(Previously posted on math.SE with no answers.)
Let $G$ be a compact Lie group and $V$ a faithful (complex, continuous, finite-dimensional) representation of it. Is it true that every (complex, continuous, finite-dimensional) irreducible representation of $G$ occurs in $V^{\otimes n} (V^{\ast})^{\otimes m}$ for some $n, m$?
The proof I know for finite groups doesn't seem to easily generalize. I want to apply Stone-Weierstrass, but can't figure out if the characters I get will always separate points (in the space of conjugacy classes). Ben Webster in his answer to this MO question seems to be suggesting that this follows from the first part of Peter-Weyl, but I don't see how this works.
Certainly the corresponding algebra of characters separates points whenever the eigenvalues of an element $g \in G$ acting on $V$ determine its conjugacy class (since one can get the eigenvalues from an examination of the exterior powers). Does this always happen?
Best Answer
You can mimic the standard finite group argument. Recall that, if $X$ and $Y$ are two finite-dimensional representations of $G$, with characters $\chi_X$ and $\chi_Y$, then $\dim \mathrm{Hom}_G(X,Y) = \int_G \overline{\chi_X} \otimes \chi_Y$, where the integral is with respect to Haar measure normalized so that $\int_G 1 =1$. The proof of this is exactly as in the finite case.
Now, let $W=1 \oplus V \oplus \overline{V}$. Your goal is to show that, for any nonzero representation $Y$, $\mathrm{Hom}(W^{\otimes N}, Y)$ is nonzero for $N$ sufficiently large. So you need to analyze $\int (1+\chi_V + \overline{\chi_V})^N \chi_W$ for $N$ large. Just as in the finite group case, we are going to split this into an integral near the identity, and an exponentially decaying term everywhere else.
I'm going to leave a lot of the analytic details to you, but here is the idea. Let $d=\dim V$. The function $f:=1+\chi_V + \overline{\chi_V}$ makes sense on the entire unitary group of $V$, of which $G$ is a subgroup. On a unitary matrix with eigenvalues $(e^{i \theta_1}, \cdots, e^{i \theta_d})$, we have $f=1+2 \sum \cos \theta_i$. In particular, for an element $g$ in the Lie algebra of $G$ near the identity, we have $$f(e^g) = (2n+1) e^{-K(g) + O(|g|^4)}$$ where $K$ is $\mathrm{Tr}(g^* g) = \sum \theta_i^2$. And, for $g$ near the identity, $\chi_X(e^g) = d + O(|g|)$. So the contribution to our integral near the identity can be approximated by $$\int_{\mathfrak{g}} \left( (2n+1) e^{-K(g) + O(|g|^4)} \right)^N (d+ O(|g|) \approx (2n+1)^N d \int_{\mathfrak{g}} e^{-K(\sqrt{N} g)} = \frac{(2n+1)^N d}{N^{\dim G/2}} C$$ where $C$ is a certain Gaussian integral, and includes some sort of a factor concerning the determinant of the quadratic form $K$. You also need to work out what the Haar measure of $G$ turns into as a volume form on $\mathfrak{g}$ near $0$, I'll leave that to you as well.
For right now, the important point is that the growth rate is like $(2n+1)^N d$ divided by a polynomial factor. In the finite group case, that polynomial is a constant, which makes life easier, but we can live with a polynomial.
Now, look at the contribution from the rest of $G$. For any point in the unitary group $U(d)$, other than the identity, we have $|f| < 2n+1$. (To get equality, all the eigenvalues must be $1$ and, in the unitary group, that forces us to be at the identity.) So, if we break our integral into the integral over a small ball around the identity, plus an integral around everything else, the contribution of every thing else will be $O(a^N)$ with $a<2n+1$.
So, just as in the finite group case, the exponential with the larger base wins, and the polynomial in the denominator is too small to effect the argument. Joel and I discussed a similar, but harder, argument over at the Secret Blogging Seminar.