Mathematics - 2025-01-21

X4J

02:21

@Thorgott Ah and then it's cyclic right?

So there is only such group

X4J

03:00

nvm forget about it

Ben Steffan

09:49

@Thorgott it's so simple math.stackexchange.com/a/5025765/681666

hbghlyj

10:14

I have a question about differential forms: Why $\omega=\frac{1}{x_{n+1}} \mathrm{~d} x_1 \wedge \mathrm{~d} x_2 \wedge \ldots \wedge \mathrm{~d} x_n$ is a non-vanishing n-form on $S^n$?

hbghlyj

10:49

For $n=1$: $x_1=\cos(\theta),x_2=\sin(\theta)$, so $\omega=\frac1{\sin(\theta)}(-\sin(\theta))\mathrm d\theta=-\mathrm d\theta$ is non-vanishing 1-form on $S^1$.

Thorgott

@X4J any product of $p$-groups over a finite set of distinct primes $p$ will do

@BenSteffan wow!

I'm surprised I've never seen this result

even tho I've taken a lecture on $4$-manifolds lol

X4J

11:07

@Thorgott In particular it implies that any abelian group has a such decomposition but I guess it’s much weaker than the fundamental theorem for classification of abelian subgroups

Thorgott

yeah, but this is nonetheless insight you can use to prove a part of it

a finite abelian group is a direct product of abelian $p$-groups by this observation, so it remains to show an abelian $p$-group is a direct product of cyclic $p$-power groups

X4J

11:21

Yes that's interesting

Bml

Hello everyone. A few minutes ago I noticed that someone unupvoted two of my questions within a few seconds. Is this legal, according to the standard, or is it still an example of targeting?

X4J

11:59

@Thorgott Is it also true that if $G$ is a finite p-group with $|G| = p^k$ then there exists $g_1, \dots, g_k \in G$ s.t $G$ is decomposed to the direct sum of $<g_i>, i=1, \dots, k$?

direct prudct*

Oh I think I just rephrased the case for abelian groups

Thorgott

this is true, but perhaps misleadingly so

you have to allow for some $g_i$ to be $0$, for example

@Ben by the way, I just went through the arguments again and I don't think dimension $4$ is actually that relevant, instead the arguments exploit small codimension

the more or less same arguments show that an $n$-manifold cannot be a $K(G,m)$ for $m\ge n-2$ (and $m\ge2$) unless $G=0$

actually, I might be lying

need to look up the cohomology of K(Z/p,n) again

X4J

12:27

@Thorgott Yes the way I wrote it is ambigious, I meant $g_i \neq 1$

However it's just the fundamental theorem, since this is abstract I sometimes get confused

Thorgott

@X4J then it's not true

one potato two potato

13:04

yesterday, by one potato two potato

I don't know in detail (or even actually true in fact) but I was told that exp(renormalized volume functional + i CS invariant functional) becomes a holomorphic function.

I just heard that it is a published paper in math (but the authors are physicists) but many experts think the paper is incomplete.

user1294729

Can we write $Z_2$ in cycle notation, if yes how?

Thorgott

@Ben oh, I just realized there's another really stupid way of finishing the argument: we know the $K(G,2)$ is an $M(G,2)$, so $H^3(K(G,2))=\mathrm{Ext}(G,\mathbb{Z})$, but this vanishes since $G$ is divisible. since we are simply connected, this suffices to apply Wall's finiteness theorem and conclude $K(G,2)$ has the homotopy type of a $2$-dimensional CW-complex, which leads to a contradiction.

one potato two potato

13:22

It's disappointing that some journals allow some authors to publish their papers in those journals due to their closeness with editors.

Thorgott

@Thorgott this is obviously complete and utter nonsense, I'm burnt

user1294729

@Thorgott Can you maybe help me with my question? Because in class we have written that $Z_2=\{e,(12)(34)\}$ but I think this is wrong or not?

Thorgott

13:44

@hbghlyj this is not even well-defined on all of $S^n$

@user1294729 what is the product of (12)(34) with itself

user1294729

(12)(34)(12)(34)=e?

@Thorgott But I mean in general I know that $Z_2=\{e,g\}$ such that $g^2=e$. But I mean (12)(12)=e?

Ben Steffan

@Thorgott that one raises an eyebrow, yeah

:)

@Thorgott yes, this has occurred to me as well

@Thorgott I'm writing a question on MO about the general case

Thorgott

14:31

@user1294729 yes to all

user1294729

perfect thanks!

