I'm more interested in his recent question about equivariant maps from manifolds with finitely many orbits

pontryagin duality says that locally compact abelian group G is isomorphic to D(D(G)) where D(G) = Hom(-,R/Z)

so much for that

Asaf's questions are very cool usually

if G is finite then G is isomorphic to D(G)

lol

Let me check it out

@BalarkaSen I thought it was Asaf Karagila

I saw that he figured something out from one of my old Frobenius Theorem answers.

They are usually too hard for me.

@Leaky, that Asaf wouldn't be doing geometry :P

I just like to see the questions, since the ideas are usually interesting.

Hi there, anyone following the Mochizuki "case"?

Yeah, I agree, Mike.

inb4 inter galactic teichmuller theory

@BalarkaSen: lol

Nevertheless, I thought this question was just silly, @MikeM. Am I missing something?

@TedShifrin Why can't $H$ be two disjoint charts in a manifold?

$H = \cup H_i$

I take manifolds to have a unique dimension, Balarka.

Unless I misunderstand you.

Ah I did not see that assumption (that $\dim H_i$ are all distinct)

Well, he didn't make it, but I do.

That is weird

this question is indeed suspekt

I agree

OK. I didn't want to be too stoooopid.

I love Whitney conditions and stratifications, but having the union be open takes out all the fun.

I am reading stratifications

Oh, wait. I'm wrong.

2hard10me

@MikeMiller check gchat

@Balarka, no, I'm wrong. You could take $H_k$ to be be missing $H_i$ in the first place, and then the closure would still be open.

I see

Still weird

Yeah, I was stooopid. He also assumed the various strata were disjoint to start with.

Take out closed submanifolds from a higher-dimensional manifold. OK, I get the question now, and the answer is NO.

Oh @Ted here is a question for you

@Balarka: Check my answer now. :P

You're sending me a sandbox?

I wrote the question in a sandbox

better than writing it in sand on a beach, I guess

I don't understand your answer. Why is the $z$-axis an open submanifold of $S$?

Why are the integrals even necessarily finite, @Balarka?

The whole union is open in $S$. The lower strata aren't open in the union.

So what are the $H_1, H_2, \cdots, H_n$?

@TedShifrin $V$ is a bounded domain.

$H_1$ is the $z$-axis, $H_2$ is the cuspidal surface minus its edge $H_1$.

The hypothesis says $H_i$ must be open in $M$

Still the integrals could diverge, even if the domain is bounded.

No, only $H_k$ is open, @Balarka. In my case $k=2$.

Oh he's denoting $k$ to be the last thing

His notation sucks

LOL ... it took me a minute to catch that at the beginning, but as you see that was only the beginning of my problems.

I see, you're correct, then. +1!

I posted a comment

What notation do you want him to use for the top stratum?

Left as an exercise :) I was troll-commenting mostly.

Don't do that! He might be hurt!

Actually, I would use a fixed letter for the top stratum and indices for the lower strata.

$H_{top}$, maybe

But I think you should remove your troll comment. He can then say $f$ is an immersion on every $H_i$, including $i=k$.

@Semicassical Your notation sucks

nah, Semiclassic.

that sounds like topological.

hmm, fair

@BalarkaSen i'd pull up my bird jpg but I probably shouldn't overuse it

I removed my troll comment

Good.

Too unprofessional

Yes, you're about to be an adult.

If the indices were indicating dimensions, for example, you wouldn't complain.

That's not unusual with stratifications.

@TedShifrin $\int_V f$ doesn't, but I guess you're speaking of $\int_{U_n} f_n$. I didn't specific how $f_n$ behaves near the boundary of $U_n$ (it could blow up)

I don't see why not, but I haven't thought seriously about the question.

So I just need to say $f_n$ is a smooth function on some neighborhood of $U_n$.

@TedShifrin How can a bounded function on a bounded domain have diverging integral?

$f$ was defined on all of $\Bbb R^2$

Right, I agree on $f$.

I'll edit the question in accordance with your complaint

Thank you

what's the usual shorthand for random variables

RVs ?

Done

@Semiclassical $X$

lol

LOL @MikeM

plural.

(so $X_i$)

$X_i$? :D

mutter

Balarka, your notation sucks.

We both went for the same joke, so either it's a good joke or we're both schlubs

@TedShifrin It is the notation of Gromov!!

Do not blame my poor soul

It looks too much like sections of the sheaf of (whatever) functions. Ugh.

@MikeMiller You're both schlubs

I think that's zhlubs, not schlubs.

