself is a set with 2 elements, and functions from $A$ to a set with two elements correspond, via the above map, with ways of writing $A$ as a disjoint union of two sets. The same thing happens in e.g. the category of topological spaces. (For the same reason that $F$ is trivial in $\text{Set}$, $F$ is trivial in any category with a zero object.)
If we look at something like $\text{Set}\times \text{Set}^{\text{op}}$, $F$ becomes a product projection. (Trying coproduct categories breaks things because there are no longer initial/final objects.)
My questions are the following:
What conditions on a category/an object of a category ensure that the collection of product decompositions is a set?
Are there "natural" categories in which the behavior is different from the 3 behaviors above? In particular, can $F$ not be a bijection?
Is there something deeper going on here?
Does the fact that this describes a similarity between the opposite of the category of rings and the category of topological spaces indicate that there could be some algebraic geometry hiding somewhere?
To my mind it's cleaner to take opposite categories and talk about coproduct decompositions of affine schemes, where $\text{Spec } \mathbb{Z} \times \mathbb{Z}$ is just "two points" (the coproduct $2 = 1 \sqcup 1$ of two copies of the terminal object). The common feature of coproducts in affine schemes, sets, and topological spaces that makes those cases work out as nicely as possible is that they are extensive categories. Intuitively this means that coproducts behave like disjoint unions and formally this means among other things that every morphism $x \to y \sqcup z$ into a disjoint union decomposes canonically into a pair of morphisms $x_1 \to y, x_2 \to y$ such that $x \cong x_1 \sqcup x_2$. Specialized to the case that $y = z = 1$ this gives that morphisms $x \to 1 \sqcup 1$ correspond to coproduct decompositions as desired. Note that extensivity also makes "coproduct decompositions" a functor.
Without extensivity of course nothing like this needs to be true. For example, in the category of groups the terminal object is the zero object and the coproduct of two copies of it is still the zero object.
You may also be interested in this blog post where I describe what can be done using only the assumption that a category is distributive (finite products and coproducts exist, and finite products distribute over finite coproducts).
|
2025-03-21T14:48:31.936337
| 2020-08-30T22:38:54 |
370484
|
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"Creator",
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|
Stack Exchange
|
When and why do we require the condition that :"a subset bounded from below and has no accumulation points?"
I have been tyring to understand the first condition given in the link https://en.wikipedia.org/wiki/Regularity_structure for quite some time now, at least a year. I have posted a similar question in https://math.stackexchange.com/questions/3524946/how-to-think-of-a-set-that-has-no-accumulation-point but did not get any reply. Please help me if possible.
As I understand the regularity structure deals with stochastic partial differential equations and this means we deal with stochastic variable or functions which are continous but not differentiable, this leads to discrete type of sets and hence we get the condition "no accumulation point" is this correct?
If so, the question is why it has to be bounded from below only? I was expecting it to be bounded from below as well as above or bounded.
Few words why this condition is necessary in regularity structure would be highly appreciated?
Edit: The comment from Nate Eldredge made it clear why it is not bounded from above, but I am still keeping this question just in case someone write in very consice terms the conection bteween the index set A and the regularity structure.
I think it may help to look at the examples in Hairer's paper. The motivating example is polynomials (section 2.2) in which case we take $A = \mathbb{N}$. This may help to explain why we do not want to require $A$ bounded above. There are other examples where we might want $A = {-2, -1, 0, 1, 2, \dots}$, or perhaps positive half-integers instead of integers, and so on.
@NateEldredge Thank you, I had a wrong undrerstanding of A.
I don't think that the Wikipedia page is very helpful in explaining why these definitions are natural. As Nate suggested, having a look at the original paper or at subsequent review papers like this one or Section 4 here is a much better start.
Anyway, a subset of $\mathbb R$ is bounded below and has no accumulation point iff it is either finite or order isomorphic to $\mathbb N$.
From the references mentioned in the comments , (section 4.4,"Renormalisation of parabolic stochastic PDEs") (RP) and (section 6.1,"Introduction to regularity structures") (IRS), we first start from the pde fixed point problem
$$\Phi = G \ast (\xi − \Phi^3 ) + G\Phi_0 ,$$
where $*$ denotes space-time convolution and where we denote by $G \Phi_0$
the harmonic extension of $\Phi_0$ and we split $G = K + \hat K$ into singular kernel $K$ and smooth $\hat{K}$. Then we consider the analogous abstract fixed point problem ((4.22) in (RP) and (6.3) in (IRS))
$$\Phi = \mathcal{K} \bigl(\Xi - \Phi^3\bigr) + \hat{\mathcal{K}} \bigl(\Xi - \Phi^3\bigr) + G \Phi_0 $$
where you see that by repeated insertion of $\Phi$, you get combinations of abstract symbols eg. $\Xi, \mathcal{I}\Xi, \mathcal{I}(\Xi) \mathcal{I}( \mathcal{I}(\Xi)^3)$ and $\mathcal{I}$ is the "abstract integration" map against that kernel.
So the question becomes: how to keep track and organize all those new abstract expressions. As you see in that section, he gives specific examples of regularity structures based on generating sets of the above symbols.
What about the index set?
A big issue with SPDEs is the presence of the white noise $\xi$ because it is a distribution and so when it shows up in an SPDE, it can make them ill-posed. This is what creates the negative regularity degree. So as in the example, we incorporate that by assigning negative numbers to $\Xi$, (the abstract symbol for noise) eg. in the previous example $deg(\Xi)=-\frac{5}{2}-\epsilon$ for some $\epsilon>0$.
On the other hand, what saves us is integrating against the kernel $\mathcal{K}$ because as you might know from the heat equation, integrating against a heat kernel immediately smoothens functions. A similar phenomenon happens here via the Schauder estimates improving the regularity of the solution pushing it to the positive side. So we assign positive increase: $|\mathcal{I} \tau| = |\tau| + 2$
So there is this battle between lowering/increasing regularity. The index set
contains all the degree of the above symbols starting from the most negative degree of $\Xi$ eg. above $\{-\frac{5}{2}-\epsilon,-\frac{5}{2}-\epsilon+2, -\frac{5}{2}-\epsilon+4,...\} $
So the requirement "bounded below" ensures that eventually with enough convolutions, we will get actual functions out of this process rather than distributions (using reconstruction operator and removal of appropriate constants).
|
2025-03-21T14:48:31.936644
| 2020-08-30T22:54:54 |
370485
|
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|
Stack Exchange
|
Concerning the Galois resolvent
Let $K$ be an algebraic number field, $f(x) = 0$ a separable algebraic equation over $K$ of degree $n \ge 2$, i.e. having only simple roots $x_1, \dotsc, x_n$, and $L \mathrel{:=} K[x_1, \dots, x_n]$ the splitting field of this equation over $K$. The basic construction on which Galois built his theory of equations was the following description of the splitting field $L$. Let $T \in K[X_1, \dots, X_n]$ be a Galois resolvent, i.e. a polynomial with the property that all the rational functions $\tau^{\sigma} := T(x_{\sigma(1)}, \dots, x_{\sigma(n)})$ in the roots of $f(x)$ are pairwise different elements of $L$ for all $n!$ permutations $\sigma \in \mathfrak{S}_n$ (such Galois reolvents exist and are in some sense not specified here any further "generic", see [1]). Then all the $\tau^{\sigma}$
are primitive elements in the sense that they generate $L$ over $K$, i.e. $L=K[\tau^{\sigma}]$; this is one of the first basic results of Galois.
In particular, put $\tau := \tau^{\textrm{id}}$. Then form the following polynomial of degree $n!$ over $K$:
$$
F(T) := \prod_{\sigma \in \mathfrak{S}_n} (T-\tau^{\sigma}) \in K[T].
$$
Factor $F(T)$ into irreducibles
$$
F(T) = \prod_{i=1}^k G_i(T),
$$
and let $G(T):=G_j(T)$ be a factor with $G_i(\tau)=0$. Then $L \cong K[T]/(G(T))$(see [1]).
Question:
Are the $G_i(T)$ pairwise different, all of the same degree?
(In fact, all of the same degree should be evident.)
[1] Edwards, H.M.,
Galois Theory
(Graduate Texts in Mathematics 101). Springer 1984
So $S_n$ acts on the $\tau^\sigma$ and the Galois group $G$ itself acts by it's embedding into $S_n$. The $G(T)$ should be a product of $(T-g\tau)$ for $g \in G$ and the other polynomials will correspond to the other orbits of $G$ on the set of $\tau^\sigma$. As such they will all be of the same degree. They have to be pairwise different because their roots of different!
|
2025-03-21T14:48:31.936923
| 2020-08-30T22:59:23 |
370486
|
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|
Stack Exchange
|
Cofinality for natural transformations
Given a diagram $D\colon\mathcal{C}\longrightarrow\mathcal{D}$, we say that a functor $J\colon\mathcal{I}\longrightarrow{C}$ is cofinal if we have a natural isomorphism
$$
\mathrm{colim}\left(\mathcal{C}\overset{D}{\longrightarrow}\mathcal{D}\right)\cong\mathrm{colim}\left(\mathcal{I}\overset{J}{\longrightarrow}\mathcal{C}\overset{D}{\longrightarrow}\mathcal{C}\right).
$$
With a number of ways to check cofinality for functors (e.g. 1, 2), this notion ends up being very useful in practice as a computational tool.
What about cofinality for natural transformations? Namely, given functors $D,D'\colon\mathcal{C}\rightrightarrows\mathcal{D}$, call a natural transformation $\eta\colon D\Rightarrow D'$ final if postcomposition with it induces a bijection¹
$$
\mathsf{Nat}\left(\Delta_{(-)},D\right)\overset{\eta}{\underset{\cong}{\dashrightarrow}}\mathsf{Nat}\left(\Delta_{(-)},D'\right),
$$
and hence a natural isomorphism $\lim(D)\cong\lim(D')$. Do we have, as in the case of cofinality for functors, easy(-ish) to check conditions implying that a natural transformations is final?
[Motivation: If so, this would similarly be useful for either computations, or for expressing co/limits of somewhat contrived functors as co/limits of nicer, more naturally-appearing, ones. For example, having such a notion in hand, one would be able to e.g. compute the $\pi_0$ of a complicated simplicial set $S_\bullet$ by first finding a “final simplicial map” $S_\bullet\to T_\bullet$ to a simpler simplicial set $T_\bullet$ and then computing the $\pi_0$ of $T_\bullet$.]
¹Here's a picture of the induced cones of $D'$ over some object $X$ of $\mathcal{D}$:
This is a right orthogonality condition. As such it is closed under composition and retracts and limits and has the 2-out-of-6 property. If $\mathcal{D}$ is, say, accessible then you only need to check the condition against a small set of objects. I don’t think there’s much more that can be said in this generality, since it depends on $\mathcal{D}$, unlike cofinality of functors.
@ZhenLin Thanks! If we impose nice conditions on $\mathcal{D}$ (say being a topos or the category of models of a finite limits theory, etc.), then would “cofinal natural transformations” admit useful characterisations?
I doubt it. If anything it may be more important to look at the diagram shape $\mathcal{C}$. For instance, if $\mathcal{C}$ has a cofinal full subcategory then you know that the components of the diagram outside that subcategory don't matter.
Robert Paré's Morphisms of colimits might be relevant?
@RoaldKoudenburg Hmm... I think the paper answers a different question, which is related to this one. Namely, referring to the very first diagram in p. 2 there, I think my question would correspond to taking $F=\mathrm{id}$ and studying when the induced morphism between co/limits is an iso. From what I understand, however, Paré is studying what would be necessary to have a morphism, not necessarily an iso, between the co/limits. In any case, the paper looks very interesting! Thanks for the pointer :)
|
2025-03-21T14:48:31.937146
| 2020-08-31T00:06:34 |
370490
|
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Stack Exchange
|
About the function $\prod_{k \in \mathbb{N}}(1-\frac{x^3}{k^3})$
I'm wondering if the function $$f(x)=\prod_{k \in \mathbb{N}}\left(1-\frac{x^3}{k^3}\right)$$ has a name, or if there are any properties (especially about derivatives of $f$) have studied so far.
I got inspired by proof of the Basel problem ($\frac{\pi^2}{6}=\sum \frac{1}{k^2}$) using the product form of $$\frac{\sin(x)}{x}=\pi \prod_{k \in \mathbb{N}} \left(1-\frac{x^2}{k^2 \pi^2}\right).$$ The function $f(x)$ above would be considered as analogue for the Apery constant and $\zeta(3k)$.
Whittaker, Watson, "Course of Modern Analysis", chapter 12.13, Example 2.
http://ramanujan.sirinudi.org/Volumes/published/ram10.html
If one starts with the Weierstrass factorisation
$$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{k}\right)^{-1} e^{z/k}$$
of the Gamma function, applied to $z = -x, -\omega x, -\omega^2 x$ (where $\omega = e^{2\pi i/3}$ is the cube root of unity), and multiplies the three together, one obtains
$$ \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^2 x) = \frac{1}{-x^3} \prod_{k=1}^\infty \left(1-\frac{x^3}{k^3}\right)^{-1}$$
and hence
$$\prod_{k=1}^\infty \left(1-\frac{x^3}{k^3}\right) = -\frac{1}{x^3 \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^2 x)}.$$
One can manipulate the right-hand side a little using the various functional equations of the Gamma function, but it does not have a dramatically simpler form (as one can already see from the zero set ${\bf N} \cup \omega {\bf N} \cup \omega^2 {\bf N}$, which is too weird to come from anything more elementary than a Gamma function). This is in contrast to the analogous identity
$$ \prod_{k=1}^\infty \left(1 - \frac{x^2}{k^2}\right) = -\frac{1}{x^2 \Gamma(-x) \Gamma(x)}$$
where the Euler reflection formula (and $\Gamma(1-x) = -x \Gamma(-x)$) applies to simplify the right-hand side to $\frac{\sin(\pi x)}{\pi x}$ as you mention, and the zero set ${\bf N} \cup -{\bf N} = {\bf Z} \backslash \{0\}$ is now simple enough to be explainable via trigonometric functions. (The different structure of the zero sets also helps explain why the zeta function evaluated at even natural numbers is much more tractable than the zeta function evaluated at odd numbers; the former relates to the Fourier analytic structure of the integers, but the latter relates to the Fourier analytic structure of the natural numbers, which is far worse.)
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2025-03-21T14:48:31.937318
| 2020-08-31T02:44:24 |
370498
|
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|
Stack Exchange
|
Information about Milnor conjecture
I'm a student of mathematics and I need know about the status of the Milnor conjecture (if there are partial results or if someone solved that). The statement is:
A complete Riemannian manifold with non-negative Ricci curvature has a finitely generated fundamental group.
If someone can help me with references to papers or anything I would be grateful.
https://www.google.com/search?q=milnor+conjecture+fundamental+group gives me that as of 2018 it is known in dimension 3 and below, but open in general.
Counterexamples have been given by Bruè–Naber–Semola first in dimension 7 https://arxiv.org/abs/2303.15347 and then 6 https://arxiv.org/abs/2311.12155.
According to David Roberts comment and the following paper it is open for dimensions $n\geq 4$.
Pan, Jiayin, A proof of Milnor conjecture in dimension 3, J. Reine Angew. Math. 758, 253-260 (2020). ZBL1432.53053.
There is a nice survey by Shen and Sormani that can be found in author homepage:
Shen, Zhongmin; Sormani, Christina, The topology of open manifolds with nonnegative Ricci curvature, Commun. Math. Anal., Conference 1, 20-34 (2008). ZBL1167.53309.
And a few related and partial attacks to the conjecture:
Paeng, Seong-Hun, On the fundamental group of manifolds with almost nonnegative Ricci curvature, Proc. Am. Math. Soc. 131, No. 8, 2577-2583 (2003). ZBL1040.53042.
Xu, Senlin; Deng, Qintao, The fundamental group of open manifolds with nonnegative Ricci curvature, Acta Math. Sin., Chin. Ser. 49, No. 2, 353-356 (2006). ZBL1120.53021.
Which is stronger than
Sormani, Christina, Nonnegative Ricci curvature, small linear diameter growth and finite generation of fundamental groups., J. Differ. Geom. 54, No. 3, 547-559 (2000). ZBL1035.53045.
Thank's a lot!!!.
To be sure you have a thorough review of what has been done, go to Milnor’s paper with the original conjecture on mathscinet and to a few of the articles mentioned above and check who has cited them. This should bring you to everything that has been published.
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2025-03-21T14:48:31.937472
| 2020-08-31T04:44:07 |
370501
|
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|
Stack Exchange
|
Can the inverse of the Riemann zeta function in $s > 1$ be expressed as a series?
In this post, we are interested in the Rimenann zeta function $\zeta(s)$ in $s > 1$ only where it is strictly decreasing rather than $s$ in the entire complex plane. We have the Stieltjes series expansion of the Riemann Zeta function. I inverted the first few terms of this series using series reversion and showed that if $s > 1$ and $\zeta(s) = a$, then,
$$
s = \zeta^{-1}(a) = 1 + \frac{1}{a - \gamma_0} - \frac{\gamma_1}{1!(a - \gamma_0)^2} + \frac{\gamma_2}{2!(a - \gamma_0)^3} - \frac{\gamma_3 - 12\gamma_1}{3!(a - \gamma_0)^4} + \mathcal O(a^{-5})
$$
It seems that $\zeta^{-1}(a)$ can be expressed in the form
$$
1 + \sum_{n=0}^{\infty} (-1)^n\frac{f(\gamma_1, \gamma_2, \ldots, \gamma_n)}{n!(a - \gamma_0)^{n+1}}
$$
where $f(\gamma_1, \gamma_2, \ldots, \gamma_n)$ is some polynomial function of the Stieltjes constants $\gamma_n$.
Question: I am looking for a closed or a recurrence formula for $f$? Also any reference in this series in literature?
Note: The question was posted in MSE. It got upvotes but no answers hence posting in MO.
This problem is not substantially different from general series reversion, so you should not have hopes for closed formulas more simple then for the reverse series itself.
I believe this problem is still more important than at first may seem. Since $\zeta(s)$ is monotone and smooth for real $s>1$ it has a smooth inverse and so is analytic, at least as a real "one-dimensional" function. The question is whether or not this inverse has an analytic continuation and in particular what its properties and limitations are. I would much like to have answers and insights to this more general question. I worry about the Stieltjes constants -- they behave terribly.
|
2025-03-21T14:48:31.937615
| 2020-08-31T05:06:53 |
370503
|
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|
Stack Exchange
|
Analytical solution to a specific differential equation
I was wondering whether there is an analytical solution to the ODE
\begin{equation}
-n\int xy(x)dx + ihy'(x) + (x^2+k)y(x) = 0,
\end{equation}
where $n=0,1,2,...$, $h \in \mathbb{R}$, and $k=+1,0$ or $-1$.
For $n=0$ this can be solved exactly, and the solution is
\begin{equation}
y(x) = C \exp\left[ih\left(\frac{x^3}{3}+kx\right)\right] \equiv f(x).
\end{equation}
However, I struggled to find a general solution for different $n$. Differentiating the ODE gives
\begin{equation}
ihy''(x)+(x^2+k)y'(x)+(2-n)xy(x)=0.
\end{equation}
I have tried the ansatz
\begin{equation}
y(x) = f(x)^{(n-2)/2} + f(x)^{(n-2)/2}\int f(x)^{1-n} dx,
\end{equation}
but it only satisfies the equation when $n=2$.
I was able to make some progress by separating the equation into its real and imaginary parts and integrate the coupled equations numerically, but I was hoping for an analytical solution.
The other possible solution is the triconfluent Heun function
\begin{equation}
y(x) = e^{-\frac{x^{3}+3 kx}{3 h}} C_{2} \text { HeunT }\left[0, \frac{-4+n}{h},-\frac{k}{h}, 0,-\frac{1}{h}, x\right]
+ C_{1} \text { HeunT }\left[0, \frac{-2+n}{h}, \frac{k}{h}, 0, \frac{1}{h}, x\right].
\end{equation}
My question is whether there is an analytical solution of a simpler form?
