Mathematics - 2020-04-24 (page 1 of 2)

Balarka Sen

00:00

Whenever you have a good class of "functions" on open subsets of $X$, which can be "restricted" from bigger open sets to smaller open sets, and if two "functions" on two open sets $U, V \subset X$, whenever they match when "restricted" to $U \cap V$, glues uniquely to a "function" on $U \cup V$, you can do the same construction as above.

These, in full generality, are known as sheaves.

A sheaf $\mathscr{F}$ on a topological space $X$ is an association to each open set $U$ an abelian group $\mathscr{F}(U)$ such that whenever $U \subset V$ there is a restriction homomorphism $\mathrm{res}^V_U : \mathscr{F}(V) \to \mathscr{F}(U)$

and whenever $f \in \mathscr{F}(U)$, $g \in \mathscr{F}(V)$ are two elements such that $\mathrm{res}^U_{U \cap V}(f) = \mathrm{res}^V_{U \cap V}(g)$, there is a unique $h \in \mathscr{F}(U \cup V)$ such that $\mathrm{res}^{U \cup V}_U(h) = f$ and $\mathrm{res}^{U \cup V}_V(h) = g$. (AKA, the pasting lemma holds)

The sheaf of continuous functions on a top. space $X$ is defined be $\mathscr{F}(U) := C^0(U)$, the restriction map is the obvious restriction of functions.

The sheaf of holomorphic functions on a domain $X \subset \Bbb C$ is defined to be $\mathscr{F}(U) := C^\omega(U)$ (the group of holomorphic functions on $U$), again restriction maps are obvious

The sheaf of locally constant functions on a top. space $X$ is defined by setting $\mathscr{F}(U)$ to be the group of $\Bbb Z$-valued constant functions on each connected component of $U$, restriction map obvious

Makes sense?

Thorgott

amazing, I never knew that this is what a sheaf is

Balarka Sen

It's really simple actually

Thorgott

it certainly makes sense. so, what's the cochain compelx thing about?

Balarka Sen

You can do the same construction I did but with a different sheaf than the sheaf of continuous functions. The resulting object $(C^\bullet(X, \mathcal{U}, \mathscr{F}), d)$ is called the Cech complex of the sheaf $\mathscr{F}$ with respect to the open cover $\mathcal{U}$.

And then you notice my earlier remark that $\ker d$ consists of tuples of functions on $k$-fold intersections which glue along $(k+1)$-fold intersections, and $\mathrm{im} d$ consists of tuples of functions on $(k+1)$-fold intersections which come from restricting tuples of functions on $k$-fold intersections and then taking that weird difference crap

So looking at the $k$-th spot in $\cdots \to C^{k-1}(X, \mathcal{U}, \mathscr{F}) \stackrel{d}{\to} C^k(X, \mathcal{U}, \mathscr{F}) \stackrel{d}{\to} C^{k+1}(X, \mathcal{U}, \mathscr{F}) \cdots \to $, compute $\ker d/\mathrm{im}\, d$ at that spot.

This makes sense, because $d^2 = 0$, so $\ker d$ contains $\mathrm{im}\, d$

anakhro

00:17

Hi all.

Rithaniel

Heya anakhro

Balarka Sen

If this group vanishes, $\ker d = \mathrm{im}\, d$, so every tuple of functions defined on $k$-fold intersections which patch on the $(k+1)$-fold intersections will arise from a tuple of functions on $(k-1)$-fold intersections by the difference stuff. These are complicated higher order scenarioes of what was happening with Simple's exercise, where we were trying to patch stuff on the double intersections

anakhro

What's up, Rithaniel?

Balarka Sen

Anyway, this group is known as the $k$-th Cech cohomology $H^k(X, \mathcal{U}, \mathscr{F})$ of the sheaf $\mathscr{F}$ with respect to the cover $\mathcal{U}$.

Rithaniel

Not much. At the end of the semester and thinking about trying to make a video game involving magmas and other algebraic objects

Also, finishing up some last homeworks

anakhro

00:20

What would the premise of this video game be?

Balarka Sen

There is a way to get rid of the dependence on $\mathcal{U}$ for nice topological spaces (nice = paracompact, Hausdorff). Then you just call the resulting group $H^k(X, \mathscr{F})$ the sheaf cohomology groups

That's all I had to say

anakhro

I am pretty sure rock paper scissors is already a game, so you missed out on magmas.

(non-associative magmas)

(non-associative, commutative magmas)

Rithaniel

The idea was that you are a deity-esque being designing the rules of different universes. Each level is like a puzzle where you have certain axioms you have to satisfy, and you design a magma to try and do so.

