Mathematics - 2015-10-23 (page 1 of 2)

Memeozuki

00:44

Any last minute GRE tips?

Lepidopterist

are there any useful bounds on the Trace$(A^2)$ in terms of Trace($A)$

Ted Shifrin

01:17

can't be, @Lepidop

Fargle

Hi, @Ted!

Ted Shifrin

((Fargle))

Fargle

How're you?

Ted Shifrin

doing pretty well, and how 'bout you?

Fargle

I'm alright. School continues, haha.

Ted Shifrin

01:18

Still too easy for you, of course

Fargle

Well, it's just probability and numerical analysis. A strain on the memory, but not much else. And my numerical professor is really good at explaining stuff simply.

Ted Shifrin

That's good ... Well, if you're too bored, get back to the stuff I told you to work on and work with me :P

Fargle

Haha, of course. I'm actually having an attack of boredom at the moment. I may take a look at that.

Ted Shifrin

Is your health good?

Fargle

Also, one of my professors gave me a researchable question that I'm looking into.

Yes, I'd say so.

Ted Shifrin

01:20

ok, good

ditto

Fargle

Good!

morphic

Hi @TedShifrin

Ted Shifrin

well, I'm happy to challenge you more, @Fargle

heya mr eyeglasses

Fargle

I'm always happy to be challenged.

skull petrol

Hi professor

Ted Shifrin

01:22

rehi @skull

anything to report of a positive nature, mr eyeglasses?

um ... so I guess not

Secret

We know that vectors in $\mathbb{C}^n$ are ordered n tuples in that $$(a,b,c)\neq (b,a,c)$$

Matrices as elements in a vector space seemed to also have some kind of ordered structure since
$$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\neq\begin{pmatrix}c & d \\ a & b\end{pmatrix}\neq \begin{pmatrix}b & a \\ d & c\end{pmatrix}$$

If we don't think of matrices as made of column vectors, then how mathematically is this order structure is being introduced?

Ted Shifrin

You just arbitrarily pick an order, @Secret. Say column by column or row by row.

Karim Mansour

@TedShifrin

Ted Shifrin

hi @Karim

Karim Mansour

I am trying to show that $|cos(z)| \leq e$

Ted Shifrin

01:29

Well, that would be very false, @Karim.

Karim Mansour

when $|z| = 1$

sorry

Ted Shifrin

Maybe believable now.

Karim Mansour

$coz(z) = \frac{1}{2} * (e^{iz} + e^{-iz})$

Ted Shifrin

ok, so?

Karim Mansour

so $|cos(z)| = |\frac{1}{2}e^{iz} + e^{-iz}| \leq 1/2|e^{iz}| + 1/2|e^{-iz}|$

by triangle inequality

Ted Shifrin

01:31

No, you threw away too much

Karim Mansour

?

I didn't distribute the 1/2

in first line

Ted Shifrin

Well, ok, that actually does it.

Karim Mansour

but second line is right

but why is $|1/2e^{iz}|$ = 1/2e ?

Ted Shifrin

And what about $|e^{iz}|$?

It isn't.

Just write things out carefully, for once.

Karim Mansour

is it true that $|e^{iz}| = e^{|z|}$?

Ted Shifrin

01:33

NO!

Write it out carefully, for once.

Karim Mansour

ok

Ted Shifrin

Goodnight @MikeM

Karim Mansour

$e^{iz} = e^{x + iy} = e^x(cos(y) + isin(y))$

Ted Shifrin

No, @Karim.

You dropped the $i$.

Mike Miller

Morning.

morphic

01:34

Hi Mike

Karim Mansour

I didn't

where?

Karl Kronenfeld

iz

Ted Shifrin

hi @Karl

Karim Mansour

ohh

Ted Shifrin

<--- getting annoyed by sloppiness

Mike Miller

01:35

Some staff are digging around in the ceiling above me which is not exactly a good environment to meet a student in.

Karl Kronenfeld

though |z|=1 iff |iz|=1

hey @Ted

Ted Shifrin

But $|z|$ is irrelephant.

Karim Mansour

$e^{iz} = e^{i(x + iy)} = e^{ix - y} = e^{-y} *(cos(x) + isin(x))$

right?