For $k=0$ the solution is a hypergeometric function,
$$y(x)=C_1 \, _1F_1\left(\frac{2}{3}-\frac{n}{3};\frac{2}{3};\frac{i x^3}{3 h}\right)-(3h)^{-1/3}(-1)^{5/6} C_2 x \, _1F_1\left(1-\frac{n}{3};\frac{4}{3};\frac{i x^3}{3 h}\right),$$
which at least for some values of $n$ can be reduced to a Bessel function and/or an incomplete gamma function.
$$n=1:\qquad y(x)=\frac{\sqrt[6]{-\frac{1}{3}} \sqrt{x} e^{\frac{i x^3}{6 h}} \left(3 \sqrt[3]{2} C_1 \Gamma \left(\frac{5}{6}\right) J_{-\frac{1}{6}}\left(-\frac{x^3}{6 h}\right)-i C_2 \Gamma \left(\frac{1}{6}\right) J_{\frac{1}{6}}\left(-\frac{x^3}{6 h}\right)\right)}{3\ 2^{2/3} \sqrt[6]{h}},$$
$$n=2:\qquad y(x)=\frac{\sqrt[3]{-1} C_2 h^{2/3} \Gamma \left(\frac{4}{3}\right) \left(-\frac{i x^3}{h}\right)^{2/3}}{x^2}-\frac{\sqrt[3]{-1} C_2 h^{2/3} \left(-\frac{i x^3}{h}\right)^{2/3} \Gamma \left(\frac{1}{3},-\frac{i x^3}{3 h}\right)}{3 x^2}+C_1,$$
$$n=3:\qquad y(x)=\frac{C_1 \Gamma \left(\frac{2}{3}\right) \sqrt[3]{-\frac{i x^3}{h}}}{\sqrt[3]{3}}+\frac{C_1 \sqrt[3]{-\frac{i x^3}{h}} \Gamma \left(-\frac{1}{3},-\frac{i x^3}{3 h}\right)}{3 \sqrt[3]{3}}-\frac{(-1)^{5/6} C_2 x}{\sqrt[3]{3} \sqrt[3]{h}}.$$
Thank you. It looks like there isn't a easy way to work around it.
|
2025-03-21T14:48:31.937753
| 2020-08-31T06:42:29 |
370506
|
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|
Stack Exchange
|
Particular Ehrenpreis factorization for covariance function
Let $f:\mathbb{R}^d\to\mathbb{R}$ be a smooth compactly supported covariance function of a stationary random fields (hence positive definite).
Is there a compactly supported function $g:\mathbb{R}^d\to \mathbb{C}$ such that $g\ast g=f$ ?
Using Fourier transform and Bochner's theorem, this question is equivalent to the construction of a entire square root for entire function of exponential type that are positive and fast decaying on the real line.
And how do you expect to get an entire square root when the Fourier transform has simple complex zeroes (which isn't hard to arrange: take any positive and fast decaying on the line entire function of exponential type and multiply it by $z^2+\mu^2$ with some real $\mu$ such that $i\mu$ is not a zero of the original function)?
Very interesting, I did not think about multiplication by such function to obtain simple zeroes. So indeed, it is not always possible. I just need to make this assumption then. I wonder if it says something interesting about the covariances functions that cannot be root squared that way.
|
2025-03-21T14:48:31.937851
| 2020-08-31T08:02:59 |
370509
|
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|
Stack Exchange
|
Multiplicity of zeros of partial sums of the Dirichlet Eta function
I am studying ways to approach the problem of the multiplicity of zeros of the partial sums of the Dirichlet Eta functions:
$$
\sum_{n=1}^{K}\frac{(-1)^{n-1}}{n^{s_o}} = 0
$$
more in particular, whether it can be proved that the above implies that its first derivative cannot vanish (i.e. multiplicity = 1):
$$
\sum_{n=1}^{K}\frac{(-1)^{n} \log n}{n^{s_o}} \neq 0
$$
considering $s_o = \sigma_o + i t_0$, and substituting $t_o \log n = a_n$, we can reformulate in a slightly more general fashion:
$$
\sum_{n=1}^{K}\frac{(-1)^{n-1}\cos a_n}{n^{\sigma_o}} = 0 \;\;\;\;\;\; AND \;\;\;\;\;\;\sum_{n=1}^{K}\frac{(-1)^{n-1}\sin a_n}{n^{\sigma_o}} = 0
$$
would it also imply that it cannot be (let us consider $t_o \neq 0$)
$$
\sum_{n=1}^{K}\frac{(-1)^{n} a_n\cos a_n}{n^{\sigma_o}} = 0 \;\;\;\;\;\; AND \;\;\;\;\;\;\sum_{n=1}^{K}\frac{(-1)^{n} a_n\sin a_n}{n^{\sigma_o}} = 0 \;\;\;\;\;\; ?
$$
whereby $a_n$ is now simply defined as a monotonically increasing sequence of positive real numbers.
I am just wondering whether this kind of more general sums have ever been studied in a systematic way. The aim is to verify whether that may help, or whether it is instead adding even more complexity to an already difficult problem.
First Edit :
I have in the meantime realised that the above conditions may also be visualized in a more intriguing way by simply switching the signs in the first one (we are considering the = 0 case anyway). Referring to the $\sin$ finite sums (in this example the sum ends at odd $k$), whereby "........." is a short notation for the corresponding finite sum of cosine terms:
$$
0\, + \, \frac{\sin a_2}{2^{\sigma_o}} \, - \, \frac{\sin a_3}{3^{\sigma_o}} \, + \, .... \, - \, \frac{\sin a_k}{k^{\sigma_o}} = 0 \;\;\;\;\;\;\;\; AND \;\;\;\;\;\;\;\; "........."
$$
would it also imply that it cannot be (let us consider $t_o \neq 0$)
$$
0\, + \, a_2\frac{\sin a_2}{2^{\sigma_o}} \, - \, a_3\frac{\sin a_3}{3^{\sigma_o}} \, + \, .... \, - \, a_k\frac{\sin a_k}{k^{\sigma_o}} = 0 \;\;\;\;\;\;\;\; AND \;\;\;\;\;\;\;\; "........." \;\;\;\;\;\; ?
$$
Of course, sufficient conditions can then be obtained by restricting to only parts of the above relatonships, such as for example limiting our examination to the the two sines sums only. Indeed, myself I am unable to imagine how there could ever exist a monotonically increasing sequence $a_n$, of positive real numbers, such that the above two finite sums of sine terms could both vanish at the same time (could not find any counterexample). Yet, I am unable to find a way to prove it.
That would also prove that zeros of the Dirichlet Eta Function (and thus of the Riemann Zeta Function) must be of multiplicity 1, through a straightforward application of Hurwitz's Theorem (this one: https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(complex_analysis) ).
I shall clarify that I am considering only positive $t_o \neq 0$ because of the symmetry of complex conjugate zeros.
|
2025-03-21T14:48:31.938038
| 2020-08-31T08:14:47 |
370510
|
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|
Stack Exchange
|
Covariance inequality with Lipschitz functions
Suppose that $X$ and $Y$ are random variables and suppose that for all Lipschitz functions $f$ and $g$ s.t. $f(X),g(Y)\in L^p$, $p>2$,
$$
|\operatorname{Cov}(f(X),g(Y))|\le \big(\operatorname{Lip}(f)\operatorname{Lip}(g)+\|f(X)\|_p\|g(Y)\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{1}\label{1}
$$
When $f$ and $g$ are the identity functions, the inequality becomes
$$
|\operatorname{Cov}(X,Y)|\le \big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{2}\label{2}
$$
Now, multiplying $X$ and $Y$ by a constant $c>0$, we get
$$
|\operatorname{Cov}(X,Y)|\le \inf_{c>0}\big(c^{-2}+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}=\|X\|_p\|Y\|_p\alpha^{1-\frac{2}{p}}.\tag{3}\label{3}
$$
This result seems weird because the inequality in \eqref{1} is clearly "scale invariant", that is, applying \eqref{1} directly (with $f(x)=g(x)=cx$) yields
$$
|\operatorname{Cov}(cX,cY)|\le c^2\big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{4}\label{4}
$$
Is the inequality in Equation \eqref{3} actually correct?
This just looks to me like an example where fixing something prematurely (or at all) gives you a sub-optimal estimate. Specifically, fixing $f=g=\mathrm{id}$ immediately leaves you unable to do much about the first term on the right hand side of (1), so it's not that surprising that doing so gives you a worse form of your estimate than if you don't. Unless I'm missing something, it looks like you've deduced that your hypotheses imply that
$$
|\mathrm{Cov}(X,Y)|\leq (c^{-2} + \Vert X \Vert_p\Vert Y \Vert_p)\alpha^{1-2/p}
$$
for any $c>0$ (i.e. (3)), discovering along the way that fixing $f=g=\mathrm{id}$ only gives you the $c=1$ case.
The hypothesis in (1) implies (2) and (4), and (2) supposedly implies (3). I don't see where my logic fails.
Re. I don't see where my logic fails, why do you think it does?
That's because (1) need not hold for $cX$ and $cY$.
Just for me (and others reading this), would you mind making your hypotheses concerning $X$, $Y$, $f$ and $g$ a little clearer? Is your hypothesis that $X$ and $Y$ are such that there is a $p>2$ such that $f(X), g(Y)\in L^p$ and (1) holds for all Lipschitz $f$ and $g$, or something else?
|
2025-03-21T14:48:31.938179
| 2020-08-31T09:39:59 |
370514
|
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|
Stack Exchange
|
Is the following function Lipschitz?
Given a vector $Q \in \mathbb{R}^{S\times A}$ where $S$ and $A$ are sets of finite cardinality, for $0<\gamma<1$ define the function $H_{w}:\mathbb{R}^{S\times A} \rightarrow \mathbb{R}^{S\times A}$ as $H_{w}Q(s,a) := w\left(r(s,a)+ \gamma \sum_{s' \in S}p(s'|s,a)\displaystyle\max_{a \in A} Q(s',a)\right)+(1-w)\displaystyle\max_{a \in A}Q(s,a).$ Here $|r(s,a)| \leq R ~ \forall ~ (s,a) \in S \times A$ and $p(\_|s,a)$ is a probability mass function on $S$ for any given $(s,a) \in S \times A$ and $0<w<w^*$ with $w^*=\displaystyle\max_{(i,a)}\left\{\frac{1}{1-p(i|i,a)}\right\}.$
Note:
$H_w$ is a contraction with respect to the max-norm/infinity-norm on $\mathbb{R}^{S\times A}$ with contraction factor $(1-w+\gamma w)$
Note the possibility of $w>1$.
Is the function $H_wQ$ jointly Lipschitz in $w$ and $Q$?
In view of your Note 1, it appears that the definition of $H_w Q(s,a)$ has to corrected as follows:
$$H_w Q(s,a):=\gamma w\left(r(s,a)+\sum_{s'\in S}p(s'|s,a)\max_{a\in A}Q(s',a)\right)
+(1-w)\max_{a\in A}Q(s,a).$$
(Your definition is missing $\gamma$.)
Now it is clear that $H_w Q$ is not Lipschitz in $(w,Q)$, because $H_w Q$ is "of degree $2$" (rather than "of degree $\le1$") in $(w,Q)$. In fact, $H_w Q$ is not Lipschitz even in $w$. Indeed, taking $Q$ to be an arbitrary real constant $q$ and letting $r=0$, we have
$$H_w Q(s,a)=h(w,q):=\gamma wq+(1-w)q=(1-(1-\gamma)w)q.$$
The partial derivative of $h(w,q)$ in $w$ is $(1-\gamma)q$, which can be however large for large $q$. So, indeed $H_w Q$ is not Lipschitz even in $w$.
Note 1: $h(w,q)$ is a polynomial in $(w,q)$ of degree $2$.
Note 2: With your current definition, too, $H_w Q$ will not in general be Lipschitz even in $w$. In that case, we can let $Q(s,a)$ take two distinct values, depending on $s$.
Thank you. You are correct, there is a typo. Edited the statement accordingly. We agree with your solution. The function is not Lipschitz with respect to $w$. We also carried out a similar calculation and obtained the same
|
2025-03-21T14:48:31.938323
| 2020-08-31T09:55:44 |
370516
|
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|
Stack Exchange
|
Is Sydler's theorem concerning Dehn invariants constructive?
Sydler proved something of a converse to Dehn's negative resolution
of Hilbert's 3rd problem. To quote Wikipedia, Sydler showed that
"every two Euclidean polyhedra with the same volumes and Dehn invariants can be cut up and reassembled into each other."
My question is: Is Sylder's proof constructive in the sense that
it implies an algorithm?
Sydler, J.-P., "Conditions nécessaires et suffisantes pour l'équivalence des polyèdres de l'espace euclidien à trois dimensions." Commentarii Mathematici Helvetici 40, pp 43–80, 1965,
doi:10.1007/BF02564364, EuDML.
A secondary question: Does anyone know if an English
translation of his paper is available?
Yes, I think so. I went through the proof awhile ago thinking I should write it up in my book. I ended up including only some of the lemmas I found especially elegant (see section 16). The proof is fundamentally algebraic, but every step can in principle be replaced by a constructive argument. It's just not particularly "effective", but neither is Bolyai–Gerwien theorem I suppose.
I thank John McCleary for pointing me to Jessen's work, and from there
I found the useful thesis of Schröder.
I have not yet assessed these papers.
Jessen, Børge. "The algebra of polyhedra and the Dehn-Sydler theorem." Mathematica Scandinavica 22, no. 2 (1969): 241-256.
JSTOR link.
Jessen, Børge, Jørgen Karpf, and Anders Thorup. "Some functional equations in groups and rings." Mathematica Scandinavica 22, no. 2 (1969): 257-265.
JSTOR link.
Felix Schröder.
"Decomposability and the Dehn Invariant."
Thesis, TU Berlin. 2015.
PDF download.
|
2025-03-21T14:48:31.938462
| 2020-08-31T10:35:29 |
370517
|
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|
Stack Exchange
|
"Universal" polynomial of bounded norm on the sphere
Consider the space $V_{d,n}=\mathbb{R}[x_1,\ldots,x_n]_d$ of homogeneous polynomials of degree $d$ in $n$ variables. I am interested in the set of bounded polynomials on the sphere $$B_{d,n}=\{f\in V_{d,n}:\, ‖f(x)‖\leq 1\textrm{ for all } ‖x‖=1\}.$$ Assume that we have fixed $d$ and $n$. I am looking for a "universal polynomial" for the set $B_{d,n}$. By that I mean a polynomial $F\in B_{d,N}$ for some possibly large $N$ such that every element $f\in B_{d,n}$ can be obtained by restricting $F$ to some suitable linear subspace $V\subset\mathbb{R}^N$ with $\dim(V)=n$. To be more precise: For each $f\in B_{d,n}$ there should be a norm preserving linear map $U:\mathbb{R}^n\to\mathbb{R}^N$ such that $f=F\circ U$.
I can find such a polynomial for $d=1,2$ (see below) and I would be very interested in the case $d=3$:
Question. Is there such a universal polynomial for $d=3$ and any $n$?
I would highly appreciate any answer to that question, positive or negative.
Sketch for $d=1,2$:
Here I want to sketch the construction of such a polynomial for $d=1,2$. For $d=1$ we claim that $N=n+1$ and $F=x_{n+1}$ does the job. Indeed, if $f\in B_{1,n}$, then after an orthogonal change of coordinates, we can assume that $f=\lambda x_1$ with $|\lambda|\leq1$. Then there is a $t\in\mathbb{R}$ such that $\lambda=\cos(t)$. Then we can apply the following orthogonal change of coordinates to $F$:
$$x_{n+1}\mapsto \cos(t)x_1+\sin(t) x_{n+1},\,\, x_1\mapsto -sin(t)x_1+\cos(t)x_{n+1}$$ and we obtain $\cos(t)x_1+\sin(t) x_{n+1}$. Restricting this to the $n$-dimensional subspace $x_{n+1}=0$ we obtain $f$. This shows the case $d=1$.
For the case $d=2$ we claim that $N=3m$ and the polynomial $F=x_{n+1}^2+\ldots+x_{2n}^2-(x_{2n+1}^2+\ldots+x_{3n}^2)$ does the job. To see that we note that any $f\in B_{2,n}$ can be written as $\lambda_1x_1^2+\ldots+\lambda_nx_n^2$, $|\lambda_i|\leq1$, after some orthogonal change of coordinates. Then we can apply orhogonal transformations to $F$ similar as in the case $d=1$ for each $x_i$, $1\leq i\leq n$, and obtain $f$ by restricting to $x_j=0$ for $j>n$.
"such that every element $f\in B_{d,n}$ can be obtained by restricting $F$ to some suitable linear subspace" Just to clarify what exactly you mean by this. Is it that there exists a norm-preserving linear mapping $U:\mathbb R^n\to\mathbb R^N$ such that $f=F\circ U$, or is it something stronger?
Yes, this is what I mean. Thanks, I will also edit the question accordingly.
Is your $F$ for $d=1$ bounded by 1?
@FedorPetrov As written, obviously not, but $x_{n+1}$ alone would do just fine.
Yes, thanks. I adjusted the case $d=1$ accordingly.
|
2025-03-21T14:48:31.938741
| 2020-08-31T10:45:15 |
370518
|
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|
Stack Exchange
|
Are all mapping classes also Dehn twists?
Let $X$ be a Riemann surface and $\Gamma$ its (pure) Mapping Class Group, then $\Gamma$ is generated by Dehn twists along simple closed curves. Is \emph{any} element of the mapping class group also a Dehn twists?
Said in another way, any element of $\Gamma$ is some complicated word in the generators associated to some collection of pairwise disjoint closed curves. Does it also exist a curve such that this element is the twist associated to it?
P.S.
I know that in general one has to take into account half-twists as well, but I am not thinking of that, and this is why I wrote "pure" Mapping Class Group
Every mapping class is a product of finitely many Dehn twists. This is Lickorish‘s Theorem.
Another no: the map on homology induced by a Dehn twist has trace 2g. But there are lots of homeomorphisms where this trace is 0; for example interchange the two circle factors of a torus.
|
2025-03-21T14:48:31.938830
| 2020-08-31T13:35:41 |
370527
|
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|
Stack Exchange
|
Connected set of vertices having large boundary in a subset?
Let $\Gamma = (V,E)$ be a connected (undirected) graph where every vertex has degree $\geq 2$. Let $E'\supset E$ be a larger set of edges between elements of $V$ such that every vertex of $\Gamma'=(V,E')$ has degree $\geq 4$.
Must there exist a subset $V'\subset V$ such that $\Gamma'|_{V'}$ is connected and $V'$ has large boundary in $\Gamma$? ("Large boundary" here means: $\geq \epsilon |V|$ elements of $V\setminus V'$ are connected by an edge to $V'$) If not, what are some counterexamples?
(For $E=E'$, the problem is not difficult (even if we just assume that many vertices of $\Gamma'$ have degree $\geq 3$, rather than that they all have degree $\geq 4$. See Existence of connected component with large boundary?)
|
2025-03-21T14:48:31.938907
| 2020-08-31T14:02:21 |
370530
|
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|
Stack Exchange
|
Elementary way to compute Hodge numbers of Grassmanian
I know that by using Hodge decomposition and the fact that Schubert cells are Hodge cycles you can compute the Hodge numbers of Grassmanian but is there a more elementary way to compute sheaf cohomology $H^i(\Omega_{Gr(m,n)}^j$)?
Whether it is more elementary depends on one's taste I guess, but the cotangent bundle $\Omega_{\mathrm{Gr}(m,n)}^1$ is completely reducible as an equivariant vector bundle (the Grassmannian is cominuscule), and hence so are its exterior powers. One can apply Borel-Weil-Bott to compute the sheaf cohomology of all of the summands that appear, and get the Hodge numbers in this way.
Are you aware that $h^{p,q}=0$ for $p\neq q$? Therefore $h^{p,p}=b_{2p}$, and the computation of the Betti numbers is well-known.
There's an arithmetic approach which is elementary to implement (but depends on hard theorems). You could calculate $#Gr(m,n)(\mathbf{F}q)$ as a function of $q$. The answer will be a polynomial in $q$. This implies that $h^{i,j} = 0$ for $i \neq j$, and moreover tells you what $h^{n,n} = b{2n}$ will be by the Weil conjectures.
@abx I know the numbets but why $h^{p,q}=0$ is there an elementry proof for this ?i already mentioned the proof by the hodge theory in my post I want a proof withoat using hodge theory
@pbelmans ,anonymous thanks these methods are very interesting but they are even more advanced than the proof by hodge theory
The point is that the Grassmannian admits a "cellular decomposition", i.e. a filtration $X\supset X_1\supset \ldots $ where each $X_i\smallsetminus X_{i+1}$ is a disjoint union of affine spaces. It is not hard then to prove that all cohomology classes are algebraic.
