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2025-03-21T14:48:29.602227
| 2020-01-08T14:38:49 |
349996
|
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"Benjamin Dickman",
"Fofi Konstantopoulou",
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|
Stack Exchange
|
Characters on Hopf algebras
For any algebra $A$, a character for $A$ is a non-zero algebra map $c:A \to \mathbb{C}$. For $H$ be a Hopf algebra, a character is given by $\epsilon:H \to \mathbb{C}$ the counit of $H$. I am looking for other examples of (co-semi-simple) Hopf algebras with characters distinct of the counit.
I am really interested in noncommutative, noncocommutative examples.
Would you consider Larson's character?
What is Larson's character?
I think that a general example is the so-called Larson's character, which in a sense ties together the trace and determinant functions.
To make the long story short: Let $C$ be a cocommutative bialgebra, $V$ a vector space and $EV$, the exterior algebra. Then, it has been shown that:
If $C\otimes V\rightarrow V$ is an action which makes $V$ a $C$-module, then there is a unique measuring $C\otimes EV\rightarrow EV$, extending the action on $V$.
In this sense, $EV$ becomes a $C$-module, with $C\cdot E^kV\subset E^kV$. If we furthermore assume that $\dim V=n$ then $E^nV$ is 1-dim. Let it be spanned by $\{z\}$. For any $c\in C$, let $\chi(c)$ be defined by $c\cdot z=\chi(c)z$. In this way, a linear map $\chi:C\rightarrow k$ is defined. It can be easily shown that this is an algebra map. It is called the Larson's character.
It can furthermore be shown that, if $g$ is a grouplike element of $C$ then $\chi(g)=\det T_g$, where $T_g:V\rightarrow V$ is explicitly given by $v\mapsto g\cdot v$; and that if $g$ is a primitive element then $\chi(g)=Trace(T_g)$.
For a detailed presentation of the above, you can see ch. VII, sect. 7.1, p.146-153, from Sweedler's book on Hopf algebras.
Furthermore, you can also take a look at Larson's paper on Characters of Hopf algebras. However, the presentation there looks quite different:
Larson adopts a dual point of view (to the usual notion of characters in group/algebra representation theory) and develops a theory of characters based on comodules of Hopf algebras. He actually considers characters as elements of the hopf algebra (instead of functionals on it) which are associated with comodules over the hopf algebra rather than modules over the hopf agebra. Furthermore, for the case of cosemisimple hopf algebras, an orthogonality relation for characters is proved.
Edit: Although i have not studied Larson's paper in detail, from what i can understand, i think that his approach is more general than Sweedler's approach (in the sense that it is not limited to the cocommutative case). In the cocommutative case, i think is essentially equivalent to the one followed in Sweedler's book; Sweedler's presentation can be recovered if we adopt Larson's approach and start from comodules of the finite dual $C^{\circ}$ hopf algebra.
Is it essential that $C$ be cocommutative? I am really interested in non-cocommutative examples.
Hmm good point. Regarding Sweddler's approach, yes it is essential. (Otherwise, the $EV$ action would not necessarily be well defined). However, in Larson's approach, the construction is given generally for cosemisimple hopf algebras.
I did some search and i think that you might find some interest in ch.3 of the thesis: https://core.ac.uk/download/pdf/48532654.pdf (in conjuction with Larson's paper mentioned above).
you are welcome
What does 'measuring' mean in "there is a unique measuring $E V \otimes V \to E V$"?
@LSpice, if $A$ is an algebra and $H$ is a bialgebra, and we have a bilinear map (not necessarily a $H$-action), $\triangleright: H \times A \to A$ satisfying $h \triangleright(ac) = (h_{1} \triangleright a)(h_{2}\triangleright c)$ and $h\triangleright 1_A=\varepsilon(h)1_A$, then we say that the bilinear map $\triangleright$ is a measuring or that $(\triangleright,H)$ measures $A$ to $A$.
The relation $EV\otimes V\rightarrow EV$ mentioned in your comment was a typo in my answer. I have edited and added the correct one (which is a measuring): $C\otimes EV\rightarrow EV$. . We can alternatively say that the action of $C$ measures $EV$. (See also the last paragraph in this: mathoverflow.net/a/284842/85967 answer).
[Sorry for what may be an ignorant question but...] Is there any relation between Larson's character and the Fredholm determinant? I know that the latter can be a trace and determinant connection (e.g. on MSE); but, I don't know if there is anything more to say in this direction.
@Benjamin Dickman, to speak the truth, i do not know. In fact i am not very familiar with Fredholm determinant. However, your remark seems very interesting to me. I will try to study a little and to think about it. Meanwhile, maybe it would be interesting to post this as a question (either here or on MSE).
I don't think I have the necessary background to parse an answer around their connection; so, I am not intending to post such a question. But please ping me if you ask any such thing on either site - now or in the future!
@Benjamin Dickman, ok i will! Thank you for your feedback.
Take an example of a finite dimensional Hopf algebra $A$, presented by generators and relations, generated by grouplikes and primitives. There are a lot of non-commutative non-cocommutative examples in the literature. Compute the group $G(A)$ of group like elements (from the presentation this should be very easy).
Now the example is $H=A^*$, the group likes in $A=A^{**}$ are the characters of $H$.
Trying to "rephrase" your description: if $A$ is a finite dimensional $k$-algebra and $A^$ its dual $k$-coalgebra, then $G(A^)=\mathcal{A}lg(A,k)$, i.e. the grouplikes of the dual coalgebra are exactly the algebra maps from $A$ to $k$, that is the characters of $A$.
And in the case that $A$ is a fin dim cocommutative bialgebra then Larson's character (mentioned in my post) is one of them.
|
2025-03-21T14:48:29.602622
| 2020-01-08T15:43:16 |
349998
|
{
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"Carlo Beenakker",
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|
Stack Exchange
|
Fourier transform derivation from Laurent series
Using Laurent Series of a function $f(z)$ around a point $a \in \mathbb{C}$
$$f(z) = \sum^{\infty}_{n=-\infty} c_n(z-a)^n \ \ \ \ (1)$$
where
$$c_n = \frac{1} {2\pi i}\int\limits_{\gamma}\frac {f(z)} {(z-a)^{n+1}} dz \ \ \ \ (2)$$
where $\gamma$ is a closed curve around $a$.
And choosing $\gamma$ such that $z$ can be parameterized as $z=a+e^{it}\ t \in [-\pi,\pi] $ in order to obtain Fourier Series
$$ f(t) = \sum^{\infty}_{n=-\infty} c_ne^{int} \ \ \ \ (3) $$
where
$$ c_n = \frac{1} {2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}dt \ \ \ \ (4) $$
we can "intuitively" prove Fourier Series by just using introductory complex calculus without any advanced mathematical background in Hilbert-Banach Spaces, Functional Analysis etc.
My question is, how can we derive other forms of Fourier Series (also called Fourier Transforms) such as Discrete Fourier Transform (DFT) and Continuous Time Fourier Transform (CTFT) by a similar manner? The problem here is that DFT equations are not closed contour integrals anymore, just discrete finite sums. I could not wrap my head around how Laurent Series may still work in this discrete case. For the CTFT case we have to let integral limits in $(4)$ to go infinity, which corresponds to infinitely many contour integrals in $(2)$. This seems to be divergent, if I am not terribly mistaken?
a continuous time Fourier transform is the limit of a Fourier series when
the period of the periodic function becomes infinitely large; is there more to say?
letting the "periodicity" of n to go infinity with redefining, say, n = n'/w and letting w to go infinity may change the nature of equation (4). Since int = in't/w after our new definition, letting w go to infinity will cause t/w to converge to 0 (remember, t is finite by definition). This means we do not describe a circular closed contour in (2), Laurent series will not work. If we change the definition of t to eliminate this issue by redefining t within an infinite interval, then we obtain a fraction of infinities t/w = inf/inf. This may yield anything.
Continuous time Fourier transform and Laurent series:
I recall equations (1) and (2), for convenience set $a=0$, and substitute $z=e^{it/T}$. The function $g(t)=f(e^{it/T})$, with $t\in(-\pi T,\pi T)$, is periodic with period $2\pi T$, given by the Laurent series
$$g(t)=\sum_{n=-\infty}^\infty c_n e^{int/T},$$
with coefficients
$$c_n=\frac{1}{2\pi T}\int_{-\pi T}^{\pi T}g(t)e^{-int/T}\,dt.$$
For $T\gg t$ the sum over $n$ may be approximated by an integral over $\omega=n/T$ with coefficients $C(\omega)=Tc_{n=\omega T}$, giving
$$g(t)=\int_{-\infty}^\infty C(\omega) e^{i\omega t}d\omega,$$
$$C(\omega) =\frac{1}{2\pi}\int_{-\infty}^\infty g(t)e^{-i\omega t}\,dt.$$
In this way the Fourier integral can be obtained as the limit of the Laurent series when the periodicity of the function tends to infinity. Notice that the exponent $nt/T$ cannot be set to zero because, even though $t/T\ll 1$, the product $nt/T=\omega t$ need not be small.
Discrete Fourier transform and Laurent series:
We now start from a discrete time signal $x_n$ and construct the Z-transform
$$X(z)=\sum_{n=-\infty}^\infty x_{n}z^{-n}=\sum_{n=-\infty}^\infty x_{-n}z^n.$$
This is a Laurent series centered at $a=0$, with inversion formula
$$x_{n}=\frac{1}{2\pi i}\oint X(z)z^{n-1}\,dz.$$
For the discrete Fourier transform one has only $N$ distinct values of $x_n$, the set $\{x_0,x_1,x_2,\ldots x_{N-1}\}$. We extend this set periodically by $x_{n+N}=x_n$. The Z-transform is a series of $N$ Dirac delta functions with coefficients $X_k$,
$$X(z=e^{2\pi iq/N})=\sum_{k=0}^{N-1}X_k\delta(q-k),\;\;X_k=\sum_{n=0}^{N-1}x_n e^{-2\pi i kn/N}.$$
The delta function converts the integral in the inversion formula into a sum,
$$x_n=\frac{1}{N}\sum_{k=0}^{N-1}X_k e^{2\pi ikn/N}.$$
n is a countable number, the sum has countably infinite elements in it. However an integral is performed on real numbers which are uncountably infinite. How can we do that transition from countably infinite numbers to uncountably infinite numbers ? Is it just an approximation ?
Isn’t that what we do when we approximate an integral by a Riemann sum? The approximation becomes more and more accurate as the discretization interval t/T becomes smaller and smaller.
That's right sir, thank you. The CTFT case seems to be closed. After the DFT case is closed I'll mark my question answered. What can we say about the DFT case ?
Thank you for your response. Isn't it z^(n-1) rather than x^(z-1) in the integral ??
I did the calculations myself and in the last identity in your answer i did not obtain 1/2π multiplier but i obtained 1/N instead. Am I mistaken ?
you are correct, $1/N$ it is.
Thank you for your help.
|
2025-03-21T14:48:29.602944
| 2020-01-08T15:48:38 |
350001
|
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"Matthew Levy",
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|
Stack Exchange
|
Braided monoidal categories
I know that it has been shown that $E_2$ algebra objects in Categories are simply braided monoidal categories.
In particular, Lurie says that an $E_2$-monoidal structure on the infinity-category $N(C)$ (the nerve of a category C) is a braided monoidal structure on C.
Is there a generalization (not sure if this is the right word) of this where instead of having braiding isomorphisms we have: $b_{V,W}:V\otimes W\to W\otimes V$ is a quasi-isomorphism?
Obviously, we need some dg-structure, so we would be talking about an $E_2$-monoidal structure on the dg-nerve of a dg-category.
What would the implications be for the composition $b_{W,V}\circ b_{V,W}$ when compared to the identity quasi-isomorphism $\mathrm{id}:V\otimes W\to V\otimes W$?
I am interested in possible new knot invariants.
Any thoughts or improvements on how I am asking the question are highly appreciated.
Any generalization will require you to choose a cellular model for the $E_2$ operad. I.e. there wont be anything as simple as what you suggest-- in this generality you need a lot more data than just the braiding maps. You could use the nerve of the groupoid of parenthesized braids, or Batanin's work (related to the category Theta_2).
What if I choose the brace operad. So there are 2 types of tensors (left/right multiplication and up/down multiplication). I realize the generality that I want requires a lot of data, I'm trying to see if it is managable in some sort of "small" model version of E2.
|
2025-03-21T14:48:29.603073
| 2020-01-08T15:49:21 |
350002
|
{
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"authors": [
"Connor Malin",
"Dmitri Pavlov",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/402"
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|
Stack Exchange
|
Generalization of familiar theorem about singular homology to general model category
I have two questions, the first one is just wether the following statement is true or not? Is there a reference for this?
The second question is maybe related, I don't know. But anyway, given $U:\Delta \rightarrow C$, we can construct
$$S:C \rightarrow \text{s}\textbf{Set}$$
given by
$$S(c)_n = \operatorname{Hom}(U_n,c)$$
Given a monomorphism $B \rightarrow Y$ in some category $D$, define $Y/B$ as the pushout of $*_D \leftarrow B \rightarrow Y $ where $*_D$ is the terminal object. My question is this: What restrictions do we have to put on $U$ so that $S(X)/S(A) \rightarrow S(X/A)$ given by the pushout diagram
(I made a typo, $S(X)$ and $S(A)$ in the top part of the diagram should change place)
is a weak equivalence of simplicial sets whenever $A \rightarrow X$ is a cofibration in $C$?
Giving a satisfactory answer would substantially generalize the result that whenever $A \rightarrow X$ is a cofibration of topological spaces, then the map $H_i(X,A) \rightarrow H_i(X/A,*)$ is an isomorphism.
The first claim is established like in Lemma 15.11.10 in Hirschhorn's Model Categories and Their Localizations (which proves a special case of cofibrant objects).
Some trivial thoughts: forgetting about the object $U$ for a second, a good class of examples where this happens just for a functor $S: C \rightarrow SSet$ is where (derived) $S$ preserves homotopy pushouts. This will happen when $S$ is a left Quillen functor. However, the motivating example of the singular set functor is not actually a left Quillen functor. It is happening because it is a (right) Quillen equivalence. So I feel like it might be rare that choosing some U actually results in homotopy pushouts being preserved.
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2025-03-21T14:48:29.603331
| 2020-01-08T15:58:10 |
350003
|
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"Dominic van der Zypen",
"Mathieu Baillif",
"Tri",
"https://mathoverflow.net/users/29491",
"https://mathoverflow.net/users/51389",
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|
Stack Exchange
|
Infinite connected $T_2$-space with fixed-cardinality fibers
What is an example of a connected $T_2$-space $(X,\tau)$ with $X$ infinite and the following property?
If $\alpha \leq |X|$ is a non-empty cardinal, then there is a continuous map $f:X\to X$ such that $|f^{-1}(\{y\})| = \alpha$ for all $y\in X$.
If the continuum hypothesis CH is true, then the circle is such a space. If CH is not true, then the circle is not such a space (a closed subset of the circle is either countable or has cardinality the continuum, any cardinal in between cannot be realized as the preimage of a point).
Wow that's amazing @MathieuBaillif - thanks!
What is the proof for a circle when alpha is infinite?
For preimages of continuum cardinality, take the projection of a space (cylinder, rather) filling curve. For countable preimages, a function as in the answer to this question: https://math.stackexchange.com/questions/1599921/countable-infinity-to-one-function/1600271#1600271
|
2025-03-21T14:48:29.603434
| 2020-01-08T16:28:37 |
350005
|
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|
Stack Exchange
|
Complexity of a proper class of extendibles
If consistent, is existence of a proper class of extendible cardinals provably equivalent to a $Σ^V_5$ statement?
Recall that in ZFC, a cardinal $κ$ is extendible iff for every $λ>κ$ there is an elementary embedding $j$ of $V_λ$ into some $V_μ$ with $\mathrm{crit}(j)=κ$. (In ZF, one also requires $j(k)>λ$, but this does not affect the above definition in ZFC.)
Extendibles are unusual in being a natural large cardinal property with high quantifier complexity (see this question for other examples of high quantifier complexity). The property of being some $V_μ$ is $Π^V_1$ (or $Π_1$ for short), being $λ$-extendible is $Σ_2$, and being extendible is $Π_3$, and so existence of an extendible cardinal is $Σ_4$, and existence of a proper class of extendibles is $Π_5$.
In the other direction, if $κ$ is extendible, then $V_κ ≺_{Σ_3} V$, and thus (assuming consistency) existence of an extendible cardinal is not equivalent in ZFC to a $Π_4$ statement. Analogously, I expect that existence of a proper class of extendibles is not provably (in ZFC) $Σ_5$, but it is easy to be confused by quantifier counting.
Offered in Q/A format as I was able to solve this question while working on its exposition.
Wait what, you figured out the answer before posting the question, but posted it anyway? Ok...
No, it is not; existence of a proper class of extendible cardinals is not provable in ZFC from any consistent $Σ^V_5$ statement.
Assume a proper class of extendibles and let $φ$ be a true $Σ_5$ statement (or even the conjunction of all true $Σ_5$ statements), and let a set $S$ witness $φ$, and let $κ$ be the least extendible above $S$. Then $V_κ$ satisfies $\mathrm{ZFC} + φ \, +$ non-existence of a proper class of extendibles. ($V_κ ≺_{Σ_3} V$, so $V_κ$ satisfies all true $Π_4$ statements with parameters in $V_κ$, and by $S∈V_κ$, $V_κ ⊨ φ$.)
$V_κ ≺_{Σ_3} V$ is a standard fact about extendibles (as an aside, despite its similarity, being supercompact is $Π_2$) and follows from existence of arbitrarily high $λ$ such that $V_κ ≺ V_λ$ (or just $V_κ ≺_{Σ_3} V_λ$): Let $λ$ with $V_κ ≺ V_λ$ be large enough. By Downward Löwenheim-Skolem, $V_λ ≺_{Σ_1} V$, so $V_λ$ satisfies all true $Π_2$ statements with parameters in $V_λ$, so (since $λ$ is large enough relative to $κ$) $V_λ$ (and thus $V_κ$) satisfies all true $Σ_3$ statements with parameters in $V_κ$, so $V_κ ≺_{Σ_3} V$.
|
2025-03-21T14:48:29.603595
| 2020-01-08T17:01:34 |
350009
|
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"Alexander Campbell",
"Tim Campion",
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|
Stack Exchange
|
When is an $\infty$-categorical localization left exact?
Let $L: \mathcal C^\to_\leftarrow L\mathcal C : i$ be an adjunction with $i$ fully faithful. In ordinary category theory, $L$ is left exact iff the class of $L$-local morphisms is stable under base change [1].
This appears to be true $\infty$-categorically, as well. Is there a proof in the literature?
But $\infty$-categorically, this is no longer true:
Example: Let $\mathcal C$ be the $\infty$-category of spaces and $L\mathcal C$ the full subcategory of $n$-truncated spaces for some fixed $n$, so that $L$ is the $n$-truncation functor and the $L$-local morphisms are those with $(n+1)$-connected fibers. Then $L$ is not left exact (failing to preserve, for example, the pullback square $K(\mathbb Z, n) \rightrightarrows \ast, \ast \rightrightarrows K(\mathbb Z, n+1)$), but the $L$-local morphisms are stable under base change.
Question: Let $L: \mathcal C^\to_\leftarrow L\mathcal C: i$ be an adjunction of finitely-complete $\infty$-categories with $i$ fully faithful. Let $\mathcal W = L^{-1}(\{\textrm{isos}\}) \subseteq \textrm{Mor} \mathcal C$ be the class of $L$-local morphisms. What are necessary and sufficient closure conditions on $\mathcal W$ ensuring that $L$ is left exact?
I'm happy to assume that $\mathcal C$ is presentable, or even an $\infty$-topos, and that $\mathcal W$ is of small generation.
[1] Here we assume that $\mathcal C$ is finitely complete. A morphism $f$ is said to be $L$-local if $L(f)$ is an isomorphism, and a class of morphisms $\mathcal W$ is stable under base change if $f \in \mathcal W$ implies $f' \in \mathcal W$ where $f'$ is any pullback of $f$ along an arbitrary morphism.
Is the class of morphisms of spaces inverted by $n$-truncation stable under pullback? Isn't your pullback square a counterexample?: $\ast \to K(\mathbb{Z},n+1)$ is inverted by $n$-truncation, but $K(\mathbb{Z},n) \to \ast$ isn't, right?
@AlexanderCampbell -- yes, I was getting confused. This criterion does hold $\infty$-categorically. I've just written up a proof over here.
Unless I misunderstand the statement, this is precisely proposition <IP_ADDRESS> in Higher Topos Theory.
|
2025-03-21T14:48:29.604013
| 2020-01-08T17:58:21 |
350011
|
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|
Stack Exchange
|
Is ETCS well-founded?
I can't find a statement about the axiom of regularity anywhere in treatments of ETCS. Perhaps this is due to the unfortunate clash of terminology with 'foundations'.
This should be relevant.
I suspect that ETCS is not well-founded in any reasonable sense. One particular way to phrase this might be the following: there are ill-founded models of ZFC - Foundation + $\neg$Foundation which "translate" in the appropriate sense to models of ETCS.
At the same time, it's not entirely clear to me what "foundation" would mean in the context of ETCS (where we're not basing everything on elementhood after all). In fact, I'd guess the following: if $M$ is a model of ETCS "generated" by a well-founded model $N$ of ZFC, there is a model of ZFC - Foundation + AFA which "generates" an isomorphic model. (Besides the obvious vagueness, I'm being a bit rude here in my mix of classical model-theoretic language and ETCS, but I think this is benign in this context.)
@NoahSchweber this is my gut feeling too. The more I look into this problem the more it becomes clear that ETCS is quite different of a beast from ZFC (at least model-wise). The question of well-foundedness then makes little sense, from a certain point of view. On the other hand, it is natural to ask how are the two theories related: in which sense ETCS sets are ZFC sets?
@MichaelGreinecker thanks, there are good remarks there. It is argued that ETCS might not be well-founded after all.
Does the pope poop in the woods?
@AsafKaragila tell us your thoughts.
I mean, the Vatican is in the middle of Rome...
As Noah suggests, the axiom of regularity doesn't make sense in the language of ETCS. What we can say is:
To construct a model of ETCS+R from a model of ZFC, one doesn't need regularity.
From a model of ETCS+R, one can construct a model of ZFC with regularity, and also separately models of ZFC with regularity replaced by various anti-foundation axioms.
The composite ETCS+R -> ZFC -> ETCS+R is the identity, as are the ill-founded versions. And the composite ZFC -> ETCS+R -> ZFC is also the identity, as are ill-founded versions whose anti-foundation axiom is sufficiently strong to characterize possible set-membership diagrams structurally.
Among other places, more details can be found in this paper of mine.
What is R here? Replacement?
@NoahSchweber Yes, R denotes any of the equivalent replacement/collection axiom schemas that can be added to ETCS.
Is there a “canonical” way of interpreting ZFC in ETCS, about which one may ask whether one ends up with a well-founded model?
@MonroeEskew Since "ZFC" by definition includes the regularity axiom, I would argue that the construction of a model with regularity is the canonical way of interpreting ZFC in ETCS. And there's similarly a canonical way of interpreting ZFC-regularity+AFA in ETCS. But no, there's no way of constructing such an interpretation without deciding in advance whether you want it to satisfy regularity or not, and it'll basically always be possible to do both -- ETCS doesn't "know" anything about regularity.
Another way to think about it is that ZFC and ZFC-regularity+AFA (AFA = anti-foundation axiom) are essentially equivalent -- one can construct a model of either one from the other -- and passing back and forth between them doesn't change the corresponding model of ETCS.
|
2025-03-21T14:48:29.604280
| 2020-01-08T18:17:32 |
350014
|
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}
|
Stack Exchange
|
The expressiveness of functions computable on trees
Motivation:
Let's define a function computable on a $k$-ary tree as a function composed with simpler computable functions defined at each node such that a function of this kind defined on a binary tree of depth $N$ receives $2^{N-1}$ inputs:
While thinking about the expressiveness of functions $F_k^N$ computable on $k$-ary trees it occurred to me that the 'combinatorial power' of such functions must grow exponentially as a function of tree depth. This would be very nice because it's natural to execute parallel algorithms with $O(N)$ time complexity on $k$-ary trees with depth $N$.
Intuitively, if there are $k^n$ functions at the nth level the expressiveness for $F_k^N$ on a tree of depth $N$ should grow on the order of:
\begin{equation}
\sim k^N \tag{1}
\end{equation}
Might there be a theorem which expresses this idea using algorithmic information theory(i.e. Kolmogorov Complexity)?
Probabilistic argument using Kolmogorov Complexity:
If $F_k^N$ is a composition of functions in $S$ where $\lvert S \rvert = \frac{k^{N}-1}{k-1}$ and $K(\cdot)$ denotes Kolmogorov Complexity then we may define:
\begin{equation}
Q = \min_{f_i \in S} K(f_i) \tag{2}
\end{equation}
and we may show that for almost all $F_k^N$ we must have:
\begin{equation}
K(F_k^N) \geq \frac{Q}{2} \cdot k^{N-1} \tag{3}
\end{equation}
Proof:
Let's suppose each $f_i \in S$ has an encoding as a binary string so $\forall i, f_i \in \{0,1\}^*$. If we compress each $f_i$ then $F_k^N$ is reduced to a program of length greater than:
\begin{equation}
n= Qk^{N -1} \tag{4}
\end{equation}
Now, the number of programs of length less than or equal to $\frac{n}{2}$ is given by:
\begin{equation}
\sum_{l=1}^{\frac{n}{2}} 2^l \leq 2^{\frac{n}{2}+1}-1 \tag{5}
\end{equation}
and so, using the principle of maximum entropy(i.e. uniform distribution) [4], we find that:
\begin{equation}
\lim_{n \to \infty} P(K(F_k^N) \geq \frac{n}{2}) \geq \lim_{n \to \infty} 1 - \frac{2^{\frac{n}{2}}}{2^n} = 1 \tag{6}
\end{equation}
Question:
My question is whether this result (3) in the setting of algorithmic information theory may be significantly improved upon and whether there might be other interesting results of this kind.
References:
Vladimir I Arnold. Representation of continuous functions of three variables by the superposition of continuous functions of two variables. Collected Works: Representations of Functions, Celestial Mechanics
and KAM Theory, 1957–1965, pages 47–133, 2009.
Roozbeh Farhoodi, Khashayar Filom, Ilenna Simone Jones, Konrad Paul Kording. On functions computed on trees. Arxiv. 2019.
M. Li and P. Vitányi. An Introduction to Kolmogorov Complexity and Its Applications. Graduate Texts in Computer Science. Springer, New York, second edition, 1997.
Edwin Jaynes. Information Theory and Statistical Mechanics. The Physical Review. Vol. 106. No 4. 620-630. May 15, 1957.
Where does your ${N\choose 2}$ term come from? That's not the count of either internal, external or total nodes of the tree, and I'm not sure what its origin is here.
@StevenStadnicki Thanks for pointing that out. This was a counting error.
I wonder whether it would be better to use the existing tag ([tag:computability-theory]) rather than creating a new tag ([tag:computable-functions]).
|
2025-03-21T14:48:29.604516
| 2020-01-08T18:32:55 |
350015
|
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"url": "https://mathoverflow.net/questions/350015"
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|
Stack Exchange
|
Is the property of being a quasi-affine morphism checkable on closed points?
Let $f:X\rightarrow Y$ be a morphism of finitely generated schemes over a $\mathbb{C}$, and $L$ be a line bundle on $X$ which is ample on all closed-fibers. I.e. for every closed point $y\in Y$, the bundle $L_y$ on $X_y$ is ample.
Does it follow that $L$ is relatively ample over $f$?
Note that I am not assuming that $f$ is a proper map, otherwise the above is true and can be proved by the theorem on formal functions.
A very related (though not exactly equivalent) question: can you chech the property of being a quasi-affine morphism on closed points?
Thank you!
Asked and answered in https://mathoverflow.net/questions/27972/relatively-ample-line-bundles.
|
2025-03-21T14:48:29.604603
| 2020-01-08T18:56:52 |
350018
|
{
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"url": "https://mathoverflow.net/questions/350018"
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|
Stack Exchange
|
Künneth theorem for G-space
Let $X$ and $Y$ be a right $G$-space and a left $G$ space, respectively, where $G=H \rtimes K$, $H$ a finite group and $K$ a compact Lie group.
Moreover, suppose that the $G$-action on $X$ is free.
Denote $H_*(-)$ the singular homology with coefficient in a field.
Is $H_*(X \times_G Y )$ isomorphic to $H_*(X) \otimes_{H_*(G)} H_*(Y)$?
Is it possible to prove it without using spectral sequences?
Crossposted on MSE.
This is very untrue. Taking $X=EG$, a contractible space with a free $G$-action, and $Y$ to be a point we have $H_*(X\times_G Y)=H_*(BG)$, which may be non-trivial in arbitrarily high degrees, while $H_*(X)\otimes_{H_*(G)} H_*(Y)$ is the homology of a point.
Thanks for the answer! What hypotheses could I add to get the result? And what if it is not true? In this case, what can I say about $H_*(X \times_G Y) $
The standard way to be compute $H_(X\times_G Y)$ when $G$ acts freely on $X$ would be to use the Serre spectral sequence of the projection $X\times_G Y\to X/G$, which is a fibre bundle with fibre $Y$ (at least when $G$ is compact). I know you said you wanted to avoid this, but in some cases (such as when $X=EG$ and you know how $G$ acts on $H_(Y)$, as in my answer) the computations can be quite tractable.
@user740379 There is one case that's simpler. If $H_\ast(X)$ or $H_\ast(Y)$ are free over $H_\ast(G)$, you do get the result desired. In addition to the useful bundle-theoretic technique Mark Grant mentions, there is also something called the Rothenberg-Steenrod spectral sequence for calculating $H_*(X \times_G Y)$, specifically measuring the deviation in terms of Tor groups. (Sorry, I know that's less useful because you've specifically asked to avoid spectral sequence techniques.)
|
2025-03-21T14:48:29.604754
| 2020-01-08T20:40:29 |
350025
|
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|
Stack Exchange
|
A weaker version of strongly graded algebras
Let $A = \oplus_{i \in \mathbb{Z}} A_i$ be a graded algebra. We say that it is strongly graded if $A_i.A_j = A_{i+j}$, for all $i,j \in \mathbb{Z}$. Can there be existing a graded algebra such that
$$
A_i.A_{-i} = A_0, ~~ \forall i \in \mathbb{Z},
$$
but which is yet not strongly graded?
If $A_i \cdot A_{-i} = A_0$ then in particular we can write
$$
1 = \sum_k a_i^{(k)} \cdot a_{-i}^{(k)},
$$
where $a_l^{(k)} \in A_l$. Now taking any $a_{i+j} \in A_{i+j}$ and multiplying it by this equality, we obtain
$$
a_{i+j} = \sum_k a_i^{(k)} \cdot (a_{-i}^{(k)} \cdot a_{i+j}) \in A_i \cdot A_j.
$$
|
2025-03-21T14:48:29.604828
| 2020-01-08T20:47:34 |
350026
|
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"url": "https://mathoverflow.net/questions/350026"
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|
Stack Exchange
|
Status of questions in "Group Actions on $\mathbb{R}$-trees"?
Culler and Morgan's "Group Actions on $\mathbb{R}$-trees" lists four questions at the end of the introduction. A few have been famously resolved by work of Rips, Bestvina–Feighn and others.
I'm curious about the status of the remaining questions?
Let me briefly recall: an $\mathbb{R}$-tree is a metric space where any two points are connected by a unique arc, which is isometric to a closed interval in $\mathbb{R}$. An action of a group $G$ on $T$ by isometries without global fixed point yields a translation length function, which I will think of as a point in the projective space $P(\mathbb{R}^\Omega)$, where $\Omega$ is the set of conjugacy classes of elements of $G$. The set of all such functions is $PLF(G) \subset P(\mathbb{R}^\Omega)$, and the subspace consisting of "small" actions (the precise definition is maybe not relevant to the question) is $SLF(G)$.
Here are the questions.
Is every pseudo-length function a translation length function?
Culler and Morgan show that translation length functions satisfy certain axioms. The question is whether these axioms completely characterize translation length functions. Edit: The axioms characterize $\Lambda$-trees more generally; the paper is "Axioms for Translation Length Functions" by Walter Parry.
Are the simplicial actions dense in $PLF(G)$ or in $SLF(G)$? If so this would give a structure theorem for those groups whose space of conjugacy classes of discrete and faithful representations into $SO(n,1)$ is non-compact. Such groups would decompose non-trivially as free products with amalgamation, or as HNN-extensions, along virtually abelian subgroups.
My understanding is that both parts of this question are resolved in the affirmative for stable actions of finitely-presented groups, (which includes actions of discrete subgroups of $SO(n,1)$) due to Rips and Bestvina-Feighn, building on work of Morgan, Shalen, Bestvina and Paulin.
What is the topology of $PLF(G)$ and $SLF(G)$? For example, when are these spaces infinite-dimensional?
Which finitely-generated groups act freely on $\mathbb{R}$-trees? Peter Shalen asks whether such a group must be a free product of abelian groups and surface groups.
My understanding is that this is also answered in the affirmative by Rips.
2 is open for finitely presented groups and general actions; 4 is proven by Rips.
In his 1987 article "Dendrology of groups: an introduction", Shalen also attributes 1 to the unpublished preprint "Pseudo-length functions on groups" by W. Parry. Presumably the Parry in question is Walter Parry, who has published extensively in the area.
@HJRW Ahh, thanks for the prompting to look at Walter Parry's papers more closely! The paper's title had changed.
As for 3: Cohen and Lustig prove in "Very small group actions on R-trees and Dehn twist automorphisms" that if $G=A*B$ is a free product of groups such that $B$ contains an element of infinite order, then the subspace of simplicial actions in $SLF(G)$ is infinite dimensional.
|
2025-03-21T14:48:29.605060
| 2020-01-08T21:38:12 |
350029
|
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|
Stack Exchange
|
Proximity in terms of characteristic functions for $n$-dimensional distributions
Let $X\in \mathbb{R}^n$ and $Y\in \mathbb{R}^n$ be random variables with characteristic functions $\phi_X(t)$ and $\phi_Y(t)$, respectively.
Suppose that
\begin{align}
\sup_{t \in \mathbb{R}^n} \frac{|\phi_X(t)-\phi_Y(t)|}{\|t\|} \le \epsilon. \quad (*)
\end{align}
Question: Can we say something about how close the distributions of $X$ and $Y$ are in some metric over probability spaces? In other words, suppose that $(*)$ small what other distance would be small too.
For example, for $n=1$. There exists the following inequality
\begin{align}
L^2(P_X,P_Y) \le 2 \sup_{t \ge 0} \frac{|\phi_X(t)-\phi_Y(t)|}{t}
\end{align}
where $L$ is the L\'evy distance.
However, I was not able to locate similar inequality for $n>1$. One distance that I have taken a look at extensively is the Levy-Prokhorov metric. However, whenever I find anything it always depends on the derivatives of characteristic functions.
There is a good reason why you cannot get anything for the standard Levy-Prokhorov distance in high dimensions. Let's consider the uniform distribution on the sphere of radius $R$ in $\mathbb R^3$ and the mixture (with weight $\frac 12$ for each) of the uniform distributions on the spheres of radii $R-r$ and $R+r$ where $R>2r$. Then the difference of the characteristic functions is $|F(R|t|)-\frac 12[F((R-r)|t|)+F((R+r)|t|)]|\le r^2|t|^2\max_{[(R-r)|t|,(R+r)|t|]}|F''|$ where $F(u)=\frac{\sin u}{u}$. However, $|F''(u)|\le \frac Cu$ for $u>0$, so to have your condition, it suffices to ensure that
$$
\frac {Cr^2|t|^2}{R|t|/2}\le\varepsilon |t|,
$$
i.e. $2Cr^2/R\le\varepsilon$, which still allows $r$ to grow without bound as $R\to+\infty$ for any fixed $\varepsilon>0$.
So, you'll have to either settle for some cruder distance that gives you less control at infinity, or find a way to get that control by some alternative means from other assumptions you may have in your problem.
Ok. Thanks. This is a lot more clear to me now. Are you aware of an alternative distance that would work?
@Boby I would look at it from the other end: what is the weakest distance you would be happy with?
Something that matrizes the weak convergence. Or is that too much to ask for?
@Boby If I understand it right, the weak convergence of probability measures (over the class of continuous bounded functions) is metrized by anything that ensures that the characteristic functions converge pointwise, so that is rather "too little" to ask for.
I don't know many probabilistic distance in high dimensions. Do you have any suggestion on which distance is small if $ \sup_{t \in \mathbb{R}^n} \frac{| \phi_X(t) - \phi_Y(t)|}{|t|}$ is small?
I know this question that I asked some time ago, but I wanted to follow and see if you can suggest any distance?
|
2025-03-21T14:48:29.605263
| 2020-01-09T00:46:13 |
350037
|
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|
Stack Exchange
|
Why are homeomorphism groups important?
For a compact metric space $X$ let $\mathcal H(X)$ denote the set of homeomorphisms in the compact-open topology (also generated by sup metric). It is known that $\mathcal H(X)$ is a Polish topological group under the composition operation. Many of the classical results in this area concern the homeomorphism type of $\mathcal H(X)$, for instance:
(0) $\mathcal H(2^\omega)\simeq \mathbb R \setminus \mathbb Q$;
(1) $\mathcal H_\partial([0,1])\simeq \ell^2$ (here $\mathcal H_\partial$ is the space of homeomorphisms which fix the boundary pointwise);
(2) $\mathcal H_\partial([0,1] ^2)\simeq \ell^2$;
(3) $\mathcal H_\partial([0,1]^n)$ is a mystery for $3\leq n<\omega$;
(4) $\mathcal H([0,1]^\omega)\simeq \ell^2$;
(5) $\mathcal H(\text{Sierpinski carpet})$ and $\mathcal H(\text{Menger curve})$ have dimension $1$. It is conjectured that these homeomorphism groups and others are homeomorphic to the $\omega$-power of $\{x\in \ell^2:x_i\notin \mathbb Q\text{ for all }i<\omega\}$;
(6) $\mathcal H(\text{Pseudo-arc})$ contains no continuum. It is unknown whether this group is zero-dimensional, connected, or something in-between.
(7) It is unknown if there is a compact space $X$ with $1<\dim(\mathcal H(X))<\infty$.
Question 1. What are some other major results and questions along these lines?
