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?)



Algebraicity of holomorphic representations of a semisimple complex linear algebraic group

> 
> "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.