I'm pretty sure it'd have to be a finite magma, and I'm currently trying to think of a mechanism that would allow you to fill out the rules of the magma without being insanely tedious (like, "fill out the entries of this Cayley Table")

Thorgott

wow, so cohomology enters the picture

I'm gonna learn my first (co)homology theories this semester, really excited for that

anakhro

I think cayley table will be the only feasible thing to code.

Thorgott

00:25

thanks for the excursion!

Balarka Sen

Yeah, the upshot is all these obstructions to getting antiderivatives to match on overlaps etc are really cohomological obstructions

No problem, glad you liked it

Rithaniel

Well, yeah, a cayley table would be the first thing, but maybe I could implement ways to fill out many entries at the same time, so that you could have 20x20 tables that don't take forever to fill out

Also, n-ary operations would be fun

Though, I'm getting distracted from HW

(what's up with you, by the way, anakhro?)

anakhro

Nothing is up with me, I am just bored today.

Enjoy your homework! I will try not to bug you. :)

geocalc33

I dropped 42 tracks

good luck Rithanial

Rithaniel

Actually, I could use some help on this one problem

I need to show that, if $V$ is a valuation domain, then $V$ is completely integrally closed if and only if every nonzero prime ideal of $V$ is maximal.

Balarka Sen

00:32

I found this conversation randomly don't ask me how

5

Can I take a moment to just appreciate how gold it is

Rithaniel

It is true: You only yoyo once

geocalc33

"enjoys are you here?"

"he's passed away" -mike

that was 6 yrs ago wow

Balarka Sen

time flies

geocalc33

I feel like I'm a human and y'all are the dinosaurs

jk sorry

Balarka Sen

Apr 11 '14 at 18:59, by Enjoys Math

Let $\Bbb{P} \times \Bbb{P}$ be the set of all prime number pairs. Let $U_k = \{(p,q) : p = q + 2k; p,q \in \Bbb{P}^2\}, k \in \Bbb{Z}$. The $U_k$ partitions the set of prime pairs, further if we define $U_k \cdot U_{\ell} = \{(a,c) : (a,b) \in U_k, (b,c) \in U_{\ell}\} \subset U_{k + \ell}$. So if you let the latter to equal $U_{k + \ell}$ then we have a group similar to $\Bbb{Z}$? It's weird. Also how would you handle identity, empty set or the diagonal?

Enjoy was thinking about twin primes even back then

what a dedicated man

Rithaniel

00:45

So, I found something claiming that, in a local valuation domain $V$, for every nonzero, nonunit $r,s\in V$, then $r\mid s^m$ and $s\mid r^n$ for some $n,m\in\mathbb{N}$. I know that in a valuation domain, either $s\mid r$ or $r\mid s$, but why does the locality give us this bit about powers?

user574847

01:06

hello. I have a basic question. if you have $\Sigma_g$ a genus $g$ orientable surface, and you want to compute its fundamental group by van kampen, then you can take the complement of any point, and a disk about the point, as your two path-connected open subsets with path connected intersection

the disk has trivial fundamental group, and the other one someone can say something about

but anyway, on the intersection of these two, you have an annulus, so that has homotopy group isomorphic to Z

so you ask about including those loops into the disk, in which they contract to a point, and for the other apparently it sends 1 to $a_1b_1a_1^{-1}b_1^{-1}\dots a_gb_ga_g^{-1}b_g^{-1}$, which is the part that confuses me

user574847

01:20

i can picture just pulling the loop onto the boundary of the standard 2g-gon that you use to construct $\Sigma_g$, but that doesn't feel very rigorous

Leaky Nun

01:40

@Rithaniel do you have the statement in context?

Rithaniel

Here you go lohar.com/mithelpdesk/hd0303.pdf

Leaky Nun

@Rithaniel you forgot "one dimensional"

and they even gave a reference

Rithaniel

I suppose I should be a bit more free with chasing down references. Though, right now I'm actually reading a thing about discrete valuation domains. It seems that in such a ring, each unit can be expressed as $ut^n$ where $u$ is a unit and $t$ generates the unique maximal ideal

Leaky Nun

in such a ring clearly this theorem is true

Rithaniel

01:59

Yeah, having everything just be associate to a power of some element makes things very easy to work with

Now, the opposite direction is still pretty elusive. Showing that if a valuation domain is completely integrally closed, then it must have krull dimension less than or equal to one

user10478

02:40

Is there a better way to write the $n$-th antiderivative of $1$ than $\int_{n\ of\ these} 1\ dx^n$?

