Mathematics - 2018-04-10 (page 1 of 4)

Zee

00:00

i dont think nonlinear theory had any progress

0celo7

@MatheinBoulomenos Interesting, I had thought it was pretty much disjoint from the rest of math

@Zee wrong

we know global existence for lots of equations, but the proofs are all ridiculous

Zee

let me ask you this

and it may sound silly but its actully something i think about often

MatheinBoulomenos

@0celo7 yeah, but as I said, I don't know how much PDE theory is actually involved, but it seems likely that there is some given the definition as eigenfunctions of the hyperbolic Laplacian

0celo7

@MatheinBoulomenos do you mean the wave operator/d'Alembertian?

MatheinBoulomenos

hmm, I have to look it up

Zee

00:02

lets say i give you this machine that answears everything you wanted to know about PDEs, now what?

0celo7

I kill myself, probably

Mission complete

Zee

thats all you are? a PDE solver?

0celo7

No, but you said everything I want to know

MatheinBoulomenos

$\Delta_k = -y^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) + i k y \frac \partial {\partial x}$ where $k \in \Bbb N$ is fixed, that's apparently called $k$th hyperbolic Laplacian on $\Bbb H$

0celo7

that's a lot of stuff

@MatheinBoulomenos um, are you sure they mean hyperbolic in the wave sense and not hyperbolic in the -1 curvature sense

Zee

00:04

why do you wanna know so much about PDEs?

MatheinBoulomenos

@0celo7 no idea, it's certainly possible that there is no connection to hyperbolic PDE at all and I got confused

0celo7

lol ok @MatheinBoulomenos

can you link me?

I should be able to tell

@Zee idk, they're cool

MatheinBoulomenos

Maass cusp form

In mathematics, a Maass cusp form, Maass wave form, Maaß form, or Maass form is a function on the upper half-plane that transforms like a modular form but need not be holomorphic. They were first studied by Hans Maass in Maass (1949). == Definition == Let k be an integer, s be a complex number, and Γ be a discrete subgroup of SL2(R). A Maass form of weight k for Γ with Laplace eigenvalue s is a smooth function from the upper half-plane to the complex numbers satisfying the following conditions: For all γ = ( ...

0celo7

Ok so I'm pretty sure they mean hyperbolic in the Poincare disk sense.

Upper half plane corresponds to the Poincare disk with a certain metric

Zee

@0celo7 ask yourself why, am not saying their not, but do that tonight in bed

user193319

00:07

I'm trying to use Lebesgue theorem to show that if $f$ is Riemann integrable, then $f^2$ is Riemann integrable. If $x$ is a point of discontinuity for $f^2$, will it be a point of discontinuity for $f^2$?....Arguing contrapositively, if $f$ is continuous at $x$, then $f^2$ is continuous at $x$, because $f^2$ is a product of functions which are continuous at $x$ (alternatively, $f^2$ is the composition of two functions which are continuous at $x$). Does this sound right?

0celo7

@Zee I've thought about it a lot and my only answer is they're cool

MatheinBoulomenos

yeah, but isn't the "diff geo Laplacian" (sorry for the weird terminology, I know no geometric analysis) on a negative-curvature manifold actually a hyperbolic operator?

0celo7

no! the Laplacian is always elliptic on a Riemannian manifold

MatheinBoulomenos

oh lol

0celo7

and the term is "Laplace-Beltrami op" if you want no confusion

MatheinBoulomenos

00:09

yeah then forget everything I said

0celo7

ok

I was going to say that hyperbolic PDE is kind of a lonely world because, AFAIK, it has no applications outside of itself

applications to GR doesn't count, that's part of the field

MatheinBoulomenos

really confusing to call an elliptic operator hyperbolic

0celo7

presumably the guy who did it wasn't a geometric analyst :)

@MatheinBoulomenos btw I am at least taking some algebra next year (based on Hungerford)

Eric Silva

I guess there's just convergence of bad terminology going on there

MatheinBoulomenos

@0celo7 sounds good! (though I haven't read the book)

0celo7

00:13

It looks pretty basic

MatheinBoulomenos

@EricSilva yeah, people just use hyperbolic/elliptic/parabolic for so many different stuff

0celo7

I'm also taking 3 analysis courses to balance out :P

and a topology class for fun

Eric Silva

I'm only taking algebra next year

Unless a topics course pops up

MatheinBoulomenos

I'm taking one number theory course which is a bit analytic, on modular forms and I'm taking 4 algebra courses to balance it out

0celo7

nice

My reading course will probably take most of my time. We're going to make an effort to understand Morgan and Tian's book in some detail

MatheinBoulomenos

00:15

@EricSilva what exactly?

