# Does every irreducible representation of a compact group occur in tensor products of a faithful representation (and its dual)?

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.

• 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-Álvarez Mar 14 '11 at 16:58

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.