Rithaniel

03:23

Phew, figured it out, I think

Suppose $V$ is a valuation domain that is not a field and that is completely integrally closed. Let $T$ be a prime ideal of $V$ and $v\in V\setminus T$. If $t\in T$ then $\frac{v}{t}\notin V$ as $v\notin T$.

However, then $\frac{t}{v}\in V$ as $V$ is a valuation ring. We assume that $v$ is not a unit and note that this implies $\frac{t}{v}\notin T$ as otherwise $\frac{1}{v}\in V$. Again, since $V$ is a valuation ring, then we know that either $\frac{t}{v}\mid v$ or $v\mid\frac{t}{v}$. If $\frac{t}{v}\mid v$ then $\frac{v^2}{t}\in V$ and therefore $\frac{1}{t}\in V$ (which is a contradiction). Thus, $\frac{t}{v^2}\in V$. Inductively, $\frac{t}{v^n}\in V$ for every $n\in\mathbb{N}$.

However, we now note that $\frac{1}{v}\in K$ and that $t(\frac{1}{v})^n\in V$ for all $n\in\mathbb{N}$. So $\frac{1}{v}$ is almost integral over $V$ and so must be an element of $V$. Thus, $v$ is a unit and $T$ must be maximal. Therefore $V$ has Krull dimension one.

(I can probably streamline some bits of this here and there)

Hawk

Are there efficient ways to compute the cyclotomic polynomials (given a root) other than by its definition or division from x^N - 1?

hchar

06:18

A basis of $\Bbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\Bbb{Q}$ would be $\{1,2^{1/2},2^{1/3},2^{2/3},2^{1/6},2^{5/6}\}$. Correct?

feynhat

@user574847 *4g-gon

@user574847 Okay, to be precise, let $U$ be the punctured $\Sigma_g$ and $V$ be a neighborhood of the puncture. You're asking why the map induced by inclusion, $\pi_1(U \cap V) \to \pi_1(U)$ takes 1 to $a_1b_1a_1^{-1}b_1^{-1}\dots a_gb_ga_g^{-1}b_g^{-1}$, right?

user574847

Yes

feynhat

07:09

hmm...

Let $q$ be the quotient map that takes the 4g-gon to $\Sigma_g$. $q$ is actually invertible at 'most' points. If you take a small loop around the puncture, its pre-image on 4g-gon will be a small circle in its interior. Does this make sense?

By 'most' I meant the interior of the 4g-gon.

Now, you deformation retract the puncture 4g-gon to its boundary. Then this small circle in its interior becomes the boundary of the 4g-gon. (The order in which this deformed circle covers the edges of the 4g-gon is important). Finally, you apply $q$.

This composition: inverting q followed by deformation retract followed by q is actually an isomorphism at $\pi_1$-level.

I don't know lol

@BalarkaSen

feynhat

07:40

@user574847 ...and if you're being pedantic, the map doesn't take 1 to that word. It takes a class of loop in $U \cap V$ to a class of loops in $U$. And how do you go from $U$ to wedge of circles? Deformation retraction. What is the image of the loop around the puncture under this deformation retraction?

Astyx

09:20

Why is everyone doing maths in terms of categories nowadays ? What's so nice about it ?

Leaky Nun

11:09

@Astyx it sounds fancy and makes you look cool

@hchar correct

Adeek

11:37

does anyone know the the bicommutant group of upper triangular matrices ?

Alessandro Codenotti

12:46

@LeakyNun The latter part is very debatable

Alessandro Codenotti

12:56

@Balarka Do hyperbolic groups that are not virtually $\Bbb Z$ always contain $F_2$?

Leaky Nun

13:12

@AlessandroCodenotti it's true

Astyx

What do the Cech cohomology groups tell you ? @Balarka

Adeek

does anyone know whether upper triangular matrices are weak operator closed ?

Astyx

What does "weak operator closed" mean ?

Alessandro Codenotti

@Astyx The same as the usual one for reasonable spaces (CW complexes)

Adeek

yeah

Astyx

13:21

So it's just a topological invariant to help classify the topology of sheafs over some space ?

Adeek

I think the commutant of 2x2 upper triangular matrices is subset of m ?

let U be the set of all upper triangular matrices

Alessandro Codenotti

Oh I don't know sheaf cohomology, I thought you were doing honest AT with dishonest spaces when you mentioned Cech cohomology

Adeek

is it true that the commutant of U is proper subset of M ?

@Astyx I can help you out I know about sheaf cohomology

@Astyx what do you want to ask about sheaf cohomology ?

