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Let $G=SL_2(k)$ considered as a linear algebraic group over an algebraically closed field of prime characteristic. Let $E$ be the natural module for $G$ and denote by $S^r (E)$ its $r-$th symmetric power. I would like to know if there is a surjective map from $$ E \otimes S^r (E) \rightarrow S^{r-1} (E).$$ Can we write this map explicitly.

Moreover Is there a general description for all symmetric power of highest weight modules for linear algebraic groups?

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    $\begingroup$ It would help to provide some motivation for this kind of map. Also, you are asking too many questions, so it would help to focus better on the most important one. Concerning the added question for other (presumably simple) algebraic groups, you are asking for a lot here given the many open questions. At least in rank one the basics are known in principle. See Jantzen's book Representations of Algebraic Groups II.2.16 and his references, but note that in higher ranks the crucial details for general or special linear groups are usually unknown. $\endgroup$ Jul 19, 2016 at 20:34
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    $\begingroup$ Do you really just want a surjective map, or do you want a split surjective map? $\endgroup$ Jul 21, 2016 at 7:37

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This group $G$ (or its close relative GL$_2$ over the same field) has been long studied and is pretty well understood. But you have not invoked the characteristic $p>0$ in your question, whose answer depends a lot on how $p$ is related to $r$: sometimes $\nabla(r)$ is simple but sometimes it's a more complicated indecomposable module. In any case the composition factors of your tensor product can be sorted out using recursion and the linkage principle.

Recall that in characteristic 0, these symmetric powers $S^r(E)$ have for almost a century been used as models of the irreducible representations of the corresponding rank 1 algebraic (or Lie) group. Such a space just consists of the homogeneous polynomials of degree $r$ in two variables, and has dimension $r+1$. Lie theory assigns to $S^r(E)$ the highest weight $r+1$ times the (unique) fundamental dominant weight $\varpi$, and tensor products are complete reducible (Weyl) with multiplicities given by a Clebsch-Gordan rule.

In the modular situation the dual Weyl module you denote $\nabla(r)$ is simple (affording an irreducible representation) for $0 \leq r < p$ but then indecomposable with an increasingly complicated (but well understood) structure. Here the simple module with highest weight $r \varpi$ is the unique simple submodule but due to Steinberg's tensor product theorem its dimension is usually $< r+1$. When $r \geq p$ other linked weights may also occur.

As an extreme example, the (first) Steinberg module $\nabla(p-1)$ is simple of dimension $p$ but has an indecomposable tensor product with $E$. This has three composition factors. In such a case your short exact sequence can't split. Indeed, this indecomposable $G$-module of dimension $2p$ is a lift of a projective/injective module for the Lie algebra (or first Frobenius kernel). For details in rank 1, including the way these modules restrict to finite Chevalley groups, look at my old 1973 J. Algebra paper (link from my homepage) or the more comprehensive treatment in Jantzen's book.

ADDED: To describe the Ext groups here precisely for all $r$ and $p$, you need a somewhat more elaborate set-up. This was done for example in Section 3 of a paper by K. Erdmann here. Rank 1 groups have been popular over the years as a laboratory for studies aimed at higher ranks, usually also aimed at the more delicate issues involved when restricting to Frobenius kernels or finite groups of Lie type.

Keep in mind that the linkage principle already implies the splitting of many extensions of your type, when the highest weights of the two dual Weyl modules fail to be linked. (This is made more precise by Donkin's block decomposition, described in Jantzen's book.) Note too that the failure of the short exact sequence to split involves a more precise assertion than existence of a surjection. (There is a surjection in my example above, but it doesn't split the sequence.)

In this very classical case, there may well be concrete ways to specify splitting when it occurs, since spaces of polynomials were a basic object of study in classical invariant theory. But most of the recent literature emphasizes what can be generalized to higher ranks.

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    $\begingroup$ 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. $\endgroup$ Jul 21, 2016 at 7:18
  • $\begingroup$ @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. $\endgroup$
    – Chunna
    Jul 21, 2016 at 13:52
  • $\begingroup$ @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. $\endgroup$ Jul 21, 2016 at 13:59
  • $\begingroup$ 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? $\endgroup$ Jul 22, 2016 at 14:01
  • $\begingroup$ @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. $\endgroup$ Jul 22, 2016 at 17:31
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Let me summarize. We take a basis $x$, $y$ of $E$ and the characteristic is $p$. There are two cases where there is a surjective map $E\otimes S^r(E)\to S^{r-1}E$.

The first case is when $r=p-1$. Then the surjective map does not split. Its kernel is spanned by $x\otimes x^r$, $y\otimes y^r$, $(x\otimes x^{r-j}y^j-y\otimes x^{r-j+1}y^{j-1})$, with $1\leq j\leq r$.

The second case is when $r+1$ is not divisible by $p$. Then the map splits and its kernel is spanned by $x\otimes x^r$, $y\otimes y^r$, $(r+1)x\otimes x^{r-j}y^j-j(x\otimes x^{r-j}y^j-y\otimes x^{r-j+1}y^{j-1})$, with $1\leq j\leq r$.

All this can be checked by tedious computations in Jantzen's framework of $G_sT$-modules. No further theory is required.

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  • $\begingroup$ Notice the homomorphism $S^{r+1}(E)\to E\otimes S^r(E)$ that sends $f$ to $x\otimes \frac{\partial f}{\partial x}+y\otimes \frac{\partial f}{\partial y}$. $\endgroup$ Jul 26, 2016 at 7:33

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