Karl Kronenfeld

Meh, not paying close enough attention to argue

Ted Shifrin

OK, @Karim, but if you're going with modulus, then $|e^{ix-y}| = |e^{ix}||e^{-y}|= e^{-y}$.

Karim Mansour

01:36

yes

yes since it has modulus 1

Ted Shifrin

it?

Karim Mansour

I mean $|e^{ix}| = 1$ because $|z| = 1$

Ted Shifrin

NO. $|e^{ix}|=1$ always when $x\in\Bbb R$.

You really need to pay attention to detail.

Karim Mansour

ohh I see

Ted Shifrin

You should be able to finish correctly now.

Dinnertime for this bonzo.

Karim Mansour

01:42

why is it true that $\frac{1}{2}e^{-y} + \frac{1}{2}e^{y} \leq e$?

anon

Chat guidelines | $\LaTeX$ in chat

13

Karl Kronenfeld

Ignores both parts

Karim Mansour

?

I am using latex

Karl Kronenfeld

The sticky-star wore out, presumably.

anon

@KarimMansour why do you think that's true? are you sure you've typed it correctly?

the LHS is unbounded

Karim Mansour

01:46

hm

Karl Kronenfeld

y is bounded in this problem

anon

what kind of y are we talking about then?

Karim Mansour

oh I see so here since we know that |z| = 1

Karl Kronenfeld

z=x+iy and |z|=1

Memeozuki

How do I get started learning PDE?

Mike Miller

01:48

What's your background?

Karim Mansour

I still don't see the last step

Karl Kronenfeld

Well, what interval must $y$ lie in?

Memeozuki

@MikeMiller I know analysis from Folland's book and some Riemannian Geometry

Karim Mansour

it must lie in [0,1]

Karl Kronenfeld

nah

Karim Mansour

01:50

why we have the modulus

Karl Kronenfeld

visualize the locus of $|z|=1$

Karim Mansour

the points will lie on a circle of radius 1

Mike Miller

@Memeozuki: I guess I'm not comfortable saying where you should start. The standard PDE text is Evans. You should probably be comfortable with measure theory and Fourier theory (know a bit about the Fourier transform).

Karl Kronenfeld

in the x,y-plane @Karim

Karim Mansour

yeah

Mike Miller

01:52

Probably there is a lot more to suggest being comfortable with but these are the obvious things to me.

skull petrol

Don't forget boundary value problems :-)

Memeozuki

@MikeMiller
Are there any tips for getting better at digesting really technical information. A lot of times I read estimate arguments and I just can't pierce through them.

Karim Mansour

I am not sure @KarlKronenfeld

Karl Kronenfeld

No one is... OH right, the |cos(z)| question. What aren't you sure about?

Mike Miller

This is something that comes with practice. (I know the feeling.) Most of these technical arguments have a main thrust - some "real idea", where the rest of it was standard manipulations. Try to find this real idea in everything you read, @Memeozuki, whether it be analysis or otherwise.

4

(This is true for most theorems in textbook mathematics. When you start reading papers things are a bit less polished and it's often harder to find this central thrust.)

Karim Mansour

01:59

if we look at the x,y plane we see that the points that have modulus 1 will be exactly the points that lie on the circle of radius 1

so

it means

Memeozuki

One thing is that for a lot of estimates they look so ugly to me I'm not sure how somebody could ever guess for it to look like that. Is that something that's picked up over time?

Karim Mansour

it is either in

[0,1]

or

[0,-1]

right ?

Karl Kronenfeld

uh huh, so just [-1,1]

Mike Miller

@Memeozuki: Yes, very much so. Stick with it. (Make sure to do lots of exercises so you're doing these ugly estimates yourself.)

Karim Mansour

yeah

Mike Miller

02:01

This is true again for every field of math. Lots of things that seem very unnatural become "the obvious idea" to someone who's been embroiled in it for a while.

Karim Mansour

well let us look at the negative part

ok ok

I get it

thanks alot @KarlKronenfeld

Karl Kronenfeld

np

Memeozuki

Ok, thanks a lot. I'll stick with it.

