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?
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asked Sep 9, 2010 at 12:09
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$\begingroup$ By the way, I am aware that for compact Lie groups, we have a Haar measure and we get an invariant bilinear form by integrating an arbitrary bilinear form. But there are things I don't like here: (1) This works only over $\mathbb R$, and we would still have to prove that all irreducible representations of $\mathrm{SL}_2\left(\mathbb C\right)$ are already defined over $\mathbb Q$. (2) $\mathrm{SL}_2$ is not compact and we would need Weyl's unitary trick to get it compact. I'm not sure whether it really preserves all representations. (3) I hate analysis and want to see it exterminated. :P $\endgroup$– darij grinbergCommented Sep 9, 2010 at 12:19
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2$\begingroup$ I'm confused. How could it be possible to have a `canonical' choice for the isomorphism? Even if V is irreducible, this can't be done because (scalar) automorphisms of V act on such choices non-trivially. $\endgroup$– t3sujiCommented Sep 9, 2010 at 12:23
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$\begingroup$ Uhm, yes. Sorry! I meant a canonical (up to scalar multiplication) choice. $\endgroup$– darij grinbergCommented Sep 9, 2010 at 12:27
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3$\begingroup$ @darij grinberg: this does not help for reducible representations: again, consider the action of automorphisms on possible choices. $\endgroup$– t3sujiCommented Sep 9, 2010 at 12:31
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1$\begingroup$ Darij, you need to be a lot more clear from the get-go about the category of representations you are interested in. For example, any automorphism $\sigma$ of $K$ induces an automorphism of $SL_2(K)$ viewed as an abstract group; composing it with a representation $\rho$ that occurs in the tensor space gives one that does not, unless $\sigma$ or $\rho$ is trivial. (Borel and Tits proved some amazing results about "abstract" homomorphisms of algebraic groups which may imply that this is the only obstruction in char 0.) Of course, this is impossible in the category of rational representations. $\endgroup$– Victor ProtsakCommented Sep 9, 2010 at 16:01
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4 Answers
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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.
answered Sep 9, 2010 at 13:37
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$\begingroup$ This is again a reduction to the irreducibles, but it gives a nice context. Thanks! $\endgroup$ Commented Sep 9, 2010 at 14:32
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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.
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$\begingroup$ Oh, the character theory approach. But using character theory to prove representations isomorphic requires knowing that the group is semisimple, right? $\endgroup$ Commented Sep 9, 2010 at 14:31
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$\begingroup$ @darij grinberg: for $\chi_V=\chi_W$ to imply $V\cong W$ we don't need the group to be semisimple (in the sense of algebraic groups, i.e. its Lie algebra ..., since $\mathbb R$ is not algebraically closed anyway); all we need is that the (continuous) representation theory of the Lie group $SU(2)$ is semisimple, which comes from the fact that one can do integral w.r.t. the Haar measure on compact Lie groups. $\endgroup$– shenghaoCommented Mar 10, 2011 at 23:53
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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.
answered Sep 9, 2010 at 15:08
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$\begingroup$ Okay, this show that $\mathrm{char}K=0$ is important. But there is still hope for synthetic methods such as the embedding $\mathrm{Sym}^n V\to \otimes^n V$ which require zero characteristic. $\endgroup$ Commented Sep 9, 2010 at 15:45
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$\begingroup$ Why is this? Do you need $p$ odd? (In the case $p=2$ it looks to me like perhaps conjugating by the matrix with 1s along the antidiagonal should exhibit self-duality, unless I've made a foolish error, which is likely). $\endgroup$ Commented Jul 19, 2013 at 14:49
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$\begingroup$ @TomLovering: If p=2, then Sym^2(V) is the adjoint representation of sl(2), and is not self-dual. I'm not completely sure what sort of answer you want to the question "why is this?". Essentially, one must check that there is a submodule $V^{[1]} \subset \operatorname{Sym}^{p+1}V$ (with $V^{[1]}$ isom to the Frobenius twist of $V$) that has no complement. $\endgroup$ Commented Jul 22, 2013 at 22:38
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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.
answered Sep 9, 2010 at 18:11