It's a zh sound.

@TedShifrin That's Gromov's fault

(Remember I'm the token Jew here.)

smacks Mike

I actually quite like the notation, though I think Gromov used $\mathcal{O}p(A)$ to stand for Open.

Yeah but the $p$ is horrible

The usual way is to say smooth on the closure.

It's nice that one can just say "an arbitrarily small open subset" with this notation instead of having to specify it

He actually uses fraktur in his PDR book though; so Yashaberg is to blame for that horribleness

In fraktur it looks good

Especially when one makes lots of choices of arbitrarily small open subsets

$\mathfrak{Op}(A)$?

Hmm not fraktur some script

$\mathscr{Op}$

aw

that is sucky

@TedShifrin Basically, my question is asking if you have a converging sequence of domains and a converging sequence of functions on those domains, do the integrals converge?

More or less

$\mathscr{Op}(A)$

(All in the C^0 norm of course)

$\mathcal p$?

wtf how do you make that p

$\mathscr{O}\!\wp$

Good enough for ya

i dont think its in tex

Gromov doesn't need TeX

no \wp, please, here.

ihes.fr/~/gromov/PDF/InequalitiesScalar.pdf

Just look at this TeX bro

Well, if your open sets were big enough to contain $V$, it should be true. This should be an exercise with uniform convergence.

@TedShifrin Oh they do not approach $V$ in a containment fashion

The limiting could be very bad

Things could wiggle wiggle wiggle wiggle in

I cannot parse anything you just said

Ignore the grey set

$V$ is a nbhd of the core unit circle

$U_n$ is the deep grey area for some $n$

I see, the new open neighborhood is the darker grey near the wiggly fella

as $n \to \infty$, the amplitude of the thing becomes less and frequency increases super fast

Yeah

Hmm ... Dunno.

You fix an open set of some size but do not take it arbitrarily small, to start corrugating inside that

@BalarkaSen I saw a really exciting paper the other day

let's compute $H^1(C_2,\Bbb Z/4\Bbb Z)$

Or rather, really exciting results

@LeakyNun Let's not computer

where the action is given by negation

Let's smartphone

@MikeMiller Mhm?

Ah, a Balarka is back in smart-aleck mode.

arxiv.org/abs/1712.06197

Ohh 0celo7 told me about that

@BalarkaSen i just realized

so $f(1) = f(1) + f(1)$, so $f(1) = 0$

and then $f(g) = f(g) + gf(1) = f(g)-f(1) = f(g)$

$f(g) = f(1) + f(g) = f(g)$

$f(1) = f(g) + gf(g) = 0$

Mostow rigidity says that if I have constant sectional curvature < 0 then Isom(X) --> Aut_{homotopy}(X) is an isomorphism in pi_0, but nothing about higher homotopy groups, right?

so $f(g)$ can be anything

so $H^1(C_2, \Bbb Z/4\Bbb Z) = \Bbb Z/4\Bbb Z$

let's compute $H^2(C_2, \Bbb Z/4\Bbb Z)$

@Balarka: I don't berember.

> berember

$\Bbb Z[C_2] \to \Bbb Z$ sending $m+ng$ to $m+n$ has kernel generated by $1-g$

@BalarkaSen It should be a natural extension of the argument to get contractibility

Hmm

I do not know the argument :3

so let's have $\Bbb Z[C_2] \to \Bbb Z[C_2]$ sending $m+ng$ to $(m+n)(1-g)$

the kernel is generated by $1-g$ again?

so our projective resolution is $\cdots \to \Bbb Z[C_2] \to \Bbb Z[C_2] \to \Bbb Z \to 0$

$0 \to \operatorname{Hom}(\Bbb Z, \Bbb Z/4\Bbb Z) \to \operatorname{Hom}(\Bbb Z[C_2], \Bbb Z/4\Bbb Z) \to \operatorname{Hom}(\Bbb Z[C_2], \Bbb Z/4\Bbb Z) \to \cdots$

hi @TobiasKildetoft

@LeakyNun Hi

I believe all of them is just $\Bbb Z/4\Bbb Z$

if $f \in \operatorname{Hom}(\Bbb Z[C_2], \Bbb Z/4\Bbb Z)$ sends $1$ to $n$ then the next level sends $1$ to $0$

ok so the maps are all zero maps

so that means $H^2(C_2, \Bbb Z/4\Bbb Z) = \Bbb Z/4\Bbb Z$

am i right @TobiasKildetoft

hi @Mr.Xcoder

hi @LeakyNun

@LeakyNun Haven't read all of that

just this

hi Tobias

@TedShifrin Hi

if so, then what are the four extensions...

