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Jul 23, 2016 at 8:25 comment added Wilberd van der Kallen @ Jim: There is nothing wrong with Erdmann's paper, but if you just want the messy concrete summand complementary to $S^{r-1}(E)$, then the linear algebra exercise based on the hyperalgebra gives the span of $x\otimes x^r$, $y\otimes y^r$, $(r+1)x\otimes x^{r-j}y^j-j(x\otimes x^{r-j}y^j-y\otimes x^{r-j+1}y^{j-1})$, with $1\leq j\leq r$ . One checks it is invariant under the generators of the hyperalgebra.
Jul 22, 2016 at 17:31 comment added Jim Humphreys @Wilberd: Yes, the short exact sequence (which I guess Chunna deleted) splits precisely when $p$ doesn't divide $r+1$. This is easy theoretically in Jantzen's framework of $G_1 T$-modules, where $G_1$ is the first Frobenius kernel and $T$ a maximal torus of $G$: in one direction, use the linkage principal, and in the other direction use my nonsplit example to get other injective/projective $G_1 T$-modules. But making it all concrete may be messier. Erdmann's paper seems fairly straightforward.
Jul 22, 2016 at 14:01 comment added Wilberd van der Kallen Using computations with the hyperalgebra I seem to get that the exact sequence splits precisely when $r+1$ is not divisible by $p$. In fact I am just guessing we are talking about the same exact sequence. Curiously the text in the answer no longer matches the question. Who caused that?
Jul 21, 2016 at 13:59 comment added Jim Humphreys @Wilberd: Yes, there is a more sophisticated general theory to invoke here, though I was trying at first to keep the focus elementary. By now there is a lot of literature which applies in particular to SL$_2$, but in higher ranks the details are usually unknown or at least very hard to work out.
Jul 21, 2016 at 13:56 history edited Jim Humphreys CC BY-SA 3.0
added 1263 characters in body
Jul 21, 2016 at 13:52 comment added Chunna @van der Kallen Is there any chance you can point me in the direction of a reference where i can find more details . I am definitely looking for a split surjection.
Jul 21, 2016 at 7:18 comment added Wilberd van der Kallen Good filtration theory tells us that there is always a unique submodule of $E \otimes S^r (E)$ that is isomorphic to $S^{r-1} (E)$. But if the characteristic is two and $r$ equals one, the surjection has this submodule in the kernel.
Jul 20, 2016 at 23:29 vote accept Chunna
Jul 21, 2016 at 12:35
Jul 20, 2016 at 21:59 history answered Jim Humphreys CC BY-SA 3.0