Eric Silva

@MatheinBoulomenos our graduate algebra sequence, it's rep theory first, then algebraic geometry, then I have no idea

Daminark

If Kato's teaching, which I think he usually does, it's a good bit of algebraic number theory/geometry

Zee

@0celo7 what kinda of material would a course called "gluing constructions of cannonical metrics" entail ?

Eric Silva

I'm down

0celo7

@Zee Not sure but I would take it. I'm thinking about various gluing constructions along black hole boundaries

MatheinBoulomenos

00:17

@EricSilva really nice! (though I like all kinds of algebra I've encountered so far)

0celo7

I regret having not done more algebra

Eric Silva

I like algebra and havent taken enough

0celo7

But I was a double major for 2.5 semesters which really limited my time

Zee

@0celo7 I mean is that stuff PDEish or global or...?

0celo7

In hindsight I should have done everything differently

@Zee do you not have a syllabus

Zee

00:19

No, nothing is posted aside from the title

0celo7

It could even be algebraic geometry, I don't know

Who is teaching it?

Zee

ill just wait and see, it seems too advanced for me though

Micheal T Anderson

0celo7

Right, he's going to be doing GR

Zee

what makes you say that?

0celo7

That sounds like a very specific course

@Zee Because I know Anderson and one of his students

Zee

00:22

lol , i mean i know him too

0celo7

Look up "Corvino-Schoen gluing", that might be something he's thinking about

Adeek

do you know the following @0celo7

Zee

@0celo7 do you know the man or the work?

Adeek

Let X be a compact Riemann surface associated to $w = \sqrt( (z - z_1) \ldots (z - z_N) )$ where $z_1,\ldots,z_N$ are distinict. Find the formula for genus in terms of N

Zee

and ill loook that up, thanks

0celo7

00:23

@Zee The work, but I know one of his students personally.

Zee

@0celo7 thas funny, i know the man but not the work

0celo7

@Zee Holy shit he solved the Nirenberg problem this year

Ah, not a full solution

Dang

Zee

he is probably the nicest professor i ever had

very humble

0celo7

@Zee Unfortunately there's two Andersons in the same field so I can't say exactly that I know this one's work

I think I know the other one better

Zee

makes alot of mistakes adding fractions lol

0celo7

00:25

@Adeek why don't you ask @BalarkaSen, he knows

Zee

whats this GR gluing

0celo7

@MatheinBoulomenos It's a little strange how useful elliptic and parabolic PDE are for math in general but hyperbolic is kind of on its own and is the most difficult of the bunch

Zee

didnt even know it exsisted

0celo7

@Zee Chapter 1 link.springer.com/book/10.1007%2F978-3-0348-7953-8

Zee

ehhh looks too complicated

so whats the essential geometric idea behind this gluing stuff

what in gods name is mathematical relativity

orbit-stabilizer

00:33

God is dead

He remains dead

And we have killed him

:(

MatheinBoulomenos

How shall we comfort ourselves, the murderers of all murderers? What was holiest and mightiest of all that the world has yet owned has bled to death under our knives: who will wipe this blood off us? What water is there for us to clean ourselves? What festivals of atonement, what sacred games shall we have to invent? Is not the greatness of this deed too great for us? Must we ourselves not become gods simply to appear worthy of it?

orbit-stabilizer

RIP Nietzsche

Zee

God does not do miracles , he does more , he turns sinners into saints

orbit-stabilizer

He had a very flamboyant writing style.