One of the ways is to use the projective bundle theorem that says that if $X = \mathbb{P}_Y(E)$ is a projectivization of a rank $r$ vector bundle $E$ over $Y$ then
$$
H^\bullet(X) = H^\bullet(Y) \oplus H^\bullet(Y)[-2] \oplus \dots \oplus H^\bullet(Y)[2-2r],
$$
where $[s]$ stands for the shift of grading.
Now consider the partial flag variety with the natural projection
$$
p \colon Fl(1,2,\dots,k;n) \to Gr(k,n).
$$
On the one hand, $Fl(1,2,\dots,k;n)$ has a structure of an iterated projective bundle, hence all of its cohomology is of $(p,p)$-type, and its generating function is
$$
h_{Fl}(t) = (1+t+\dots+t^{n-1})(1+t+\dots+t^{n-2})\cdots(1+t+\dots+t^{n-k}).
$$
On the other hand, the map $p$ is also an iterated projective bundle, hence $H^\bullet(Gr(k,n))$ is a direct summand of $H^\bullet(Fl(1,2,\dots,k;n))$, hence it is also of $(p,p)$-type, and moreover its generating function is
$$
h_{Gr}(t) = \frac{h_{Fl}(t)}{(1+t)(1+t+t^2)\cdots(1+t+ \dots+t^{k-1})}.
$$
|
2025-03-21T14:48:31.939099
| 2020-08-31T15:20:28 |
370532
|
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|
Stack Exchange
|
Relationship between continuous vector fields and divergence measure fields in dimension $\ge 2$
Let $\Omega \subset \mathbb R^d$ with $d \geq 2$ (I am mostly interested in the case when $\Omega$ is the unit ball). A vector field in $L^p(\Omega,\mathbb R^d)$ is called a divergence measure field if its distributional divergence is a Radon measure. The space of divergence measure fields over $\Omega$ is denoted by $\mathcal D \mathcal M^p(\Omega, \mathbb R^d)$.
My question seems trivial, but I couldn't find an answer anywhere.
Question: Let $\mathcal C(\Omega,\mathbb R^d)$ be the space of continuous vector fields over $\Omega$. Is it true that $\mathcal C(\Omega,\mathbb R^d) \subset \mathcal D \mathcal M^p(\Omega)$?
Discussion: The result clearly fails in one dimension, since then $\mathcal D \mathcal M^p(\Omega) = BV(\Omega) \cap L^p(\Omega)$ (with a ${}_{loc}$ if $\Omega$ is unbounded) and there exist continuous functions that are not in $BV$. However, in more than two dimensions the inclusion $BV(\Omega,\mathbb R^d) \cap L^p(\Omega,\mathbb R^d) \subset \mathcal D \mathcal M^p(\Omega,\mathbb R^d)$ is proper. For example,
$$
g(x,y) = \Big(\sin\big(\frac{1}{x-y}\big), \sin\big(\frac{1}{x-y}\big)\Big) \in \mathcal D\mathcal M^\infty(\mathbb R^2,\mathbb R^2) \setminus BV_{\mathrm{loc}}(\mathbb R^2,\mathbb R^2).
$$
There are quite a few papers on continuous solutions of
\begin{equation}\label{eq2}
\text{div}\, v = F, \tag{1}
\end{equation}
where $F$ is a distibution. The closest I came to some sort of an answer is Theorem 3.7 (see also Definitions 2.1 & 2.3) in de Pauw and Pfeffer, Distributions for which div v = F has a continuous solution, 2007. It says that a continuous solution of \eqref{eq2} exists if and only if for any compactly supported sequence of test functions $\varphi_i \in \mathcal D(\Omega, \mathcal R^d)$ it holds
\begin{equation}\label{eq3}
\lim_i |\varphi_i|_1 = 0 \quad \text{and} \quad \sup_i |\nabla \varphi_i|_1 < \infty \quad \implies \quad
\lim_i F(\varphi_i) = 0. \tag{2}
\end{equation}
A consequence would be that if there exists a distribution satisfying \eqref{eq3} which is not a Radon measure, then the inclusion I'm after would be false. However, I don't know if \eqref{eq3} implies that $F$ is a Radon measure (perhaps under additional assumptions on $\Omega$, such as compactness).
Any help will be appreciated.
Identifying $\mathbb R^d\simeq\mathbb R\times\mathbb R^{d-1}$, to get a counterexample for $d\ge 2$, why don't you just take $(x,y)\mapsto(f(x),\chi(y),0)$, where the smooth $\chi$ has compact support and $f$ is "your favourite" continuous function not having bounded variation?
@TaQ true. I guess you even don't need to have a $\chi$, you can extend any scalar $f(x)$ to a vector field $(f(x),0)$. I'm trying to think how to exclude this trivial case.
@TaQ to exclude your counterexample, one might restrict fields in $\mathcal C(\Omega,\mathbb R^n)$ to gradient fields, i.e. to consider the quotient space $\mathcal C(\Omega,\mathbb R^n)/{g \colon \text{div }g \equiv 0}$. This doesn't exclude the trivial case $(x,y) \mapsto (f(x),0)$, though
|
2025-03-21T14:48:31.939288
| 2020-08-31T15:24:55 |
370533
|
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|
Stack Exchange
|
Where to read about the toric variety coming from a principal nilpotent element of a (semi)simple algebraic group?
Given a principal (regular) nilpotent element $e$ in the Lie algebra $\mathfrak g$ of a complex semisimple algebraic group $G$, let $\mathfrak s=(e,f,h)$ be an $\mathfrak{sl}_2$-triple for $e$. Then the stabilizer of $h$ in $G$ is a maximal torus $T$ acting on the $2$-eigenspace $\mathfrak g_2$ of $\operatorname{ad}h$ with an open orbit (orbit of $e$).
This must be one of the very basic examples of an affine (in fact, linear) toric variety, and it must be doubtlessly studied in much detail somewhere. What text would you recommend to read about it?
@MikhailBorovoi Thanks for the useful edits, Misha!
|
2025-03-21T14:48:31.939366
| 2020-08-31T15:39:30 |
370534
|
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|
Stack Exchange
|
Weighted blowup of a Cohen-Macaulay ring along a regular sequence is Cohen-Macaulay?
Let $(R, \mathfrak{m})$ be a Cohen-Macaulay ring, let $f_1, \dotsc, f_d \in \mathfrak{m}$ be a regular sequence, and let $n_1, \dotsc, n_d > 0$ be weights (feel free to assume that $n_1 = 1$ if it helps). I would like to consider the graded $R$-subalgebra $R' \subset R[T]$ generated by the $f_iT^{n_i}$ and form the "weighted blowup" $X = Proj(R')$. Is $X$ Cohen-Macaulay? If $n_1 = 1$ (the only case I care about), is at least the locus of $X$ where ``$f_1T$'' does not vanish, i.e., the last (first?) affine coordinate patch, Cohen-Macaulay?
I suspect that the answer may be contained in the commutative algebra literature and that maybe even $R'$ itself is Cohen-Macaulay--this would be even better--but I do not know where. For instance, in the case $n_1 = ... = n_d = 1$, when we are talking about the usual blowup / Rees algebra, the Cohen-Macaulayness of $R'$ was proved by Barshay in 1973.
I would be very grateful if someone could point me to the relevant literature, I know that Cohen-Macaulayness of various graded rings has been studied extensively, but I don't know where to look for this particular question. I tried using Hyry's paper on Cohen-Macaulayness of multigraded rings implying that of the diagonal subring, but its results did not seem to help directly (because of generation in degree $1$ assumptions).
Yes. Here's a proof when all the weights are 1 (so we are looking at a usual blowup. For further details (and some further references) you might be interested in Prop. 5.5 of https://arxiv.org/abs/1703.02269.
Let $Y = \mathrm{Spec}R$, let $I = (f_1, \dots, f_d)$, let $Z = V(I) \subset Y$, let $\pi : X \to Y$ be the projection and let $E = \pi^{-1}(Z)$. By the universal property of blowing up, $E$ is a Cartier divisor, hence if $E$ is Cohen-Macaulay so is $X$. Since $f_1, \dots, f_d$ is a regular sequence, the co-normal bundle $I/I^2$ of $Z$ is locally free, and so $E$ is a $\mathbb{P}^{d-1}$-bundle over $Z$. So if $Z$ is Cohen-Macaulay, so is $E$. Finally, $Z$ is Cohen-Macaulay since $R$ is Cohen-Macaulay and $f_1, \dots, f_d$ is a regular sequence.
Perhaps the same strategy works for weighted blowups — one would have to verify:
$E$ is a Cartier divisor,
$E \to Z$ is a flat family of weighted projective spaces (It would then follow that since $Z$ is Cohen-Macaulay, so is $E$ - the key point here is that weighted projective spaces are Cohen-Macaulay, since finite quotient singularities are Cohen-Macaulay).
Thank you for your response. I know the case when all the weights are $1$; some other references for it are Lemma A.6.1 in Fulton's Intersection theory or, for a stronger statement, 1.3--1.4 in expose VII of SGA6. Could you elaborate on the argument in the general weighted case? For one thing, is $E$ still a Cartier divisor then?
Ah, good point. I'm not sure (edited accordingly). Sorry I can't elaborate further off the top of my head, since I don't have much working experience with weighted-blowups.
|
2025-03-21T14:48:31.939564
| 2020-08-31T15:41:21 |
370535
|
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|
Stack Exchange
|
When is an algebra derived indecomposable?
Call a finite dimensional (acyclic) quiver $K$-algebra A derived indecomposable in case $A$ is not derived equivalent to an algebra of the form $B \otimes_K C$.
For example when the number of simples or the vector space dimension of $A$ is a prime number $A$ is derived indecomposble. But $K D_4$ is derived equivalent to $K A_2 \otimes_K KA_2$ and thus not derived indecomposable.
$KE_6$ is derived equivalent to $K A_2 \otimes_K K A_3$. $KE_8$ is derived equivalent to $K A_2 \otimes K A_4$.
Question 1: Is there an (easy) criterion when a given algebra is derived indecomposable?
Easy could mean that it can be decided by some discrete data that are computable for example using QPA.
Question 2: Is a derived equivalence like $K E_6$ to $K A_2 \otimes_K K A_3$ pure coincidence or is there more behind it? Can we also "build" $E_6$ up to some equivalences from $A_2$ and $A_3$ in other situations like Lie algebras?
Question 3: For which acyclic graphs $Q$ is the path algebra $KQ$ derived indecomposable?
In singularity theory, one can make the simple singularity E6 ($x^3+y^4$) from A2 ($x^3$) and A3 ($y^4$). This sum-with-disjoint-variables is named the Thom–Sebastiani sum.
There is a necessary condition, that the Coxeter polynomial is a tensor product of two polynomials. This can be checked for instance on the set of roots.
|
2025-03-21T14:48:31.939681
| 2020-08-31T16:59:43 |
370537
|
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|
Stack Exchange
|
Examples of (infinite) graphs which cannot be embedded into 3d space?
I was thinking about the concept of embedding graphs into Euclidean spaces. Specifically, i was looking for examples of infinite graphs which cannot be embedded in $\mathbb{R}^3$ but can be embedded into some higher-dimension Euclidean space.
This came about when I was thinking about planarity, where in $\mathbb{R}^2$ whether or not a graph can be embedded is an interesting question. Of course in $\mathbb{R}^3$ all finite graphs can be embedded (as can any graphs with cardinality $\aleph_0$ I believe, by the same construction). Obviously any graph with cardinality greater than $\mathfrak{c}$ can't be embedded in any finite-dimension Euclidean space. So I believe any examples would have to have $\aleph_0 < |V| \le \mathfrak{c}$.
For reference, I'm using this definition of embedding. I'd particularly enjoy graphs that are directly constructed or that I can "understand", but a proof of existence would be better than nothing.
Any graph with at most continuum many vertices can be embedded in $\mathbb R^3$. Consider points $p_t=(t,t^2,t^3)$ for $t\in\mathbb R$. No four of these points are coplanar, so no two line segments spanned by pairs of these points can intersect except at an endpoint. As you note, no graph with more vertices can be embedded.
Perhaps a better notion of embedding is representing every point by a sphere, requiring adjacent nodes to correspond to touching spheres, and other spheres to be disjoint. For finite planar graphs, Koebe's (-Andreev-Thurston) theorem gives such an embedding
@Wojowu that construction is much simpler than the one I came up with for finite/countable graphs.
@YuvalPeres that was interesting to read about, although some parts of the proof were probably over my head. Thanks!
@Wojowu Even simpler, if you don't demand a straight-line embedding: put all the vertices on a straight line, and draw each edge on a different plane through that line. This works even for graphs with multiple edges, provided that the number of edges is at most the continuum. . . . But which (infinite) graphs are homeomorphic to a subset of $\mathbb R^3$ or $\mathbb R^2$? Are those interesting questions, or not? Have they been studied?
|
2025-03-21T14:48:31.939859
| 2020-09-02T16:09:24 |
370690
|
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|
Stack Exchange
|
Comparing sizes of sets of integers
Is there a total preorder $\lesssim$ on the power set of $\mathbb Z$ such that:
$A<B$ if $A\subset B$ (proper subsets are smaller)
$1+A\lesssim 1+B$ iff $A\lesssim B$ (where $1+C = \{1+c:c\in C\})$ (shift invariance)
if $A\cap C=B\cap C=\varnothing$, then $A\lesssim B$ iff $A\cup C\lesssim B\cup C$ (additivity)?
The answer is positive if (3) is dropped or if (2) is dropped (easiest way for me to see it is by using an ultrafilter to create a hyperreal-valued finitely additive strictly positive measure on $\mathbb Z$). The answer is trivially positive with $\mathbb N$ in place of $\mathbb Z$: just use lexicographic ordering on the indicator functions.
If one adds reflection invariance ($-A\lesssim -B$ iff $A\lesssim B$), the answer is easily seen to be negative.
It's easy to show that such a comparison would have various weird properties, such as that it says that there are more positive odd numbers than positive even numbers, and that either: (a) $(-\infty,a]\cap\mathbb Z < [b,\infty)\cap\mathbb Z$ for all $a,b$ (it is biased to the right), or (b) $(-\infty,a]\cap\mathbb Z > [b,\infty)\cap\mathbb Z$ for all $a,b$ (it is biased to the left).
If I don't miss something, writing $f\lesssim g$ for all $f,g$ fulfills the condition. Do you have an additional non-triviality requirement?
Yeah: I meant $A<B$ in (1).
What about: $A\lesssim B$ iff $A=B$ or $\min(A\smallsetminus B)>\min(B\smallsetminus A)$ (with $\min(\emptyset)=+\infty$)?
I assume $\min$ is $\inf$, or else it's not defined for sets not bounded below. But then it's not a total (linear) order. Consider the evens and the odds. It becomes total if you replace $>$ with $\ge$, but I think last time I tried something like this, transitivity failed.
Yes, I think it's a total preorder. Indeed min is defined (since every nonempty subset of N has a min and for $A\neq B$ at least one of $A-B$, $B-A$ is nonempty). Assuming $0\in N$, for odd and even, min(ODD-EVEN)=1 and min(EVEN-ODD)=0 so EVEN$\le$ODD. Assuming $0\notin N$ (I don't know your convention), it yields ODD$\le$EVEN.
But the question is about subsets of $\mathbb Z$ not just of $\mathbb N$.
Sorry, I couldn't guess this. One is called "relative" integers and the other "natural" integers, if you don't want to use standard notation Z/N.
Interesting: I've never heard of "integers" as used for anything but $\mathbb Z$. I always say "naturals" for $\mathbb N$. I never heard "relative integers" before, but there are 6660 google hits for it, so it's a real phrase.
I've certainly heard "integers" for both.
I'm probably missing or misreading something, but if you can do it for $\mathbb N$, why can't you do it for every set which is in bijection with $\mathbb N$ (such as $\mathbb Z$) ?
@Vepir: Condition 2 is not preserved by bijections.
what about the approach with Zorns lemma on the set of preorders satisfying these conditions. If we say one preorder is larger than another, if it extends it, we can easily show that any totally ordered chain of preorders has an upper bound given by $A\le B$, if there is some preorder $\le'$ in the chain for which $A\le'B$ holds. It remains to show that any non-total preorder satisying your condition is not maximal, i.e. If we have two uncomparable sets, we can somehow construct a larger preorder in which these sets are comparable. I dont know whether the last step works.
That's a likely strategy: it's how one proves Szpilrajn's theorem, and how I proved the version without the additivity condition, but I don't know how to do the "last step" with it in place.
Interesting question. I have no idea about the answer, but I think condition (3) could be stated more neatly: $A\lesssim B$ iff $A\setminus B\lesssim B\setminus A$.
Without (2) I think the easiest way is to take an ultrafilter $\mathcal U$ on the set $S$ of all finite subsets of $\mathbb Z$ with the property that ${F\in S:a\in F}\in\mathcal U$ for every $a\in\mathbb Z$, and decree that $A\lesssim B$ iff ${F\in S:|A\cap F|\le|B\cap F|}\in\mathcal U$. Maybe that's the same as what you said, which I didn't actually understand.
Your way of proving this without 2 is a little more direct than what I had in mind. Nice! And you're right about 3. My formulation comes from qualitative probability theory, where 3 is a common axiom mirroring Kolmogorov's finite additivity axiom.
I was getting ready to make a trivial comment. Now that the question is answered my comment seems redundant as well as trivial but I won't let that stop me. By a standard compactness argument the preorder you want on $\mathcal P(\mathbb Z)$ exists iff every finite $\mathcal B\subset\mathcal P(\mathbb Z)$ admits a total preorder which satisfies your conditions as far as they are defined, e.g., (2) holds whenever $A,B,1+A,1+B\in\mathcal B$.
Moreover we may assume that $\mathcal B$ is a finite Boolean subalgebra of $\mathcal P(\mathbb Z)$, and then so is $\mathcal B_0={A\in\mathcal B:1+A\in\mathcal B}$. So it seems to me that the question is about a finite Boolean algebra $\mathcal B$ with given subalgebras $\mathcal B_0,\mathcal B_1$ and an isomorphism $f:\mathcal B_0\to\mathcal B_1$.
Yes, there is such a preorder. I will argue that there is a preorder on the space of bounded functions $\mathbb Z\to\mathbb R$ so that comparing indicator functions in this space does the job. A vector space preorder can be constructed from a suitable "positive cone", the set of non-negative elements, so the main task is to construct this cone.
Let $M$ be the set of non-negative, not identically zero, finitely-supported functions $\mathbb Z\to\mathbb R.$ Let $B$ be the real vector space of bounded functions $\mathbb Z\to\mathbb R.$ Let $a*\phi$ denote convolution of a function $a\in B$ by a function $\phi\in M.$ Define $a\sim b$ for $a,b\in B$ to mean that $a*\phi=b*\psi$ for some $\phi,\psi\in M.$ This is an equivalence relation because $M$ is a (commutative) monoid under convolution. The $\sim$-equivalence class $[0]$ of zero is a linear subspace of $B.$
Define a good cone to be a set $C\subset B$ such that
C1. $y\in C\iff z\in C$ whenever $y\sim z,$ and
C2. $C$ is a convex cone ($x,y\in C\implies \lambda x+\mu y\in C$ for $\lambda,\mu\geq 0$), and
C3. $C\cap (-C)=[0].$
Define $C_0$ to be the set of $x\in B$ such that $x\sim y$ for some non-negative function $y\in B.$ Because non-negative functions are closed under convolution by any $\phi\in M,$ the definition of $C_0$ simplifies slightly to $x*\phi$ being non-negative for some $\phi\in M.$ The set $C_0$ satisfies the good cone conditions: (C1) is obvious, for (C2) if $x*\phi$ and $y*\psi$ are non-negative and $\lambda,\mu\geq 0$ then $(\lambda x+\mu y)*\psi*\psi$ is non-negative, and for (C3) if $x*\phi$ is non-negative and $x*\psi$ is non-positive, then $x*\phi*\psi$ is identically zero so $x\sim 0.$ By Zorn's lemma there is a maximal good cone $C$ containing $C_0.$
Consider $x\in B\setminus C.$ Define $C_x$ to be the set of $y\in B$ such that $y*\phi= x*\psi+c$ for some $\phi\in M$ and $\psi\in M\cup\{0\}$ and $c\in C.$ By maximality of $C,$ the set $C_x$ is not good.
