Mathematics - 2019-06-25 (page 1 of 3)

Karl Kronenfeld

12:03 AM

@rschwieb Heh, what's the context?

Ultradark

Hey @anakhro everyone

anakhro

Hey ultradark.

Ultradark

I just couldn’t solve the conjecture

anakhro

Conjecture of what?

Ultradark

A conjecture about @anakhro; -; the: flow of information from place to place being necessary for the universe to exist. I’ve given up

And I’m never returning to it

anakhro

12:10 AM

what does that even mean?

Ultradark

It’s

Related to the holographic principle in Physics

according to the H lophic

Principal

The universe is constructed only from the flow of info

anakhro

Yeah I agree, throw this idea in the trash

Ultradark

Yeah

Whoever came up with that idea

Is a crank

There are about 500 of them

anakhro

ya

Ultradark

Leaders in their field

They’re not cranks lol.

anakhro

12:16 AM

Cranks are allowed to have their own subfield of crankery.

Ultradark

Okay

Yeah maybe they are a bit crank

Like 30 percent

anakhro

Can you link me to something on the arxiv?

Ultradark

Sure

GFauxPas

12:31 AM

What's a group presentation for $SO(3)$?

Ted Shifrin

I have no idea.

Is it even finitely presented?

GFauxPas

I should have asked that first

Ted Shifrin

Why do you want a group presentation?

GFauxPas

Idle curiosity

Ted Shifrin

Oh.

I would be shocked if it could be finitely presented. Hard to see how composing a finite number of rotations could give rotations through arbitrary angles around arbitrary axes.

GFauxPas

12:35 AM

The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.

Ted Shifrin

Yes, I know that result.

GFauxPas

I didn't but it makes sense

Ted Shifrin

Compute it in the plane first.

anakhro

It's a great theorem.

Ted Shifrin

But using reflections, my complaint still stands.

I actually don't know the Cartan-Dieudonné Theorem. What sayeth it?

GFauxPas

12:37 AM

Yeah it does seem unreasonable to expect a finite presentation

Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.

Ted Shifrin

I would hazard a guess that it's not even countably presented, but someone might show me I'm wrong.

Oh, cool.

I see: That's the generalization of the physicists' Euler angles in $\Bbb R^3$.

Oh, no, never mind. That isn't right.

That would say you can write every rotation as a composition of ($n$) rotations about the coordinate axes.

I doubt that's true for $n>3$.

These are all good questions for an industrious undergraduate to investigate.

Daminark

12:54 AM

Hello nerds

Ted Shifrin

Hi nerd Demonark.

rschwieb

@KarlKronenfeld Just filling a gap in my database , and I couldn’t believe it hadn’t been filled earlier.

Daminark

How's everything going?

N. Maneesh

1:28 AM

1

Q: How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?

How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself? Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...

differential-geometry

anakhro

I can't even parse that title.

N. Maneesh

I got this question from a lecture video.

Differential Geometry | Math History | NJ Wildberger

►

17:04

He is not explaining completely.

anakhro

1:44 AM

Did he actually say the title anywhere

Because I still can't parse it.

N. Maneesh

yes. !7:04

evolute of the involute is the curve itself

anakhro

Do you mean to ask "how is the evolute of an involute of a curve $\Gamma$ itself?"

N. Maneesh

yes

for evolute, we have a fixed point on the curve right?

then we consider tangent to each point right of that

Involute is the curve that is perpendicular to each tangent line.

sorry, for involute we have fixed point.

then we consider tangent to each point right of that
Involute is the curve that is perpendicular to each tangent line.

For finding the Evolute of this Involute, we need to find the centre of curvature at each point.

anakhro

I have never heard of an evolute or involute, sorry.

N. Maneesh

okay

Kenkar

6:57 AM

Z

Thanks @TedShifrin for answering

ÍgjøgnumMeg

7:34 AM

Mooornin'

Grešnik

Good morning, we are trying to solve some games on infinite chessboard in one room, I will post them here also, it could be that someone finds them interesting.