Karim Mansour

@KarlKronenfeld sorry I have one quick question

about something else in complex analysis

Karl Kronenfeld

ok, if I can't answer it, surely somebody else can

Sanic Hodgeheg

02:04

Hello!

skull petrol

Hi pal

Memeozuki

Hi @SanicHodgeheg

Karim Mansour

consider the following integral $\int_{\tau}cos(cos(sin(z))) dz$, where $\tau$ is given by $\tau(t) = e^{it}$, for $0 \leq t \leq \pi$ and by $\tau(t) = e^{-it}$ for $\pi \leq t \leq 2\pi$

Mike Miller

@Memeozuki: I know the frustration; I used to feel the same way. Now I study a part of math which is very firmly analytic (before first thinking I was going to be an algebraist because I couldn't deal with all these inequalities :P )

Karim Mansour

so the first one is a circle that goes counter clockwise

from 0 to pi

the second one is a clock wise curve that goes from $2\pi$ to $\pi$

I was thinking to use cauchy integral formula

but its not a closed path

since we don't return to our initial point right ?

Memeozuki

02:07

@MikeMiller That's very encouraging. I really want to learn about geometric analysis and PDE.

Karim Mansour

actually no

we do right ?

anon

can you explain in words what the path is Karim?

Karim Mansour

ok so for first path its easy to describe

we go trace a curve from $0$ to $\pi$ that goes counterclockwise

right ?

Mike Miller

@Memeozuki Very cool. (I study gauge theory, which is related to geometric analysis. Lots of the PDE involved is more classical though, much of it being elliptic PDE. I stay mostly in that realm, though some people do gauge theory with some parabolic PDEs.)

Karim Mansour

do you agree ?

anon

02:09

yes

well, from 1 to -1 on the unit circle is what you mean

Karim Mansour

yeah

Memeozuki

Ah, awesome. It seems like the geometric evolution equations that I've seen are parabolic but I'll learn more about what I want to do as I learn more PDE things.

Karim Mansour

now for the second path we go from $2\pi$ to $\pi$ but instead of going counter clockwise we go clockwise

since we added the negative

I am sorry

I meant to say

For the second path we start at $\pi$ and then we go back to $2\pi$ and we go clockwise, so we do indeed go to our initial spot, so we can use cauchy integral theorem.

correct @anon?

anon

pi and 2pi are not points on the circle. refer to points on the circle.

Karim Mansour

oh ok

yesss

Mike Miller

02:12

@Memeozuki: I think pretty much all modern study is on parabolic and hyperbolic PDE. The elliptic equations are the easy stuff.

Karim Mansour

I guess one way to do it is to directly compute stuff

Mike Miller

I gotta go for now. Good luck!

Karim Mansour

$\tau(2\pi) = cos(2\pi) - isin(2\pi) = 1$, while $\tau(0) = 1$

so it is indeed a closed path

anon

here is what I was looking for: tau goes from +1 to -1 around the unit circle counterclockwise, then goes back from -1 to +1 around the unit circle clockwise.

Karim Mansour

yes

anon

02:17

in real variable calculus, do you know the relationship between $\int_a^b f(x)dx$ and $\int_b^a f(x) dx$? One goes on a path from $a$ to $b$, the other on a path from $b$ to $a$.

Karim Mansour

it is the negative

going from a to b is the negative of going from b to a

anon

so what do you think will happen with your integral if you split it into two parts, each integral over the same path but in opposite directions?

Karim Mansour

they will evaluate to zero

anon

try it and see what happens

Karim Mansour

alright

Karim Mansour

02:34

yeah

it evaluates to zero

is it true that $|\int_{\tau}f(z)| \leq \int_{\tau}|f(z)|$?

anon

yes

well, you have to write it correctly

$\int_\tau |f(z)|dz$ needn't even be a real number

one should write $\int_\tau |f(z)dz|$ (can you interpret that?)

Karim Mansour

ohh I see

coz I am using the problem I solved before to show that $|\int_{\tau} cos(z) /z dz| \leq 2\pi*e$

anon

$|\int_\gamma f(z)dz|$ is bounded by ${\rm len}(\gamma)\cdot \sup_\gamma |f(z)|$

Karim Mansour

$|\int_{\tau} \frac{cos(z)}{z} dz| \leq \int_{\tau} |\frac{cos(z)}{z}dz| $

$\leq \int_{\tau} \frac{e}{|z|}|dz| \leq e * 2\pi$

right ?