(C1) holds: whenever $y*\eta=z*\zeta$ and $y*\phi= x*\psi+c$ we have $z*\zeta*\phi=x*\psi*\eta+c*\eta,$ which implies $z\in C_x.$
(C2) holds: if $y*\phi= x*\psi+c$ and $y'*\phi'= x'*\psi'+c'$ and $\lambda,\mu\geq 0$ then $(\lambda y+\mu y')*\phi*\phi'=x*(\psi*\phi'+\psi'*\phi)+(c*\phi'+c'*\phi).$
So (C3) fails: some $y\not\sim 0$ satisfies $y*\phi= x*\psi+c$ and $y*\phi'=-x*\psi'-c'.$ But then $$-x*\psi'*\phi-c'*\phi=y*\phi*\phi'=x*\psi*\phi'+c*\phi'$$
which implies $x*(\psi*\phi'+\psi'*\phi)+(c*\phi'+c'*\phi)=0.$ If $\psi$ and $\psi'$ are both zero, then $c*\phi=-c'*\phi$ is in $C\cap (-C)$ contradicting $y\not\sim 0.$ So $\psi$ and $\psi'$ are not both zero, which means $-x\sim c*\phi'+c'*\phi\in C.$
In other words $x\in -C.$
We have shown that $C\cup (-C)=B.$ The cone $C$ defines a total vector order on $B/[0],$ but to answer the question we just need to define $S\lesssim T\iff 1_T-1_S\in C.$ Your condition 1 comes from $C_0\subset C$ and (C3). Your condition 2 comes from (C1) - shifting is convolution by a delta function. Your condition 3 comes from $1_{T\cup U}-1_{S\cup U}=1_T-1_S$ whenever $S\cap U=T\cap U=\emptyset.$
Am I right in thinking that this generalizes to any abelian group $G$ in place of $\mathbb Z$, with (2) replaced by $x+A\lesssim x+B$ iff $A\lesssim B$ for all $x\in G$? Or is there some step that fails?
@AlexanderPruss: yes, it generalizes to give a total preorder on the powerset of any abelian group $G.$
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2025-03-21T14:48:31.940721
| 2020-09-02T16:53:11 |
370694
|
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Stack Exchange
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Is the Frog game solvable in the root of a full binary tree?
This is a cross-post from math.stackexchange.com$^{[1]}$, since the bounty there didn't lead to any new insights.
For reference,
The Frog game is the generalization of the Frog Jumping (see it on Numberphile) that can be played on any graph, but by convention, we restrict the game to tree graphs. The Frog Jumping is equivalent to the case on "paths (tree graphs with exactly 2 leaves)" of the Frog game, which was resolved in the linked video.
The mechanics of the Frog game are in short,
Given a tree graph and one frog on each vertex, the goal of the game is for all frogs to host a party on a single vertex of the graph. All of the $f$ frogs on some vertex can "jump" to some other vertex if and only if both vertices have at least one frog on them and both vertices are exactly $f$ edges apart from each other.
If a sequence of "jumps" exists such that all frogs end up on a single vertex, then the party is successfully hosted on that vertex and we call that vertex a "lazy toad" (or just "lazy") because the frog that started there never jumped during the game.
Trivially, if $|V|=n$ is the number of vertices of some graph $G=(V,E)$, then every game on $G$ must end in $n-1$ moves (or less if the party isn't hosted, e.g. if the sequence of jumps is suboptimal and we can no longer make moves or if none of the vertices are lazy).
Here, I'm trying to resolve the Frog game on "full binary trees".
Let $T_h=(V,E)$ be a full binary tree of height $h$. Denote its layers with $0,1,2,\dots,h$ where the root $r\in V$ is on the layer $0$ ("bottom layer"), where layer $i=1,2,\dots,h$ contains vertices $v_{i,j}$ labeled with $j=1,2,\dots,2^i$.
That is, we have $|V|=2^{h+1}-1$. Let $h\in \mathbb N$ because $h=0$ is a single vertex which is trivial.
Conjecture. For all $h\ge 4$, every vertex of $T_h$ is a "lazy toad" (is "lazy").
I couldn't prove the above, and for simplicity, my question is just about the root vertex:
Let $h\in\mathbb N, h\ne 2$. Can we prove that the root $r$ of every such $T_h$ is a "lazy toad" (is "lazy")?
However, if the approach that can solve the root can be generalized to all other vertices, that's a big bonus.
Solving the root
Reduction. We say that $T_h$ can be reduced to $T_s,s\le h-2$, if there exists a sequence of moves that can bring all frogs from the top $s$ layers $h,h-1,\dots,h-s+1$ to the root vertex.
I've managed to reduce $T_h$'s for all $h\in(2,20)$ by hand pretty easily. At most, I needed to consider only $31$ vertices or less to find a pattern that allows a reduction of entire layers, which is very efficient if you consider that $T_{19}$ has over a million vertices. It looks like that maybe an inductive argument would be able to solve the problem, but I did not manage to find a sufficient set of move sequences.
I'll summarize some immediate results below.
Trivial reduction. If $h=2^t+t-2,t\ge 2$ and $T_{t-1}$ is solvable in root, then $T_h$ can be reduced to $T_{h-t}$.
In the above reduction, we are simply (at the top layers) solving $T_{t-1}$ subtrees in root and then jumping from that root to the root of $T_h$.
Necessary minimal reduction condition. Let $h\gt2,a\ge2$. If reduction from $T_h$ to $T_{h-a}$ is possible, then there exists $b\in[0,h-a+1]$ and a partition of the number $(2^a-1)2^b$ into the layers $H=\{h,h-1,h-2,\dots,h-a+1\}$ where $(h-l)\in H$ is used at most $(2^{a-l-1})2^b$ times.
In the above condition, we say that "a partition is solvable" if we can bring the corresponding amounts of frogs to the corresponding layers. I.e. the above condition simply says that we must be able to partition the total number of frogs from the top layers (that we are reducing) into the corresponding vertices from which they can jump to the root.
It remains to show (or find an explicit set of move sequences that will imply) that at least one partition from the above condition is solvable. Surprisingly, for all $h\in(2,20)$ that I've solved so far, the solution was always available in the smallest or one of the smallest partitions ($a\le 5,b\le 3$) of no more than $31$. The $h=20$ is the first case that requires considering more than $31$ vertices.
I still do not see how to construct explicit move sequences for larger $h$ without tackling each one individually. Alternatively, could there be a way to prove that the root is lazy, without finding explicit move sequences?
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2025-03-21T14:48:31.941162
| 2020-09-02T17:39:25 |
370696
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Stack Exchange
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Are injective analytic maps between non-archimedean spaces open?
Let $\Omega$ be a non-archimedean complete field, $n\in\mathbb N$ and $f:\Omega^n\to\Omega^n$ be an injective analytic map.
Is the application $f$ open?
In the complex case, this is a consequence of a Remmert theorem, but in the non-archimedean case, I do not know.
Thanks in advance for any answer.
Let $p\equiv 2\pmod{3}$ so that $\mathbb{Q}_p$ contains no cube root of unity other than $1$. The map $x \mapsto x^3$ from $\mathbb{Q}_p$ to itself is injective, yet its image contains no neighborhood of $0$ (because only points whose valuation is multiple of $3$).
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2025-03-21T14:48:31.941242
| 2020-09-02T18:12:57 |
370698
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Stack Exchange
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Work on triply periodic minimal surfaces
I have seen in some engineering departments that they manufacture models of periodic minimal forms (characterised by equal and opposite curvature at every points on the surface). In pure mathematics, they are known as triply periodic minimal surfaces.
If I understand rightly, these have been observed experimentally in crystallography and polymer chemistry but I assume they must have been studied in differential geometry as well. The Wikipedia page mentions the classification of these surfaces as an open problem: has there been any recent progress on this? The
physics literature also mentions the possibility of constructing minimal surfaces with the properties of a quasicrystal (ie. minimal surfaces with a quasicrystalline order). Again, has there been any further geometric work on this construction?
This is an active research topic. I'm currently working on the front line towards a classification of TPMSs of genus 3 (TPMSg3s). My collaborators include Weber and Traizet. I also know a Japanese team working on the moduli space.
Recent progress include surprising discoveries of new examples:
https://arxiv.org/abs/1804.01442 (j/w Weber, to appear in Trans. AMS)
https://arxiv.org/abs/1807.10631 (j/w Weber, to appear in Trans. AMS)
and rigorous proof of some deformations of the gyroid.
https://arxiv.org/abs/1901.04006
These examples are important because they reveal concrete singularities in the moduli space of TPMSg3s. So their existence basically proves that the classification is very complicated.
However, an ultimate classification of TPMSg3s is not completely hopeless. We are making big progress on the boundary of the moduli space, and we expect to see more new examples in the near future. I estimate about 5 years of hard works before we can finally evaluate the feasibility of the classification. I'm optimistic.
We also constructed uncountably many examples of non-periodic minimal surfaces that of great interest for material scientists and crystallographers.
https://arxiv.org/abs/1908.06276 (j/w Traizet, to appear in SIAM J. Math. Anal.)
Some of them can be regarded as "quasi-periodic", but not "quasi-crystallographic". I have been thinking about quasi-crystallographic minimal surfaces, but currently have no progress at all.
My involvement in the topic is mostly motivated by physics, and I'm actively collaborating with many experimental teams.
Matthias Weber has been active in this area (classifying TPMSs) recently. For example. I suggest you contact him.
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2025-03-21T14:48:31.941421
| 2020-09-02T18:28:21 |
370699
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Stack Exchange
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Classifying spaces of amalgamated topological monoids
Let $\mathsf{Top}_*$ be the category of well-based spaces and $\mathsf{TopMon}$ the category of topological monoids. Recall the James construction $\mathcal{J}:\mathsf{Top}_*\to \mathsf{TopMon}$ which is left adjoint to the forgetful functor.
Now let $A\hookrightarrow X$ be a cofibration of based spaces and let $M$ be a topological monoid as well as $f:A\to M$ a based map. I want to “attach” the free topological monoid $\mathcal{J}X$ to $M$ along $\mathcal{J}A$. More precisely, I want to consider the adjoint map $\mathcal{J}A\to M$ and form the pushout of $(\mathcal{J}X\leftarrow \mathcal{J}A\to M)$ in the category $\mathsf{TopMon}$.
Call this pushout $M'$. I am interested in the classifying space $BM'$. The naïve hope would be of course something like
$$BM' = \mathrm{hocolim}_{\mathsf{Top}_*}\left(B\mathcal{J} X\leftarrow B\mathcal{J} A\to BM\right),$$
and then one would identify up to homotopy $B\mathcal{J}\simeq \Sigma$, where the occuring map $\Sigma A\to BM$ is just the composition $\Sigma A\to \Sigma M\to BM$ with $\Sigma M\to BM$ being induced by the inclusion $\Delta^1\times M\to |NM|=BM$ of the $1$-simplices (or in other words, $\Sigma M\to BM$ is the adjoint of the group completion $M\to \Omega B M$).
However, I guess that this is too nice to wish for. I found the following paper by Fiedorowicz in which he gives explicit conditions under which the classifying space of a discrete amalgamated monoid is a (homotopy) pushout of classifying spaces.
Is there anything known for the non-discrete case? What would we additionally have to assume for the inclusion $A\hookrightarrow X$ or the attaching map $f:A\to M$ to hold?
The paper of Fiedorowicz has some issues. See https://mathoverflow.net/questions/297583/a-flatness-result-of-fiedorwicz-for-amalgamated-free-products-of-monoids-in-conn
Possibly of interest: https://arxiv.org/abs/1203.4978
Thank you, @GustavoGranja! I think that this should help: Vogt proves that B preserves homotopy colimits up to homotopy equivalence. Now there is a left proper model structure on $\mathsf{TopMon}$ such that $\mathcal{J}$ is left Quillen, so $\mathcal{J}(A\hookrightarrow X)$ is a cofibration, whence $M'$ is indeed a homotopy pushout. Or am I missing something?
Well … one would have to argue why $M'$ is not only a homotopy pushout with respect to the model structure on $\mathsf{TopMon}$, but also a homotopy pushout in Vogt’s sense. Or is this clear?
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2025-03-21T14:48:31.941615
| 2020-09-02T19:31:02 |
370703
|
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"Andrej Bauer",
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|
Stack Exchange
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Can primitive recursive functions be simulated in the smallest reasonable primitive recursive group?
Second Edition, completely rewritten with unchanged questions.
The said questions are motivated by the bizarre wording of the concluding § in A Class of
Reversible Primitive Recursive Functions by L. Paolini, M. Piccolo & L. Roversi (Electronic Notes in Theoretical Computer Science 322 (2016) 227–242, doi:10.1016/j.entcs.2016.03.016, author pdf).
Kudos to the 2 upvoters for making sense of the 1st Ed.: 2 days after posting it, I couldn't & came up with this. Apologies and thanks to the 100+ viewers for your efforts & special thanks to Andrej Bauer for your sharp questioning.
The paper itself is not problematic. It purports to simulate the usual set of primitive recursive functions (PRF) by means of carefully chosen bijections from $\mathbb{Z}_∗$, the free monoid over the signed integers, to itself. Then, they proceed the other way around, to simulate those bijections using PRF's; which, it turns out, amounts to simulate all the members of the group they generate.
I attempt here to make sense of their concluding remark by rephrasing it in group-theoretic terms. Sorry if it is old moons to you, and any comments to that effect are welcome: all this is way beyond the fields I know.
So, I shall explain my terminology at some tedious length, before I can restate the results of the Paolini-Piccolo-Roversi paper and only then, ask my questions. Meaning you might find the post more entertaining by reading it from the bottom up; you're welcome to try.
The locus dramatis is the group $U_p$ of bijective functions from $\mathbb{Z}^p$ to itself, with group law the composition of functions and unit the identity $I_p$; and the disjoint union thereof $U_*=\bigsqcup_{p \in N} U_p$.
In my ignorance of established terminology, I chose the following; again, your comments are welcome if you know any better.
Identifying tuples from $\mathbb{Z}^p$ with words from the free monoid $\mathbb{Z}_*$, we let $|w|$ denote the size of the tuple $w$ & $(w_1, w_2)$ denote $w_1$ concatenated with, i. e. followed by, $w_2$, and identify $\mathbb{Z}$ with $\mathbb{Z}^1$;
a denizen of $U_p$ has arity at most $q$ if, up to conjugation by some reordering of its $p$ arguments, it is equal to the Cartesian product $f_q \times I_{p-q}$ for some $f_q \in U_q$;
lifting concatenation to an operation over $U_*$, we rewrite $f_q \times I_{p-q}$ as $(f_q, I_{p-q})$ and extend every bijection $ f \in U^p$ to a bijection $ (f, I_\infty) : \mathbb{Z}[X] \to \mathbb{Z}[X]$.
This switch, from the free monoid $\mathbb{Z}_*$ to polynomials $\mathbb{Z}[X]$, disposes of technicalities about how to feed a function with a word too short for its taste: just zero-pad it the way you pad a polynomial with null terms to match degree requirements;
this promotes $U_*$ to the status of a group acting on $\mathbb{Z}[X]$: the group of (bijections of) fixed arity with values in $\mathbb{Z}$;
From now on, the subscript $q$ in $f_q \in U_p$ will systematically denote the arity of $f$: the smallest $q$ such that $f$ has arity at most $q$. By exception, all the $I_q$'s remain available for tuple juggling, as aliases of the identity function: $I_0$, the unit element of $U_*$ and the only one of arity $0$.
So much for the preliminaries! Finally,
groups acting on $\mathbb{Z}_*$ have a unique feature: an exponentiation defined over the whole group. As usual, $f^k$ in $U_p$ denotes $f$ composed $k$ times with itself for all $k \in \mathbb{Z}$ : $f^0 = I_0$, $f^{-z} = (f^{-1})^z$. Then, the exponentiated $f_q$ is the bijection $f_q^* : (z, w_q) \to (z, f_q^z(w_q))$ for all $(z, w_q) \in \mathbb{Z}^1 \times \mathbb{Z}^q$; it has arity $q+1$.
With $U_*$ closed under exponentiation, a minimally recursive group is any subgroup of $U_*$ that is closed under exponentiation and under permutations of (finite subsets of) function arguments; and,
for any subset $S$ of $U_*$, the primitive recursive group generated by $S$, denoted $R_*(S)$, is the smallest subgroup of $U_*$ containing $S$ that is minimally recursive. $R_p(S)$ is $R_*(S) \cap U_p$.
Beware, gentle reader!!! My choice of terms is far from ideal. It suggests $R_*(S)$ is the natural representative of the usual PRF's in reversible computing; however, it is only true if the set of generators is neither too small, nor too rich. More on this in a moment.
In the above setting, the primitive recursive group chosen by Paolini–Piccolo–Roversi has 4 generators: the unit translation I denote $++ : \mathbb{Z}^1 \to \mathbb{Z}^1$; a zero-testing bijection: $\mathbb{Z}^2 \to \mathbb{Z}^2$ that maps $(0, 0)$ to $(0, 1)$ and $(z, 0)$ to itself, $z \ne 0$; a pairing bijection $\mathbb{Z}^3 \to \mathbb{Z}^3$ that maps $\mathbb{Z}^2 \times \{0\}$, bijectively, to $\mathbb{Z}^1 \times \{0\}^2$; an unpairing bijection that provides the inverse mapping, $\mathbb{Z}^1 \times \{0\}^2 \to \mathbb{Z}^2 \times \{0\}$.
That this $R_*(S)$ contains simulations of all the PRF's and conversely, can be simulated by PRF's, is not in question; I take it for granted.
The bizarre conclusion is, they regard an "open issue" whether the pairing function is "independent from the remaining functions" (sic); said functions being, I surmise, the other generators of their group.
At first glance, the issue seems to me closed and tightly barred with a negative answer: there are PRF's that map bijectively $\mathbb{N}^2$ to $\mathbb{N}$, and it looks like an easy programming exercice to extend one to a pairing fucntion in $R_*(S)$, as soon as it features a zero-testing bijection.
In other words: if $S$ is restricted to the fist 2 generators, $R_*(S)$ still contains the other 2; it just makes it all the easier to simulate $R_*(S)$ with PRF's.
In view of this, let me suggest a tougher issue: what if we also remove the zero-testing capability? In other, hopefully unambiguous, words:
Q1: if $S$ contains only translations over $\mathbb{Z}^1$, does the resulting group,
$R_*(\{++\})$, contain a zero-testing bijection ? Such
a function is any bijection from $\mathbb{Z}^p$ to itself that maps $(z, 0, 0_{p-2})$ to some $(z, b, w_{p-2}(z))$, with $b=1$ if $z=0$, else $0$.
As soon as one such bijection is found, it is a straightforward matter to build a cleaner one upon it, one of arity $p+1$ that maps $(z, 0, 0_{p-1})$ to $(z, b, 0_{p-1})$.
Given that concatenation & exponentiation behave well with respect to the Euclidean norm over $\mathbb{Z}^p$ and that, for $S$ as above, $R_p(S)$ contains the linear group over $\mathbb{Z}^p$, the next question cries out to be asked:
Q2: if $S$ is the orthogonal group over $\mathbb{Z}^{p-1}$, does $R_p(S)$ contain a zero-testing isometry? This one is a bijection from $\mathbb{Z}^p$ to itself, tweaked to
preserve Euclidean norms: it maps $(z, 1, 0, a_{p-3})$, for some
constant $a_{p-3}$ and any $z \in \mathbb{Z}$, to $(0, 1, 0, w_{p-3}(0))$ if
$z=0$, else to $(z, 0, 1, w_{p-3}(z))$; with $w_{p-3}$ of size $p-3$, having the same
norm as $a_{p-3}$ and otherwise unspecified.
The answer to Q1 is likely negative, which turns the group over 1 generator $R_*(\{++\})$ into a fairly exotic fellow: the smallest group with reasonable claims to be closed under some sort of recursion, yet too small to simulate the PRF"s.
It offers addition and iteration, hence a sizeable bit of linear algebra, even of polynomial algebra with substitution... only equality testing is wanting. Hence the title of this post, and my choice of the term "minimally recursive" .
I added Q2 out of curiosity and a personal distaste for undecidable questions: if my life depended on it, I would bet the answer to Q1 is "no or maybe undecidable" and the answer to Q2 "decidably no".