3

Q: $6$ bishops and a knight on an infinite chessboard

Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants. Then $A$ makes a move, then $B$, and so on... The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...

combinatorics game-theory chessboard infinite-games

11

Q: A simple game on infinite chessboard

Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...

combinatorics game-theory chessboard infinite-games

Grešnik

7:57 AM

Here is a room:

https://chat.stackexchange.com/rooms/95333/discussion-between-peter-and-gresnik

Adam

11:43 AM

I get a 6 day ban for telling someone to shut up?

fair enough I am socially inept in the ways everyone else I've met are not, but they are also socially inept in the ways I am not

Albas

11:56 AM

The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms

Balarka Sen

@Albas What is the "invariant formula for the exterior product"

Albas

That should be exterior derivative. en.wikipedia.org/wiki/Exterior_derivative look under in terms of the invariant formula

Balarka Sen

Exterior derivative, not product!

So this is just an infinitisimal Stokes theorem

Think about what is happening for $n = 2$. $d\omega(X, Y) = X\omega(Y) - Y\omega(X) - \omega([X, Y])$ for a $1$-form $\omega$

Albas

@BalarkaSen Yea sorry I always confuse that the naming. I guess its because of the interior product

Balarka Sen

If you have two vector fields $X$ and $Y$ on your manifold, then you have the following formula relating the flows of $X$, $Y$ and $[X, Y]$:

Call $\Phi_t$ the flow of $X$ for time $t$ and $\Psi_t$ the flow of $Y$ for time $t$. Then $[X, Y] = d/dt \Phi_{\sqrt{t}} \Psi_{\sqrt{t}}\Phi_{-\sqrt{t}} \Psi_{-\sqrt{t}}$.

Albas

12:12 PM

Yeah

Balarka Sen

This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.

Albas

Yea measures the noncommutativity of the flows

The lie bracket that is

Balarka Sen

So there's an infinitisimal truncated square spanned by $X$ and $Y$ which is really a pentagon with an edge really small that represents $[X, Y]$

Albas

Yup

Balarka Sen

Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$

Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$

So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$

Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$

Albas

12:18 PM

Thats very nice. I have a doubt, what do you mean evaluating $\omega$ on the edge. As in $\omega$ takes in the vector field along that edge?

Balarka Sen

More or less, yeah. Think intuitively instead of too rigorously though

Albas

Cool cool

Balarka Sen

But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$

Didn't try it out but it should work out

Albas

Okay will try it out.

Balarka Sen

For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube

Albas

12:22 PM

A "multidimensional" box

Balarka Sen

@Albas Your analogy with the covariant derivative is apt. What does $X\omega(Y) - Y\omega(X) - \omega([X, Y])$ remind you of in Riemannian geometry?

Albas

Something like torsion?

Balarka Sen

Exactly.

Albas

Ohhhhhhhhh

Awesome man

Balarka Sen

Actually you can make this purely formal

I can tell you if you're interested

Albas

12:26 PM

Sure

Balarka Sen

Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.

Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$

Albas

Makes sense

Balarka Sen

You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point

Albas

Yeah that's how a connection when defined for a riemannian manifold behaves. So I guess a generality should also follow that

user193319

Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).

Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...

Balarka Sen

12:38 PM

@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.

This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.

You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.

I'll use the latter notation consistently if that's what you're comfortable with

Albas

Yea kind of makes sense. So you're fixing your s so that the map now varies with $X$. This is somewhat similar like saying that $\nabla Y$ is $(1,1)$ tensor

Balarka Sen

Yeah precisely

$\nabla s$ is a bundle homomorphism $TM \to E$, sending $X$ to $\nabla_X s$. That's the (1, 1) tensor in your context.

And then I also "contract $s$" to just make a map $d^\nabla : \Gamma(E) \to \Gamma(\text{Hom}(TM, E))$, defined by $d^\nabla(s) = \nabla s$

Albas

Yeah

Balarka Sen

(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)

Albas

Ah yes.

Balarka Sen

12:46 PM

@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)

Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms

So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.

Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.

Makes sense so far?

Albas

Yeah. It will take me some time to take all this in. There's a lot of notation to swift through

Balarka Sen

Let me know when you're ok so we can continue :)

Albas

Cool

user193319

It seems that $orb(u) \neq orb(v)$ is equivalent to there existing a $g \in G$ such that $g v \in T'$. Freeness of the action needs to be used somehow, but i don't see it.

Balarka Sen

My next plan is to define higher exterior derivatives $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ for $k > 0$, but only after this formalism is clear.

Alessandro Codenotti

12:51 PM

@user193319 if the orbits were equal then $gu=v$ and $hv=u$ for some $g,h\in G$. Then $hgu=u$ and $ghv=v$, since the action is free $h=g^{-1}$. Set $w=gv$. Then $hw=u$, thus $w=v$, a contradiction

user193319

@AlessandroCodenotti Ah, very nice. Thanks!