Astyx

Yesterday Balarka introduced me to the Cech complex of sheafs w.r.t. an open cover of the space, then stated you could abstract the cover out for "nice" space

Just what it's general use can be

Leaky Nun

Apr 19 at 14:36, by Leaky Nun

how about the generalized MV: Stacks 01ES: the Čech-to-cohomology spectral sequence

Adeek

13:26

you can compute sheaf cohomology for instance

Sheaf cohomology is very useful

Astyx

Could you give an example ?

Adeek

@Astyx here this is where I learned about all this stuff

www-users.math.umn.edu/~garrett/m/algebra/cech.pdf

Astyx

Thank you !

Leaky Nun

@BalarkaSen hey

Balarka Sen

@AlessandroCodenotti Yeah

Alessandro Codenotti

13:35

Is that trivial and I'm missing something obvious or?

Balarka Sen

No I don't think it's obvious at all. It's ping-pong on the boundary of the hyperbolic group

Alessandro Codenotti

Do you know a reference where I can read a proof?

Balarka Sen

Haha no clue really

these are all factoids from the void i overheard

@Astyx Here are some classical problems that sheaf cohomology tells you the answer to: First and second Cousin problems

Astyx

Cheers

Balarka Sen

If you use the sheaf $\mathscr{F}$ of locally constant functions and $X$ is a very nice space, then the sheaf cohomology groups are interesting topological invariants, also known as singular cohomology groups

Alessandro Codenotti

13:42

In Löh's book there's a result that if $x$ and $y$ are infinite order elements of an hyperbolic group with $\langle x\rangle\cap\langle y\rangle=\{e\}$, then $\langle x,y\rangle$ is isomorphic to $F_2$ but it's not proved in detail

Balarka Sen

Do you want to try the ping pong game

I have no idea how it works but I know it should :P

Alessandro Codenotti

I cannot right now, but we can think about it later

I don't know how the Gromov boundary works very well though

Balarka Sen

Me neither

I know less GGT than you Alessandro no need to get intimidated

Alessandro Codenotti

But you understand the geometry part much better than I do

Balarka Sen

Who's spamming the star panel

Cut it out

Alessandro Codenotti

13:47

Oh no again

MEcho

14:04

I am wondering if someone can help me with a problem

??

????!!!!!!!!!!!!!!!!!!!!!!!!!!

Leaky Nun

if you don't ask it then nobody can help you

no matter how many punctuation marks you type

MEcho

I understand that

but I dont feel comfortable just spamming it here

Balarka Sen

Ask; don't ask to ask

Leaky Nun

looks like you feel perfectly comfortable spamming exclamation marks here though

MEcho

Oh look seriously dude, I have anxiety, I am struggling with some things at the moment

I asked for help, but be as jerky as you like

Or as judgemental as you like

Leaky Nun

14:10

you still haven't asked your question

MEcho

Because I want to do it in private chat

as was obviously implied

Leaky Nun

there is no private chat

MEcho

No private messaging?

Balarka Sen

The chats in chat.stackexchange are all public yeah

MEcho

Oh well

I guess that is that then

Mike Miller

14:13

The way it works here is that people ask questions and then if someone is around who knows how to help and is interested they respond

No one's gonna judge if they don't want to answer your question, the worst case is they just don't answer it

Balarka Sen

the Riemann curvature tensor is totally determined by sectional curvature

this is the formula

where does the curly braces close lol

Balarka Sen

14:42

Say $R(X, Y, Z, W) = R(X, Y)Z \cdot W$. Then $R(X, Z, Y, Z) = R(Y, Z, X, Z)$ so $R(X + Y, Z, X + Y, Z) = R(X, Z, X, Z) + R(Y, Z, Y, Z) + 2R(X, Z, Y, Z)$

Cool this is what I want

Papa

15:38

Hi
Can I get some help on graph theory?

Balarka Sen

@LeakyNun Apparently $\ell$-adic cohomology behaves well under etale descents

Leaky Nun

@BalarkaSen XD

Balarka Sen

Rekt

Leaky Nun

@BalarkaSen why do I feel like this piece was written for the sole purpose of having people show off?

Balarka Sen

yeah nuts right

Leaky Nun

15:51

I don't like the competitiveness

Sujal Motagi

Can someone please help me with this: math.stackexchange.com/questions/3641762/…

Can you guys tell me how to get better at contest math? Right now i suck at it.

Leaky Nun

@BalarkaSen Groups acting acylindrically on hyperbolic spaces

Balarka Sen

Lol nice

Leaky Nun

16:09

@BalarkaSen étale behaves cohomology under descents $\ell$-adic well

Balarka Sen

Lmao

geocalc33

1

Q: Mathematical music theory text that treats the concept of the "freestyle"

This question relates to mathematical music theory and linguistics. I am investigating and trying to understand a phenomena I call "pseudo-musical speech." That is, the artist just let's the words "roll off the tongue," or "freestyles on the track." Note: The words might not make perfect linguist...