Can you please write down explicitl what $w_{p-2}$ is in the first question, without resorting to "ad lib"?
Please don't say "unspecified" or "arbitrary". Give one particular choice, and then say later that the choice did not matter. I am immediately suspicious about "arbitrary". How arbnitrary? Can I make it a non-bijection? Perhap I can start choosing things so that I sneak in some non-computability somewhere? I should not be worrying about these things before I understand the definitions.
Ah, I see. Ok, how about something non-problematic? How would one represent addition in this setting? I am not asking these question to answer your question, but to understand your question.
@AndrejBauer Addition is w/o zero testing: by way of what I called "signed recursion". Shall we move to chat? Thanks for your patience anyway.
Thank you for your patience. Yes, let's move to chat.
Let us continue this discussion in chat.
@YCor thanks for tidying up. I doubt very much $\mathbb{U}$ is standard for such a peculiar group, so I'll change back only that one.
@FrançoisJurain sure I just coined it, just using mathbb to distinguish from a certained fixed explicit group vs $G$ which is varying. Of course choice is yours and $U$ is OK.
Why, shiver mey timbers! Q1 has a positive answer. I cannot believe I am the 1st to notice it, although I do not recall even hearing of it, as a classroom exercise or otherwise; if you know it in published form, kindly provide a link & I will be glad to credit the inventor.
So, we are looking for an algorithm that compares the integer $z$ with zero, using only incrementation and exponentiation. First, note those are enough to square integers, which reduces the problem to comparing the unsigned $z^2$ with $0$.
The key observation is, they are also enough to compute the quotient and remainder of the division of $n \ge 0$ by $2$: defining $Add_{1/2} : (n, b, c) \to (n+b, c, b)$, the exponential ${Add_{1/2}}^*$ maps $(2 \cdot n+b, 0, 0, 1)$ to $(2 \cdot n+b, n, b, 1-b)$.
Then, replace $2 \cdot n + b$ with $n$ and reiterate at least $\log_2 (n + 1)$ times, adding the successive remainders. This computes $L_2(n_0)$, the sum of the digits in the base 2 expansion of $n_0$, in ${} \le n_0$ iterations: $L_2(\sum_{k \ge 0} {b_k \cdot 2^k}) := \sum_{k \ge 0} b_k$. The function $L_2(n_0)$ only takes on the value 0 when $n_0 = 0$, otherwise it is ${} \ge 1$: almost the one we were looking for.
(Edited §, found a simplification). So, define the sequence $n_{k+1} := L_2(n_k)$; from any $n_0 \ge 1$ it falls in a few steps to $1$, where it remains stationary. This is because $L_2$ is strictly decreasing when $n_k \ge 2$. Whereas it is 0 whenever $k \ge n_0 = 0$.
Et voilà, your zero-testing function: $n_{n_0 + 1}$. It is a not-so-easy exercise, then, to make the computation reversible; however it is pure programming, not math.
(Edited on second thoughts); not so pure. The usual programming trick to reversibly replace $2 \cdot n + b$ with $n$ by means of a pairing function is not available yet, and we must do with an arithmetic ersatz; it may help answer Q2 to expose it.
So: we have bijections $ Swap: (m, n) \to (n, m)$, $ Add(z): (n) \to (n + z)$ and $Halve(2 \cdot n + b): (z, 0, 1) \to (z+n, b, 1-b)$; we work with a 9-tuple $((2 \cdot n + b)_0, 0_1, 0_2, 1_3, l2_4, c_5, 0_6, 0_7, 1_8)$ where $c$ acts as a bit bucket, or rather a bit dump to throw used binary digits after reclaiming their component in the tuple.
Now, compose the following in sequence:
compute quotient & remainder: $Halve((2 \cdot n + b)_0): (0_1, 0_2, 1_3) \to (n_1, b_2, (1-b)_3)$;
add the remainder: $Add(b_2): (l2_4) \to ((l2+b)_4)$;
reclaim component 0: $Add^{-2}(n_1): ((2 \cdot n + b)_0) \to (b_0)$; $Add^{-1}(b_2): (b_0) \to (0_0)$;
lay $b$ on top of the dumped bits: $Add^2(c_5): (0_6) \to ((2 \cdot c)_6)$; $Halve^{-1}((2 \cdot c)_6): (c_5, 0_7, 1_8) \to (0_5, 0_7, 1_8)$; $Add(b_2): ((2 \cdot c)_6) \to ((2 \cdot c + b)_6)$;
reclaim components 2 and 3: $Swap^{(2 \cdot c + b)_6}: (b_2, (1-b)_3) \to (0_2, 1_3)$.
The final tuple is $(0_0, n_1, 0_2, 1_3, (l2+b)_4, 0_5, (2 \cdot c + b)_6, 0_7, 1_8)$: swap components 0-1 and 5-6, and you are ready to iterate. In the end, component 4 will have grown from $0$ to $L_2(2 \cdot n + b)$ and component 0 dwindled from $(2 \cdot n + b)$ to $0$, swap them and you are ready to go for $n_{k+2}$.
I do not claim authorship for separating fixed from changing function arguments and for the notation $value_{index}$ and in the preceding §§: they are so natural, I probably reinvented the wheel as is my wont. Again, just provide a link to published material & I will gladly credit the inventor.
This leave Q2 open, although without computability-theoretical added value. Still, if you care to provide an answer, I will rather accept it than this one.
@Andrej-Bauer Thanks; LaTeX is not my kink, really.
Very nice! Who would have thought. So this is now enough to get all p.r. functions, yes?
@AndrejBauer I feel confident I can get them; let's move to chat if you want to discuss it. Hardly any chance I'll find a way to remove the last generator as well, though.
|
2025-03-21T14:48:31.942503
| 2020-09-02T19:45:40 |
370704
|
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|
Stack Exchange
|
A group where the Weil topology induced by the Haar measure does not coincide with the original topology
Let $(G,\tau)$ be a locally compact Hausdorff topological group that is $\sigma$-finite with respect to the Haar measure $\mu:\mathcal{B}(G)\to[0,\infty]$ ($\mathcal{B}(G)$ is the Borel $\sigma$-algebra for $G$). Define $\mathcal{B}\boldsymbol{a}(G)\subseteq \mathcal{B}(G)$ to be the Baire $\sigma$-ring in $G$ (the $\sigma$-ring generated by the compact $G_\delta$'s), and furthermore assume that $G\in\mathcal{B}\boldsymbol{a}(G)$ (i.e. $\mathcal{B}\boldsymbol{a}(G)$ is a $\sigma$-algebra). Let
$$\mathcal{A}=\{EE^{-1} \mid E\in \mathcal{B}\boldsymbol{a}(G), 0<\mu(E)<\infty\}.$$ Now forget about the topology $\tau$. It is well known that $\mathcal{A}$ forms a system of neighborhoods for $e$, which induces a topology $\tau_\mu$ in $G$ which makes it a Hausdorff topological group. This topology is called Weil's topology (see [1]). Under this topology $G$ is densely embeddable in a Hausdorff locally compact group $\overline{G}$, and the Haar integral in $\overline{G}$ coincides with the integral with respect to $\mu$ for all continuous functions of compact support contained in $G$.
It can be easily shown that $\tau \subseteq \tau_\mu$, and it was shown in [2] that adding the assumption that $\mathcal{B}\boldsymbol{a}({G})$ is analytic, $\tau_\mu\subseteq \tau$.
I am trying to come up with a simple example where $\tau_\mu\not\subseteq \tau$ (evidently in the case where $\mathcal{B}\boldsymbol{a}({G})$ is not analytic), but I have not been successful. Any ideas?
Refs:
[1] Halmos, Paul R., Measure theory. 2nd printing, Graduate Texts in Mathematics. 18. New York - Heidelberg- - Berlin: Springer-Verlag. XI, 304 p. DM 26.90 (1974). ZBL0283.28001.
[2] Mackey, George W., Borel structure in groups and their duals, Trans. Am. Math. Soc. 85, 134-165 (1957). ZBL0082.11201.
It seems that what needs to be done is to take a locally compact group $G$, and find a subgroup $H$ of full outer measure, and a distinct topology on $H$ that both a group topology and has the same Baire sets as the subspace topology on $H$. Unfortunately I don't know how to do this.
This seems like a good path, although the distinct topology on $H$ needs to be also locally compact. Halmos gives an example of a thick subgroup in $G=\mathbb{R}^2$ in [1] p. 276 exercise (6), using some basic linear algebra techniques. But I would not know either how to give this subgroup a distinct locally compact topology so the Baire sets coincide.
If $G$ is a locally compact group, then the topology on $G$ agrees with the Weil topology. One way follows from local compactness and continuity of $(g,h) \mapsto gh^{-1}$, and the other is Weil's extension of Steinhaus's theorem to locally compact groups (that if $E$ has strictly positive Haar measure, $EE^{-1}$ is a neighbourhood of the identity). I assumed that you knew this, and thought you were looking for a (necessarily non-locally compact) topological group that had a Haar measure such that the topology differed from the Weil topology.
You assumed wrong! Thank you so much, I completely missed this.
Do you want me to post my comment as an answer? Then you can accept it and the question will be considered "answered". Otherwise the question recirculates to the front page in search of answers for ever and ever.
Sure, go ahead!
There are no such locally compact groups, because if $G$ is a locally compact group under the topology $\tau$, then the Weil topology $\tau_\mu$ defined by the Haar measure $\mu$ is the same as the original topology $\tau$.
To show $\tau_\mu$ is finer than $\tau$, let $N$ be a $\tau$-neighbourhood of $e$. Since the mapping $g \mapsto gg^{-1}$ is continuous $G \rightarrow G$, there is a neighbourhood $M$ of $e$ such that $MM^{-1} \subseteq N$. Since $G$ is locally compact, we can find a compact $G_\delta$ neighbourhood $K$ of $e$ such that $K \subseteq M$ and therefore $KK^{-1} \subseteq N$. Since $K$ is compact, $\mu(K) < \infty$, and since it contains an open set, $\mu(K) > 0$ and therefore $KK^{-1} \in \mathcal{A}$ and so $N$ is a $\tau_\mu$-neighbourhood of $e$.
The other direction holds by Weil's extension of Steinhaus's theorem, which states that if $\mu(E) > 0$ then $EE^{-1}$ is a $\tau$-neighbourhood of $e$. Weil proved this by what is now the standard argument that convolving $\chi_E$ by $\chi_{E^{-1}}$ produces a continuous function vanishing outside $EE^{-1}$ but taking the nonzero value $\mu(E)\mu(E^{-1})$ at $e$.
For the more general question of topological groups with Haar measures, I do not know an example of a topological group $G$ with a left-invariant Radon measure $\mu$ such that the original topology $\tau$ differs from $\tau_\mu$. However, if we drop the requirement that $\mu$ be Radon there is a simple example. Take $G = \mathbb{Q}$, and let $\tau$ be its subspace topology in $\mathbb{R}$. The counting measure $\mu$ is an invariant measure on this group. However, the Weil topology $\tau_\mu$ defined by the counting measure on $\mathbb{Q}$ is easily seen to be the discrete topology, which is strictly finer than $\tau$.
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2025-03-21T14:48:31.942848
| 2020-09-02T22:42:46 |
370709
|
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|
Stack Exchange
|
Fundamental regions in convex programming
In linear programming, the fundamental regions are polyhedra, since those are the intersection of half-spaces defined by linear inequalities. In semidefinite programming, the fundamental regions are spectrahedra$^\color{magenta}{\dagger}$:
A spectrahedron is the intersection of the cone of positive semidefinite matrices with an affine-linear space.
What are the fundamental regions in convex programming in general?
$\color{magenta}{\dagger}$ Bernd Sturmfels, GAeL Lecture I on Convex Algebraic Geometry — Spectrahedra, Coimbra, June 2010.
Would "feasible" be better than "fundamental"?
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2025-03-21T14:48:31.942928
| 2020-09-02T22:44:11 |
370710
|
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|
Stack Exchange
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$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
Let $\mathbb{Z}/p^{\infty}$ denote the Prufer group. By $p$-completion properties, it follows that $(B\mathbb Z/p^{\infty})^{\wedge}_p\simeq K(\mathbb{Z}^{\wedge}_p,2)\simeq(BS^1)^{\wedge}_p$. But, why does the inclusion $\mathbb{Z}/p^{\infty}\hookrightarrow S^1$ induce such a homotopy equivalence?.
Think of the Prüfer group as $\mathbb{Q}/\mathbb{Z}$ and of $S^1$ as $\mathbb{R}/\mathbb{Z}$. Here $\mathbb{Q}$ is discrete, $\mathbb{R}$ has its usual topology and is contracible. The map $\mathbb{Q} \to\mathbb{R} $ becomes an equivalence after $p$-completion.
Thank you!, do you mean $B\mathbb{Q}^{\wedge}_p\rightarrow B\mathbb{R}^{\wedge}_p$ is a homotopy equivalence?
Yes, I think if you want to turn this idea into an actual proof you look at the induced map of fiber sequences $B\mathbb{Q}\to B\mathbb{Q/Z} \to K(Z,2)$ and $B\mathbb{R}\to B\mathbb{S^1}\to K(Z,2)$, and observe that the map on the fibers becomes an equivalence after $p$-completion.
but $B\mathbb{R}$ is contractible, and $B\mathbb{Q}$ is $p$-good. So far, their $p$-completions cannot be homotopy equivalent.
I'm not sure I correctly remember what $p$-good means, but note that the mod $p$ homology of $B\mathbb{Q}$ is of course zero.
Fine!, I was wrong, sorry. Thank you again.
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2025-03-21T14:48:31.943160
| 2020-09-02T23:18:32 |
370714
|
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|
Stack Exchange
|
A possible error in Elliot's book "Probabilistic Number Theory"
In Elliot's book "Probabilistic Number Theory", there seems to be an inaccuracy. The author defines, for any sequence $a_n$, the quantity
$$V(p)=\sum_{r=0}^{p-1}\left|\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(p)}}^N a_n-p^{-1}\sum_{n=1}^N a_n
\right|^2$$
He then asserts that, if $a_n$ assumes only the values 0,1, then
$$\sum_{p\leq Q}pV(p)\leq c_1 Q^2 \log(Q)\sum_{n=1}^N|a_n|^2$$
where $c_1$ is some absolute constant. The issue is, this would imply that
$$\limsup_{N\to\infty}\sum_{p\leq Q}p\frac{V(p)}{N^2}\leq c_1 Q^2 \log(Q)\limsup_{N\to\infty}\frac{1}{N^2}\sum_{n=1}^N|a_n|^2=0$$
which isn't always true. An easy counter-example is $a_n$ defined as $0$ when $n$ is even and $1$ when $n$ is odd, and $Q=2$. Namely, we have that
\begin{align*}
\sum_{p\leq Q}p\frac{V(p)}{N^2} &= 2\frac{V(2)}{N^2}\\
&=\frac{2}{N^2}\sum_{r=0}^{1}\left|\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(2)}}^N a_n-\frac{1}{2}\sum_{n=1}^N a_n
\right|^2\\
&=2\sum_{r=0}^{1}\left|\frac{1}{N}\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(2)}}^N a_n-\frac{1}{2N}\sum_{n=1}^N a_n\right|^2\\
&=2\left|\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 0\mathrm{mod}(2)}}^N a_n-\frac{1}{2N}\sum_{n=1}^N a_n\right|^2+2\left|\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 1\mathrm{mod}(2)}}^N a_n-\frac{1}{2N}\sum_{n=1}^N a_n
\right|^2\\
\end{align*}
Since
$$\lim_{N\to\infty}\frac{1}{2N}\sum_{n=1}^Na_n=\frac{1}{4}$$
$$\lim_{N\to\infty}\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 1\mathrm{mod}(2)}}^N a_n=\frac{1}{2}$$
$$\lim_{N\to\infty}\frac{1}{N}\sum_{\substack{n=1 \\n \equiv 0\mathrm{mod}(2)}}^N a_n=0$$
We see that
$$\lim_{N\to\infty}\sum_{p\leq Q}\frac{V(p)}{N^2}=2\left(\frac{1}{4}\right)^2+2\left(\frac{1}{4}\right)^2=\frac{1}{8}$$
Numerical computations show that $\frac{1}{8} \neq 0$ and thus this is a contradiction. The paper cited for this result is locked behind a paywall so I cannot access it and see what the true theorem is. Does anyone know what the actual result should have been? Where is the typo?
The paper cited with the result is
Roth, Klaus F., On the large sieves of Linnik and Renyi, Mathematika, Lond. 12, 1-9 (1965). ZBL0137.25904.
SIDE QUESTION:
In the book, there are many inequalities given for the sum $\sum_{p<Q}pV(p)$, but if you think about $V(p)$ as being on the order of $\frac{N^2}{p^2}$ for large $N,p$ (which is the worst-case scenario), then the sum $\sum_{p<Q}p^2V(p)$ feels much more natural to study, and $\sum_{p<Q}pV(p)$ feels like a logarithmically weighted version. Does anyone know of any inequalities for $\sum_{p<Q}p^2V(p)$?
Can you give the page number in Elliot's book, and a reference to the cited paper?
I got a good chuckle out of "Numerical computations show that $\frac{1}{8} \ne 0$"...
The pages are 183-184, and the cited work is "On the Large Sieves of Linnik and Renyi"
There's an "Insert Citation" button when you edit that will let you insert the full citation (title, authors, journal reference, DOI), pulled from Zbmath. Anyway it's better to have this info in the question instead of a comment which is not meant to be permanent.
In Roth's paper one has the condition $Q \ge \frac {\sqrt N}{\sqrt {\log N}}$, otherwise the inequality is more complicated
@Conrad what is the more complicated inequality?
Did you see my response? I gave you the correct general form of the inequality with a reference. For $Q\geq\sqrt{N}/\sqrt{\log N}$, it implies Elliott's version, since in that case $Q^2\log Q\gg N+Q^2$.
@GHfromMO Yes, I did. Thank you. I was wondering if Conrad was referring to a different inequality to the general one you had stated.
The large sieve inequality has several variants. Read Montgomery's survey, for instance.
@GHfromMO I already did, a few minutes ago.
Sorry, and thanks!
The result in Roth is as folows (actually he allows the sum to be on a subset of the primes less than $Q$, using the appropriate cardinality instead of $P(Q)$) but ignoring that and taking the full sum one gets: $\sum_{p\leq Q}pV(p)\leq c(ZN+ZQ^2\log(R)+Z^2P(Q)R^{-2}), c$ absolute constant, where $Z$ is the cardinality of the set $n \le N, a_n=1$ (so $\sum a_n=\sum a_n^2$ etc in this special case), $R \ge 2$ parameter, $P(Q)$ the number of primes $\le Q$; choosing $R=Q \ge \frac{\sqrt N}{\sqrt {\log N}}$ gives $\sum_{p\leq Q}pV(p)\leq c_1ZQ^2\log Q$ which is what Elliott quotes
You are right that the second display is false in general (Elliott might impose some conditions). The following version is well-known, and a consequence of Selberg's optimized large sieve inequality:
$$\sum_{p\leq Q}pV(p)\leq (N+Q^2-1)\sum_{n=1}^N|a_n|^2.$$
This holds for any complex numbers $a_n$. See (23) in Montgomery: The analytic principle of the large sieve. Well, Montgomery has $N+Q^2$ instead of $N+Q^2-1$, but the latter is also valid in the light of Theorem 3 in Montgomery's survey.
Is there a similar inequality to this one but with $p^2$, i.e $\sum_{p\leq Q}p^2V(p)$?
@MiloMoses: I don't know. On the other hand, $\sum_{p\leq Q}p^2V(p)$ is at most $Q$ times $\sum_{p\leq Q}pV(p)$, and this is practically optimal. Indeed, in practice one would localize $p$ between $Q/2$ and $Q$, and then the new sum is essentially $Q$ times the old sum.
I don't think that this is entirely correct. If $\lim_{N\to\infty}\frac{p^2V(p)}{N^2}=V'(p)$ exists for every prime $p$, which is true for many bounded sequences $a_n$, then one can relatively easily get that $\frac{1}{\pi(Q)} \sum_{p\leq Q}p^2V(p)\to0$, whereas the "practically optimal" bound would only imply boundedness. My main question is whether or not this implied rate of convergence to $0$ is global or if there are some sequences for which the convergence to $0$ is arbitrarily slow.
