Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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 — 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
2  
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

3 Answers 3

up vote 15 down vote accepted

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.

share|improve this answer
    
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).

share|improve this answer
    
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

The result is true for linear algebraic reductive groups over C. The sum is over all (isomorphism classes of) irreducible regular finite dimensional representations and the isomorphism is an isomorphism of G\times G-modules.

See Theorem 12.1.4 of Goodman and Wallach Representations and Invariants of the Classical Groups.

share|improve this answer
    
The proof does not use much more than Schur's Lemma and the fact that k[G] is a locally regular repn of G\times G so I guess it goes through for any algebraically closed k. –  Fran Burstall Oct 31 '09 at 0:15
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.