Balarka Sen

17:30

Should I sleep today

Thorgott

why not

Mathphile

I have a conjecture

Anonymous

In a short exact sequence like $1 \to A \to G \to B \to 1$ I'm a bit confused about the definition of a "section". Does a section $s: B \to G$ necessarily have to induce an identity on $B$ or can it be any homomorphism $B \to G$?

Balarka Sen

It's a homomorphism $B \to G$ such that if you compose it with $G \to B$ (the last map in the sequence), you get identity on $B$

i mean if it was a random homomorphism it wouldnt have anything to do with the sequence

Mathphile

${}^{\infty}x=x^n$ where $n\in \Bbb{N}$ only when $x=1$

How do we prove this?

Anonymous

17:36

@BalarkaSen Ah, right. Thanks!

Anonymous

So in the example, $0 \to \Bbb Z/p \to \Bbb Z/p^2 \to \Bbb Z/p \to 0$ (let's represent this as $0 \to A \to G \to B \to 0$), I'm basically trying to compute the number of possible sections $B \to G$. Say $\sigma$ is the surjective homomorphism $G \to B$ and we know $\sigma \circ s = \mathrm{id}_B$. And we know if a composition is injective the first map is injective. This means the element (generator) $1 \in B$ must get mapped to an element of order $p$ in $G$.

Anonymous

And since there are only $p-1$ elements of order $p$ in $G = \Bbb Z/p^2$, the number of possible sections $s$ should be $p-1$. I'm not sure if this is right though.

Leaky Nun

@BalarkaSen yes

Mathphile

9 mins ago, by Mathphile

${}^{\infty}x=x^n$ where $n\in \Bbb{N}$ only when $x=1$

I feel that it is trivial to prove this but I don't see how

Astyx

17:51

What is $^\infty x$ ?

Mathphile

The infinite tetration of $x$ or $x^{x^{x^{x...}}}$

it converges for $e^{-e}\le x \le e^{1/e}$

@Astyx

@Astyx any idea how to prove this?

user10478

Do matrices apply to systems of inequalities?

Rithik Kapoor

18:17

@Mathphile Interesting conjecture

Astyx

I think we have $x^{^\infty x} = ^\infty x$, so you get $x^{x^n} =x^n$ which rewrites as $y^{y/n} =y$ ie $y \log y = ny$, so $\log y = n$

Rithik Kapoor

I though have no idea on how to prove this one

Astyx

or y=0

(y is $x^n$)

Hmm something's wrong

Mathphile

$y^{y/n}=y$?

oh wait that seems fine

Astyx

$x^{x^n} = (y^{1/n})^{y}$

Ah yes of course

It's not $y\log y = ny$, it's $y\log y = n\log y$

Mathphile

18:22

yes

Astyx

So $y=n$ or $\log y =0$ ie $y=1$

So $x = n^{1/n}$ or $1$

Now you just need to see which $n^{1/n}$ are valid for infinite tetration

Which I have no will to do :)

But since it converges to 1, probably a lot

Mathphile

Apparently all $n^{1/n}$ are valid for infinite tetraton

Alessandro Codenotti

@Balarka still around by any chance?

Astyx

Yes because $x\mapsto x^{1/x}$ has a maximum at $e$

geocalc33

@Mathphile nice conjecture

I have a conjecture that boosted hyper-Kolgomorov entropy is square integrable on $L^2$

Stan Shunpike

18:38

If I have a finite set S, would intersections of all its subsets also be in S?

Mathphile

@geocalc33 thank you

@geocalc33 I have no idea what that means :D

I have a stronger conjecture for this

geocalc33

what is it?

Mathphile

${}^{\infty}x=a^n$ where $a$ is some rational number and $n$ is some natural number only when $x=1$

geocalc33

neat

Balarka Sen

@Alessandro Ya

Alessandro Codenotti

18:50

Do you want to help me sort out some topology confusion?

Balarka Sen

OK

Mathphile

@geocalc33 even stronger conjecture

Mad Spaces

$\int \! \vec{v^2} \, \mathrm{d}t$ how to integrate this integral ? where as $v$ is the velocity.

geocalc33

@Mathphile do tell :)

Alessandro Codenotti

@BalarkaSen Do you remember how for nice spaces $X$ there is a correspondence between compactifications of $X$ and closed subrings of $C_b(X)$ that generate the topology of $X$?

