Mathematics - 2025-01-22

Thorgott

00:02

@BenSteffan what does $p$-completely equivalent mean?

Ben Steffan

equivalent after $p$-completion

Thorgott

there is a dimension shift, probably something about the fibration induced by $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}[1/p]\rightarrow\mathbb{Z}_{p^{\infty}}\rightarrow0$?

@BenSteffan is that something I don't know or synonymous with localization at $p$?

leslie townes

psie: if K is compact it is bounded and everything in K_eta will be within eta or so of something in K. this is a triangle inequality thing.

Ben Steffan

It is not synonymous with localization at $p$

I'm not too familiar with how this works for spaces, but for spectra it is Bousfield localization at the Moore spectrum $M(\mathbb{Z} / p)$

psie

@leslietownes ok, I will try triangle inequality. The K_eta seem to be extensions of K, so it wasn't immediately obvious to me.

Ben Steffan

00:05

In spectra it can also be described by $X^\wedge_p := \lim X / p^n$, but for spaces you'd have to do something else I believe

Sine of the Time

draw a picture in $\Bbb R^2$ and you'll understand the geometric interpretation of $K_{\eta}$

Ben Steffan

@Thorgott what is this supposed to refer to?

Thorgott

I only know Bousfield localization for (model) categories

@BenSteffan he says $K(\mathbb{Z}_{p_{\infty}},2)$ is $p$-completely equivalent to $K(\mathbb{Z},3)$, not $K(\mathbb{Z},2)$

Ben Steffan

ah, hmm

Thorgott

sounds like that would come from a connecting map in this fiber sequence after $p$-completion

Ben Steffan

00:07

yes, then maybe more is going on

Thorgott

but since I don't really know what $p$-completion is, that's just a guess

leslie townes

psie: if K is contained in a ball with radius R at 0, try to convince yourself that K_etas is contained in a ball with radius R + eta at 0. hint |z| <= |z - k| + |k| for any z and k

psie

ok 👍

Thorgott

his answer is interesting, I was thinking earlier about making a CW-complex into a manifold by taking a nice nbhd in Euclidean space, but for some reason I tossed the idea again...

Ben Steffan

it's in Bousfield's original paper sciencedirect.com/science/article/pii/0040938375900233

Thorgott

00:09

@BenSteffan it's a bit subtler cause it also asks for countability, but this is still in Wall's paper

Ben Steffan

so I have to go read that then, huh :')

alright, alright

thanks again :)

Thorgott

yeah, it's Theorem E that summarizes anything

in the simply connected case, it comes down to asking that homology is countable and bounded

I think rational homotopy theory also gives you a very concrete model for $K(\mathbb{Q},m)$, $m$ odd that does the job

Ben Steffan

I just found out that completion of spaces is the topic of part 3 of more concise

that's probably a better exposition than Bousfield's paper

we should all read more of more concise

Thorgott

part 2 already kills me

Ben Steffan

00:26

study some stable theory and then come back to it, at least you'll have a positive outlook :^)

Thorgott

by the way, the above is clearly too complicated, we know $K(\mathbb{Q},m)$ is a Moore space, so we can obviously construct it as a countable finite-dimensional complex

Ben Steffan

oh, right

might not be true for other $K(G, m)$ to which that answer applies, however, so that's still good to know

Thorgott

yeah, in the general case, I don't see how CW-approximation alone would imply it

handan_toddler

07:47

±7 hours later...

Soumik Mukherjee

08:44

What do those symbols mean?

psie

09:41

Suppose $f:U\to\mathbb R^n$, where $U\subset\mathbb R^n$. If $f$ is $C^1$ (meaning continuous partials and Jacobian matrix existing), does then say, the fundamental theorem of calculus apply and $$f(x)-f(y)=\int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}f(y+t(x-y))\,\mathrm{d}t=\int_0^1 f'(y+t(x-y))(x-y)\,\mathrm{d}t?$$Do these equalities make sense (for say, the line segment between $x,y\in U$ entirely contained in $U$)? It doesn't look like the fundamental theorem of calculus to me anymore when $n>1$.

So I'm not sure if these equalities make sense.