Question 2. I suppose the dimension/connectedness of $\mathcal H(X)$ says something about generalized homotopy between homeomorphisms. But specifically why is knowing the homeomorphism type of $\mathcal H(X)$ important/useful? Also are there applications and/or interpretations of $\mathcal H(X)$ which lead to a greater appreciation?
Your claim that the homeomorphism group of $\mathbf{R}^2$ is connected sounds suspicious. I'd bet it has at least 2 connected components. Also, most of your examples are with $X$ locally compact, not always compact.
@YCor I will double-check my sources. It may be that the spaces should be products of $[0,1]$ and not $\mathbb R$.
What is $\ell^2$?
@YCor Indeed $\mathcal H(\Bbb R^2)$ is homotopy equivalent to $O(2)$, combining a result of Smale with some fiber-sequence fiddling. Perhaps OP meant the group of homeomorphisms of $[0,1]^2$ which fix the boundary pointwise; Smale's result is that this is contractible.
@ARG Hilbert space (equivalent to $\mathbb R ^\omega$).
@MikeMiller That is correct, I will edit original post.
My memory had briefly failed me: Smale's result is on the diffeomorphism group. That the homeomorphism group is contractible is older and is called the Alexander trick.
Often knowing the homeomorphism type is a test case for characterisation theorems: we know how to recognise $\ell^2$, or the irrationals, or irrational Erdös space, by checking a list of properties. Maybe we can find ‘new types” or new characterisation theorems by studying these homeomorphism groups.
@HennoBrandsma Can you expound on this or give an example? How would studying homeomorphism groups lead to topological characterization theorems?
I think homeomorphism groups of compact metric spaces are important, but especially as groups, or topological groups, rather than just topological spaces.
This comment is not quite on topic because it is not "along the lines" of the listed properties, as requested in Question 1. Nonetheless one has to note that $\mathcal H(X)$ can be important because of its quotient modulo the component of the identity, called the mapping class group of $X$. This does wash out huge swaths of interesting local topological structure, and in fact the quotient group becomes discrete in many cases. But the mapping class group of $X$ is very important in certain cases, particularly when $X$ is a surface.
Of course the automorphism group of any mathematical object is interesting if the object itself is interesting. But that doesn't single out the homeomorphism group . Let me give some examples.
Let $X$ be a space and $H=H(X)$ the homeomorphism group.
The homeomorphism group is a space. Knowing its homotopy type is much weaker than knowing its homeomorphism type, but still can give interesting results. Let me give two examples.
Fiber bundles over the sphere $S^n$ with fiber $X$ are in 1 to 1 correspondence with elements of $\pi_{n-1}(H)$. To classify the fiber bundles of any reasonable space $Y$, you need to understand all homotopy classes $[Y,BH]$, where $BH$ is the classifying space of $H$. Sometimes it is possible to partially compute these sets. This allows you to construct new bundles, with interesting properties.
Here is another nice theorem of Reeb, which depends on the fact that the group $H_\partial(D^n)$ is connected.
Let $f:M\rightarrow \mathbb{R}$ be a smooth function on a smooth compact manifold $M$. Suppose that $f$ has only two critical points. Then $M$ is homeomorphic to a sphere.
This theorem is false if we want to construct a diffeomorphism. Indeed, the group of diffeomorphisms $\mathrm{Diff}_{\partial}(D^n)$ is not always connected. Milnor famously found the first examples of manifolds that are homeomorphic but not diffeomorphic.
|
2025-03-21T14:48:29.605719
| 2020-01-09T01:42:31 |
350038
|
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|
Stack Exchange
|
Minimize sum of unique elements in columns of a matrix by moving entries within rows
Suppose we have a matrix with $R$ rows and $C$ columns.
1.An element appears no more than once in the same row.
2.The set of elements in each row has intersection.
3.Elements in the matrix can only be moved in the same row.
Objective:
Through the movements of the elements in the same row, get the minimum sum of the numbers of unique elements in each column.
$\min(\sum_{i=1}^Cunique(column_i))$
Example:
$$\begin
{bmatrix}
{a}&{d}&{f}&{b}\\
{d}&{e}&{c}&{g}\\
{b}&{a}&{e}&{s}\\
{s}&{b}&{a}&{h}\\
\end{bmatrix}$$
Objective = 4+4+4+4=16
After movements:
$$\begin
{bmatrix}
{a}&{b}&{d}&{f}\\
{c}&{g}&{d}&{e}\\
{a}&{b}&{s}&{e}\\
{a}&{b}&{s}&{f}\\
\end{bmatrix}$$
Objective = 2+2+2+2 = 8
This is a example only with 4 columns and 4 rows. What about a matrix with many columns and rows? How to find the optimal movements policy and prove it?
Thank you!
The title is very uninformative.
I edited the title, but am not sure to understand the meaning of "minimum sum of the numbers of unique elements in each column" In a column, one sums all entries occurring only once?
You can solve the problem via integer linear programming as follows. For $i\in\{1,\dots,R\}$, let $S_i$ be the set of $C$ elements in row $i$. Let binary decision variable $x_{i,s,j}$ indicate whether row $i$ element $s$ is assigned to column $j$, and let binary decision variable $y_{s,j}$ indicate whether element $s$ appears at least once in column $j$. The problem is to minimize $\sum_{s,j} y_{s,j}$ subject to linear constraints:
\begin{align}
\sum_{j=1}^C x_{i,s,j} &=1\\
\sum_{s\in S_i} x_{i,s,j} &=1\\
x_{i,s,j} &\le y_{s,j}
\end{align}
|
2025-03-21T14:48:29.605848
| 2020-01-09T02:09:19 |
350039
|
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|
Stack Exchange
|
Linear formal series in positive characteristic
Let $K$ be an infinite field positive characteristic and $F(X,Y)\in K[[X,Y]]$. Assume that $F(Z_1+U_1,Z_2+U_2)=F(Z_1,Z_2)+F(U_1,U_2)$ where $Z_1,Z_2,U_1,U_2$ are four indeterminates. Can one assert that $F(X,Y)=G(X)+H(Y)$ with $G,H\in K[[X]]$?
Yes. It suffices to consider the case that $F$ is a homogeneous polynomial. Write $F(X,Y) = aX^d + Y^eP(X,Y)$, where $d = \deg(F)$ and $P$ is homogeneous of degree $d - e$, $e \geq 1$.
Claim: if $a \neq 0$, then $d$ is a power of the characteristic $p$ of $K$.
Proof: Otherwise the coefficient of $Z_1^{d-1}U_1$ in $F(Z_1+U_1, Z_2+U_2)$ is nonzero, whereas this coefficient is zero in $F(Z_1,Z_2) + F(U_1, U_2)$.
Now write $F = aX^d + bY^d + Q(X,Y)$, where both $X$ and $Y$ divide $Q$. We will show that $Q = 0$. Indeed, by the claim, $Q$ also satisfies: $Q(Z_1+U_1, Z_2+U_2) = Q(Z_1,Z_2) + Q(U_1, U_2)$. Plugging in $Z_2 = -U_2$ in the preceding identity yields that $Q(Z_1,-U_2) + Q(U_1, -U_2) = Q(Z_1+U_1, 0) = 0$. If $Q$ is a nonzero polynomial, this gives a contradiction.
Thanks for the answer. But I do not see why it sufficient to consider the homogenous case. Can you explain?
if $F(Z_1 + U_1, Z_2 + U_2) = F(Z_1, Z_2) + F(U_1, U_2)$, then for each $d$, the homogeneous components of degree $d$ of LHS and RHS must be equal.
|
2025-03-21T14:48:29.605967
| 2020-01-09T02:34:27 |
350040
|
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|
Stack Exchange
|
Mathematical meaning for the (continuous) Sine-Gordon transformation
I've been trying to understand the so-called Sine-Gordon Transformation which occurs in both classical and quantum statistical mechanics. One of the most cited references on this topic seems to be Fröhlich's article, which is my main reference at the moment. We consider a function $V: \mathbb{R}^{n}\times \mathbb{R}^{n}$ which is continuously differentiable, satisfies $\sup_{x,y \in \mathbb{R}^{n}}|V(x,y)| \le K$ and
$$ \langle f,Vg \rangle := \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f(x)V(x,y)g(y)ddy \ge 0 \tag{1} $$
for every $f,g \in L^{2}(\mathbb{R}^{n})$. If we define $B: \mathcal{S}(\mathbb{R}^{n})\times \mathcal{S}(\mathbb{R}^{n})$ to be $B(f,g) \equiv \langle f, Vg\rangle$, the associate quadratic form $f \mapsto B(f,f)$ is non-negative, so that, by Minlos' Theorem, there exists some (Gaussian) measure $\mu_{V}$ on $\mathcal{S}'(\mathbb{R}^{n})$ such that
$$ W(f) := e^{-\frac{1}{2}B(f,f)} = \int_{\mathcal{S}'(\mathbb{R}^{n})}d\mu_{V}(T)e^{iT(f)}$$
Because $\mathcal{S}(\mathbb{R}^{n})\subset \mathcal{S}'(\mathbb{R}^{n})$, each $f \in \mathcal{S}(\mathbb{R}^{n})$ induces a distribution in $\mathcal{S}'(\mathbb{R}^{n})$. Thus, if we fix $\epsilon_{1},...,\epsilon_{N}\in \mathbb{R}$ and $x_{1},...,x_{N}\in \mathbb{R}^{n}$, we can choose sequences $\{f_{l}^{(j)}\}_{l\in \mathbb{N}}$ such that $f_{l}^{(j)} \to \epsilon_{j}\delta_{x_{j}}$, for each $j=1,...,N$. Fröhlich proves that
$$\lim_{l\to \infty}\int_{\mathcal{S}'(\mathbb{R}^{n})}d\mu_{V}(T)\prod_{j=1}^{N}:e^{iT(f_{l}^{(j)})}:_{V} = e^{-\sum_{1\le i< j \le N}\epsilon_{i}\epsilon_{j}V(x_{i},x_{j})}$$
where $:e^{iT(f)}:_{V} := e^{iT(f)}e^{\frac{1}{2}B(f,f)}$. Everything looks fine till now. Fröhlich introduces the notation:
$$lim_{l\to \infty}\int_{\mathcal{S}'(\mathbb{R}^{n})}d\mu_{V}(T)\prod_{j=1}^{N}:e^{iT(f_{l}^{(j)})}:_{V} \equiv \bigg{\langle}\prod_{j=1}^{N}:e^{i\epsilon_{j}T(x_{j})}:_V\bigg{\rangle}_{V} \tag{2} $$
The right hand side of (2) does not make sense, once $T$ cannot be evaluated pointwise. However, this seems to be defined only as a notation, which is fine to me. The problem is that the Sine-Gordon transformation, which follows these calculations, is obtained by means of manipulations of (2). For instance, one can write the partition function of the system as a Gaussian integral with respect to the measure $\mu_{V}$, but this seems to be ill-defined to me.
Question: Is it possible to give mathematical meaning to this Sine-Gordon transformation? Am I missing something? Or this version of the Sine-Gordon transformation can only be defined in a formal way? What is the purpose of defining the partition function in terms of a Gaussian measure if it is formal and does not have mathematical meaning?
EDIT: I should clarify what I mean by 'Sine-Gordon Transformation' once Fröhlich does not use this term in his article. The Sine-Gordon Transformation is equation (2.24) in Fröhlich's work, which is a way of writing the partition function $\Xi_{V}(z)$ in terms of a Gaussian measure.
I'm not sure what exactly the "Sine-Gordon transformation" is (Frölich's article doesn't use that terminology), but I guess your question is specifically about the meaning of the symbol ${:} e^{i \epsilon T(x)} {:}_V$, when not inside the expectation value $\langle - \rangle_V$, and what algebraic manipulations are allowed with it.
Ideally, it would be an element of the algebra of functions on $\mathcal{S}'(\mathbb{R})$, so that its expectation value $\langle - \rangle_V$ could be taken by integrating against the Gaussian measure $d\mu_V$. Of course, as you point out, that is not possible. But it is very easy to take the algebra of functions on $\mathcal{S}'(\mathbb{R})$ and extend it to formal series in the new formal variables ${:} T^k(x) {:}_V$, for various $x$ and $k$. Then $${:} e^{i\epsilon T(x)} {:}_V := \sum_{k=0}^\infty \frac{(i\epsilon)^k}{k!} {:} T^k(x) {:}_V$$ is a well-defined element of this extended algebra, which can be thought of as a generating function for the elements ${:} T^k(x) {:}_V$. So expressions involving ${:} e^{i\epsilon T(x)} {:}_V$ can be manipulated using the usual algebraic operations. The expectation value $\langle - \rangle_V$ can be uniquely extended to the larger algebra by requiring that your (2) holds as an identity of corresponding generating functions, where the limit on the left-hand side has already been computed.
Strictly speaking, at this stage, applying $\langle - \rangle_V$ to an element of the formal series algebra only gives a formal series of real (or complex) numbers. But formal numerical series can also be manipulated using the usual algebraic operations. Moreover, if manipulating absolutely convergent series in such a way, the result is again absolutely convergent. And indeed, the left-hand side of (2) is given by an absolutely convergent series. So you can presume that Frölich implicitly sums all such formal numerical series. You will notice that only such (implicitly summed) expectation values are subject to inequalities in Frölich's article.
Frölich is also using smeared generating functions, which can be defined by $$\int dx\, g(x) {:} e^{i\epsilon T(x)} {:}_V := \sum_{k=0}^\infty \frac{(i\epsilon)^k}{k!} {:} T^k(g) {:}_V,$$ where ${:} T^k(g) {:}_V$ are yet new formal variables, whose expectation values are required to satisfy $$ \langle {:} T^k(g) {:}_V \rangle_V = \int dx \, g(x) \langle {:} T^k(x) {:}_V \rangle_V, $$ so I think you get the idea.
So, why use formal series? First of all, they absolutely have a precise and rigorous mathematical meaning, just as I have explained above. Second, the use of generating functions, even formal ones, can dramatically simplify and shorten various formulas. As an exercise, try to imagine how much more complicated the formulas in Frölich's article would have been without resorting to generating functions.
Thanks so much for your answer! It helped me a lot! Just to clarify: you new variables $:T^{k}(x):_{V}$ are, again, formal and do not have mathematical meaning right? But the formulas we get from it have meaning?
@Willy.K, essentially Yes. But I have a quibble with your terminology: formal variable is a mathematical meaning in itself. When one talks about the polynomial ring $\mathbb{R}[x]$, it's sure true that the variable $x$ is not by itself a real number, but I see no-one going around complaining that it has no mathematical meaning.
I see. I always treated formal power series as objects without mathematical meaning but I've been wrong this hole time and I now realize it! Do you know any good reference on this topic (formal power series)?
@Willy.K, formal power series are basic construction in algebra. There's not much more to the definition than meet eye. You can already find some useful general info on Wikipedia.
|
2025-03-21T14:48:29.606345
| 2020-01-09T03:27:15 |
350043
|
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|
Stack Exchange
|
Another application of Borel-Cantelli Lemma
I ask this question on math stackexchange, but there is no answer, so please forgive me I ask it here again.
Let $c>0$ and $P(x)$ be a polynomial. Then there exists a $p>1$ (e.g. we can take $p$ to be greater than the degree of $P$) such that for almost all $x=e^{2\pi i\theta},\theta\in[0,1]$ on the unit circle, $∣P(x^N)∣<\frac{c}{N^p}$ occurs only for finitely many $N\in \mathbb{N}$.
We can show this for $P(x)=x−1$. $|P(x^N)|=|x^N−1|=|e^{2\pi iN\theta}−1|=|e^{2\pi i(N\theta−m)}−1|∼|N\theta−m|$ where $m$ is the closest integer to $N\theta$. For any $p>1$, applying this result proved by Borel-Cantelli Lemma gives the result.
For general $P(x)$, we only need to factorize it over $\mathbb{C}$ and analyse the factors similarly.
I believe similar result also holds for multivariate polynomial $P(x_1,x_2,\dots,x_n)$. Namely, there exists a $p>1$ such that for almost all $x_1=e^{2\pi i\theta_1},\dots,x_n=e^{2\pi i\theta_n}$ on the $n$-dimentional torus, $∣P(x_1^N,\dots,x_n^N)∣<\frac{c}{N^p}$ occurs only for finitely many $N\in \mathbb{N}$.
Moreover I guess this holds for several variable analytic functions.
Could you prove this or give a reference please? Thanks.
Math.SE link: Another application of Borel-Cantelli Lemma.
|
2025-03-21T14:48:29.606449
| 2020-01-09T04:29:31 |
350044
|
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|
Stack Exchange
|
Degree of $\mathbb{Z}_p[\zeta_n]$
Milne’s note says that the degree of $\mathbb{Z}_p[\zeta_n]$, where $p$ is a prime, $p$ doesn’t divide $n$, $\zeta_n$ is the primitive n-th root of unity, is $f$, where $f$ is the smallest integer such that $n$ divides $p^f-1$. Can anyone tell me the reason?
Thank you very much!
That is also the degree of $\mathbf F_p(\overline{\zeta_n})/\mathbf F_p$ when $n$ is not divisible by $p$, and the $\mathbf Q_p$-conjugates of $\zeta_n$ are its $p$-th power iterates $\zeta_n, \zeta_n^p, \zeta_n^{p^2}, \ldots$ (this is from Hensel's lemma, since $x^n - 1$ is separable in characteristic $p$). In short, the situation in the $p$-adics can be regarded as a "lifting" of the situation mod $p$.
Sorry I don’t understand why the conjugates are its $p$-th power iterates?
Read about unramified extensions of $\mathbf Q_p$.
Ohh I see. Thanks!
|
2025-03-21T14:48:29.606540
| 2020-01-09T06:16:16 |
350047
|
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|
Stack Exchange
|
Completed cohomology and variants
I am interested in the following set up:
I have an ind-sequence of curves $\dots X_2\to X_1$ defined over a finite field of characteristic $p$ such that $X_n/X_{n-1}$ is a Galois degree $\ell$ cover and the galois group of the tower is $\mathbb Z_\ell$ for $\ell \neq p$.
I would like to study the variation in the cohomology $H^1_{et}(X_n,\mathbb Z_{\ell'})$ with $n$ for $\ell' \neq p$ (but possibly equal to $\ell$). This seems like exactly the set up of completed cohomology (for instance from here) except that continuous cohomology seems to be defined using singular cohomology.
I expect there should be a straightforward variant of completed cohomology using etale cohomology and the major theorems should hold in this case too. Is this true? Is this stuff written down anywhere?
And just in general, what would be the best place to start learning about completed cohomology?
There is no reason why you shouldn't consider completed etale cohomology, instead of completed singular cohomology. If you look at Emerton's 2006 Inventiones paper which started the whole theory, he allows etale cohomology as well, in order to get a Galois action on completed cohomology; and there is a huge industry of studying completed cohomology spaces as Galois modules (this is how one formulates and proves "local-global compatibility" in the p-adic Langlands programme).
Emerton doesn't consider towers of function fields, but it's totally clear that the definitions go over to that case. I suspect that shouldn't be too hard to establish the key finiteness property, that $\widetilde{H}^*$ is an admissible Banach space representation, in the function field case as well. (I don't know if this is written down in the literature explicitly, though.)
PS: As for where to start reading about completed cohomology: Emerton's 2006 paper is actually amazingly readable, miraculously so considering the density of completely new ideas in it. I got a lot out of reading it myself as a PhD student and I would strongly recommend starting with that as your reading material.
|
2025-03-21T14:48:29.606700
| 2020-01-09T06:37:10 |
350048
|
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"Terry Tao",
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|
Stack Exchange
|
frequence of block of digits in Mobius sequence
Let $\mu$ be the Mobius function from $\mathbb{N}$ to $\{-1, 0, 1\}$. It is well known for the frequency of $-1, 1$, and $0$ for the sequence $(\mu(1), \mu(2), \mu(2), \dots, )$.
For any $k\in \mathbb{N}$, it is natural to ask what is the frequency of any given block of $k$-digits in $\{-1, 0, 1\}^{k}$ . I do not know whether this is known in the literature.
Any comments and remarks will be appreciated.
Terry Tao has a blog post on this here
The Chowla conjecture asserts that all $k-$ blocks are equidistributed.
Matomaki, Radziwill and Tao (MRT) have shown that each of the sign patterns in
$\{-1,0,+1\}^k$
is attained by the Möbius function for a set ${n}$ of positive lower
density for $k\leq 4.$
What this means is that for all $(a_1,\ldots, a_k)\in \{-1,0,+1\}^k,$ if $k\leq 4,$
there is a subset $I$ of $\mathbb{N}$ such that there is some $\varepsilon>0,$
with $$\lim_{N\rightarrow \infty} \frac{I \cap \{1,2,\ldots,N\}}{N}>\varepsilon,$$
and
$$
\{(\mu(n),\mu(n+k-1)=(a_1,\ldots,a_k):\forall n \in I\},
$$
Edit: Edited to update the value of $k$ for which the MRT result holds.
The Chowla conjecture asserts uniform equidistribution for sign patterns of the Liouville function, but for the Mobius function the distribution is a bit more complicated due to local factors (e.g., Mobius vanishing at every multiple of 4). The result of MRT that you mention has now been extended to $k \leq 4$ (with the correct logarithmic density): see Corollary 1.10 of https://arxiv.org/abs/1708.02610 .
|
2025-03-21T14:48:29.606828
| 2020-01-09T06:43:08 |
350049
|
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|
Stack Exchange
|
Recover the field from its Milnor K-groups
For every field $F$, consider $K_n^M(F)$ the $n$-th Milnor K-group of $F$ for each $n \in \Bbb N$, and form the Milnor K-ring $K^M(F)=\oplus_{n \geq 0}K^M_n(F)$. For instance, $K_1(F)=F^{\times}$.
What information of $F$ can be recovered from the graded ring $K^M(F)$? Can it determine $F$ uniquely? For instance, we can consider $F$ among local fields or number fields.
Is there any example of two non-isomorphic fields $F_i$ such that $K^M(F_1) \cong K^M(F_2)$? There is no such example if $F_i$ are both finite or algebraically closed.
Motivations:
(1) Number fields can be determined by their absolute Galois groups.
(2)This paper proves that the function field of an algebraic variety
of dimension ≥ 2 over an algebraically closed field is completely
determined by its first and second Milnor K-groups;
(3) There are different $p$-adic fields with isomorphic $K_1$.
(4) Norm residue isomorphism theorem.
Very interesting question! First of all, what happens if we ask the same question with algebraic $K$-theory instead of Milnor $K$-theory? The only thing I can think of is that if $K(F_1) \cong K(F_2)$ in a way compatible with regulators, and we believe in the conjectures on the special values of $L$-functions, then $\zeta_{F_1} = \zeta_{F_2}$, hence $F_1$ and $F_2$ would be equivalent number fields.
|
2025-03-21T14:48:29.606940
| 2020-01-09T07:10:14 |
350050
|
{
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|
Stack Exchange
|
Finite quotients of surface braid groups
Let $\Sigma_b$ be a closed orientable surface of genus $b \geq 2$, and denote by $\mathsf{P}_2(\Sigma_b)$ the pure braid group with two strands on $\Sigma_b$.
There is a braid $A_{12} \in \Sigma_b$ corresponding to the classical braid generator of the disk, together with $2b$ further generators coming from the usual representation of $\Sigma_b$ as an identification space of a polygon with $4b$ sides, see for instance [B04].
Question. How can we construct finite quotients $\pi \colon \mathsf{P}_2(\Sigma_b) \to G$ such that $\pi(A_{12})$ is non-trivial?
In [CP19] it is remarked that some finite Heisenberg groups do the job, and similar proofs work for other extra-special $p$-groups.
Are there different examples? Is it possible to classify them in some way? For instance, what is the smallest order of such a quotient in terms of $b$?
Note that $A_{12}$ is a commutator in $\Sigma_b$, so our condition $\pi(A_{12}) \neq 1$ implies that $G$ must be non-abelian. For $b=0$, some results are contained in [CKLP19].
Edit. Notation changed in order to be consistent with the one in [CP19].
References.
[B04] P. Bellingeri: On presentations of surface braid groups, J. Algebra 274, No. 2, 543-563 (2004).
[CP19] A. Causin, F. Polizzi: Surface braid groups, finite Heisenberg covers and double Kodaira fibrations, arXiv:1905.03170.
[CKLP19] A. Chudnovsky, K. Kordek, Q. Li, C. Partin: Finite quotients of braid groups, arXiv:1910.07177.
Just as an aside, the smallest finite quotient for the full surface braid group on any number of strands (for $g>0$) is $\mathbb{Z}/(2)$; the “classical” generators collapse to an order two element in the abelianization.
Just to be sure: the group in question is $\mathbb{Z} \rtimes \pi_1(\Sigma_g)$ where the action is by monodromy, right? So shouldn't this be about determining the kernel of the mondromy action in $\pi_1(\Sigma_g)$?
@StefanWitzel: I am not sure of understanding your comment. The subgroup $\langle A_{12} \rangle$ is not normal in $\mathsf{P}2(\Sigma_b)$. In fact, from the presentation (that you can find in my paper with Causin, see p. 10) we have for instance, when $j > k$, $$\tau{2k} A_{12} \tau_{2k}^{-1} = A_{12} [\rho_{1j}, , \tau_{2k}].$$
In order to be consistent with the notation in [CP19], I am denoting by $A_{12}$ (instead of $\sigma$) the braid generator of the disk and by $\rho_{ij}$, $\tau_{ij}$ the geometric generators coming from the polygon.
Mapping class groups are residually finite, by Grossman's theorem, so there are lots of finite quotients like you want. Grossman proves her theorem in a way that shows that you can take the quotients to be congruence quotients, meaning that you can find finite quotients by taking characteristic subgroups $C$ of the fundamental group $\pi_1(S)$ and considering the map from the Mapping class group to the finite group $\mathrm{Out}(\pi_1(S)/C)$. Conjecturally these are essentially the only kind of quotients of the mapping class group.
|
2025-03-21T14:48:29.607159
| 2020-01-09T09:28:01 |
350056
|
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|
Stack Exchange
|
Continuous-time extension of a discrete dynamical system
It is clear that one can obtain a discrete dynamical system from a continuous one, but is the converse possible if the system is "nice"?
Define the discrete-time dynamical system on $\mathbb{R}^d$ by
$$
x_{n+1} = f(x_n);\, x_0\triangleq x
$$
where
$f \in C^2(\mathbb{R}^d;\mathbb{R}^d)$ and $x \in \mathbb{R}^d$.
Fix a (large) positive integer $N$, is there a function $F:\mathbb{R}^d\rightarrow \mathbb{R}^d$ such that the solution to the continuous-time dynamical system
$$
\partial_t X_t = F(X_t) ; \qquad X_0=x,
$$
and $X_n = x_n$ for every $n \in \left\{1,\dots,N\right\}$?
If you mean "for all $x$", then no, it is not possible. For every $F$, the Jacobian determinant of the corresponding flow is non-zero, and hence it remains positive. Thus, $f(x) = (-x_1,x_2,x_3,\ldots,x_d)$ provides a simple counter-example. On the other hand, if $x$ is fixed, then I do not see the answer right away.
@MateuszKwaśnicki No, $x$ should be specified (so the "anti-discritization" is dependent on the choice of x); It would only need to hold for an (arbitrarily large but fixed) number of iterations.
Then this is essentially a question whether there is a smooth simple curve passing through given $N$ points $x_1, \ldots, x_N$ in a given order. Or, more precisely, we require this curve to pass through $M$ points $x_1, \ldots, x_M$ in a given order and have no self-intersections except possibly at $x_M$, in which case it must be tangent to itself; here $M$ is the least number such that $x_M \in {x_1, \ldots, x_{M-1}}$, or $M = N$ if no such number exists. This seems to be fairly straightforward, no?
|
2025-03-21T14:48:29.607334
| 2020-01-09T10:27:06 |
350060
|
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|
Stack Exchange
|
Why are we interested in permutahedra, associahedra, cyclohedra, ...?
The following families of polytopes have received a lot of attention:
permutahedra,
associahedra,
cyclohedra,
...
My question is simple: Why?
As I understand, at least the latter two were initially constructed by their face lattice representing certain combinatorial objects (e.g. ways to insert parentheses into a string).
So I assumed that representing these structures as a face lattice was of some use.
But then people got interested in realizing these objects geometrically, and it turns out that, e.g. the associahedron can be realized in many ways.
Was this surprising?
Is there something to be learned from that fact?
On the other hand, for the permutahedron the realization came probably first, so is there anything deep to learn from its combinatorial structure?
Further, there seem to exist connections to algebra, e.g. homotopy theory. I cannot wrap my head around these connections.
For me, these polytopes are just further examples of polytopes, nothing else.
So what's up?
Do they have some extremal properties?
Are they especially symmetric (i.e. are they interesting for their symmetries)?
Does the geometric point of view make apparent some hidden combinatorial properties of the underlying structures (e.g. the cyclohedron is said to be "useful in studying knot invariants")?
What justifies this interest?
As an example, a recent result (https://arxiv.org/abs/1902.08059) uses the Loday realization of the associahedra to construct a cellular operad structure and diagonal on the associahedra. This allows one to endow the tensor product of two $A_\infty$-algebras with an $A_\infty$-structure.
Unrelated to my answer below, but an interesting fact about the regular permutohedron is that it tiles space: see https://en.wikipedia.org/wiki/Permutohedron#Tessellation_of_the_space.
Against the unlikely possibility of arXiv link rot, the paper referenced by @BertramArnold is Masuda, Thomas, Tonks, Vallette - The diagonal of the associahedra [sic].
I despair of writing down the majority of the diverse, beautiful connections of permutohedra and associahedra to algebra, analysis, combinatorics, and geometry/topology and am constantly surprised at how "surprised" other mathematicians have been in the past few years at first learning their diversity, given the existence of the OEIS entries related to the topics.
Philosophical questions deserve philosophical answers, so I am afraid no amount of references and specific results will probably satisfy you. Let me try to explain it in a somewhat generic way.
Think about it this way - why care about sequences like $\{n!\}$, Fibonacci or Catalan numbers? The honest answer is "because they come up all the time". Now, once you know these sequences, you may want to understand the underlying structures (permutations, trees, Dyck paths, triangulations, etc.) You may then want to understand connections between structures (e.g. bijections), algebraic or geometric interpretations (e.g. group representations, volumes of polytopes), etc. Once you have developed some kind of structures you may want to understand the relations between different structures, whether your bijections are structure-preserving, etc. That's how you develop the theory starting with just numbers!
In general, basic objects in combinatorics tend to lack structure. Adding structures is always welcome as they present a deeper understanding of the underlying objects (and sometimes even just numbers). It's what allows to employ and further develop tools from other parts of Combinatorics and other fields. This is the setup in which one can understand results such as Kuperberg's proof of the number of ASMs or the Adiprasito-Huh-Katz theorem, but it doesn't have to be so spectacular. Sometimes even a weak structure can lead to unexpected connections and generalizations unforeseen otherwise.
In summary, "these polytopes are just further examples of polytopes" is a misunderstanding of the context in the same way as Fibonacci and Catalan numbers are not "just numbers". Viewed in context, permutahedra and associahedra exhibit structures of combinatorial objects invisible otherwise.
In "Lessons I Learned from Richard Stanley" (https://arxiv.org/abs/1501.00719) Jim Propp starts with "Two big ideas": "The biggest lesson I learned from Richard Stanley’s work is, combinatorial objects want to be partially ordered! ... A related lesson that Stanley has taught me is, combinatorial objects want
to belong to polytopes!"
Thank you! This answer really convinced me and I totally agree that "adding structure" is a good thing, especially in combinatorics.
See also the slew of refs in https://oeis.org/A133314 for the permutahedrs and https://oeis.org/A133437 for the associahedra.
The origins of associahedra go back to the thesis work in homotopy theory of Jim Stasheff in the early 1960's.
He did graduate work at Oxford, working with Ioan James, who in the mid 1950's had proved lots of beautiful theorems about the free associative topological monoid $JX$ generated by a based topological space $X$; in particular, showing that, when $X$ is connected, that $JX$ is homotopy equivalent to the free loopspace $\Omega \Sigma X$ generated by $X$.
He also did graduate work with John Moore at Princeton. Moore was part of the first generation to grow up thinking comfortably about both algebra and topology through the lens of category theory. For example, thinking hard about the associativity of the tensor product operation was already in the air, thanks to work of MacLane.
Stasheff's work channels the vibe of both of his advisors. He is answering the question: when is a topological space homotopy equivalent to the space of based loops on another space? Two examples are topological groups, and the spaces $JX$. These have strict associative multiplications, whereas, as everyone learns when they learn about the fundamental group, one makes some arbitrary choices when `adding' loops in a space, but then shows the choices are associative up to homotopy. Taking this last line seriously - remembering all basic choices of n-fold multiplications, and homotopies, and assembling them into (contractible) geometric objects - led Stasheff to his answer of the question. The family of objects are the associahedra.
Continuing through history, homotopy theorists fit the associahedra into their general theory of operads (named by J. Peter May) in the 60's and 70's. In the mid 1990's, these ideas became fashionable in algebra too, and have continued ever since.
Not surprising, objects arising from such a universal source (the associative law!) end up being interesting to study from many points of view, including combinatorics. But original and continuing interest comes from homotopy theory.
IIRC, at the time of his thesis defense Stasheff had shown that the associahedron has the structure of a regular CW complex, and then Milnor (?) who was on his thesis defense panel asked about polytopiality, and then later showed Stasheff that it is in fact a polytope. Also, earlier Tamari had studied the 1-skeleton of the associahedron and its lattice structure.
The history of the associahedra is discussed by Stasheff in the article "How I 'met' Dov Tamari" from the Tamari Memorial Volume mentioned in Joseph O'Rourke's answer. Stasheff gives credit for the independent discovery of the associahedra to Tamari, and in fact the latter included graphical depictions of the associahedra as convex polytopes already in his 1951 thesis. (Alas, this was in the unpublished part of his thesis.)
In my opinion there are two answers to this question.
The first is that these particular classes of polytopes have fascinating combinatorial properties and structure. Presumably you're aware of the work of Postnikov and others in this direction. In my view, and the view of many others, these properties make the polytopes worth studying in their own right.
The second is that these polytopes arise naturally in other contexts, e.g., as moment polytopes (images of certain manifolds/varieties under the moment map) and as weight polytopes (convex hulls of subsets of the weight lattice of certain Lie groups). Various geometric properties of the manifolds, and representation-theoretic properties of the group, can be reduced to combinatorial properties of the polytopes. For example, GKM theory tells you that (provided certain hypotheses are satisfied) you can define and calculate equivariant cohomology directly on the moment graph.
This of course is a very high-level answer. If you are wondering about whether certain specific combinatorial properties have geometric or representation-theoretic significance, then maybe you can state a more specific question.
Good answer, yes, it’s worth pointing out the appearance of these polytopes in e.g. the book by Gelfand, Kapranov, Zelevinsky.
Can you provide some particularly good links to open access (no logins required) papers on the Web on moment maps and weight lattices of Lie groups related to the permutohedra and associahedra?
@TomCopeland : Hmm...I'm not sure. Maybe you'll find Hausel's thesis to be a useful introduction to moment maps. The permutohedron is the image under the moment map of something called the "permutohedral variety", but I'm not sure of a good reference where this is spelled out in detail. The description of the permutohedron as a weight polytope is given in one of Postnikov's papers.
I initially mistook moment map for moment curve rather than momentum map. I was unfamiliar with moment maps in relation to polytopes. Thanks for the intro.
There are remarkable combinatorial formulas for the face numbers and the volumes (of certain geometric realizations of) of these polytopes and a more general family ("generalized permutohedra" a.k.a. "polymatroids") to which they belong. These numbers include classical sequences like the Eulerian numbers, Catalan numbers, $(n+1)^{n-1}$, etc. This is a major combinatorial interest in these polytopes.
Regarding the question about why geometric realizations are interesting in particular, note that to meaningfully talk about volume (and its relatives like discrete volume, i.e., number of lattice points, etc.) requires a particular geometric realization.
(Warning: "generalized permutohedra" and "generalized associahedra" are two different families of polytopes, despite the very similar names. "Generalized permutohedra" are obtained by deforming the normal fan of the regular permutohedron, i.e., sliding facets in and out along a normal vector. "Generalized associahedra" have to do with analogs of the associahedron in other Lie types.)
I'm not an expert, but perhaps this could help:
Hohlweg, Christophe. "Permutahedra and associahedra: Generalized associahedra from the geometry of finite reflection groups." arXiv:1112.3255 (2011):
"Permutahedra are a class of convex polytopes arising naturally from the study of finite reflection groups, while generalized associahedra are a class of polytopes indexed by finite reflection groups. We present the intimate links those two classes of polytopes share."
Fig.5.
This is one chapter in the following collection, which includes other possibly illuminating chapters:
Müller-Hoissen, Folkert, Jean Marcel Pallo, and Jim Stasheff, eds. Associahedra, Tamari lattices and related structures: Tamari memorial Festschrift. Vol. 299. Springer Science & Business Media, 2012. List of chapters with links.
Especially: Cesar Ceballos and Günter M. Ziegler:
Realizing the Associahedron: Mysteries and Questions.
arXiv. 2011.
They quote Haiman from 1984: "The associahedron is a mythical object"!
I think it’s important to note that the major interest in generalized associahedra is their connection to cluster algebras.
Edit Dec. 2024:
For a concrete physical reason in line with my arguments below for an interest in associahedra, see the discussion on associahedra in the YouTube vid posted by Quantum Magazine
"2024's Biggest Breakthroughs in Physics"
featuring work by Arkani-Hamed and Figueiredo on quantum geometry (timestamp 11:11).
See "Hidden zeros for particle/string amplitudes and the unity of colored scalars, pions and gluons" by Arkani-Hamed, Cao, Dong, Figueiredo, & He.
If I were an orthodox Bourbakist, I would say the polytopes per se are of no interest, but I'm not. The geometric enumerative combinatorics of the faces of the permuto/permutahedra and the associahedra respectively reflect the algebra of multiplicative and compositional inversion of formal power series, and I find that alone fascinating.
But, how useful are the geometric realizations of the combinatorics of these relations?
First consider the algebraic equation of a circle and the geometric rep. Through which rep. is it easier to discern the possible number of intersections of two circles of different radii, not given their locations in a plane?
Now consider two particles traveling on two circles, one on each. The geometry of the intersections tells us a lot about the possible trajectories of collisions of the particles and would guide us in developing a program to calculate and depict them.