Balarka Sen

18:52

Yeah

geocalc33

@MadSpaces there's a relationship between velocity position and acceleration based on a differential equation

Mathphile

${}^{\infty}x=a^n$ where $a$ is any algebraic number not equal to $m^{1/m}$ for inegral $m$ and $n$ is some natural number only when $x=1$

geocalc33

derivative of position is velocity. derivative of velocity is acceleration

Mad Spaces

Yea i know that.

I just dont know how to integrate this integral at this moment. my brain is dead

Alessandro Codenotti

Ok so the setting is this, I have $f:X\to Y$, and I have two compactifications $cX$ and $cY$ associated to the rings $R_x\subseteq C_b(X)$ and $R_y\subseteq C_b(Y)$ respectively. I know that $g\mapsto g\circ f$ is a well defined map $R_y\to R_x$. Is that enough to get a map $cX\to cY$ induced by $f$?

Mathphile

18:55

I should phrase my last conjecture better

geocalc33

@MadSpaces what's the explicit function for $v(t)$?

Mad Spaces

There is no function. Its just this integral. Theoretical mechanics.

Oh wow nevermind i dont even need this integral.

Balarka Sen

I forget the correspondence. Given a closed subring of $C_b(X)$ generating the topology of $X$, how do you recover the compactification? Embed $X \to [0, 1]^{C_b(X)}$ by $x \mapsto (f(x))_f$, and then take the closure?

Mad Spaces

The velocity is constant. thus its not a function of t!

geocalc33

haha good catch

Balarka Sen

18:59

I guess you need functions to $[0, 1]$ which separate points for my thing to work

Alessandro Codenotti

@BalarkaSen rather than $[0,1]$ to get the compactification associated to $R$ you need $\prod_{f\in R}I_f$ where $I_f$ is a closed interval containing the image of $f$

Balarka Sen

Ah yeah

Nice

Alessandro Codenotti

Ah actually my question is much nicer to think about by considering the complex valued $C_b(X)$ and its C*-subalgebras, because then the associated compactifications are the spectra of the subalgebras and Gelfand-Naimark gives me the map I want

Yet it should work with the real valued rings too

Balarka Sen

Ah I don't know that story at all. I vaguely remember Stone-Cech is the spectrum of the $C^*$-algebra $C_b(X)$

Whatever spectrum of a $C^*$-algebra is

Alessandro Codenotti

The space of functionals

It gets a topology as a subspace of the dual with the weak* topology

Mathphile

19:05

@geocalc I have a question

If $x\ne 1$ is a rational number then will $x^{1/x)$ always be irrational?

geocalc33

I may or may not have an answer

Ted Shifrin

howdy, a @Balarka, demonic @Alessandro, et al.

Alessandro Codenotti

Hi Ted

Balarka Sen

Hi @Ted

geocalc33

howdy Ted

Balarka Sen

19:07

@Alessandro I might ask you to explain this story to me someday

Seems useful to know

Alessandro Codenotti

Sure, I'd be happy to

geocalc33

@Mathphile if a,b are rational and not equal to 1 then a^b is algebraic

if however a,b are algebraic...

then look at this en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem

Ted Shifrin

@Mad What does $\vec{v^2}$ even mean?

user714630

takes me back to last quarter

Balarka Sen

Lol @TedShifrin

Mathphile

19:11

alright

user76284

3

Q: What captures our intuitive notion of faces, edges, and vertices?

This answer suggests that laypeople's intuitive notion of the meaning of these words is consistent with the following claims: A cube has 6 faces, 12 edges, 8 vertices. A cylinder has 3 faces, 2 edges, 0 vertices. A cone has 2 faces, 1 edge, 1 vertex. A sphere has 1 face, 0 edges, 0 vertices. ...

geometry manifolds differential-topology tangent-spaces orbifolds

If anyone wants to share any thoughts.

Balarka Sen

Look up simplicial complexes

Ted Shifrin

I don't agree that a cylinder has $3$ faces, $2$ edges, $0$ vertices, etc. You have to give me a particular triangulation of the cylinder. And how can there be $0$ vertices?

user76284

Aren't those discrete? Or can you talk about them "under homeomorphism" or something like that?

Ted Shifrin

A sphere has $1$ face, $0$ edges, $0$ vertices? Huh?

Balarka Sen

19:16

Lol Ted you're great at spotting these stuff

I should pay attention

Ted Shifrin

nods in Balarka's general direction :D

user76284

Hmm well that's what I'm asking about: What's a good formal definition that can capture this intuition?

Ted Shifrin

But whatever intuition you're referring to ... it's wrong.

Balarka Sen

0 vertices does not make sense, so some of those claims are baloney, is Ted's point

user76284

Qiaochu suggested the tangent spaces definition.