Thorgott

10:17

@Ben what I suggested yesterday cannot work, the dimensions in the LES go up, so $K(\mathbb{Z}(p^{\infty}),2)$ and $K(\mathbb{Z},3)$ are actually $3$ steps apart

not sure how to compare their (co)homology

Ben Steffan

I think the 3 might have been a typo

but I'm reading about $p$-completion now

not a typo. this is essentially Example 10.3.3. in more concise

Thorgott

10:46

well, I think p-completion is only relevant here insifar as to say they have the same mod p (co?)homology

Ill check the example

ok this seems beyond me

perhaps there isnt a sttraightforward computation that demonstrates this

Ben Steffan

@Thorgott but this should be enough

since $K(\mathbb{Z}, 3)$ is known to have $p$-torsion in homology in infinitely many degrees

Thorgott

11:14

no it is obvious, look at K(Z[1/p],2)->K(Z[1/p]/Z,2)->K(Z,3)

fiber has vanishing mod p cohomology, so run the Serre SS

Xander Henderson

13:01

"Obvious".

Sine of the Time

13:39

"Trivial".

Ben Steffan

15:13

the holy trinity: Obvious, Trivial, (removed)

2

Thorgott

but do you agree it's obvious now? :)

Ben Steffan

yes I do

Thorgott

nice, I've also made some more observations whilst not paying attention to the non-archimedean geometry lecture I was sitting in earlier

first of all, I think the relevant version of Wall's finiteness theorem is actually not much harder than CW-approximation (as claimed in Achim's answer)

That is, suppose $X$ has the homotopy type of a CW-complex, is simply connected and has vanishing homology in dimension $>n$. I claim we can construct an $(n+2)$-dim. CW-complex $Y$ with a homotopy equivalence $Y\rightarrow X$. To do so, apply standard CW-approximation techniques to construct an $(n+1)$-dim. CW-complex $Y$ with a map $f\colon Y\rightarrow X$ that is $(n+1)$-connected.
then Hurewicz yields $\pi_{n+2}(f)=H_{n+2}(f)$ and the pair sequence yields $H_{n+2}(f)=H_{n+1}(X)$. the latter group is free abelian by cellular homology. choosing a basis, this means we can attach $(n+2)$-ce…

If your $X$ is a $K(G,n)$ with $G$ countable, Serre mod C theory implies $K(G,n)$ has countable homology groups, in which case CW-approximation only attaches countably many cells at every step and the resulting complex is countable.

second point is that I was looking at the answer about EML spaces that are Moore spaces again and it turns out their argument that $G$ is necessarily divisible is also flawed. this led me to look at the divisibility argument in your question again and spot an error.
you argue using the SES $0\rightarrow G\rightarrow G\rightarrow G/pG\rightarrow 0$, but that's only a SES if you know $G$ has no $p$-torsion. to get a correct argument, we have to *first* use the answer to get $G$ torsion-free and then make either your divisibility argument or my rationalization argument.

psie

15:39

Let $\det(\varphi'(x))=J_\varphi(x)$ be the Jacobian determinant of a $C^1$-diffeomorphism $\varphi:U\subset\mathbb R^n\to\mathbb R^n$ at $x$, where $U$ is open. I'm paraphrasing from my textbook. If $K\subset U$, where $K$ is compact, then there exists a $\delta>0$ for all points $u,v\in K$ such that $|u-v|<\delta$ and $$\left(1-\epsilon \right)\left|J_{\varphi }\left(u\right)\right|\le \left|J_{\varphi }\left(v\right)\right|\le \left(1+\epsilon \right)\left|J_{\varphi }\left(u\right)\right|.$$

Why is this inequality true? It looks like some kind of (uniform?) continuity statement with epsilon in the definition of uniform continuity equal to $\epsilon |J_\varphi(u)|$. Is that allowed? Does it make sense?

Ben Steffan

@Thorgott that's neat

@Thorgott very surprising that nobody caught that!

Thorgott

yeah lol

Ben Steffan

Do you maybe want to write a second answer under my 4-fold question to give the correct argument?

And then what do we do about the other question

Thorgott

sure, I'll write an "addendum" answer in a bit

@BenSteffan I'm trying to fix the divisibility argument right now

I should be writing an expository section for my thesis right now, but this is more fun lol

Ben Steffan

15:54

lol

psie

16:23

@psie There's this lemma that precedes this claim. Let $\varphi, U$ be as above and $\lambda_d$ Lebesgue measure on $\mathbb R^n$.

> Lemma 7.3 Let $K$ be a compact subset of $U\subset\mathbb R^n$ open and let $\epsilon>0$. Then we can choose $\delta>0$ sufficiently small so that $n\delta<\mathrm{dist}(K,U^c)$ and, for every cube $C$ with faces parallel to the coordinate axes, with center $u_0\in K$ and sidelength smaller than $\delta$, we have $$(1-\epsilon)|J_{\varphi}(u_0)|\lambda_d(C)\leq \lambda_d(\varphi(C))\leq (1+\epsilon)|J_{\varphi}(u_0)|\lambda_d(C).$$

But I feel like the claim above is unrelated to this lemma.

