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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?

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You can apply Stone-Weierstrass not to characters but to matrix coefficients (and for functions on $G$ not for functions on conjugacy classes). –  Torsten Ekedahl Mar 16 '11 at 16:45
    
@Torsten: right. That implies that the matrix coefficients I get are dense, and then... oh, then I apply orthogonality for matrix coefficients. Okay, now I have to understand this result (I've only used orthogonality of characters in the past). –  Qiaochu Yuan Mar 16 '11 at 17:25
    
See GTM98 (Representations of Compact Lie groups), p.137 for a proof using Peter-Weyl. –  shenghao Mar 16 '11 at 17:48
    
@shenghao: I think Peter-Weyl is stronger than what I need. As Torsten indicates, the fact that $G$ is already assumed to have a faithful representation means I can use Stone-Weierstrass to prove the part of Peter-Weyl that is relevant. –  Qiaochu Yuan Mar 16 '11 at 17:57
    
what about "for all sufficiently large m,n" ? –  Vivek Shende Mar 17 '11 at 9:38
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2 Answers 2

up vote 11 down vote accepted

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.

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Very nice! I tried to mimic this argument but it is much nicer when you add in that copy of the trivial rep. –  Qiaochu Yuan Mar 16 '11 at 15:59
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The answer is "yes". This is (the compact version of) Proposition 2.20 b on page 139 of the Deligne-Milne article on Tannakian categories in Hodge cycles, Shimura Varieties, and Motives (Springer LNM 900). You can find a scan on Milne's web page.

Edit: You can reconstruct a self-contained proof of the algebraic tensor generation statement by mixing Proposition 3.1 from Deligne's Absolute Hodge article with the lemma in section 3.5 of Waterhouse's Introduction to Affine Group Schemes (GTM 66):

Let $G$ be a linear algebraic group, and $V$ a faithful representation. If $W$ is an $n$-dimensional representation of $G = \operatorname{Spec} A$, then $W$ embeds by the comodule structure map into $A^{\oplus n}$, where $A^{\oplus n} = W \otimes A$ has comodule structure given by $id_W \otimes \Delta$. The comodule structure map is an $A$-comodule homomorphism by the axiom relating coactions with comultiplication, and the injectivity follows from the counit axiom. If $W$ is irreducible, then its embedding in $A^{\oplus n}$ lies in one copy of $A$, so $W$ can be viewed as a sub-$A$-comodule of $A$.

$G$ is a closed subvariety of $GL(V)$, which is in turn a closed subvariety of $\operatorname{Spec}(B)$ by the map $g \mapsto (g, g^{-1})$, where $B = \operatorname{Sym}(\operatorname{End}(V) \oplus \operatorname{End}(V^\vee))$. This yields a surjection $\pi: B \twoheadrightarrow A$ of $A$-comodule algebras. $B$ is inductively filtered by sub-$GL(V)$-modules $B_{k,\ell}$ of polynomials whose degree in $\operatorname{End}(V)$ is at most $k$ and whose degree in $\operatorname{End}(V^\vee)$ is at most $\ell$. Since $W$ is a finite dimensional sub-$A$-comodule of $A$, it is contained in the image of some $B_{k,\ell}$ under $\pi$.

If $C_{k,\ell} \subset B$ is the space of bihomogeneous polynomials of degree $(k,\ell)$, then $B_{k,\ell} \cong \bigoplus_{r\leq k, s \leq \ell} C_{r,s}$ as $GL(V)$-modules. Also $C_{r,s}$ is a $GL(V)$-module quotient of $C_{1,0}^{\otimes r} \otimes C_{0,1}^{\otimes s}$. $C_{0,1}$ is a sum of copies of $V$, and $C_{1,0}$ is a sum of copies of $V^\vee$. This yields $W$ as a quotient of $V^{\otimes k} \otimes (V^\vee)^{\otimes \ell}$ for some $k$ and $\ell$.

This reduces your question to two problems:

  1. Algebrization of compact groups takes complex representations to complex representations.
  2. Finite dimensional representations of compact groups are completely reducible.
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Great! The proposition refers to I.3.1a. I assume that is the first part of the paper? Is there a reasonably self-contained proof of this? –  Qiaochu Yuan Mar 16 '11 at 14:29
    
You can look at page 40 of jmilne.org/math/articles/1982a.pdf to see this proposition. I'm afraid this material is too foreign to me to tell if it's easy to reconstruct a self-contained proof... –  Matthew Daws Mar 16 '11 at 14:43
    
This is interesting, but algebrization of compact groups seems like a little too much work to prove this result. Or is this easier than I think it is? –  Qiaochu Yuan Mar 18 '11 at 21:00
    
I think there is an unavoidable analytic input. –  S. Carnahan Mar 18 '11 at 21:04
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