@anon ?

no wait I think that last line is wrong

Huy

05:27

@KarimMansour: are you in $\mathbb{R}$ or $\mathbb{C}$?

Karim Mansour

C

Huy

@KarimMansour: how do you get $\cos(z) \leq e$?

(or was your idea something different from the first to the second line?)

Karim Mansour

$|cos(z)| \leq e$

@Huy I made a mistake

Huy

@KarimMansour: for which $z$?

Karim Mansour

for |z| = 1

Huy

05:30

ah

Functional Analysis

05:52

someone will discuss functional analysis things with me?

Weak convergence of $A_n\rightharpoonup A$ means $\forall x\in X, \forall f\in X^*, f(A_nx)\to f(Ax)$

Functional Analysis

06:30

why does people call the norm a function? isn't it a functional since it maps from the linear space to the scalar field?

sajjad

again this one is just like one before "expanding a non-periodic" function (even and odd) , Can anyone help me to confirm my solution(?) :

0

Q: Expanding Fourier Series of $f(x)=x^2$ where $0<x<1$ (even and odd)

I tried to solve Fourier series (which appeared on title) and ended up to below solution : on even state : $ \phi(x)= \begin{cases} x^2 & 0<x<1 \\ x^2 & -1<x<0 \end{cases} $ $a_{0}=\dfrac{2}{3}$ and also $a_{n}=\dfrac{4(-1)^n}{n^2\pi^2}$ on odd state I reached below answer : $ \phi(x)= \be...

Balarka Sen

06:53

Figured the topology problem I gave you, @JulianRachman?

Julian Rachman

07:34

Sorry haven't had time @BalarkaSen can yoi give me a jist because I am really busy right at this moment

Balarka Sen

07:47

@JulianRachman What do you mean by "gist"?

The problem was to show that $\Bbb R$ and $\Bbb R^2$ are not homeomorphic, both given the standard topology.

user61230

08:02

Anyone here have a moment to help with a hopefully straightforward covariant derivative question?

Julian Rachman

Sorry autocorrect. @Balarka

user61230

I'm trying to show that, supposing $W$ has $||W||=C$ (constant), then $(\forall V)(\nabla_VW\perp W)$.

Balarka Sen

@JulianRachman Then what did you mean?

user61230

I have a vague intuitive understanding of why this should be, but I'm not sure how to approach proving it. I've started with $(\nabla_VW)\cdot W=\sum V[w_i]w_iU_i$, but I'm not sure how to show this quantity is zero (or even if this is the right approach).

Julian Rachman

Gist

Balarka Sen

08:06

Yeah, I guessed, but gist of what?

Tobias Kildetoft

@Emrakul Take the inner product of $W$ with itself and differentiate

Julian Rachman

The solution

user61230

...that is glaringly obvious in retrospect. Thank you, @Tobias!

Tobias Kildetoft

@Emrakul I only knew how to do it because I happened to have done a similar exercise (like 10 years ago)

Balarka Sen

@JulianRachman Are you sure? It has a one-line solution.

There is a trick involved, but I guess you need to do some thinking for that.

Tobias Kildetoft

08:09

@BalarkaSen And the trick involved is one that it useful for a large number of exercises on showing spaces not to be homeomorphic

Balarka Sen

Right.

Tobias Kildetoft

(especially subspaces of the reals and the plane)

Julian Rachman

I just need time to do everything wnat first

user61230

Glorious. That's a nifty little proof.

Balarka Sen

@TobiasKildetoft I gave it to Julian because essentially the solution comes from looking at $\pi_0$. This generalizes to $\Bbb R^m$ and $\Bbb R^n$ where one looks at $\pi_k$ for higher $k$'s instead.

So it's a neat introduction to the concepts of algebraic topology, I think.

One can explain why $\pi_1$ is interesting/relevant by looking at the case $m = 1$, $n = 2$.

That provides some motivation.

Julian Rachman

08:15

Ya well Ted has been telling me to focus on the basics and not care about other stuff yet

Balarka Sen

He's quite right.

Julian Rachman

And is Munkres the best text for Algebraic Topology?

Balarka Sen

"best" of course not. It's good as an introduction, in my opinion.

