Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a compact (Hausdorff) group and $V$ a faithful (complex, continuous, finite-dimensional) representation of it. (Hence $G$ is a Lie group.) Is it true that every irreducible representation of $G$ occurs as a summand of $V^{\otimes n} \otimes (V^{\ast})^{\otimes m}$ for some $m, n$? (The original question only asked for summands of $V^{\otimes n}$ and, as anon mentions below, $\text{U}(1)$ is an easy counterexample.)

I know that the corresponding result is true for finite groups, but the proof I know doesn't seem to easily generalize. It seems we ought to be able to apply Stone-Weierstrass: the characters you get from summands of $V^{\otimes n} (V^{\ast})^{\otimes m}$ form an algebra of class functions closed under addition, multiplication, and complex conjugate, so if we know that they separate points, they ought to be dense in the space of class functions. But

1) I'm not sure if we can show that these functions separate points, and

2) I'm not sure if the space of conjugacy classes (with the quotient topology from $G$) is even Hausdorff.

Motivation: I was looking for cheap ways to set up the representation theory of $\text{SU}(2)$. In this rather special case the character of the defining representation $V$, which is self-dual, already separates conjugacy classes, and I think the above argument works. Then Clebsch-Gordan allows me to quickly classify the irreducible representations of $\text{SU}(2)$ without using Lie algebras.

share|improve this question
2  
For (2): the quotient space of your group by its action on itself by conjugation is Hausdorff, because the group is Compact and Hausdorff, so that the action is proper (In general, a compact Hausdorff space acting on a Hausdorff space has Hausdorff quotient) –  Mariano Suárez-Alvarez Mar 14 '11 at 16:58

3 Answers 3

It doesn't work for $U(1)$: take $V$ to be the standard representation, you only get the "nonnegative" representations. Indeed to use Stone Weierstrass in the complex case, the algebra has to be stable under complex conjugation. Maybe consider the $V^n \otimes (V^*)^m$ instead.

share|improve this answer
    
Very good point. I'll edit the question. –  Qiaochu Yuan Mar 14 '11 at 17:25

Yes all irreducible rep. of compact groups can be found in such tensor products of a faithful rep. and its dual. The idea is to use the so called representative functions. One way to define this algebra, is that it's the sum of matrix elements of finite dim. representations. One then shows that this is dense in the algebra of continuous functions, and that it is the algebra generated by the matrix elements of a faithful matrix representation and their conjugate elements. Now you can use this density to show that all irreducible rep. must sit in such tensor products. It's also true that the conjugacy class space is Hausdorff; actually, it's an important fact towards the understanding of compact group representations. This space is homeomorphic to the quotient of the compact group by a maximal torus. You should have a look to "Brocker, tom Dieck, Representations of compact Lie groups", page 137 (applications of the theorem of Peter Weyl), and p.166 (consequences of the conjugacy theorem).

Cheers, Amin

share|improve this answer
    
Thanks. I forgot to mention that this question was answered on MathOverflow already. –  Qiaochu Yuan Apr 6 '11 at 16:29
up vote 0 down vote accepted

Asked and answered on MathOverflow.

share|improve this answer

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.