Ramp up your knowledge of algebra, analysis, geometry, and physics, and now you are prepared to understand "Combinatorics and Topology of Kawai–Lewellen–Tye Relations" by Sebastian Mizera. The abstract:
We revisit the relations between open and closed string scattering amplitudes discovered by Kawai, Lewellen, and Tye (KLT). We show that they emerge from the underlying algebro-topological identities known as the twisted period relations. In order to do so, we formulate tree-level string theory amplitudes in the language of twisted de Rham theory. There, open string amplitudes are understood as pairings between twisted cycles and cocycles. Similarly, closed string amplitudes are given as a pairing between two twisted cocycles. Finally, objects relating the two types of string amplitudes are the $\alpha_0$-corrected
bi-adjoint scalar amplitudes recently defined by the author 3. We show that they naturally arise as intersection numbers of twisted cycles. In this work we focus on the combinatorial and topological description of twisted cycles relevant for string theory amplitudes. In this setting, each twisted cycle is a polytope, known in combinatorics as the associahedron, together with an additional structure encoding monodromy properties of string integrals. In fact, this additional structure is given by higher-dimensional generalizations of the Pochhammer contour. An open string amplitude is then
computed as an integral of a logarithmic form over an associahedron. We show that the inverse of the KLT kernel can be calculated from the knowledge of how pairs of associahedra intersect one another in the moduli space. In the field theory limit, contributions from these intersections localize to vertices of the associahedra, giving rise to the bi-adjoint scalar partial amplitudes.
(For more on associahedra and scattering amplitudes in quantum physics, just google the terms. See also this MO-Q and the attendant comments and links.)
Conceivably, the permutohedra might prove useful since they can be deformed into associahedra. Analogously, conic sections other than the circle are important in understanding gravity and collisions of asteroids with planets as well as SL2 transformations. Good to have diverse reps shedding different light on mathematical and physical relations, in general.
For connections to extremal properties, see the Aquiar and Ardila ref in https://oeis.org/A133314 for permutahedra (also for multiplicative inversion of power series, surjective mappings, the inversion of modified Pascal matrices and other matrices derived from infinitesimal generators, lattice paths, the formalism of Appell polynomials related to Weyl algebras, quantum field theory) and https://oeis.org/A133437 for the associahedra (compositional inversion, quantum physics, lattice paths, and a slew of connections to other diverse geometric / combinatorial structures).
By the way, Newton was the first I have found to have written down the first few partition polynomials for the comp. inversion of a power series long before Tamari, Stasheff, and Loday developed the convex associahedra (or Lagrange, his inversion algorithm).
Three papers that discuss the all three families of convex polytopes listed by the OP: "Algebraic structures on graph associahedra" by S. Forcey an M. Ronco; "Differential algebra of polytopes and inversion polytopes" by V. M. Buchstaber and A. P. Veselov; and "Toric Topology" by Buchstaber and Panov.
Some of these polyhedra are of current interest to physicists—see for instance Nima Arkani-Hamed in the arXiv.
It looked like "Bruno Harris" was part of a signature rather than part of the answer; since your name is automatically appended, I edited it out when I edited in a link to the arXiv search. Can you recommend any particular paper(s) by Arkani-Hamed?
@LSpice, see https://mathoverflow.net/questions/182622/an-intriguing-tapestry-number-triangles-polytopes-grassmannians-and-scatteri
|
2025-03-21T14:48:29.608662
| 2020-01-09T11:15:30 |
350066
|
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|
Stack Exchange
|
Hauptvermutung for non-manifolds
The Hauptvermutung proposes the following: if two finite simplicial complexes are homeomorphic then they are PL-homeomorphic, meaning that they have a common refinement.
People are mostly interested in the manifold case but I am asking about the general case. According to Ranicki's Hauptvermutung book the 2-dimensional case was proven by Papakyriakopoulos in a paper written in Greek in Bull. Soc. Math. Grèce. My first question is: is there another more accessible reference for this?
In this paper Edward Brown proves the 3-dimensional case, but the paper is not mentioned in Ranicki's book (and hardly at all in general). My second question is: is there a particular reason for that?
Context: the Hauptvermutung for combinatorial manifolds is known to hold in dimension $\le 3$ by work of Radó and Moise. The Hauptvermutung was shown to be false by Milnor. Today obstructions to being PL-homeomorphic are well understood.
Incidentally, Dugundji's MathSciNet report (MR0024619) does not help to gain confidence in Papakyriakopoulos's paper: "the lemma [...] cannot be correct", "the proof [...] contains a serious gap", "this leads him [...] to hastily assume, incorrectly".
That's a great question. The seemingly inaccessible 153 page paper by Papakyriakopoulos (or does anybody have a digital copy?) is a pretty scary reference. I don't know about the status of Brown's paper. We really need a system for flagging papers, otherwise we have no idea which of the older papers are reliable and which ones are not. Perhaps people back in the 1960s knew which papers are shaky, but how we now supposed to know?
|
2025-03-21T14:48:29.608806
| 2020-01-09T13:04:53 |
350074
|
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|
Stack Exchange
|
The Poincaré Lemma
Let me consider an $L^1(\mathbb R^N)$ function $f$ such that $$
\int_{\mathbb R^N} f(x) dx =0.
$$
Then I claim that the $N$-form $f(x) dx_1\wedge\dots\wedge dx_N$ is closed, i.e. there exists a vector field $X$ such that
$$
\text{div} X=f.
$$
I would like to pick-up an $X$ in $W^{1,1}(\mathbb R^N)$. Is it possible? Is there a somewhat constructive proof?
No, there is no reason why $X$ would be integrable, too. For example $f(x)=1/x^2$ for $x\ge 1$, $f(x)=-1/x^2$ for $x\le -1$, and interpolate linearly, say. So $X=\pm 1/x$ for $|x|\ge 1$.
@Christian Remling Thanks, you are absolutely right. I could ask more to the function $f$, say $f\in L^p$ so that $X=\Delta^{-1}\nabla f\in W^{1,p}$. Then if $p>N$, it should be true that the continuous function $X$ goes to zero at infinity and thus I could apply Green's Theorem to get an equivalence between null integral for $f$ and the existence of a vector field $X$ going to zero at infinity such that $f=\text{div} X$.
|
2025-03-21T14:48:29.608904
| 2020-01-09T14:07:10 |
350081
|
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|
Stack Exchange
|
What information about the lattice $\Lambda$ can be recovered from the metric space $\mathbb{R}^n/\Lambda$?
Let $\Lambda\subset\mathbb{R}^n$ a lattice, i.e., a discrete subgroup that spans $\mathbb{R}^n$. Now we can look at the torus $T=\mathbb{R}^n/\Lambda$ which naturally carries the metric $d_T$ induced by the euclidean metric $d$ on $\mathbb{R}^n$: $$d_T(x+\Lambda,y+\Lambda):=\min_{a,b\in\Lambda}(d(x+a,y+b)).$$ Now my question is the following: What information about $\Lambda$ can be recovered from the metric space $T$? Can we completely recover $\Lambda$ (up to some orthogonal equivalence)?
Yes, we can. Consider the universal cover $U$ of your torus $T$. One can easily show that $U=R^n$ equipped with a Euclidean metric.
So we have $f:R^n\to T$,
The $f$-preimage of a point is your lattice, up to the shift of the origin (to one point of this preimage) and an orthogonal transformation (an isometry of the Euclidean metric).
A missing detail: $T$ is considered as metric space. So $U$ is a topological space and hence "locally" a metric space. One then has to endow $U$ with corresponding length distance, i.e., the largest distance which is locally bounded above by the original bounded distance. This is indeed isometric to the original Euclidean distance.
|
2025-03-21T14:48:29.609008
| 2020-01-09T14:10:29 |
350082
|
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"Dima Pasechnik",
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|
Stack Exchange
|
Finding a cycle of a specific length in an edge-weighted graph
I'm looking for some suggestions on how we might calculate cycles of a specific length in an edge-weighted graph.
For example, imagine my phone tells me that I need to walk three miles today. It might be nice if the phone could calculate a few tours of appropriate length, ideally avoiding things like walking up and down the same streets repeatedly.
A previous posting reviewed some algorithms for finding fixed-length cycles in unweighted graphs. However, I'm not aware of anything concerning the edge-weighted case.
A simple algorithm for a road network might work as follows. Assume that I want a cycle of length $x$ that starts and ends at a source vertex $s$. Now use Dijkstra's algorithm to produce a shortest path tree. Next, find the node $v$ in this tree whose distance from $s$ is close to $x/2$. Our solution then involves walking from $s$ to $v$ and then back to $s$ again. While this algorithm is polynomial, the solution is not great as we will end up walking up and down the same streets (so its not really a cycle at all...).
Another option might be to convert a weighted graph into an unweighted one. Assuming all edge-weights are integer, an edge of length $x$ is replaced by a chain of $x-1$ vertices, as the following demonstrates:
How to convert an edge weighted graph into an unweighted graph
We can then apply methods used for the unweighted graphs. An issue with this approach is that it could massively increase the size of the graph. Given that existing approaches for simple graphs can be quite expensive, this doesn't seem like a good idea.
Does anyone have any ideas/thoughts?
if your cycles must not re-visit vertices, then it's a hard problem, as hard as TSP.
@DimaPasechnik If you just do not want to revisit edges, there is a trivial realization of the knapsack problem in the given setup: just consider the graph with 2 vertices and multiple edges of lengths $0,w_1,\dots,w_n$ between them. Then just ask if you can make a path of length exactly $W$. However, as the setup suggests, we do not really need to achieve exactly $W$: a multiplicative factor close to $1$ should be allowed, in which case knapsack is just polynomial, so there is some hope. Of course, the minimal length computation is a red herring if we want just to stroll, not to get anywhere.
To help answer my own question, here's a paper that I recently published on this topic. Thanks also for the suggestions of Manfred Weis above.
Lewis, R. and P. Corcoran (2022) 'Finding Fixed-Length Circuits and Cycles in Undirected Edge-Weighted Graphs: An Application with Street Networks'. Journal of Heuristics, vol. 28, pp. 259-285.
https://link.springer.com/content/pdf/10.1007/s10732-022-09493-5.pdf
There are the so-called diverse routing algorithms that manage to calclulate the edge-disjoint or vertex-disjoint pair of paths with minimal lengths-sum between a pair of vertices. With such an algorithm you can check all pairs of points and filter out the ones, that have path-sum that comes close enough to the desired cycle length.
|
2025-03-21T14:48:29.609247
| 2020-01-09T14:44:18 |
350085
|
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Stack Exchange
|
How to describe the power operation on Lie groups?
Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$, or its compact form over $\mathbb{R}$. Recall that the automorphism group $\operatorname{Aut}(\mathfrak{g})$ is of the form $G^{\mathrm{adj}}\rtimes \operatorname{Out}(\mathfrak{g})$, where $G^{\mathrm{adj}}$ is the adjoint form of $G$ and $\operatorname{Out}(\mathfrak{g})$ is the group of Dynkin diagram automorphisms.
A theorem of Kac parameterizes the conjugacy classes of finite-order elements of $\operatorname{Aut}(\mathfrak{g})$. It is quoted for example in Theorem <IP_ADDRESS> of these lecture notes.
Briefly, you write down the affine Dynkin diagram of $\mathfrak{g}$, with its numbering (meaning numbers $m_i$ attached to each root, so that the affine root is given the number $m_0 = 1$, and the vector $\vec m$ is in the kernel of the Cartan matrix); then a finite-order element is, up to conjugacy, a pair $\rho \in \operatorname{Out}(\mathfrak{g})$ of order $r$ and some nonnegative rational numbers $a_i$ assigned to the roots of the $\rho$-folded Dynkin diagram (the twisted affine Dynkin diagram of the $\rho$-folded diagram of $\mathfrak{g}$), so that $\sum a_i m_i = 1/r$. (The order is the common denominator of the $a_i$.) These data are considered up to Dynkin diagram automorphism.
It is pretty easy to see why these data parameterize elements of $\operatorname{Aut}(\mathfrak{g})$. Indeed, the idea is to describe the maximal torus therein as $(\mathbb Q/\mathbb Z) \otimes (\text{weight lattice})$, and then to track any further Weyl group actions.
Is there a similarly simple description of the power operations $x \mapsto x^n$ on this set, for $n \in \mathbb Z$?
The easiest case is when $\rho = \mathrm{id}$, in which case the power operation is "$a_i \mapsto na_i$," more or less. But that map quickly destroys any bounds on $\sum a_i m_i$, so you need to subtract off some integers and then act by the Weyl group.
Do you mean "some nonnegative rational numbers $a_i$" (rather than $n_i$)?
Do you mean "the power operations $x\mapsto x^n$ on this set, for $n\in\mathbb Z$" rather than "for $a\in \mathbb Z$"?
@MikhailBorovoi Oops, yes, I will correct it.
Your description of finite order elements of ${\rm Aut}(\mathfrak g)$seems to be not quite correct. First, I suspect that you should assume that $\mathfrak g$ is simple. Second, in the case $\rho\neq 1$ you should attach numbers $m_i$ to the vertices of the $\rho$-twisted affine Dynkin diagram of $\mathfrak g$ rather than to the vertices of the affine Dynkin diagram of $\mathfrak g$.
@MikhailBorovoi I think all I need is that $\mathfrak{g}$ be semisimple. If it is not simple, then the Dynkin diagram may be disconnected, and there may be outer automorphisms that permute the components of the diagram. I think that what I'm calling "the twisted affine Dynkin diagram of the $\rho$-folded diagram of $\mathfrak{g}$" is what you are calling the "$\rho$-twisted affine Dynkin diagram", but perhaps I am using the terminology wrong?
I think that there is no standard terminology, and I find your terminology to be better than mine. Anyway, to vertices of what do you attach the numbers $m_i$, and how do you define them?
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2025-03-21T14:48:29.609749
| 2020-01-09T15:26:19 |
350088
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Stack Exchange
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Can one Gershgorin circle (only) contain all eigenvalues, when the other circles are not contained in it
In short, following a question from my students, I am trying to find a special case where all the eigenvalues of a matrix lie within only one circle, but not in the others, and the other circles are not completely contained in it.
Reminder: Gershgorin circle theorem
Given a matrix $A\in\mathbb{C}^{n\times n}$, define the disks $D_1,\ldots,D_n$ as follows:
$$D_i = \Bigl\{ z : |z-a_{ii}|\le \sum_{j\ne i} |a_{ij}|\Bigr\}.$$
The theorem states that all eigenvalues of $A$ lie within (at least) one of the disks. Moreover, if a connected component of the union of the disks contains $k$ disks, then exactly $k$ eigenvalues of $A$ lie in that union.
I am trying to find an example where all eigenvalues lie within only one circle (of course, all circles must partially overlap for this).
For $n=2$
Such an example is easy to find.
For instance,
$
A=\begin{bmatrix}
7 & 9 \\
-5 & -5
\end{bmatrix}
$ has two eigenvalues $\left\{1+3j, 1-3j\right\}$, which are only inside one Gershgorin circle, as illustrated below:
For $n\ge 3$
I could find many examples (by simulation) where one circle completely contains all other circles:
But I am looking for an example where the circles overlap, but are not all contained inside one of the circles. Like in following (hand made) graphs:
I already randomized billions of such graphs without any success, so I am beginning wonder:
Is there an interesting mathematical property making this case impossible?
I guess a related question would be: can a Gershgorin circle not intersect with any other circle that contains an eigenvalue.
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2025-03-21T14:48:29.610020
| 2020-01-09T16:19:08 |
350089
|
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Stack Exchange
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Diophantine representation of the set of prime numbers of the form $n²+1$
A polynomial formula for the primes (with 26 variables) was presented by Jones, J., Sato, D., Wada, H. and Wiens, D. (1976). Diophantine representation of the set of prime numbers. American Mathematical Monthly, 83, 449-464.
The set of prime numbers is identical with the set of positive values taken on by the polynomial
$(k+2)(1-(wz+h+j-q)^2-((gk+2g+k+1)\cdot(h+j)+h-z)^2-(2n+p+q+z-e)^2-(16(k+1)^3\cdot(k+2)\cdot(n+1)^2+1-f^2)^2-(e^3\cdot(e+2)(a+1)^2+1-o^2)^2-((a^2-1)y^2+1-x^2)^2-(16r^2y^4(a^2-1)+1-u^2)^2-(((a+u^2(u^2-a))^2-1)\cdot(n+4dy)^2+1-(x+cu)^2)^2-(n+l+v-y)^2-((a^2-1)l^2+1-m^2)^2-(ai+k+1-l-i)^2-(p+l(a-n-1)+b(2an+2a-n^2-2n-2)-m)^2-(q+y(a-p-1)+s(2ap+2a-p^2-2p-2)-x)^2-(z+pl(a-p)+t(2ap-p^2-1)-pm)^2)$
as the variables range over the nonnegative integers.
I am asking if there exist a similar result for the primes of the form $n²+1$, i.e., this set of prime numbers is identical with the set of positive values taken on by certain polynomial.
The existence of infinitely many such primes is a conjecture of Landau, and is still open. If one had a polynomial representation, I would expect it to resolve the conjecture.
@Stopple: But, there exist a such polynomial for twin primes and Mersenne primes and no one can solve these conjectures.
@Stopple: A Note on Diophantine Representations
Author(s): Christoph Baxa
Source: The American Mathematical Monthly, Vol. 100, No. 2 (Feb., 1993), pp. 138-143
If we call your polynomial $P$, how about $-c(X^2+1)(P - X^2 - 1)^2 + X^2 + 1$ for some $c>1$? This will be negative if $P \neq X^2+1$, but if there is equality it will be equal to the prime $P = X^2+1$. Or am I making some silly mistake here?
@RP_: This set of prime numbers is identical with the set of positive values taken on by certain polynomial.
@Stopple The resulting polynomial will probably be at least as complicated as the one in the body of the question. Polynomials like this are almost tautological (well, it uses Wilson's Theorem IIRC) so they are completely useless for actually studying primes.
Call your polynomial $P$. I propose the following polynomial:
$$
P' = (\xi^2+1)(1 - (\xi^2+1-P)^2)
$$
Proof (that the positive values of $P'$ are exactly the primes of the form $N^2+1$):
Let $P_0$ be one of the values of $P$, and let $\xi_0$ be any integer.
Case (i). Suppose $P_0 = \xi_0^2+1$. Then the value of the above polynomial (with the appropriate substitutions made) is $P_0 = \xi_0^2+1$, since the second factor evaluates to unity.
Case (ii). Suppose $P_0 \neq \xi_0^2+1$. Now the second factor evaluates to some non-positive number, and hence the value of the polynomial itself is non-positive.
Now the first case gives us all primes of the form $N^2+1$ as values of $P'$, and no other values (since $P_0=\xi_0^2+1$ is positive), whereas the second case gives us zero or negative numbers exclusively. So the positive values of $P'$ are exactly the primes of the form $N^2+1$.
Note that this is a template for making polynomials for some other subsets of primes. One needs a little more tweaking to make it detect primes of the form 8n+5 for example, but that is a simple exercise given this template. Gerhard "May Do Some Simple Exercise" Paseman, 2020.01.11.
Yes, in that case you need to explicitly "enforce" $P$ to be positive, which I didn't need to do since $P=\xi^2+1$ already did it for me. But of course, one can accomplish this by using Lagrange's theorem. (Which goes to show that, while the existence of these kinds of polynomials is unlikely to yield non-trivial number-theoretical results, one sometimes needs to use non-trivial number theoretical results to construct them... :-))
Putnam (1960) proved that a set is Diophantine if and only if it can be described as the set of positive values of a suitable polynomial with integer coefficients. Matiyasevich (1970) proved that a set is Diophantine if and only if it is recursively enumerable. It follows that every recursively enumerable set, such as the primes of the form $n^2+1$, can be described as the set of positive values of a suitable polynomial with integer coefficients.
I'm sorry if you don't want to answer next doubt that I've after I known the theorem in the second paragraph of your linked Wikipedia. It is well known that there exists an algorithm to find Mersenne primes (an also Fermat primes), as you know the Lucas-Lehmer primality test (respectively Pépin's test). Is the set of Mersenne primes computably enumerable? I suspect that some fails if I answer it as yes, but from the definition of the Wikipedia Recursively enumerable set, I don't get easily the idea why the set of Mersenne primes isn't computably enumerable
@user142929: The sequence of Mersenne primes is not only recursively enumerable (i.e. you can generate it by a computer program) but even recursive (i.e. you can recognize its elements by a computer program). Basically every sequence you know is recursive (hence also recursively enumerable).
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2025-03-21T14:48:29.610352
| 2020-01-09T16:34:14 |
350091
|
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Stack Exchange
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weierstrass points on hyperelliptic curves of genus 3
Let $C$ be a complex projective smooth hyperelliptic curve of genus $3$ and
$A_1, A_2, A_3$ three distinct Weierstrass points on $C$.
Consider the divisor $D=A_1+A_2+A_3$ and $L$ the line bundle associated to
$D$. Question: is $h^0(C,L)$ bigger than one?
To be more precise, $C$ is a Galois cover of degree two of a curve of genus 2.
If $A_1$, $A_2$ and $A_3$ are in distinct fibers of the double cover $C \to \mathbb P^1$, then $\mathrm h^0(C, L) = 1$. Otherwise you would get a map $C \to \mathbb P^1$ of degree 3; together with the double cover this would yield a map $C \to \mathbb P^1 \times \mathbb P^1$, whose image $X \subseteq \mathbb P^1 \times \mathbb P^1$ is a divisor of degree $(2,3)$, and such that the map $C \to X$ is birational. Since the arithmetic genus of $X$ is 2, this gives a contradiction.
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2025-03-21T14:48:29.610472
| 2020-01-09T16:47:01 |
350093
|
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Stack Exchange
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Gently changing measure
This question was asked and bountied on MSE without answer, so I'm porting it here:
There's an easy way to change the measure of a set of reals by moving to a larger universe: simply make $\mathbb{R}$ of the ground model be null. Conversely, it's not hard to show that if $M\subseteq N$ are transitive models of ZFC and $A\in\mathcal{P}(\mathbb{R})^M$ is measurable in $M$ and has positive measure in $N$, then $m(A)^M=m(A)^N$, so at least for "nice" sets that's the only way measure can change via this process. The answer there is stated for forcing extensions only, but that's unnecessary.
For non-measurable sets things are more complicated. However, the construction given there still somewhat fits the "nullify-or-leave-unchanged" pattern: the anomolous set is built from two pieces such that we can make one null while not changing the outer measure of the other. I'm interested in whether this is optimal:
Question 1: Can there be a pair $M\subseteq N$ of transitive models of ZFC and a set of reals $A\in M$ such that $(i)$ $\mu^*(A)^M>\mu^*(A)^N$ but $(ii)$ there is no partition $A=B\sqcup C$ with $B,C\in M$ such that $\mu^*(B)^N=0$ and $\mu^*(C)^N=\mu^*(C)^M$?
I suspect the answer is no. Annoyingly, I haven't been able to make any progress on this, and in particular I can't even rule out the following extreme (and "obviously" ridiculous) possibility:
Question 2: Can there be a pair $M\subseteq N$ of transitive models of ZFC such that no non-null $A$ in $M$ is null in $N$ but some $A$ in $M$ has $\mu^*(A)^M\not=\mu^*(A)^N$?
The "forcing" and "inner-models" tags are because those are the primary ways we know how to build transitive models "to order." By "can there be" I mean "is it consistent with ZFC + large cardinals." I'm also interested in the same question nontransitive models (and demanding that $N$ be an end-extension of $M$), but I'm primarily interested in the transitive case. Meanwhile, I'm not very intereted in going below ZFC and I'm not interested at all (at the moment) in dropping below ZF + DC, since the question is less exciting if measure is nastier.
Do you have the measure inequality backwards in Question 1?
@JamesHanson Frack, yes. Fixed!
If $N$ is a generic extension of $M$ obtained by forcing with a weakly homogeneous forcing, then the asnwer to Question 2 is negative (Lemma 6.3.10 in the Bartoszynski-Judah book). I don't know if we can drop weakly homogeneous here.
@Ashutosh Ah neat, I didn't know that! But the general situation (esp. non-generic extensions) is still unclear, right?
Yes. Some related questions are discussed in Kellner, Shelah, Preserving preservation, JSL, Vol. 70 No. 5, 2005 (See section 3). But I didn't see your question being addressed there.
To question 1, there is such a pair. This is a minor reworking of Ashutosh's example you linked.
Start with $L,$ add an $\omega_1$-sequence of random reals $X=\langle r_{\alpha}: \alpha<\omega_1 \rangle,$ then a Cohen real $c,$ then another $\omega_1$-sequence of random reals $Y=\langle s_{\alpha}: \alpha<\omega_1 \rangle,$ then another random real $t.$ Let $N=L[X][c][Y][t]$ be the final model. Identify the reals in $X$ and $Y$ as elements of $[0,1],$ and $t$ as a subset of $\omega.$ In $N,$ $\mu^*(X)=0$ and $\mu^*(Y)=1.$
We now work in $L[X][Y][t],$ noting that $X,$ $Y,$ and $t$ are mutually random over $L.$ Let $\langle b_n \rangle$ be the increasing enumeration of $t,$ and $\langle c_n \rangle$ the increasing enumeration of $\omega \setminus t.$
We use $t$ to interweave $X$ and $Y.$ Precisely, define $\langle u_{\alpha}: \alpha<\omega_1 \rangle$ by letting $r_{\omega \alpha + n} = u_{\omega \alpha + b_n}$ and $s_{\omega \alpha + n} = u_{\omega \alpha + c_n}.$ Applying permutation invariance of random forcing in $L[t],$ we have that $u$ is a random $\omega_1$-sequence of random reals over $L[t]$ and $t \not \in L[u].$
Let $M=L[u] \subset L[X][Y][t] \subset N,$ and let $A=\{2^{-n-1}(1+u_{\omega \alpha +n}): n<\omega, \alpha<\omega_1\}.$ Then $\mu^*(A)^M=1$ and $\mu^*(A)^N = \sum_{n \in \omega \setminus t} 2^{-n-1} \in (0,1).$
Suppose there is a partition $A = B \sqcup C$ in $M$ as in the description of question 1. Then $t = \{n<\omega: \mu^*( C \cap [2^{-n-1}, 2^{-n}])^M = 0\} \in M,$ contradiction.
I think "morally" this counterexample doesn't really violate your intuition, in that we still have in $N$ a partition of $A$ into one set of "preserved points" and another set of "nullified points." A natural follow-up question (which I don't know the answer to) is whether there can be $A \in M \subset N$ such that (i) $\mu^*(A)^M > \mu^*(A)^N$ and (ii) in $N,$ every point in $A$ is a Lebesgue density point of a minimal measurable cover of $A.$
Here's a sketch of an attempt at negative answer for #2. First, we may assume that our measure space is the Cantor space $2^{\omega}$, and that $\mu^{*}(A)^{M} = 1$. Working in $M$, $(2^{\omega})^{\omega}$ is isomorphic to $2^{\omega}$ and $\mu^{*}_{\omega}(A^{\omega}) = 1$, letting $\mu^{*}_{\omega}$ be the corresponding version of outer measure.
I would think that this last claim is standard, but I haven't been able to find it anywhere. One argument for it would be to take the construction of the $\omega$-size product in Theorem 254F of Fremlin (https://www1.essex.ac.uk/maths/people/fremlin/chap25.pdf) and apply it to measure space on $A$ induced by its being a subspace of $2^{\omega}$ of full outer measure. Another argument would be to take a supposed closed set $C$ of positive measure in $(2^{\omega})^{\omega}$ disjoint from $A^{\omega}$ and, using Fubini's Theorem, iteratively choose the members of a sequence $\langle a_{n} : n \in \omega \rangle \in A^{\omega} \cap C$.
If this is right then we seem to be done: if $\mu^{*}(A)^{N} < 1$, then $\mu^{*}_{\omega}(A^{\omega})^{N} = 0$.
Why can we assume that $\mu^*(A)^M=1$?
First you can assume that it's finite by considering infinitely many disjoint subsets of finite measure and what happens to them in the outer model. Then you can scale it by the appropriate constant.
Or, in the second step, find a Borel set in which it has full measure, and map everything over to the Cantor space.
Or, if $A$ is a subset of the Cantor space, and $B$ is a Borel set containing $A$ with the same outer measure, use $A \cup (2^{\omega} \setminus B)$ instead.
Ah, yes, I was being thick. I'm still interested in Q1.
The claim $\mu_{\omega}^*(A^{\omega})=1$ is right, and is one of the few things in analysis which seems to require $\text{DC}{\mathbb{R}}$ and not just $\text{CC}{\mathbb{R}}.$ (I haven't confirmed $\text{CC}_{\mathbb{R}}$ doesn't imply this principle but it does fail in Feferman-Levy model).
|
2025-03-21T14:48:29.610918
| 2020-01-09T17:39:52 |
350098
|
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|
Stack Exchange
|
Is there a version of algebraic de Rham cohomology that can be used to calculate torsion classes?
Much work has gone into the construction of cohomology theories which are defined on algebraic varieties (étale, crystalline, etc.) and comparison isomorphisms between them.
Say $X$ is an algebraic variety over $\mathbb Z$. I am interested in computing $H^*_{\rm sing}(X_{\mathbb C}, \mathbb{Z}_p)$ using a variant of the de Rham complex. However, the obvious integral version of de Rham cohomology, gives the completely wrong answer even for $X = \mathbb A^1$.
This is surprising to me, because (if I am reading the literature correctly, as a non-expert) the de Rham complex of $X_{\mathbb Z_p}$ computes the crystalline cohomology of $X_{\mathbb F_p}$. And crystalline cohomology is often claimed to be the "correct" $p$-adic replacement for the etale cohomology groups $H^*_{et}(X_{\overline{\mathbb F_p}}, \mathbb Z_l)$. However, it seems that crystalline cohomology/de Rham cohomology groups of affine space are "wrong", in the moral sense of not lining up with what I would naively expect.
Is there a correction/explanation for this discrepancy? Is there a variant of de Rham/crystalline cohomology which computes the "right" cohomology for $\mathbb A^1$? Morally, is there a reason why we would want integral crystalline cohomology to carry so much extraneous torsion? Is there something I'm missing?
This claim about crystalline cohomology is only for smooth proper varieties. Berthelot's rigid cohomolgy extends crystalline cohomology to more general varieties, and it is $\mathbb A^1$-invariant, but it has rational coefficients. A reasonable $\mathbb A^1$-invariant theory with mod $p$ coefficients for $\mathbb F_p$-varieties is logarithmic de Rham cohomology (= motivic cohomology).
@MarcHoyois Thanks, now I see that what I was objecting to really is the failure of $\mathbb A^1$ invariance.
You should read the introduction to Bhargav Bhatt's lecture notes on prismatic cohomology: available here. This is a new cohomology theory introduced by Bhatt-Scholze (closely related to prior work by Bhatt-Morrow-Scholze) for explaining these torsion phenomena.
The presence of extra torsion in de Rham (or crystalline etc.) cohomology is unavoidable, as SashaP's answer demonstrates. These cohomology theories are only supposed to be reasonable analogues of singular cohomology (i.e. Weil cohomology theories) when you use torsion free coefficients.
However, we can explain where the extra torsion "comes from" (relative to the "true" etale/singular cohomology with e.g. $\mathbf F_p$ coefficients) with a certain long exact sequence.
Let $X$ be a reasonable variety over (e.g.) $\mathbf Z$. Then we can consider its etale cohomology $H^i(X_{{\overline{\mathbf{Q}}}}, \mathbf Z_p)$. This agrees as a module with the singular cohomology of $X_{\mathbf{C}}$, so it's the "right answer" (in particular it's $\mathbf A_1$-invariant).
On the other hand, we can consider the algebraic de Rham cohomology of $X_{\mathbf Z_p}$ (this is essentially the algebraic de Rham cohomology of $X$ tensored with $\mathbf Z_p$), $H^i_{\mathrm{dR}}(X/\mathbf Z_p)$. This agrees with the crystalline cohomology of the special fiber - in particular it only depends on the special fiber, so we shouldn't expect it to capture the same information as the etale/singular cohomology of the generic fiber.
These two $\mathbf Z_p$-modules have the same rank by Fontaine's comparison theorem (at least in the proper case, but if you use some sort of log theory you can extend to the open case), but in general the de Rham cohomology has extra $p$-torsion.
To explain this, Bhatt-Scholze construct a third cohomology theory for schemes over $\mathbf Z_p$ (it's essentially a variant of the construction of crystalline cohomology), valued in modules over the ring $\mathbf Z_p[[u]]$. Let's call this $\mathcal{H}(X)$. Then:
$\mathcal{H}(X)/u$ is isomorphic to de Rham cohomology of $X_{\mathbf{Z}_p}$ (equivalently, it's isomorphic to the crystalline cohomology of $X_{\mathbf{F}_p}$).
$\mathcal{H}(X)[1/u]$ is non-canonically isomorphic to the etale cohomology with $\mathbf Z_p$ coefficients of $X_{{\overline{\mathbf Q}}}$ (equivalently to singular cohomology of $X_{\mathbf C}$), tensored with $\mathbf Z_p((u))$.
$\mathcal{H}(X)$ (at least after some scalar extension) can in principle be computed using a "q-de Rham complex"
(I'm sweeping some homological algebra consideration, about e.g. exactness of modding out by u, under the rug - see the Bhatt-Scholze paper for precise statements. You really need to use derived categories everywhere.)
Let $k$ be a field of characteristics $p$ and $R$ be any ring where $p$ is not invertible.
Asuume that $F:Var_{k}\to D(R-mod)$ is a cohomology theory of smooth algebraic varieties over the field $k$ with values in $R$-modules that satisfies etale descent and has $F^0(\mathrm{Spec}\,k)=R[0]$ such that for any geometrically connected variety $X$ the structure morphism induces an isomorphism $R=F^0(\mathrm{Spec}\,k)\simeq F^0(X)$(these assumptions are satisfied for crystalline cohomology with torsion-free coefficients, e.g. over $W(\bar{k})$). Then at least one of the modules $F^1(\mathbb{A}^1_k)$ and $F^2(\mathbb{A}^1_k)$ must be non-zero.
Indeed, consider the etale Artin-Schreier $\mathbb{Z}/p$-cover given by $\mathbb{A}^1\to\mathbb{A}^1,t\mapsto t^p-t$ (a generator of the cyclic group acts by $t\mapsto t+1$). The Hoschschild-Serre spectral sequence looks like(group cohomology is taken in the category of $R$-modules)$$E_2^{i,j}=H^i(\mathbb{Z}/p,F^j(\mathbb{A}^1))\Rightarrow F^{i+j}(\mathbb{A}^1)$$
If $F^1(\mathbb{A}^1)=0$ then there is an injection(there are no differentials that could touch this term) $H^2(\mathbb{Z}/p,F^0(\mathbb{A}^1))=E^{2,0}_2\hookrightarrow F^2(\mathbb{A}^1)$. But we've assumed that $F^0(\mathbb{A}^1)=R$ with any automorphism acting trivially, so $H^2(\mathbb{Z}/p,F^0(\mathbb{A}^1))=R/p\neq 0$ hence $F^2(\mathbb{A}^1)$ is non-zero.
Thanks, this is a nice observation. However, imposing the requirement of etale descent in char $p$ seems to me be a bad condition, since we already know that etale cohomology with $\mathbb F_p$ coefficients (in some sense the universal theory satsifying etale descent) is poorly behaved, and that the etale site of $\mathbb A^1_{\mathbb F_p}$ is very complicated.
@user1092847 Indeed $p$ is automatically inverted in any $\mathbb A^1$-invariant étale sheaf of spectra on smooth $\mathbb F_p$-schemes. So if you care about $p$-torsion you have to give up either étale descent or $\mathbb A^1$-invariance.
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2025-03-21T14:48:29.611338
| 2020-01-09T18:42:37 |
350101
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{
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"authors": [
"H A Helfgott",
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|
Stack Exchange
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Fully explicit version of Atkinson's formula?
Let $$I(T)=\int_0^T \left|\zeta\left(\frac{1}{2} + i t\right)\right|^2 dt$$
and let $E(T)$ be $I(T)$ minus what turn out to be its main terms:
$$E(T) = I(T)- T \log \frac{T}{2 \pi} - (2 \gamma - 1) T.$$
Atkinson (1949) gave an explicit formula for $E(T)$. As it happens, his formula (see the original paper, or Chapter 15 of Ivić's book) has a small error term (namely, $O(\log^2 T)$) whose constant is not made explicit. Working the constant out looks like something that would involve plenty of time and ink, but should be doable.
Has anybody done it?
(And what about the case of the integral over a line with $\Re s=\sigma$, $0<\sigma<1$, $\sigma$ other than $1/2$? I know Matsumoto and Meurman have a version of Atkinson's formula in that case, but they also have a small, non-explicit error term, similar to the above.)
I should perhaps add that I do not understand why Atkinson or Ivić do not use a version of Voronoï with a smoothing ("given by nature" in this case). There has to be a catch, since such a version can be found as (3.2) in Ivić's book.
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2025-03-21T14:48:29.611480
| 2020-01-09T19:23:34 |
350102
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{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Alejandro Tolcachier",
"David Handelman",
"Derek Holt",
"Luc Guyot",
"YCor",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350102"
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|
Stack Exchange
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Isomorphism of $\mathbb{Z}\ltimes_A \mathbb{Z}^m$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^m$
here it's a question that I've posted in MSE but unfortunately got no answers:
Let $A$ and $B$ be matrices of finite order with integer coefficients.
Let $n\in\mathbb{N}$ and let $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ be the semidirect product, where the action is $\varphi(n)\cdot (m_1,\ldots,m_n)=A^n (m_1,\ldots,m_n)$, and similarly with $B$.
It is easy to construct an isomorphism between $G_A$ and $G_B$ if $A$ is conjugate in $\mathrm{GL}(n,\mathbb{Z})$ to $B$ or $B^{-1}$.
But, this is also a necessary condition? I mean, does $G_A\cong G_B$ implies $A\cong B$ or $A\cong B^{-1}$ in $\mathrm{GL}(n,\mathbb{Z})$ or is there a counterexample?
I've seen at this MSE question that it is true if $A$ and $B$ are hyperbolic, i.e none of their eigenvalues have module 1, but it isn't the case.
Thank you very much!
I think it is a necessary condition, but I haven't had time to write down a proof. If the isomorphism $\phi:G_A \to G_B$ maps the normal subgroup $Z^n$ in the semidirect product to the $Z^n$ in $G_B$ then $\phi$ on $Z^n$ is essentially the required conjugating element of ${\rm GL}(n,Z)$. Otherwise $\phi(Z^n)$ must be a normal subgroup that intersects the $Z^n$ in $G_B$ in a subgroup $Z^{n-1}$. But then that subgroup would be centralized by $B$ and since $B$ has finite order, that would force $B=I_n$. Similarly $A=I_n$, so they are conjugate as required.