I know that there is the semidirect product corresponding to 0

that's just D8

there can't be four extensions, there's only D8 and Q8

@BalarkaSen something about quasiconformal maps on the boundary of hyperbolic space

Right, but cohomology tends to be bad at counting things exactly

Though in this case it it probably right, since you ended up with a free module of rank $1$, which is the correct number of non-split extensions

@MikeMiller Huh

woah, 400 new messages

chat is busy today

How do you tell if an abelian group is interesting

hi @mercio

hi ted

think about whether other people would be interested in it, it's not an objective answer

I'm going to try and read that cohomology computation

@geocalc33 "interesting" is rarely a precisely defined term

i didnt mean "interesting" as any math codeword. I meant, would other people be interested in hearing you talk about it?

$\Bbb Q/\Bbb Z$ is a classic imo

nominate it for the abelian group hall of fame

never seen that one talked about that I can recall, Mike

what about the trivial group

might be just me, but I find the trivial group kinda trivial

only so many times you can write $x = xxxxxxxxxxxxx^{-1}xxxxxxx$ before you get bored

@geocalc33 I already told you

@GFauxPas: It's famous because it's infinite but every element has finite order.

if you date people around to tell if they are interesting, boy are you going through a tough love life

Ted whoa

that's neat

Rehi @Eric

@TedShifrin And of course it is also a key ingredient in Pontryargin duality

Hlo

what's a good example of an abelian group with $\Bbb Z^\infty \subset G$ but no homomorphism $G \to \Bbb Z$

$\Bbb Q^\infty$ works, but somehow I would like a 'smaller example'

whatever that means

does it have a nice presentation?

$\mathbb Q / \mathbb Z$

@GFauxPas Not really. It is not even finitely generated

There's no $\Bbb Z$ sitting inside $\Bbb Q/\Bbb Z$, @GFauxPas

@TedShifrin He wasn't answering me, he was asking if that had a nice presentation

Oh, sowwee.

@MikeMiller By ${}^{\infty}$ you mean the infinite direct sum?

Take $\langle 1/p^i \mid p \cdot 1/p^i = 1/p^{i-1}, p \cdot 1/p = 0 \rangle$

yeah

where $p \cdot$ is read as "sum of p copies"

and p varies over primes, i over nonnegative integers

yeah I guess that isn't "nice"

I can write it down in the span of a minute so it can't be that bad

What is an infinite abelian group

Z

infinitely many things inside it

the order of an element is how many times you add/multiply it to itself before getting the identity

Okay so infinite elements

so, take $\mathbb Z / 2\mathbb Z$

the integers mod 2

everything in there has order two, because if you add two integers you get $0 \pmod 2$

except we don't think of it as infinite because the quotient gives you two sets, $1 \pmod 2$ and $0 \pmod 2$

I can't figure out my identity

well you're multiplying things, right?

Yeah

remember, the identity $e$ in a group satisfies $c \circ c^{-1} = c^{-1} \circ c = e$, so it's implicit when you define an inverse that you already have an identity

what's the inverse of an element in your group?

@MikeMiller does $\Bbb Q^\ast$ work?

a @Balarka: I'm thinking about your question. There are two issues. The first is, do we have a uniform bound on the volume of the $U_n$'s? Yours look like they could get big. Second, it seems to me that you need a $C^1$ estimate on the diffeotopy in order to use change of variables to estimate $\int_{U_n} f$ as close to $\int_V f$.

@LeakyNun I believe that's isomorphic to $\Bbb Z^\infty \oplus \Bbb Z/2$, with generators $1/p$ and $-1$

By prime decomposition

@TedShifrin Negative to both! Great, I really do want a negative answer.

Yeah I do think the volumes could get big due to wiggle

@MikeMiller so $\Bbb Q_{>0}$ is what you want

that should be small enough

No, that's $\Bbb Z^\infty$, which has many homomorphisms to $\Bbb Z$

in fact it has $\prod_{i=0}^\infty \Bbb Z$ homomorphisms

ok

Rekt Leaky

actually @geocalc33 "infinitely many elements" isn't a good description because of things like what I said

:(

So, the obvious way to do the estimate is to write $$\left|\int_{U_n} f_n - \int_V f\right| \le \int_{U_n} |f_n-f| + \left|\int_{U_n} f - \int_V f\right|.$$ You'd want reverse triangle inequality to get badness. @Balarka

I suspect I should stop being silly and just be happy with $\Bbb Q^\infty$ hehe

let's say that an infinite abelian group is an abelian group $G$ with an injection $\Bbb Z \to G$

is that good enough?

