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

feynhat

17:01

man I am allergic to patches. I find it hard to read anything that doesn't uses charts.

how do I get over my patch-phobia.

Balarka Sen

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

Thorgott

I still don't get the argument

this is kind of subtle, I think

Fargle

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

Thorgott

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

Fargle

you can get the injective nbhd by my argument

and the local diffeo nbhd by Balarka's

intersect, bada bing

Balarka Sen

17:08

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

Fargle

wiggles of linear isos are linear isos

Thorgott

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?

Balarka Sen

yeah local compactness

Fargle

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

then yeah, pass to convergent subsequences

Balarka Sen

or that

Thorgott

17:11

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

but it lies in finitely many

Balarka Sen

thats what, so its not an issue

Thorgott

and finite union of compact sets is compact

Balarka Sen

this is a general topology argument

Thorgott

so ok

Balarka Sen

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

Thorgott

17:12

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

Balarka Sen

so you can use completeness of R^n

Fargle

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 $z$

Thorgott

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

Fargle

because $f$ injective on $Z$

Balarka Sen

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

Thorgott

17:14

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

Balarka Sen

bing bang boom

Thorgott

wahoo

Balarka Sen

as Hikaru would say @LeakyNun

Mike Miller

@BalarkaSen Anything interesting here

Balarka Sen

@MikeMiller Nope

just good old G&P exercises

Fargle

17:19

I'm a widdle baby, basically

Thorgott

and I can't topology

robjohn

Test

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

Balarka Sen

do you guys know the Busemann boundary

Fargle

wuzzat

robjohn

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

Balarka Sen

17:22

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?

Thorgott

this feels weirdly asymmetric

Balarka Sen

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

Alessandro Codenotti

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

Thorgott

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

Balarka Sen

17:26

Oh wait that's not your question

Alessandro Codenotti

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

feynhat

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

Thorgott

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

Balarka Sen

@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

Alessandro Codenotti

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

Thorgott

17:28

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

Balarka Sen

@AlessandroCodenotti First is polynomial growth

feynhat

ah.

Alessandro Codenotti

@Thorgott Uh ok weird

Balarka Sen

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

Fargle

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

Alessandro Codenotti

17:29

@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

Thorgott

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

growth should only care about asymptotics

Alessandro Codenotti

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

Thorgott

ah, you're right

Balarka Sen

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

Alessandro Codenotti

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

17:33

Thank you!

Balarka Sen

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

Thorgott

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

Balarka Sen

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

Alessandro Codenotti

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

Balarka Sen

Yeah

Alessandro Codenotti

17:36

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

Balarka Sen

yeah Gromov said it obviously converges lmao

Mahan told me you need ultralimits or some shit

using filters

Thorgott

clearly obvious that its obviously clear

Alessandro Codenotti

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

Balarka Sen

What a fucked up idea though

Mad genius

Fargle

trivially clearly obviously transparent

Balarka Sen

17:38

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

Alessandro Codenotti

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

Balarka Sen

I see Los' theorem

I bail out

this is clearly your cup of tea

Alessandro Codenotti

Haha yes it's written by logicians

Los's theorem is really useful though

Balarka Sen

You've read the proof in Kapovich-Drutu?

I might try sometime

Alessandro Codenotti

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

Balarka Sen

17:42

Gotcha

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

Alessandro Codenotti

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

Balarka Sen

Ah nice

Let me see

Alessandro Codenotti

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

Balarka Sen

Thanks!

Alessandro Codenotti

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

Balarka Sen

17:44

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

Thats probably more relevant to me

Alessandro Codenotti

Fair

Balarka Sen

Although this seems like good analysis

Alessandro Codenotti

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

Balarka Sen

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

Alessandro Codenotti

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

Balarka Sen

17:49

How do you see that without Gromov's theorem

Alessandro Codenotti

Ah there's a remark at the end

Balarka Sen

Oh that remark also contains my comment

Poisson boundary is Gromov boundary by a famous theorem

Alessandro Codenotti

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

Balarka Sen

ahh

thats real neat

Alessandro Codenotti

Yeah I think I read this argument on MSE somewhere

Balarka Sen

17:54

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

Alessandro Codenotti

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

Balarka Sen

Very nice

Alessandro Codenotti

Unfortunately the missing cases are the most interesting ones

I have to leave for a while, bye

Balarka Sen

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

18:05

@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

Thorgott

18:58

@Balarka tell me how to integrate $\omega$

Ted Shifrin

19:08

@Thor: Who is $\omega$?