Albas

Okay make sense let's continue @BalarkaSen. There's this itch in me which makes me think that can we define some crazy generalization of De Rham cohomology here.

Balarka Sen

@Albas Alas, we won't have any de Rham cohomology. But not having one would precisely be the point of interest.

Ryan Unger

Oh no what is happening in here

Albas

Ah I guess $(d^{\nabla})^2$ won't be $0$

Ryan Unger

1:00 PM

You get a curvature term

Balarka Sen

It would be the Riemann curvature tensor, yep

user193319

@AlessandroCodenotti Wait, why does $hw = u$?

Balarka Sen

@Albas Okay! So here is a definition of the higher exterior derivatives $d^\nabla : \Omega^k(M, E) \to \Omega^{k+1}(M, E)$ takes a $E$-valued $k$-form $\omega \in \text{Hom}(TM^{\wedge k}, E)$ to $d^\nabla \omega \in \text{Hom}(EM^{\wedge k+1}, E)$ defined by $d^\nabla \omega(X_1, \cdots, X_{k+1}) = \sum_i (-1)^{i+1} \nabla_{X_i} \omega(X_1, \cdots, \hat{X_i}, \cdots, X_n) + \sum_{i < j}(-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_n)$

Broke the chat but you get my point

This is exactly the invariant formula for higher exterior derivatives, modified to fit the scenario

Albas

Yeah

Alessandro Codenotti

@user193319 Ah I got confused $hw=v$

Whatever, assume the orbits intersect and play around with the action until you get an edge inversion of $e$ I think

user193319

1:06 PM

What is an edge inversion?

Balarka Sen

@Albas Note how $\omega(X_1, \cdots, \hat{X_i}, \cdots, X_n)$ is a section of $E$, so I am doing $\nabla_{X_i}$ of that

Alessandro Codenotti

An elelement swapping $u$ and $v$, thus fixing an edge

Albas

@BalarkaSen Yea its well defined like that

Balarka Sen

That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection

Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$

Voila, Riemann curvature tensor

Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature

Albas

This is super neat

Balarka Sen

1:11 PM

Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?

Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.

Albas

Umm let me get this wait

Alessandro Codenotti

Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.

Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?

Balarka Sen

Just like $ds = 0$ for a section $s \in \Gamma(E)$ would mean $\nabla_X s = 0$ for all vector fields $X$ on $M$ would mean $s$ is a "parallel section". Think in Riemannian geometry, for example.

Albas

Oh okay, yeah

Balarka Sen

If $ds(X) = 0$ then $\nabla_X s = 0$ i.e., $s$ doesn't change if you parallel transport along $X$.

Albas

1:15 PM

Yeah.

Balarka Sen

OK, let's wrap this up and interpret the torsion tensor in this formalism

Albas

Yeah, this is a lot to think about

Balarka Sen

Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.

You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form

Albas

Yeah on the cotangent bundle

Balarka Sen

(The cotangent bundle is naturally a symplectic manifold)

Yeah

So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.

But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!

Albas

1:21 PM

Wow

Where did you learn all this

Balarka Sen

$d\theta = 0$, or vanishing of torsion, is exactly the condition that the soldering form is parallel with respect to the connection

Albas

Like torsion free connection

Balarka Sen

Which is natural, I mean: you'd want the identity homomorphism $TM \to TM$ to be parallel in whatever geometry you're working on. Roughly speaking it means the space doesn't "twist in itself"

@Albas I dunno, I saw it somewhere on MO once. Picked up bits and pieces of bundle-valued differential forms by working it out by hand. This is all language game at the end of the day

But a powerful one I think

Albas

Yea kind of enforces Cartan's thing that differential forms are supreme

They are everywhere

Balarka Sen

Lol well I am very bad at differential forms to be honest

This is just something cool without having to do any local computations by hand. That's what bundle geometry helps you to do, avoid local computations as long as possible

Albas

1:25 PM

Yea, computations especially here will kill you

I remember trying to find killing vector fields for various kinds of metrics. It was horrendous. But Killing Vector fields are nice

Balarka Sen

I don't know any explicit calculations man

Albas

They are nice man. If you find the Killing vector fields for the Minkowski space, I think they give you the generators of the Poincare algebra.

