Timeline for Decomposition into Weyl modules
Current License: CC BY-SA 3.0
13 events
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| Mar 15, 2018 at 10:04 | comment | added | Tobias Kildetoft | @MichielVanCouwenberghe The Weyl modules are indeed the reductions mod $p$ of suitable modules over $\mathbb{Z}$. But a decomposition over $\mathbb{C}$ will rarely lead to one over $\mathbb{Z}$. | |
| Mar 15, 2018 at 8:51 | comment | added | Michiel Van Couwenberghe | @JimHumphreys So my mistake is probably the following. I thought that the decomposition of $S^2(V \times \mathbb{C})$ leads to a decomposition of $S^2(V)$ over $\mathbb{Z}$ and then, by tensoring with $k$ would lead to a decomposition of $S^2(V \otimes k)$. But probably the Weyl modules over $\mathbb{Z}$ are not as nice as the ones over $\mathbb{C}$. | |
| Mar 15, 2018 at 8:47 | comment | added | Michiel Van Couwenberghe | @JimHumphreys I am aware of the fact that most indecomposable modules fail to be simple in prime characteristic so maybe I should be more precise. I am interested in the following problem (coming from S. Garibaldi and R.M. Guralnick's article on Simple Groups Stabilizing Polynomials Lemma 7.1). Let $\mathcal{E}_8$ be the split semisimple group scheme of type $E_8$ over $\mathbb{Z}$ and $V$ the Weyl module of $\mathcal{E}_8$ over $\mathbb{Z}$ of some highest weight $\lambda$. Is it true that $V \otimes k$ is the Weyl module of highest weight $\lambda$ of $\mathcal{E}_8 \times k$? | |
| Mar 14, 2018 at 20:40 | comment | added | Tobias Kildetoft | On the other hand, if we pick $\lambda$ small enough relative to the group and $p$ (for example we could let $\lambda$ be any fundamental weight and assume that $p\geq h+1$ where $h$ is the Coxeter number), then the tensor square itself will be a sum of Weyl modules and hence so will the symmetric square. | |
| Mar 14, 2018 at 20:38 | comment | added | Tobias Kildetoft | However, if we take a weight of the form $\lambda = (p^r-1)\rho$ where $\rho$ is the halfsum of the positive roots (so $V(\lambda$ is the $r$'th Steinberg module), then the tensor square is tilting and hence so is the symmetric square. And since the highest weight lives in the symmetric part, it will have the corresponding indecomposable tilting module as a summand, and this will not be a Weyl module, since it has the trivial module in its head. | |
| Mar 14, 2018 at 20:36 | comment | added | Tobias Kildetoft | @MichielVanCouwenberghe Let me give a few more precise statements here: The example I gave actually shows that in characteristic $2$, the symmetric square need not have a Weyl filtration, since in that example it becomes the induced module $H^0(2)$, being the quotient of that tilting module by its socle. This is a specific thing to characteristic $2$ though, since otherwise the symmetric square will be a direct summand of the tensor square, so it has a Weyl filtration by standard results. | |
| Mar 14, 2018 at 14:30 | comment | added | Jim Humphreys | @Michiel: To reinforce what Tobias says, I'd emphasize that most (though certainly not all) indecomposable $G$-modules in prime characteristic fail to be simple. Also, "symmetric square" needs precision, while for special (or general) linear groups the second symmetric power of the standard module is itself a Weyl module. Offhand I don't know whether symmetric squares in other cases (or other types) are direct sums of Weyl modules, or not. But your sentence "Basically ..." is seriously out of focus anyway. | |
| Mar 14, 2018 at 11:38 | vote | accept | Michiel Van Couwenberghe | ||
| Mar 14, 2018 at 8:41 | comment | added | Tobias Kildetoft | @MichielVanCouwenberghe It should have a Weyl filtration, but why would you expect it to decompose? That the modules exist in all characteristics does not really say anything about splitting. I mean, the tensor square need not even split into the symmetric and exterior parts. For a concrete example of this, in $SL_2$ with $p=2$ we have that $V(1)\otimes V(1)$ is indecomposable (being isomorphic to the tilting module $T(2)$). | |
| Mar 13, 2018 at 9:14 | comment | added | Michiel Van Couwenberghe | Everything starts to make sense... However, I still feel that it is possible to decompose the symmetric square of a Weyl module into Weyl modules. Basically because it is possible in characteristic zero and these Weyl modules exist in every characteristic. Do you know a good reference for this statement? | |
| Mar 12, 2018 at 20:07 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Mar 12, 2018 at 18:21 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
added 11 characters in body
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| Mar 12, 2018 at 18:07 | history | answered | Jim Humphreys | CC BY-SA 3.0 |