Algebraicity of holomorphic representations of a semisimple complex linear algebraic group

Question

Let $G$ be a complex linear algebraic group, given to us as a closed subgroup of some $\mathrm{GL}(n,\mathbb{C})$. Suppose moreover that $G$ is semisimple. Then it's a fact that every finite-dimensional holomorphic representation of the complex Lie group $G(\mathbb{C})$ is actually an algebraic representation (i.e., given by polynomials in the matrix entries, together with $\mathrm{det}^{-1}$).

One can certainly deduce this from the highest-weight theory (that is, by provably constructing all the representations, and noting that everything you've constructed is in fact algebraic). But this isn't remotely satisfying.

In another direction, I was led by notes of Milne to a book and some papers by Dong Hoon Lee, and from those to a series of papers by Hochschild and Mostow. But those authors want to do something harder: classify the Lie groups (not necessarily semisimple!) that can be given an algebraic group structure such that all the holomorphic representations of the Lie group are algebraic representations of the algebraic group. It seems to me that if you start out with an algebraic group, and then assume that the group is semisimple, then most of the complications should go away.

So my question is, is there a satisfying and reasonably elementary proof of the fact in the first paragraph? I should say something about my motivation. I'll be teaching a Lie groups/algebras course next year, and when we talk about representation theory we'll observe this phenomenon; so it would be nice to explain it if there's a reasonable way to do so. Given that I can't expect my students to have had an algebraic geometry course, I'd want to minimize the algebraic geometry in the argument, possibly at the cost of making more serious use of the structure theory of Lie groups/algebras.

Here's an approach that I would find especially clarifying if it can be made to work. We're handed a faithful representation of $G(\mathbb{C})$ (the inclusion into $\mathrm{GL}(n,\mathbb{C})$) which is certainly algebraic. Its tensor powers are algebraic. Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one. How might one prove the latter? (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)

Answer 1

"Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one. How might one prove the latter? (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)"

Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}_K(W,V^{\otimes N}) \cong (W^* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int_K \overline{\chi_W} \cdot \chi_V^{N}$, where $\chi_V$ and $\chi_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d_V$ and $d_W$ be the dimensions of $V$ and $W$.

We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi_V(g)| < d_V$. We first present the proof in the setting that $Z = \{ e \}$.

Choose a neighborhood $U$ of $\{ e \}$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi_V(g) = d_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi_W(g) = d_W + O(g-e)$. Manipulating $\int_U \overline{\chi_W} \chi_V^N$ should give you $$\frac{d_W \pi^{\dim K/2}}{\det Q} \cdot d_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$ Meanwhile, there is some $D<d_V$ such that $|\chi_V(g)| < D$ for $g \in K \setminus U$. So the integral of $\overline{\chi_W} \chi_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d_V^N$ term in the $U$ integral.

We deduce that, unless $d_W=0$, we have $\mathrm{Hom}_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.

If $Z$ is greater than $\{e \}$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V_N:=V^{k+Nd_V} (\det \ )^{-N}$. $V_N$ is constructed so that $\overline{\chi_W} \chi_{V_N}$ is identically $d_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.

Answer 2

I also want to add another, much more elementary, answer in the case $G=\mathbb{C}^*$. Let $\rho(t)$ be a holomorphic map $\mathbb{C}^* \to \mathrm{GL}_n(\mathbb{C})$. Then we can write $\rho$ as a convergent sum $\sum_{i=-\infty}^{\infty} P_i t^i$. Write out the equation $\rho(tu) = \rho(t) \rho(u)$ and look at the $t^i u^i$ term to deduce that $P_i^2=P_i$. So $\mathrm{Tr} \ P_i$ is a nonnegative integer, equal to its rank. But $\sum P_i = \rho(1) = \mathrm{Id}_n$, so $\sum \mathrm{Tr} P_i =n$. We deduce that all but at most $n$ of the $P_i$ must be $0$, so $\rho(t)$ is a polynomial.

I've been trying to think of a clever way to extend this to the general case, using that a generic matrix is diagonalizable, but I haven't found one yet.

Answer 3

(Edit: removed application of Peter-Weyl's --- thanks to Victor Protsak for pointing out it is unnecessary.)

As long as you don't mind passing to compact subgroups, it can be done using Weyl's unitary trick. Here's a sketch: let $K\subset G({\mathbb C})$ be the maximal compact. Consider in $L^2(K)$ the matrix elements of irreducible representations. They form an orthogonal set. On the other hand, the subset consisting of matrix elements of irreducible algebraic representations spans $L^2(K)$ (by the Stone-Weierstrass Theorem --- any polynomial in matrix elements and $det^{-1}$ is a linear combination of those). Hence the two sets are the same.

P.S. If you prefer, you can work with characters instead of matrix elements and with class functions instead of all functions.

Answer 4

I can think of a funny shortcut but you will be there on your own in the mathematical jungle (it is not written anywhere as far as I know) and you may lose your students to tigers and anakondas:-))

Write your group by generators and relations (using Tits-Steinberg relations and unipotent root subgroups as generators). Now start with a holomorphic representation, differentiate it to Lie algebra, and verify that nilpotent elements act nilpotently. Now you are home: exponents of your generators are polynomials and, so will be for any element...

BTW, your idea of tensoring things around could be fun but you will have to do it case by case and you won't get far in exceptional cases anyway. You may read a light account of it in Fulton-Harris and a much heavier one in Weyl's Invariant Theory.