Did you try the simplest possible example: $n = 2$, $A = {\rm diag} (1,-1)$ and $B$ the other conjugacy class for a matrix of order two and size two (the two by two permutation matrix)? The fact that one of them is diagonalizable and the other isn't, probably allows the semidirect products to be distinguished.
@DavidHandelman I've checked this example: the abelianizations are then not isomorphic!
It is known to be a necessary and sufficient condition when $n = 2$ (elementary).
Another easy case is when $1$ is not an eigenvalue of $A$ (which means that the abelianization of $G_A$ has rank 1) — not assuming $A$ has finite order by the way. In this case the normal copy of $\mathbf{Z}^m$ in $G_A$ is characterized as the inverse image of the torsion from the abelianization.
I believe now that David Speyer's example can be adapted to provide a counterexample to the original question. (So I retract my earlier comment on the question and will delete it soon.)
In David's example, $A$ is a degree $\phi(m)$ matrix of order $m$ defining the action by multiplication of $\zeta_m$ on the ideal $I$ of the number field ${\mathbb Q}[\zeta_m]$, and $B$ is the action on the ideal $\sigma(I)$, and $A$ and $B$ are not conjugate to each other or to their inverses in ${\rm GL}_{\phi(m)}({\mathbb Z})$. A specific example is $m=37$, $\phi(m)=36$.
We define degree $n:=\phi(m)+1$ matrices $A'$ and $B'$
as the diagonal joins of $A$ and $B$ with the identity matrix $I_1$. So the corresponding ${\mathbb Z}$-modules can be thought of as $I \oplus \langle y \rangle$ and $\sigma(I) \oplus \langle z \rangle$, with trivial action on the second factors. These modules cannot be isomorphic, because an isomorphism would have to map the fixed points submodule $\langle y \rangle$ onto $\langle z \rangle$ and then their quotients $I$ and $\sigma(I)$ would be isomorphic, which they are not. So $A'$ and $B'$ are not conjugate in ${\rm GL}_{n}({\mathbb Z})$.
I claim (at least in some cases) that we can choose $A$ and $B$ such that the corresponding semidirect products $\langle \alpha \rangle \ltimes_{A'} {\mathbb Z}^n$ and $\langle \beta \rangle \ltimes_{B'} {\mathbb Z}^n$ are isomorphic, where $\alpha$ and $\beta$ generate infinite cyclic groups. We can (in some cases?) choose $A = B^a$ with $a$ coprime to $m$ and $2 \le a < \phi(m)-1$ such that $B$ is not conjugate in ${\rm GL}_{\phi(m)}({\mathbb Z})$ to $A$ or to $A^{-1}$, and choose integers $r,s$ with $ra-sm=1$.
Then we can define a isomorphism from $\langle \alpha \rangle \ltimes_{A'} {\mathbb Z}^n$ to $\langle \beta \rangle \ltimes_{B'} {\mathbb Z}^n$ by mapping $I$ to $\sigma(I)$ as in David's example, $y$ to $\beta^m z^r$ and $\alpha$ to $\beta^a z^s$. Note that this induces an isomorphism from the free abelian group $\langle \alpha, y \rangle$ to $\langle \beta, z \rangle$, such that the image of $y$ centralizes $\sigma(I)$.
I did some calculations in Magma in the case $m=37$, and found a degree 36 integer matrix $A$ that is not conjugate to $A^a$ for any $a$ with $2 \le a \le 36$.
For completeness, here is the matrix $A$ in machine readable format. I used the Magma function $\mathsf{AreGLConjugate}$ to test $A$ for conjugacy with $A^i$. This uses a fairly new algorithm published in Bettina Eick, Tommy Hofmann, and E.A. O'Brien.
The conjugacy problem in ${\rm GL}(n,{\mathbb Z})$.
J. London Math. Soc., 2019.
[
[-4,149,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-1,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,133,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-2,64,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-2,42,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,130,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-3,76,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,143,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-1,24,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-2,53,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-3,86,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-3,103,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-1,35,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-1,9,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,113,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,144,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-1,20,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-2,69,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-1,22,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-2,61,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-2,54,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-3,82,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,119,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],
[-4,120,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0],
[-4,116,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0],
[-4,132,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0],
[-2,68,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0],
[-1,26,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0],
[-2,45,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0],
[-4,118,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0],
[-4,124,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0],
[-3,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0],
[-2,47,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0],
[-3,110,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0],
[-1,7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
[15,120,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1]
]
Do you remember how you defined or found the matrix $A$? I would have had to asked this before, sorry. I'd like to know at least if $A$ has det 1.
I am afraid I cannot find any record of the matrix $A$ that I found, but since it has order $37$ it must have determinant $1$.
What do you call "degree" of a matrix?
By a matrix of degree $36$, I mean a $36 \times 36$ matrix. (I guess I am taking this terminology from the degree of a corresponding group representation.)
It is strange.. I put the matrix in Magma and the command IsGLZConjugate(E,E^2) returns "true"
@AlejandroTolcachier I can't explain that. For me $\mathtt{AreGLConjugate}$ is returning false, and $\mathtt{IsGLZConjugate}$ is crashing with an error message about an assertion failing.
$\mathtt{} $ also returns a conjugating element $C$. Have you checked directly whether $C^{-1}AC = A^2$?
$\newcommand{\IZ}{\mathbb{Z}}$
One can easily verify that $G_A' = \{0\}\times \operatorname{im}(A-1_{m\times m})$. Moreover $G_A$ acts on $G_A'$ by conjugation. The elements of $\IZ^m$ act trivially and the extra $\IZ$ acts by multiplication with $A$. The normal subgroup $K_A:=\operatorname{ord}(A)\IZ \times \IZ^m$ is the kernel of this action, i.e. the subgroup of all elements that act trivially on $G_A'$.
Therefore any isomorphism $G_A \to G_B$ must map $K_A$ to $K_B$. In particular $ord(A)=|G_A/K_A| = |G_B/K_B|=\operatorname{ord}(B)$, let's call that $n$, and $G_A/K_A \cong G_B/K_B \cong \IZ/n\IZ$.
Now consider the conjugation action of $G_A$ on $K_A$ instead of $G_A'$. Since $K_A$ is abelian, this is really an action of $G_A/K_A\cong \IZ/n\IZ$ on $K_A\cong \IZ \times\IZ^m$ given by multiplication with the block matrix $A':=\begin{pmatrix}1&\\&A\end{pmatrix}$.
By considering the induced action on $K_A \otimes \mathbb{Q}$, we find that the two $\mathbb{Q}[\IZ/n]$-modules $K_A \otimes \mathbb{Q}$ and $K_B\otimes \mathbb{Q}$ must be isomorphic. That means that $A'$ and $B'$ are $\mathrm{GL}_{1+m}(\mathbb{Q})$-conjugated at the very least. I'm not sure how one would go from there.
Since $A'$ and $B'$ have the same canonical Jordan form, the matrices $A$ and $B$ have also the same canonical Jordan form. Therefore $A$ and $B$ are $GL_m(\mathbb{C})$-conjugated and hence $GL_m(\mathbb{Q})$-conjugated.
$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$I misread the question as asking about $C_m \ltimes_A \ZZ^n$ and $C_m \ltimes_B \ZZ^n$, where $m$ is the order of $A$ and $B$. If we work with $\ZZ \ltimes_A \ZZ^n$ and $\ZZ \ltimes_B \ZZ^n$, I'm not sure what happens.
Working with $C_m \ltimes_A \ZZ^n$, this is not true. Let $m$ be the order of $A$ and $B$, let $\zeta_m$ be a primitive $m$-th root of unity, let $K$ be the cylotomic field $\QQ(\zeta_m)$. Let $G$ be the Galois group of $K$ over $\QQ$, so $G \cong (\ZZ/m \ZZ)^{\times}$. Let $H$ be the class group of $K$. Suppose that $H$ contains a class $h$ whose $G$-orbit is larger than $h^{\pm 1}$; say $\sigma(h) \neq h^{\pm 1}$.
Let $I$ be an ideal representing the class $h$, so $I$ is a free $\ZZ$-module of rank $\phi(m)$. Let $A$ be the matrix of multiplication by $\zeta_m$ on $I$, and let $B$ be the matrix of multiplication by $\zeta_m$ on $\sigma(I)$. Since $I^{\pm 1}$ and $\sigma(I)$ are not isomorphic as $\ZZ[\zeta_m]$ modules, $A^{\pm 1}$ and $B$ are not conjugate.
However, $C_m \ltimes_A \ZZ^{\phi(m)} \cong \langle \zeta \rangle \ltimes I$ and $C_m \ltimes_B \ZZ^{\phi(m)} \cong \langle \zeta \rangle \ltimes \sigma(I)$, and these are isomorphic by $(\zeta^j, x) \mapsto (\sigma(\zeta)^j, \sigma(x))$.
This occurs for $m=37$, where $H \cong \ZZ/37 \ZZ$. If I recall correctly, if $\sigma(\zeta) = \zeta^a$ then $\sigma(h) = h^{a^{21}}$. Since $\mathrm{GCD(21,36)} = 3$, the monomial $a^{21}$ takes $12$ different values modulo $37$ so, taking $h$ a generator of the class group, there are values of than $h^{\pm 1}$ in the $G$ orbit of $h$.
Of course the significant difference between $C_m$ and ${\mathbb Z}$ is that the only isomorphisms of ${\mathbb Z}$ are the identity and inversion.
Why $\sigma(h)=h^{a^{21}}$? I couldn't understand the following either. Can you give me a reference to read more about these number theory (I suppose) things?
This is a complement to Johannes Hahn's answer.
Corrigendum. In the previous version of this answer, I have made an erroneous claim, allowing $\omega$, the order of $A$ and $B$, to be any positive number.
The claim below is valid only if $$\omega \in \{1, 2, 3, 4, 6 \},$$ which is sufficient to address OP's subsequent examples.
Following Johannes Hahn's approach, we can prove the following:
Claim. Assume that $G_A$ and $G_B$ are isomorphic. Then
$\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$ is a conjugate of $\begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 \\ 0 & B^{-1} \end{pmatrix}$ in $\text{GL}_{n + 1}(\mathbb{Z})$. In particular $A$ is a conjugate of $B$ or $B^{-1}$ in
$\text{GL}_{n}(\mathbb{Q})$.
Proof. Let $K_A$ be the centraliser of the derived subgroup
$G_A' = [G_A, G_A]$ of $G_A$. It is clearly a characteristic subgroup of $G_A$. Let $C_A$ be the infinite cyclic subgroup of $G_A$ generated by $a \Doteq (1, (0, \dots, 0))$. The conjugation by $a$, or equivalently, the multiplication by $\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$ induces a structure of $\mathbb{Z}[C_A]$-module on $K_A$. This structure is almost invariant under isomorphism in the following sense: if $\phi:G_A \rightarrow G_B$ is a group isomorphism, and if we identify $C_A$ with $C_B$ via $a \mapsto b = (1, (0, \dots, 0)) \in G_B$ then $K_A$ is isomorphic to $K_B$ or to $K_{B^{-1}}$ as a $\mathbb{Z}[C]$-module with $C = C_A \simeq C_B$, depending on whether $\phi(a) = bk$ or $b^{-1}k$ for some $k \in K_B$. This is so because conjugation by $bz$ induces a group action on $K_B$ which is independent of $k$. Now the claimed result immediately follows.
Thus the pair of modules $\{K_A, K_{A^{-1}}\}$ is a group isomorphism invariant of $G_A$ It turns out to be useful for this example and this one.
Addendum. Here are some details on the module $K_A$.
An element of $\mathbb{Z}[C_A]$ is a Laurent polynomial with coefficients in $\mathbb{Z}$ of the form $P(a) = \sum_{i = 0}^d c_i a^{e_i}$ where $e_i \in \mathbb{Z}$ for every $i$. The structure of $\mathbb{Z}[C_A]$-module of $K_A$ is defined in the following way:
$$P(a) \cdot k = (a^{e_0}k^{c_0}a^{-e_0}) \cdots (a^{e_d}k^{c_d}a^{-e_d})$$ for $k \in K_A$. Assume now that there is an isomorphism $\phi: G_A \rightarrow G_B$. As $\phi$ is surjective and $\phi(K_A) = K_B$, there is $f \in \mathbb{Z}$ coprime with $\omega$ and $z \in \mathbb{Z}^n \triangleleft G_B$, such that $\phi(a) = b^f z$. Since $\omega \in \{1, 2, 3, 4, 6\}$, we infer that $\phi(a) = b^{\epsilon}k'$ for some $\epsilon \in \{\pm 1\}$ and some $k' \in K_B$. Thus $\phi(a^e) = b^{\epsilon e}k''$ where $k'' \in K_B $ depends on $e$, $k$ and $\epsilon$. The image of $P(a) \cdot k$ by $\phi$, after substituting $\phi(a^{e_i})$ with $b^{\epsilon e_i}k_i''$, and after simplification ($K_B$ is Abelian), results in $$(b^{\epsilon e_0}\phi(k)^{c_0}b^{- \epsilon e_0}) \cdots (b^{\epsilon e_d}\phi(k)^{c_d}b^{- \epsilon e_d}) = P(b^{\epsilon}) \cdot \phi(k).$$
Therefore $\phi$ induces an isomorphism of $\mathbb{Z}[C]$-module if $\epsilon = 1$, where $C = C_A \simeq C_B$. Let $e_0 \Doteq (\omega, (0, \dots, 0))$. Let $(e_1, \dots, e_n)$ denote the canonical basis of $\mathbb{Z}^n$ and let $C \in \text{GL}_{n + 1}(\mathbb{Z})$ be the matrix of $\phi$ with respect to $(e_0, e_1, \dots, e_n)$. If $\epsilon = 1$, then the following identity $\phi(a \cdot k) = b \cdot \phi(k)$ holds true and translates into
$$C \begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}
k = \begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix}C k$$ simply because of the way we defined the action of $a$ and $b$ on $K_A$ and $K_B$ respectively. The claimed result on the matrix conjugation follows.
Dear Luc, I tried to understand every line in your re-proof. I have two questions or point where I need a little bit of explanation if it is possible:
(1) Why $K_A$ is isomorphic to $K_B$ or to $K_{B^{-1}}$ as $\mathbb{Z}[C]$-module? I can't see that $\phi:K_A\to K_B$ is a $\mathbb{Z}[C]$-module homomorphism (in the case of $\phi(a)=(1,z)$.
(2) How follows that $1\oplus A$ is conjugate to $1\oplus B$ or $1\oplus B^{-1}$ from the isomorphism of the corresponding $\mathbb{Z}[C]$-modules? Thanks
@AleTolcachier Dear Ale, I added some details in an Addendum.
Great! Now that part is very clear. Yesterday I noticed that I didn't know why $\phi(a)$ can be only $(\pm 1, z)$. It isn't true that $\phi(\mathbb{Z}^n)=\mathbb{Z}^n$, is it?
@AleTolcachier Dear Ale, this is good that you check so carefully. I have made a mistake: it is not true that $\phi(a) = (\pm 1, z)$. The claim needs to be downgraded to something much less general.
Very interesting. It is a pain the result is less general but still it is useful because I have examples of matrices of order 4 and 6. I don't see why the order $\omega$ is important, what is its role in the proof?
@AleTolcachier If $\omega \in {1, 2, 3, 4, 6}$, then $f \in \mathbb{Z}$ is coprime with $\omega$ if and only if $f \equiv \pm 1 \text{ mod } \omega \mathbb{Z}$.
Excellent, now I understand every line of the proof. Thank you again for all the (very clear) explanations. Thinking about the role of the order $\omega$, I wonder if the result can be extended to say "$G_A\cong G_B$ implies $1\oplus A$ is conjugate to some of the matrices $1\oplus B^\varepsilon$ where $\varepsilon$ is coprime with $\omega=\operatorname{ord} A=\operatorname{ord} B$". Do you agree?
|
2025-03-21T14:48:29.612708
| 2020-01-09T20:03:11 |
350104
|
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|
Stack Exchange
|
Morphism in a Verdier quotient
Let $\mathcal{T}$ be a triangulated category and take $\mathcal{S}$ a triangulated subcategory. Consider the Verdier quotient $\mathcal{T} \left/ \mathcal{S} \right.$, morphisms in this category are constructed as roofs
$E \leftarrow H \rightarrow F$ where $\text{cone}(H \rightarrow E) \in \mathcal{S}$ modulo a certain equivalence relation. In general it's not easy to compute morphisms in a general Verdier quotient because one needs to replace either E or F with a "resolution" in terms of objects which are in the left, respectively right, orthogonal to $\mathcal{S}$. An example to keep in mind is the derived category of an abelian category, in which case we replace $E$ with a projective resolution or $F$ with an injective resolution. However, in the case $D(\mathcal{A})$, where $\mathcal{A}$ is an abelian category, one has some spectral sequences converging to Homs in $D(\mathcal{A})$ (I'm thinking of $Hom_{D(\mathcal{A})} ( E, H^q(F)[p]) \implies Hom_{D(\mathcal{A})}(E,F[p+q])$ for example).
My question is whether there exist tools like such (spectral sequences or any kind of approximation) to compute morphisms in $\mathcal{T} / \mathcal{S}$. Even a way to relate them to morphisms in $\mathcal{T}$ would be fine. I hardly think such a thing exists in general, but I am mainly interested in the case $\mathcal{T} = D(\mathcal{A})$.
Thank you in advance for the help.
I agree that this kind of computations may be tricky but, if you are happy to restrict a little bit your setting, then computations can be done. In particular, two assumptions may help: 1) that your triangulated category is the homotopy category of a (stable, simplicial, maybe even combinatorial) model category, and 2) that the Verdier quotient functor has an adjoint, so that you have a Bousfield localization. In this case, the quotient $\mathcal T/\mathcal S$ should be the homotopy category of a Bousfield localization of the original model category, so you have descriptions for homs.
@Simone Virili Can you give me a reference? Those conditions should be satisfied as I'm considering $T = D(A)$ and thus the quotient factor has a right adjoint by Brown representability.
I think the standard reference for Bousfield localization of Model categories is "Model Categories and Their Localizations" by Philip S. Hirschhorn. To get the idea and to understand if that is what you are looking for, you may have a look at the nLab page about "Bousfield localization of model categories".
|
2025-03-21T14:48:29.612896
| 2020-01-09T20:25:51 |
350105
|
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|
Stack Exchange
|
Does any real projective plane incidence theorem follow from axioms?
Is it known whether any projective geometry statement that holds true in the real projective plane (equivalently, can be deduced from Hilbert axioms) follows from the standard projective axiomatics?
By projective geometry statement I mean any statement it terms of incidence of points and lines, that is a statement of the form:
Let $I\subset \mathbb{N}\times\mathbb{N}$ be a finite subset and $(i_0,j_0)\in\mathbb{N}\times\mathbb{N}$.
For all sequences $(p_i)$ of points and sequences $(l_j)$ of lines,
if $p_i\in l_j$, for all $(i,j)\in I$ then $p_{i_0}\in l_{j_0}$.
By standard projective axiomatics I mean four incidence axioms
plus Desargues, Pappus and Fano.
No. Your axioms hold in the projective plane over any field of characteristic $\neq2$. So they don't imply, for example, the existence of $\sqrt2$ in the underlying field. But that existence can be expressed as an incidence statement. Unfortunately, I don't have time right now to reconstruct that incidence statement or to search for a reference, so this is only a comment, not an answer.
This is false. I don't see how to get a counterexample from Andreas Blass's comment (the only uses for the existence of $\sqrt2$ which are obvious to me require a more flexible notion of incidence statement), so I am posting this as an answer although it is probably more complicated than necessary.
For fixed prime $p$, one can take a point and line configuration corresponding to the rank 3 Dowling geometry of $\mathbb{Z}/p$, in which a certain equality must occur unless the underlying field has a primitive p-th root of unity. The presence of a root of unity of prime order $p \ge 3$ is not ruled out by these axioms, which can therefore never prove the equality, although it holds over $\mathbb{R}$.
Explicitly, what you do is the following: take 3 points and the 3 lines connecting them (think of this as a triangle). On each of these three lines, place $p$ points, labelled $0,1,\ldots,p-1$. Label the lines by $a,b,c$, and denote by $i_a, i_b, i_c$ the point labelled $i$ on the corresponding line. Define a new line on which $i_a, j_b, k_c$ are present whenever $i+j+k = 0 \mod p$.
Any realization of this point and line arrangement over a field, in which the corners of the triangle are all distinct, corresponds to a representation of $\mathbb{Z}/p$ in the multiplicative group of the field. Hence if $p\ge 3$, either some two corners coincide or the representation is trivial, so all points $\{i_a\}_i$ are identical, as are all points $\{i_b\}_i$ and all points $\{i_c\}_i$.
The many lines connecting various triples of points also imply that if two corners of the triangle coincide, so do all points $i_a, j_b, k_c$ for all $i,j,k$.
Hence in $\mathbb{R}$, one has that $1_a$ lies on the line through $0_a, 0_b, 0_c$. This incidence statement is not true over $\mathbb{C}$, hence it is not implied by the axioms given.
Amazing. Thank you. The statements you give seem to be independent for different $p$'s. Is there some sensible description of real projective incidence theorems? E.g. is there finite axiomatics?
What would be a good reference to educate myself about these matters?
I learned these things by studying some matroid theory. I'm fairly sure there is no finite description in terms of incidence statements only: see Vamos, "The Missing Axiom of Matroid Theory is Lost Forever" (unfortunately not free for access as far as I can tell.) The real-representable matroids also have the property of having infinitely many excluded minors, and one can translate between your incidence statements and the language used by Vamos (if you allow conjugation and disjunction).
If you do not have access to Vámos's paper, you might want to look at a later paper by Mayhew, Newman and Whittle on the same topic.
|
2025-03-21T14:48:29.613167
| 2020-01-09T20:40:02 |
350106
|
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|
Stack Exchange
|
Lattices from quaternion algebras (MAGMA software)
I am studying the paper "Lattice Packing from Quaternion Algebras" from 2012 about the construction of ideal lattices.
In Section 3.3 the authors construct very interesting examples of lattices using quaternion algebras over number fields. In Example 3.3, they choose an ideal of the maximal quaternion order, but I could not understand how to do this using Magma (or even other software). I am really puzzled about how to get this ideal (with N (N (I)) = 7).
I understand the construction and the way it happens, but I do not understand how I can select an ideal as they do.
I'm not sure what you're asking. The paper gives generators for the lattice, and my recollection is that Magma will give you the ideal generated by a set of elements in a quaternion algebra. Are you saying you can't get the ideal function to work?
Yes, Magma gives the ideal generated by a set in a quaternion algebra. My question is how to find an ideal with a given norm. In Example 3.3 they select an ideal of a maximal quaternion order such that $N(N(I)) = 7$. I'm confused about how to obtain that using Magma (or another software) or if there is some theoretical result that leads me to obtain such ideal.
Given an order $O$ in a quaternion algebra $B$ over a number field $K$, I believe you want to construct an integral ideal $I$ in $O$ with some given norm $n$ (an ideal in $K$). I don't believe Magma has any built-in function to do this directly, but you should be able to do this in one of the following ways using various built-in Magma commands:
(1) If $n$ is principal and $N(O)= \mathfrak o_K$, look for an element $\alpha$ of norm $n$ in $B$. If you find one, then you can take $I=\alpha O$.
(2) In general, given a (pseudo)basis for $O$, construct all "small" linear combinations whose norms lie in $n$. Then construct the ideal generated these elements and compute the norm. If the norm is $n$, you're done. Otherwise, try adding more linear combinations to the ideal.
|
2025-03-21T14:48:29.613318
| 2020-01-09T21:02:33 |
350107
|
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|
Stack Exchange
|
Counting binary vectors that satisfy given distance constraints
Let's start with a warm up problem.
Suppose I am given two binary vectors $x, y \in \{0, 1\}^n$ of length $n$ that differ in exactly $r$ places, i.e. $||x-y||_0=r$, i.e. their hamming distance is $r$.
Now I want to compute the size of following set:
$$ S_{a,b} = \{ z ~~\vert~~ ||x-z||_0 = a, ||y-z||_0 = b \} $$
Intuitively, the set $S_{a,b}$ contains all vectors that are exactly $a$ bits apart from $x$ and at the same time $b$ bits apart from $y$.
First, we notice that we can re-order the dimensions of $x$ and $y$ such that they disagree in the first $r$ dimensions and agree in the remaining $(n-r)$ dimensions without loss of generality.
To compute $\vert S_{a,b}\vert$ imagine the we look at the vectors $z$ where we "flip" $i$ zeros in the first $r$ bits where $x$ and $y$ disagree, and we flip the rest $a-i$ bits in the remaining $(n-r)$ dimensions. There are $\binom{r}{i}\binom{n-r}{a-i}$ such vectors. But in order to satisfy the other conditions it has to hold
$$(r-i) + (a-i) = b$$
since $y$ should be $(r-i)$ apart from $z$ in the first $r$ dimensions (in order to be the complement of $x$) and $(a-i)$ bits apart in the remaining $(n-r)$ dimensions (in order to match $x$).
Solving for $i$ we get $i=\frac{a-b+r}{2}$ and plugging it in we get that the size of the region in question for even $(a+b-r)$ is
$$
\vert S_{a,b} \vert=\binom{r}{\frac{a-b+r}{2}}\binom{n-r}{\frac{a+b-r}{2}}
$$
Now onto the real question.
Suppose we can partition our $x, y, z$ into subspaces of equal size, i.e. $x=[x_1, \dots, x_m]$, where each $x_i$ is of size b, making the total size $n=b\cdot m$. Similarly we partition $y$ and $z$.
Now I want to compute the size of the following region:
$$ S_{a,b,c} = \{ z ~~\vert~~ ||x-z||_0 = a, ||y-z||_0 = b, \\
\sum_{i=1}^m\mathbb{I} ( ||x_i - z_i||_0 > 0 ) <= c, \\\sum_{i=1}^m\mathbb{I} ( ||y_i - z_i||_0 > 0 ) <= c \} $$
Intuitively, this region contains the same vectors as before with the additional constraint that when $x$ and $z$ differ they can do so in at most $c$ partitions.
I tried to approach this in a similar manner as before without any luck. We can do a similar "trick" with re-ordering the dimensions per partition. One idea would be to compute the size of some other smaller sets and combine them to obtain the size of $S_{a,b,c}$.
A variant of the problem would be to specify how much $x$ and $y$ differ per partition i.e. specify $r_i=||x_i - y_i||_0$ for all $i=1,\dots, m$ and compute the size of $S_{a,b,c}$ then.
Yet another variant would be to add specify $\sum_{i=1}^m\mathbb{I} ( ||x_i - y_i||_0 > 0 ) <= d$, i.e. that $x$ and $y$ themselves differ in at most $d$ partitions.
For both of these variants I can conclude that if $2c<d$, or for more than $2c$ partitions $r_i>0$'s then the size of the region $S_{a,b,c}$ must be $0$.
Any thoughts or helpful references or different ideas on how to approach this problem? What if only have 2 partitions for $x, y, z$ rather then $m$?
A brute-force solution would be to enumerate all $z$'s and simply count how many satisfy the given constraints but that would not scale since I need to compute the size of $S_{a,b,c}$ for many different combinations of $a$ and $b$.
|
2025-03-21T14:48:29.613537
| 2020-01-09T21:20:54 |
350109
|
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|
Stack Exchange
|
Octonion algebras over number fields
Is there any textbook or paper about Arithmetic of Octonion Algebras or Octonion Algebras constructed over number fields? I know J. Voight book and K. Martin notes about quaternion algebras but I was thinking about a generalization using Octonions.
More than that: Is it possible to define Orders and Maximal Orders over an octonion algebra? Or these concepts just make sense over associative algebras?
Possibly you'll find information here: https://arxiv.org/abs/1810.09979 (Composition algebras by A. Elduque)
I'm voting to close this question as off-topic because it was simultaneously asked at Mathematics (no cross-posting, please). If after a week there is no answer there, it may be asked here. (Update: there was an answer at math.se: see https://math.stackexchange.com/questions/3503269/octonion-algebras-over-number-fields)
|
2025-03-21T14:48:29.613625
| 2020-01-09T22:36:04 |
350114
|
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|
Stack Exchange
|
The generalized word problem on groups
Given some group $G$ that is generated by $a$ and $b$, each of which has infinite order, and some free subgroup $N$ generated by $a^k$ and $b^k$, is there any algorithm that tells me if some $x \in G$ is also in $N$? If not generally, what about for $k=2,3,4$?
No, you can assume that $a$ and $b$ have order $k$ and $G$ has unsolvable word problem. Hence $N$ is trivial (a free group on $0$ generators) and there is no way to tell if $x\in G$ is in $N$.
What if $a$ and $b$ have infinite order?
The problem in the wording of your question is that you want to require $(a^k,b^k)$ to be a free family, but you say something weaker "$a^k$ and $b^k$ generate a free subgroup" which is not what you want. Fortunately Mark's answer interpreted it correctly.
The special linear group is generated by two elements:
Trott, S. M., A pair of generators for the unimodular group, Can. Math. Bull. 5, 245-252 (1962). ZBL0107.02503.
And the generalized word problem is known to be undecidable for $SL(4, \mathbb{Z}).$
Notice that in the question, the subgroup $N$ is assumed to be free and, moreover, generated by small powers of generators of $G$.
True. On the other hand, is it clear that the subgroup generated by small powers of the Trott generators (or some generators obtained from them by Whitehead moves) is not free?
Igor: What I had in mind is that Mikhailova subgroups are never free (they are also not 2-generated). I am unaware of any examples of f.g. free subgroups of $SL(n,Z)$ with undecidable membership problem. Do you know any? The worst distortion I know (for free subgroups) is given by hydra subgroups, where one can realize arbitrary Ackermann functions as distortion functions.
@Misha Agreed. On the other hand, even deciding whether $k$-th powers of generators generate a free group might be impossible :(
|
2025-03-21T14:48:29.613769
| 2020-01-09T22:54:49 |
350116
|
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|
Stack Exchange
|
History of the name "subexponential distribution" in probability
In probability theory, the term subexponential distribution has historically been used for a distribution whose CDF $F(x)$ satisfies the relation
$$
n(1-F(x)) \sim 1 - F^{*n}(x)
$$ for any $n \ge 1$ as $x \to \pm \infty$ (depending on the context), where $F^{*n}(x)$ is the CDF of the $n$-fold additive convolution. The interpretation is that for large $x$, $F(x) \approx 1-\epsilon$ and so $$
n(1-(1-\epsilon)) = n\epsilon \approx 1 - (1-\epsilon)^n
$$ and the latter represents the maximum of $n$ independent copies of the distribution, so we can say that $$
\mathbb{P}[X_1 + X_2 + \ldots + X_n > x] \sim \mathbb{P}[\max\{X_1,X_2,\ldots,X_n\}>x]
$$ for large $x$, or specifically that it is a distribution such that if the sum of $n$ independent copies is large, it is likely due to the contribution of one particularly large jump. So subexponential is a type of large-tailed distribution, subject to extreme events.
On the other hand, a subgaussian distribution is one whose moment generating function does not grow faster than the Gaussian-like function, so
$$
\mathbb{E}[e^{\lambda(X-\mathbb{E}[X])}] \le e^{\lambda^2\sigma^2/2}
$$ for all $\lambda$ and some $\sigma^2 >0$. This guarantees tails that are no larger than Gaussian. If we relax this so that it only holds in a neighborhood of zero, specifically that
$$
\mathbb{E}[e^{\lambda(X-\mathbb{E}[X])}] \le e^{\lambda^2\sigma^2/2}, \ \ \ \ \ |\lambda| < 1/\alpha
$$ then the tail bounds become
$$
\mathbb{P}[|X-\mathbb{E}[X]|>t] \le 2e^{-t/2\alpha}, \ \ \ \ \ t> \sigma^2/\alpha
$$ It seems that in some circles, these distributions are now being referred to as subexponential. See, for instance, https://www.stat.berkeley.edu/~bartlett/courses/2013spring-stat210b/notes/4notes.pdf, and other top hits on "subexponential distribution".
Frustratingly, these conditions are quite opposite of one another: one implies heavy tails, the other light tails. The latter seems more appropriate, meaning "tails no heavier than exponential". My question is, what is the historical background that led to calling the former subexponential?
In the known survey Subexponential distributions by Goldie and Klüppelberg, we find this:
"The name arises from one of their properties, that their tails decrease more slowly than any exponential tail".
So, the logic behind this term seems to be that the rate of decrease of the tails of such distributions is less than the rate of decrease of the tail of any exponential distribution.
I am not sure if this logic is compelling. I guess people did not think much about the choice of the term.
So far, there seems to be no written record like this: "Let us refer to the distributions satisfying this condition as subexponential." One of the earliest papers in the field, by Teugels, says just this: "The distribution $F$ is said to belong to the subexponential class $\mathscr S$ if [...]". I guess the term first appeared during a rather casual presentation or conversation, and then took root.
|
2025-03-21T14:48:29.614106
| 2020-01-09T23:54:28 |
350120
|
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}
|
Stack Exchange
|
Hochschild cohomology of (generalizations) of Khovanov's arc algebra
Backgroud: In his seminal paper A functor-valued invariant of tangles, Khovanov (among many other things) introduced the arc algebra $H^{n}$ and several functors between $H^{n}$ and $H^{m}$ related to tangles. In particular, Khovanov homology of links could be recovered from these functors/bimodules.
Stroppel defined a generalization of Khovanov's arc algebra in Parabolic category O, perverse sheaves on Grassmannians, Springer fibres and Khovanov homology by considering arcs with "semi-infinite" ends and the resulting algebra is denoted by $\mathcal{K}^n$. Brundan and Stroppel generalized these algebras further in Highest weight categories arising from Khovanov's diagram algebra I: cellularity, such as $K_{m}^{n}$.
As these papers suggest, the algebras $K_{m}^{n}$ are important for geometric representation theory. Some more recent work shows the connection between such algebras and annular Khovanov homology.
Question: Do we have some concrete computations of Hochschild cohomologies of the algebras $H^n$ and $K_{m}^n$? Certainly we can compute everything out by writing some computer programs, so the question is really about patterns/geometric intepretations of the Hochschild cohomologies in these cases.
Is it possible to generalize the technique of Diogo-Lisi and compute symplectic cohomologies instead?
That's possibly true, but it is know that the total rank of Hochschild cohomology of the ordinary arc algebra grows exponentially w.r.t. $n$. To compute the differentials, one needs to compute GW invariants inside some Hilbert scheme, and actually we need some precise information about curves avoiding a divisor. The extended version is some kind of FS category, whose HH* may not have nice "closed-string" cousins.
|
2025-03-21T14:48:29.614248
| 2020-01-10T00:41:04 |
350124
|
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|
Stack Exchange
|
Definition of Lie derivatives of sections of natural vector bundles - product preservation needed?
Section 6.15 of Natural Operations by Kolár, Michor, and Slovak defines the Lie derivative of a section of a natural vector bundle along a vector field. Set-theoretically, the definition is clear. Just after, the authors assert the Lie derivative to be a smooth section. To see this for myself, I have tried to write it as a composite of smooth maps. However, there seems to be an obstacle without assuming the natural vector bundle is product-preserving.
Question. Is the Lie derivative of a section itself a smooth section also for non-product preserving natural vector bundles?
Below I give my description of the Lie derivative as a composite of smooth maps. First, a couple of details.
If $\vec v$ is a vector field I will write $X\times \mathbb R\supset \mathfrak D(\vec v)\overset{\exp \vec v}{\longrightarrow} X$ for its flow map. The map $\mathfrak D(\vec v)\to X\times \mathbb R$ given by $(x,t)\mapsto (\exp \vec v(x,t),t)$ is a $C^\infty$ isomorphism $\mathfrak D(\vec v)\cong \mathfrak D(-\vec v)$.
If we have a morphism $f$ of submersions over $Z$ (commutative triangle with top side $f:X\times (a,b)\to Y$) whose domain is $X\times (a,b)$ then its (relative) time-derivative is the smooth lift $X\times (a,b)\to \mathrm T_{Y/Z}$ defined by the formula $\partial_tf(x_0,t_0)=\mathrm T(x_0,0,t_0,1)$. Here $\mathrm T_{Y/Z}$ denotes the vertical bundle.
In particular, if $Y\to Z$ is a vector bundle, then its vertical bundle is canonically isomorphic to the first projection $Y\times_ZY\to Y$, so composing with the second projection allows to interpret the time-derivative as valued in $Y$.
Now the diagrams defining the Lie derivative. $F$ denotes a natural vector bundle functor. The problem lies in the first diagram. Using the notation of Natural Operations, I want to define $ (\exp\vec v)^\ast s$ as the composite by "fattening up" the section $s$. However, without product preservation it seems this "fattening up" is not available as e.g in the case of $F=\mathrm T$ the tangent bundle.
If the above diagram is granted, the second one is as follows. The lengthy composite applies remarks (2,3) to interpret the time-derivative as taking values in $FX$.
Finally, since $X\times \left\{ 0 \right\} \subset \mathfrak D(\vec v)$, the time-derivative at zero is defined on $X$ and is smooth as the restriction of a smooth map. This is the Lie derivative, exhibited as a smooth section $X\to FX$.
Am I missing something? Is the "fattened section" unnecessary?
|
2025-03-21T14:48:29.614452
| 2020-01-10T02:04:08 |
350126
|
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|
Stack Exchange
|
Any good references on the decay rate of Legendre coefficient?
Let $P_n:[-1,1]\rightarrow\mathbb{R}$ be the $n$-th Legendre Polynomial. and, let
$$a_n:=\int_{-1}^1 f(t) P_n(t)\, dt$$
for some $f:[-1,1]\rightarrow\mathbb{R}$.
Are there any good references on the decay rate of $\vert a_n \vert$? I am not familiar with this kind of problem, but I guess there must be a lot of methods.
From the following similar mathoverflow question: Reference for the exponential decay of Legendre coefficients, I found one paper. Also, I could find a book "Spherical Harmonics and Approximations on the Unit Sphere" by Atkinson. After skimming those references, it seems that showing smoothness of $f$ is one method. But the application in my mind is the case when $f=\arccos^2(t)$, which is not smooth enough.
So, I'm wondering are there any other references explaniing various methods to compute decay rate of $\vert a_n \vert$. Especially, if there are some techniques one can use for non-smooth $f$, I really want to know.
You would have to assume some kind of smoothness. Otherwise the decay could be arbitrarily slow (e.g. $f(t)=\sum_{n>0}2^{-n}P_{n!!!!!}$).