Thorgott

any product of disjoint transpositions in a symmetric group is an element of order $2$, so it generates a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$

user1294729

ah I see

Thorgott

@BenSteffan though it's still interesting to note that Wall's finiteness theorem does imply something like $K(\mathbb{Z}[p^{\infty}],2)$ has a $3$-dimensional model

@BenSteffan nice, I was gonna ask if you were

the only thing I'm unsure about regarding my first observation is whether $H^{n+2}(K(\mathbb{Z}/p\mathbb{Z},n))\neq0$

Steenrod operations are the bane of my existence

Ben Steffan

@Thorgott Sure: $H_{n + 1}(K(\mathbb{Z} / p, n)) = 0$ and $H^{n + 2}(K(\mathbb{Z} / p, n)) \cong H_{n + 1}(K(\mathbb{Z} / p, n))$

The first by Hurewicz and the second from UCT + the fact that the co/homology of $K(\mathbb{Z} / p, n)$ consists of f.g. $p$-power torsion groups entirely

Thorgott

14:37

so it would vanish

Ben Steffan

yes

Thorgott

which is also what I thought at some point, but doesn't that contradict the claim in your divisibility argument?

Ben Steffan

@Thorgott I'm being a little slow on the uptake: how does this follow from Wall's finiteness theorem?

@Thorgott which claim?

the one with the cup square is with $\mathbb{Z} / p$-coefficients

Thorgott

in the simply connected case, it reduces to: given $X$ has the homotopy type of a simply connected CW-complex, it has the homotopy type of a simply connected CW-complex of dimension $\le n$ if and only if the homology in degrees above $n$ and the cohomology in degree $n+1$ vanish

which follows in particular if all homology in degrees above and including $n$ vanish

Ben Steffan

huh, do you have a reference for this? I'm not very familiar with Wall's theorem

Thorgott

14:42

it's in Wall's original paper, Finiteness Conditions for CW-complexes

@BenSteffan ah, your point is that H^4 is zero, but H_4 isn't, right?

with Z-coefficients

Ben Steffan

yes

if $p = 2$ it is $\mathbb{Z} / 4$, for instance

Thorgott

ok, so the generalization we would need is $H_{n+2}(K(\mathbb{Z}/p\mathbb{Z},n))\neq0$

Ben Steffan

@Thorgott thanks

@Thorgott I'd suspect this is false for large $n$

because for $n \gg 0$ that group will be in the "stable range", and for large $p$ there is no stable operation that only raises degree by 2

but I'm generalizing arguments I'm only 100% sure are valid for $p = 2$

Thorgott

this does sadly sound convincing

so your divisibility argument might break down to rule out an $n$-manifold cannot be a $K(G,n-2)$

Ben Steffan

yeah

Thorgott

14:56

but Michael's argument that $G$ is torsion-free does generalize

Ben Steffan

does it?

oh, for $n - 2$, yeah

Thorgott

actually, perhaps torsion-free is enough half the time? $K(G,n-2)$ has the same rational (co)homology as its rationalization $K(G_{\mathbb{Q}},n-2)$ and if $G$ is torsion-free non-trivial, then $G_{\mathbb{Q}}$ is a non-trivial $\mathbb{Q}$-vector space and we have a contradiction for even $n$, right?

actually, the argument was only ever gonna work for even $n$ anyway

Ben Steffan

Yes, that would work

Thorgott

but I believe this skips the divisibility step?

though I suppose odd $n$ are wide open

like, can a $5$-manifold model $K(\mathbb{Q},3)$

Ben Steffan

also what about $2 \leq k < n - 2$

hbghlyj

15:06

@Thorgott I think it can be extended to all of $S^n$. By changing coordinates.

On the chart $\{x_{n+1}>0\}$
with local coordinates $(x_1,\dots,x_n)$ define $\omega=\frac{1}{x_{n+1}} \mathrm{~d} x_1 \wedge \mathrm{~d} x_2 \wedge \ldots \wedge \mathrm{~d} x_n$.

On the chart $\{x_n>0\}$
with local coordinates $(x_{n+1},x_1,\dots,x_{n-1})$ we have $\frac{\partial x_n}{\partial x_{n+1}}=-\frac{x_{n+1}}{x_n}$, so $\omega=\frac{1}{x_{n+1}} \mathrm{~d} x_1 \wedge \ldots \wedge \mathrm{~d} x_{n-1}\wedge(-\frac{x_{n+1}}{x_n}\mathrm{~d} x_{n+1})=-\frac1{x_n}\mathrm{~d} x_1 \wedge \ldots \wedge \mathrm{~d} x_{n-1}\wedge\mathrm{~d} x_{n+1}$.

Ben Steffan

Your charts don't cover $S^n$.

hbghlyj

I think we can do the same for $\{x_i>0\}$

and for $\{x_i<0\}$

for each $i\in\{1,\dots,n+1\}$

But $\omega$ seems to be different from the volume form on $S^n$

I guess they should differ by factor of a non-vanishing smooth function

Thorgott

how are you getting $\frac{\partial x_n}{\partial x_{n+1}}=-\frac{x_{n+1}}{x_n}$?

hbghlyj

$\frac\partial{\partial x_{n+1}}(x_n^2)=\frac\partial{\partial x_{n+1}}(1-x_1^2-\dots-x_{n-1}^2-x_{n+1}^2)=\frac\partial{\partial x_{n+1}}(-x_{n+1}^2)$

so $2x_n\frac{\partial x_n}{\partial x_{n+1}}=-2x_{n+1}$

on the chart $\{x_n>0\}$.

