# Decomposition of k[G]

There's a well-known decomposition of $L^2(G)$, a regular representation of compact complex group Lie $G$, called Peter-Weyl theorem.

Turns out for some reason I automatically think that there is a similar theorem that decomposes regular representation $k[G]$ of algebraic group $G$:

$$k[G] = \bigoplus_R \ R^* \otimes R$$

where sum goes over representations to $GL(n, k)$. For this to work I think we need $G$ to be a linear reductive group over, say, algebraically closed field $k$ of characteristic 0. Also, perhaps we need $\pi_1(G) = 1$.

But perhaps this is not true — the search hasn't given me a reference yet, but I wasn't able to provide a counterexample either.

Consider, for example, the multiplicative group $\mathbb G_m$. Then $k[\mathbb G_m] = k[x, x^{-1}]$ where each summand $k\cdot x^n$ is a separate representation of $\mathbb G_m$ into $\mathbb G_m = GL(1, k)$, specifically the one given by $a \mapsto a^n$. So the identity works.

So, is there such a theorem? What's a reference or a counterexample?

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I also found books.google.com/… –  Ilya Nikokoshev Oct 30 '09 at 22:04
What happens for SL(2,k)? When k is the complex numbers, this has no non-trivial finite-dimensional unitary representations, so one can't get decomposition of the left-reg rep of SL(2,C) as a sum of fin-dim reps. But your question is slightly different. Nonetheless, SL(n,k) or its universal cover seems an obvious first case to consider –  Yemon Choi Oct 30 '09 at 22:24
I tried checked it before posting an answer &mdash; it feels like it works and passes some checks, e.g. you find the 3- and 5- dimensional reps. But not sure yet. –  Ilya Nikokoshev Oct 30 '09 at 23:03
Indeed, SL(2) algebraic <---> SU(2) complex (for the purposes of this question at least). –  Ilya Nikokoshev Oct 30 '09 at 23:04
Yemon- algebraic functions on SL(2,R) aren't L^2. They don't form a Hilbert space, and the action isn't unitary. –  Ben Webster Oct 31 '09 at 0:55
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This is true for reductive groups, more or less by definition. An algebraic representation of an algebraic group is a comodule V over the algebra of functions O(G) of the group. Therefore, every representation V induces a map V -> V ⊗ O(G), or equivalently V^* ⊗ V --> O(G) (call the source of this map C(V) for coefficient space of V). It is not hard to see that the latter is a map of G x G modules. If G is reductive, then its representation category is semi-simple, and thus so is the representation category of G x G. In this case the simples of G x G are external tensor product of simples of V, and Hom(A ⊗' B, C ⊗' D) = d(A,C) ⊗ d(B,D) where d(V,W)=0 if v \cong W, C else. Here ⊗' means external tensor product. There doesn't appear to be a ⊠

For non-reductive groups, you can still form O(G) in an analogous way:

Let A = ⊕V V^* ⊗' V, where here the sum is over ALL finite dimensional modules V (not just isoclass representatives, and not just simples), and again the tensor product is external, so this lives in a completion of Rep(G) ⊗' Rep(G), and ⊗' means Deligne tensor product of categories.

Well this A is way too big, but now let's quotient A by the images of f^* ⊗' id - id ⊗' f, for all f:V-->W. This cuts A back down, for instance it identifies C(V) and C(V') whenever V and V' are isomorphic. If the category Rep(G) is semi-simple, you can similarly use the projectors and inclusions of simple objects to reduce to a Peter-Weyl type decomposition.

One nice thing about this construction (even in the semi-simple case) is that it is basis free because you don't choose representatives of simple objects, and also it makes the multiplication structure completely trivial: V^* ⊗' V ⊗2 W^* ⊗' W = V^* ⊗ W^* ⊗' V ⊗ W --> W^* ⊗ V^* ⊗' V ⊗ W, using the braiding (tensor swap). It also works in braided tensor categories and explains the multiplication structure on the "covariantized" quantum group.

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Thanks David! It's great to meet again, funny how MathOverflow connects people. By the way, I changed &boxtimes to its Unicode code, since it appears Safari doesn't like the former one. –  Ilya Nikokoshev Nov 1 '09 at 8:30
Good to see you too, and read all your questions! –  David Jordan Nov 1 '09 at 14:19
I'd like to add Chuck's comment from below to here, lest anyone be confused by what I wrote above. As Chuck pointed out, "reductive" doesn't mean "the category of its modules is semi-simple" in non-zero characteristic. So the first sentence above should have read "In the case your category is semi-simple"... However, the general construction starting in paragraph 2 should hold in any characteristic. Probably Jantzen's filtration is related to what is discussed in paragraph 4 above. –  David Jordan Nov 12 '09 at 15:30

This statement is false in general for algebraic groups. It's true in characteristic 0, but it is not in general true in positive characteristic. Instead, one has a weaker statement in positive characteristic (cf Proposition 4.20 on page 213 in Jantzen's "Algebraic Groups"):

Let $G$ be a reductive linear algebraic group over an algebraically closed field of positive characteristic $k$. Then $k[G]$ has an increasing filtration whose subquotients are of the form $H(\lambda) \otimes H(-w_0 \lambda)$, where $\lambda$ runs over the dominant weights for $G$ and the $H(\lambda)$ are the modules arising as global sections of line bundles on the flag variety of $G$ (the so-called costandard modules for $G$).

Moreover, this is true when $k[G]$ is considered as a $G\times G$-module.

Note that unlike in characteristic 0, these modules $V$ are not in general irreducible. (It's worth noting that the category of modules over a reductive algebraic group is not in general a semisimple category — this is only true in characteristic 0).

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Good to know! The question was restricted to char 0 since I suspected something will not work, thanks for an explanation of what breaks down! –  Ilya Nikokoshev Nov 6 '09 at 16:52
Absolutely! I just fixed a small problem in what I'd written, but it wasn't a big thing. You can search inside Jantzen's book on Google Books if you're interested in the proof -- it's on page 213. –  Chuck Hague Nov 6 '09 at 17:49
Your answer is very informative: I returned to change formatting and expand it a bit; feel free to revert! –  Ilya Nikokoshev Feb 3 '10 at 22:35
That's what I naively think, but then how to explain the fact that textbooks and papers almost always consider only the case of k=C? What if there are some important nuances? –  Ilya Nikokoshev Oct 31 '09 at 10:32