@TedShifrin What is wrong?

Ted Shifrin

19:17

You need a triangulation, or a simplicial complex, or a $\Delta$-complex.

Qiaochu made some reference to manifolds and manifolds with boundary. He didn't begin to justify this nonsense.

geocalc33

what captures our intuitive notion of faces edges and vertices? You absolutely must define the "idea" mathematically and precisely like ted's saying @user76284 you can speak all day about "intuitive notion" but at the end of the day it must be mathematically justified

user76284

Why are you calling it nonsense? That seems rude to me. They were brainstorming ideas for what kind of formal definition would capture this intuition.

@geocalc33 It's pretty clear what I'm asking.

Ted Shifrin

You can think of a sphere, for example, as a blown-up tetrahedron (make it out of balloon and then blow air into it), or as a blown-up cube, or as a blown-up octahedron. All of these are valid triangulations or simplicial structures.

I don't see any way to make a sphere out of one face and nothing else.

user76284

Please don't derail the conversation, geocalc. I made it very clear what I'm looking for: What formal, mathematical definition best captures this intuition and is consistent with the above claims? A lot of math starts like this: You start with some intuitive notion (the natural numbers, space, distances, sets/collections, smooth manifolds, etc.) and you put it on rigorous footing.

Ted Shifrin

So go read about triangulations and simplicial complexes. This was done over a hundred years ago.

Mike Miller

19:21

@SanchayanDutta You must have $f\sigma(1) = 1$, and because $\sigma$ is a homomorphism you have $p\sigma(1) = 0$. But what does the map $f: \Bbb Z/p^2 \to \Bbb Z/p$ do to elements of order $p$ in the domain?

Balarka Sen

@user76284 I looked at the question you linked carefully now. I see what JDH's answer is trying to capture; these are not called "faces", "edges", "vertices" in mathematics.

They have different meaning. What they are trying to capture is the notion of boundaries, corners, singularities on manifolds.

I understand what you mean now.

user76284

@TedShifrin What do you think of Qiaochu's comment "I believe it's true that the naive count of faces, edges, etc. counts connected components of the subspaces of points whose tangent spaces have the relevant dimensions"?

Alessandro Codenotti

@BalarkaSen I think you meant $\mathsf{Set}^{\Delta^{\mathsf{Op}}}$

Balarka Sen

Lol

Anonymous

@MikeMiller I'm using the natural homomorphisms as detailed here...

Leaky Nun

19:25

@AlessandroCodenotti oh no

Ted Shifrin

So it's purely descriptive and nothing to do with topology. That's the first observation. Topologically, a cube is the same as a sphere. Now, if we talk about smooth structures, then we are forced to talk about stratifications of these objects by various lower-dimensional smooth objects. That's what Qiaochu is referring to.

Mike Miller

I know what they are.

I'm asking you to tell me the answer.

Ted Shifrin

I wonder what the kid's teacher would have said to a smart-aleck kid who said he could blow air into a cube and turn it into a sphere.

Balarka Sen

LOL

user76284

Right, it seems to me like a promising start. Intuitively it seems to match what I think of when I think of "faces", "edges", etc. in this context.

Mike Miller

19:27

I think that's unfair. Smooth topology is still topology.

user76284

(Where we're no longer dealing with polytopes only.)

Ted Shifrin

It's bad, though, even with little kids, because Euler's Theorem $V-E+F=2$ does show up in the curriculum as a fun thing ... and there it means actual triangulations.

@MikeM: Stratifications aren't topologically invariant.

Mike Miller

I know, and don't care. They're still topology.

Balarka Sen

@user76284 So just to hit the word home, the right word here is "stratification by singularity type".

user76284

I'll look into that, thanks.

Balarka Sen

19:29

You might enjoy looking up the formal definition of a stratification

Mike Miller

@BalarkaSen It's interesting to ask how to get the "dimension" of a singularity type from its tangent cone

Is that totally obvious? You take the set of extreme points of the cone and ask for its dimension?

Anonymous

@MikeMiller I guess the map $f$ takes the elements of order $p$ in $\Bbb Z/p^2$ (i.e., $G$) to the identity $0$ in $\Bbb Z/p$ (i.e., $B$).

Balarka Sen

The cone will have various extreme rays of different dimensions, it seems to me

Ted Shifrin

Hmmm ... so I'm confused. If I look at an ordinary double point or a tacnode, the tangent cones look different. But how are they detecting the point?

Mike Miller

On the vertex of the compact cone I see that the tangent cone is a noncompact cone, with a single extreme point

Ted Shifrin

19:33

I guess it depends if we're doing algebraic or semialgebraic. I.e., is a cone a true cone or one nappe thereof?