Julian Rachman

and how about your situation? Arent you the same age as me?

Balarka Sen

Yes. What about my situation?

Julian Rachman

08:17

then what is the strongest base introduction and what is the best structure book to completely read out of

Well you are like so good and trail right behind you... gg

Balarka Sen

@JulianRachman What do you mean by "strongest base introduction" and "best structure book"?

I have read algebraic topology from 2 book - Munkres and Hatcher. Hatcher is just too hard for beginners.

Julian Rachman

Ok that is all I need

Balarka Sen

@JulianRachman Well, I'm studying the basics too.

I could read and understand algebraic topology because (I think) I have a firm background on some algebra and topology. But I am lacking on analysis.

Julian Rachman

Ok

Balarka Sen

So, e.g., if I want to read about differential topology, I'm going to have to learn multivariable calc. That's what I am doing right now.

Julian Rachman

08:20

I am learning multivariable next year as a formal class

in college classroom

Balarka Sen

Cool. :)

Julian Rachman

so that is covered

Balarka Sen

and I think you also have more background on real analysis than me?

Julian Rachman

I think I going to read munkres part 2 right after I finish homework then do a few problems from my problem set

ya I technically do

I studied it for a year before topology

Balarka Sen

If you want. But do this $\Bbb R$ and $\Bbb R^2$ problem first.

Julian Rachman

08:22

Ok

Balarka Sen

@JulianRachman Yeah, I thought so. You know measure theory?

Because I don't know anything about it.

Julian Rachman

Not familiar with it however it was mentioned in my text for real analysis

Balarka Sen

I attended a lecture in ergodic theory but came out confused. They're applying so much real analysis to prove completely number theoretic things.

Julian Rachman

Oh my.

that must suck because you sit there not knowing what or how to do anything

Balarka Sen

well, not completely. I understood what the statement and the theorems are. I just don't have enough background to understand the proofs.

The point of being in lectures is to get an idea of which branch you would want to study, really :P

Tobias Kildetoft

08:25

@JulianRachman That is the default state of about half the audience at most seminars

Balarka Sen

haha.

Julian Rachman

Ah I see. Honestly so far for me I just love category theory. I want find that gap in the spectrum where category theory could change it all

lol I want to attend some seminars

I think I might try some at UCLA

Tobias Kildetoft

@JulianRachman Then do so, they are usually open

Balarka Sen

I don't see the point of studying category theory for the sake of it, as I had said earlier.

Julian Rachman

do you guys take notes

?

In the seminars I mean

Balarka Sen

08:27

In lectures? Yes.

Tobias Kildetoft

@JulianRachman Sometimes, though not that often

Julian Rachman

I am not learning it just for the sake of it. I want to apply it to some other mathematics discipline

Balarka Sen

Most of the time, things go the other way around. You learn some branch where some ideas of category theory are introduced, and you pick them up along the way.

Julian Rachman

and ya. I would figure that it would be pretty hard to take notes during a seminar that like 3/5 of you don't understand

Balarka Sen

That's how I learnt the category theory I know, from studying algebraic topology.

Julian Rachman

08:29

ok

Well it is also a huge part of algebra, then like homology theory, homotopy theory, etc.

Balarka Sen

@JulianRachman But if you have the notes with you, then you can come back to it later on to see which things you need to study to do understand it.

Julian Rachman

ok i see

Balarka Sen

@JulianRachman It's not a part of any three of the things you mention. In homology theory, one studies homological algebra - this might be categorical, but is a far-reaching branch of math. I don't know what kind of algebra you have in mind when you say that.

Tobias Kildetoft

@BalarkaSen homological algebra is best understood in the language of category theory

Balarka Sen

In homotopy theory (something I don't know much), they do have an abstract setting, or so I have heard. But that has concrete applications, like computing homotopy groups of spheres.

Julian Rachman

08:31

Aluffi's categorical algebra

Tobias Kildetoft

Then there are all the applications of $2$-categories and their $2$-representations that are starting to become apparent in some areas of representation theory

Balarka Sen

Yeah, n-categories are useful in bordism theory. But that's all I know about them, haha.

