Mathematics - 2020-07-17 (page 2 of 3)

chart is just a small patch on your manifold

read Thurston's little book on 3-manifolds

if you are serious about wanting to up your geometric intuition

not the big book, the little book

I honestly can't think without a picture in my head. Symbols are too hard for me. Maybe that's bad who knows, but my point is I don't understand the full calculations perspective

I still don't get the argument

this is kind of subtle, I think

there's an intermediate theorem that if a function is an injective local diffeo, then it's a diffeo onto its image

sure, that's clear

I don't see how we are guaranteed to find a neighborhood of $Z$ on which $f$ is injective and a local diffeo yet

you can get the injective nbhd by my argument

and the local diffeo nbhd by Balarka's

intersect, bada bing

5:08 PM

if it's not injective on any nbhd you'll find two sequences which converge to two distinct points on Z which map to the same thing

like Fargle said

shrink the nbhd

wiggles of linear isos are linear isos

why can you guarantee convergence? does the compact set necessarily possess a compact neighborhood in which the sequences eventually lie so you can pass to a convergent subsequence?

yeah local compactness

say these manifolds lie in some $\Bbb R^N$

then yeah, pass to convergent subsequences

or that

5:11 PM

local compactness, but $Z$ needn't lie in a single chart

but it lies in finitely many

thats what, so its not an issue

and finite union of compact sets is compact

this is a general topology argument

so ok

G&P's manifolds lie in R^n anyway

5:12 PM

yeah, you should know by now that I love getting hung up on general topology

so you can use completeness of R^n

passing to conv subsequences, we have $a_i \to a$, $b_i \to b$, and by construction I'm making $a$ and $b$ lie in $Z$ so $a = b = $ some $z$

but why can't it happen that your two sequences have the same limit?

because $f$ injective on $Z$

because take a small chart

around the same convergent point

and $df$ at that point cant be an iso

because you have folding around that chart at that same limit

5:14 PM

ofc, it's injective on a neighborhood of any point of Z by IFT, so the limits have to be distinct, but then f isn't injective on Z

ok, I get it

bing bang boom

wahoo

as Hikaru would say @LeakyNun

@BalarkaSen Anything interesting here

@MikeMiller Nope

just good old G&P exercises

5:19 PM

I'm a widdle baby, basically

and I can't topology

robjohn

Test

Interesting. I was on another chat and the back ticks did not work as they do here

do you guys know the Busemann boundary

wuzzat

robjohn

@BalarkaSen does it divide the front from the back of the Busemann?

5:22 PM

Say $(X, d)$ is a nice metric space, consider for any $x, y, z$ the quantity $b(x, y, z) = d(x, z) - d(x, y)$, the difference between the distance from $x$ to $z$ and $x$ to $y$

Let $C(X)$ be the space of continuous functions on $X$, you have a natural map $b_y : X \to C(X)$, $b_y(x) = b(x, y, -)$. $b_y(z)$ is indeed continuous by triangle inequality (in fact Lipschitz) and $b_y$ itself is a continuous function by triangle inequality again. You can also check it's an embedding

Konformist Liberal

Is there a CW-complex for every homotopy type?

this feels weirdly asymmetric

@KonformistLiberal Yes, take a product of $K(\pi_n, n)$'s

I'm confused by something which is likely extremely stupid. I have a function $f:\Bbb N\to\Bbb N$ (it is the growth function of a group but that is irrelevant). What's the difference between there exist constants $c,d$ such that $f(n)\leq cn^d$ for all $n>0$ vs for infinitely many $n>0$ and which one is polynomial growth

$x\mapsto d(x,-)$ also gives you an embedding $X\rightarrow C(X)$, but I assume you haven't gotten to the punchline yet

5:26 PM

Oh wait that's not your question

I don't understand the question, but I guess hawaiian earring is the answer

feynhat

$\pi_n$'s don't uniquely determine homotopy type?

@Alessandro I mean, one is clearly weaker than the other, no?

@KonformistLiberal For every space $X$ there's a CW complex which is weak homotopy equivalent to $X$, by CW-approximation. There's no CW complex homotopy equivalent to the Cantor set

Sure, but what's a function satisfying the second but not the first

5:28 PM

$f(n)=e^n$ for odd $n$ and $f(n)=n$ for even $n$?

@AlessandroCodenotti First is polynomial growth

feynhat

ah.