Balarka Sen

@MikeMiller OK, so I am not entirely sure how to phrase this question yet, but if $X$ is stratified of top dimension $n$ and $S$ is a stratum of dimension $k$, it will locally look like $c(A) \times \Bbb R^k$ where $A$ is an $n-k-1$-dimensional stratified set

aka $A$ is the link of $S$ in $X$

The extreme points of the tangent cone are that $\Bbb R^k$ direction, right?

so that does read off the dimension of the stratum

@TedShifrin Tacnodes seem like interesting examples. They're apparently one in Arnold's list of classification for germs of smooth functions R^2 -> R

user76284

Regarding my question, is there a connection to orbifolds?

Balarka Sen

the $A_k$-type singularities

Anonymous

@MikeMiller I suppose my basic confusion is about the definition of a section of a surjection in the category of sets. I initially thought a section $s: B \to G$ is necessarily a group homomorphism, but that's apparently not the case (because in that sense a non-split extension can't have a section at all). In other words, I think a section can be any set of one representative element from each coset of $A$ in $G$.

Mike Miller

Yes, non-split extensions don't have sections.

user76284

19:41

@BalarkaSen arxiv.org/pdf/1409.0501.pdf "Note this also encodes the singularity type found at a corner of the surface of a cube." Is this what you were referring to?

Balarka Sen

@user76284 Orbifold singularities are a class of singularities, sure. They capture more information than just "topological"; in the following sense. For example, you can cut out a small sector of the plane with vertex angle $2\pi/n$ and then glue to a cone.

For different $n$'s, they all look the same: they are cones. But in the orbifold world you keep track of this angle at the singularity

Mike Miller

@BalarkaSen I guess I can't get the statement about extreme points from this unless I want to exclude eg the hourglass singularity

If we just work in convex smooth sets I think maybe you can do my extreme point definition

Balarka Sen

Ah fair I see

Ted Shifrin

@Balarka: Various of my joint papers are full of these singularity classifications ... :P

Anonymous

@MikeMiller Indeed. However, my professor seems to define sections in the category of sets. And since 2-cocycles can be cohomologously varied by choosing different "sections" I'm trying to compute the number of elements in an equivalence class of a 2-cocycle in this context. Basically, I'm trying to see if there is any direct method to compute $|B^2(B; A)|$ from first principles.

Balarka Sen

19:44

@Ted I have been struggling so hard to understand these things. Can't make an actual headway without billions of germinal commutative diagrams

user76284

@BalarkaSen If I'm understanding correctly, we want to "quotient over" (in the sense of equivalence classes) those angles so that cones with different angle deficits are treated the same. (Of course the cone isn't infinite so you might say something like the unit disc instead of the plane.)

Ted Shifrin

We have all sorts of fun pictures in one of the JDG articles.

user76284

Is that what you mean?

Balarka Sen

@user76284 In orbifold world you do keep track of the angle deficits, though, is my point.

Two cones with different angle deficits are not the same orbifolds, formally speaking

user76284

Yeah, that's why I said we want to forget that information.

Balarka Sen

19:46

Who's "we"? :)

It depends on what is the right framework for singularities for you

user76284

I mean in the context of my question :P

Balarka Sen

Ah, I suppose, yeah.

Anonymous

@MikeMiller Well, since there are $p$ possible representatives of each coset corresponding to $1, 2, 3, \ldots, p-1$ in $\Bbb Z/p$ (i.e., $B$), would it be sensible to say that there are $p^{p-1}$ possible 2-coboundaries or $|B^2(B; A)| = p^{p-1}$. And for some reason I guess $0 \bmod p$ can only be mapped to $0 \bmod p^2$, but I'm not sure about this.

Balarka Sen

@TedShifrin I wanted to talk about orbifold Gauss Bonnet for surfaces in the math camp talk thing but thought it might be too difficult

So I stopped at combinatorial Gauss Bonnet

Mike Miller

@SanchayanDutta I don't much like using the word section of a group homomorphism to mean in the sense of set map without explicitly stating it, but I understand your point now.

Anonymous

19:54

@MikeMiller I agree with you, and that is precisely what confused me while reading the lecture notes. Nevertheless, do you have any idea about whether my computation of $|B^2(B; A)|$ makes sense (above)?

user76284

This sounds like "differentiable manifolds with kinks".

Balarka Sen

That's exactly what Whitney stratified spaces are :)

Kinks of various dimensions, mind you

And maybe there are kinks inside kinks

user76284

math.berkeley.edu/~qchu/Notes/274-2/Lecture2.pdf

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