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It seems that, for $GL_n$, and possibly for something like complex reductive groups $G$ in general, there's an algebraic version of the Peter-Weyl theorem, which might say that the coordinate ring of $G$ decomposes as a direct sum of endomorphisms of all the irreducible algebraic representations. That is, that $$\mathbb C[G] = \bigoplus V^* \boxtimes V$$ as $G \times G$-representations (sum is over all irreps of $G$).

Does anyone happen to know of a reference for this such an algebraic kind of Peter-Weyl?

Grounds for suspicions: this blog post by David Speyer, this comment by Ben Wieland on mathoverflow, and the first few lines of notes from lecture 5b of an MIT seminar on quantum groups that one can find by searching for "peter-weyl algebraic" and that I'm not allowed to link to.

A clean and true statement for when such a theorem might hold would be a lovely start as well, I suppose...

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I found Procesi's book very helpful. –  Akhil Mathew Oct 22 '11 at 5:46
Yes! It looks like what I was looking for is the theorem of section 3.1 in Chapter 7. It looks like Matt E's argument below with a by-hand argument that $V^* \boxtimes V$ is the $V$-isotypic component of $\mathbb C[G]$ instead of the general Frobenius reciprocity statement. Thanks! –  sibilant Oct 23 '11 at 5:37

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This will be true for any complex reductive group. A general Frobenius reciprocity argument shows that $\mathrm{Hom}_G(V,\mathbb C[G]) \cong V^{\vee}$ as $G$-representations. On the other hand, since $G$ is reductive, $\mathbb C[G]$ is a direct sum of irreducible reps. Putting these two observations together proves that indeed $\mathbb C[G] \cong \bigoplus_{V \text{ irred.} } V\boxtimes V^{\vee}$.

It is a good exercise to check this concretely when e.g. $G = \mathrm{SL}_2$.

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So is the fact that ${\rm Hom}_G(V, \mathbb C[G]) \cong V^\vee$ used to conclude that $V \boxtimes V^\vee$ is the $V$-isotypic component of $\mathbb C[G]$? Or is there another way to combine the two observations? Thanks! –  sibilant Oct 23 '11 at 5:50
@sibilant: Dear sibilant, Yes, the first computation exactly say that $V\boxtimes V^{\vee}$ is the $V$-isotypic component for the left action of $G$ on $\mathbb C[G]$, and that the right action on this isotypic component is via the $G$-action on $V^{\vee}$. Regards, –  Matt E Oct 23 '11 at 9:38

There is a proof of this statement for any complex reductive group, more or less along the lines of Matt's sketch, in Chapter 12 of Goodman and Wallach's book "Representations and invariants of the classical groups".

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Thanks! It looks like in the current edition (the one called "Symmetry, Representations, and Invariants") this result appears on p.199 as Theorem 4.2.7 in Chapter 4. You can see it here by searching inside for "199". I can't quite follow the argument there, nor really Matt E.'s sketch above, but it seems that they might indeed be similar. Chapter 12 is more mysterious still. –  sibilant Oct 23 '11 at 4:34
Ok, I think I'm finally starting to understand what's going on. Goodman & Wallach's argument is actually the same as Procesi's: $V^* \boxtimes V$ maps into $\mathbb C[G]$ as the functions given by the matrix coefficients of the action of $G$ on $V$, and then one checks that that's the entire $V$-isotypic component of $\mathbb C[G]$. –  sibilant Oct 23 '11 at 5:43

A good refernce in my opinion is Theorem 27.3.9 in the book by Patrice Tauvel and Rupert Yu named Lie Algebras and Algebraic Groups.

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