Zee

no i actully like him

although i dont agree with him

i dont think he agrees with himself

MatheinBoulomenos

00:38

@Zee yeah, he was a fool who on a bright morning lighted a latern and ran to marketplace

Zee

is there an honost man?

orbit-stabilizer

@Math, what's that a reference to?

Thus Spoke Zarathustra?

MatheinBoulomenos

no, it's a reference to the beginning of "the madman" in "the gay science" aka the text where he wrote "God is dead etc." for the first time

Zee

and that is obviously a refrence to diogenes

orbit-stabilizer

Diogenes was the man.

Zee

00:42

ya, he is almost transcendental in a way, not part of the world but beyond it

like a monk, without the hippie bullcrap

orbit-stabilizer

How are monks hippies?

Zee

well not monks but am thinking gurus

you know, if you like Nietzche

you should read Dostoyevsky

even Nietzche said he leanred from Dostyoevsky

orbit-stabilizer

I'm not too interested in Nietzsche atm

Zee

what are you intrested in?

Orbits ?

orbit-stabilizer

Buddhism atm

nbro

00:47

Hi!

Can anyone here explain to me the following step?

It's from the following Wiki article: en.wikipedia.org/wiki/Conjugate_prior

I don't understand how the denominator becomes what it becomes in the last expression

I was trying to do it by myself

However, I am not sure how to proceed to obtain the desired Gamma function at the denominator

orbit-stabilizer

Don't do it by hand. Simply note that we end up with what looks like a gamma distribution multiplied by some constant.

That constant being the beta function acting as the normalizer

I don't think it's worthwhile doing the computation

nbro

Btw, I am using a different notation, where s = H and f = T.

Akiva Weinberger

Good stuff happening to the family that I probably can't legally share

orbit-stabilizer

immigration related?

nbro

It would be nice to perform the calculations, but not now because I am tired

:D

But I think what I did is correct

I just need to show that the denominator is equal to $B(\alpha + H, \beta + T)$

Sha Vuklia

01:18

@Akiva are you an algebra fan?

o wait, there's @Dami. would you mind looking at an algebra ex?

it's about finite groups

:p

Akiva Weinberger

@orbit-stabilizer Not immigration related, but career related

Daminark

Oh shit let's go

Akiva Weinberger

All finite numbers are small and thus approximately equal to 3

Sha Vuklia

@Daminark was that.. directed at me?

Akiva Weinberger

Yes

Sha Vuklia

01:24

hm, alright, I guess I'll see that as a green light: let $G$ be a finite group with exactly two conjugacy classes. I want to show that #$G=2$. So let $Gx$ and $Gy$ be the two conjugacy classes. Let $N_x$ and $N_y$ be the associated normalisers. Then we know that #$G=\operatorname{index}[G:N_x]+\operatorname{index}[G:N_y]$. We would basically have to show that $G=N_x=\{gxg^{-1}\vert g\in G\}$, but.. not sure how?

Akiva Weinberger

@Silent Consider the circle with diameter AA' (A' being opposite to A). We want to show that the angle ABA' is 90 degrees.

Consider B', the point on the circle opposite B, and think about the quadrilateral ABA'B'.

@ShaVuklia Exactly two conjugacy classes… including $\{e\}$?

So the two conjugacy classes would be the identity, and everything else

Sha Vuklia

oh right, of course

did not think of that

Leaky Nun

@ShaVuklia and then use the orbit stabilizer theorem

Sha Vuklia

omg leaky, do you ever sleep:p

I don't know if I've had that theorem

I had another approach tho

Leaky Nun

it's 09:34 AM now

Sha Vuklia

01:35

03.35 here

ohh

so you just woke up, I hope:p

Leaky Nun

@ShaVuklia whoever teaches conjugacy classes without the orbit-stabilizer theorem should be hanged imo

right

Sha Vuklia

let me check

oh right

yea, indeed, I wanted to use that

didn't know it was called like that

Leaky Nun

my book notes: "The case for infinite groups is entirely different. Any infinite group in which each nontrivial element has infinite order may be embedded in a group with just two conjugacy classes: {1} and the set of all nontrivial elements."