@Thorgott Uh ok weird

But actually the second can't happen without the first for growths can it

@KonformistLiberal The closed topologist's sine curve has no CW-complex with the same homotopy type

5:29 PM

@BalarkaSen Apparently Gromov's theorem for groups of polynomial growth actually holds for groups of growth satisfying the second too

But I don't see how a group can satisfy the second but not the first

wouldn't polynomial growth be this for all but finitely many $n$?

growth should only care about asymptotics

If you have a big enough $c$ you can get this for all $n$, no?

ah, you're right

@AlessandroCodenotti yeah I think it should be easy to see the second => first

I don't know, see the bottom of page 1 here, as it's phrased it doesn't seem that they are equivalent

Konformist Liberal

5:33 PM

Thank you!

Hm, maybe I'm wrong that it's easy to see then.

Seems to say they are equivalent because Gromov's theorem

virtually nilpotent are polynomially growing in the strong sense

it's certainly not equivalent for any old function $f$, but idk how restrictive being the growth rate of a group is

virtually nilpotent groups have scaling limits which converge to nilpotent Lie groups by Pansu's observation, and from that you can deduce it means they have polynomial growth

this was Gromov's original line of attack iirc

It's possible that it is a slight generalization in the sense that it has weaker hypothesis, but they just turn out to be equivalent to Gromov's

Yeah

5:36 PM

@BalarkaSen Yeah the paper I linked above follows Gromov's approach but uses some nonstandard machinery to build Gromov's limit of metric spaces

yeah Gromov said it obviously converges lmao

Mahan told me you need ultralimits or some shit

using filters

clearly obvious that its obviously clear

Yes, the proof in the book by Kapovich and Drutu used ultralimits of metric spaces

What a fucked up idea though

Mad genius

trivially clearly obviously transparent

5:38 PM

It would never ever ever have made sense to me if I hadn't encountered the Heisenberg group

I'm quite curious about the nonstandard approach, and the paper is only 26 pages, I'll see if it's readable though

I see Los' theorem

I bail out

this is clearly your cup of tea

Haha yes it's written by logicians

Los's theorem is really useful though

You've read the proof in Kapovich-Drutu?

I might try sometime

No I only skimmed it, I had to pick between going through that and doing asymptotic dimension stuff for my thesis

5:42 PM

Gotcha

I'm not sure if it's immediately useful for me but I will try sometime

There's also a blog post by Tao doing a simplified proof iirc

Ah nice

Let me see

terrytao.wordpress.com/2010/02/18/a-proof-of-gromovs-theorem

Thanks!

You probably won't like the flavour of this approach though, it seems more analytic than I remembered

5:44 PM

Yeah I really want to understand how the scaling limit is constructed

Thats probably more relevant to me

Fair

Although this seems like good analysis

Yeah I think theorem 2 and 3 in Tao's post are interesting on their own

Yeah I do need to understand this

L^2 functions on the Gromov boundary of a hyperbolic group gives rise to natural harmonic functions on the group I think

I don't understand why Tao splits the proof of Thm 2 into amenable and nonamenable cases, groups of polynomial growth are always amenable

5:49 PM

How do you see that without Gromov's theorem

Ah there's a remark at the end

Oh that remark also contains my comment

Poisson boundary is Gromov boundary by a famous theorem

$\liminf |B_{n+1}|/|B_n|=1$ for subexponential growth groups, so you can extract a Følner sequence from the $B_n$

ahh

thats real neat

Yeah I think I read this argument on MSE somewhere

5:54 PM

Of course, neato idea

$\{B_n\}$ is literally a Folner sequence

$gB_n \Delta B_n$ is like at most $\partial B_n$

and $\lim_n |\partial B_n|/|B_n| = 0$ for polynomially growing guys

Oh yeah, Ycor wrote it as an answer to an old question of mine, he's an expert on this kind of stuff

Very nice

Unfortunately the missing cases are the most interesting ones

I have to leave for a while, bye

Thanks, see ya!

I need to get dinner as well

Edward Evans

@Leaky "so all the ramification occurs in $E_\sigma/E_\sigma \cap K^{\text{ur}}$" is the point, but I'm failing to make it formal lol

Leaky Nun

6:05 PM

@EdwardEvans let $F$ be the maximally unramified subextension of $E_\sigma / E_\sigma \cap K^{ur}$

then $F/K$ is unramified

so $F \subseteq K^{ur}$

Edward Evans

lol

6:58 PM

@Balarka tell me how to integrate $\omega$

7:08 PM

@Thor: Who is $\omega$?

I was gonna ask

@Thorgott $\int \omega$

connection form

I'm still trying to figure out your comments from yday

Jacob Bumgarner

Can someone help me with a stats question?

Well, I'm totally out of context.