Thorgott

15:36

oh yeah, sorry, I was a bit distracted

you can do this on all pairs of charts to get that $\omega$ is actually well-defined

but that it's nowhere vanishing is obvious once you have that it's well-defined

hbghlyj

16:00

Yes. Thanks for confirmation.

user1294729

16:18

@Thorgott If I have two vector spaces $V,W$ and denote by $V^\star$ their dual space, is then $(V\oplus W)^*=V^*\oplus W^*$ and what about the tensor product?

I think yes if $V,W$ are finite dimensional because one could take a basis ${v_i}$ of $V$ and ${w_j}$ of $W$. Then ${v^i}$ and ${w^j}$ is a basis for $V^*$ respectivey $W^*$ and then define the map $\phi:(V\oplus W)^*\rightarrow V^*\oplus W^*$ by $\phi((v\oplus w)^{ij})=v^i\oplus w^j$. For the tensor product, one can exchange $\oplus$ by $\otimes$.No?

Thorgott

16:34

you can always define a canonical (i.e. without any choice of basis) homomorphisms $(V\oplus W)^{\ast}\rightarrow V^{\ast}\oplus W^{\ast}$ and $V^{\ast}\otimes W^{\ast}\rightarrow(V\otimes W)^{\ast}$ (note the directions), try this

once you've written down the correct maps, you can show (without choosing bases still) that the first map is always an isomorphism

the second map will not be an isomorphism in general (though always injective), so at this point you can choose bases and check that it's an isomorphism at least in the case that both are finite-dimensional

if my cardinal arithmetic isn't failing me right now, $V^{\ast}\otimes W^{\ast}$ and $(V\otimes W)^{\ast}$ are always non-canonically isomorphic, but that's not a particularly helpful statement

user1294729

ah okey so my statement is true but the argument not right?

Thorgott

16:53

your argument does not make it clear that the map is canonical

and one should never write "=" for a non-canonical isomorphism

psie

17:10

Let $f:\mathbb R^d\to\mathbb R^d$ and let $f$ be differentiable on an open subset $U\subset\mathbb R^d$. We then have $$\lim_{h\to0}\frac{|f(h+a)-f(a)-Df(a)(h)|}{|h|}=0,\quad \forall a\in U.$$ Let now $K\subset U$ be compact. Under what conditions on $f$ does the fraction above converge uniformly to $0$ on $K$? In other words, if we express the limit definition as an $\epsilon$-$\delta$ definition, under what conditions can we choose $\delta>0$ independent of $a$?

I'm partly reading this answer (and also my own textbook, where this claim seems to be made) and it seems to hold for $d=1$ with the help of the mean value theorem, but I have no clue how one would go about it for $d>1$.

In that answer I linked, the assumption is that $f$ is continuously differentiable, so my guess is this is something one would require for $d>2$ too.

leslie townes

17:31

smoothness of f would probably be enough. a closer-to-minimal hypothesis would probably require something like existence/boundedness/continuity of something like second partial derivatives on K.

if a textbook is raising this question i would see how they phrase "taylor's theorem" (which again is an enormous family of similarly spirited results and not one thing) and see if their version of that suggests an argument.

psie

17:48

hmm, is there a Taylor's theorem for vector-valued functions?

leslie townes

yes there are many such things, depending on the hypotheses made by the authors and the effort they want to spend proving them.

there is a pretty obvious way of thinking of a function taking values in R^n as a tuple of functions taking values in R. so depending on the textbook and treatment they may lean on the scalar valued case pretty heavily.

there are like thirty million almost arbitrary choices that go into formulating a "taylor's theorem" even from R^1 to R^1. i think rudin PMA gives at least two forms, for example (maybe one is in the exercises).

psie

@leslietownes so you're saying the uniform convergence of the difference quotient above might just be an application of a multivariable generalization of Taylor's theorem in R?

leslie townes

18:09

if you have an appropriately formulated theorem of that type, you might be able to prove something like that. yes. in saying this i am not excluding other approaches. just saying the first thing it made me think of.

i've said before that at least in my view, analysis is full of sufficient conditions that are usually not strictly necessary. i would basically never suggest that there is only one route to a result like that.

in the same spirit, i can also imagine that for a lot of applications, whatever one was hoping to deduce from the uniform convergence of the difference quotients might be deducible from some weaker hypothesis.