You seem to think of $V(p)$ as a function of $N$, with a fixed sequence $(a_n)$. That is fine, but then I don't know what you mean by $\frac{1}{\pi(Q)} \sum_{p\leq Q}p^2V(p)\to0$. Under what condition? There are two variables going to infinity in your setup: $N$ and $Q$.
Sorry, that was a typo. I meant to say that $\frac{1}{\pi(Q)}\sum_{p\leq Q}p^2V'(p)\to 0$, since $V'$ is purely a function of $p$.
OK, but there is a difference between $\sum_{p\leq Q}p^2V'(p)$ that you talk about now, and $\sum_{p\leq Q}p^2V(p)$ that you asked about originally.
From what I understand $V(p)$ is well studied, unlike $V'(p)$, and so I intended on taking any inequality given on $\sum_{p<Q}p^2V(p)$ and applying it to get bounds on $\sum_{p<Q}p^2V'(p)$.
Does this function $V(p)$ have a standard name? No name is given in Elliot's book. I would like to ask this question about $V'(p)$ here on MO but I do not know what to call the question.
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2025-03-21T14:48:31.943589
| 2020-09-02T23:44:26 |
370720
|
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"Benjamin Steinberg",
"https://mathoverflow.net/users/15934"
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}
|
Stack Exchange
|
Reference request concerning splitting fields for groups that are related to special symmetric groups
Denote the symmetric group of order $n!$ by $S_n$. Let $H:=S_p$ for an odd prime $p$.
Every finite field $k$ is a splitting field $(^*)$ for $kH$, in particular $k:=\mathbb{F}_p$.
Questions:
Is $k:=\mathbb{F}_p$ also a splitting field for $kG$ where
a) $G:=H\times H$ ?
b) $G:=H \wr C_2$ ?
I would be interested in references appearing in the literature which deal with these (similar) questions.
Thank you in advance for the help.
Remark 1: $(^*)$ $k$ is a splitting field of $S_n$ if the $k$-algebra $kS_n$ splits over $k$, i.e. if for every simple $kS_n$-module $M$, we have End$_{kSn}(M)\cong k$. (cf. https://math.stackexchange.com/questions/422979/splitting-fields-of-symmetric-groups)
Remark 2: I looked at https://ncatlab.org/nlab/show/direct+product+group and due to remark 2.2 there my questions might not have an affirmative answer for arbitrary groups, but I was interested, if the statement is nevertheless true in these special cases.
Remark 3: I posted the same question at math.stackexchange.com (see https://math.stackexchange.com/questions/3800244/reference-request-concerning-splitting-fields-for-groups-that-are-related-to-spe) more then one week ago, but I did not receive an answer. I hope that it is ok to ask it here now.
Thanks in advance for the help.
If k splits groups G,H then it splits $G\times H$. More generally the tensor product of split finite dimensional algebras is split.
The result mentioned in the answer to this question https://mathoverflow.net/questions/369932/looking-for-citable-reference-for-a-well-known-fact-about-tensor-product-of-fini certainly covers the case of HxH. The proof given there will handle the general case. It all boils down to the tensor product over k of matrix algebras over k if a matrix algebra over k.
|
2025-03-21T14:48:31.943742
| 2020-09-03T00:49:55 |
370722
|
{
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/370722"
}
|
Stack Exchange
|
Understanding the wrapping criterion in percolation theory
Context:
When studying percolation in finite sized systems, there exist various definitions and criteria for determining when a given system is percolating, i.e., given a definition for connectivity, it contains a system-spanning cluster which mimics that of an infinite cluster in the limit of infinite system sizes. Examples of percolative systems might be physical, such as molecular systems, or more mathematical such as bond/site percolation in 2D lattice domains.
Two commonly used definitions for qualifying a cluster as percolating are
The side-to-side spanning clusters, where a cluster is found to connect two opposing sides/walls of the system together. This is generally used when the system has free boundaries (no periodic conditions).
The wrapping criterion is another one, where a cluster wraps around the system (box, domain, etc). This definition is used when the system domain is endowed with periodic boundary conditions. Wrapping is usually described as follows: all constituent bonds/sites in the wrapping cluster are connected by a contiguous path to their own periodic image.
More formally, below are two excerpts from Newman and Ziff 2001 (also relevant is Fig 7.):
Cluster spanning: In many calculations one would like to detect the onset of percolation in the system as sites or bonds are occupied.
One way of doing this is to look for a cluster of occupied sites or
bonds which spans the lattice from one side to the other...
Cluster wrapping: An alternative criterion for percolation is to use periodic boundary conditions and look for a cluster which wraps
all the way around the lattice...
However, at least to me, the latter is still a very counter-intuitive image of what such cluster might look like, and how it differs from the more conventional spanning definition. And naively, on which level is having a cluster comprised of constituents that are connected to their respective periodic images equivalent to the wall-to-wall spanning definition of percolation.
Questions:
Is there an intuitive way of seeing what the wrapping criterion entails? I am struggling in two particular senses: first, simply understanding what it means for a particle/bond/site to be connected to its periodic image, does that mean that if we draw neighbouring periodic images of our system, we see the cluster continuing in these images? (I have not found images that somehow visualise this idea).
And secondly, how does the wrapping definition relate to the more conventional spanning criterion used in percolation? For instance, is wrapping always a stronger condition? (namely, a wrapping cluster is also spanning in the usual sense if we were to remove periodic boundaries?...).
Any pictures/examples, or references where such questions might be tackled would be much appreciated. Unfortunately, I still cannot wrap my mind around the wrapping percolation criterion (no pun intended), so any conceptual or intuitive insights would definitely be very helpful.
Q1: Here is an image that shows a wrapping cluster [source]. So yes, the wrapping condition means that the cluster would extend out to infinity if the lattice is repeated periodically in all directions. Just imagine tiling the plane with the image, and you would find a band of colored sites extending from the lower left to the upper right.
Q2: This example also shows that wrapping is a weaker condition than spanning: the image does not have a cluster that connects opposite edges, so there is no spanning cluster while there is a wrapping cluster.
Thank you for the prompt reply! About Q2, I am still somewhat confused as to how the two are related. In your example it is indeed the case that it is wrapping but not spanning, and vice versa with other examples. See e.g. the last paragraph of the first column of the intro in this paper, starting from "In all these studies of percolation,..." which suggests the opposite, that wrapping always implies spanning.
I was referring to a situation where we apply the open boundary conditions to the unit cell of the periodic lattice; if you allow me to introduce open boundaries at an angle of $45^\circ$ with the boundaries of the unit cell, then the wrapping cluster spans opposite boundaries.
I had to admit Carlo, I have lost you... I will think more about this to understand. Thanks again.
|
2025-03-21T14:48:31.944046
| 2020-09-03T04:45:11 |
370724
|
{
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"Evgeny Shinder",
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|
Stack Exchange
|
$K_0$ of configuration of hyperplanes
Let $\ell_n$ where $n\geq 3$ be the configuration of $n$ lines in a plane, such that $n-1$ of them pass through a single point and the last one does not and it intersects rest of the $n-1$ lines. I'm interested in computing the $K_0$ of algebraic vector bundles on these lines. More generally for every $m$ I'd like to calculate $K_0$ of $\mathbb{A}^m\times \ell_n$. I expect this to be $\mathbb{A}^1$-invariant for $m>n-2$. I'd appreciate any ideas about how to approach such a computation.
There is paper by "BARRY H. DAYTON AND CHARLES A. WEIBEL1". They compute $K$-theory of hyperplanes in general position which satisfy some conditions. None of my cases above satisfy it except $\ell_3$ which gives $K_0(\ell_3)\cong K_0(k)\oplus K_1(k)$ where $k$ is the base field.
For singular curves, the nontrivial part of $K_0(X)$ should come from the Picard group $Pic(X)$ which can be computed in terms of the local structure of singular points. One reference about Picard group of singular curves is https://link.springer.com/content/pdf/10.1007/BF01440949.pdf.
|
2025-03-21T14:48:31.944143
| 2020-09-03T05:57:55 |
370726
|
{
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"Iosif Pinelis",
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"url": "https://mathoverflow.net/questions/370726"
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|
Stack Exchange
|
Emergence of non-power-law behaviour under infinite summing
Suppose $X_1,X_2,...$ is a sequence of random vectors in $\mathbb{R}^n$ s.t for all $k \in \mathbb{Z}^+$ and $u \in \mathbb{R}^n$ we have that $E [ \langle u, X_i \rangle ^k]$ is finite. (The $X_i$s could thus be sub-exponentially distributed too and they are not necessarily independent)
Now is it possible that the above kind of moment finiteness is not true for some $u$ and $k$ for the sum $\sum_i X_i$ ?
It would be helpful to know what theorems are known about when can this happen or not happen.
First of all, it is easy to see that without loss of generality the dimension $n$ is $1$.
Next, of course $S:=\sum_i X_i$ can equal $\infty$ everywhere (if e.g. $X_i=1$ for all $i$), and then we will have $ES^k=\infty$ for all natural $k$.
If the $X_i$'s are independent, then $ES^k$ will be finite for all natural $k$ if and only if $\sum_i EX_i$ is finite and $\sum_i E|X_i|^k<\infty$ for all natural $k\ge2$.
This follows from Rosenthal's inequalities
$$c_1(p)(A_p+B^p)\le E\Big|\sum_i Y_i\Big|^p\le c_2(p)(A_p+B^p),$$
where $p\in[2,\infty)$, $c_1(p)$ and $c_2(p)$ are positive real constants depending only on $p$, the $Y_i$'s are independent zero-mean random variables, $A_p:=\sum_i E|Y_i|^p$, and $B:=A_2^{1/2}$.
In general, if the $X_i$'s are not independent, anything can happen. To get anything specific in the "dependent" case, you would need to specify dependence conditions on the $X_i$'s. One thing that can be said in the "dependent" case is as follows.
By Jensen's inequality, for any real $p\ge2$ and any natural $n$
$$(E|S_n|^p)^{2/p}\ge ES_n^2=(ES_n)^2+Var\, S_n,$$
where $S_n:=\sum_1^n X_i$.
So, we will have $E|S_n|^p\to\infty$ whenever $|\sum_1^n EX_i|=|ES_n|\to\infty$ (as $n\to\infty$). Also, if the $X_i$'s are nonnegatively correlated and $\sum_1^n Var\, X_i\to\infty$, then $Var S_n\ge\sum_1^n Var\, X_i\to\infty$ and hence again $E|S_n|^p\to\infty$.
Thanks for the pointers! So is there a Rosenthal inequality when there is dependency between the $X_i$s?
@gradstudent : I have added a bit on the "dependent" case. I am not aware of lower Rosenthal-type bounds for the "dependent' case.
Thanks! Theorem 1.8 (more generally Theorem 4.3) of this paper https://arxiv.org/pdf/0907.2261.pdf seems to be a general result in this direction : as to how for rercursively defined sequences of random variables a heavy tail can emerge?
|
2025-03-21T14:48:31.944324
| 2020-09-03T05:59:45 |
370727
|
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|
Stack Exchange
|
On cycloids and other roulettes
It is well known that the cycloid is the curve traced by a point on a circle as the circle rolls along a line without slipping.
Consider wheels with smooth convex shapes (not necessarily circular) all of the same perimeter being rolled along a line. Is the area of an arch of the resulting roulette curve an extremum when the wheel is a circle - and the roulette is a cycloid?
Note 1: If the wheel is not a circle, the trajectories of different points on the boundary would be different although each trajectory consists of arches of same base length (equal to perimeter of the wheel) - eg: if an ellipse is rolled, the ends of the major axis would trace arches of height 2a and no other point rises so high. So, above question should be be about the average among the area under the arches traced by all the points.
Is there any smooth convex shape for a wheel such that when rolled along a straight line, a point on its rim traces out the arc of a circle?
Note 2: question 2 could be whether any one point on the rim of some smooth wheel traces arches that are perfect circular arcs OR whether there is some smooth wheel such that every point on its rim traces arches that are circular arcs.
Note 3: Question 2 could be extended by seeking wheel shapes for which each arch of the roulette is segment of a catenary, parabola, etc. instead of circular arc..
As far as the second question, one can't have the entire roulette be an arc of a circle because, where the point touches down on the line, the roulette is parametrized in the form $$\bigl(x(s),y(s)\bigr) = \left(\frac{\kappa^2}6,s^3+\cdots,\ \frac{\kappa}2, s^2+ \cdots\ \right), $$ (where $\kappa$ is the curvature of the convex curve at the tracing point), and this implies that the curvature of the roulette goes to infinity at $s=0$. One can arrange (by solving a differential equation) that part of the roulette be an arc of a circle, but just how large an arc is not entirely clear to me.
Will at least a semicircle be possible as a roulette? For a sequence of semicircular arches, at the cusps, the point of the wheel in contact with the line seems to be totally at rest (unlike for other arcs where the velocity of the contact point along the line of motion seems nonzero at cusps).
A whole semicircle is not possible with a smooth (strictly) convex curve. Whenever the roulette touches the line, it will look like the formula above for some $\kappa$. Notice that $\bigl(x'(0),y'(0)\bigr) = (0,0)$, so the point does actually 'stop' when it gets down to the line, i.e., when $y(s)=0$, but $x(s)$ does not 'turn around' there; it just pauses and then goes on. I expect that you can get quite close to a semi-circle, maybe except on an arbitrarily small neighborhood of the 'touchdown' points.
@R Nandakumar: For the second question the kinematical relation $s^2+R^2=a^2$ (s=arc, R= radius of curvature) yields spiral roulettes (on numerical integration) for circular arc traces of curvature center. The center of a rolling circular arc traces a straight line anyhow.
@Narasimham, I hadn't thought about points not on the rim of the wheel and the trochoids. As you point out, the center of a circular wheel will trace a straight line. Based on Prof. Bryant's answer, I suspect even if we examine curtate trochoids (https://en.wikipedia.org/wiki/Trochoid) resulting from noncircular wheels, we wouldn't get a sequence of circular arcs.
Choosing a different base line (instead of straight) as a circle, there seems to be a possibility. When a moving involute gear wheel M rolls externally on another fixed involute gear wheel tooth F without sliding, any point on M would trace out a circular arc as roulette locus.This is a start point but can however be tweaked a bit towards a rack.
|
2025-03-21T14:48:31.944603
| 2020-09-03T07:24:17 |
370729
|
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|
Stack Exchange
|
Non-ccc and non-locally ccc
What is an example of a category $\mathcal C$ which is not cartesian closed or locally cartesian closed, but for which there exist interesting examples of objects $X,Y \in \mathcal C$ such that an exponential object $Y^X$ exists?
The category of (small) semicategories also called non-unital categories (https://ncatlab.org/nlab/show/semicategory) should answer your question (I am not sure to understand the end of your sentence).
Note that if a category with a terminal object is not cartesian-closed, it is also not locally cartesian-closed.
@varkor Good point. the category of general topological spaces is a much simpler example. A Hausdorff topological space is exponentiable if and only if it is locally compact.
The category of smooth manifolds. The category of classes and class functions.
@DavidRoberts And the category of diffeological spaces is cartesian closed. I am wondering whether diffeological spaces are a closure of some sort of the category of smooth manifolds ? I can't find the answer in the nLab. It is only explained why the inclusion functor is full and faithful.
I've just edited the question with a guess at what the mysterious clause "except x, of all x" might mean. It seems to align with the sorts of examples people were giving in the comments anyway.
@TimCampion Ah thanks: now I understand the question.
Another example: in any category with finite limits, if $1$ denotes the terminal object then $X^1$ always exists and is $X$. in an extensive category, if n denotes the coproduct of n-copies of $1$, then $X^n$ exists and is the usual cartesian power of $X$
The category of topological spaces and continuous maps is like that, see for instance Topologies on spaces of continuous functions by Martín Escardó and Reinhold Heckmann.
|
2025-03-21T14:48:31.944756
| 2020-09-03T07:40:04 |
370731
|
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|
Stack Exchange
|
Colimits in cohomology of profinite arithmetic groups
Let $G\subset \operatorname{GL}_n$ be a linear algebraic group over $\mathbb{Q}$ and let $\Gamma\subset G\cap \operatorname{GL}_n(\mathbb{Z})$ be an arithmeric subgroup without torsion. Using a result of Borel-Serre, one shows that $\mathbb{Z}$ has a bounded resolution with finite free terms as $\mathbb{Z}[\Gamma]$-module (see for example section 5.1). This implies that $\Gamma$ has finite cohomological dimension. Moreover, this also shows that taking group cohomology commutes with filtered colimits.
The questions are the following:
Does the pro-completion $\hat{\Gamma}$ of $\Gamma$ have either finite
cohomological dimension or the compatibility with filtered colimits?
Assume that $\Gamma$ is a congruence subgroup. Does the ''congruence'' pro-completion (i.e. the completion with respect to congruence subgroups of finite index) have finite cohomological dimension or compatibility with filtered colimits?
In particular, I would like to know if filtered colimits commute with continuous cohomology of the Iwahori subgroup of $\operatorname{GL}_2(\mathbb{Z}_p)$.
|
2025-03-21T14:48:31.944854
| 2020-09-03T08:33:17 |
370733
|
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"url": "https://mathoverflow.net/questions/370733"
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|
Stack Exchange
|
Lower bound on the cardinality of set $A\in \{1,2,3,\dotsc,n\}$ with $\operatorname{lcm}(A)>\phi^n$ and asymptotic of number of such subsets
There have been a lot of information published about bounds on lcm of polynomials and other types of sequences evaluated at consecutive naturals.
Moreover it’s known that $\operatorname{lcm}(1,2,3,\dotsc, n)>2^n$, and it behaves asymptotically as $e^n$.
However I wanted to ask two questions.
Does there exist an effective lower bound on cardinality of set $A\in\{1,2,3, \dotsc, n\}$ such that the lcm of the sequence is greater than $\phi^n$ where $\phi$ is the golden ratio?
Is there any rough asymptotic of the number of subsets $A$ such that the lcm of the members of sequence is greater than $\phi^n$?
For 1, the trivial lower bound $(n\log\phi)/\log n$ is asymptotically tight.
In the paper The least common multiple of random sets of positive integers, Cilleruelo, Rué, Šarka, Zumalacárregui prove that if each subset of
$\{1,\dotsc,n\}$ is chosen with equal probability (equivalently each point $a\in \{1,\dotsc,n\}$ is included in $A$ with probability $\delta=1/2$) then for almost all sets $A\in \{1,\dotsc,n\}$ one has
$$
\operatorname{lcm}\{a: a \in A\}=2^{n(1+o(1))}.
$$
This would seem to indicate the answer to question 2 is asymptotically $2^n$.
|
2025-03-21T14:48:31.944972
| 2020-09-03T09:42:36 |
370736
|
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"url": "https://mathoverflow.net/questions/370736"
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|
Stack Exchange
|
Two questions on counterexamples to Borsuk's conjecture and ball-packings
In 1933 Karol Borsuk conjectured the following
Can every bounded subset $E$ of $\mathbb{R}^d$ be partitioned into $(d+1)$ sets, each of which has a smaller diameter than $E$?
Whilst new to this field of geometry I still have some open/unclear questions on this topic.
Early in game (to my knowledge even before any counter-example were known) Larman suggested investigating the problem when $E$ is a two-distance set. As a result of this suggestion many counter-examples emerged. Most recently the counter-example of Bondarenko in Dimension 64 using strongly regular graphs which form a two distance set on a sphere.
My question to this is:
How did Larmann come up with two-distance set? Why is it reasonable to think that those kind of sets contain many difficulties intrinsic to the general problem? What are those difficulties?
Another question emerges from a result by Hao Chen on Ball packings with high chromatic numbers from strongly regular graphs.
Most likely it seems that this publication emerged from another MathOverflow question by Cantwell. Namely:
Chromatic number of graphs of tangent closed balls .
Just as Bondarenko, Chen uses strongly regular graphs to form a spherical two-distance set to construct a ball-packing whose tangency graph is highly chromatic.
My question regards the part where Chen wants to link Borsuk's conjecture to the ball packing problem. As written:
The finite version of the Borsuk conjecture can be formulated as follows: the chromatic number of the unit-distance graph for a set of points with maximum distance $1$ is at most $d+1$. So the chromatic number problem for unit ball packings is the “opposite” of the Borsuk conjecture.
Why can Borsuk's conjecture be regarded as Chen names it. Somehow I don't get the connection between the "classic" conjecture by Borsuk and Chens version.