Balarka Sen

I was gonna ask

Alessandro Codenotti

@Thorgott $\int \omega$

Thorgott

connection form

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

Jacob Bumgarner

Can someone help me with a stats question?

Ted Shifrin

Well, I'm totally out of context.

Balarka Sen

19:12

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

Ted Shifrin

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

Balarka Sen

Yeah.

Ted Shifrin

I would just stick to surfaces :P

Balarka Sen

I specialized to surfaces

Ted Shifrin

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

Balarka Sen

19:14

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

Mike Miller

Ambrose-Singer is great

Deane Yang has a good writeup

Balarka Sen

Deane Yang is a good geometer

I learnt the Weyl tensor shenanigans from his papers

Mike Miller

7

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

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}\\...

Thorgott

wtf is ambrose-singer

this was just supposed to be the geometric interpretation of curvature

Mike Miller

It is

Balarka Sen

19:18

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

Ted Shifrin

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

Balarka Sen

@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?

Ted Shifrin

notes deathly quiet

Balarka Sen

Lol

Thorgott

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

Balarka Sen

19:25

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.

Ted Shifrin

and probably the negative.

Balarka Sen

Yes, indeed, thanks.

Ted Shifrin

And maybe a coefficient. But ... blah.

Balarka Sen

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

Thorgott

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 $\varepsilon\mapsto\varepsilon^2$

Balarka Sen

19:29

Right.

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

Thorgott

anyway, I don't know that limit

Balarka Sen

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?

Thorgott

right, our picture pushes forward under a chart

Balarka Sen

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

Thorgott

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

Fargle

19:49

Howdy, nerds

Balarka Sen

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.

Ted Shifrin

Hi @Fargle

Fargle

How goes it @Ted?

Ted Shifrin

Still alive :)

Fargle

Good to hear

Ted Shifrin

19:55

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

Balarka Sen

I'm waiting for Thorgott.

Ted Shifrin

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

Fargle

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

Ted Shifrin

puts Fargle in humorless mode

Balarka Sen

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

Ted Shifrin

19:57

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

Balarka Sen

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

Fargle

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

Ted Shifrin

Oh, you're back to G&P?

Fargle

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

Ted Shifrin

Well, cool.

Fargle

20:02

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

Balarka Sen

Oh yeah I have seen those flying around

Fargle

They're quite good. Remarkably clear

Balarka Sen

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?

Ted Shifrin

Milnor is a great lecturer.

Fargle

Cool beard, too

Balarka Sen

20:05

Encyclopedia Britannica has this following picture:

Very religious interpretation

Only Gromov deserves such a background

Mike Miller

@BalarkaSen Does this even make sense

Ted Shifrin

LOL ...

Balarka Sen

Yes, for sure, I think.

Mike Miller

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

Balarka Sen

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

Mike Miller

20:08

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

Balarka Sen

I don't know what a path integral is man

Mike Miller

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

2

Balarka Sen

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

Mike Miller

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

Ted Shifrin

Wiener :P

Balarka Sen

20:11

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

Ted Shifrin

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

Fargle

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

Balarka Sen

have you heard of Wiener sausages, Fargle

Ted Shifrin

They don't like Vienna?

Balarka Sen

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...

Mike Miller

20:12

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

Ted Shifrin

I think those are weeners. :)

Fargle

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

Ted Shifrin

LOL, oh @Balarka

OK, lunchtime. BBIAB.

Fargle

@MikeMiller killing_in_the_name_outro.flac

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

Mike Miller

Well that's why you have to define them

Fargle

20:19

:|

Balarka Sen

This is like one of the worst things Thom did

Mike Miller

Which is not to say it's bad

Balarka Sen

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

Mike Miller

Unfortunately it was pure algebra

Balarka Sen

:(

Fargle

20:22

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

Balarka Sen

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

Mike Miller

Your first approach was better

Balarka Sen

send fargle the exposition mike

Fargle

eh, 5 seconds of my life vs 30

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

taps temple

Mike Miller

Yeah I just respect this approach less

Fargle

20:24

understandable, so do I

Balarka Sen

@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

Mike Miller

Just give up man

Balarka Sen

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

Ted Shifrin

20:45

Well, two are still alive.

Balarka Sen

True

Misha Gromov - October 24, 2016 (Part 2 of 2)

►

I never got to part 2, I'll watch it today

6 minutes in he drops a joke about involutions

Ted Shifrin

33 involutions per minute?

Transcript for

$Mathematics$ Mathematics