Thorgott

@Ben it would really helpful to know in which degree above $n$ the homology of $K(\mathbb{Z}/p,n)$ is first non-vanishing again

or even the cohomology

this value is known, but I'm too stupid to parse the descriptions of stuff in terms of cohomology operations

leslie townes

perhaps you can get it out of the [uniform] continuity of (u,v) -> |J(u)/J(v)| on KxK

Ben Steffan

you should be able to work this out using the Serre spectral sequence

let me try

Thorgott

the cohomology algebra is freely generated by the Bockstein and Steenrod operations (I think), so the cohomology is non-zero in degree n, n+1 (Bockstein of the universal element) and then 0 until n+2(p-1) (Steenrod power of universal element), right?

cause Bockstein twice is 0

(this is odd p case)

Ben Steffan

16:39

@Thorgott Is it? Isn't this the case only stably?

It will definitely be non-zero in degree $2n$ when $n$ is even, independent of $p$

Thorgott

sciencedirect.com/science/article/pii/…

Theorem 3.4

I might be missing something stupid

Ben Steffan

16:53

that sparks joy but also not

I don't immediately see how to get the square, but it has to be there

At $p = 2$ that's obviously no problem

psie

@leslietownes yes! I think you are right :) Thanks a lot. If we consider (u,v) -> |J(v)/J(u)| as a map on KxK, which is compact, then |(u,v)-(0,0)|<d should imply |J(v)/J(u)-1|<e, which is the inequality I'm looking for. Note, since varphi is a C1-diffeomorphism, the Jacobian determinant is never 0. Thanks again leslie.

leslie townes

psie well, maybe take some amount of care to specify the metric you are using on KxK. :) if you use the sum of the usual metrics on R^n you might want to compare (u,v) to (u,u) or (v,v) which are both in KxK and have distance |u-v| from it in that sum metric. i do not see any guarantee that 0 is in K but maybe that is part of the setup that you did not include in the summary above.

Thorgott

yeah, I'm always confused by cohomology ring versus Steenrod algebra versus dual Steenrod algebra

the other bad news is that I'm not managing to get a spectral sequence argument working that proves G divisible

psie

yeah good points. Yes, my book constantly uses just simple absolute value for any norm. Sometimes I don't know either which norm they are currently using.

but yes, 0 might not be in K 👍

Thorgott

17:21

another approach might be to show that G is torsion-free and then do a rationalization argument, but a) I can't show this either, b) for n odd, it still remains to exclude proper subgroups of Q

perhaps we should re-ask the EML = Moore space question on MO

Ben Steffan

that might be a good idea

this is turning into a saga

Thorgott

I also considered asking a question about whether K(Q,3) can be an open 5-manifold lol

or more generally what the minimal dimension it can be is, the argument we have gives a 9-manifold

Ben Steffan

yeah, I've wondered as well

one question begets the next

Thorgott

18:51

@Ben mathoverflow.net/questions/486439/…

Ben Steffan

nice

Thorgott

20:06

@Ben Will Sawin made a good point in the comments

Ben Steffan

yeah, I saw

Thorgott

20:23

its sounds reasonable, but may be difficult to work out (I don't wanna do the entire Serre calculations)

Ben Steffan

I'll consider trying it, but perhaps not tonight

I do at some point need to study :)

oh, classification of lens spaces, sweet

that's a nice topic

the lecture notes claim that the map $\{(z, w) \in \mathbb{C}^2 \mid |z|^2 + |w|^2 = 1, |z| \geq 1 / 2\} \to \{(z, w) \in \mathbb{C}^2 \mid |z| = 1, |w| \leq 1\}$ given by $(z, w) \mapsto (z / |z|, w / |w|)$ is a diffeomorphism

really makes you think

Thorgott

20:48

@BenSteffan nice

I never did the non-trivial parts of that

Ben Steffan

actually the proof might not have been part of the course after all :(

but it proves that the 3-manifolds of Heegaard genus 1 are exactly the lens spaces

Thorgott

yeah, I think the non-trivial part needs torsion

Xander Henderson

21:46

Ugh... what should I make for dinner tonight?

Ben Steffan

22:05

@XanderHenderson That question is trivial. Or is it obvious?