@LeakyNun No

because $\mathbb Z / 2 \mathbb Z$ is isomorphic to a group with 2 elements

@GFauxPas And it only has finitely many elements

@TobiasKildetoft you're right, I'm stupid

there's no injection $\Bbb Z \to \Bbb Q/\Bbb Z$

An infinite abelian group is by definition one with infinitely many elements

well "all even integers" and "all odd integers" can be viewed as infinite or finite so I'm trying to think of how to be more precise

@GFauxPas no, they really can't

those are two sets, and two is finite

$\ldots \equiv -6 \equiv -4 \equiv -2 \equiv 0 \equiv 2 \equiv 4 \equiv \ldots$ ?

ah I see where my mistake it

is*

the fact that the sets themselves are infinite is completely irrelevant

@TedShifrin Hmmm

right, that was my mistake Tobias

anyway @geocalc33 learning group theory is a worthwhile endeavor if you can handle the abstraction of it

It is unclear to me why $\int_{U_n} |f_n - f|$ doesn't poof as $n \to \infty$

I don't particularly enjoy abstract algebra but I recognize that it's worth knowing

$\|f_n - f|_{U_n}\|_{C^0} \to 0$ by hypothesis

I guess the point is changing domains again

That is the problematic bit you're right

Yeah, but does the volume of $U_n$ increase to cancel out or worsen that decrease?

Well, the other term is problematic, too, without $C^1$ control. Write down the COV.

@GFauxPas: I liked it a lot more when I taught it and wrote a book on it :)

@TedShifrin Thank you, your insights are very helpful! I will think about this and hopefully come up with a example pointing to the negative.

my professor for it was pretty awful, I felt bad for him because it was his first job as a teacher and his greenness was apparent

hopefully he'll get better with practice

got better*

Keep me posted, @Balarka.

Will do. Thanks for confirming my suspicion that this was the problematic bit.

@GFauxPas I was about to say I hoped I was not that teacher :)

nah he wasn't Kildetoft,his name escapes me atm but it wasnt that

@GFauxPas I figured given your change to "got".

lel

What's the intuition behind Gauss-Bonnet? Especially the local version? It feels like black magic to me

I think of it in terms of the Poincare-Hopf battery.

I don't think about the boundary term though, which is bad of me

Battery because it powers machines instead of being a machine?

@BalarkaSen I know nothing about that :/ Yeah the boundary term is just annoying

Yeah I don't think of it as a theorem; it's just a cycle of different theorems

If you give me two of those I'll probably use one to prove the other

It's an endless loopy loop

The boundary term is difficult to get at through differential topology machinery

@Alessandro: It's Stokes's Theorem.

There is a sense in which it is the Atiyah Patodi Singer theorem: A formula relating Euler characteristic to an integral, with a correction term that depends on metric data on the boundary

You either do Green's Theorem with coordinate yuck or you use differential forms and it's beautiful.

Damnit I was going to say Stokes too so I don't look like a totla asshole

@TedShifrin That's another reason why I should learn differential forms

You are indeed a totla asshole

I do think it's interesting that all of the terms are topological and geometric except for the boundary which is intrinsically metric

The key classical computation is on p. 82 of my notes. See p. 105 for the forms.

This is a surprisingly challenging question.

@Alessandro OK, here goes my view of things. $\Sigma \subset \Bbb R^3$ be a (oriented) embedded surface. Consider the map $G : \Sigma \to S^2$ given by sending $p \in \Sigma$ to $G(p)$, the normalized normal vector of $\Sigma$ at $p$, pointing in the outward direction.

Say $v \in S^2$ is a regular value of $G$ and consider the projection of $n$ to each of the tangent spaces of $\Sigma$ (we can do that; we're in $\Bbb R^3$) to get a vector field $X$ on $\Sigma$ which vanishes whenever $v$ is normal to $\Sigma$. That happens for points $p \in \Sigma$ such that $G(p) = v$ or $-v$. So the "total index" of the zeroes of the vector field $X$ of $\Sigma$ is total sum of local degrees of points in $G^{-1}(v)$ and $G^{-1}(v)$ - which is twice the degree of $G$.

@Balarka: In all fairness, the global version is just Poincaré-Hopf. It's the local version where you see differential geometry.