I really have to thank my physics background for providing me motivation to do calculations

Balarka Sen

I have 0 clue about semiRiemannian geometry. Killing fields are the ones which integrate to geodesics, right? So, what, the flows are isometries and for R^{3, 1} the pointwise isometry group is O(3,1)

The global isometry group is that semidirect product R^4 or whatever

You'd get the Lie algebra of R^4 $\rtimes$ O(3, 1)

Albas

So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up

Balarka Sen

The space of Killing fields on $M$ is just the tangent space at identity (or the Lie algebra) of $\text{Isom}(M)$ (which is a Lie group by a famous theorem)

at least when $M$ is Riemannian

Albas

1:33 PM

@BalarkaSen Yup exactly, thats why when you find the vector fields you see that they are the generators of the Poincare algebra which is just the Lie algebra of O(3,1) the lorentz group

@BalarkaSen Oh did not know that

Gtg then, thanks for this @BalarkaSen

Balarka Sen

I don't know much about the Poisson bracket, seems interesting

Cya

@Albas $S^2$ is not a symplectic manifold!

Try torus, maybe, if you want model examples

The standard symplectic structure on $\Bbb R^{2n}$ descends to $T^{2n}$

Albas

@BalarkaSen $S^2$ is, its the only even dimensional sphere to satisfy the cohomology condition

Balarka Sen

Oh yeah sorry rip the volume form of course lol

Albas

Yeah

Balarka Sen

Mentally replaced that with $S^{2n}$, smh

Albas

1:44 PM

Lol. S^2 even has the almost complex structure which originally made me think about symplectic manifolds without Kahler structure.

Balarka Sen

It has a complex structure. It's CP^1

CP^n are the prime examples of Kahler manifolds

In particular it means any smooth projective algebraic variety are also Kahler, because the Fubini-Study metric on CP^n restricts

@Albas You should ask that question to Ted, he would know an example

Albas

Oh cool, I will. The question has actually peaked my interest, let me see if I can construct some example on my own. Most probably won't, I don't even know the theory of complex manifolds properly

Balarka Sen

Idk any of this

just barely understand out of folkloric knowledge

Albas

You know more than I do man. I mean you have the language, now you just have to sit with the material.

Ryan Unger

@ÉricoMeloSilva its not possible to ask Alice Chang a question

The lecture must go on uninterrupted

Albas

1:55 PM

And the point is not to compare how much one knows. Just have to learn I guess

Balarka Sen

sure its whatever just learn because its fun

ill have to read symplectic geometry seriously at some point

Albas

Exactly.

You at TIFR now?

Balarka Sen

just need to read these 50 things before that

@Albas not yet

i leave on 30th

Albas

Hmm for a month?

Balarka Sen

a little less than that

3 weeks more or less

Ryan Unger

1:58 PM

@BalarkaSen come to Princeton. Pardon is teaching it next semester

Balarka Sen

o shit

Albas

I have a couple of friends who are doing something with elliptic curves there. Went through VSRP.

Balarka Sen

aha nice

Ultradark

Yo what’s up chat. Anyone know an inter sing result in symplectic geometry? Like geodesic and the metric on the manifold maybe torus or that possibly convex thing?

*interesting

Ryan Unger

I actually have no idea what it’s about

Ultradark

2:09 PM

Oh :( that’s okay though

You know a lot lol

Ryan Unger

Not enough

Ultradark

Spoken like a true champ

Érico Melo Silva

@RyanUnger y

Adam

2:27 PM

@BalarkaSen lol @ bullshit generality.

Ultradark

Say hwat’ Adam @Adam

Say something

Arcane

Something modern.

Secret

> There exists an infinite geodesic that is geodesically complete

Ultradark

Really?

That is indeed something modern

you modernist

@Secret how do you know so much math if you’re a PhD chemistry student

I’m perplexed. I really am. My brother barely knows basic cell biology and his IQ is nearly 106!

Secret

You just learn along the way. Passion makes a difference

Ultradark

I think I’m going to purchase a text book about that geodesic thing you said. Where did you read that fact?

Passion!

Yes you are right.

That’s awesome though

Not many people are that adventurous

Ryan Unger

2:46 PM

@ÉricoMeloSilva there’s a good chance she’ll just ignore you

Or she’ll say something like “did I make a mistake? Sorry” and just carry on

Balarka Sen

lol

Ryan Unger

Paul Yang didn’t bring the notes so I guess I just won’t understand anything

Daminark

3:38 PM

Hey everyone!