How is the coloring of tangency graphs of ball-packings connected to Borsuk's conjecture? Why can the chromatic number problem for unit packings be regarded as the "opposite" of Borsuk conjecture? What opposite?
I will be thankful for every kind of advice or tip to my questions.
I have no specific knowledge about this problem or its history, but this is my guess: you have a problem about distances between points, and what could be a simpler test case than a set in which only two distances exist. If you find no counterexamples among them, then maybe you might be able to prove that the conjecture is true for these kinds of sets. It is worth to take a look.
|
2025-03-21T14:48:31.945274
| 2020-09-03T09:50:38 |
370737
|
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|
Stack Exchange
|
Random matrix invertible
I am trying to figure out why the following random matrix is invertible:
\begin{align*}
A_j = I_d + J_{\mu}(\hat{X}_{t_{j-1}})(t_j - t_{j-1}) +
\left( \begin{array}{rrr}
B_1^T \\
\vdots \\
B_d^T \\
\end{array}\right) \in \mathbb{R}^{d \times d}
\end{align*}
with
\begin{align*}
B_l = \Delta W_{t_{j}}^T J_{\sigma_l}(\hat{X}_{t_{j-1}}) \in \mathbb{R}^{1 \times d},
\end{align*}
whereas
\begin{split}
\hat{ X}_{t_{j+1}}&:= \hat{X}_{t_j} + \mu (\hat{X}_{t_j}) \Delta t_i + \sigma (\hat{X}_{t_j}) \Delta W_{t_{j+1}},
\end{split}
and
\begin{align*}
\hat{ X}_{t_{0}} = x_0 \in \mathbb{R}^d
\end{align*}
the Euler discretization of a SDE X is. For
\begin{align*}
\mu : \mathbb{R}^d \rightarrow \mathbb{R}^d
\end{align*}
and
\begin{align*}
\sigma : \mathbb{R}^d \rightarrow \mathbb{R}^{d \times m}
\end{align*}
we assume that they are globally lipschitz continuous and in continuously differentiable and with
\begin{align*}
\Delta W_{t_i} := (W_{t_i} - W_{t_{i-1}})
\end{align*}
I mean the increment of a Brownian motion with dimension $m$. With $J_{\sigma_l}$ I denote the Jacobian matrix.
I know that under the lebesgue measure the measure is zero that $A_j$ is not invertible. Now I want to show that the matrix $A_j$ is invertible under the probability measure with probability 1. My idea would be to show that the terms in the matrix are absolutely continuous with respect to the lebesgue measure. For a gaussian distributed random variable I understand that this is true. My problem is that I do not see how I can proof that the Jacobian of $ \mu$ and $ \sigma$ and the Euler discretisation are absolutely continuous.
I would appreciate any references or hint!
|
2025-03-21T14:48:31.945393
| 2020-09-03T09:57:14 |
370738
|
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|
Stack Exchange
|
Shape derivative at manifold $M$ in direction $v$ is equal to the shape derivative at $\partial M$ in drection $\langle v,n\rangle n$
Let $\tau>0$ and $d\in\mathbb N$.
Definiton 1$\:\:\:$If $v:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ with $v(\;\cdot\;,x)\in C^0([0,\tau],\mathbb R^d)$ and $$\sup_{t\in[0,\:\tau]}\left\|v(t,x)-v(t,y)\right\|\le c_1\left\|x-y\right\|\tag1$$ for some $c_1\ge0$, then there is a unique $X^x\in C^0([0,\tau],\mathbb R^d)$ with $$X^x(t)=x+\int_0^tv(s,X^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau].$$ Moreover, $$T^{(v)}_t(x):=X^x(t)\;\;\;\text{for }x\in\mathbb R^d$$ is a $C^1$-diffeomorphism for all $t\in[0,\tau]$. Let $\mathcal V$ denote the set of all such $v$.
Now let $\mathcal F:2^{\mathbb R^d}\to\mathbb R$.
Definiton 2$\:\:\:$If $\Omega\subseteq\mathbb R^d$ and $v\in\mathcal V$, then $\mathcal F$ is called shape differentiable at $\Omega$ in direction $v$ if $$[0,\tau]\ni t\mapsto\mathcal F\left(T^{(v)}_t(\Omega)\right)\tag2$$ is differentiable at $0$. In that case, $${\rm d}\mathcal F(\Omega;v):=\lim_{t\to0+}\frac{\mathcal F\left(T^{(v)}_t(\Omega)\right)-\mathcal F(\Omega)}t.$$
Let $M$ be a $d$-dimensional properly embedded $C^1$-submanifold of $\mathbb R^d$ with boundary and $\nu_{\partial M}$ denote the unique outward pointing unit normal field on $\partial M$. We can show that if $v\in\mathcal V$ with $$\langle\left.v\right|_{[0,\:\tau]\times\partial M},\nu_{\partial M}\rangle=0\tag3,$$ then $$T_t^{(v)}(\partial M)=\partial M\tag4.$$ From this result we can infer that $$T_t^{(v)}(M)=M\tag5.$$
Now let $v\in\mathcal V$ be arbitrary and $\tilde\nu\in C^0(\mathbb R^d,\mathbb R^d)$ be any extension of $\nu_{\partial M}$. I would like to show that, if $\mathcal F$ is shape differentiable at $M$ in direction $v$, then $\mathcal F$ is shape differentiable at $\partial M$ in direction $u:=\langle v,\tilde\nu\rangle\tilde\nu$ and $${\rm d}\mathcal F(M,v)={\rm d}\mathcal F(\partial M,u)\tag6.$$
I'm not sure what the best approach is. We may note that $$T^{(u)}_t(\langle x,\tilde\nu(x)\rangle\tilde\nu(x))=\langle T^{(v)}_t(x),\tilde\nu(x)\rangle\tilde\nu(x)\;\;\;\text{for all }(t,x)\in[0,\tau]\times\mathbb R^d\tag7.$$ Or we may consider $w:=v-u$ and note $$\langle\left.w\right|_{[0,\:\tau]\times\partial M},\nu_{\partial M}\rangle=0\tag8,$$ which implies $T_t^{(w)}(\partial M)=\partial M$ and $T_t^{(w)}(M)=M$ by the former result.
EDIT
The following example might clarify what I'm trying to achieve. Assume $$\mathcal F(\Omega)=\lambda^{\otimes d}(\Omega)\;\;\;\text{for all }\mathcal B(\mathbb R^d)\tag9,$$ where $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R^d)$. Let $\Omega\in\mathcal B(\mathbb R^d)$, $v\in\mathcal V$ and assume that $$\det{\rm DT^{(v)}_t}>0\;\;\;\text{for all }t\in[0,\delta)\tag{10}$$ for some $\delta>0$. We then obtain $$\left.\frac{{\rm d}}{{\rm d}t}\mathcal F\left(T^{(v)}_t(\Omega)\right)\right|_{t=0}=\int_\Omega\left.\frac{{\rm d}}{{\rm d}t}\det{\rm D}T^{(v)}_t\right|_{t=0}{\rm d}\lambda^{\otimes d}=\int_\Omega\nabla\cdot v_0\:{\rm d}\lambda^{\otimes d}\tag{11},$$ where I wrote $v_0:=v(0,\;\cdot\;)$. If now $\Omega$ is a $d$-dimensional properly embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, we can write $$\int_\Omega\nabla\cdot v_0\:{\rm d}\lambda^{\otimes d}=\int\langle v_0,\nu_{\partial\Omega}\rangle\:{\rm d}\sigma_{\partial\Omega}\tag{12},$$ where $\sigma_{\partial\Omega}$ denotes the surface measure on $\mathcal B(\partial\Omega)$. What I'm trying to obtain is a generalization of this result as described in the question.
If I'm understanding your setup correctly, (8) can't be true in general. For example, if $\mathcal F$ is $d$-dimensional volume, then $\mathcal F(\partial M)$ is always zero, and thus the right-hand side of (8) is always zero. (Of course, volume is not defined for arbitrary subsets of $\mathbb R^d$. But you could replace it by inner measure or outer measure, and the same comment would apply.)
@JackLee (8) does not depend on $\mathcal F$. Did you really mean (8)?
Sorry, that should have been (6).
@JackLee Well, from your example, it seems like something is wrong here. This is supposed to be a simplified version of "Hadamard's theorem". Please take a look at Propostion 2.26 here. The claim there should be more or less the same as in the quesiton.
@JackLee I've added the example you've mentioned to the question. Hopefully it's clearer now what I'm trying to achieve.
@JackLee Maybe I'm misunderstanding the claim made in Proposition 2.26. Maybe it's only claiming that there exists a functional $\varphi(\partial M,;\cdot;)$ on a space of functions $\partial M\to\mathbb R$ such that $${\rm d}\mathcal F(M,v)=\varphi(\partial M,u)\tag{6'}.$$ To be honest, this is the claim literally made, but denoting $\varphi(\partial M,;\cdot;)$ by ${\rm d}\mathcal F(\partial M,;\cdot;)$ is highly confusing and obviously ambigiuous.
|
2025-03-21T14:48:31.945658
| 2020-09-03T10:16:43 |
370739
|
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"Henri Cohen",
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|
Stack Exchange
|
I have a question on the definition of 'good' primes in the paper of Cohen and Martinet
I'm reading the paper of Cohen and Martinet 'Etude heuristique des groups de classes'.
In the section 6, for an central idempotent $e$ of $\mathbb{Q}[\Gamma]$ and a prime $p$, the 'goodness' of $p$ is defined.
The second condition says for each irreducible component $e'$ of $e$, $e'\mathbb{Z}_{(p)}[\Gamma]$ is a maximal $\mathbb{Z}_{(p)}$-order of $\mathbb{Q}[\Gamma]$.
But the $\mathbb{Q}$-dimension of $\mathbb{Q}\otimes_{\mathbb{Z}_{(p)}}e'\mathbb{Z}_{(p)}[\Gamma]$ is always strictly less than $|\Gamma|$. Thus the second condition does not make sense to me.
Does the second condition actually mean that $e'\mathbb{Z}_{(p)}[\Gamma]$ is a maximal order in $e'\mathbb{Q}[\Gamma]$?
Thank you very much!
I just asked Martinet, and although he says it is very old in his memory he believes the answer is yes, and this sentence should have been rephrased.
@HenriCohen Thank you very much for the clarification!
|
2025-03-21T14:48:31.945752
| 2020-09-03T12:52:17 |
370742
|
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|
Stack Exchange
|
Is the degree a sufficient condition such that a measure is the pullback of another one?
Consider $M$ a smooth compact connected manifold with $w$ a volum form. Take for example $M=\mathbb{S}^n$ with the uniform measure. For any smooth map $f:M\rightarrow M$ and its pullback measure $\nu = f^* w$ it is well known that we have $$\int_M \nu = \text{deg}(f)\int_M w $$with $\deg(f)\in \mathbb{Z}$. Is this condition sufficient?
For any $\nu$ volum form on $M$ such that $\int_M \nu / \int_M w \in \mathbb{Z} $ does there exist $f:M\rightarrow M$ such that $\nu = f^*w$ ? Can we construct such an $f$ explicitly ?
$S^n$ does not have a Haar measure unless $n=0,1,3$. Maybe $7$ if you stretch the meaning of Haar measure a bit. Ignoring this, most manifolds do not have self-maps of arbitrary degree; and if you want $\nu$ to also be nonvanishing then $f$ should be a covering map (even less common). It is a theorem of Moser that if $\int \omega = \int \nu$ for two nonvanishing top forms, then there's an oriented diffeomorphism $f$ with $f^*\nu = \omega$. The proof is what's usually called the Moser trick.
a note on terminology: usually measures are only pushed forward, and differential forms are pulled back. Defining pullback measures is tricky https://mathoverflow.net/q/122704/3948 .
No. Consider the $2$-sphere, and the standard volume $\omega$ form with volume $4\pi.$ Consider $\nu=2\omega.$ If $f\colon S^2\to S^2$ with $f^*\omega=\nu$ would exist, it would be a local diffeomorphism and by compactness a covering map. As $S^2$ is simply connected and $f$ would have degree 2, this is not possible.
|
2025-03-21T14:48:31.945879
| 2020-09-03T13:29:34 |
370747
|
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|
Stack Exchange
|
Condtions for a stochastic process to be locally non-factorizable
Given a stochastic process $X=(X_t)_{t\in I}$ on $\mathbb{R}^d$ with continuous sample paths supported on a prob. space $(\Omega, \mathscr{F}, \mathbb{P})$ and such that each pair $(X_s, X_t)$, with $(s,t)\in \mathcal{I}:=\{(s',t')\in I^{\times 2}\mid s' < t' \}$, admits a continuous joint Lebesgue density $\chi_{s,t}\in C(\mathbb{R}^{2d})$.
Suppose that for each $(\emptyset\neq)\,U\subset\mathbb{R}^d$ open there is $\Omega_U\in\mathscr{F}$ with $\mathbb{P}(\Omega_U)>0$ as well as $(\emptyset\neq)\,I_U\subseteq I$ open such that
$$\tag{1}(X_t(\omega))_{t\in I_U}\subset U\qquad\text{for each }\quad \omega\in\Omega_U.$$
We call a function $\chi\in C(U\times U)$ ($U\subseteq\mathbb{R}^d$) factorizable if $\chi\equiv\chi(x,y)=\chi_1(x)\cdot\chi_2(y)$ for some $\chi_1, \chi_2\in C(U)$.
Question: Under which 'minimal' additional assumptions can we infer that $X$ is locally non-factorizable in the sense that
$$\tag{2}\forall\, (\emptyset\neq)\,U \text{ open } \ : \ \exists\,(s,t)\in \mathcal{I} \ \quad \text{ such that }\quad \ \left.\chi_{s,t}\right|_{U^{\times 2}} \ \text{ is not factorizable}? $$
Remark: Condition (2) should hold for instance if $X$ is a (sufficiently regular) diffusion, as the joint density $\chi\equiv\chi_{s,t}$ then adheres to the boundary condition $\lim_{s\rightarrow t}\chi_{s,t}(x,y) = \delta(x-y)$ of the Kolmogorov equations associated to $X$.
Also, by the continuity of $X$, Kolmogorov's zero-one law implies that (2) holds for the fixed open set $U=\mathbb{R}^d$ (as otherwise each $X_t$ would be constant almost surely, defying the existence of $\chi_{s,t}$). Surely the regularity condition (2) will be satisfied under some more general circumstances..
|
2025-03-21T14:48:31.946015
| 2020-09-03T13:31:33 |
370748
|
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|
Stack Exchange
|
Multiplicities of the Laplacian eigenvalues of a graph
Let $\lambda^G_1 > \lambda^G_2 > \dots$ be the eigenvalues of the Laplacian matrix $G$ of a graph on $n$ vertices.
Let $\mu(G)$ be the composition $a_1,\dots,a_k$ of $n$ where $a_i$ is the multiplicity of $\lambda^G_i$.
Is $\mu$ surjective as a map from (finite) simple graphs to integer compositions?
It barely works for $n=3$!
We can actually construct a graph with the desired composition of multiplicities by adding isolated vertices and taking complements:
As an initial remark, note that the smallest Laplacian eigenvalue of a graph $G$ is $0$ and that its multiplicity is $1$ if and only if $G$ is connected.
Let $(a_1,\dots,a_k)$ be the desired composition. If $a_k > 1$, we can choose a graph $G$ with $\mu(G)=(a_1,\dots,a_{k-1},1)$ and add $a_k-1$ isolated vertices.
If $a_k = 1$, choose a graph $G$ with $\mu(G) = (a_{k-1},\dots,a_2, a_1+1)$. Then the complement of $G$ has the desired multiplicities.
(For people like me who are a bit slow, regarding the last paragraph: since $a_1+1 > 1$, the graph $G$ is disconnected; hence its complement is connected; hence $\lambda_1^{G} < n$; and so indeed the complement of $G$ has the desired $\mu$.)
I believe it also follows from your argument that to get all the compositions we only need to consider threshold graphs: https://en.wikipedia.org/wiki/Threshold_graph
Note, interestingly, that the number of isomorphism classes of threshold graphs on $n$ vertices is $2^{n-1}$, the same as the number of compositions of $n$. I guess your construction gives a bijection between these sets.
This reminds me of "The critical group of a threshold graph" Hans Christianson and Victor Reiner. Connection?
Theorem 5 of that paper (https://doi.org/10.1016/S0024-3795(02)00252-5) says "The eigenvalues of L(G) for G a threshold graph are the column lengths of the Ferrers diagram of the degree sequence of G," which is directly related to your eigenvalues multiplicity question. They attribute that result to Merris (https://doi.org/10.1016/0024-3795(94)90361-1). But I don't know if there's a direct connection to critical groups.
There's a simple bijection between (unlabeled) threshold graphs on $n$ vertices and binary sequences of length $n-1$: let $v_1, \ldots, v_n$ be the vertices in increasing order of degree; then the $i$th letter is 1 if $v_{i+1}$ is adjacent to $v_i$ and 0 otherwise. Of course there's also a canonical bijection between compositions of $n$ and binary sequences of length $n-1$ (via "stars and bars"). But note that your eigenvalue multiplicity bijection is not the composition of these (think about $11\ldots 1$). But the aforementioned Theorem 5 still explains what's going on, I think.
|
2025-03-21T14:48:31.946467
| 2020-09-03T14:31:43 |
370753
|
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|
Stack Exchange
|
Finding minimum operations to move ants through connected graph
I am working on a project that requires to find the minimum number of steps to move ants from source to sink in a graph; one step is the movement of all ants from one node to the next of the graph. Only one ant per node
I already solved the problem of finding all possible disjoint paths. The issue I have is selecting the number of ants to send per path.
Here is an example:
every colour is a disjoint path. Given 12 ants, the best solution is to send 4 along gray, 3 along black, 3 along red, 1 along yellow and 1 along blue.
The only way that I found for now is a greedy solution but this will not scale well at all. I am not too good at math and I am sure there must be some smart formula to help me solve this problem. I tried to use the total length of the paths to help me solve it but I can't figure it out... Any help or resource is greatly appreciated
My current take:
I am calculating solutions by seeing how many ants go through the shortest path (grey) while one goes through a second one. So for "black" is 2 as len(n) / len(shortest). This helps me create proportions. Moreover I have the paths stored in increasing order. I do not send ants through a edge with increased len unless all previous edges are at capacity. if more edges have the same len I treat them as one (send same amount through them). This method seems to work but only with small tweaks in numbers when I find a solution. I feel like there is some sort of mistake but can't really pinpoint it.
This may come under the heading of "network flows" although it differs from the netwrok flow problems I'm familiar with in that only one unit of flow can be in any given arc at any given time.
This is more of a queueing theory flavour. Once the paths are identified, it's like saying each one is a queue (a processor) and the grey processor completes a job in 1 step and the red processor completes in 2 steps and the yellow processor completes in 3 steps, etc.
If there are many ways to decompose the graph into a set of disjoint paths, it may not be easy to find the best way.
The question is supremely unclear. What's the meaning of the numbers zero to nine (with the number three missing) in the vertices? Which vertex is the source, and which is the sink? How can there be "only one ant per node" if you start with $12$ ants, and only nine nodes? What is the capacity of each edge? A real dog's breakfast.
I'm thinking of this from the perspective of how many ants can you accommodate in $s$ steps. Suppose your disjoint source-to-sink paths have lengths $(p_1, \ldots, p_n)$ listed in nondecreasing order. The number of ants you can handle in $s$ steps is then, using the integer floor function,
$$ \left\lfloor \frac{s}{p_1} \right\rfloor + \cdots + \left\lfloor \frac{s}{p_n} \right\rfloor.$$
Your example has path lengths $(1,2,2,3,4)$. In five steps, you can handle
$$ \left\lfloor \frac{5}{1} \right\rfloor + \left\lfloor \frac{5}{2} \right\rfloor + \left\lfloor \frac{5}{2} \right\rfloor + \left\lfloor \frac{5}{3} \right\rfloor + \left\lfloor \frac{5}{4} \right\rfloor = 5+2+2+1+1 = 11$$
ants, while with $s=6$ steps you can handle
$$ \left\lfloor \frac{6}{1} \right\rfloor + \left\lfloor \frac{6}{2} \right\rfloor + \left\lfloor \frac{6}{2} \right\rfloor + \left\lfloor \frac{6}{3} \right\rfloor + \left\lfloor \frac{6}{4} \right\rfloor = 6+3+3+2+1 = 15.$$
So for 12 ants, you need 6 steps. To make something algorithmic, set up the plan for the optimal 15 ants in 6 steps and then ignore, say, the first 3, giving here (3 gray, 3 black, 3 red, 2 yellow, 1 blue). Or the last 3, giving (6 gray, 3 black, 3 red).