You might find an article by Guillemot-Tessier in the Ann. Scuola Norm. Sup. Pisa, 25 (1971) 519-573 interesting (available online). She is interested in the extreme cases (smooth test functions, with rapidly decreasing coefficients, and distributions, with slowly increasing ones) but gives estimates which could be useful for other situations.
@Anthony Quas Thank you. So, $f$ should be smooth on whole $[-1,1]$? If $f$ is smooth only interior $(-1,1)$ Is there any way to control the decay rate?
@user131781 Thank you. I will look into it.
|
2025-03-21T14:48:29.614590
| 2020-01-10T08:31:02 |
350131
|
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|
Stack Exchange
|
Approximation of $\sigma$-finite Borel measures by equivalent finite measures
Let $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\mu)$ be a $\sigma$-finite Borel measure on $d$-dimensional Euclidean space. Can one always construct a sequence of finite equivalent measures $\left\{\mu_n\right\}_{n \in \mathbb{N}}$ on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ such that
$$
\lim_{n \rightarrow \infty} \frac{d\mu_n}{d\mu} \uparrow 1.
$$
Motivation:
This is true for both the Lebesgue measure and the counting measure on $\mathbb{R}^d$ by taking
$\frac{d\mu_n}{d\mu}(x)\triangleq \begin{cases}
e^{-q_n\min\{\|x\|,\frac1{\|x\|}\}}&:\, x\neq 0\\
1 &: \, x =0
\end{cases}
$, where $q_n$ runs over $\mathbb{Q}\cap(0,\infty)$.
Since $\mu$ is $\sigma$-finite there is a $\mu$-integrable function $f$ with $0<f\le 1$. For an increasing sequence $A_n$ with $\bigcup_{n\in\mathbb N} A_n=\mathbb R^d$ define densities $f_n=I_{A_n}+fI_{A_n^c}$. Then $\mu_n=f_n\cdot \mu$ are equivalent finite measures whose densities converge to $1$.
But how to show that such a function exists (ie the bounded away from 0 part)?
If $A_n$ are as in the answer set $B_1=A_1$, $B_n=A_n\setminus A_{n-1}$, and $f=\sum\limits_{n=1}^\infty c_n 1_{B_n}$ with suitable $c_n>0$, so that $\int f d \mu= \sum\limits_{n=1}^\infty c_n \mu(B_n)<\infty$.
Note that this has nothing to do with the topology of $\mathbb R^d$. It holds for every $\sigma$-finite measure space.
Yes I had noticed that; thanks :)
Let $(E,\cal{E},\mu)$ be a measure space and $\mu$ a not finite $\sigma$-finite measure on $\cal{E}$. Then there is a sequence $E_n$ of $E_n \in \cal{E}$ with $E_n \uparrow E$ and $0 < \mu(E_n) < \mu(E_{n+1}) < \infty$. Let $a_1 := \mu(E_1)$ and $a_n := \mu(E_n \setminus E_{n-1})$ for $n > 1$. Let $\mu_n(A) := \mu(A \cap E_n) + \sum_{k=n+1}^\infty 2^{-k} / \mu(E_k \setminus E_{k-1}) \cdot \mu(A \cap (E_k \setminus E_{k-1}))$ for $A \in \cal{E}$. Then each $\mu_n$ is finite, equivalent to $\mu$ and $\mu_n \uparrow \mu$.
Edited after reading Jochen Wengenroths answer. I ommited the topological assumtions.
|
2025-03-21T14:48:29.614752
| 2020-01-10T08:52:05 |
350133
|
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|
Stack Exchange
|
Coin flipping game
Motivation. My elder son played the following game. He had a bunch of coins, all with heads up, arranged in a circle. He flipped one coin, so that it showed tails, then he moved $1$ position clockwise, flipped that coin, then moved $2$ positions clockwise, flipped that coin, then moved $3$ positions clockwise etc. He stopped whenever he reached a coin that was already showing tails.
Informal version of question. For what values of $n$ can you flip the coins in the manner described above so that you end up with all coins showing tails?
Formal version. For any positive integer $n$, let $[n] = \{0,\ldots,n-1\}$. For what values of $n$ is the map $j:[n]\to [n]$ defined by $$k \mapsto \big(k(k+1)/2 \mod n\big)$$ injective (and therefore bijective)?
0 and -1 are both mapped to 0.
for even $n$ this map $f$ is not well-defined: $f(0)=0$ while $f(n)=n/2\pmod n$
Apologies for the bad formal version - it didn't correspond to the informal one! I hope I fixed things now.
A quick experiment suggests that the map is bijective if and only if $n$ is a power of $2$.
See https://wordplay.blogs.nytimes.com/2014/11/03/toss/.
I think it may be worth noting that a very similar problem came up in another MO question (though it may be tooting my own horn as well): https://mathoverflow.net/questions/344401/divisibility-of-certain-polynomials/344407#344407
Only a partial answer: Assume that $n$ is not a power of $2$. Then there is a prime $p | n$ with $p > 2$. First we consider the case $n = p$. Then for $p = 3$ $\{(k (k+1)/2 \mod 3 \colon k = 0,1,2\} = \{0,1\}$ and $f_p(k) := k(k+1)/2 \mod p$ is not surjective. For $p > 3$ we have $f_p(k) \equiv f(p-k-1) \mod p$, in particular $f_p(1) \equiv f(p-2) \mod p$, hence $f_p$ is not injective. Now for $n$ as above with prime factor $p > 2$ consider the natural map $\pi \colon \mathbb{Z}/(n) \to \mathbb{Z}/(p)$ ($(n)$ the ideal generated by $n$). If $f_n$ is surjective, then $f_p = \pi \circ f_n$ must be surjective, which is false. Hence if $n$ is not a power of $2$ the map $f_n$ is not bijective.
To complete the proof of Dieter Kadlka:
For $n=2^m$, we are looking for $a\neq b< n$ such that $$ b^2+b=a^2+a+l 2^{m+1} $$ for some $l\in \mathbb{N}$ and then $$l 2^{m+1} = (b-a)(b+a+1)$$
Because $b-a$ and $b+a+1$ have different parity, we have (Gauss Lemma) either $2^{m+1}|(b-a)$ or $2^{m+1}|(b+a+1)$ which is impossible since $2^{m+1}>|a|+|b|$
Lovely, thanks both of you - and I am sorry that I can only accept one answer. Since Dieter "planted the seed", I am going to accept his answer - but I would have liked to accept both!
|
2025-03-21T14:48:29.614961
| 2020-01-10T11:14:57 |
350138
|
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|
Stack Exchange
|
Incorrect information in an old article about the Kervaire invariant
In the Soviet times there was a famous Encyclopedia of Mathematics. I think it is still familiar to every Russian mathematician maybe except very young ones, and yours truly is in possession of all 5 volumes. Browsing it recently (with no real purpose) I came across a certain peculiarity. In the article "Kervaire invariant" by M.A.Shtan'ko there was a claim that the Kervaire invariant is nontrivial in all dimensions $2^k-2$ for $2\le k\le 7$ - yes, including 126. Which, for what I know, is still an open problem
Kervaire invariant: Why dimension 126 especially difficult? . In this article, the credit for the $k=6$ and $k=7$ cases (lumped together) was given to M. Barratt, M. Mahowald, and A. Milgram, but with no actual reference.
To be fair, the absence of references is understandable (in the original article, that is) because it was written in 1978 while a complete proof for dimension 62 was only published six years later
https://web.math.rochester.edu/people/faculty/doug/otherpapers/barjoma.pdf
It is possible that back in 1978 the result was just announced. But, what happened to the 126? And to Milgram? The simplest possible explanation is that a proof for 126 was also announced but later retracted. However this is by no means the only possibility, so I am curious what really happened. Besides, those MO folks who know more about the subject then myself might wonder what the attempted proof was like.
After a bit of search I found a reference which may be relevant (hopefully). In "Some remarks on the Kervaire invariant problem from the homotopy point of view" by M.E.Mahowald (1971) there is Theorem 8 attributed to Milgram and after it the following Remark: "It can be shown that $\theta_4^2=0$ and thus Milgram's theorem implies $\theta_6$ exists". If I get it right this indeed means a nontrivial Kervaire invariant in dimension 126, so probably there is a mistake somewhere in this argument. (But even if it is so, damned if I have a clue who has made it: Milgram, Mahowald, or somebody else.)
I have to admit that a few things about this story look suspicious. To begin with,
A. Milgram died in 1961 so it should probably be R. J. Milgram if any. In the introduction to "The Kervaire invariant of extended power manifolds " J. Jones stated explicitly that the 62 case is solved by Barratt and Mahowald but not published yet while in the higher dimensions the problem is open, in contradiction to what Shtan'ko wrote the same year. In a couple of papers between 1978 and 1981 I spotted references like [Barratt M. G., Mahowald M., The Arf invariant in dimension 62, to appear] but no traces whatsoever of 126 and Milgram. (Besides this article of Mahowald from almost a decade before.) I am at a loss what to make of all this. It would be nice if someone can set it straight - at the very least, I want to know if Shtan'ko made it up.
By the way, an English translation of this Encyclopedia article can be found here
https://www.encyclopediaofmath.org/index.php/Kervaire_invariant
Only, the year is written 1989 instead of 1978 (a second edition, apparently).
A second edition is one possibility, but a more likely scenario is that a technical typist moved fingers too far when typing the year.
I found the following remark in Zhouli Xu's paper "The strong Kervaire invariant problem in dimension 62":
In [19], R. J. Milgram claims to show that under the same condition as in Theorem 1.1, one has $θ_{n+2}$ exists. If this were true, then we would have that $\theta_6$ exists. However, Milgram’s argument fails because of a computational mistake
[8].
The paper containing the mistake is
R. J. Milgram, "Symmetries and operations in homotopy theory" Amer. Math. Soc. Proc.
Symposia Pure Math., 22(1971), 203-211
and the other reference is private communication with Robert Bruner.
Thank you. So this is where the mistake was. By the way, in the Remark I mentioned Mahowald have really said "it can be shown that $\theta_4^2=0$". And yes it can, except this was actually done 40+ years later! (Theorem 1.2 in this paper of Xu.) Amazing. I am still curious though why Shtan'ko counted the problem as solved in 1970s. Was there a time when experts were optimistic about it?
@RobertBruner It would be great if you could add more details
There seems to be some confusion about the difference between "A. Milgram" and "R.J. Milgram". The latter used a version of his middle name "James" and was a prolific algebraic topologist at Stanford (now retired).
I'm feeling mischievous. The history of the Kervaire invariant problem is strewn with false proofs, Jim Milgram's is only one of many. A student of mine (who I will leave nameless) had a preprint (around 1980?) that solved the problem but that Mark Mahowald quickly shot down. The most recent example I know of is that of a Russian mathematician (who I will also leave nameless). See Math Reviews MR2590025 (2010k:55031)
Differentials of the Adams spectral sequence and the Kervaire invariant (Russian) Dokl. Akad. Nauk 427 (2009), no. 5, 601–604; translation in Dokl. Math. 80 (2009), no. 1, 573–576. From the text (translated from the Russian): "In this paper, we study the differentials of the Adams spectral sequence for stable homotopy groups of spheres and solve the Kervaire invariant one problem for n-dimensional manifolds when $n=2^i-2$, $i\geq 6$." That is a four page paper. Would that it were so simple!
I'll try to give more precise detail soon, but here's my understanding of this history. The 'proof' Peter May mentions is the 'standard mistake' in the subject.
Consider elements in a spectral sequence coming from the homotopy exact couple of a tower. If you have a geometric construction that the boundary of $x$ is $y$, and you can show that $y$ is itself null-homotopic, it is tempting to think you have shown that $x$ survives to a non-zero class. However, $x$ is one of the reasons that $y$ is null-homotopic, so you haven't really observed anything about $x$ from knowing only that $y$ is null-homotopic. What you need is that $y$ was already null-homotopic before $x$ got there to kill it. In other words, you need that $y$ is null-homotopic in an appropriately high stage of the tower, not just in the $0^{th}$ term. For an example related to this case, see pp. 38-39 of
http://www.rrb.wayne.edu/papers/fin_conj_handout.pdf
The mistake Milgram made in "Symmetries and operations" was just a miscalculation in the $\Sigma_4$ extended power of the $30$-sphere, or perhaps of $S^{30} \cup_2 e^{31}$. It did seem like this would give $\theta_6$ in the $126$-stem, given what was then known about $\theta_4$, until the mistake was noticed, and apparently Shtan'ko wrote his report during this burst of enthusiasm.
Shtan'ko's 'A. Milgram' was just a mistake. Surely he meant R. J. Milgram'.
|
2025-03-21T14:48:29.615722
| 2020-01-10T12:01:59 |
350139
|
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|
Stack Exchange
|
Why is $\Delta - p_0 - p_{2}$ a projector?
I apologize in advance, since I am probably doing a very naive mistake in my computation. I am learning about pure (Chow / Grothendieck) motives. One of the first steps is to consider the category where objects are smooth projective varieties and morphisms from $X$ to $Y$ are defined to be correspondences of degree $0$ from $X$ to $Y$, i.e., algebraic cycles on $X \times Y$ of codimension $\dim X$ (up to some adequate equivalence relation). Composition works as follows: given a cycle on $X \times Y$ and one on $Y \times Z$, I pull them back to $X \times Y \times Z$ and then intersect them and push forward the result to $X \times Z$. The identity on $X$ is the graph $\Delta$ of the diagonal morphism. A projector is a correspondence whose square is equal to itself. Now, I can prove that given a curve $X$ and a point $P$ on $X$, both $p_0 := P \times X$ and $p_{2} := X \times P$ are projectors on $X$. But I am stuck proving that $p_1 := \Delta - p_0 - p_{2}$ is also a projector. Can someone please point out where I am making a mistake? I am computing
$$p_1^2 = \Delta^2 + p_0^2+p_2^2 -\Delta \circ p_0 - \Delta \circ p_2-p_0 \circ \Delta - p_2 \circ \Delta + p_0 \circ p_2 + p_2 \circ p_0$$
$$= \Delta +p_0 + p_2 -p_0-p_2-p_0-p_2+X \times X +P\times P $$
$$ = p_1 + X \times X+P\times P. $$
But I can really not see why the last two summands should be $0$. The mistake is probably in the computation of $p_0 \circ p_2$ and $p_2 \circ p_0$, since I do not even obtain cycles of the right codimension. But it seems to me that, when I compute $p_2 \circ p_0$, pulling back $p_0$ (first two factors) gives $P \times X \times X$, pulling back $p_2$ (last two factors) gives $X \times X \times P$, their intersection is $P \times X \times P$ and pushing forward to the first and third factor gives $P \times P$. And similarly for $p_0 \circ p_2$. What am I doing wrong?
Your equality is in $CH^1(X\times X)$, how could you get $X\times X$ or $P\times P$???
@abx I remarked by myself in the question that this cannot possibly make sense. But I don’t see why. So the question is, how can I correct it? How does the correct argument go?
Indeed, $p_0 \circ p_2$ and $p_2 \circ p_0$ are $0$. For $p_0 \circ p_2$, we have to cap $P \times X \times X$ with $X \times X \times P$ and pushforward onto the first and third factor. The cap product is just the intersection $P \times X \times P$. However, the projection of $P \times X \times P$ onto the first and third factor has positive dimensional fibers, so the pushforward of $[P \times X \times P]$ is $0$. (continued)
For $p_0 \circ p_2$, we have to cap $[X \times P \times X]$ with itself. But you can't compute a self cap-product by literally intersecting the subvariety with itself! You have to either look at the normal bundle, or else replace one of the two copies with an equivalent cycle. If your equivalence relation is algebraic equivalence (or anything coarser), then the second is easier. Let $Q$ be a different point than $P$, then $[X \times P \times X] = [X \times Q \times X]$ and $(X \times P \times X) \cap (X \times Q \times X)=0$, so the self intersection of $[X \times P \times X]$ is zero.
Thanks a lot, I was indeed using a very naive notion of intersection and push forward. Now everything is clear
|
2025-03-21T14:48:29.615948
| 2020-01-10T12:10:18 |
350140
|
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|
Stack Exchange
|
Compute lower bound on $\min_{E} \mathcal N(0,\sigma^2 I_n)(E)$ subject to $vol(E \cap H_n(r)) / vol(H_n(r)) \ge p$ where $H_n(r)$ is $n$-hemisphere
Let $n \ge 2$ be an integer, which may be assumed to be very large. For $r > 0$, consider the hemi-sphere $H_n(r) := S_n(r) \cap (\mathbb R^+ \times \mathbb R^{n-1})$, where
$$
S_n(r):= \{x \in \mathbb R^n \mid \|x\|_2 \le r\}
$$
is the $n$-sphere of radius $r$.
and consider the measure $\lambda_n$ define on borel-measurable subsets of $\mathbb R^n$ by
$$
\lambda_n(E; r) := \frac{vol(E \cap H_n(r))}{vol(H_n(r))}.
$$
Finally, let $\sigma > 0$ and $\mathcal N(0,\sigma^2 I_n)$ be the standard $m$-dimensional gaussian distribution with variance $\sigma^2$.
Question 1. For $p \in (0, 1)$, what is the borel-subset $\hat{E}$ of $\mathbb R^n$ which solves
$$
\min_{E} \mathcal N(0,\sigma^2 I_n)(E)\text{ subject to } \lambda_n(E; r)\ge p. \tag{1}
$$
Question 2. Same question with the hemisphere $H_n(r)$ replaced with the sphere $S_n(r)$.
Question 4. Same questions, but just provide the optimal objective value in (1), or a good lower-bound thereof.
Notes
Of course, any helpful pointers / refs for addressing such problems are more than welcome
Edit
Question 3. Cancelled as it would make sense in general.
Let $\nu:=\mathcal N(0,\sigma^2 I_n)$ and $p\in[0,1]$.
Concerning Question 1: For the minimizing $E$, without loss of generality (wlog) we clearly have $E\subseteq H_n(r)$ and $\lambda_n(E;r)=p$. It is then clear that, subject to these conditions, wlog the set $E$ must consist of the points in $H_n(r)$ with the smallest density of $\nu$ wrt the Lebesgue measure; here, one might want to recall the Neyman–Pearson (NP) lemma. So, wlog
$$E=E_*:=H_n(r)\setminus H_n(s), \tag{1}
$$
where $s$ is the solution to the equation $\lambda_n(H_n(r)\setminus H_n(s);r)=p$, which can be rewritten as
$$1-(s/r)^n=p.
$$
(See details on (1) below.)
The answer to Question 2 is quite similar. This also answers Question 4.
As for Question 3, it does not quite make sense as stated, because the limits $\lambda_n(E;0)$ and $\lambda_n(E;\infty)$ will in general not exist, and also in view of the answers to Questions 1 and 2. Perhaps, Question 3 was intended as a relaxed, limit version of Questions 1 and 2.
Details on (1): Basically, here I will give a proof of the NP lemma, which I think is better than the one given in the linked Wikipedia article. Let $\lambda$ denote the Lebesgue measure on $\mathbb R^n$, and let $g:=\frac{d\nu}{d\lambda}$. Let real $c>0$ be such that for all $x\in H_n(r)$ we have $g(x)<c\iff x\in E_*$. Then for any Borel $E\subseteq H_n(r)$
$$(1_E-1_{E_*})(g-c)\ge0. \tag{2}
$$
Assuming now that $\lambda(E)=\lambda(E_*)$, expanding the left-hand side of (2), and then integrating there wrt $\lambda$, we get $\nu(E)\ge\nu(E_*)$, as claimed.
Great, thanks for the answer. Indeed Question 3 was asked in a hurry. Deleted. Do you mind adding a one-line sentence explaining how you've applied NP here ?
Intuitively, by symmetry arguments, the set which solves the set can be taken a horse-shoe $H_n(r)\setminus H_n(s)$ as you're proposed (then one would chose $s$ to meet the volume constraint). What I'm missing is the NP link.
@dohmatob : I have added the details on the NP lemma.
OK great, thanks for the details.
On second thought, one thing which seems mysterious in this solution is that the gaussian distribution $\nu$ has barely been used in the arguments. Q: what property of $\nu$ has been used by the construction of the solution, apart (perhaps) the fact that $\nu$ has density w.r.t Lebesgue ?
@dohmatob : The following two properties of the density $g$ were used in the answer: (i) that $g(x)$ depends only on $|x|_2$ and (ii) that $g(x)$ is a decreasing continuous function of $|x|2$. These two properties were used in the sentence "Let real $c>0$ be such that for all $x\in H_n(r)$ we have $g(x)<c\iff x\in E*$."
OK, makes sense. Thanks. Then your construction should hold for densities of the form $\varphi(-|x|_p)d\mathcal L(x)$, where $\varphi$ is increasing nonnegative and $p \in [1,\infty]$, and the $\ell_2$ hemisphere replaced with the $\ell_p$ hemisphere.
|
2025-03-21T14:48:29.616240
| 2020-01-10T12:19:41 |
350142
|
{
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"R. N. Marley",
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}
|
Stack Exchange
|
Vlasov Poisson: linear momentum conservation
The 3-dimensional Vlassov -Poisson equation I am studying at university is
$$ \partial_t f (t,x,v) + v\cdot \nabla_x f (t,x,v) - \nabla_x \phi (t,x) \nabla_v f (t,x,v) =0,$$
where $$\Delta \phi = 4\pi\gamma \rho (t,x,v) \text{ and } \rho (t,x) = \int_{\mathbb{R}^3} f(t,x,v)\ dv$$
I am trying to prove the momentum is conserved:
$$q(t) = \int_{\mathbb{R}^6} vf(t,x,v)\ dxdv \Rightarrow q' (t)=0$$
I need some help to prove that This is my attempt,
I first substitute the equation of $f$ into the expression:
$$ q'(t) = \int \partial_t [vf(t,x,v)]\ dxdv = \int v [-v\cdot \nabla_x f (t,x,v) + \nabla_x \phi (t,x) \nabla_v f (t,x,v)]\ dxdv$$
Secondly, I noticed that the first term is equal to zero after an integration by parts (passing the $x$-derivative to $|v|^2$ which does not depende on $x$), therefore
$$ q'(t)= \int v\nabla_x\phi(t,x)\nabla_v f(t,x,v)\ dxdv $$
What can I do next to show that $q' =0$? Thanks in advance!
It helps to consider components of $q$, in order to keep the notations clear. From what you have we can write
$$ q_i'(t) = \int v_i \sum_{j} \nabla_{x_j} \phi \nabla_{v_j} f ~\mathrm{d}x ~\mathrm{d}v $$
integrate by parts in $v$, and using that $\phi$ is independent of $v$ and $\nabla_{v_j} v_i = \delta_{ij}$ we get
$$ q_i'(t) = - \int \nabla_{x_i} \phi f ~\mathrm{d}x ~\mathrm{d}v $$
perform the $v$ integral first since $\phi$ is $v$-independent
$$ q_i'(t) = - \int \nabla_{x_i} \phi \rho ~\mathrm{d}x $$
by definition, and using the definition of the gravitational potential you get
$$ q_i'(t) = - \frac{1}{4\pi\gamma} \int \nabla_{x_i} \phi \Delta \phi ~\mathrm{d}x $$
Integrating by parts the Laplacian you get
$$ q_i'(t) = \frac{1}{8\pi\gamma} \int \nabla_{x_i} ( |\nabla\phi|^2 ) ~\mathrm{d}x = 0 $$
Nice! Thank you very much Willie +1
|
2025-03-21T14:48:29.616495
| 2020-01-10T12:57:28 |
350144
|
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"Maxime",
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|
Stack Exchange
|
Convergence of Chebyshev interpolation in L^1
Let $f\in C^0([-1,1])$ and $P_n(f)$ its interpolation polynomial at the Chebyshev nodes.
I would be interested to know about any existing results (positive or negative) about the convergence of $P_n(f)$ to $f$ in $L^1([-1,1])$ (only assuming that $f$ is continuous).
The negative results I'm aware of (existence of a continuous function $f$ for which $P_n(f)(x)$ does not converge for any $x$ in $[-1,1]$, from Marcinkiewicz and Grünwald (1936)) do not seem to leave much hope, but still they are not strong enough to rule out convergence in $L^1$. Any additional insight would be welcome!
If I understand it right what the Chebyshev nodes are, you are just interpolating continuous functions on the unit circle with the property $f(\bar z)=f(z)$ by trigonometric polynomials of degree $n$ over $2n+2$ equidistant nodes. This creates a bounded operator from $C$ to $L^2$ on the circle, so you have convergence in $L^2(\frac{dx}{\sqrt{1-x^2}})$ in the original problem, which is stronger than what you are asking for. Am I misunderstanding something?
Dear fedja, thanks for your answer!
Chebyshev nodes are usually defined as $\cos((2k+1)\pi/(2n))$, $k=0,\ldots,n$, or as $\cos(k\pi/n)$, $k=0,\ldots,n$, and in both cases you can indeed view the problem on the circle as you described (but in the second case, which is the one I'm mostly interested in, you only have $2n$ different points on the circle).
Could you please elaborate (or give me a reference) regarding the rest of your arguement? (I agree that once we have convergence in $L^2(\frac{dx}{\sqrt{1-x^2}})$ we have convergence in $L^1$)
Here are the details (for the second definition of Chebyshev nodes). We can define the new continuous function $g$ on the circle by $g(z)=f(\frac{z+z^{-1}}2)$ and consider the interpolation by the trigonometric polynomials $Q_n(z)=\sum_{k=0}^n b_k\frac{z^k+z^{-k}}{2}$ (so that $P_n(\frac{z+z^{-1}}{2})=Q_n(z)$) on the $2n$-th roots of unity.
The square of the $L^2$ norm of $Q_n$ on the circle is (up to a normalization factor) just $\sum_{k=-n}^n|c_k|^2$ where $c_0=b_0$ and $c_k=\frac {b_{|k|}}2$ for $1\le k\le n$ (so $Q_n(z)=\sum_{k=-n}^n c_kz^k$.
Notice that $z^n= z^{-n}$ on every node $z$, so we can just as well consider the modified polynomial $\widetilde Q_n(z)=\sum_{k=-(n-1)}^{n-1}c_kz^k+b_nz^n$. It will have the same values on the nodes but now all the participating powers of $z$ will be orthogonal with respect to the counting measure on the nodes, so we will have
$$
\sum_{k=-(n-1)}^{n-1}|c_k|^2+2(|c_n|^2+|c_{-n}|^2)=\frac 1{2n}\sum_{z:z^{2n}=1}|\widetilde Q_n(z)|^2\\ =\frac1{2n}\sum_{z:z^{2n}=1}|g(z)|^2\le\|g\|_C^2=\|f\|_C^2
$$
and the declared boundedness of the interpolation operator from $C$ to $L^2$ on the circle follows immediately.
The rest is the usual mumbo-jumbo. We have a sequence of linear interpolation operators $I_n:g\mapsto Q_n$ whose norms from $C$ to $L^2$ are uniformly bounded by some constant $M$. Now decompose $g$ into $g_m+h_m$ where $g_m$ is a trigonometric polynomial of degree $m$ and $\|h_m\|\to 0$ as $m\to\infty$ and notice that $I_ng_m=g_m$ for $n>m$, say, so
$$
\|g-I_ng\|_{L^2}=\|h_m-I_nh_m\|_{L^2}\le (1+M)\|h_m\|_C.
$$
The rest should be clear, but feel free to ask questions if needed :-)
Thanks for the clarification! Now that I know the answer, and that you made me think about going via the $L^2$ norm, I think this can also be proved by using the $L^2$ convergence of the Chebyshev series (the Chebyshev polynomials form a Hilbert basis for the weigth $\frac{1}{\sqrt{1-x^2}}$), and then the "aliasing" relation between the Chebyshev series and the Chebyshev interpolant. (I'm not saying that this way is any better, maybe just closer to the type of arguments I'm used to)
@Maxime You should be a bit careful with that idea because we have boundedness $C$ to $L^2$, not $L^2$ to $L^2$, so pure $L^2$ convergence of the series is insufficient and you'll have to invoke some additional ideas as well. On the other hand, you can, of course, rewrite the argument without the trigonometric substitution if you feel like you would prefer it that way.
|
2025-03-21T14:48:29.617074
| 2020-01-10T15:02:26 |
350154
|
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|
Stack Exchange
|
Triangles with a given outer Soddy circle of the Malfatti circles
I did a JavaScript interactive picture of the Malfatti circles of a triangle. The user can drag the vertices of the triangle and the Malfatti circles are updated accordingly.
Now, I would like to restrict the transformations of the original triangle in order that they preserve the outer Soddy circle of the Malfatti circles (the goal is to do an interactive Apollonian gasket with a fixed exterior circle).
In other words, let's say that the outer Soddy circle of the Malfatti circles of a triangle is the Soddy-Malfatti circle of this triangle; what are all the triangles which have a given Soddy-Malfatti circle ?
EDIT
Here is the interactive picture (click and drag). How to move the triangle in such a way that the outer circle remains fixed?
This is a suggestion on how to compute the required (two parameter) family of triangles which have a given $C$ as Soddy-Malfatti circle (a comment since I haven‘t done the computations). Wlog $C$ is the unit circle with centre at the origin of a coordinate plane. Each triangle is directly similar to one of the form $ABC$ with $A=(0,0)$, $B=(1,0)$ and $C=(p,q)$. We now compute the centre $M(p,q)$ and radius $r(p,q)$ of its Soddy-Malfatti circle. A translation by $–M(p,q)$ and dilation by $\frac 1 {r(p,q)}$ maps $ABC$ into a suitable triangle. Doing this for each $p$ and $q$ suffices.
@user131781 Thank you, this works! See the animation :-)
The new JS picture. The two sliders give the values of $p$ and $q$. @user131781 Maybe you can convert your comment to an answer?
nice graphics. will do
At the request of the poster, I am expanding my comment to an answer. We are given a circle $\cal C$ which we take to be the unit one with centre at the origin of the coordinate plane. Now if $PQR$ is a triangle in the plane, there is a unique triangle $ABC$ with $A=(0,0)$, $B=(1,0)$ and $C=(p,q)$ to which it is directly similar. (This is a common ploy in triangle geometry which seems to have been discovered and forgotten several times but goes back at least to Euler). We compute the Soddy-Malfatti circle of $ABC$
and denote its centre by $M(p,q)$, its radius by $r(p,q)$. If we translate the original triangle by $-M(p,q)$, then dilate it by $\frac 1 {r(p,q)}$, we obtain one with $\cal C$ as Soddy-Malfatti circle. Repeating this for all values of $p$ and $q$ gives a two parameter family of triangles with the required properties.
Thanks. Here is the Apollonian gasket.
It has just occurred to me that I missed one degree of freedom—one actually gets a three parameter family (as one would expect—the family of triangles has dimension 6 and one has three restraints, radius and centre of the S-M circle). The missing dimension can be restored by rotating about the centre of the S-M circle. You see this in your graphics—all the triangle bases are horizontal.
I added a degree of freedom to $B$ like this: $B=(1,y)$. Graphics, with $y$ corresponding to the third slider. Does that sound correct?
Yes— looks great. you now see the triangles rotating. By the way, the same method could be used with respect to other circles associated with a triangle—the nine point circle springs as first to mind (circum- in- and ex-circles are probably not too interesting), but there are many others.
|
2025-03-21T14:48:29.617336
| 2020-01-10T16:45:08 |
350163
|
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"Alexandre Eremenko",
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|
Stack Exchange
|
Is there an easy way to solve an "almost quadratic" equation?
I have an equation of the form:
$ax^{k_1} + bx^{k_2} + cx^{k_3} + dx^{k_4} + e = 0$
where
$a$, $b$, $c$ and $d$ are arbitrary real numbers; $k_1$ and $k_2$ are positive reals in the range of 1.92...2.08; $k_3$ and $k_4$ are positive reals in the range of 0.96...1.04.
It is known that there exist two real $x$ that satisfy the equation. We are interested in finding both. At least one of these $x$ is positive.
Is there a relatively lightweight way to solve this equation numerically?
By "relatively lightweight" I mean that an operation as complex as a logarithm should be taken no more than just a handful of times (say, 20 or so).
Right now the best idea I have is to assume that $k_1$ and $k_2$ are strictly equal to 2 and $k_3$ and $k_4$ are strictly equal to 1. This way the equation becomes quadratic and I can solve it by taking just a single square root, but I wonder if I can have more precise solution.
Would the task be easier if we can get another set(s) of coefficients {$a,b,c,d,k_1,k_2,k_3,k_4$} having approximately the same solution?
Yes, and it is called Newton method. First you localize the roots by plugging certain values, then run Newton method, until you find the roots.
@AlexandreEremenko I usually run Newton simultaneously with bijection (to have some sanity check at each step), which doubles the running time in the case when everything is nice but eliminates problems in ugly scenarios. However localizing the roots may present some problem in the case when crossings are near touches, so that part may require some thought unless some a priori information is known from other considerations.
I made up an algorithm in 1984 which worked for an arbitrary number of variables, however, all coefficients and exponents were positive, except that the free term was on the other side of the equation (i.e. this one was negative after all :) ). I believe that my algorithm was much faster than Newton's algorithm in the given case - I didn't do a solid error analysis (it was just a 1-page handwritten memo which I've written for my colleagues at a company).
@fedja: "bijection" --> bisection?
If you just plot your function with any software to see where the roots are, and then run the Newton method if you need high accuracy. Since you know the number of roots in advance, no problems can arise.
@AlexandreEremenko 1) Yes, of course :-) 2) My understanding of the problem was that you need a quick and reliable solver for dynamic inputs of $a,\dots, k_4$ under the assumption that there are always exactly $2$ roots for some external reason, in which case "plotting with any software" is totally out of question.
I suppose they mean mean positive roots (since the exponents are not integers). Their number can be estimated by Decartes rule in any case.
|
2025-03-21T14:48:29.617572
| 2020-01-10T17:59:20 |
350172
|
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"Martin Brandenburg",
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"url": "https://mathoverflow.net/questions/350172"
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|
Stack Exchange
|
"discrete" objects of a $2$-category
Let $\mathcal{K}$ be a $2$-category. Is there a special name of those objects $B \in \mathcal{K}$ which have the property that the category $\mathrm{Hom}_{\mathcal{K}}(B,C)$ is essentially discrete for all $C \in \mathcal{K}$? This means that for every two morphisms $f,g : B \to C$ any $2$-morphism $f \to g$ is an isomorphism, and it is unique if it exists.
Also, is there some intuition what this property actually means, intuitively?
Assuming that $\mathcal{K}$ has more properties (such as the existence of certain limits), can we simplify the property perhaps to some "internal" information of $B$?
Here is the example which motivates my question: Consider the $2$-category $\mathrm{Cat}_{c\otimes/R}$ of cocomplete $R$-linear tensor categories and let $A$ be a commutative $R$-algebra. Then $\mathrm{Mod}(A) \in \mathrm{Cat}_{c\otimes/R}$ has this property, since $\mathrm{Hom}_{c\otimes/R}(\mathrm{Mod}(A),\mathcal{C}) \simeq \mathrm{Hom}_R(A,\mathrm{End}(1_\mathcal{C}))$ and the latter is a set. I have recently proven that $\mathrm{Qcoh}(X)$ has this property as well, where $X$ is any quasi-compact quasi-separated $R$-scheme. When $X$ is an algebraic stack, usually $\mathrm{Qcoh}(X)$ doesn't have this property.
I would call those objects "codiscrete" or "co-0-truncated", since "discrete" and "0-truncated" are used for the dual property, e.g. here and here. It's equivalent to saying that $B$ is equivalent to the copower $B \odot (\cdot \rightrightarrows \cdot)$.
Thanks, Mike! In fact I looked for "discrete object" at the nlab, but only saw the "wrong" page: https://ncatlab.org/nlab/show/discrete+object
@MartinBrandenburg Note that that page links to the correct page in the second paragrah of the "Idea" section.
|
2025-03-21T14:48:29.617747
| 2020-01-10T18:14:51 |
350173
|
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|
Stack Exchange
|
Relationship between $q$-Weyl dimension formula and $q$-analog of weight multiplicity?
$\DeclareMathOperator\dim{dim}$For a dominant (integral) weight $\lambda$ and any (integral) weight $\mu$ of a simple Lie algebra $\mathfrak{g}$, Lusztig's $q$-analog of weight multiplicty $K_{\lambda,\mu}(q)$ is a $q$-analog of the dimension $\dim V^{\lambda}_{\mu}$ of the $\mu$-weight space $V^{\lambda}_{\mu}$ of the highest weight $\mathfrak{g}$-irrep $V^{\lambda}$ with highest weight $\lambda$. There are a few ways to define it, but probably the simplest is to think of it as the Hilbert series associated to a filtration on $V^{\lambda}_{\mu}$ induced by the action of any regular nilpotent element of $\mathfrak{g}$. See for instance Lusztig's original paper "Singularities, character formulas, and a $q$-analog of weight multiplicities" and Joseph, Letzter and Zelikson - "On the Brylinski–Kostant filtration". In Type A, the $K_{\lambda,\mu}(q)$ are called Kostka–Foulkes polynomials and sometimes this name is used in other types too.
(Other ways to define $K_{\lambda,\mu}(q)$: via a $q$-analog of Kostant's partition function; in terms of intersection cohomology of Schubert varieties; as certain affine Kazhdan–Lusztig polynomials.)
On the other hand, let us define the $q$-dimension of $V^{\lambda}$ to be $\dim_q(V^{\lambda}) \mathrel{:=} \sum_{\mu} (\dim V^{\lambda}_{\mu}) q^{\langle \mu, \rho^{\vee}\rangle}$, where $\rho^{\vee}$ is the dual of the Weyl vector $\rho$, which is the sum of the fundamental weights. I believe the Weyl character formula tells us that $\dim_q(V^{\lambda}) = \prod_{\alpha \in \Phi^+} \frac{[\langle\lambda+\rho,\alpha\rangle]_q}{[\langle\lambda,\alpha\rangle]_q}$, where $\Phi^+$ are the positive roots of $\mathfrak{g}$ and $[m]_q \mathrel{:=} \frac{(q^{1/2}-q^{-1/2})^m}{(q^{1/2}-q^{-1/2})}$. Maybe I'm slightly off here, but something more-or-less like this should be true, and I think the notion of $q$-Weyl dimension formula is anyways an established thing.
Question: what is the relationship between the $K_{\lambda,\mu}(q)$ and $\dim_q(V^{\lambda})$? In particular, is there some way to write $\dim_q(V^{\lambda}) = \sum_{\mu} c_{\lambda,\mu}(q) K_{\lambda,\mu}(q)$ for some "simple" coefficients $c_{\lambda,\mu}(q)$?