Tobias Kildetoft

@BalarkaSen I never go beyond $2$. Plenty of stuff to do there

Balarka Sen

@JulianRachman You mean Aluffi chapter 0? Bleh. That unnecessarily complicates simple, concrete topics in algebra.

Julian Rachman

Well Idk I like it so far

Balarka Sen

08:35

How much have you read it?

Julian Rachman

Half way through groups

Balarka Sen

OK. Then you haven't started on the concrete stuff yet, I guess.

Julian Rachman

I guess so

So why do you say that category theory is not something to learn?

Tobias Kildetoft

There are categories involved in the group theory stuff in Aluffi?

Balarka Sen

I have never said it's not something to learn.

@TobiasKildetoft You know, wrapping everything in a fancy language.

Tobias Kildetoft

08:39

@BalarkaSen I agree that using categories there is going too far

(not that there are not things in group theory one can use categories for, it just comes later)

Balarka Sen

@JulianRachman I'm just saying you're better off learning concrete things (have a look at Artin) and pick up categorical stuff as you go ahead. The real application of categories come way later.

Julian Rachman

Ok

Tobias Kildetoft

arxiv.org/abs/1412.4494 is a neat example

Julian Rachman

So should I keep with all aluffi or just stop where I am and go artin

Balarka Sen

I am unclear what representations of G \wreath S_n has to do with classification of simple modules, although I admit I don't know much about this stuff.

Julian Rachman

08:43

?

Balarka Sen

@JulianRachman Have a go at Artin. It's a beautiful book.

Julian Rachman

Ok. Then a hat should I do with category theory?

Tobias Kildetoft

@BalarkaSen representation = module

Balarka Sen

@Julian Learn bits of it in your leisure if you want. But I'd say you'd find algebra much better than category theory when you really learn it.

Julian Rachman

Oh and @Balarka isn't since \mathbb{R} is compact that means that it is not homeopathic to \mathbb{R}^2 which is not compact?

Balarka Sen

08:45

Both are noncompact.

Julian Rachman

Oh

Balarka Sen

Why would $\Bbb R$ be compact?

Julian Rachman

Ugh I don't want to give up on category theory because that is where lots of the big research is

Tobias Kildetoft

@JulianRachman Category theory is a lot easier to learn once you have learned all the stuff that motivated it in the first place

Balarka Sen

The big research is on applying category theory to prove concrete stuff, last time I heard.

Tobias Kildetoft

08:47

i.e. pretty much all other math

Julian Rachman

Like I was even thinking to other day on how we can prove the ABC conjecture using category theory instead of the original 500 page proof

Balarka Sen

So you need to know the concrete stuff.

Tobias Kildetoft

@JulianRachman ABC conjecture is still not proven

Julian Rachman

Oh. Wait didn't take a Crack at it and the paper was wrong?

Balarka Sen

To be frank, I think it's silly to hope ABC conjecture can be proved using category theory.

@JulianRachman Mochizuki has claimed he has proved it. But nobody bothered to check it because it's so badly written.

Julian Rachman

08:49

Ok well ill go artin

and p2 munkres

Julian Rachman

09:03

I have decided to finish the basic categorical part of Aluffi then jump into Artin and Munkres Part 2

Balarka Sen

@TobiasKildetoft So, what have you been thinking about lately?

Tobias Kildetoft

@BalarkaSen still trying to understand the Kazhdan-Lusztig cells in type $B$

though I have also been trying to understand the implications of a new description of so-called special representations, which allows the concept to be extended to a much broader generality

Balarka Sen

Must be fun. I have no idea about the kind of math you do :)

Tobias Kildetoft

@BalarkaSen The Kazhdan-Lusztig cell stuff is really just combinatorics (since someone has worked out previously how it should be). It just happens to not be described very well, in a paper that is not very well written.

Balarka Sen

09:18

Can you give me a rough idea of what it is?

I'm curious.

Tobias Kildetoft

@BalarkaSen Are you familiar with Weyl or Coxeter groups?

Balarka Sen

I know the definition of a Coxeter group.

Tobias Kildetoft

actually, let's just take the "easy" case of type $A_{n-1}$, which is just the symmetric group $S_n$

Balarka Sen

mmkay

Tobias Kildetoft

The KL cells come from three different partial preorders on the group, the left-, right- and two-sided preorders

let's just focus on the left for now (the right is just the same with inverses and the two-sided is generated by the other two)

Balarka Sen

09:22

okay.

