Discussion between nigelvr and Qiaochu Yuan

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01:27

Let $G=SL(2,\mathbb{R})$ and let $K=SO(2)$ be our maximal compact subgroup. Let $(\pi,V)$ be a real irreducible representation of dimension $d$. Apparently one has that the set of $K$-fixed vectors $V^K$ is $1$-dimensional if and only if $d$ is even. In the other case, the dimension of $V^K$ is...

Qiaochu Yuan

$G$ has a unique (up to isomorphism) real irreducible representation of each dimension $d$. Do you know how to construct it?

nigelvr

Yes, I am familiar with it. It's the representation on the space of degree $d$ homogeneous polynomials in $2$ variables, with highest weight $x^d$ (depending on exactly how one defines the action).

Qiaochu Yuan

Cool. Now, do you know how to compute the action of $K$ on this representation?

nigelvr

It doesn't appear to be anything special... just the usual rule of $g.f(x,y) = f(g^{-1}(x,y))$.

Qiaochu Yuan

I mean, for example, do you understand the action of $K$ in enough detail to know how $V$ decomposes as a representation of $K$?

nigelvr

01:27

It's the circle group, which is abelian... perhaps characters somehow come into play here. Sorry, I am really guessing.

Are you there?

peter a g

02:23

hi @nigelvr - one moment please...

nigelvr

peter a g

as QY isn't here I'll put my two cents in - hopefully correct cents. The diagonal elements diag (t, t^{-1) have as eigen spaces the X^j Y{d-j}, with eigen-value t^{d -2j} (or the multiplicative inverse)... yes?

nigelvr

let me think...

peter a g

never...

that was a joke!

nigelvr

I know :)

i'll take it for granted at the moment

yes I see that it is true

peter a g

02:31

(tX)^j(t^{-1}Y^{d-j}= t^(2j - d) x^j Y^{d-j}

nigelvr

of course.

peter a g

one moment again!

nigelvr

no problem

peter a g

OK - so t has 1 has an eigen-value iff d is even

because of the 2j- d in the exponent

nigelvr

by t surely you mean \pi(diag(t,t^{-1}) has 1 as an eigenvalue

peter a g

02:34

yes I mean that

nigelvr

very good. i'm following you so far

peter a g

If one complexifies the representation one ends up with an irreducible representation of SL_2 C - OK?

nigelvr

correct.

peter a g

Great - one can use instead of the diagonal torus, SO_2 to do the decomposition into eigen spaces

OK?

nigelvr

SO_2 or SU_2? Since now the max cpt subgroup of SL_2 C is SU_2

peter a g

02:38

In fact, I think I understand this better using lie algebras - are you ok with this?

nigelvr

yes I have some familiarity with lie algebras. we can work with those

peter a g

OK the H = diag{1,-1} has an eigen value of zero iff d is even

OK?

nigelvr

correct. i suppose once you differentiate pi that's what you get.

peter a g

Rats hold on a second

nigelvr

well right from the get-go you know that SO(2) is a commuting family of matricies, hence you write V as a direct sum of eigenspaces

so then the goal is to show that one of them has eigenvalue 1 if and only if d is even

peter a g

02:48

yes - since SO_2 is compact, the characters are of the form e^{i k \theta} - the point is that the k are the same as that show up for the diag {t, t^{-1}}

so - if you buy that - I think it is correct! that was my rats, I wanted to show it - k= 0 shows up iff d is even -

nigelvr

i think i'm almost getting there ... but I am still confused about why diag(t,t^{-1}) comes up.

peter a g

because you can see the weight spaces easily using the diagonal torus..

nigelvr

OK. I'm new to Lie theory but I think I see what you're doing

instead of using the torus SO(2) in SL(2,R) you're using a conjugate one, that looks like diag(t,t^{-1})

peter a g

Right - but the conjugation has to be in SL_2 C

I am sure that QY would have done a way better job of this. I 'used to' know this and I am shocked I am chocking and not giving you a good explanation. Sigh.

nigelvr

it's OK. I'm pretty sure I will have it completely if I mull over it for a while now, thanks to your help

I never learned Lie theory properly. I learned basic things about Lie algebras, and Fourier analysis more in depth, and now my MSc supervisor seems to expect me to be some type of expert

peter a g

03:02

OK - for what it's worth! I hope I didn't screw up. The relevant words are 'unitary trick' You might look at knapp's representation theory of semisimple groups chapter 2

ciao!

nigelvr

ok, i'll look it up. thanks. take care

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