Note that in Type A we have $s_{\lambda}(\mathbf{x}) = \sum_{\text{dominant $\mu$}} K_{\lambda,\mu}(q) P_{\lambda}(\mathbf{x})$, where $s_{\lambda}(\mathbf{x})$ is the Schur polynomial and $P_{\lambda}(\mathbf{x})$ is the Hall–Littlewood polynomial. And the $q$-Weyl dimension is essentially a principal specialization of $s_{\lambda}(\mathbf{x})$, and I believe the corresponding specialization of $P_{\lambda}(\mathbf{x})$ should give something like a $q$-binomial although I didn't fully work it out.
My main conceptual difficulty in linking $K_{\lambda,\mu}(q)$ and $\dim_q(V^{\lambda})$ is that, for $\dim_q(V^{\lambda})$, each weight space contributes a single power of $q$, while for the $K_{\lambda,\mu}(q)$ the weight spaces contribute many different powers of $q$.
But I still suspect something like what I'm asking for should be true and probably well-known, although I'm having trouble googling for the answer.
EDIT: I just noticed that in a comment to this MathOverflow answer of Jim Humphreys, Victor Protsak says "There is an interesting $q$-analogue of the character of a finite-dimensional module that involves the notion of $q$-multiplicity of weight due to Lusztig, and it specializes into a natural $q$-analogue of the Weyl dimension formula." An expansion of Victor's comment would likely answer my question.
For the coefficients $c_{\lambda,\mu}(q)$: I am happy to either sum over all $\mu$ or all dominant $\mu$. When $\mu$ is not dominant there are two different things $K_{\lambda,\mu}(q)$ could mean: one is defined in terms of the $q$-Kostant partition function and is "bad" because it can have negative coefficients; the one I want is the one coming from the Brylinski-Kostant filtration. It is known that this second one is equal to $q^{\textrm{some explicit power}} \cdot K_{\lambda,\nu}(q)$ where $\nu$ is the dominant representative of $W\mu$ (see the Joseph-Letzter-Zelikson paper).
I think that the best reference for what I mention here is Stembridge's notes "Kostka-Foulkes Polynomials of General Type". The picture you describe for type A actually generalizes nicely to all types.
Let's denote by $\Lambda$ the weight lattice, by $\Lambda^{+}$ the dominant weights, $W$ the Weyl group, and by $\chi_{\lambda}=\sum_{\mu} \dim(V^{\lambda}_{\mu})e^{\mu}$ the Weyl characters that live in the ring $\mathbb Z[\Lambda]^W$. There are Hall-Littlewood polynomials $P(\lambda,q)$ defined in the notes above, that satisfy the important identity
$$\chi_{\lambda}=\sum_{\mu\in \Lambda^+}K_{\lambda, \mu}(q)P(\mu,q) \tag{1}$$
In Lusztig's paper that you refer to, he proves this for one of the definitions of the $K_{\lambda,\mu}$ (Kazhdan-Lusztig for affine Weyl group) and states that conjecturally this is the same if one defines them by taking the q-analog of the Kostant's partition function (conjecture 9.6). This conjecture was proved by Kato in "Spherical Functions and a q-Analogue of Kostant's Weight Multiplicity Formula" using methods from spherical analysis on p-adic groups. A more elementary proof was given by R. K. Brylinski in "Characters and the q-analog of weight multiplicity".
In your question you want to take the principal specialization of $(1)$, so that $\chi_{\lambda}$ becomes $\operatorname{dim}_q(V^{\lambda})$. As a result, $c_{\lambda,\mu}(q)$ becomes the principal specialization of the Hall-Littlewood polynomial $P(\mu,q)$. Just like in type A, this is always a nice q-product. In fact, this was one of the motivating examples behind the evaluation conjecture for Macdonald Polynomials (See conjecture 12.10 in Macdonald's "Orthogonal polynomials associated with root systems", and the following discussion in (v) where it is explained that the principal specialization is known for the Hall-Littlewood case). The general principal specialization of Macdonald polynomials was proven later by Cherednik in "Macdonald's Evaluation Conjectures and Difference Fourier Transform".
Thanks, especially for all the references; this is very illuminating.
Is it possible that the principal specialization of the $P(\mu,q)$ gives exactly the sum of all the $q^{\textrm{some explicit power}}$ (from my comment under the question) for the Weyl orbit of $\mu$? So that one could simply sum all the $K_{\lambda,\mu}(q)$ over all including non-dominant $\mu$ to get the $q$-Weyl dimension?
Disclosure of interest in this: Rhoades (https://arxiv.org/abs/1005.2568) proved cyclic sieving results for both the Kostka-Foulkes polynomials (see also Fontaine-Kamnitzer https://arxiv.org/abs/1212.1314), and the $q$-Weyl dimension polynomials. But his method of relating the two results is kind of subtle. I was wondering if the Kostak-Foulkes result formally implies the $q$-Weyl dimension result.
@SamHopkins I haven't looked at Rhoades paper in a long time, but here are some superficial comments: In type A there are only so many natural candidates for q-analogs, so you will see $\frac{\text{q-thingy}}{\prod [h]_q}$ a lot. As Rhoades remarks, such expressions are $q$-dimensions but also Kostka-Foulkes polynomials. When you upgrade to the $q,t$-analogs you will see some expressions start to diverge, for example some $[h]_q$ will become $(1-q^at^{l+1})$ and some will be $(1-q^{a+1}t^l)$, similarly some expressions will start to diverge once you generalize outside of type A. [continued]
If anything, the main idea in these papers is that the Kazhdan-Lusztig theory is quite crucial to get these cyclic sieving results.
@GjergjiZaimi Well, so far... :)
@PerAlexandersson Yup, so far :) I think it would be very cool to have cyclic sieving of q-dimensions (especially in other root systems) that didnt rely on geometric interpretation of Kostka-Foulkes polynomials.
what you want is $(8.10)$ of
Lusztig, George (1-MIT)
Singularities, character formulas, and a q-analog of weight multiplicities. Analysis and topology on singular spaces, II, III (Luminy, 1981), 208–229,
Astérisque, 101-102, Soc. Math. France, Paris, 1983.
available here
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2025-03-21T14:48:29.618297
| 2020-01-10T18:53:56 |
350175
|
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Stack Exchange
|
W H Lin's thesis and Hopf subalgebras of the Steenrod algebra
If $B$ is a subalgebra of $A$, you can ask whether the $B$-module structure on $B$ can be extended to give an $A$-module structure on $B$.
W H Lin, in his 1973 PhD thesis at Northwestern, showed that the only Hopf subalgebras of the mod 2 Steenrod algebra for which this can be done are the algebras $A(n)$ — this is the algebra generated by $\text{Sq}^{2^i}$ for $i\leq n$. Are there any electronic copies of the thesis, or at least this particular proof, available?
this is the library entry --- no digital copy I'm afraid.
Using @CarloBeenakker's answer, our librarian found an electronic version, produced from the microfilm copy of the original: https://search.proquest.com/docview/302701183 (full text may require accessing through a university library or similar). At a quick glimpse, it looks complete.
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2025-03-21T14:48:29.618513
| 2020-01-10T19:02:15 |
350177
|
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Stack Exchange
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Monoidal categories from the projective modules of a ring
Let $R$ be a not necessarily commutative ring, and denote by $_R\mathrm{lp}_R$ the category of $R$-bimodules, which are finitely generated projective as left modules, with morphism $R$-bimodule maps, denoted by $_R\mathrm{Hom}_R(M,N)$, for any two objects $M,N$.
i) Is $_R\mathrm{lp}_R$ a monoidal category? In other words is $M \otimes_R N$ again projective as a left $R$-module?
ii) Denote by $_R\mathrm{Hom}(M,R)$ the left $R$-module maps from $M$ to $R$. Is $M^* := _R\mathrm{Hom}(M,R)$, endowed with its usual bimodule structure, projective as a left $R$-module?
iii) If $M^*$ is projective, then is it a dual for $M^*$, that is, is $_R\mathrm{lp}_R$ a rigid monoidal category?
Yes to (i).
It's easier to think about if you weaken/generalize.
Suppose that $M$ is an $(R,S)$-bimodule and $N$ is a left $S$-module. If $N$ is projective then the $R$-module $M\otimes_SN$ is a summand of a direct sum of copies of $M\otimes_SS\cong M$, so that it is a projective module if $M$ is.
No to (ii).
$_RHom(M,R)$ gets its left module structure module from the right module structure of $M$. You can have an example where $R$ is a domain and where $M$ as a right module is a torsion module and $_RHom(M,R)$ as a left module is a torsion module. For example, let $R=k[X]$, $k$ a field, take $M$ to be $R$ with the usual left module structure but right module structure given by $f(X)g(X)=f(X)g(0)$. (Bad notation, but I think you get the idea.)
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2025-03-21T14:48:29.618637
| 2020-01-10T20:37:12 |
350183
|
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Stack Exchange
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Confusion regarding a definition of cycles
For a projective variety $X$ over $\mathbb{C}$, let us denote by $CH_k(X)$ the Chow group of $k$-cycles of $X$, modulo rational equivalence. Also, let $CH_k(X)_{hom}$ denote $k$-cycles modulo homological equivalence.
I know that $CH_k(X,\mathbb{Q}) = CH_k(X)\otimes \mathbb{Q}$ (follows from flatness of $\mathbb{Q}$ over $\mathbb{Z}$). My question is:
In certain papers they use the notation $CH_k(X,\mathbb{Q})_{hom}$ without the definition; is it also same as $CH_k(X)_{hom}\otimes \mathbb{Q}?$ I just want to be sure without getting into trouble later.
Thanks in advance!
Yes, they are same. By definition $ CH_k(X)_{hom} :$= kernel of the cycle class map , $CH_k(X)\to H_{2k}(X, \mathbb{Z})$ also $CH_k(X , \mathbb{Q})_{hom}$ := kernel of the cycle class map $CH_k(X,\mathbb{Q})\to H_{2k}(X, \mathbb{Q})$. Now the assertion follows as $\mathbb{Q}$ is flat over $\mathbb{Z}$.
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2025-03-21T14:48:29.618734
| 2020-01-10T21:20:20 |
350188
|
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Stack Exchange
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Smallest size of a non-measurable set of reals
The question is pretty much the title. I'm wondering if anything is known about the smallest size $\kappa$ of a non-measurable subset of the real numbers (regarding the Lebesgue measure). Since we have $\kappa\geq\aleph_0$ and $\kappa\leq\mathfrak{c}$ with $\kappa=\mathfrak{c}$ at least being consistent (under CH or MA), it might be an interesting cardinal invariant to look at.
If I look at the Wikipedia articles Cardinal characteristic of the continuum (current revision) and
Cichon's diagram (current revision), it seems that this is the cardinal denoted $\operatorname{non}(\mathcal N)$ and $\operatorname{non}(\mathcal L)$.
Knowing this notation might help when searching for results about this cardinal. (And perhaps also some of the references given in those Wikipedia articles might contains some pointers.)
I wasnt aware of the connection to $\text{non}(\mathcal{N})$. So does this just follow, because we can carry out the Vitali construction on any set of positive and finite measure and therefore obtain, for every non-nullset $M$, a non-measurable set $N\subseteq M$?
Well, a measurable set of positive measure necessarily has cardinality $\mathfrak{c}$. So if it has cardinality less than $\mathfrak{c}$ and isn't null, it must already be non-measurable.
Two precise questions (to start with): (a) is it consistent under ZFC that every subset of reals of cardinal $\aleph_1$ is measurable (hence null)? (b) Is it consistent under ZFC that CH fails and there exists a non-measurable subset of cardinal $\aleph_1$?
@YCor Yes to both.
Here is a link to a post about "set of positive measure has cardinality $\mathfrak c$": Cardinality of a set of positive Lebesgue measure.
This post seems also related to the above comments: Do sets with positive Lebesgue measure have same cardinality as R?
I am just going to compile the comments into an answer so i can close this question.
Claim: The smallest size of a non-measurable set is $\text{non}(\mathcal{L})$:
$\geq$: If $A$ is non-measurable, then $A$ is not null.
$\leq$: If $A$ is not null and not of size continuum, then $A$ has to be non-measurable, because the difference set of any set of positive measure has to contain an interval around $0$ (https://en.wikipedia.org/wiki/Steinhaus_theorem) and therefore be of cardinality $\mathfrak{c}$. Because furthermore, the cardinality of the difference set is less than or equal to the cardinality of $A\times A$ which is equinumerous with $A$, $A$ is of cardinality $\mathfrak{c}$.
Therefore either $\text{non}(\mathcal{L})=\mathfrak{c}$ and $\leq$ is trivial or there is a non-null set of cardinality $<\mathfrak{c}$ which, by the argument above, also is a non-measurable set.
|
2025-03-21T14:48:29.618944
| 2020-01-10T23:55:29 |
350195
|
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Stack Exchange
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Hausdorff dimension between $(1,2)$
Is there a number $c \in (1,2)$ for which there exist some interesting geometric property/properties which hold for every set of Hausdorff dimension in $(1,c)$ and does not hold for any set of Hausdorff dimension in $(c,2)$? As is stated this question is somehow vaque but I will welcome any examples of such properties, especially those which can be see ,,with the naked eye''.
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2025-03-21T14:48:29.619000
| 2020-01-11T00:07:16 |
350196
|
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"Mateusz Kwaśnicki",
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Stack Exchange
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Reference (foundamental sol. and grad estimate, etc.): a particular elliptic PDE
In $\mathbb{R}^d$, consider the following equation
$$\Delta u -x\cdot \nabla u = f $$
where $f$ can be $C^\infty$ and decay like $e^{-\frac{c|x|^2}{2}}$.
I would like to know fundamental sol. to this equation or some gradient estimate, preferably pointwise. In particular I'm interested in the following quantity
\begin{align*}
\int_{\mathbb{R}^d}|\nabla u|^2e^{-\frac{|x|^2}{2}}dx.
\end{align*}
One useful observation might be
$$\nabla \cdot (e^{-\frac{|x|^2}{2}}\nabla u)=f e^{-\frac{|x|^2}{2}} .$$
Thank you!
This is called Ornstein–Uhlenbeck evolution, and it has been studied a lot. In particular, an explicit solution is known in terms of the usual heat semigroup.
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2025-03-21T14:48:29.619078
| 2020-01-11T00:23:13 |
350197
|
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Stack Exchange
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Numerical method for simultaneous computation of eigenvalues of a family of commuting matrices
I have a problem where I have $n$ commuting matrices $M_1,\dots,M_n$. It is a well-known fact that commuting matrices are simultaneously diagonalizable/triangularizable. I need to find the eigenvalues of these matrices, but I need to know the eigenvalues grouped up by the common eigenspaces.
In exact arithmetic, this would be as easy as Schur-factorizing $M_1 = UT_1U^*$, and then computing $T_i = U^*M_iU$. However, in floating-point arithmetic, it is my understanding that U may be computed inaccurately when the eigenspaces are poorly conditioned (i.e. when the eigenvalues are clustered).
Is there a stable numerical method for performing this computation?
For small Hermitian (or real symmetric) matrices, yes, but really this is a hard problem not fully solved. See [1,2] for algorithms. The Cardoso paper [2] looks at the non-commuting case, but in the commuting case should minimize the off diagonal errors with respect to the Frobeneius norm.
I don't know about about getting matrices simultaneously into upper triangular form. I would look at papers that cite these two papers.
[1] Bunse-Gerstner, Angelika, Ralph Byers, and Volker Mehrmann. "Numerical methods for simultaneous diagonalization." SIAM journal on matrix analysis and applications 14.4 (1993): 927-949.
[2] Cardoso, Jean-François, and Antoine Souloumiac. "Jacobi angles for simultaneous diagonalization." SIAM journal on matrix analysis and applications 17.1 (1996): 161-164.
Perhaps this will work. I have not read it.
Oseledets, Ivan V., Dmitry V. Savostyanov, and Eugene E. Tyrtyshnikov. "Fast simultaneous orthogonal reduction to triangular matrices." SIAM Journal on Matrix Analysis and Applications 31.2 (2009): 316-330.
Didn't realize that it was such an open problem. Thank you for the references. In the meantime, I'm going to rely on the fact that the matrices at hand are generally quite small, and hope that a heuristic sort based on comparing the eigenvalues of $M_i$ to the diagonal of $U_1 M_i U_1^*$ will suffice.
Just to be clear, 1000-by-1000 should be small enough unless you have lots of multiplicity (or lots of near points) in the spectrum. The paper by Bunse-Gerstner explains how this quickly turns into a problem with almost commuting matrices. That's what makes it hard, I think.
A cheap trick is to do the Schur decomposition for a random linear combination of the matrices, then hope that unitary works rather well for all. Defeated by close point in the spectrum, but might work in your case.
|
2025-03-21T14:48:29.619274
| 2020-01-11T00:30:25 |
350198
|
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Stack Exchange
|
submodules of the exterior algebra
Let $A_{n,q}$ be the exterior algebra of a vector space of dimension $n$ over the finite field $F_q$.
Let $a_{n,q}$ be the number of submodules of $A_{n,q}$ (meaning submodules of the $A_{n,q}$-module $A_{n,q}$).
Question: What is $a_{n,q}$?
For $q=2$, $a_{n,q}$ starts with 3,7,47,...
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2025-03-21T14:48:29.619327
| 2020-01-11T02:39:24 |
350202
|
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Stack Exchange
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Properness of reductive group actions on smooth varieties
Suppose that $G$ is a reductive algebraic group acting on a smooth variety $X$, and that the action has finite stabilizers. When is the action of $G$ on $X$ proper? What is an example where the action is not proper?
I am aware of a similar statement which is Proposition 0.8 in Mumford's GIT, which says that the action is proper if the geometric quotient $\phi : X \to X / G$ exists and $\phi$ is affine. I would like to know in what situations the action can be assumed proper without this assumption.
Actions of reductive groups with finite stabilizers on quasi-projective varieties are often not proper. The simplest example I know is given by he action of $\mathrm{PGL}_2$ on the projective space $\mathbb P^4$ of effective divisors of degree $4$ on $\mathbb P^1$. Consider the open subset $X \subseteq \mathbb P^4$ of points whose stabilizer is finite. These are of two types.
(1) four distinct points in $\mathbb P^1$, and
(2) three distinct points, one of them double.
The stabilizer of a point of $X$ of type (1) is well-known to be of type $\mathbb Z/2 \times \mathbb Z/2$, while one of type (2) has $\mathbb Z/2$ as a stabilizer. Thus the generic stabilizer is larger than the stabilizer at a special point: this means that the inertia can not be finite, and the action can not be proper.
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2025-03-21T14:48:29.619441
| 2020-01-11T03:53:46 |
350203
|
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Stack Exchange
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On time-dependent invariant Riemannian metrics
Let $X$ be a $G$-homogeneous space that admits a $C^\infty$ family of invariant Riemannian metrics $(\alpha,\beta)\ni t\mapsto g(t)$. Then there exists a family $(\alpha,\beta)\ni t\mapsto\Phi(t)\in\mathrm{Diff}(X)$ of diffeomorphisms of $X$ such that the map $(t,x)\mapsto \Phi(t)x$ is $C^\infty((\alpha,\beta)\times X,X)$ and $d\Phi(t)^*g(t)=g(0)$ for all $t\in(\alpha,\beta)$.
This can be proven by first constructing the map locally through the exponential map, then using the $G$-equivariance of the exponential map to glue local maps etc. But I am pretty sure that this can be found somewhere in the literature.
Question: Would you, please, provide a reference where a proof of this statement can be found or at least easily read off from a stronger statement.
Thank you.
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2025-03-21T14:48:29.619518
| 2020-01-11T04:07:31 |
350204
|
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Stack Exchange
|
Writing rational number as $\frac{a^k+b^k}{c^k+d^k}$
Let $k$ be an odd positive integer. Can every positive rational number $n$ be written as $\frac{a^k+b^k}{c^k+d^k}$ where $a,b,c,d$ are positive rational numbers/ rational numbers?
The answer is true for $k=1$, and probably also true for $k=3$ by explicit construction (see http://mathworld.wolfram.com/DiophantineEquation3rdPowers.html). How about the case $k=5$ and $k=7$? For general $k$?
Motivation: By general philosophy of arithemetic for surfaces, there will be little rational points on a surface of general type. And for $k=3$ and fixed rational number $n$, we are considering rational points on cubic surfaces which are del Pezzo so shall contain rational curves.
The k=3 case has an affirmative solution. This appeared on the shortlist of the 1999 IMO.
Note that if 3 can be written in this form with $k=7$ then the map $(x,y)\mapsto x^7+3y^7$ from $\mathbb{Q}^2$ to $\mathbb{Q}$ is not injective. However, this map actually believed to be injective, see https://mathoverflow.net/questions/21003/polynomial-bijection-from-mathbb-q-times-mathbb-q-to-mathbb-q?noredirect=1&lq=1
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2025-03-21T14:48:29.619621
| 2020-01-11T05:44:27 |
350207
|
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Stack Exchange
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Need help in proving an inclusion between some subspaces of operators
The following question was first posted on Math.Stackexchange.com but unfortunately I didn’t get any answer. This might be obvious for many researchers but I can’t see how this is so, thus I am asking it here:
Let $V$ be a closed subspace of $B(H,K)$ such that $xy^*z \in V$ for all $x,y,z \in V$. Let $I$ be an ideal $(IV^*V+VV^*I \subset I)$ of $V$. Let $C(I)$ denotes the $C^{\ast}$-algebra generated by $II^{\ast}$. In a paper the following inclusion is stated without proof:
$VI^{\ast} \subset C(I)$
I don’t see how this can be true, since $C(I)$ is generated by $II^{\ast}$ therefore how could $VI^*$ be included inside $C(I)$?
What paper is this from? I don't immediately see the answer.
@NikWeaver: This inclusion is stated in proposition $2.2.4$ in Manmohan Kaur’s PhD thesis. She has used this result crucially in showing that $A(I)$ is ideal of $A(V)$, where $A(V)$ denotes linking $C^{\star}$- algebra of TRO $V$. Do you think this result is true?
Why don't you just email her for an explanation? Assuming it isn't explained in the thesis.
@NikWeaver: I found a counter example on math.stackexchange.com. Thank you.
My guess is that there are extra assumptions in this part of Kaur's thesis, which are used to justify this inclusion and the subsequent result that $A(I)$ is an ideal in $A(V)$. It would help if you quote the whole statement of Prop 2.2.4
By the way, it it is good practice to link to the MSE question, so that people here can see what has been said over there: https://math.stackexchange.com/questions/3504653/need-help-in-proving-an-inclusion-between-some-subspaces-of-operators Likewise, at MSE you should link to this MO post, so that the people responding at MSE can see what has been said in comments here
@YemonChoi: Here is the statement: “Let $I$ be closed ideal of $V$ then $V/I$ is TRO”.
@MathLover think about how your last comment would be received by someone asking for more details and context. Does the thesis not specify what $V$ is? Yemon wonders if there are extra assumptions in the thesis and your comment does not answer this.
@NikWeaver (and MathLover) looking at the JFA 2002 paper of Kaur and Ruan which seems to be the only publication based on her thesis, the basic facts about the Cstar algebras associated to TROs seem to predate her thesis, and if you look at the references in that paper, then it seems like you might find further explanation in this paper of Hamana: http://doi.org/10.15099/00003027
@NikWeaver BTW, Nik, perhaps you could leave a comment for Martin A over on MSE so that we are not getting duplicated effort?
@NikWeaver: As far as I have read her thesis there is no extra assumption on $V$, it was just assumed to be a TRO.
@YemonChoi: I will check out this paper of Hamana. Btw, I was mainly interested in a fact which I have posted recently here on MSE: https://math.stackexchange.com/questions/3505790/trying-to-prove-a-subset-is-ideal-of-a-c-ast-algebra. You may like to see it.
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2025-03-21T14:48:29.619848
| 2020-01-11T06:24:50 |
350208
|
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Stack Exchange
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Convexity of the Voronoi cells of higher-dimensional polyhedra
let a convex polytope $\mathcal{P}$ in $E^n$ be defined as in the tag-description with the additional requirement that their volume be strictly positive.
let further the Voronoi Cells $VC(f)$ of $\mathcal{P}$ be defined as the set of points of $\mathcal{P}$ that attain one of their minimal distances to the boundary of $\mathcal{P}$ in facet $f$.
Question:
is it true that the Voronoi cells $VC(f)$ are in turn convex polytopes with strictly positive volume, resp. what are counter examples?
Yes. Denote $H_1,\ldots,H_m$ hyperplanes of facets $f_1,\ldots,f_m$ of $\mathcal{P}$, denote by $\partial \mathcal{P}$ the union of $f_1,\ldots,f_m$ (the boundary of $\mathcal{P}$). Let $d(x,A)$ denote the distance from point $x$ to set $A$.
Proposition. For $x\in \mathcal{P}$ we have $x\in VC(f_j)$ if and only if $d(x,H_j)\leqslant d(x,H_i)$ for all $i=1,\ldots,m$.
Proof. Denote $r=d(x,\partial \mathcal{P})$. Consider the ball centered in $x$ with radius $r$. It contains a point from $f_j$ if and only if it touches $H_j$, otherwise $H_j$ does not intersect this ball. Therefore $d(x,H_i)\geqslant r$ for all $i$, with equality if and only if $x\in VC(f_i)$.
Therefore $VC(f_j)$ is an intersection of finitely many half-spaces (bounded by $H_i$'s and the bisector planes between $H_j$ and $H_i$). So it is a convex polytope. It has positive volume, since for fixed interior point $p$ of $f_j$ all points of $\mathcal{P}$ sufficiently close to $p$ belong to $VC(f_j)$.
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2025-03-21T14:48:29.619962
| 2020-01-11T06:35:10 |
350210
|
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Stack Exchange
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Applications of character sheaves
There are many important recent works (for example, by Lusztig, Bezrukavnikov-Finkelberg-Ostrik, Ben-Zvi-Nadler, Boyarchenko-Drinfeld, Lusztig-Yun, Vilonen-Xue) on character sheaves (which are certain perverse sheaves on an algebraic group $G$ over an algebraically closed field that were originally introduced by Lusztig), where the structure and properties of various categories of character sheaves are studied. What are the established or conjectural applications (for example, to algebraic or geometric representation theory or link homology) of the theory of character sheaves (and in particular, the results on character sheaves mentioned above) other than Lusztig's original motivation of determining the characters of finite groups of Lie type $G(\mathbb F_q)$?
You can probably find some insights in the Math Reviews of Lusztig's papers by Bhama Srinivasan, a former student of J.A. Green; she got heavily involved with the finite groups of Lie type in particular.
Relation of character sheaves to physics is discussed in Section 9.1.3 of https://arxiv.org/abs/1810.10652
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2025-03-21T14:48:29.620056
| 2020-01-11T07:52:57 |
350213
|
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"url": "https://mathoverflow.net/questions/350213"
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Stack Exchange
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Superlevel sets of a parametrized quadratic forms
Let $N$ be an odd integer, $n\in\mathbb{N}$, and $-\frac{2T}{NR^2}\leq a\leq0$ for given $R,T\in\mathbb{R}$ with $\frac{T}{NR^2}\leq\frac{\pi}{2}$.
Now consider the quadratic form $\Omega(a)=\sum_{l\in\mathbb{Z}/N\mathbb{Z}}(\frac{1}{\tan{a}}q_l^2-\frac{1}{\sin{a}}q_lq_{l+1})$ on $(\mathbb{R}^n)^N$ where $q_l\in\mathbb{R}^n$ for $l\in\mathbb{Z}/N\mathbb{Z}$ when $a\neq 0$, and $\Omega(0)=0$ defined only on the diagonal $\{ (q,\cdots,q)|q\in\mathbb{R}^n\}$. Then for nonzero $a$ it has eigenvalues $\lambda_0=\cot{a}-\csc{a}$ with multiplicity $n$ and $\lambda_l=\lambda_{N-l}=\cot{a}-\csc{a}\cos{\frac{2\pi l}{N}}$ for $l=1,\cdots,\frac{N-1}{2}$ with multiplicity $2n$.
Let $\rho:\mathbb{R}\times(\mathbb{R}^n)^N\rightarrow \mathbb{R}$ be the canonical projection given by $(a,\{q_l\})\mapsto a$ and let $W=\{(a,\{q_l\})|\Omega(a)\geq 0\}\subset \mathbb{R}\times(\mathbb{R}^n)^N$.
In Chiu's paper "Non-squeezing property of contact balls", he says that
$W'=\rho^{-1}(\{0\leq -a\leq \frac{2T}{NR^2}\})\cap W$ is homotopic to the Euclidean space $\mathbb{R}^{D+1}$ where $D+1$ is the number of positive eigenvalues of $\Omega(-\frac{2T}{NR^2})$, or equivalently $D+1=2nI+n$ where $I$ is the number of solutions $\lambda_l>0$ among $l=1,\cdots,\frac{N-1}{2}$. Moreover their compactly supported cohomologies are isomorphic.
In addition, he says that $\rho^{-1}(\{0< -a\leq \frac{2T}{NR^2}\})\cap W$ is homotopic to $S^{D-n}\times\mathbb{R}^n\times\mathbb{R}_{>0}$.
I wanted to check all Chiu's assertions but I could't. Please tell me if you know about that.
I am happy to see the question which bothered me too for a long time. But now I know what he did, hope it could help.
Actually, Chiu want to compute $R\Gamma_c(W\cap\rho^{-1}(\{0\leq -a\leq \frac{2T}{NR^2}\}),\mathbb{K})$.
Let's take $F=\mathbb{K}_{W\cap\rho^{-1}(\{0\leq -a\leq \frac{2T}{NR^2}\})}$, and $\pi(a,\{q_l\})=(q_l)$.
The cohomology Fubini theorem shows $$R\Gamma(\mathbb{R}_a\times \mathbb{R}^{nN},F)\cong R\Gamma(\mathbb{R}^{nN},R\pi_!F).$$
The Begle-Vietoris theorem tell us $R\pi_!F\cong \mathbb{K}_{\pi\left(W\cap\rho^{-1}(\{0\leq -a\leq \frac{2T}{NR^2}\})\right)}$. In this case, you can find
$$\pi\left(W\cap\rho^{-1}(\{0\leq -a\leq \frac{2T}{NR^2}\})\right)=\{(q_l): \Omega(a)\geq 0, a\in [-\frac{2T}{NR^2},0] \}.$$
Now, like what Chiu said, you can choose $a=-\frac{2T}{NR^2}$, and compute cohomology of
$$\{(q_l): \Omega(-\frac{2T}{NR^2})\geq 0\}.$$
This space is a quadratic cone, which is properly homotopic to the maximal positive defined subspace of the quadratic form $\Omega(-\frac{2T}{NR^2})$. In particular, it is $\mathbb{R}^{D+1}$.
(In fact, if $Q(x,y)=|x|^2-|y|^2$, then $f_t(x,y)=(x,(1-t)y)$ is the homotopy you want.)
Actually, it is still possible to write down a homotopy directly to compute the cohomology without the Begle-Vietoris theorem, but I don't think it is a good point of view. In my work (arXiv 2103.05143) on generalizing Chiu's result to toric domains, it looks simpler to use Begle-Vietoris.
The second part works similarly. A quadratic cone without a center axis will homotopy to some sphere, right?
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2025-03-21T14:48:29.620235
| 2020-01-11T07:57:45 |
350214
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|
Stack Exchange
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Conformal map from a 7-sided polyhedron to a square pyramid
I have a right-angled square pyramid, $A$, whose height and base-length are $l$. By 'right-angled', I mean that the apex of $A$ lies vertically above one of the vertices in its base. Now supposed I form a new polyhedron, $B$, by gluing a cube with side-length $l$ to the base of $A$ ($B$ now has a base-length $l$ and height $2l$). Additionally, suppose that I have a new pyramid $A^{'}$ that is similar to $A$, but with a height and base-length of $2l$.
Does anyone know how to derive a conformal map from the interior of $B$ to that of $A^{'}$?
P.s. I have also cross-posted this on MathStackExchange: https://math.stackexchange.com/questions/3503921/conformal-map-from-a-7-sided-polyhedron-to-a-square-pyramid
Duplication of https://math.stackexchange.com/questions/3503921/conformal-map-from-a-7-sided-polyhedron-to-a-square-pyramid
I figured I was going to get a different calibre of responses on MO as opposed to MSE. Is it not common to post the same question on both? Sorry I'm relatively new to SE.
Your conformal map is just the similarity.
niran90, I would delete the duplicate at math.se since it's been answered here. Then, if I were you, I would post the smooth bijection question over there first, and wait some days for a response. If there is none, you can try asking here (although it might easily be closed as off-topic for this site).
No such conformal map exists.
Conformal mapping in dimensions above 2 is very different from conformal mapping in dimension 2. In dimensions above 2, any conformal mapping is a (finite) composition of rigid motions, dilations, and inversions. In particular, such a mapping carries planes and spheres to planes and spheres and preserves the intersection angles between them. Your pyramid has 5 boundary planes, and so any conformal image of it will have 5 faces that are parts of either planes or spheres. In particular, the image cannot look like the pyramid+box that you describe as your polyhedron $B$.
Okay I see. Suppose I relax the requirement for conformality, and simply require that the mapping is smooth and bijectve, would that be feasible? I imagine there are infinitely many such mappings. But what's the most straightforward way to derive one of them? Actually for my specific application, conformality is not crucial. I just thought it would be a sufficient condition for a smooth bijective map.
Actually, I assumed that you wanted to map the interior of $B$ to the interior of $A'$, and that's what my answer addressed. Did you instead want to map the actual polyhedrons (i.e., the surfaces) conformally?
No, you assumed correctly - I want to map the interiors to each other. I've now clarified the ambiguity in the post. Thanks.
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2025-03-21T14:48:29.620548
| 2020-01-11T10:54:42 |
350219
|
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"Dieter Kadelka",
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Stack Exchange
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On tensor products of (essentially) bounded measurable functions
Let $F:\mathbb R^2\rightarrow\mathbb R$ be an essentially bounded measurable function ($\mathbb R^d$ is equipped with its standard Lebesgue measure) and assume that
$F(x,y)=F(y,x)$. I would like to write
$$
F(x,y)=\sum_{\nu, \mu} a_{\mu,\nu}\mathbf 1_{A_\mu}(x)\mathbf 1_{A_\nu}(y),
\text{ with $a_{\mu,\nu}=a_{\nu,\mu}$ and $\bigcup_\mu A_\mu=\mathbb R$, measure$(A_\mu\cap A_\nu)=\delta_{\mu, \nu}$.}
$$
Is it possible?
What are the underlying function spaces?
Let $\mu$ be a probability measure on $\mathbb R$ equivalent to the Lebesgue measure (for instance, the Gaussian measure). Then $A_\mu$ is a partition of $(\mathbb R,\mu)$ modulo null sets, and is therefore countable; so $\mu,\nu$ range over a countable set. Furthermore, throwing away intersections of zero measure, one can assume that the sets $A_\mu$ are disjoint. But then $F$ is equal in $L^\infty$ to a function which has the following property: for almost all $y\in\mathbb R$, the function
$$
x\mapsto F(x,y)
$$
is a step function (there will be only one $\nu_0$ which contributes to the sum). This is, of course, quite a severe restriction, so by far not all functions will have this property.
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2025-03-21T14:48:29.620660
| 2020-01-11T11:02:15 |
350221
|
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"JCM",
"Wlod AA",
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"https://mathoverflow.net/users/33849"
],
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"url": "https://mathoverflow.net/questions/350221"
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|
Stack Exchange
|
Roughness of a differentiable function
If a function $f: [0, 1] \to \mathbb R$ is differentiable at a point $p$, then there exists some unique linear function $L_p$, and some unique function $r_p$ such that $f(x) = f(p) + L_p (x - p) + r_p (x - p)$, where $r_p$ is $o(|x-p|)$ as $x$ approaches $p$.
Given $e > 0$, and a function $f$ differentiable at $p$, define $t(e, f, p)$ to be
$$\sup \{d:\ \min(p, 1-p) > d > 0;\ r_p(x-p) < e|x-p| \text{ whenever } |x-p| < d\} $$
Now for $0 < p < 1$, define the quantity $N(f, p)$ to be
$$\limsup_{e \to 0+} \frac{1}{2et} \int |r_p (y)/(y - p)| \ dy$$
where $t$ is short for $t(e, f, p)$, and the integral is taken over all $y \in [p - t, p + t]\setminus p$ with respect to Lebesgue measure.
We call $N(f, p)$ the roughness of $f$ at $p$. Note that $N(f, p)$ is between $0$ and $1$ inclusive.
Are there any differentiable functions $f: [0, 1] \to R$ such that $N(f, p) = 1$ for every $ p \in (0, 1)$? What about continuously differentiable functions?
You could make your definition of $,t(e,f,p) :=\ldots,$ readable. (?)
I'm not certain but a continuously differentiable function, such that at each $p\in (0,1)$, the derivative of the function is Dini continuous, but not Holder continuous, seems like it will be enough (maybe further details are required). I should emphasise the first 3 words of my comment though.
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2025-03-21T14:48:29.620772
| 2020-01-11T11:06:33 |
350222
|
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"Gerald Edgar",
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|
Stack Exchange
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addition theorems for hypersine
I learned from Wolfram MathWorld about hypersine, as being a dimensional analog trig function for hypersolid angles. There it is being defined by
The hypersine ($n$-dimensional sine function) is a function of a vertex angle of an $n$-dimensional parallelotope or simplex. If the content of the parallelotope is $P$ and the contents of the $n$ facets of the parallelotope that meet at vertex $v_0$ are $P_k$, then the value of the $n$-dimensional sine of that vertex is $$\sin(v_0)=\frac {P^{n-1}}{\prod^n_{k=1}P_k}$$
[…] The vertex simplex of a vertex of an $n$-dimensional parallelotope is the simplex that has as its vertices that vertex and the $n$ adjacent vertices of the parallelotope. Its content (and the content of all the other vertex simplices) is the content of the parallelotope divided by $n!$. If the content of an $n$-dimensional simplex is $S$, and the contents of the $n$ facets that meet at vertex $v_0$ are $S_1, S_2,..., S_n$, the simplex can be considered a vertex simplex of a parallelotope, and the facets also vertex simplices of the facets of the parallelotope with respect to the same vertex. Substituting in [the above] equation gives $$\sin(v_0)=\frac {(n!\ S)^{n-1}}{\prod^n_{k=1}(n-1)!\ S_k}$$
Confering the first defining equation to the area formula of a parallelogram, it becomes obvious that this definition is nothing but the usual sine function for $n=2$.