Sha Vuklia

right, I'm too sleepy to process that unfort.

Akiva Weinberger

Remind me how @orbit-stabilizer works?

Daminark

01:38

Okay lemme get on a computer because TeX

Sha Vuklia

order of the orbit equals index of the associated stabilisor @Akiva

Leaky Nun

@AkivaWeinberger you mean the theorem itself or how it is applied here?

@ShaVuklia so you know how to do it now?

Sha Vuklia

@leaky yea, I think so. Let $\operatorname{ord}G=n$. And let $q=[G:N_x]$. Then we have $q\mid n$, so write $n=kq$. The orbit-stabilizer thm gives: $kq=1+q$, so $q=1/(k-1)$. Since $q$ is an integer, the only possibility is $q=1$

Leaky Nun

nice

Zee

I love the smell of steak in the morning

Leaky Nun

01:46

I found a proof of a similar theorem online: groupprops.subwiki.org/wiki/…

Sha Vuklia

anyhow, I'm off the bed. I appreciate your help a lot leaky!

cya

Nicholas Roberts

01:58

@Zee

Do your hw

Hahaha

Anyone know how existence of a nowhere-vanishing vector field on S^n implies there exists a continuous map $V: S^n /rightarrow S^n$ such that $<V(x) , x> = 0

For all $x \in S^n$?

$V: S^n /rightarrow S^n$

Sorry, I messed up the Latex, but i think its clear what Im asking

0celo7

@NicholasRoberts If you view the sphere in the standard way in $\Bbb R^{n+1}$, then tangent vectors correspond to vectors satisfying $\langle v,x\rangle=0$

Nicholas Roberts

@0

@0celo7 Can you explain a little more in detail? Thanks

0celo7

think geometrically

Nicholas Roberts

Wouldnt we need to define the vector field initially as the normal vector field on S^n?

0celo7

$S^1\subset\Bbb R^2$

Nicholas Roberts

02:07

And then all tangent vectors will be Perpindicular implying the dot product will vanish?

0celo7

there's nothing normal here?

$v\in T_xS^1$ is orthogonal to $x\in S^1$

draw a picture

Nicholas Roberts

Yes, I see that

So is the function simply the vector field?

The function we seek to show existence of

0celo7

yes

Are you defining manifolds abstractly or via embeddings?

Nicholas Roberts

This is respect to do the dot product on $R^{n+1}$

0celo7

You need to somehow show that the geometric interpretation of $TS^n$ corresponds to whatever definition you're using

It's kind of obvious but a rigorous proof takes some thought

Nicholas Roberts

02:12

I believe this via embedding as the question says to take the dot product using R^n+1

0celo7

ok then the geometric picture is sufficient

Nicholas Roberts

So when writing this up, how would does this sound? "Indeed, existence of a non-vanish vector field V on S^n gives us that $<V(x) , x> = 0 \forall x \in S^n$. This is because tangent vectors are orthogonal to vector on the sphere"

Sorry, few typos.

0celo7

orthogonal to the position vector on the sphere

(this can actually be proved easily using the tangent vector to curve definition of tangent vectors)

Nicholas Roberts

Yes, thats better wording. Interestingly enough, this question im working on is an outline of the proof of the hairy ball theorem.

It makes you prove a chain of equivalences that imply it

0celo7

yeah this is how everyone proves the HBT :)

what book are you using

Nicholas Roberts

02:17

Lee's intro to smooth manifolds. Ever read it? Whats your opinion on it

0celo7

Depending on the time period I'm sent back in time to, I would take it with me and become a legendary mathematican

great book

Nicholas Roberts

Lol! My friend Zee hates it

0celo7

Of course he does

Nicholas Roberts

But its ok. The bigger problem is that our Prof doesnt follow along with it but yet he assigns problems in the book that reference material we never covered in class

Of course as grad students we should be expected to deal with this sorta thing, but its a tad annoying

0celo7

I don't know why anyone would have a problem with it. I never used it for learning, but it's nice to have a detailed reference for such things

Nicholas Roberts

02:22

I dislike how he explains tensor fields as sections of the tensor bundle. Very un-intutitive explanation in my opinion