Xander Henderson

@BenSteffan Neither, unfortunately.

Ben Steffan

hmm

well, what do you have, in terms of ingredients?

Xander Henderson

That's part of the problem. I have no idea, but I do need to stop by the store on the way home, so anything available at the store is fair game.

I do know that I need more rice. I can't believe that I've run out.

Ben Steffan

rice is good. how about risotto?

if you can get risotto rice

Xander Henderson

@BenSteffan Unlikely. I can almost always get basmati, and I can sometimes get the sushi rice in the bag with a flower on it. I'm maybe thinking of throwing some kind of soup together.

I have duck stock at home, and could throw in a bunch of veggies, and maybe some chicken...

Ben Steffan

22:12

duck stock? oh, how I envy you...

Xander Henderson

@BenSteffan I made it from a duck.

Ben Steffan

that's generally how it's done I suppose :)

Xander Henderson

Or maybe I'll get leeks and taters, and make a blendy soup.

Ben Steffan

I don't have duck stock. I could make some, if I felt like splurging on a duck, and spending the time it takes to make some, and was happy making an unroasted stock since I don't have an oven.

Xander Henderson

I can usually get leeks...

But I think that I am committed to soup...

copper.hat

22:26

@XanderHenderson regarding the closed question, the user has been around for 8 years, perhaps they are gaming, but seems unlikely?

Xander Henderson

@copper.hat It is not so much that the user is gaming the system. It is that their actions give the impression that they are trying to do something that they know they shouldn't be doing. It is the appearance of impropriety that bothers me.

copper.hat

hmmm...

i suspect that their original question would have been closed as well, so from their perspective they might as well try again?

my memories of duck are biting on something hard and discovering that it was not a bone.

Xander Henderson

@copper.hat I have no idea. But my vote is based on the reposting, which looks like shenanigans to me.

copper.hat

some folks think the word shenanigans originated from an Irish saying.

Xander Henderson

@copper.hat Oh? I have no idea of the etymology.

Ben Steffan

22:39

Well, there once was a guy called Shen Anigans...

copper.hat

:-)

Xander Henderson

@BenSteffan Was he from Nantucket?

copper.hat

'twas on the good ship Venus...

Xander Henderson

Ah!

Got it.

'Cause women are from Venus, right?

copper.hat

:-)

Ben Steffan

22:40

@XanderHenderson yes, and extremely lewd as a corollary

Xander Henderson

Nice.

copper.hat

the figurehead was a nude in a bed

i'll skip the rest for now...

psie

Consider in $\mathbb R^d$ so-called elementary cubes: $$C=\prod_{j=1}^d (k_j2^{-n},(k_j+1)2^{-n}],\quad k_j\in\mathbb Z$$ and $n\in\mathbb N$. Let $U\subset \mathbb R^d$ be open. It is claimed that the collection of all elementary cubes whose closure is contained in $U$ generates the Borel $\sigma$-field on $U$. What are some ways I can show this?

I know that the closed or open rectangles in $\mathbb R^d$ generate the Borel $\sigma$-field on $\mathbb R^d$, and the elementary cubes are definitely a subset of either one of these collections. But we are considering a subset $U$ here, so I probably need to consider the appropriate collections intersected with $U$?

psie

23:01

What I'm doubting is if $U$ intersected with say all closed rectangles on $\mathbb R^d$ would give me a collection that contains all elementary cubes whose closure is contained in $U$.

leslie townes

i suggest trying d = 1 and U = an interval before the more general case

the general way of showing two sigma algebras specified by generating sets are equal to one another is to show that each member of a generating set for one is also in the other

psie

indeed, ok, I'll try d=1 and U = an interval

leslie townes

as you note, one of the inclusions is maybe already known or at least easier here, the other one might be deduced from an underlying fact that any open ball is a countable union of dyadic rectangles

i guess maybe you're also struggling with how 'open in U' relates to 'open in R^n' which a review of the definition of the subspace topology might help with [? it might come up if you are distinguishing U as a topological space from R^d as one, depending on your definition of borel sigma algebra]

but a lot of the features of the general problem are present in the d = 1 case

psie

@leslietownes yes, in particular, I'm staring at this, i.e. how one generating set of a sigma algebra on a starting set can induce another generating set for a subset of the starting set.

leslie townes

this feature of the general problem is present in the d = 1 case

it is nice/significant that U is an open set here, and not some random subset of R^n. "subset of U that is open in U with the subspace topology" and "subset of U that is open in R^n" are the same thing

psie

23:15

yes

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