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

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Jul 22, 2013 at 22:38	comment	added	George McNinch		@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.
Jul 19, 2013 at 14:49	comment	added	Tom Lovering		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).
Sep 9, 2010 at 15:45	comment	added	darij grinberg		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.
Sep 9, 2010 at 15:08	history	answered	George McNinch	CC BY-SA 2.5