Tobias Kildetoft

we say that $x \to_L y$ if two things are true: There is some element in the generating set (which are the transpositions of the form $(i\ i+1)$ here) which reduces the length of $y$ when multiplied on the left, but which does not reduce the length of $x$ when multiplied on the left

Balarka Sen

OK, that makes sense.

Tobias Kildetoft

(so for example if $x = e$ is the identity, then this part holds for any $y$

Balarka Sen

yeah.

Tobias Kildetoft

Now, the second thing is somewhat technical. It requires that the KL-polynomial $K_{x,y}$ or $K_{y,x}$ (whichever is non-zero if one of them is) has largest possible degree given the lengths of $x$ and $y$ (there is a bound for any KL polynomial in terms of these lengths)

Balarka Sen

09:26

what's a KL-polynomial?

Tobias Kildetoft

and don't ask me why anyone would consider this condition, as I have no intuition for that

the KL-polynomials come from a distinguished basis of the Hecke algebra of the Coxeter group

Balarka Sen

ah, ok, I'll just take it as granted that there is such a thing for now, then. what do you mean by "largest possible degree"?

Tobias Kildetoft

@BalarkaSen There is a general bound on the degree of $K_{x,y}$ in terms of the lengths of $x$ and $y$ (which I can never remember)

so largest possible means that the degree equals this bound

Balarka Sen

ah.

Tobias Kildetoft

anyway, one extends this to a partial preorder on the group in the obvious way

Balarka Sen

09:30

the first condition can be reproduced in any group by defining length to be the Gromov metric of the element and $1$, but this condition is too technical, it seems.

Tobias Kildetoft

@BalarkaSen yeah, the second condition is very technical

Balarka Sen

plus, those KL-polynomial things only come from Coxeter groups, you say.

ok, please continue.

Tobias Kildetoft

@BalarkaSen Yeah, not sure if they can be defined in higher generality

Balarka Sen

nod.

Tobias Kildetoft

anyway, ths gives of a partial preorder denoted $\leq_L$

Balarka Sen

09:31

yep.

Tobias Kildetoft

and as with any preorder, we get an equivalence relation denoted $\sim_L$ (so $x\sim_L y$ if both $x\leq_L y$ and $y\leq_L x$)

Balarka Sen

true.

Tobias Kildetoft

the equivalence classes of this are called left cells

Balarka Sen

and the equivalence class is called the left cell?

ok, fair enough.

Tobias Kildetoft

similarly, one defines $x\leq_R y$ iff $x^{-1}\leq_L y^{-1}$

Balarka Sen

09:33

ah.

Tobias Kildetoft

and $x\leq_{LR} y$ as the union of the preorders

and these give right- and twosided cells

Balarka Sen

alright, I get it.

Tobias Kildetoft

for $S_n$ there is a really nice description of the cells and of the two-sided order (not sure if there is a nice one for the left or right order)

Balarka Sen

why are these things of general interest?

@TobiasKildetoft oh?

Tobias Kildetoft

Well, given a left cell, one can define a representation of the group, and the the symmetric group, one gets precisely the irreducible representations this way

plus, they turn out to encode a bunch of important stuff

Balarka Sen

09:35

How can one define a representation of the group from a left-cell?

Tobias Kildetoft

@BalarkaSen in a technical way that I can never recall

Balarka Sen

hah, fair enough.

Tobias Kildetoft

anyway, the nice description of the cells is via the Robinson-Schensted correspondence if you are familiar with that

Balarka Sen

no, I am not really familiar with it.

Tobias Kildetoft

ok, the important thing is that it is easy to compute and associates to each permutation a pair of standard young tableaux of the same shape

Balarka Sen

09:38

hm, ok

Tobias Kildetoft

elements are then in the same left cell iff they have the same left tableau and in the same twosided cell iff their tableaux have the same shape

Balarka Sen

ah.

Tobias Kildetoft

and the twosided order (on the twosided cells) coincides with the dominance order on the corresponding partitions

though that last is a theorem from 2006, despite the rest being known since like the 80's

Balarka Sen

ok, interesting.

Transcript for

$Mathematics$ Mathematics