In characteristic zero one can use the Clebsch-Gordan rule to decompose tensor products of SL(2)-modules. In characteristic $p$ things are more complicated.

I am interested in the special case $S^dV\otimes V$ (where $V$ is the 2-dimesional standard representation) for fields $k$ of characteristic $p>0$. In fact, I mainly want to know about $d=3$.

If one computes the Clebsch-Gordan isomorphism explicitly, one can see that the denominator is $(d+1)$. So there will be a problem for $p|(d+1)$.

What is known in this case? I'd be happy just to know the case $d=3$, especially an explicit composition series and whether one still has some nice direct sum decompositions into representations of smaller dimension (I realize that these will no longer be simple modules as in characteristic 0). I'd also like to know references about how the invariant theory of SL(2) works in positive characteristic.

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    $\begingroup$ Could you be a bit more specific about what you mean by decompose? The modules will not decompose as a direct sum of simples in positive characteristic. Do you want the composition factors? Or a specific composition series? A good way to start is probably to do the usual decomposition to get a good filtration of the tensor product, and then use that we actually do know the characters of the modules in such a good filtration in terms of the characters of the simple modules, since this is $SL_2$. $\endgroup$ – Tobias Kildetoft Feb 25 '14 at 8:00
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    $\begingroup$ And what do you mean by `the invariant theory'? Just the invariants in this module? (They are the same as predicted by Clebsch-Gordan.) $\endgroup$ – Wilberd van der Kallen Feb 25 '14 at 8:21
  • $\begingroup$ @TobiasKildetoft: Sorry for the ambiguity (I have edited the question to remove it). And thank you for your answer. I want to be explicit as possible, so I want a composition series as you give below. I would also like to know whether $S^3V\otimes V\cong H^0(4)\oplus H^0(2)$. $\endgroup$ – Lloyd Yu-West Feb 25 '14 at 15:16
  • $\begingroup$ @WilberdvanderKallen: Thank you for your comment. I am naive about invariant theory in positive characteristic. How much carries over from the classical case? I'd be grateful of any pointers/references. $\endgroup$ – Lloyd Yu-West Feb 25 '14 at 15:25
  • $\begingroup$ After some more thought I realized that this module is indecomposable. I am on a tablet now, so I will elaborate later. $\endgroup$ – Tobias Kildetoft Feb 25 '14 at 15:28

$\newcommand{\Hom}{\operatorname{Hom}}$Here is an elaboration of my comment with what happens in characteristic $2$ when $d=3$:

The usual decomposition rule gives us a filtration of $S^dV\otimes V$ with two factors: $H^0(4)$ and $H^0(2)$ (I use the notation from Jantzen's Representations of Algebraic Groups and write all weights in terms of the fundamental weight).

Applying the Jantzen sum formula, we see that $H^0(4)$ has a composition series consisting of $L(4)$, $L(2)$ and $L(0)$.

We also see that $H^0(2)$ has a composition series consisting of $L(2)$ and $L(0)$.

All this gives us a composition series $$0 \subseteq M_1 \subseteq M_2 \subseteq M_3 \subseteq M_4 \subseteq M_5 = S^3V\otimes V$$ where $M_1\cong L(2)$, $M_2/M_1\cong L(0)$, $M_3/M_2\cong L(4)$ and $\{M_4/M_3,M_5/M_4\}\cong \{L(2),L(0)\}$. Which order the two top factors come in is less obvious (I will need to think a bit about it), and whether we actually have $S^3V\otimes V\cong H^0(4)\oplus H^0(2)$ I will also need to think a bit more to figure out.

Added: So, after some further thought, we can actually say a bit more.

First note that as mentioned by Jim Humphreys, we have $S^3V\otimes V\cong L(1)\otimes L(1)\otimes L(1)^{(1)}$ which means that it is self-dual. In particular, we see that our composition series can be chosen to be "symmetric", so we get $M_4/M_3\cong L(0)$ and $M_5/M_4\cong L(2)$ (it is also good to notice that we actually have $M_2\cong H^0(2)$ and $M_5/M_2\cong H^0(4)$ as these are sometimes easier to work with).

We can also show that $S^3V\otimes V$ is indecomposable. In fact, we have $\operatorname{soc}_{SL_2}(S^3V\otimes V) = L(2)$.

To see this, we need a bit more machinery (it might be possible to do this in a more elementary way). Let $G = SL_2$ and let $G_1$ be the first Frobenius kernel of $G$. We let $\lambda = \lambda_0 + p\lambda_1$ be a dominant weight with $\lambda_0 < p$ and use that $L(\lambda) \cong L(\lambda_0)\otimes L(\lambda_1)^{(1)}$. Now we note that $$\Hom_G(L(\lambda),L(1)\otimes L(1)\otimes L(1)^{(1)})$$ $$\cong \Hom_{G/G_1}(L(\lambda_1)^{(1)},\Hom_{G_1}(L(\lambda_0),L(1)\otimes L(1))\otimes L(1)^{(1)})$$ so it is sufficient to show that $\operatorname{soc}_{G_1}(L(1)\otimes L(1)) = L(0)$.