if you look at the arxiv, particularly the PDE section, you will see a lot of papers where people invent conditions that are weaker than textbook conditions, but also more specific than necessary, in that they are 'tuned' to get some result in some range of circumstances of interest to the author.

there are times where even basic things like continuity and compactness are "only" used to ensure that some quantity of interest is bounded, and you can just say, well, under any circumstances where this thing is bounded, i'd still get X Y and Z.

psie

leslie townes

well sure, but what would you be using that result for? deducing some regularity property of Df? or what? that's what i meant by "whatever one was hoping to deduce from the uniform convergence of the difference quotients might be deducible from some weaker hypothesis."

it is rare to care too much about something converging uniformly to something else just in the abstract. you have some reason for wanting to deduce [who knows what] and things like uniform convergence are tools that get you there.

e.g. if i have a series of functions and i want to be sure that its sum is continuous, one way i can arrange that is to assume that the summands are continuous and that the convergence is uniform. but that is far from the only way that such a series can sum to a continuous function.

Ben Steffan

18:41

@Thorgott mathoverflow.net/questions/486388/…

Thorgott

18:55

nice

should I leave a comment adding that $m\ge n-1$ can be excluded and $m=n-2$ can be excluded for even $n$?

Ben Steffan

sure :)

Thorgott

19:22

@Ben just saw that the question has also been asked before (you may have found this already): mathoverflow.net/questions/233379/…

but the answer only addresses what we already knew, of course

Ben Steffan

I've in fact not seen this

yeah, that answer is not very useful

It's a little funny in the context of this recent question by Qiaochu math.stackexchange.com/questions/5025181/…

Ben Steffan

19:58

once again it turns out the MO people are not stingy with their upvotes :)

Thorgott

20:26

yeah, I should also migrate my latest MSE question to MO tbh

@Ben I just realized another thing: if $V$ is a $\mathbb{Q}$-vector space of dimension $>1$, then $K(V,n)$, $n$ odd, has rational cohomology in degree $n^2$ by Künneth, so we can actually exclude that too in the $m=n-2$ case for $n$ odd, so the only case left for $m=n-2$, $n$ odd is $K(G,n-2)$, where $G$ is a subgroup of $\mathbb{Q}$ (and these are probably no more or less difficult than the case of $\mathbb{Q}$ itself, which is still completely unclear to me)

Ben Steffan

Yes

I think I was aware of this in the back of my head but didn't bother to differentiate

for the $n = 4$ case rational stuff was out anyhow

Thorgott

yeah, it's kind of obvious, but I hadn't explicitly considered it before

it's just funny that it really comes down to this one annoying case

handan_toddler

20:47

\o @AlexanderGruber

Ben Steffan

21:18

mathoverflow.net/a/486394/142596 ?!?!

psie

I'm considering asking on main about the uniform convergence of the difference quotient to the derivative, when we are in $\mathbb R^n$ for $n>1$. I've spent a couple of hours pondering on this.

I just don't know which "tool" to use to prove this. As far as I can see, Wikipedia doesn't list any mean value theorem or Taylor's theorem for functions $\mathbb R^n\to\mathbb R^n$.

handan_toddler

Wikipedia is not really meant to ponder upon.

Thorgott

21:40

@BenSteffan very interesting, though I don't quite grasp the last sentence

Ben Steffan

yeah, I feel unsure about a number of things there

it's manifold talk

but the author is Greg Arone, so

Thorgott

well, he does wonder if it works, but it's very much worth thinking about in detail

Ben Steffan

anyways, if it is true it is horrible

my life is in shambles, etc.

Thorgott

21:56

to be honest, I don't think it's that bad after seeing the construction

Ben Steffan

I'm exaggerating of course, but it does feel very strange that the answer to that question should be "yes"

I guess there's no good description of that 5-fold you could give that qualifies it as an "explicitly known model for a $K(\mathbb{Z}_{p^\infty}, 2)$

Ben Steffan

23:48

well so much about $K(\mathbb{Z}_{p^\infty}, n)$

"it's $p$-completely equivalent to $K(\mathbb{Z}, n)$ so has $p$-torsion in infinitely many degrees" of course

can you remind me of why e.g. $K(\mathbb{Q}, 3)$ should admit a finite-dimensional CW-structure on countably many cells before I go ask that question to Achim Krause and embarass myself? @Thorgott

psie

Let $D\subset\mathbb R^d$ be open and suppose $K \subset D$ is compact. Now, define for $\eta > 0$ small enough the following set: $K_{\eta}=\{z \in D: \mathrm{dist}(z,K) \leq \eta \}$, where it is understood that $\mathrm{dist}(z,K)=\inf\{|z-k|:k \in K\}$. Show that the set $K_{\eta}$ is compact.

I know that $z\mapsto \mathrm{dist}(z,K)$ is continuous, so the set is closed. However, how can I argue that it is bounded?

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