Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am not sure of the right terminology, but here goes. Let $G$ be a compact, connected, simply connected, non-abelian Lie group.

For any choice of one-dimensional torus $S\subset G$, and any finite-dimensional representation $\pi:G \to U_{d(\pi)}({\mathbb C})$, we can define $$ m_S(\pi) = \max\{ |n| \colon \pi\vert_S \text{ has a component of degree $n$, $n\in{\mathbb Z}$} \} $$

Question 1. Can we always choose $S$ and find a constant $C=C(S,G)>0$ such that $$ m_S(\pi)\leq C d(\pi) \quad\hbox{for all irreps $\pi\in \widehat{G}$?} $$

For instance, this is easily seen to be the case when $G=$SU(2) and $S$ is any choice of maximal torus, with $C=1$; it also seems to be true when we take $G=$SU(3) and $S$ to be an appropriate subgroup of a maximal torus, using the well-known formula relating degrees of irreps of SU(3) to their highest weights.

My (limited) understanding is that for SU(3) this formula is obtained using the Weyl character formula, so it seems likely that if Q1 has a positive answer then this could be found using the WCF and known classification of such groups $G$.

However, the motivation for Q1 is that if true it would give an alternative proof of some known results about the Fourier algebra of $G$, which are usually proved by observing that such $G$ contains a closed subgroup isomorphic either to SO(3) or SU(2), cf. this Wikipedia entry. That observation seems rather more basic than the usual incarnation of the Weyl character formula. So my next question, which is admittedly vague, is:

Question 2. If Q1 has a positive answer, is there some way to see this which is "global" in flavour and does not need much theory of weights and roots, let alone classification?

I have in mind something along the lines of David Speyer's answer to another question. Perhaps one could show that when inducing a linear character of "degree" $n$ from $S$ up to $G$, the resulting representation of $G$ contains no irreducible component of dimension $\ll n$ -- but I admit I have no idea if this is a sensible or feasible approach.

share|improve this question
    
Is there a chance that there might be such a constant independent of $S$? Given a representation $G\to SU(d)$, pull back the Killing form of $SU(d)$ to get a multiple of the Killing form for $G$. What multiple is it? Is that positive number bounded above by $Cd$ for some constant $C$ (depending on $G$)? –  Tom Goodwillie Dec 27 '12 at 15:27
1  
I think that the constant must depend on $S$. For example take $G=SU(3)$, $\pi$ to be the usual 3 dimensional representation, and $S$ consisting of diagonal matrices $diag(z^N,z^{-N-1},z)$. –  Victor Ostrik Dec 27 '12 at 19:46

1 Answer 1

The answer to Question 1 is positive. Choose $S$ which is a maximal torus in $SU(2)\subset G$ or $SO(3)\subset G$)(it is well known that such embedding always exists). Then the inequality in Question 1 holds with $C=1$.

share|improve this answer
1  
Yes. The OP says that a positive answer to Q1 would give an alternative proof of some known results which are usually proved by observing that such $G$ contains an $SO(3)$ or $SU(2); but Q1 can easily be answered using exactly that observation. –  Tom Goodwillie Dec 27 '12 at 18:23
    
Oops, my bad. I did not read the question carefully.. –  Victor Ostrik Dec 27 '12 at 18:53
    
Thanks anyway, Victor: I admit that I had overlooked this simple argument. I would still be interested in a proof which doesn't use these embedded copies of SO(3) or SU(2) –  Yemon Choi Dec 28 '12 at 1:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.