Now you can localize the Poincaré-Hopf argument with integral geometry and "averaged Morse theory." I sent you that paper.

Total index of $X$ is exactly the Euler characteristic $\chi(\Sigma)$ by Poincare-Hopf (think: total index of a vector field on $S^2$ is $2 = \chi(S^2)$), so $2 \deg(G) = \chi(\Sigma)$

@TedShifrin Yeah I remember that one

It was very intriguing

@Alessandro Here is the relation to curvature. Consider $dG$. That keeps track of how the normal vectors bend and stuff when you move away from a point infinitisimally. Take for example a sphere; at any point $p \in S^2$, if you move away from $p$ in any direction, the unit normal curves in exactly the way $S^2$ is curved. If you are on the saddle point of a saddle, the normal vector curves oppositely on two different directions.

Indeed, $\det(dG)$ is exactly the Gaussian curvature of $\Sigma$.

It measures how much the image of an infinitisimal square on $\Sigma$ gets distorted by $G$

Ok, that's the Gauss map and Weingarten operator, right?

Correct!

I think I know it by the shape operator

Now, you also have a global formula for the degree of $G$. I wonder if I have a concrete way to write that

Here is the point.

Write $n = \text{deg}(G)$. The way you should think about $G : \Sigma \to S^2$ is as an $n$-sheeted covering map, but possibly with singularities. If you are willing to think of it as such, you can convince yourself that "$G\text{vol}(\Sigma) = n \text{vol}(S^2)$" as $\Sigma$ sort of "wraps around" $S^2$ n times, so multiplies the volume by $n$.

But what should be this mystical quantity "$G\text{vol}(\Sigma)$"? Locally we know what $G$ does to volume; it stretches a unit square by $\text{det}(dG)$.

@BalarkaSen Wait who is $n$ here?

So it should just be the case that $"G\text{vol}(\Sigma)" = \int_{\Sigma} \text{det}(dG) dA$ where $dA$ is the volume (by which I mean area, 2-dimensional volume) element on $\Sigma$

@Alessandro $n = \text{deg}(G)$, as I wrote. The degree of $G$

In the previous message, you're projecting $n$

Oh lol

Sorry about that

I meant projection of $v$

Oh ok

Once you are convinced of this mystical equivalence I wrote, you have $$\int_\Sigma \det(dG) dA = \int_\Sigma K dA = n \text{vol}(S^2) = \deg(G) 4\pi = \chi(\Sigma)/2 \cdot 4\pi = 2\pi\chi(\Sigma)$$

Somebody give me a medal for this shite

hiss sorcery

Well that makes sense intuitively even though it takes a couple of acts of faith :P @Balarka

Thanks a lot for your explanation!

To make that last step I made rigorous you'd have to pass through differential forms

But yeah it is a mystery that $\int_{\Sigma} K dA$ does not depend on the embedding of $\Sigma$ that I choose in $\Bbb R^3$ (or rather, the metric I choose on $\Sigma$), only it's topology

Like @MikeMiller said these kind of things generalize

You get some geometric quantity coming from choice of geometric data and you integrate it to obtain topological invariants

Sounds fascinating

I'm surely going to take algebraic topology next year, I have to check if I have enough free credits to fit some differential geometry into my study plan as well :/

That'd be a great combination

Topological invariants that come from algebraic geometry are bollocks though

how so

They come from homological algebra

Total bollocks

I'm going to focus on logic and set theory of course, but I'll also do as much geometry as I can

No reason why they do what they do

I have been especially mad at spectral sequences for the past few days

Are they haunting you?

I should have seen that coming

very angery

@BalarkaSen Nobody sees ghosts coming!

Stop

Please stop

They say two bad jokes in a row will instantaneously summon @Daminark to the room

Go forth and study hauntology

Boom

Oh my fucking god

Here he is!

Hi @Dami

what have we done

In his house at Chicago missing Daminark waits meming.

@Semiclassical can confirm personally that @Daminark is an eldritch being of pure meme

Sounds right.

whereas I'm more like a Lethargian out of The Phantom Tollbooth

How do you do fellow memers?

Pretty well

You are on summer break now, right?

'"You see," continued another in a more conciliatory tone, "it's really quite strenuous doing nothing all day, so once a week we take a holiday and go nowhere, which was just where we were going when you came along. Would you care to join us?"'

Yup!

Got about half of week of just messing around before my next schtick starts

i havent read that book in years

me neither, if i'm honest

It's scary how bad humo(u)r immediately summons Demonark. It makes him jealous.

Bad humor? What is this nonsense?