7:12 PM

I was telling Thorgott why $d\omega(X, Y) = X \omega(Y) - Y \omega(X) - \omega([X, Y])$ follows from Stokes' theorem on an infinitisimal flow pentagon given by $X, Y, X^{-1}, Y^{-1}, [X, Y]$

$\omega$ being the connection form is irrelevant here

This is true generally

Oh, presumably trying to motivate Ambrose-Singer for getting holonomy from curvature?

Yeah.

I would just stick to surfaces :P

I specialized to surfaces

Then it's just the structure equation $d\omega_{12} = \pm K\,dA$.

7:14 PM

Easier to say $\Omega^1_2(X, Y) = d\omega^1_2(X, Y)$.

Ambrose-Singer is great

Deane Yang has a good writeup

Deane Yang is a good geometer

I learnt the Weyl tensor shenanigans from his papers

A: Alternative (easier) Proof of Ambrose Singer Holonomy theorem

7

I always find it easier to work with the vector bundle induced by a linear representation of the structure group. I believe this theorem is a consequence of the following loop formula (a terse proof can be found in Holonomy Equals Curvature): If \begin{align*} M &= \text{smooth manifold}\\...

wtf is ambrose-singer

this was just supposed to be the geometric interpretation of curvature

It is

7:18 PM

That's what it is, I'm going backwards to explain curvature to you through holonomy

Deane was a Griffiths student. I've known him since his grad student days. He actually took my complex manifolds course at MIT.

I actually haven't seen his Ambrose-Singer write-up.

We communicate almost exclusively by Facebook posts :P

@Thorgott Call the pentagon given by $\varepsilon$-flows of $X$ and $Y$ to be $P_{\varepsilon}$. Then $d\omega(X, Y) = \lim_{\varepsilon \to 0} \frac{1}{\text{Area}(P_\varepsilon)} \int_{P_\varepsilon} d\omega$, agree?

notes deathly quiet

Lol

as in, we flow $X,Y-X,-Y$ for $\varepsilon$ time and then $[X,Y]$ for $\varepsilon^2$?

7:25 PM

Yes, that's the pentagon. Well, the missing edge isn't quite flow of $[X, Y]$ for time $\varepsilon^2$, it is so upto first order.

You close it anyway.

and probably the negative.

Yes, indeed, thanks.

And maybe a coefficient. But ... blah.

The $1/2$ appears only if you write it as 2nd derivative of $[\Phi_{\varepsilon^2}^X, \Phi_{\varepsilon^2}^Y]$, I believe.

sure, all up to first order

yeah, negative sign is missing, but coefficient is fine

if you flow $X,Y,-X,-Y$ for $\varepsilon$ time, there is no missing edge up to first order and it's $-1/2[X,Y]$ in second order, whereas if you flow $X,Y,-X,-Y$ for $\sqrt{\varepsilon}$ time, you get $-[X,Y]$ up to first order

this is the second picture after substituting $\varepsilon\mapsto\varepsilon^2$

7:29 PM

Right.

The missing edge doesn't matter upto first order in any case.

anyway, I don't know that limit

It is "intuitively clear" but why don't you give me an argument instead of me thinking about how to write one

The pentagon goes to the parallelogram spanned by $X$ and $Y$ in the tangent space, and $d\omega$ eats that in the limit.

It should be definition-chase

Just work in $\Bbb R^2$, yeah?

right, our picture pushes forward under a chart

Approximate $X$ and $Y$ by their linearizations.

this should come down to some sort of uniform continuity

it's essentially saying that $d\omega$ converges uniformly to $d\omega(X,Y)$ as $P_{\varepsilon}$ closes in

now how do I make this precise

7:49 PM

Howdy, nerds

There's nothing special about $d\omega$, just take any 2-form on $\Bbb R^2$, $f dx\wedge dy$. Write it out, and notice that only the linearized vector fields $\hat{X} = \sum X(p)^i \partial_i$, $\hat{Y} = \sum Y(p)^i \partial_i$ matters.

Maybe you do need 1st order terms, I haven't thought about it. Because you're dividing by area. Higher order terms in the vector field gives $O(\varepsilon^3)$ terms, which cancel with $O(\varepsilon^2)$ term in the denominator.

Hi @Fargle

How goes it @Ted?

Still alive :)

Good to hear

7:55 PM

It's lunchtime. I figure Balarka will get to his punchline around dinnertime :P

I'm waiting for Thorgott.

I think you should have just illustrated holonomy on a spherical cap.

I think Balarka prefers to go "no cap", as I'm told the kids say

puts Fargle in humorless mode

I thought about giving the example later, but I already started the elaborate explanation of why $\Omega^1_2$ obviously measures holonomy around the flow pentagon, so.

