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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?)

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Even if a proof using tensor algebra can be given, it can hardly be called illuminating. If you are willing to do a bit more legwork, you can prove Borel-Weil theorem and observe that $G/B$ is algebraic. That has extra payoff of explicitly realizing all irreducible representations. I seem to remember that Pressley and Segal, "Loop groups", had a nice self-contained approach. –  Victor Protsak Jun 11 '10 at 17:12
David, if you're going to teach max. compacts and real Lie gps arising from linear alg. groups over $\mathbf{R}$, may want to follow development of max. compacts in Ch. XV of Hochschild's book "The structure of Lie groups" since handles case when comp. gp is nontrivial but finite (most references assume connectedness at the outset, and seems not obvious how to get theory of max. compacts to work for finite comp. gp as formal consequence of theory in conn'd case). Also, Ch. XVII gives good discussion of "complexification" of compact Lie gps and relation with max. compacts of complex Lie gps. –  BCnrd Jun 11 '10 at 20:58
This is definitely a tricky area for exposition, since older work relied on passing from complex semisimple groups to compact real forms (following Weyl) and then studying representations via the complexified Lie algebra. Algebraic geometry and algebraic groups come into the picture later, historically. Shortcuts are essential in a course, but may not be "transparent". One graduate text to be aware of Several Complex Variables with Connections to Algebraic Geometry and Lie Groups - Joseph L. Taylor, AMS, 2002. Ignoring the title, the last chapters may be relevant for you (or not). –  Jim Humphreys Jun 11 '10 at 21:05
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4 Answers

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"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.

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Of course, to use this approach, you still have to show that your complex group $G$ contains a compact subgroup $K$ of half dimension. What definition of reductive/semisimple you are using? –  David Speyer Jun 11 '10 at 19:01
Thanks! I haven't read your answer carefully yet (I certainly will do so) but for now let me respond to your comment. I don't know if I will get far enough to prove the existence of compact real forms, but I'm perfectly happy to check it on a case by case basis. That is, if I can explain how the observed (with serious difficulty!) phenomenon that all the reps are algebraic follows from the observed (with not much difficulty) phenomenon that there's a compact real form, I think that's satisfying. But I do have a full year, so I can hope to get to the existence of compact real forms. –  D. Savitt Jun 11 '10 at 20:11
One can also use Stone-Weierstrass directly: it implies that character of $W$ can be well-approximated by a polynomial in the matrix entries of $V$ and their conjugates. So $\chi_W$ is not orthogonal to some matrix entry for the representation $V^{\otimes k} \otimes (V^*)^{\otimes l}$. This implies that there is a nonzero homomorphism from $W$ to $V^{\otimes k} \otimes (V^*)^{\otimes l})$. –  moonface Jun 11 '10 at 21:12
Existence of compact real forms for Lie algebras is relatively straightforward. Passing to groups requires nontrivial structure theory. In particular, an honest (hah!) unitary trick requires showing that the corresponding $K$ has finite center. To me, it's just unsatisfying that in order to prove purely complex/algebraic result, you need to develop all this real theory that's not going to be used anywhere (unless you head int the direction of representation theory of real reductive groups). –  Victor Protsak Jun 11 '10 at 22:32
Thanks, David & moonface. –  D. Savitt Jun 12 '10 at 9:20
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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.

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Earlier today I asked my question to someone here at MPIM and I think we worked out something along these lines. The group G is generated by finitely many tori; in fact you have a dominant map P \to G where P is a product of finitely many tori, and the composite map to GL_n is algebraic. So you're done if you can prove that if X \to Y is a dominant algebraic map and Y \to Z is an holomorphic map such that the composite X \to Z is algebraic, then Y \to Z is algebraic. I think we sketched this (saying so, I hope it's true!!) but it was a bit much to give to my students. Thoughts? –  D. Savitt Jun 11 '10 at 20:23
D. Savitt, since not assuming students have experience in alg. geom, can they follow arguments using "dominant algebraic map", etc.? I recommend not proving these things -- point out algebraicity of many examples and tell them that after they learn some alg. geom. they can learn the purely algebraic approach to the theory (i.e., "linear alg. groups") which explains the algebraicity of the examples in a systematic manner. Tell them w/o proof that (over C) image of adj. repn is "algebraic" when Lie alg. is ss; that clarifies why "algebraic" objects arise naturally in the analytic theory. –  BCnrd Jun 11 '10 at 21:30
BCnrd -- indeed, I certainly would not give my class the argument that's sketched in my comment. –  D. Savitt Jun 11 '10 at 21:41
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(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.

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There are many technicalities to keep track of in this approach (due to Weyl), starting with the fact that $GL(n,\mathbb{C})$ is not closed in $n\times n$ complex matrices (consequently, not all its representations are polynomial). Assuming only basic facts about $K$, how do you show that there is only one holomorphic representation of $G$ restricting to a given irreducible representation of $K$? Or that an irreducible holomorphic representation of $G$ remains irreducible upon restriction to $$K$? –  Victor Protsak Jun 11 '10 at 22:11
First comment: I am not sure I understand the problem. Real polynomials are dense in L^2(K), therefore, complex polynomials and their conjugates span everything. On K, the conjugate of a matrix element is a matrix element of the dual representation. So you get that every irrep is either polynomial or its dual is polynomial. Other comments: What exactly are `basic facts'? I need to know that the real dimension of K equals to the complex dimension of G. Do you want to forget about this assumption? –  t3suji Jun 11 '10 at 22:40
I think what's needed is the fact that $K$ is Zariski dense in $G$, not just that $\mathrm{dim}_R K=\mathrm{dim}_C G.$ I guess I don't understand what you mean by "complex polynomials and their conjugates span everything": when we start out, we don't know that all holomorphic representations are polynomial (or rational), and there is no "Peter-Weyl theorem" on G (indeed, harmonic analysis on $L^2(G)$ involves unitary reps of G none of which is fin-dim). So how are you going to conclude something about holomorphic f.d. representations of $G$ from your knowledge of representations of $K$? –  Victor Protsak Jun 12 '10 at 3:07
I am probably missing something simple. Take for granted that the Lie algebra of $G$ is the complexification of the Lie algebra of $K$ (this easily follows from $dim_RK=dim_CG$). Then any holomorphic function on $G$ that vanishes on $K$ vanishes identically. In particular, if a matrix representation is upper-triangular on $K$, it is upper-triangular on $G$, and if two matrix representations are equal on $K$, they are equal on $G$. Thus irreducible representations of $G$ form a subset of irreducible representations of $K$. –  t3suji Jun 12 '10 at 4:29
Yes, you are right, irreducibility follows. I've removed the previous comment in which I was thinking of a different meaning of "holomorphic" that doesn't apply here. So the argument, as I understand it, goes as follows. Restrict a matrix coefficient of a f.d. holomorphic rep of G to K: it becomes a bi-K-finite function on K, hence a polynomial on K using Peter-Weyl when G is semisimple (not true for C* and the circle). Restriction map is injective by polar decomposition G=KP ("Zariski density of K in G"), so the original matrix coefficient is a polynomial. –  Victor Protsak Jun 12 '10 at 5:50
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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.

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