This article moreover continues on how to calculate the latter formula when knowing the respective dihedral angles $\alpha_{jk}$ between $S_j$ and $S_k$.
Applying this makes it easy to get the hypersine of the corner angle of any hypercube generally: $$\sin(v_0)=1$$
Even the hypersine of the corner angle of any regular simplex could be calculated to: $$\sin^2(v_0)=\frac{(n+1)^{n-1}}{n^n}=\frac1{n+1}\ \left(1+\frac1n\right)^n$$
But in order to calculate correspondingly the hypersine of the corner angle of the orthoplex one needs to extrapolate this function to non-simplicial corners as well. One clearly could easily subdivide the orthoplex corner symmetrically into $2^{n-1}$ equal (mostly right-angled) simplices, but then again one needs to know about addition theorems for this hypersine function.
Is there anything being known in this direction or could anything be derived for that purpose? - I would be very grateful to learn about it.
--- rk
That's not Wolfram Alpha, that is Wolfram Mathworld.
From the paper mentioned in the other answer it looks to me, that addition theorems are kind out of near reach.
But at least we could come up with a result for the crosspolytopes (orthoplexes) none the less. This can be done by using the already mentioned dissection of the vertex corner angle into according subsimplices: Just consider $x3o3o...o3o4o$ (orthoplex) as being the vertex figure of $x4o3o3o...o3o4o$ (hypercubical honeycomb). That is, each looked for subsimplex is nothing but the vertex corner pyramid of an hypercube $-$ just that the vertex of consideration, i.e. which contributes to the vertex corner of the orthoplex, would be one of its base vertices.
Thus the contributing subsimplex corner angle could be derived as
$$\sin_n^2(v_{0,\ subsimplex})=\left|\begin{array}{ccccc}
1 & 0 & … & 0 & -\frac1{\sqrt n}\\
0 & 1 & … & 0 & -\frac1{\sqrt n}\\
… & … & … & … & …\\
0 & 0 & … & 1 & -\frac1{\sqrt n}\\
-\frac1{\sqrt n} & -\frac1{\sqrt n} & … & -\frac1{\sqrt n} & 1
\end{array}\right| = \frac1n$$
And therefrom we clearly get at least
$$\begin{array}{rcl}
v_{0,\ orthoplex} & = &2^{n-1}\cdot v_{0,\ subsimplex}=\\
& = & 2^{n-1}\ \arcsin_n(\frac1{\sqrt n})
\end{array}$$
(Note that I attached the index $n$ to both the sine and the inverse sine functions in order to remind that we are dealing with hypersine instead of usual 2D sine only.)
--- rk
This is the only reference found by MathSciNet:
Lerman, Gilad; Whitehouse, J. Tyler, On (d)-dimensional (d)-semimetrics and simplex-type inequalities for high-dimensional sine functions, J. Approx. Theory 156, No. 1, 52-81 (2009). ZBL1170.46025.
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2025-03-21T14:48:29.621039
| 2020-01-11T11:12:49 |
350223
|
{
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"ABIM",
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|
Stack Exchange
|
$L^1_{\mu}$ as limit
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space. Does there exist a countable set of finite measures $\{\mu_n\}_{n \in \mathbb{N}}$ on $(X,\Sigma)$ such that $L^1_{\mu}(\Sigma)$ can be written as the projective-limit in the category of LCS
$$
L^1_{\mu}(\Sigma) = \projlim\, L^1_{\mu_n}(\Sigma),
$$
for some suitable restriction maps $\pi_n^m:L^1_{\mu_m}(\Sigma) \rightarrow L^1_{\mu_n}(\Sigma)$.
Related: Can $L^1_{loc}$ be represented as colimit?
Taking projective limits of Banach spaces in the sense of lcs’s will not usually produce a Banach space, as pointed out below. However, the category of Banach spaces with contractions as morphisms does have projective limits and I would suggest that this might be what you want. As a simple example, if $\mu$ is Lebesgue measure on the line, and $\mu_n$ is its restriction to $[-n,n]$, the sequence $(L^1(\mu_m))$ has the corresponding $L^1$-space as projective limit in the second sense, but $L^1_{loc}$, the Fréchet space of locally integrable functions, in the first.
This is very interesting actually. Could you possibly provide a reference to this; maybe some book?
I can‘t think of one for this explicit situation but both follow directly from the general construction of such projective limits as compatible threads, in this case as compatible sequences $(f_n)$ with $f_n \in L^1(\mu_n)$. In the lcs case there is no growth condition so they combine to form a locally integrable function, in the Banach space one, the $L^1$ norms are bounded so you get a globally $L^1$ function.
@user131781 would you happen to have a reference on $L^1$ described this way in the Category of Banach spaces and (linear?) contractions?
Well, no but it is so elementary that I don‘t think that one needs a reference. It is simply the fact that if you have a compatible, bounded sequence $(f_n)$ where $f_n$ is in $L^1([-n,n])$, then they combine to form an integrable function on the line. The only possible finesse comes from the fact that the functions are only defined a.e. but the countability condition takes care of that. Of course, I only chose the real line as an example—-this works for any $\sigma$-finite meaure. A useful toy example to make everything transparent would be the positive integers with counting measure.
Perfect! I'll work this out then; thanks.
There is a strictly positive integrable function $f\in L^1_\mu(\Sigma)$, hence $\nu=f\cdot \mu$ is a finite measure and $\Phi:L^1_\nu(\Sigma)\to L^1_\mu(\Sigma)$, $g\mapsto gf$ is an isomorphism. In particular, $L^1_\mu(\Sigma)$ is isomorphic to a projective limit of $L^1_{\mu_n}(\Sigma)$ with finite measures (and if you define a projective limit in the categorial sense by universal properties isomorphic to and equal to a projective limit is the same).
Let me add a general remark: Projective limits in the category of locally convex spaces are extremely useful to represent general spaces by simpler ones, in particular Frechet spaces as countable projective limits of Banach spaces. Useful results usually require an additional property called reducedness, i.e., the connecting maps $\pi_m^n:X_m\to X_n$ have dense range for $m\ge n$ or variants of that, like the Mittag-Leffler condition for all $n$ there is $m\ge n$ such that for all $k\ge m$ we have $\pi_m^n(X_m)\subseteq \overline{\pi_k^n(X_k)}$, the abstract Mittag-Leffler theorem then implies that this also holds for $k=\infty$ where $\pi_\infty^n$ is the map from the projective limit into the $n$th step. Your question asks for a representation of a Banach space by other Banach spaces which is somewhat queer to the theory because the limit of a reduced spectrum of Banach spaces is again a Banach space only if for all but finitely many $n$ the linking maps $\pi_m^n$ are isomorphisms.
Thank you, I'll definitely keep the Mittag-Leffler condition in mind; I think it will be useful for my applications. Would you happen to have a recommendation on the subject?
My Springer Lecture Note "Derived Functors in Functional Analysis" contain some information about that.
Thank you very much for the reference I'm looking at it now. However, now I'm curious about the dual question; namely, can $L^1_{\mu}$ be represented as a colimit of $L^1_{\mu_n}(\Sigma)$ spaces for aptly chosen $\mu_n$?
As it is isomorphic to such a space it is isomorphic to a (trivial) colimit.
|
2025-03-21T14:48:29.621677
| 2020-01-11T13:56:56 |
350227
|
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|
Stack Exchange
|
For a manifold of positive curvature, can we lower bound the distance between unit normals?
Suppose $M \subset \mathbb R^d$ is a $C^2$ $(d-1)$-manifold. In particular I am interested in when $M$ is the set $\{x \in \mathbb R^d: f(x) \le 1\}$ for some $C^2$ function $f:\mathbb R^d \to \mathbb R$. Suppose also the curvature at each point at least $m \ge 0$. There are loads of equivalent definitions of this. For example
For every twice-differentiable path $\gamma:(0,1) \to M$ with all $\|\gamma'(t)\| =1 $ we have $\|\gamma''(t)\| \ge m$ for all and $t \in (0,1)$. Here $\|\cdot\| $ is the Euclidean norm.
For each point $x \in M$ if we locally represent the manifold as the graph of a function $F:\mathbb R^{d-1} \to \mathbb R$ that is minimised at $x$ then the Hessian $H$ of $F$ at $x$ has $v^T H v \ge m \|v\|^2$ for all $v$ in the tangent space at $x$.
The curvature at $x$ describes how quickly the unit normal $N_y$ changes as we vary the basepoint $y$ in a small region about $x$. Thus one would expect we could bound $\|N_a-N_b\|$ from below by integrating the infinitesimal change along a path from $a$ to $b$.
Suppose $M \subset \mathbb R^d$ is a $C^2$ $(d-1)$-manifold with curvature at each point at least $m \ge 0$. Is there $C \ge 0$ such that $\|N_x-N_y\| \ge Cm \|x-y\|$ for all $x,y \in M$ with unit normals $N_x,N_y$?
I suspect if there is no uniform bound on the curvature this is impossible, as $N_x$ might vary in one direction and then back the other direction, resulting in $N_a =N_b$. But if we have $m \ge 0$ then $M$ is the boundary of a convex set and it's hard to imagine this happening.
For the special case when $M =\{x \in \mathbb R^d: f(x) \le 1\}$
with $\|\nabla f(x)\|=1$ for $x \in M$ then we have $N_x = \nabla f(x)$ and I think a proof might look something like this:
For any $x,y$ let $\gamma:[0,1]$ be a geodesic from $y$ to $x$ with constant speed $s$ equal to the geodesic distance between the points. Since $\nabla f(x)$ has derivative $\nabla ^2f(x)$ some version of the fundamental theorem of algebra gives
$$\nabla f(x) - \nabla f(x) = \int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt. $$
Take squares to get $ \|\nabla f(x) - \nabla f(x)\|^2 $ equal to
$$ \left(\int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt \right) \cdot \left( \int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt \right) $$
what I would like to do now is say the right-hand side is at least $$ \int_{0}^1 \Big(\nabla^2 f(\gamma(t)) \gamma'(t) \Big) \cdot \Big (\nabla^2 f(\gamma(t)) \gamma'(t) \Big)dt $$
$$ = \int_{0}^1 \gamma'(t) \nabla^2 f(\gamma(t)) \nabla^2 f(\gamma(t)) \gamma'(t) dt.$$
By the Hessian condition the integrand should be at least $m^2\|\gamma'(t)\|^2 = m^2 s^2 \ge m^2 \|x-y\|^2$. Now take square roots to get $\|\nabla f(x)-\nabla f(y)\| \ge m \|x-y\|$.
This only works in a very special case, and the inequality step is just wishful thinking. But I have been staring at this for far too long and am making no progress. Would someone with more geometry experience mind taking a look?
|
2025-03-21T14:48:29.621935
| 2020-01-11T15:36:51 |
350228
|
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|
Stack Exchange
|
How useful is differential geometry and topology to deep learning?
After seeing this article https://www.quantamagazine.org/an-idea-from-physics-helps-ai-see-in-higher-dimensions-20200109/ I wanted to ask myself how useful of an endeavor would it be if one goes through the process of learning differential geometry to deep learning? Perhaps there are other tools as well?
Can concepts such as homotopy be 'appropriately' encodable in deep learning? I feel notions of homotopy invariance if properly defined could be much more valuable than the simple attempt in quantamagazine article. A basic attempt (away from deep learning framework) is in https://arxiv.org/abs/1408.2584.
Day-to-day application of machine learning generally uses simple statistical models, such as GLMs. Most of the work is in modeling and cleaning data and figuring out how to communicate results to non-technical people (whether through reports and data viz or embedding in a production application). So I'd expect differential geometry/topology are not immediately useful in industry jobs outside of big tech companies' research labs.
@Neal I doubt it will still be that way in the future if progress is made. Right now it is not the case because not much is known to architecture the divide and conquer process. If properly framed ML (as it is envisioned today) could benefit from modern mathematics (we can always sell the product nicely to lay people with nice words as currently done with quantum computing).
I don't think there is a definitive answer to your question at this point. Presumably the answer will eventually be a strong "yes, very useful". But as far as I am aware, at present it sounds like "maybe in some special cases" is more appropriate.
Learning differential geometry certainly can't hurt, as the answers below show. The only real downside is opportunity cost: depending on your objective and your current knowledge, there could easily be another subject that's even more helpful. Can you say more about your goal? (E.g., practical applications, personal intuition, new research results, etc.)
I suspect there's at least one quality PhD dissertation in ML that will be written about employing differential geometry in ML. Even if the topic is of interest to only academia for quite a long time, it would still set you up nicely for a career in both academia and industry.
Also, re @Neal, while most of the "data science" jobs out there really just involve data cleaning and using previously developed methods, the interesting (and higher paying) problems (and jobs) involve a lot more than just that.
Differential geometry can take a long time to learn but the benefits are that you can consider curvature for manifolds of arbitrary dimension, which is worth the effort imo.
A "roadmap type" introduction is given by Roger Grosse in
Differential geometry for machine learning.
Differential geometry is all about constructing things which are
independent of the representation. You treat the space of objects
(e.g. distributions) as a manifold, and describe your algorithm in
terms of things that are intrinsic to the manifold itself. While you
ultimately need to use some coordinate system to do the actual
computations, the higher-level abstractions make it easier to check
that the objects you're working with are intrinsically meaningful.
This roadmap is intended to highlight some examples of models and
algorithms from machine learning which can be interpreted in terms of
differential geometry. Most of the content in this roadmap belongs to
information geometry, the study of manifolds of probability
distributions. The best reference on this topic is probably Amari and
Nagaoka's Methods of Information Geometry.
One should note that this kind of application of differential geometry to ML is different to the one described in the quanta article. The former is based on the infinite-dimensional manifold of (probability) distributions while the work of Cohen et al (https://arxiv.org/abs/1902.04615) is concerned with data and features that life on a finite-dimensional manifold, e.g. vector fields on a sphere.
Also is it relevant to deep learning mathematical architecture and intuition?
This answer does not seem to be about deep learning.
You can read this paper:
https://arxiv.org/abs/1805.10451
Although what they say is by no means new - Coifman had had this point of view for the last thirty years at least.
Topological data analysis is already pretty popular in "data science", so much so that there are companies built on this idea. See:
1). https://en.wikipedia.org/wiki/Topological_data_analysis
2).https://web.stanford.edu/class/archive/ee/ee392n/ee392n.1146/lecture/may13/EE392n_TDA_online.pdf
3). https://towardsdatascience.com/a-concrete-application-of-topological-data-analysis-86b89aa27586
TDA as of now how it stands does not appear very interesting for many fields such as NLP or vision let alone invariance detection in general.
@VS I am not sure what the success stories of TDA are. I am sure that's due to my own ignorance.
@IgorRivin I think TDA as of now is too simple to have any applications useful to industry. The right way seems to be deep learning (which is some glorified non-uniform divide and conquer by design of architecture) with some elaborate context specific structural information and intuition from mathematics that specifies the architecture. That is why I think differential geometry as of how it is applied now is not the end of story.
Well link #2 is presentation by a industrial firm on their use of TDA. So atleast they claim it is useful
@ Piyush Grove Link #2 ls not by an industrial firm. It is by Ayasdi, which is a company which produces and sells Topological Data Analysis software..In your parlance, one of the companies built on this idea.
TDA can be useful in a number of settings, especially when the number of data points is small but the dimension is large (and so is complementary to machine learning). Consider the following in medicine, psychology, materials science and neuroscience respectively.
https://www.pnas.org/content/108/17/7265
https://www.pnas.org/content/114/40/10767
https://www.ncbi.nlm.nih.gov/pubmed/28534490
https://www.pnas.org/content/112/44/13455
Quite a few people myself included expect TDA and ML to play well together. Carlsson has started a second company, and Leygonie, Oudot and Tillmann have started setting the stage for gradient descent on persistence diagrams.
http://front.math.ucdavis.edu/1910.00960
Adding to what @DevSinha said... There's also been related, interesting work representing the persistent homology of data as barcodes.
Taco Cohen has written some papers that use Differential Geometry, Topology, Guage theory, etc. in Machine learning.
This is about deep learning.
Yes, Cohen does apply it on deep learning.
That is what I mean contrasting CarloBeenakker's answer.
Colah gives a very interesting perspective about deep learning and neural networks in the context of topology. He discusses the "Manifold Hypothesis" which, in short, tries to explain why deep learning is so effective. To read more about the Manifold Hypothesis, Goodfellow has a chapter on it.
On another note, there are interesting articles being published in the context of topological analysis of some "image spaces". For example, Carlsson et al.:
In this study we concentrate on qualitative topological analysis of the local behavior of the space of natural images. To this end, we use a space of 3 by 3 high-contrast patches ℳ. We develop a theoretical model for the high-density 2-dimensional submanifold of ℳ showing that it has the topology of the Klein bottle.
I think learning basic topological ideas and especially all the work that has already been done on the connection of data analysis and topology can be very fruitful (Mind that I am no expert though).
Is "manifold hypothesis" supposed to be glossed as "low-dimensional subspace hypothesis", or is smoothness really an aspect of it? This reminds me of the "T" in "TQFT", which stands for "topological" but typically means "smooth".
As far as I can remember it is usually only meant as "low-dimensional subspace hypothesis", yes.
Heat kernels are based on ideas from differential geometry:
http://www.jmlr.org/papers/volume6/lafferty05a/lafferty05a.pdf
I also found this paper which gives a topological characterization of a classification neural network to be intuitive :
https://arxiv.org/abs/2008.13697
They defined data as a topological space and defined labels as closed subsets of this space. They also defined a notion of "separable data" which is topologically equivalent to "correctly classifying the data". Urysohn lemma is then used to prove that the data, considered as a topological space, can be "separated", if you can find embedded disks that are mutually disjoint and separate the labeled parts of the data (after being mapped by the neural network to the final space).
This can be seen with the disentanglement figure at the end of the paper:
which shows how the above neural network acts on the input topological space, deforms it in order to achieve a given classification task (in this case separating the two labeled linked spaces). Some hints to the general position theorem are also made there. There are also some links to "topological moves" in the paper :a neural network acts on the input topological space by a sequence of topological moves (similar to Reidemeister moves in knot theory or other moves in topology) in order to achieve the given task : which is deforming the input space to the final space, in this case the Voronoi diagram that represents the classes of the input data, such that every labeled region in the input space maps to the correct cell in the final Voronoi diagram.
To me, from this perspective, a classification neural network can be interpreted as if it is acting on the input space as a "discrete colored homotopy" : discrete because of the layers (time is the index of the layer) and colored because the class label which can be considered as "coloring" on the input space.
Since you mentioned homotopies, you might find interest in A homotopy training algorithm for fully connected neural networks. I haven't read the paper, but from the abstract I gather that the neural network (which is a function) is first trained and becomes $f(x)$. Then its architecture's complexity is increased and trained some more. The complexity increase (and training) is done multiple times until the final model, which is $g(x)$. The function $h$ from $f$ to $g$ is a homotopy.
|
2025-03-21T14:48:29.622677
| 2020-01-11T16:58:15 |
350230
|
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|
Stack Exchange
|
Is the representation of $GL_n(\mathcal{O})$ in functions on Grassmannian multiplicity free?
Let $\mathbb{F}$ be a local non-Archimedean field. Let $\mathcal{O}\subset \mathbb{F}$ be its ring of integers. Let $GL_n(\mathcal{O})$ be the (compact) group of $n\times n$ invertible matrices with entries in $\mathcal{O}$ such the inverse matrix also has entries in $\mathcal{O}$. Let $Gr_{i,n}$ be the Grassmannnian of linear $i$-dimensional subspaces in $\mathbb{F}^n$.
Is it true that the natural representation of $GL_n(\mathcal{O})$ in the space of locally constant functions on $Gr_{i,n}$ is multiplicity free? In other words is it true that any irreducible representation of $GL_n(\mathcal{O})$ appears with multiplicity at most 1?
A reference would be helpful.
Remark. Archimedean analogue of the above statement, i.e. $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, is true; here $GL_n(\mathcal{O})$ should be replaced by $O(n)$ or $U(n)$ respectively. (That follows from the fact that the Grassmannian is a symmetric space.)
Perhaps it would help to give some examples, e.g. what happens when $i=n-1$? Or for small $n$ such as $n=2$? A classification of irreducible representations of $\operatorname{GL}_2(\mathcal O)$ seems to be known.
It is indeed known, see https://www.sciencedirect.com/science/article/pii/S0001870808002260
Yes, this is due to Hill:
Hill, Gregory, On the nilpotent representations of (GL_ n({\mathcal O})), Manuscr. Math. 82, No. 3-4, 293-311 (1994). See especially Corollary 3.2.
This was generalised and extended by Bader and Onn in
Bader, Uri; Onn, Uri, Geometric representations of (\text{GL}(n,R)), cellular Hecke algebras and the embedding problem., J. Pure Appl. Algebra 208, No. 3, 905-922 (2007).
and
Bader, Uri; Onn, Uri, On some geometric representations of (GL_N(\mathfrak{o})), Commun. Algebra 40, No. 9, 3169-3191 (2012).
This may be redundant as a complete answer with references has already been posted.
Just in case this is still useful: I think multiplicity one can be proved by the usual Gefland trick: we need to check that endomorphisms of the induced representation is a commutative ring, this follows once we check that the identity automorphism of that algebra is an anti-involution. The endomorphism algebra coincides with the Hecke algebra ${\mathbb{C}}[P(O)\backslash G(O)/P(O)]$ where $P$ is the group of block upper triangular matrices with blocks of sizes $i, \, n-i$. This can also be realized as the space of $G(O)$-invariant functions on pairs of rank $i$ summands in $O^n$. Switching the two elements in the pair induces an anti-involution.
Looking at relative positions of two direct summands in $O^n$ one sees that the relative position of a pair $(N,M)$ is the same as the relative position of $(M,N)$, which yields the statement.
|
2025-03-21T14:48:29.622995
| 2020-01-11T19:37:52 |
350238
|
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|
Stack Exchange
|
How to (easily) obtain the splitting field for dihedral extensions
Let $f(x) \in {\mathbb Q}[x]$ be a polynomial that is irreducible over ${\mathbb Q}$ with $D_{n}$ (the dihedral group of order $2n$) as its Galois group. Let $\alpha$ be a root of $f(x)$ and put $K={\mathbb Q}(\alpha)$. The splitting field of $f(x)$, denote it by $L$, will be a quadratic extension of $K$.
Is there an ''nice'' element, $\beta \in K$ such that $L=K(\sqrt{\beta})$?
Ideally ''nice'' here means that $\beta$ has a simple expression in terms of $\alpha$.
The discriminant of $f$ works, at least if $n$ is odd, and I think in general.
@Ari Shnidman thanks for that. Would you have a reference, please? And do we know what happens for $n$ even too?
Sorry, I was being too confident. Even when $n$ is odd the discriminant does not always work. The square root of disc($f$) lives in $L$, since it is the product of the difference of the roots of $f$. So if disc($f$) is not a square, then its squareroot gives a quadratic extension, which is not contained in $K$ (since $n$ is odd). The problem is that disc($f$) may be a square. If $n$ is an odd prime, then disc($f$) is a square if and only if $n$ is 1 mod 4 (see page 2 here: arxiv.org/pdf/1609.09153.pdf). So when $n$ is a prime that is 3 mod 4, disc(f) does what you want, but not for all $n$.
@Ari Shnidman thanks for your follow-up comment, Ari. And for the reference too. That is very helpful to me. I do appreciate you taking your time to help.
The formula in Cohen-Thorne only says that the ideal of the discriminant is a square, not that the discriminant is a square. I am not convinced that your first remark is wrong.
Here's an easier way to see it. The discriminant is a square iff $D_n$ is a subgroup of $A_n$, which happens iff $n$ is odd (so rotation is an even cycle) and congruent to 1 mod 4 (so reflection is a product of an even number of transpositions). If $n$ is even, then the discriminant is never a square, but the square root may already live in $K$ (from some examples it looks like this happens iff $n$ is 2 mod 4).
@Ari: thank you!
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2025-03-21T14:48:29.623170
| 2020-01-11T21:10:18 |
350240
|
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|
Stack Exchange
|
On the permanent dominance conjecture for symmetric group
The Lieb's permanent dominance conjecture states that the expression $$\frac{d_{\chi}^HA}{\chi(e)}\le per(A)$$ holds for all positive semidefinite matrices $A$, where $d_{\chi}^HA=\sum_\limits{\sigma\in H}\chi(\sigma)\prod_\limits{i=1}^na_{i\sigma(i)}\ \ ,\{a_{ij}\}= A\ \ ,H\le S_n$ is the immanent function of the matrix $A$ and $per(A)$ is the permanent of $A$.
Now, I think for $H=S_n$, the symmetric group of order $n$, it is sufficient to show that $\chi(e)$, the irreducible character corresponding to the trivial conjugacy class, or the dimension of the representation, dominates all the other irreducible characters; i.e $\chi(e)\ge \chi(\sigma)$ for any $\sigma\in S_n$. Now, $\chi(e)$ is straightly given by hook length formula. The other irreducible characters can be found by the Murnaghan-Nakayama rule, or the determinantal formula. Since the Murnaghan-Nakayama formula involves the recurrence on the number of rim hooks(Border-strip Tableau) and the length of each of which do not exceed the proper hook in the hook's formula, therefore, it seems intuitive to think the conjecture could be proven in the case $H=S_n$. Any way to rigorously look at the difference between the counting of characters in the Murnagahan-Nakayama rule and the hook length formula? Thanks beforehand.
For any finite group and any character $\chi$ it is clear that $\chi(e)\geq \chi(\sigma)$, since $\chi(\sigma)$ is a sum of $\chi(e)$ roots of unity. I don't see how this implies Lieb's conjecture since we could have $\prod_{i=1}^n a_{i\sigma(i)}<0$.
@RichardStanley ok, so then the conjecture is trivial for matrices with positive entries, right?
That is correct.
One can refer for more details from Pate papers and the link https://arxiv.org/pdf/1608.02844.pdf
|
2025-03-21T14:48:29.623327
| 2020-01-11T21:21:43 |
350241
|
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|
Stack Exchange
|
Polynomial approximation (in $L^1$ norm) at minimal cost
Define the cost of a polynomial $\sum_{i=0}^N a_i x^n$ to be $\sum_{i=0}^N |a_i|$. Let $g:[0,1]\to \mathbb{R}$ be a function to be approximated — say, $g(x)=0$ if $0\leq x < e^{-1}$, $g(x)= 1/x$ if $e^{-1}\leq x\leq 1$. (That function comes up in practical contexts.) We are interested in polynomials $P_+$, $P_-$ such that $P_-(x)\leq g(x)\leq P_+(x)$. We define the tightness $\epsilon(P)$ of $P$ to be $\epsilon(P) = \epsilon = \int_0^1 |P(x)-g(x)| dx$.
For given $\epsilon>0$ and $N$ what are the polynomials $P_+$, $P_-$ of tightness $\leq \epsilon$ and minimal cost?
What are those minimal costs $c_+(\epsilon,N)$, $c_-(\epsilon,N)$?
What if we allow the degree $N$ to be arbitrary? In other words, what are $c_-(\epsilon) = \inf_N c_-(\epsilon,N)$ and $c_+(\epsilon) = \inf_N c_+(\epsilon,N)$?
Is there a simple way to see the right order of magnitude of $c_+(\epsilon)$ and $c_-(\epsilon)$?
As an alternative, we could allow $P_+(x)$, $P_-(x)$ to be a linear combination of fractional powers $x^r$, $r\geq 1$.
The back of envelope computation suggests that the approximation problem is solvable, roughly speaking, for $N>\varepsilon^{-1}$ and the associated cost is about $\exp(\varepsilon^{-1})$ (in the sense that changing $\varepsilon$ to $100\varepsilon$ influences the answer more than its inherent inaccuracies). If it is to your liking, I'll try to figure out the details; if not, just let me know and we'll forget about it. What was your hope, BTW?
Uf, $exp(\epsilon^{-1})$ is a disaster. Why can't one do better? Where is that exponential coming from?
Analyticity in the unit disk. Basically you ask whether one can approximate the sign function in $L^1$ on $[-1,1]$ up to $\varepsilon$ by an analytic function in the unit disk bounded by $M=c(\varepsilon)$. That is terribly costly.
Aha. And why is that terribly costly? (I have a nagging feeling that I once knew why...)
BTW - I see that $\exp(\epsilon^{-1})$ can be reached by Beurling-Selberg. Is there another, possibly simpler example?
And why is that terribly costly?
@AlexandreEremenko will, probably, have a simpler explanation, but this is how I see it if no pen and paper is allowed.
Let $\varphi$ be the conformal mapping of the horizontal strip $\{|\Im z|<1\}$ to the unit disk such that $\varphi(0)=0$ and the real axis is mapped to $[-1,1]$ so that $-\infty$ goes to $-1$ and $+\infty$ goes to $1$. Suppose $f$ is an analytic function in the unit disk bounded by $M$ and such that $\int_{[-1,1]}|f-H|\le\varepsilon$ where $H(x)=0$ on $(-\infty,0)$ and $H(x)=1$ on $[0,+\infty)$. Consider $F=(f\circ \varphi)\cdot\varphi'$. We have $F$ analytic and dominated by $M|\varphi'|$ in the strip and $\int_{-\infty}^\infty |F-H\varphi'|\le\varepsilon$, whence for every $y>0$, we must have $\left|\int_{-\infty}^\infty F(x)e^{iyx}dx-\int_{0}^\infty \varphi'(x)e^{iyx}dx\right|\le\varepsilon$. Shifting the contour in the first integral to $i+\mathbb R$ and taking into account that $\int_{-\infty}^{\infty}|\varphi'(i+x)|dx=\pi$ (the top line is mapped to the top half-circle), we see that the first integral is at most $\pi Me^{-y}$. On the other hand, the second one is about $i\frac{\phi'(0)}{y}$ for decent size $y$ (shift the contour to the vertical interval $[0,i]$ followed by $i+[1,\infty)$ and use the trivial version of the "Laplace formula"). Thus, we get
$$
\frac cy\le \varepsilon+\pi Me^{-y}
$$
Now choose $y=\frac c{2\varepsilon}$, say.
Is there another, possibly simpler example?
Yes, if you want just an $L^1$ approximation but adjusting it to your requirement that the polynomial should stay on one side of the function is a bit cumbersome exercise, so I'll not bother going into it.
I'm confused by the notation $F = (F\circ \phi) \phi'$. Do you mean $F = (f\circ \phi)\cdot \phi'$?
@HAHelfgott Yes, I'll edit
|
2025-03-21T14:48:29.623603
| 2020-01-12T02:39:21 |
350245
|
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|
Stack Exchange
|
Pseudo-Abelian Completion in the constrution of Motifs (by Y. Manin)
I reading Yuri Manin's famous paper on "CORRESPONDENCES, MOTIFS AND MONOIDAL TRANSFORMATIONS" and struggle with his definition for so called pseudo-abelian completion given on page 453 by a reason I would like to explain below. first of all the setting:
Definition 1: An additive category $\mathscr{D}$ is called pseudo-abelian if the following condition holds in
it:
(P)For any projector $p \in Hom(X,X)$ (recall projector = idempotent map satisfying $p=p^2$) there exists a kernel $Ker p$, and the canonical
isomorphism $Ker(p) \oplus Ker(id_X-p) \cong X$.
Definition 2: Let $\mathscr{D}$ be an additive category. Its pseudo-abelian completion is the category $\tilde{\mathscr{D}}$ defined
by the following data:
-Objects $Ob(\tilde{\mathscr{D}})$ are pairs $(X,p)$ with $X \in Ob(\mathscr{D})$ and $p$ projector
-Morphisms are defined by
$$Hom_{\tilde{\mathscr{D}}}((X,p),(Y,q)):= \\ \Big\{\text{group of } f \in Hom_{\mathscr{D}}(X,Y) \text{ with } f \circ p=q \circ f \Big\} \Big / \Big\{\text{ the subgroup of } f \text{ for } \\ \text{ which } f \circ p=q \circ f=0 \Big\}$$
Then it is claimed that the "category" $\tilde{\mathscr{D}}$ is pseudo-abelian.
Problem 1: I'm not satisfied with the definition of morphisms. Essentially what is here done one associate to a category $\mathscr{D}$ a Karoubian hull. See here: https://ncatlab.org/nlab/show/Karoubian+category#properties or here: https://en.wikipedia.org/wiki/Pseudo-abelian_category#Pseudo-abelian_completion for details.
And I think that it not suffice to demand that the morphisms in $\tilde{\mathscr{D}}$ are $f \in Hom_{\mathscr{D}}(X,Y) \text{ with } f \circ p=q \circ f$ modulo certain relation.
I think it's neccessary to require that these morphism have to satisfy $f \circ p=q \circ f= f$ as well.
Why I think so: Let $(X, p)$ an object in this category and the question is what is the identity $id_{(X,p)}$ of it? If we only demand that the morphisms have to satisfy $f \circ p=q \circ f$, then we have a uniqueness problem of the identity:
$id_X$ and $p$ behave both well as identities of $(X,p)$ as one can easily check. And if $p \neq id_X$ that is a problem. I don't also see how the moduled out relations can repair this leck of uniqueness.
But if we require instead the extended relation $f \circ p=q \circ f= f$, we see that $p$ and not $id_X$ is unique identity $id_{(X,p)}$ on $(X,p)$.
Problem 2: What is the reason for taking quotient by $\{\text{ the subgroup of } f \text{ for which } f \circ p=q \circ f=0 \}$? The classical description of Karoubian completion also not require it.
This is an equivalent way of describing the same thing. To see that, notice that for any $f$ such that $q\circ f=f\circ p$, $$( f\circ p - q\circ f\circ p)\circ p =q\circ( f\circ p - q\circ f\circ p)=0$$ Therefore $[f\circ p]=[q\circ f\circ p]$ in our homomorphism group. Similarly, $[f]=[f\circ p]$ in our homomorphism group. Hence every homomorphism is represented by an element of the form $g=q\circ f\circ p$( and hence $q\circ g=g\circ p=g$).
On the other hand, if two maps $f,g:X\to Y$ are such that $q\circ f= f\circ p=f$, $q\circ g= g\circ p=g$, and represent the same homomorphism, then by definition $q\circ (f-g)=(f-g)\circ p=0$. Since composition is bilinear, this forces $f=g$.
So, the answer to your questions are 1.) this is a different way of describing the same thing, and 2.) to make the problems you described in 1. go away.
|
2025-03-21T14:48:29.623855
| 2020-01-12T04:02:19 |
350246
|
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"Pierre PC",
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"https://mathoverflow.net/users/129074",
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|
Stack Exchange
|
Construction of an infinitely Fréchet differentiable function with given set of zeros in a Banach space
After looking at this question, I am now wondering if the following is true.
Let $X$ be a separable Banach space over $\mathbb R$ or $\mathbb C$, and $A\subseteq X$ a closed set. Then there exists a function $f: X\to\mathbb R$ (Cannot be $\mathbb C$ here; always $\mathbb R$ even if the previous one is $\mathbb C$) such that $A=\{x\in X:f(x)=0\}$ and $f$ is infinitely differentiable in the sense of Fréchet (see Fréchet derivative).
Or can it be true after some minor modifications?
Here is my attempt to prove it.
Let $\phi:\mathbb R\to[0,\infty)$ be a $C^\infty$ function satisfying $\phi(t)=1$ for $|t|\leq1/2$ and $\phi(t)=0$ for $|t|\geq 1$. Such $\phi$ can be constructed by using functions like $\exp(-1/t)$.
Since $X$ is separable and $X\backslash A$ is open, $X\backslash A$ can be expressed as a countable union of open balls,
$$
X\backslash A=\bigcup_{k=1}^\infty B(x_k,r_k).
$$
Define, for $x\in X$,
$$
f_k(x)=\phi\left(\frac{\|x_k-x\|}{r_k}\right).
$$
Each $f_k$ is infinitely differentiable, since for $x\neq x_k$, this follows from the chain rule, and if $x=x_k$, obviously all orders of derivatives are zero since it is constant in a neighborhood of it.
Now, define $f(x)=\sum_k a_k f_k(x)$, where
$$
a_k=\frac{1}{2^k} \left(\sup_{x\in B(x_k,r_k)\\ m\le k}\|D^m f_k(x)\|\right)^{-1},
$$
any if the thing in the bracket is zero, we can choose any positive value for $a_k$ as we like.
Does my proof work?
The norm may be non-smooth, even in finite dimension. This is easily corrected in the finite dimensional setting, but the existence of smooth function with compact support is not a given in general Banach spaces. This phenomenon is particularly annoying when one is hoping partitions of unity would exist.
@PierrePC What does it mean by "non-smooth"? I know that the norm in a Hilbert space is (at least first order) differentiable.
Sorry, I meant not smooth, like the $L^\infty$ norm in $\mathbb R^n$. In a Hilbert space, the norm is indeed smooth.
@PierrePC So my proof will work in a Hilbert space?
I think so; see my answer below.
If there is a non-zero smooth function with bounded support, there are smooth partitions of unity, and any closed set is a the zero set of some smooth function. But a Banach space may fail to have non-zero differentiable function with bounded support.
@PietroMajer I knew I had seen this somewhere!... I just remembered a reference for that statement, by R. Bonic and J. Frampton: https://www.jstor.org/stable/24901438 . The result about the zero set is not explicitly stated, but a simple tweaking of the proof of Theorem 1 (about the partitions of unity) does the trick. Of course it means that the existence of such a "bump function" is equivalent to the characterization of zero sets as stated by the OP.
I should add that the definition of $V_x$ in the proof given is terrible; I assume it should read $V_x = {y : f_{n(x)}(y)>\alpha(x)}$.
@Pierre PC thank you for this reference!
In a few words, I believe there is not much to add to you proof to make it work, but it nevertheless highly decreases the breadth of the result.
Edit: As Pietro Majer said in the comment, the theorem below actually hold without the uniform bound for the derivatives. In other words, every open set of a fixed Banach space $B$ is the zero set of a $\mathcal C^k$ function if and only if there exists a $\mathcal C^k$ bump function, i.e. a non-zero function with bounded support. A self-contained reference is R. Bonic and J. Frampton, Smooth Functions on Banach Manifolds, 1966. The proof of Theorem 1 contains everything needed to prove the result I state here, up to a notational typo (maybe I am the only one to be terribly confused); the definition of $V_x$ should read:
$$ V_x = \{y:f_{n(x)}(y)>\alpha(x)\}. $$
Notations
I write $L^k(B;B')$ for the Banach space of continuous $k$-linear functions with arguments in $B$ and image in $B'$; also $L(B,B')$ will mean $L^1(B;B')$. By definition, I say that $f:B\to B'$ is differentiable at $x$ only if there exists $A\in L(B,B')$ such that for any $\varepsilon>0$, we have
$$ |f(x+h) - f(x) - A(h)|_{B'} \leq \varepsilon|h|_B $$
for all $h$ small enough. In particular, I always ask for the derivative at some point to be continuous. I define the $\mathcal C^k$ classes as usual, and the higher derivatives $D^k\!f$ of a function $f$ are functions from $B$ to $L^k(B;B')$.
