(Previously posted on [math.SE](https://math.stackexchange.com/questions/26909/does-every-irreducible-representation-of-a-compact-group-occur-in-tensor-products) 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](https://mathoverflow.net/questions/40926/a-faithful-representation-generates-all-the-other-representations) 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? 



Does every irreducible representation of a compact group occur in tensor products of a faithful representation and its dual?
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 $\dim \mathrm{Hom}\_G(X,Y) = \int\_G \overline{\chi\_X} \chi\_Y$, 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](http://sbseminar.wordpress.com/2010/11/25/passage-from-compact-lie-groups-to-complex-reductive-groups/).