0celo7

I learned tensors from physicists first

I don't know how I think about them, I just do

Nicholas Roberts

The more i deal with them, the less mysterious they become

0celo7

they're not mysterious at all

they're just matrices

Nicholas Roberts

Anyhow, can you quickly explain how the dot product vanish is easy to prove using the tangent vector to curve defintion

im looking at it now and dont understand

0celo7

Ok so you have a curve in $S^n$, say $x(t)$, with $x(0)=x$

and you have that $x'(0)=v\in T_xS^n$

But since $x(t)\in S^n$, $|x(t)|^2=1$

now differentiate this at $t=0$

exercise for you

where of course $|x(t)|^2=x(t)\cdot x(t)$, Eucl. dot product

Nicholas Roberts

02:30

Why differentiate the square? I thought you would differentiate the curve and evaluate that at t = 0?

Indeed, x'(0) = v, our tangent vector

0celo7

@NicholasRoberts Did you differentiate the square or not

Nicholas Roberts

having trouble with differentiating the norm. With respect to which variable? All of them?

0celo7

$t$

Nicholas Roberts

Before plugging in 0, i got the square root of the norm of x(t)

is this correct?

@0celo7

0celo7

no

take the derivative of $\sum_i x^i(t) x^i(t)$

Nicholas Roberts

02:42

$2x^i(t)$

and let i range from 1 to n

0celo7

dude, do you need to review calculus?

chain rule

Semiclassical

That's a bit blunt, but uh

yeah

if you're differentiating position w/r/t time then you'd better not end up with position again

Nicholas Roberts

Lol, but isnt that just $x_i^2$?

Semiclassical

No. The $t$-derivative of $\sum_i x^i(t)x^i(t)$ is not $\sum_i 2x^i(t)$

Nicholas Roberts

I feel stupid but what is it?

Semiclassical

02:50

Simpler question: What's the $t$-derivative of $x^i(t)$?

Nicholas Roberts

To me, that would just be $\frac{d}{dt} \space x^i(t)$

Semiclassical

Sure.

So why are you omitting that when you differentiate $\sum_i x^i(t)x^i(t)$?

Nicholas Roberts

So its $2x^i(t) \cdot \frac{d}{dt} x^i(t)$

Semiclassical

Right.

You get a factor from the chain rule.

Nicholas Roberts

And let i run of 1 to n

Anyway @0celo7, I got 0

0celo7

02:59

so you got $\sum x^i v^i=0$

that's what you wanted

Nicholas Roberts

Im just not seeing where we have one being a tangent vector and another a position vector in this calculation

Ah perhaps the x^i is the position vector.

I see

Nameless

03:30

Let $V$ be a vector space and GL(V) be the associated General Linear Group. What is the name of the map $\psi : GL(V) \to V$? Does anyone know?

Leaky Nun

@Nameless but what is the map?

I mean, in some sense $GL(V) = \operatorname{Aut}(V)$

and we don't usually have a map from Aut(-) to -

Nameless

Well never mind then, I thougt there was a name for it.

Leaky Nun

or is it the evaluation map that sends $A$ to $A(v)$ for every $v \in V$?

i mean, for every $v \in v$ there is an evaluation map $\psi_v:GL(V) \to V$ sending $T$ to $T(v)$

but, you see, $GL(V)$ is a group, not a vector space

Nicholas Roberts

03:47

anyone like representation theory of finite groups?

Leaky Nun

03:58

@NicholasRoberts not too familiar with it, but hopefully I know some of it

orbit-stabilizer

Question.

Leaky Nun

@orbit-stabilizer Answer.

orbit-stabilizer

I have integral of $\bar z$ over a circle centered at $i$ with radius $1$. I thought that $\bar z$ had a removable singularity at $z = 0$, turns it... it doesn't. Why?

Alexander Gruber

@NicholasRoberts yuh

Nicholas Roberts

@AlexanderGruber awesome!

Leaky Nun

03:59

@orbit-stabilizer because it's perfectly continuous at $z=0$?

Transcript for

$Mathematics$ Mathematics