You can tell him for me

7:57 PM

Nah. I'm going to eat lunch. I always start with illustrative examples :P

I really like how you immediately get the spherical cap from the cone

Okay, fine, I'll keep doing G&P. grumble grumble

Oh, you're back to G&P?

It was only a matter of time until I got stir-crazy enough to do it, given all the, uh, everything.

But yes, I am. I even did like two exercises

Well, cool.

8:02 PM

I also found some really old Milnor lectures on YouTube that have been interesting to watch alongside, for my own historical edification

Oh yeah I have seen those flying around

They're quite good. Remarkably clear

Off-topic: What would be a random Riemannian metric on a manifold? The space of Riemannian metrics is the positive cone on $T^*M \otimes T^*M$; equip a fiberwise Gaussian measure?

Milnor is a great lecturer.

Cool beard, too

8:05 PM

Encyclopedia Britannica has this following picture:

Very religious interpretation

Only Gromov deserves such a background

@BalarkaSen Does this even make sense

LOL ...

Yes, for sure, I think.

I feel like if you can cook up a good measure on the space of sections of something you've solved the path integral.

I don't think the measure needs to be canonical. Most random models don't have a canonical measure

Gaussian is the best for most things

8:08 PM

The path integral doesn't need to be canonical either

I'm pretty sure you're saying you know how to invent the path integral

I don't know what a path integral is man

Nobody does that's why I'm calling you a fraud

2

Oh I see what you mean, the shit physicists do

They did that to cook up the Wiener measure as well

Ok yeah that's the Gaussian

So you're saying what you're doing does exist already, it's the weiner measure

Wiener :P

8:11 PM

Wiener measure is a measure on the space of paths in something

solves the problem of "how to pick a random path"

aka a Brownian motion

@Fargle: When you wanna discuss something interesting in G&P, let me know.

Everyone in my math program has been calling it the "Winer" measure out loud, presumably because they can't fathom saying "Wiener".

have you heard of Wiener sausages, Fargle

They don't like Vienna?

Wiener sausage

In the mathematical field of probability, the Wiener sausage is a neighborhood of the trace of a Brownian motion up to a time t, given by taking all points within a fixed distance of Brownian motion. It can be visualized as a sausage of fixed radius whose centerline is Brownian motion. The Wiener sausage was named after Norbert Wiener by M. D. Donsker and S. R. Srinivasa Varadhan (1975) because of its relation to the Wiener process; the name is also a pun on Vienna sausage, as "Wiener" is German for "Viennese". The Wiener sausage is one of the simplest non-Markovian functionals of Brownian motion...

8:12 PM

Fargle, define cobordism groups and prove that $\bigoplus_{k \geq 0} \Omega_k(Z)$ is a ring

I think those are weeners. :)

My grandparents are Kentuckian, so they're "Vye-innies" to them

LOL, oh @Balarka

OK, lunchtime. BBIAB.

@MikeMiller killing_in_the_name_outro.flac

(which is my way of saying "I dunno what any of that means")

Well that's why you have to define them

8:19 PM

:|

This is like one of the worst things Thom did

Which is not to say it's bad

Dude are you sure Thom didn't compute Omega^*(pt) by geometry

This is so unlike Thom

Maybe the Bourbaki took over him during that era

Unfortunately it was pure algebra

:(

8:22 PM

jsyk Mike, instead of doing what you asked, I'm just going to cheat by looking up what cobordism groups are and copy-pasting a proof of the result you said

I recently read an English translation of Thom's stratified space and stability of maps paper it was 100% impressionism

Your first approach was better

send fargle the exposition mike

eh, 5 seconds of my life vs 30

didn't say I'd try to understand what I read

taps temple

Yeah I just respect this approach less

8:24 PM

understandable, so do I

@MikeMiller But wait doesn't Thom-Pontryagin say $\Omega^{SO}_*(pt) \cong \pi_*(MSO)$, and rational homotopy groups of $MSO$ are rational homology groups of $MSO$ by Sullivan theory i.e., $H_*(BSO; \Bbb Q)$ which is generated by Pontryagin classes

Oh Sullivan theory wasn't available to him

Just give up man

Impossible man you're just trying to paint Thom black

I can't believe it

Wow actually it's all rational homotopy theory, $H_\ast(BSO; \Bbb Q) \cong U(\pi_*(BSO) \otimes \Bbb Q)$ and $\pi_\ast(BSO) \cong \pi_{\ast-1}(\Omega BSO) \cong \pi_{\ast-1}(SO)$ which has rational homotopy $\Bbb Q$ in $4k$ degrees by google and Bott periodicity

I don't even need to know what Pontryagin classes are

Sullivan, my savior

To Gromov, Thurston and Sullivan I pray, every day

8:45 PM

Well, two are still alive.