In general, compute the $\lfloor s/p_i \rfloor$ sum for increasing $s$ until you first reach or exceed the number of ants you need, make the allocation from the formula for the optimal number that can be handled, and then ignore the appropriate number of ants.
|
2025-03-21T14:48:31.946743
| 2020-09-03T15:14:41 |
370755
|
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|
Stack Exchange
|
Identity map minus Cremona transformation
Let $ \delta $ be the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ in $\mathbb R^3$. It's a face of the standard octahedron. The Cremona transformation
$$\mathcal C: (x, y, z) \mapsto - \frac {(yz, zx, xy)} {yz + zx + xy}$$
maps $ \delta $ to the opposite face $-\delta$ of the octahedron. And the half-sum of $\mathcal C $ with the identity transformation
$$\mathcal F: (x, y, z) \mapsto \frac {(x, y, z) + \mathcal C (x, y, z)} {2}$$
maps $\delta$ to the hexagon $ h $, which is the section of the octahedron by the plane $ x + y + z = 0 $.
Question. Is the map $ \mathcal F: \delta \to h $ a diffeomorphism from the interior of $ \delta $ to the interior of $ h $? Is there a simple closed formula for the inverse map $ \mathcal F^{- 1}: h \to \delta $?
The answer is yes and yes: the map is a diffeomorphism from the interior of $\delta$ to the interior of $h$, and there is a rather simple formula for the inverse map.
Let $s:=(s_1,s_2,s_3):=\mathcal F$, so that $s_1+s_2+s_3=0$, let $t:=(s_1,s_2)$, and let $S:=\{(x,y)\colon0<x<1,0<y<1-x\}$.
We have to
(i) show that the map
$$S\ni(x,y)\mapsto s(x,y)\in h$$
is a bijection or, equivalently, that the map
$$S\ni(x,y)\mapsto t(x,y)$$
is a bijection;
(ii) show that the Jacobian determinant of $t$ is nonzero everywhere on $S$;
(iii) find the inverse of $s$.
All this is done in a Mathematica notebook, whose image is here (click on the image to enlarge it):
Many thanks for this answer, Iosif!
|
2025-03-21T14:48:31.946869
| 2020-09-03T15:38:08 |
370760
|
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|
Stack Exchange
|
Which cases of Beilinson-Bloch-Kato for elliptic motives are known?
Let $V$ be a semisimple geometric Galois representation of a number field. Then the Bloch-Kato conjectures state that
$$
\operatorname{ord}_{s=0}{L(V^*(1),s)} = \operatorname{dim}{H^1_f(G_k,V)}-\operatorname{dim}{H^0(G_k,V)}.
$$
Beilinson has similar conjectures relating the LHS to algebraic K-theory rather than Selmer groups.
If $E$ is an elliptic curve, and we set $V=h_1(E)=h^1(E)(1)$, then the conjecture above is equivalent to the statement that the analytic rank is the same as the rank of the $p$-adic Selmer group. Beilinson's conjecture in this case is equivalent to the assertion that the analytic rank equals the Mordell-Weil rank.
For $V$ of non-negative weight, the conjecture simply asserts that the Selmer group vanishes. Assuming the conjectured properties of $L$-functions, the conjectures for $V$ and $V^*(1)$ are equivalent (in particular, the case of weight $\le -2$ follows from the case of weight $\ge 0$).
My question is: which cases of this conjecture are known for $V=\operatorname{Sym}^k{h^1(E)}(n)$ for $E$ an elliptic curve? I know that many cases are known when $k=n=1$ due to the theory of Heegner points, Gross-Zagier, etc, but I'd like to know what's known outside that range. (I'm especially interested in cases where $w=k-2n=-2, -3, -4$).
Feel free to give reference, or even better, specific elliptic curves in LMFDB.
One way to construct elements in $H^1_f(V)$ is to use the $p$-adic realization map from motivic cohomology. In you particular situation, elements in motivic cohomology have been constructed by Deninger for $n=k+1$ using torsion points of $E$ (see Higher regulators and Hecke L-series of imaginary quadratic fields I & II). One can show that these elements are linearly independent by computing the archimedean regulator (but the relation with the $L$-function is not known). I don't know if computing the $p$-adic regulator has ever been implemented.
Recarding the case $k=1$, the motivic elements have been constructed by Beilinson. Gealy has computed their $p$-adic realisation in his PhD thesis. He also shows a result of the form ''$H^1(V)$ has rank $1$'' (as expected) but it's conditional on the Leopoldt-type conjecture that $H^2(V)$ is finite, and on Kato's main conjecture for $E$.
You probably want some assumptions on the base number field. If it's not totally real or CM, then we know basically nothing.
To be honest, I was really interested in elliptic curves over $\mathbb{Q}$ or maybe a quadratic imaginary field. So that's no issue
There are three approaches I know of to studying $H^1_{\mathrm{f}}(K, V)$, where $V = Sym^k(h^1(E))(n))$. All rely on $E$ being modular, so let me assume this henceforth (of course, this is no assumption if $K = \mathbf{Q}$, or for some other small-degree fields).
Via "anticyclotomic" Euler systems, such as Heegner points (and the closely-related method of "arithmetic level-raising"). This works extremely well when $k = n = 1$, and $K$ is totally real (or $K$ is CM and $E$ is base-extended from $K^+$); under these hypotheses we know the BK conjecture holds, for any $p$, whenever the analytic rank is 0 or 1 (Zhang, Nekovar). More generally, this might potentially be accessible for any $n$ and $k = 2n-1$, although huge amounts of work would be needed to carry that out. However, it's entirely impossible to generalise this approach beyond the case of motivic weight $w = -1$.
Via modularity-lifting theorems. This gives a way of studying Selmer groups of representations that have the shape $W \otimes W^*$, where $W$ is irreducible. More generally, if $W$ has some extra structure (e.g. self-duality) which forces $W \otimes W^*$ to be reducible, then you can get some information about the cohomology of the pieces. This gives you very nice control over $Sym^k(h^1(E))(n)$ for $k = 2$ and $n = 1$ (or by duality $n = 2$) (Diamond--Flach--Guo). More generally, one should be able to get some information about general $n$ and $k = 2n$ or $2n-2$ using the recent work of Newton--Thorne; Theorem 5.6 of this paper tells you something about $W\otimes W^*$ where $W = Sym^k(h^1(E))$, and this representation breaks up as a sum of $Sym^{2m}(h^1(E))(m)$ for $0 \le m \le k$. [Caveat: I'm not sure exactly what their method gives; it's possible that you need to twist by an odd quadratic character at some point.] However, this is again restricted to specific values of $w$; it won't tell you anything unless $w = 0$ or $w = -2$.
Via "cyclotomic" Euler systems, such as Kato's Euler system. This method has the advantage that it can tell you something about general motivic weights (i.e. a fixed $k$ and any $n \in \mathbf{Z}$). However, it gives you a criterion for vanishing of the $H^1$ in terms of p-adic L-functions, and these are only indirectly related to complex $L$-functions outside the critical range (i.e. away from $w = -1$ in your case). Nonetheless, these p-adic L-functions are computable, so you can check explicitly whether they vanish in examples. For elliptic curves over $\mathbf{Q}$, an Euler system for $Sym^k E$ exists for $k = 1$ due to Kato, and for $k = 2$ [*] and $k = 3$ due to Zerbes and myself (building on work of lots of other people). So, for example, if $K = \mathbf{Q}$, and $k = 1, 2, 3$, this would give an approach to proving the vanishing of $H^1_{\mathrm{f}}(\mathbf{Q}, Sym^k(h^1(E))(n))$ for your favourite elliptic curve and a specific but arbitrary value of $n$ (and $p$), using only a finite amount of computation.
[*] Actually there is a caveat here -- embarrassingly, I forgot the statement of my own theorem! -- so the result as published only applies to $Sym^2(E)$ twisted by a non-trivial Dirichlet character. But the un-twisted case might also be accessible with some extra work.
Thanks! What about over an imaginary quadratic field? In one case, I have a specific curve of rank $0$, and I'm interested in it over imaginary quadratic fields over which it has rank $1$. If I can show it explicitly for $k=2,3$, that's really all I need.
(But I'm also interested in other cases, e.g., for certain rank $1$ curves over $\mathbb{Q}$.)
I'd be happy to discuss this further but MO isn't really the place for extended conversations -- feel free to get in touch via my warwick.ac.uk email.
To add to the second bullet-point, it seems that $\operatorname{Sym}^2{h^1(E)}(1)$ is covered by arxiv.org/abs/1411.7661 under some mild assumptions when the curve is modular (known in general e.g., over $\mathbb{Q}$ or a real quadratic field).
Is it fair to say that $k=3$ would use something like Theorem D of https://arxiv.org/pdf/2003.05960.pdf?
Morally yes, although there are some mildly fiddly technical issues meaning we don't have a full proof written out yet for Sym^3 of an elliptic curve. See https://arxiv.org/abs/2005.04786 for Sym^3 of modular forms of level 1 and big weight.
|
2025-03-21T14:48:31.947414
| 2020-09-03T15:56:26 |
370763
|
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|
Stack Exchange
|
It is possible that $ X \simeq ΩX $? and that $ X \simeq Ω^ 2X $?
Original post: https://math.stackexchange.com/questions/3810423/it-is-possible-that-x-simeq-%ce%a9x-and-that-x-simeq-%ce%a9-2x
I am studying J. Strom's Modern Classical Homotopy Theory. In chapter 4 he proposes the following exercise
Let $ X $ be a path-connected non-contractible space.
(1) It is possible that $ X \simeq ΩX $ ?
(2) It is possible that $ X \simeq Ω^ 2X $ ?
My answers would be
(1) No, because otherwise $ \pi_n X \simeq \pi_{ n - 1 } \Omega X \simeq \pi_{ n - 1 } X \simeq \cdots \simeq \pi_0 X \simeq *$ and by Whitehead $X \simeq *$ (assuming $X$ CW)
(2) Yes, for example $X = \prod_{n=0}^\infty K (\mathbb{Z}, 2n+1)$ (or $X = U$ by Bott periodicity)
The problem is that at that chapter no Whitehead theorem nor Eilenberg-MacLane spaces (nor Bott periodicity) are available, so there would be a more down-to-earth answer!
What is available up to chapter 4?
Chapter 1: categories and functors.
Chapter 2: limits and colimits.
Chapter 3: convenient categories of spaces, i.e. with CW complexes, their limits and something more; smash product, smash-map adjuction, suspension, loospace.
Chapter 4: homotopies, contractible spaces, nullhomotopies, abstract homotopies with cylinder and path objects, (homotopy) groups and cogroups, homotopy groups; mapping spaces; maps over and under a space; CW structure on loospace.
What is $U$ in question (2) ?
@Joël Isn't that the underlying space of the infinite unitary group? I think Bott periodicity gives you deloopings Ω(BU×Z) ~ U and ΩU ~ BU×Z
You can try asking the author himself: https://wmich.edu/math/directory/strom Anyway, I might get ahead of myself but maybe the book is simply not meant to be read 100% linearly. Maybe when you first read the chapter your answer should be "uh, I can't find any obstruction to that, weird" and when you read the rest you should think "ahah, now I know!". (Though I'd be happy to learn of an answer to the question as asked.) The question is "is it possible", not "find an example", after all. "With what I've learned so far, it's possible" is a fine answer...
@Joël The infinite unitary group
@NajibIdrissi Coulde be! I have seen that the author is a quite active user on MO, so maybe it is not necessary to contact him to get an answer!
For (1) you only need to prove a CW-complex with no homotopy retracts onto its basepoint, which is easier than Whitehead's theorem. For (2) maybe you have the bar construction?
@JohnGreenwood Thank you! I saw your comment only today. I will follow your hint for (1). For (2), no, there is no bar construction.
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2025-03-21T14:48:31.947617
| 2020-09-03T15:58:14 |
370764
|
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|
Stack Exchange
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Name for "Category without nontrivial automorphisms"?
It's all in the title: Is there really no name for categories in which all automorphisms are trivial? I've encountered some examples of these, the most prominent being the simplex category.
In other places, objects without nontrivial automorphisms are called rigid. There is already a different notion of rigid category, but I'm of the opinion that that's the wrong adjective and one shouldn't be using this terminology.
@JohannesHahn In this terminology, should'nt "Rigid category" mean a category with no non-trivial self equivalences.
I think Gaunt category (https://ncatlab.org/nlab/show/gaunt+category) is not too far off. It usually refers to the stronger property of having no non-trivial isomorphisms. But a category has no non-trivial automorphism if and only if it is equivalent to a Gaunt category. So maybe "Essentially gaunt category" would be a good name ? (or use "gaunt" and "strictly gaunt")
I like "essentially gaunt". Even in the absence of the existing notion of "rigid category", I would expect a "rigid category" to be one that has no automorphisms itself as an object of Cat, rather than one whose objects have no automorphisms.
Thank you, all. Maybe my notion of cat without nontrivial automorphisms is too specific to deserve a name (but yes, "essentially gaunt" is good): it seems cats without nontrivial endomorphisms are more common, and called "one-way"; even more common, it seems, are "direct" categories which have no infinite descending chains of non-id morphisms. See https://ncatlab.org/nlab/show/direct+category
To sum up the comments & discussion: no established terminology seems to exist, but essentially gaunt would be a good name for categories without nontrivial automorphisms.
For a related notion, categories without notrivial isomorphisms are called gaunt, see https://ncatlab.org/nlab/show/gaunt+category
On the other side, categories without nontrial endomorphisms are called one-way, and those without infinite descending chains of non-id morphisms are called direct, see https://ncatlab.org/nlab/show/direct+category
|
2025-03-21T14:48:31.947763
| 2020-09-03T16:15:41 |
370765
|
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|
Stack Exchange
|
How to compute the following probability involving 4 normal random variables?
$\alpha, \alpha', \beta$ and $\beta'$ are four independent standard normal random variables, I am wondering how to compute the probability of the following two events:
$\alpha>\alpha'>0, \ \ \beta<\beta', \ \ |\alpha-\alpha'+\beta-\beta'|\geq c_1,$
$\alpha>0>\alpha', \ \ \beta<\beta', \ \ |\alpha-\alpha'+\beta-\beta'|\geq c_1, \ \ |\frac{\alpha}{\alpha'}|>c_2$
where $c_1$ and $c_2$ are positive constants. I know that the sum and difference of normal random variables are still normal random variables, but I'm not sure how to use this with the inequality relation here.
@MattF. Thank you for your answer! Is it possible to simplify the formula so that we can have a lower bound of the probability?
@MattF. I am analyzing some machine learning algorithms and I want to say that parameters initialized as standard normal variables have the properties above with certain probability. So I guess any lower bound would be good for me.
This might work, the pdf might have the answer or at least part of the question but could not find the rest. https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter4.pdf
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2025-03-21T14:48:31.947861
| 2020-09-03T17:19:50 |
370768
|
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|
Stack Exchange
|
ramification locus for finite morphism and Abhyankar's Lemma
I want to ask given a finite morphism between projective varieties $f:X\rightarrow Y$.
What is exactly the ramification locus $\Delta(X/Y)$. If $X$, $Y$, $f$ are smooth, then I can more or less understand it.
Q1: There should be strict definition of $\Delta(X/Y)$ (Any reference?). Also, is there a good example to see $\Delta(X/Y)$ with $X,Y$ singular variety.
Q2: I saw a theorem, which says that if $f:X\rightarrow Y$ is finite morphism,$X$ normal, $Y$ smooth and $\Delta(X/Y)$ simple normal crossing, then $X$ has quotient singularity. The proof is refered to Abhyankar's Lemma (in Germany), I wonder how the Abhyankar's Lemma can be applied here. Can I get a rough explanation! (Reference in English is also helpful)
Thanks!
The ramification locus is just the support of the sheaf $\Omega_{X/Y}$ on $X$. This is in many texts. To get examples with $X,Y$ singular, all you need to do is to compute what $\Omega_{X/Y}$ for a map of singular varieties $X\to Y$ and see where it's supported.
|
2025-03-21T14:48:31.947957
| 2020-09-03T17:26:31 |
370769
|
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"Ben W",
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|
Stack Exchange
|
What are the complemented subspaces of $(\bigoplus\ell_q^n)_p$?
Bourgain/Casazza/Lindenstrauss/Tzafriri proved in their unconditional basis UTAP book (1985) that $\ell_1$ is the only nontrivial complemented subspace of $(\bigoplus\ell_2^n)_1$, and hence by duality $c_0$ is the only nontrivial complemented subspace of $(\bigoplus\ell_2^n)_0$. And of course there are no nontrivial complemented subspaces of $(\bigoplus\ell_2^n)_p$ for $1<p<\infty$ due to the fact that it's just $\ell_p$ itself.
I am wondering about the general case where $p\neq q\neq 2$. Is $\ell_p$ then the only nontrivial complemented subspace of $(\bigoplus\ell_q^n)_p$?
I guess you mean $p \not= 1$. I think your question is still open.
@BillJohnson Ah okay, that's unexpected. I thought for sure this would have been solved.
But are you also saying that the question is closed for $(\oplus\ell_q^n)_1$ for $q$ other than just 2?
I just meant that when $p=1$ and $1<q \not=2 < \infty$ the space contains a complemented subspace isomorphic to the $1$ sum of $\ell_2^n$, which is not isomorphic either to $\ell_1$ or to the whole space.
@BillJohnson Oh, yes, the complemented version of Dvoretsky.
|
2025-03-21T14:48:31.948061
| 2020-09-03T17:28:18 |
370770
|
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|
Stack Exchange
|
Can we define geodesic in the space of compactly supported functions?
From Wikepedia, the definition of geodesic is stated as:
A curve $\gamma: I\to M$ from an interval $I$ of the reals to the metric space $M$ is a geodesic if there is a constant $v\geq 0$ such that for any $t\in I$ there is a neighborhood $J$ of $t$ in $I$ such that for any $t_1, t_2 \in J$ we have $d(\gamma(t_1), \gamma(t_2)) = v|t_1- t_2|$.
It is a known fact that a metric space, in general, may have no geodesics, except constant curves.
I wonder whether there is a well-defined geodesic in the space of compactly supported functions. If there is, how can we compute it? Is the geodesic unique?
I'm specifically interested in the cases of:
$L^p([0, 1])$ with $L^p$-norm;
$\mathcal{P}([0, 1])$ with sup-norm, where $\mathcal{P}([0, 1])$ denotes the space of piecewise constant functions over $[0, 1]$.
I'm a newbie to the topic of geometry. So any direct answer or reference to this question would be extremely helpful to me!
These are vector spaces, so the geodesic is just a straight line, is it not? That is, the geodesic from $f_1$ to $f_2$ is simply $\gamma(t) = (1-t)f_1 + t f_2$. You can verify this from properties of the norm.
@NateEldridge: for homogenous metrics that is a geodesic, but it is not necessarily unique.
@NateEldredge That's a good catch. I was too involved thinking about the `advanced' geometric ideas... Is there any way to justify the uniqueness?
@mw19930312 I think for $L^p$ it's true and for sup false. At least $l^p$ on $\mathbb{R}^d$ is (Google it), and for sup there's a counterexample on $\mathbb{R}^2$ and you can modify that a bit.
@VilleSalo Thanks for the comments! I'll take a look.
Sorry about the "Google it", I wanted to link https://math.stackexchange.com/questions/477438/uniquely-geodesic-spaces but it opened in the app and I don't know how to link from that and had to run.
A good reference for geodesic spaces (which includes a chapter dedicated to normed vector spaces) is Athanase Papadopoulos, Metric Spaces, Convexity, and Nonpositive Curvature (google books has a preview).
As Nate Eldredge pointed out in the comments, in a normed space you can always form a geodesic between two points, say $f$ and $g$, as the affine line $\gamma(t) = (1-t)f + tg$. However, this geodesic is not always unique. In particular, the $L^p$ spaces are uniquely geodesic when $1<p<\infty$, but not for $p=1$ or $p=\infty$. In fact, any function space with the sup-norm is not uniquely geodesic, as the sup-norm is not strictly convex.
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