"Why" is every polynomial representation of SL(2) selfdual? - MathOverflow

"Why" is every polynomial representation of SL(2) selfdual?

2010-09-09T12:09:52Z

Given a field $K$ of characteristic $0$. It seems to me that every finite-dimensional polynomial representation of $\mathrm{SL}_2\left(K\right)$ is self-dual (i. e., isomorphic to its dual). In fact, every representation of $\mathrm{SL}_2\left(K\right)$ is a direct sum of irreducible representations (since $\mathrm{SL}_2\left(K\right)$ is semisimple), and the irreducible representations are the canonical representations on $K\left[x,y\right]_n$ which are known to be self-dual. But is there a simpler proof without subdividing into irreducibles?

Answer by Richard Borcherds for "Why" is every polynomial representation of SL(2) selfdual?

2010-09-09T13:37:09Z

Because its Weyl group contains -1.

For split semisimple groups in char 0, taking duals corresponds to acting by -1 on the weight lattice, where irreducible polynomial representations correspond to weights modulo the action of the Weyl group. So if -1 is in the Weyl group (acting on the weight lattice), then any (irreducible) representation is isomorphic to its dual. This includes the groups with Dynkin diagrams A1, Bn, Cn, Dn for n even, E7, E8, F4, G2 but not An for n>1, Dn for n odd and E6.

Answer by Joel Kamnitzer for "Why" is every polynomial representation of SL(2) selfdual?

2010-09-09T13:59:02Z

Here is a variant of Richard's answer. In the group $ SU(2) $ every element is conjugate to its inverse. Hence the characters of a representation and its dual are the same. (Richard's answer is essentially this answer restricted to the torus, which is sufficient.)

To alleviate your concerns, smooth complex representations of $SU(2) $ and algebraic complex representations of $SL_2(\mathbb{C}) $ are the same thing.

Answer by George McNinch for "Why" is every polynomial representation of SL(2) selfdual?

2010-09-09T15:08:13Z

I'm not sure that you can expect to avoid "subdividing into irreducibles", since the result is not true without semisimplicity. If $K$ has characteristic $p>0$, the $SL_2(K) = SL(V)$-representation on the $p+1$-dimensional space $\operatorname{Sym}^p(V)$ (the $p$th symmetric power of the natural 2 dim'l representation) is not self-dual.

Answer by Jim Humphreys for "Why" is every polynomial representation of SL(2) selfdual?

2010-09-09T18:11:32Z

The question itself and some of the comments seem out of focus to me, so let me add to what Richard and George write the following summary version of an answer. I'd stress that nothing here is really complicated or subtle to prove apart from the basic Cartan-Weyl classification and (in characteristic 0) complete reducibility for finite dimensional representations.

First, the group itself is defined and split over the prime field (here $\mathbb{Q}$ ), hence over any larger field. Chevalley's theory implies that the representations discussed here are absolutely irreducible over $K$ . (For a semisimple group defined but not split over a field, more analysis is needed of representations which require a field extension to become absolutely irreducdible.)

Anyway, for a connected semisimple group the "rational" and "polynomial" representations are the same, unlike the reductive group GL $(n,K)$ . The group also being simply connected in this case, the rational/polynomial representations are essentially those of the Lie algebra and are more easily classified by dominant integral highest weights in that setting. So each irreducible representation or simple module in question has a unique highest weight $\lambda$ . The easy textbook criterion for such a module to be self-dual is just that $\lambda = -w_0 \lambda$ where $w_0$ is the longest element of the Weyl group. As Richard Borcherds points out, this is -1 just for simple types listed, including type $A_1$ .

So far nothing really depends on characteristic 0. But as George McNinch observes, there are plenty of cases where nonsimple modules in prime characteristic fail to be completely reducible and are typically not self-dual. So you do need to invoke complete reducibility (and non-canonical direct sum decompositions) to dispose of the characteristic 0 question.

P.S. It's certainly possible to treat the rank 1 case here by direct ad hoc methods in characteristic 0, including the needed proof of complete reducibility (using the easily computed Casimir operator). For irreducible representations, self-duality is a trivial consequence of the fact that these representations are uniquely classified (up to isomorphism) by their dimensions 1, 2, 3, .... But such an ad hoc argument fails to provide much enlightenment. And the general theory allows one to see that the group representations and Lie algebra representations are essentially the same, whether the groups are regarded as Lie groups or algebraic groups (or just as abstract groups). Of course, finite dimensionality is a key point throughout, since the infinite dimensional representation theory involves harder questions.