Recall the following classic fact from metric topology.
Fact.
Let $B$ and $B'$ be two Banach spaces. The space of bounded continuous functions $f:B\to B'$ is complete, with respect to the norm
$$ |f|_{B\to B'} := \sup_{x\in B}|f(x)|_{B'}. $$
It is actually true for $B$ a topological space, and $B'$ a complete metric space.
Positive results
I will show, using your proof, the following result.
Theorem.
Let $B$ be a separable Banach space and $0\leq k\leq\infty$. Suppose that there exists some non-zero function $f:B\to\mathbb R$ of class $\mathcal C^k$ with bounded support, such that
$$ \sup_{x\in B}|D^\ell\!f(x)|_{L^\ell(B;\mathbb R)}<\infty $$
for all $0\leq\ell\leq k$. Then any closed set of $B$ is the zero set of some function of class $\mathcal C^k$.
The requirement seems absurd from a finite-dimensional perspective, but actually two things can go bad in infinite dimensions. First off, it might not be easy to find smooth functions to begin with — I am not an expert, but I seem to recall smooth functions with bounded support are not a given in an arbitrary Banach space. Second, a smooth function with bounded support can actually be unbounded. I give an example at the end of this answer.
Before giving the proof, here is a proof of existence in a particular case.
Proposition.
If $H$ is a Hilbert space, the squared norm $x\mapsto |x|_H^2$ is smooth, and all its derivatives are bounded on bounded sets.
Of course $|x|^2_H$ is continuous and quadratic, so its derivatives are $|x|_H^2$ (of order 0), $h\mapsto2\langle x,h\rangle$, $h,h'\mapsto2\langle h,h'\rangle$ and zero (of order $k\geq3$), and we see that they are bounded on bounded sets. Hence for instance $\phi(|x|_H^2)$, with $\phi$ as in your question, will work for the above theorem (we can choose $k=\infty$).
Proof
The proof of the Theorem is basically the one you gave, where you implicitly proceeded by induction using the following elementary lemmas.
Lemma 1.
Suppose $f_n:B\to B'$ is a sequence of functions of class $\mathcal C^1$. Assume moreover that $f_n$ and $Df_n:B\to L(B,B')$ are Cauchy sequences with respect to $|\cdot|_{B\to B'}$ and $|\cdot|_{B\to L(B,B')}$ respectively.
Then $f_n$ and $Df_n$ converge to some continuous functions $f$ and $g$, and $f$ is $\mathcal C^1$ with $Df=g$.
$~$
Lemma 2.
Let $B$ be a separable Banach space, $U_0$ a bounded open neighbourhood of $0$ in $B$, and $(x_n)_{n\geq 0}$ a dense sequence in $B$.
Then for any open subset $U$ of $B$ and any $x\in B$, $x+r(x)U_0\subset U$ for
$$0<r(x):=\sup\left\{r>0\text{ such that }x+r U_0\subset U\right\},$$
and $U=\bigcup_{n\geq0}\big(x_n+r(x_0)U_0\big)$.
To prove the first result, you just have to use the above Fact, and take the limit in
$$ f(x+h) = f(x) + \int_0^1Df(x+th)(h)\mathrm dt. $$
The second one is easy, and I imagine you know it already.
A ‘negative’ result
I am giving here an explicit example of a smooth function $B\to\mathbb R$ with bounded support, but which is itself unbounded.
Suppose you have a continuous inclusion of Hilbert spaces $H\hookrightarrow K$. Then $|x|_K\leq C|x|_H$ for $C>0$ large enough. However, maybe $|x|_K$ can be arbitrary small for $|x|_H=1$. Both $|x|_H^2$ and $|x|_K^2$ are smooth functions on $H$, as discussed in the Proposition, so
$$ f:x\mapsto \frac{\psi\big(1/|x|_H^2\big)}{|x|_K^2} $$
is smooth provided $\psi:\mathbb R\to\mathbb R$ is smooth and vanishes on a neighbourhood of zero.
Now if $x_n$ is a sequence in $H$ such that $|x_n|_H=1$ but $|x_n|_K$ goes to zero, fix some $\psi$ with support in $(1/2,2)$ and $\psi(1)=1$. Then for $f$ defined as above, $f(x_n)$ is $1/|x_n|_K^2$ and diverges, even though $f$ is smooth and has support in the ball of radius $2$ centred at zero.
For an actual concrete example, take the inclusion of Lebesgue spaces $\mathrm L^2\big([0,1],\mathrm dx\big)\subset\mathrm L^2\big([0,1],x\mathrm dx\big)$ and
$$x_n:t\mapsto\begin{cases}
n\text{ if }t\leq 1/n^2,\\
0\text{ else.}\end{cases}$$
As a side remark note that of there is a nonzero function with bounded support on X, one can make it bounded (in uniform norm) e.g. composing with $\arctan$
@PietroMajer Maybe I am missing your point, but 1) I agree that my example was artificial and that the unboundedness is easily corrected. I just think it is a neat example that speaks against something that we often take for granted. 2) Ideally, one would want to change the function so that the uniform norm but also a (say fixed but arbitrary large) number of its derivatives as well. Are you saying that there is a way to do that with methods similar to what you describe?
no, that was only a side remark. As far as I know (i.e. $\epsilon$) there may be a hierarchy of situations, of Banach spaces where there are non-zero functions with bounded support in some regularity classes, but not in others (e.g. $C^k$ but not $C^k_b$ or $C^k_b$ but not $C^{k+1}$)
For instance, I think one can make a function $f$ as in the hypothesis of your Theorem out of some $C^k$ function $g$ defined on some $U\subset B$ that has $D^2g(x_0)>0$ at least in one point (as a symmetric bilinear form). Indeed up to adding a linear form, $x_0$ is a local minimum, and a composition $f=\phi\circ g$ with a suitable smooth bump function $\phi:\mathbb{R}\to\mathbb{R}$ supported around $g(x_0)$ should do. (Boundedness of differentials of $f$ should follow from the fact that differentials of $g$ are continuous, hence locally boundeed)
I think you know the answer to my question: https://mathoverflow.net/q/410368/121665.
|
2025-03-21T14:48:29.624476
| 2020-01-12T06:27:42 |
350250
|
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|
Stack Exchange
|
Convex combination of semi-algebraic sets
Suppose we are given two semi-algebraic sets $S_1$ and $S_2$. Define
$$
S=\{s=ps_1+(1-p)s_2: s_1\in S_1, s_2\in S_2, 0\leq p\leq 1.\}
$$
$S$ is semi-algebraic.
Can we bound the degree of $S$?
If we have more than two semi-algebraic sets $S_1,\cdots,S_n$, can be bound the degree of their convex combination?
|
2025-03-21T14:48:29.624529
| 2020-01-14T14:13:00 |
350404
|
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|
Stack Exchange
|
Completeness of the field of fractions of a ring of formal power series
Let $k$ be a field and let $k[[X,Y]]$ be the ring of formal power series with coefficients in $k$. Let $k((X,Y))$ be its field of fractions. For $F\in k[[X,Y]]$, $F\neq 0$ define $v(F)$ as the least degree of a monomial appearing in $F$, and extend $v$ to $k((X,Y))$ by $v(F/G)=v(F)-v(G)$. Then $v$ is a valuation in $k((X,Y))$ that induces an absolute value (namely $|F|=2^{-v(F)}$) and it is well-known that $k[[X,Y]]$ is a complete ring, but:
is the field of fractions $k((X,Y))$ also complete?
This is a particular case of this question.
With just one indeterminate the answer is positive.
I think I have found the answer to my own question. Since I cannot post it here, I have posted it as an answer to the related question cited above. I hope it will be useful, since I have not been able to find it in any book I have had at hand.
|
2025-03-21T14:48:29.624623
| 2020-01-14T15:33:45 |
350409
|
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"Ali Taghavi",
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|
Stack Exchange
|
A vector field $X$ on $\mathrm{GL}(n,\mathbb{R})$ with $\begin{cases} X.\mathrm{trace}=\mathrm{Det} \\X.\mathrm{Det}=-\mathrm{trace} \end{cases}$
Is there a vector field $X$ on $\operatorname{M}_n(\mathbb{R})$ or $\operatorname{GL}(n,\mathbb{R})$ with the following condition:
$$\begin{cases} X\cdot \operatorname{trace}=\operatorname{Det} \\X\cdot \operatorname{Det}=-\operatorname{trace} \end{cases}$$
where $\operatorname{Det}$ is determinant?
What do you mean with $X.trace$ and $X.Det$?
@Wojowu It is the derivative of Det along solution curves of vector field $X$.
$\DeclareMathOperator\trace{trace}\DeclareMathOperator\Det{Det}$No. Let $\sum_{i,j}x_{ij}(M)\frac\partial{\partial m_{ij}}$ be this vector field. These conditions amount to writing
$$\sum_ix_{ii}(M)=\det M,\qquad\sum_{ij}x_{ij}(M)\hat m_{ij}=-{\rm Tr}M,$$
where $\hat M$ is the cofactor matrix. Take $M=a I_n$, for which $\hat M=a^{n-1}I_n$. One must have
$$\sum_ix_{ii}(a I_n)=a^n,\qquad a^{n-1}\sum_{i}x_{ii}(a I_n)=-na,$$
which is inconsistent when $a^{2n-1}\ne-an$.
Edit. Since both the trace and the determinant are homogeneous polynomial, you could be interested in a vector field in which the conditions are consistent with homogeneity. Then its coordinates would be themselves homogeneous polynomials in the entries of $M$. For instance, looking for a vector field $X$ such that
$$X\cdot \trace=\Det,\qquad X\cdot \Det=-(\trace)^{2n-1}$$
makes sense. The last righ-hand side could be replaced by $- \trace^{n-1}\Det$.
Does a vector field satisfying your modified condition exist?
Thank you for your answer and the very interesting modification of the question
@LSpice that is a very good point to think. Thanks!
|
2025-03-21T14:48:29.624756
| 2020-01-14T15:51:42 |
350414
|
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|
Stack Exchange
|
When is there a unique perfect group of order $n$?
For which $n$ is there a unique perfect group of order $n$? Are there infinitely many such $n$?
Some guesses for infinite sequences of such $n$: $|\mathrm{PSL}(2,p)|$, $|\mathrm{SL}(2,p)|$, $|A_m|$, $|S_m|$.
If you replace "perfect" with "simple", a complete understanding follows from the Classification. Apart from a few small coincidences and one well-understood infinite family, there are no coincidences. See this previous question for instance. But the picture for perfect groups is less clear to me.
A complete list of the orders of perfect groups up to $10^6$ is contained in Section 5.4 of the book by Holt and Plesken (Holt, Derek F.; Plesken, Wilhelm, Perfect groups. (1989). ZBL0691.20001.). There seem to be many $n$ for which there is a unique perfect group of order $n$, but there are also many $n$ (e.g. $2^{14} \cdot 60$) for which the number of perfect groups is very large. There are precisely five $n < 10^6$ for which there are both simple and non-simple perfect $G$ of order $n$:
20160: $A_8$, $L_3(4)$, $2.(A_5 \times L_3(2))$
181440: $A_9$, $A_6 \times L_2(8)$, $L_3(2) \times 3.A_6$
262080: $L_2(64)$, $2^2.(A_5 \times L_2(13))$
443520: $M_{22}$, $2^2.(L_3(2)\times L_2(11))$
604800: $J_2$, $A_5^2 \times L_3(2)$, $2^2.(A_5 \times A_7)$
$|S_m|$ being the order of the Schur cover of $A_m$, obviously.
Is it obvious ( or at least known) when we can have $|S|^{a} = |T|^{b}$ for non-Abelian simple groups $S$ and $T$ and positive integers $a,b$?
@GeoffRobinson I don't know about that, but if you look at just $\mathrm{PSL}_2(q)$ for small $q$ ($q \leq 17$ is enough) then eventually you have more groups than prime divisors, so you will have many coincidences between direct products of these groups. I'm more interested in the other direction: can you be sure of some $n$, say $n = p(p-1)(p+1)$, that there is only one perfect group of order $n$?
@GeoffRobinson Your question about $|S|^a=|T|^b$ is known -- it only happens for $a=b$. Reference: W. Kimmerle et al., Proc. London Math. Soc. (3) 60 (1990), no. 1, 89–122.
@RichardLyons :Thanks for that reference Richard.
For the case $n = p(p-1)(p+1)$ for $p >3$ a prime, a theorem of M. Herzog (to be found in the article "Finite groups with a large cyclic Sylow subgroup" (MSN) in the conference proceedings "Finite Simple Groups" (Oxford, editors M. Powell and G. Higman, published 1971 by Academic Press (MSN))) answers your question postively. The Theorem of Herzog uses a 1958 Theorem of Brauer and Reynolds ("On a problem of E. Artin", Annals of Mathematics, 68 (MSN)) which probably suffices to cover the cases you are interested in.
Note that if $n = p(p-1)(p+1)$ with $p >3$ a prime, and $G$ is a perfect group of order $n$, then $G$ is not $p$-solvable (for if $G$ is $p$-solvable of order $n$, then $G$ either has a factor group of order $p$ or a non-trivial cyclic factor group of order dividing $p-1$)-(later edit: more generally, if $G$ is any $p$-solvable group with a Sylow $p$-subgroup of order $p$, then $G$ is not perfect).
The Theorem of Herzog/Brauer–Reynolds proves that a finite group $G$ of order divisible by $p$, but of order less than $p^{3}$ (for a prime $p >3$) which is not $p$-solvable is one of the following groups: $\operatorname{PSL}(2,p)$, $\operatorname{PSL}(2,p-1)$
(for $p >5$ a Fermat prime), $\operatorname{SL}(2,p)$, $\operatorname{PGL}(2,p)$, $\operatorname{PSL}(2,p) \times \mathbb{Z}/2\mathbb{Z}$.
The only perfect group of order $n$ on the list is ${\rm SL}(2.p)$ and the only perfect group of order $\frac{n}{2}$ is ${\rm PSL}(2,p)$, showing that ${\rm SL}(2,p)$ and ${\rm PSL}(2,p)$ ( for $p >3$ a prime) are the unique perfect groups of their orders.
Nice, thank you!
Details of second paragraph (please correct me if you had something simpler in mind): If $G$ is $p$-solvable and has order $n$ then some quotient $G/N$ has a normal subgroup of order $p$, and this extension is split by Schur--Zassenhaus. Your two cases are whether this split product is direct or not.
I think that works, but the argument I had in mind was the following: G/Op′(G) has a self centralizing normal Sylow p-subgroup of order p (eg by Hall-Higman centralizer Lemma), so is isomorphic o a subgroup of order divisible by $p$ of a Frobenius group of order p(p−1), so is itself either a Frobenius group or cyclic of order $p$. If cyclic of order $p$, then G has a factor group of order p, while if the Frobenius complement is non-trivial, it is a cyclic subgroup of order dividing p−1 which is a homomorphic image of G.
|
2025-03-21T14:48:29.625164
| 2020-01-14T15:53:35 |
350415
|
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|
Stack Exchange
|
Log-concavity of function
Consider the function
$$f_{n}(x)=e^{-x^2}x^n.$$
My goal is to show that
$$ G(y):=\frac{(f_2*f_0)(y)}{(f_0*f_0)(y)}- \left(\frac{(f_1*f_0)(y) }{(f_0*f_0)(y)}\right)^2$$
is log-concave.
Let us first observe that indeed $G(y) \ge 0.$
This just follows from a Cauchy-Schwarz
$$(f_1*f_0)(y) \le \sqrt{(f_2*f_0)(y)(f_0*f_0)(y)}$$
so everything is well-defined.
Usually one can say a lot when convolutions are involved about log-concavity due to standard theorems see wikipedia
but this combination looks a bit tricky.
Addendum I should add that I am in particular very interested in theoretical insights why this particular expression has to be log-concave.
This can be done by a straight-forward calculation with help of Maple ( see the code and results here https://www.dropbox.com/s/zwf335sfkxm10nx/log-concave.pdf?dl=0 ).
Direct calculations show that
$$(f_2*f_0)(y)=\frac{1}{4} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} \left(y^2+1\right),
$$
$$(f_1*f_0)(y)=\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} y,
$$
$$(f_0*f_0)(y)=\sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}}
$$
for all real $y$,
so that $G$ is the constant $1/4$ and hence log concave.
thank you. although i must admit i was curious whether there is something behind my expression that explains why it has to be log-concave.
@Martinique : I doubt that a "behind-the-expression" reasoning exists, in particular given that $G$ is a constant and thus just "barely" log concave.
I should say that- although not related to log-concavity-there is some interesting structure that I noticed which I ask about in this post https://mathoverflow.net/questions/350524/phase-transition-in-convolution
|
2025-03-21T14:48:29.625290
| 2020-01-14T16:37:08 |
350417
|
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|
Stack Exchange
|
Nilpotency of topological groups
A group $G$ is said to be nilpotent if $G$ has a central series of finite length, that is, a series of normal subgroups
$$
\{1\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G
$$
where $G_{i+1}/G_i$ is a subgroup of $Z(G/G_i)$.
A (based) topological space $X$ is said to be nilpotent if the fundamental group $\pi_1 X$ is a nilpotent group, and $\pi_1 X$ acts nilpotently on the higher homotopy groups, $\pi_i X$ for $i \geq 2$, i.e., there is a central series
$$
\{1\} \triangleleft G_0^i \triangleleft G_1^i \triangleleft \cdots G_n^i = \pi_i X
$$
such that the induced action of $\pi_1X$ on the quotient $G_{k+1}^i/G_k^i$ is trivial for all $k$.
Here are the two questions.
Q1. Is there an example of a topological group $G$ which is nilpotent as a group, but not nilpotent as a topological space?
Q2. Is there an example of a topological group $G$ which is nilpotent as a space, but not nilpotent as a group?
Aren't all topological groups nilpotent as spaces? (Because $G = \Omega BG$?)
In any case, Q2 is trivial: take a simply connected semi-simple group, e.g. $SU(n)$.
|
2025-03-21T14:48:29.625393
| 2020-01-14T17:18:32 |
350420
|
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|
Stack Exchange
|
Refinement: Can $L^1_{loc}$ be represented as colimit?
Let $\mu$ be a $\sigma$-finite measure on a measure space $(\mathbb{R}^d,\Sigma)$. Can $L^1_{\mu,loc}$ be represented as an injective-limit in the category of LCS (locally convex spaces and continuous linear maps) of the form:
$$
\injlim L^{p_n}_{\mu_n} = L^1_{\mu,loc}
$$
for some finite-measures $\mu_n$ on $(\mathbb{R}^d,\Sigma)$ and for some $p_n \in [1,\infty)$;
with the additional constraint that:
The resulting universal maps $\iota^n_{\infty}:L^1_{\mu_n}\rightarrow L^1_{\mu,loc}$ are the inclusions (ie: $f \mapsto f$)?
This is a refinement of this post.
What do you mean by the injective-limit? Here is my suggestion, for the case of Lebesgue measure. We cover the space by balls of radius $n$ and denote by $\mu_n$ its corresponding restrictions. The the Banach spaces $(E_n)$ form both an inductive and a projective system, where these are the corresponding $L^1$-spaces. The projective limit is naturally identifiable with the Fréchet space of (equivalent classes of) locally integrable functiins. The inductive limit is the strict $LF$-space of integrable functions with compact support. I am not sure which one you mean—hope this helps.
As an additional remark, these are not just any old inductive or projective limits but have partitions of unity in the sense of de Wilde which makes them particularly tractable.
Grubb treats these LF-spaces in the case where $\mu$ is Lebesgue and $d=1$, can the analogous construction still be identified with $L^p_{\mu,comp}(\mathbb{R}^d;\mathbb{R}^D)$ in general (ie for $\mathbb{R}^D$-valued functions with respect to any $\sigma$-finite measure)? That is, do you have a reference?
Also, I'm assuming that these spaces are not metrizable since they're a strict injective limit of infinite-dimensional Fr/'{e}chet spaces.
Again, do you mean inductive or projective limits (I am not familiar with the term „injective limit“)? As I wrote, the projective limit is a Fréchet space, the inductive limit an $LF$- ( even $LB$-) space. The same holds for any value of $p$ and for vector-valued (even Banach space valued) functions. I haven‘t really thought about the case of a general measure but imagine it will hold, say, for a locally finite measure on a $\sigma$-compact, locally compact space.
Indeed, I meant inductive (sorry I didn't notice my modified terminology). Also if you could provide a reference I'd accept this as a complete answer since its has been so helpful in directing me.
Again, I don‘t have an explicit reference but would emphasise that the inductive limi in the lcs sense gives the space of functions with compact support, not the locally integrable ones. The inductive limit in the sense of Banach spaces gives precisely $L^1$. Glad to have been of help.
Perfect, you've helped so much. Thanks a million :))
@user131781 Was this the article you were referring to:
https://link.springer.com/content/pdf/10.1007%2FBF01397608.pdf
|
2025-03-21T14:48:29.625593
| 2020-01-14T17:32:16 |
350425
|
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|
Stack Exchange
|
Recovering a $G$-valued representation/parameter
Number theoretic phrasing
Let $G$ be a connected reductive group over a characteristic $0$ field $F$. Associated to $G$ is its Langlands dual group $^{L}G$. For every dominant cocharacter $\mu$ of $G_{\overline{F}}$ Kottwitz constructs in [Kot1,(2.1.1)] a representation $r_{-\mu}:\, ^{L}(G_E)\to \mathrm{GL}(V)$ where $E$ is the reflex field of $\mu$. Given two $L$-parameters $\psi_1,\psi_2:W_F\to {^L}G$ I believe that the equivalence $r_{-\mu}\circ \psi_1\sim r_{-\mu}\circ\psi_2$ for all $\mu$ does not imply the equivalence of $\psi_1$ and $\psi_2$ (e.g. this is true for some $G$ but examples for Spin groups are given in [Lar]). I am under the impression that if one considers not just a single $\mu$ but tuples of $\mu$'s simultaneously that this fixes the issue. I think this is supposed to be pivotal to the philosophy of why V. Lafforgue and his predecessors consider moduli spaces of Shtukas with multiple legs (but I am really interested in the result I asked--not how Lafforgue deals with it which I understand is slightly different).
Can anyone provide a precise statement? Is it that if for all tuples $(\mu_1,\ldots,\mu_n)$ one has that the representations $r_{-\mu_i}\circ\psi_1$ and $r_{-\mu_i}\circ\psi_2$ are simultaneously isomorphic (in some sense?) then $\psi_1\sim\psi_2$?
Lie group theoretic statement
Suppose that $G$ is a complex reductive Lie group and that $\psi_1,\psi_2:\Gamma\to G$ are two continuous homomorphisms where $\Gamma$ is some topological group (in my setting it's something like a Galois group, but not compact). I believe it's known that if $\rho\circ\psi_1$ is conjugate to $\rho\circ\psi_2$ for all irreducible algebraic representations $\rho$ of $G$ then it needn't be true that $\psi_1$ and $\psi_2$ are $G$-conjugate (cf. again the paper [Lar]). I believe that one can remedy this if one considers tuples of representations.
Can anyone provide a precise statement? Is it true that if for all tuples $(\rho_1,\ldots,\rho_n)$ one has that the representations $\rho_i\circ\psi_1$ and $\rho_i\circ\psi_2$ are simultaneously isomorphic (in some sense?) then $\psi_1$ and $\psi_2$ are $G$-conjugate?
[Lar] Larsen, M., 1994. On the conjugacy of element-conjugate homomorphisms. Israel Journal of Mathematics, 88(1-3), pp.253-277.
[Kot1] Kottwitz, R.E., 1984. Shimura varieties and twisted orbital integrals. Mathematische Annalen, 269(3), pp.287-300.
|
2025-03-21T14:48:29.625779
| 2020-01-14T17:57:32 |
350427
|
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|
Stack Exchange
|
Geometric models for 2-gerbes
One can think of a complex line bundle as a geometric model for an integral cohomology class of degree 2. Similarly, a locally-trivial bundle of $C^*$-algebras with fiber B(H) (the $C^*$-algebra of bounded operators on an infinite-dimensional Hilbert space) can be thought of as a geometric model for an integral cohomology class of degree 3. One can say that such a bundle is a 1-gerbe, while a complex line bundle is a 0-gerbe. Are there similarly nice models for integral cohomology classes of degree 4 (that is, for 2-gerbes)?
$E_8$ principal bundles? :)
One silly answer is: your description of 0 and 1 gerbes is equivalent to the observation that $K(\mathbb{Z},2)=BU(1)$ and $K(\mathbb{Z},3)=BPU(H)$. So we want to identify $K(\mathbb{Z},4)$ as $BG$ for some topological group $G$. But of course we could take $G=K(\mathbb{Z},3)$ and find that $2$-gerbes are principal $K(\mathbb{Z},3)$ bundles...
It might help to understand what it is you’re looking to do. A four form is just a map to a $K(Z,4)$, and that is fairly geometric. As John said, that’s the same as a $K(Z,3)$-bundle, which you can model as $E_8$ in reasonably low dimensions or use some other example of a $K(Z,3)$ with a group structure. I think you can even work directly with bundles of homotopy types in $\infty$-land. You could also model it as a differential cohomology class. Or you can look up the definition of a 2-gerbe in Breen’s work. I’m not sure any of these is better or worse than any of the others.
One requirement is that the model is geometric, i.e. does not involve objects which are only defined up to homotopy equivalence. Modeling a degree-$n$ cohomology class as a map to $K(Z,n)$ is not geometric, because $K(Z,n)$ is a homotopy type, not a concrete space. Similarly, modeling a degree-2 class as a map to $BU(1)$ is not geometric, because the classifying space of a group is defined only up to homotopy equivalence. A further more vague requirement is that I want a model which has a physics flavor. Roughly, it should involve Hilbert spaces in in some way.
@AntonKapustin in case you are still interested, I have a paper coming out soon that has a much easier geometric model than bundle 2-gerbes, as defined by Stevenson, and which suffices to capture all examples.
Sure, David, I would be happy to have a copy.
One fairly concrete way to view these things is via the Cech model (= transition functions). But maybe this isn't what you were looking for...
Say the base space $X$ is a manifold or finite $CW$-complex. Choose a cover by contractible open sets $\amalg U_{i}\rightarrow X$.
A line bundle on $X$ is the data of functions $f_{ij}:U_{ij}\rightarrow U(1)$ on double overlaps, satisfying certain conditions on triple overlaps $U_{ijk}$.
A 1-gerble on $X$ is the data of line bundles $L_{ij}\rightarrow U_{ij}$ on double overlaps, isomorphisms $\phi_{ijk}:L_{ij}\otimes L_{jk}\rightarrow L_{ik}$ on triple overlaps, satisfying certain conditions on quadruple overlaps $U_{ijkl}$. Assuming the cover is nice enough, this is equivalent to functions $f_{ijk}:U_{ijk}\rightarrow U(1)$ satisfying conditions on $U_{ijkl}$.
So a 2-gerble would be either: assign a 1-gerble to each double overlap, as well as the appropriate coherence data on higher overlaps (which seems baffling to me), or: repeat the 1-gerbe recipe but start at triple overlaps, or: repeat the 0-gerbe recipe but start at quadruple overlaps.
I guess I should have made clear that I want a description which does not involve a choice of a cover. The descriptions of 0-gerbes and 1-gerbes that I gave satisfy this requirement.
Two possible ideas. One is that you can realize $K(\mathbb{Z},3)$ as the quotient $U(HS)/PU(\infty)$ where $U(HS)$ is the unitary group on the Hilbert space of Hilbert-Schmidt operators. However, I don’t know if you can see the group structure in this model to make a principal bundle. This construction is in
Alan L. Carey, Diarmuid Crowley, Michael K. Murray, Principal Bundles and the Dixmier Douady Class, Commun.Math.Phys. 193 (1998) pp 171-196, doi:10.1007/s002200050323, arXiv:hep-th/9702147.
There is also Andre Henriques’s conjecture that the outer automorphisms of a hyperfinite type III factor model $K(\mathbb{Z},3)$. See `Naturally occuring' $K(\pi, n)$ spaces, for $n \geq 2$.
The second option sounds good to me! I need to watch this lecture again (I already did once, years ago, but forgot most of it).
Bundle $n$-gerbes seem to be what you are looking for.
Bundle gerbes can be defined w.r.t. an open cover (then it is what John Greenwood wrote), but don't have to. Instead of a cover, any surjective submersion is ok. $k$-fold intersections are then replaced by $k$-fold fibre products.
A nice example are basic bundle gerbes over compact Lie groups. For the groups $SU(n)$ and $Sp(n)$, one can construct them using an open cover of the Lie group, with a canonical line bundle over the double intersections. For the other simply-connected Lie groups, one needs to "resolve" the open sets by principal bundles over them, related to certain stabilizer subgroups. The union of their total space then gives a surjective submersion mapping to $G$, over whose 2-fold fibre product then again a canonical line bundle can be defined.
A further example are so-called lifting bundle gerbes, which are constructed from the problem of lifting the structure group of a principal bundle along a central extension. In this case, the surjective submersion is the bundle projection.
For bundle 2-gerbes, there is the Chern-Simons bundle 2-gerbe. Its surjective submersions is the bundle projection of a principal $G$-bundle. Over its 2-fold fibre product, it has a bundle gerbe, over its three-fold fibre product, a bundle gerbe isomorphism, over its 4-fold fibre product, a 2-isomorphism, and over its 5-fold fibre product, a condition. The Chern-Simons bundle 2-gerbe represents a class in $H^4(M,\mathbb{Z})$, the level of a Chern-Simons theory over $M$. It also carries a canonical connection, whose 3-holonomy is the value of the Chern-Simons theory on closed oriented 3-manifolds. This works for arbitrary Lie groups $G$, with no need to assume simply-connectedness.
I think bundle 2-gerbes will work for my purposes, thanks.
|
2025-03-21T14:48:29.626171
| 2020-01-14T18:22:03 |
350431
|
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|
Stack Exchange
|
Second moment estimates for $\zeta(s)$: different methods?
What are some different ways to achieve the bound $$\int_0^T \left|\zeta\left(\frac{1}{2} + i t\right)\right|^2 dt = T \log \frac{T}{2 \pi} + (2 \gamma - 1) T + E(T)$$
with an error term $E(T) = O(T^{\alpha+\epsilon})$ with $\alpha=1/2$ or $\alpha=1/3$?
Note I am not asking for the best $\alpha$; I am asking for different ways - the simpler the better - with the intention of choosing one of them to make explicit.
Methods I know:
Proofs based on Atkinson's 1949 formula. This formula yields $\alpha=1/2$ automatically and $\alpha=1/3$ after some smoothing and some work. The problem is that there is an asymptotically tiny error term that seems hard to make explicit without applying the saddle-point method explicitly (twice) - something that I know to be time-consuming and sometimes painful.
Balasubramanian (1978). Again, this doesn't look simple.
Ingham/Atkinson (1939). Short and relatively simple. Yields $E(T) = O(T^{1/2} \log^2 T)$.
I'm also curious about the same question (in the same way) for
$$\int_0^T \left|\zeta\left(\sigma + i t\right)\right|^2 dt,$$
where $0\leq \sigma <1$.
Theorem 7.2(A) of Titchmarsh's 'The Theory of the Riemann zeta function' uses a simplified form of the special case $x=t$ of the approximate functional equation of the Riemann zeta function which holds for $1/2<\sigma\leq 1$,
$$
\zeta(s)=\sum_{n<t} n^{-s} + O(t^{-\sigma}).
$$
The result is
$$
\int_1^T|\zeta(\sigma+it)|^2 \ dt< AT \min\left(\log T, \frac1{\sigma-\frac12}\right)
$$
uniformly for $1/2\leq \sigma \leq 2$.
For $0<\sigma<1/2$, the above result and the functional equation may help.
More precisely, for $\sigma>1/2$, Theorem 7.2 of the same book gives
$$
\int_1^T |\zeta(\sigma+it)|^2 \ dt \sim \zeta(2\sigma) T.
$$
That's not what I would call an approximate functional equation, and I was looking for more than just an upper bound.
If that's the case, Theorem 7.2 gives an asymptotic formula, not 7.2(A).
Again, see my post for the rough strength I am aiming for. Incidentally, Theorem 7.2 is valid only for sigma>1/2.
|
2025-03-21T14:48:29.626330
| 2020-01-14T18:54:43 |
350434
|
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|
Stack Exchange
|
Complete invariant of filtered chain complexes under chain homotopy equivalence
Betti numbers are a complete invariant of chain complexes of vector spaces modulo chain homotopy equivalence.
Can we similarly find complete invariants for (say, finite dimensional) filtered chain complexes? In particular, do the dimensions of the spaces appearing in the spectral sequence give a complete invariant?
Every chain complex of vector spaces is quasi-isomorphic to its homology, so homology is a complete invariant (every chain complex here is bi-fibrant so no issues worrying about chain homotopy equivalent versus quasi-isomorphic).
About the second question: You might want to add the assumptions that each chain complex is complete with respect to the filtrations. Otherwise you get the following counterexample. Let $C_$ be $\bigoplus_\mathbb{N} k$ concentrated in degree $0$ and let $D_$ similarly be the direct product, both equipped with the canonical decreasing filtrations given by the submodules where the first $k$-coordinates are zero. You get the same spectral sequences, but these two chain complexes cannot be homotopy equivalent, since $H_0$ is different.
@ConnorMalin thanks, edited the question accordingly.
@HenrikRüping thanks, added finiteness requirement to make things simple
|
2025-03-21T14:48:29.626448
| 2020-01-14T20:17:14 |
350439
|
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|
Stack Exchange
|
Sphere inversion in Riesz potential
I am reading the paper: ``ON THE DISTRIBUTION OF FIRST HITS FOR THE SYMMETRIC STABLE PROCESSES" by Blumenthal, Getoor and Ray, (Trans. Amer. Math. Soc. 99 (1961), 540-554).
On page 546, the authors talk about the idea of Riesz regarding spherical inversions. In particular they say, for the sphere $\{u:|x-u| =r\}$, inversion is the change of coordinates
$$
u \mapsto v = x+r^2\frac{(u-x)}{|u-x|^2}.
$$
Then the authors explain
``Riesz noted that if $f(u)$ is a potential of exponent $\alpha$ in $R^N$, then after inversion in a sphere with center $x$, $(x —v)^{\alpha-N}f(v)$ is a potential."
I dont understand this sentence. Also later they use this idea to write some identities which I do not understand at all. For example, in page 547, they write ``if $|y| >1$, inverting along the sphere $\{u: |y-u|^2 = |y|^2-1\}$ yields
$$
\int _{|u| \le 1} (1-|u^2|)^{-\alpha/2} |u-y|^{\alpha-N}du = (|y|^2 -1)^{\alpha/2} \int_{|v| \le 1} (1-|v|^2)^{-\alpha/2} |v-y|^{-N}dv "
$$
I would be obliged if anyone can shed some light on these mysterious identities.
Well, you may like to have a look at the original M. Riesz's 1938 paper: it is a fantastic read!
In the language of these papers, a "potential" of exponent $\alpha$ is a function $f$ of the form
$$ f(x) = \int_{\mathbb{R}^N} |y - x|^{\alpha - N} \mu(dy) , $$
where $\mu$ is a non-negative measure. M. Riesz observed that if $f$ is a potential, then its Kelvin transform
$$ \mathcal{K} f(x) = |x|^{\alpha - N} u(x / |x|^2) $$
is again a potential. This corresponds to inversion in the unit sphere $B(0, 1)$, that is, the change of coordinates:
$$ x \mapsto y = x / |x|^2 . $$
Obviously, the above observation extends (by translation and change of scale) to arbitrary spheres, and this is precisely the result that Blumenthal, Getoor and Ray refer to.
The above observation is particularly useful when inversion in the corresponding sphere preserves the unit ball. Such a sphere have a form given in the second part of your question, and the identity you are asking about is proved by the corresponding change of variable.
All authors mentioned above (M. Riesz, Blumenthal, Getoor and Ray) used these kind of calculations a lot, and rarely provided details. I believe, mostly because these are direct extensions of the well-studied case corresponding to $\alpha = 2$.
I do not know good references which give all details. A book by Bliedtner and Hansen, and perhaps a book by Landkoff, have them, if I remember correctly, but I find them both rather unfriendly.
Thank you, that helps a lot! I have now checked the computation for N=1 and the identity is clearer now. Unfortunately, the paper by Riesz is written in French which makes it inaccessible to me. If you happen to know of any English translation, it would help me a lot. Thanks again!
|
2025-03-21T14:48:29.626658
| 2020-01-14T20:18:38 |
350440
|
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|
Stack Exchange
|
Connes Embedding Conjecture is false
This preprint from yesterday claims to prove that Connes Embedding Conjecture fails.
Since the paper is from outside Operator Algebras (Computer Science/Quantum Computing) and they actually work on Tsirelson's Problem, it would be nice to have some feedback from experts from the Operator Algebra side on both the validity and the consequences of this.
I'm sure it would be nice. This is not the place for it though. If you have a specific question about the preprint, you might get an expert to weigh in on that point. Gerhard "Discussing Preprints Is Off-Topic Here" Paseman, 2020.01.14.
The authors are certainly very pedigreed and have been working on the problem for a long time and have a history of making partial progress. Alas, I am also an outside but I did find this blog post by one of the authors extremely enjoyable: https://mycqstate.wordpress.com/2020/01/14/a-masters-project/
There is some discussion on Scott Aaronson's blog.
Martin: maybe fire off an email to Vern or Taka, as operator algebraists and operator-space experts who have worked on the links between what-I-shall-always-refer-to-as-Kirchberg's-QWEP-conjecture and Tsirelson's problem?
there is equivalences proved by Klep and Schweighofer (with later corrections by more people) between Connes Embedding Conj. and existence of certain decompositions of "positive polynomials in matrix variables" (https://arxiv.org/abs/math/0607615, https://www.fmf.uni-lj.si/~klep/rcec-19jul13.pdf), which are sort of "pure algebra". It would also be interesting to know what kind of said "positive polynomials" arise from that preprint that do not admit these decompositions.
|
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