Decomposition of k[G] - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:50:30Z http://mathoverflow.net/feeds/question/3474 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3474/decomposition-of-kg Decomposition of k[G] Ilya Nikokoshev 2009-10-30T21:55:07Z 2010-02-03T22:41:05Z <p>There's a well-known decomposition of $L^2(G)$, a regular representation of compact complex group Lie $G$, called <strong><a href="http://en.wikipedia.org/wiki/Peter%E2%80%93Weyl%5Ftheorem" rel="nofollow">Peter-Weyl theorem</a></strong>.</p> <p>Turns out for some reason I <a href="http://mathoverflow.net/questions/3446/tannakian-formalism/3466#3466" rel="nofollow">automatically think</a> that there is a similar theorem that decomposes regular representation $k[G]$ of <em>algebraic</em> group $G$:</p> <p>$$k[G] = \bigoplus_R \ R^* \otimes R$$</p> <p>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 $\pi_1(G) = 1$. </p> <p>But perhaps this is not true &mdash; the search hasn't given me a reference yet, but I wasn't able to provide a counterexample either.</p> <p>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.</p> <p>So, is there such a theorem? What's a reference or a counterexample?</p> http://mathoverflow.net/questions/3474/decomposition-of-kg/3493#3493 Answer by Fran Burstall for Decomposition of k[G] Fran Burstall 2009-10-30T23:47:57Z 2009-10-30T23:47:57Z <p>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.</p> <p>See Theorem 12.1.4 of Goodman and Wallach <a href="http://books.google.co.uk/books?id=MYFepb2yq1wC&amp;lpg=PP1&amp;ots=J8xg-38RVi&amp;dq=goodman%20wallach&amp;pg=PP1#v=onepage&amp;q=&amp;f=false" rel="nofollow">Representations and Invariants of the Classical Groups</a>. </p> http://mathoverflow.net/questions/3474/decomposition-of-kg/3580#3580 Answer by David Jordan for Decomposition of k[G] David Jordan 2009-10-31T18:45:10Z 2009-11-01T08:23:19Z <p>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 &otimes; O(G), or equivalently V^* &otimes; 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 &otimes;' B, C &otimes;' D) = d(A,C) &otimes; d(B,D) where d(V,W)=0 if v \cong W, C else. Here &otimes;' means external tensor product. There doesn't appear to be a &#x022A0;</p> <p>For non-reductive groups, you can still form O(G) in an analogous way:</p> <p>Let A = &oplus;<sub>V</sub> V^* &otimes;' 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) &otimes;' Rep(G), and &otimes;' means Deligne tensor product of categories.</p> <p>Well this A is way too big, but now let's quotient A by the images of f^* &otimes;' id - id &otimes;' 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.</p> <p>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^* &otimes;' V &otimes;<sub>2</sub> W^* &otimes;' W = V^* &otimes; W^* &otimes;' V &otimes; W --> W^* &otimes; V^* &otimes;' V &otimes; W, using the braiding (tensor swap). It also works in braided tensor categories and explains the multiplication structure on the "covariantized" quantum group.</p> http://mathoverflow.net/questions/3474/decomposition-of-kg/4322#4322 Answer by Chuck Hague for Decomposition of k[G] Chuck Hague 2009-11-06T00:36:33Z 2010-02-03T22:34:59Z <p>This statement is <strong>false in general</strong> 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 <a href="http://books.google.com/books?id=Liqas0afjR0C&amp;lpg=PP1&amp;dq=Jantzen%20%22Algebraic%20Groups%22&amp;pg=PA213#v=onepage&amp;q=&amp;f=false" rel="nofollow">page 213 in Jantzen's "Algebraic Groups"</a>): </p> <blockquote> <p>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 <em>costandard modules</em> for $G$). </p> </blockquote> <p>Moreover, this is true when $k[G]$ is considered as a $G\times G$-module.</p> <p>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 &mdash; this is only true in characteristic 0).</p>