To see this we further note that it will suffice to show that $\operatorname{soc}_G(L(1)\otimes L(1)) = L(0)$ since the $G_1$-socle is a $G$-submodule. But this final part is a simple calculation, as we clearly just need to check that neither $L(1)$ nor $L(2)$ are submodules. That $L(1)$ is not a submodule is clear by parity (all highest weights of composition factors in $L(1)\otimes L(1)$ must be even), and that $L(2)$ is not a submodule is seen by noting that $$\Hom_G(L(2),L(1)\otimes L(1))\cong \Hom_G(L(1),L(1)\otimes L(2))\cong \Hom_G(L(1),L(3))$$ and $L(3)$ is simple (it is the 2'nd Steinberg module as also mentioned by Jim Humphreys).

A few final notes: The above actually shows that as a $G_1$-module, $L(1)\otimes L(1)$ is the injective hull of the trivial module. This is a general fact about $SL_2$ in characteristic $2$, ie, that for all $r$, $St_r\otimes St_r$ is the injective hull of the trivial module as a $G_r$-module (this does not generalize to other groups, nor to other primes).

Also, the conclusion about the module $S^3V\otimes V$ is in fact that it is indecomposable tilting (in the notation from Jantzen, it is denoted $T(4)$).

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  • $\begingroup$ I can't follow your notation, which is nonstandard, so I've written down my own version. $\endgroup$ – Jim Humphreys Feb 25 '14 at 18:33
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    $\begingroup$ @JimHumphreys Which part of the notation? $\endgroup$ – Tobias Kildetoft Feb 25 '14 at 19:13
  • $\begingroup$ Sorry, I got confused by the later part of your answer (probably expecting to see some twisted tensor proaducts). The basic notation is not a problem. $\endgroup$ – Jim Humphreys Feb 25 '14 at 21:01
  • $\begingroup$ One other comment: your expanded answer is helpful, but there's no need to invoke Jantzen's sum formula here to find composition factors (that much is very easy). for the module structure of such tensor products, methods of Doty and Henke are systematic. $\endgroup$ – Jim Humphreys Feb 28 '14 at 15:57
  • $\begingroup$ @JimHumphreys I had a feeling that would be overkill, but my knowledge of the theory for $SL_2$ is a bit lacking (something I will need to fix at some point). Do you know any good reference for the $SL_2$ specifics? $\endgroup$ – Tobias Kildetoft Feb 28 '14 at 18:34

There is unfortunately no "formula" for tensor products in prime characteristic. Instead you can derive a list of composition factors $L(\lambda)$ (with multiplicity) by recursion. When $p=2$ there are only two simple modules with restricted highest weights (abbreviated by non-negative integers), namely the trivial module $L(0)$ of dimension 1 and the natural module $L(1)$ of dimension 2. After this you need to rely on Steinberg's twisted tensor product theorem relative to a $p$-adic expansion of the highest weight. For instance, $L(2) \cong L(1)^{(1)}$, the first Frobenius twist of the natural module (still having dimension 2).

In your specific example, the recursion is easy to carry out: peel off the composition factor of highest weight and see what weights remain. Here yuu are looking at the tensor product of the natural module $L(1)^{(1)}$ with the "induced" module $H^0(3)$ of dimension 4 as in Jnntzen's book (polynomials in two variables of homogeneous degree 3) which is actually simple when $p=2$, isomorphic to $L(1) \otimes L(1)^{(1)}$. (These factors are the respective Steinberg modules for the first and second Frobenius kernels.)

From the recursion one arrives at a list of composition factors (having total dimension 8): $L(1)^{(2)}, \: L(1)^{(1)} \text{ twice }, L(0)\: \text{ twice}.$

Here at least there is a recursive method, but getting the precise module structure can be quite tricky. In this special case, you might take advantage of the fact that you are tensoring with a projective module for a certain Frobenius kernel. But in general it's complicated even in rank 1.

ADDED: For some recent work on decomposition of tensor products into indecomposables, see for example a paper by Doty and Henke, Decomposition of tensor products of modular irreducibles for SL$_2$. Q. J. Math. 56 (2005), no. 2, 189-207 (preprint here). This involves tilting modules and their Frobenius twists.

FURTHER COMMENTS: I intended to mention something about the extra question raised by Lloyd on invariant theory in prime characteristic. There is a classical result describing the ring of invariants for the general linear groups over a finite field, or for $\mathrm{SL}_n(\mathbb{F}_q)$, which goes back to L.E. Dickson in 1911. This has been reworked a number of times, for instance in an article by R. Steinberg, On Dickson’s theorem on invariants, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 34 (1987), no. 3, 699–707. (It's especially fitting to recall Steinberg's diverse contributions, since he died very recently on his 92nd birthday.)

In the special case dicussed here, Jantzen (unpublished) showed how to recover Dickson's theorem from the well-understood representation theory of the groups. This led me to investigate the general case in a similar spirit, though it remains to be seen whether we will know enough about representation theory to recover the full theorem this way. Anyway, my short paper contains an assortment of references to the literature (including work of Donkin on tilting modules and a paper by Wilkerson motivated by algebraic topology): Another look at Dickson's invariants for finite linear groups, Comm. Algebra 22 (1994), no. 12, 4773-4779.

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  • $\begingroup$ Thank you very much for the reference, which does indeed answer the decomposition part of my question. $\endgroup$ – Lloyd Yu-West Feb 28 '14 at 20:03

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