Balarka Sen

Nerd alert

Daminark

Where?

Alessandro Codenotti

Hi

Balarka Sen

There ^

Alessandro Codenotti

D:

Kenkar

4:03 PM

If someone has the time to quickly check my result, I would appreciate.
Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$
Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?

$\frac{1}{n\lambda^2}+\frac{2}{\lambda^2}$ *

Alessandro Codenotti

Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty

Adam

@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method

anakhro

4:24 PM

@Ultradark the non-squeezing theorem is one of the defining problems of symplectic geometry: en.wikipedia.org/wiki/Non-squeezing_theorem

It's one I always like to tell people who don't know anything about math who wonder about symplectic geometry and what it means.

Balarka Sen

Oh cool this is nice

This is very nice lmao. Gromov is life

How the heck did he think of proving something like this

anakhro

The linear one is a little bit easier to fathom.

Balarka Sen

Yeah I believe that

anakhro

From Mcduff & Salamon, by the way.

Balarka Sen

I need to read that someday

anakhro

4:33 PM

I need to finish it some day. :(

looks at his bookcase sadly.

user193319

I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.

My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?

Balarka Sen

$d = 6$ shouldn't be hard, it's an embedded $S_3$ in $S_4$!

Maybe it's easier if you think about $S_4$ as the orientation preserving symmetry group of the cube.

user193319

which is isomorphic to the subgroup generated by $(1,2)$ and $(1,2,3)$

That's why I said "semblance of trouble"

Balarka Sen

Yeah but I mean you shouldn't have to think more than 2 seconds or even write an explicit generating set

It's just a stabilizer subgroup of the usual action of $S_4$ on a set of size $4$ by permutation

user193319

"orientation preserving symmetry group of the cube" that doesn't sound much easier.

Balarka Sen

4:44 PM

It's immediate from that what the $d = 8$ guy is. Take the equatorial cube, it's symmetry group is a $D_8$ embedded in $S_4$.

No Sylow theory required

The $d = 12$ guy is the symmetry group of an embedded tetrahedra in the cube, which is an $A_4$ in $S_4$

You can list down all the groups purely geometrically from this

user193319

Well, the only reason why I used sylow is because this problem is part 3 of a problem, where in part 2 I showed that S4 has a sylow 2 subgroup of order 8 (three actually).

Balarka Sen

Sure

user193319

What is an equatorial cube?

Balarka Sen

Equatorial square I meant, whoops

Think of the cube as $[-1, 1]^3$ and slice by $z = 0$

user193319

Ah, okay.

Thanks!

Balarka Sen

4:48 PM

You can also slice by $x = 0$ and $y = 0$, which gives three squares inside the cube, which correspond to those three nonconjugate 2-Sylow $D_8$'s in $S_4$ that you speak of!

Semiclassical

equatorial square is a nice phrase, i may have to steal it for my own purposes

as an example of that kind of usage: the elliptope, defined by $x^2+y^2+z^2\leq 2xyz+1$ in the [-1,1]^3 cube, looks like this

images.app.goo.gl/PAQDXBvMWjJt25uA9

and, amusingly, the three equatorial cross-sections are all circles

e.g. $z=0\implies x^2+y^2 \leq 1$

Balarka Sen

In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.

Semiclassical

(the non-equatorial cross-sections are all ellipses, hence why it's called theh elliptope)

Balarka Sen

Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group

Everything about $S_4$ is encoded in the cube, in a way

The same can be said of $A_5$ and the dodecahedron, say

Ted Shifrin

Hi, a @Balarka. You're channeling Artin :)

Semiclassical

4:54 PM

and A_4 for the tetrahedron, etc

Balarka Sen

Hi @Ted! Yup

Ted Shifrin

This was one of my favorite parts of my algebra course when I used to teach it :P

Balarka Sen

Haha I love it a lot

Semiclassical

hmm. Now I'm wondering if every finite group is the symmetry group of some polyhedron (of appropriate dimension)

Balarka Sen

certainly not regular

Semiclassical

4:57 PM

Not seeing why that restriction is obvious

Balarka Sen

if $G$ is symmetry group of a regular polyhedron in $\Bbb R^{n+1}$ say then $G$ acts on $S^n$ freely I think

Because embed the regular polyhedron in a sphere

but groups acting freely on spheres have nontrivial cohomological restrictions

Hm no reason for the action to be free I suppose

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