Let $G$ be a split reductive group over an arbitrary field $k$. By definition, see Jantzen (*), an ascending chain $$0 = V_0 \subset V_1 \subset V_2 \subset \dots$$ of submodules of a $G$-module $V$ is called a Weyl filtration of $V$ if $V = \bigcup_{i \geq 0} V_i$ and if each $V_i/V_{i-1}$ is isomorphic to some Weyl module $V(\lambda_i)$ of highest weight $\lambda_i$.

It seems that if a module admits such a Weyl filtration, we have $V \cong \bigoplus_{i} V(\lambda_i)$. Indeed, according to remark II.4.19 in Jantzen, we can order the $\lambda_i$ such that $\lambda_i < \lambda_j$ implies $i > j$. The decomposition $V \cong \bigoplus_{i} V(\lambda_i)$ then follows by induction from $\operatorname{Ext}^1_G(V(\lambda),V)=0$ if $V$ has no weight $\mu > \lambda$ (see Jantzen II.2.14 Remark 2).

In particular I would like to use that we can write the symmetric square of a Weyl module as a direct sum of Weyl modules and that we can derive the terms from Weyl's character formula. This should follow from Proposition II.4.21 in Jantzen which basically says (the dual of) if $V$ admits a Weyl filtration then so does $V \otimes V$.

However, this results seems quite standard and I am quite surprised that Jantzen doesn't mention it (as far as I know). So my question is whether there is a more direct approach towards this problem and whether anyone knows a good reference.

EDIT. The reasoning in the second paragraph is not correct. We would want the $\lambda_i$ to be ordered such that if $\lambda_i < \lambda_j$ then $i < j$. I am not sure whether this is possible. Probably it is not, which explains why this isn't in Jantzen.

(*) Jantzen, Representations of Algebraic Groups (AMS, 2003)

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    $\begingroup$ I don’t understand your second paragraph. The results you quote don’t rule out a module $V$ with a Weyl filtration $0=V_0<V_1<V_2=V$ with $V_1/V_0\cong V(\mu)$, $V_2/V_1\cong V(\lambda)$, $\mu>\lambda$ and $V\not\cong V(\mu)\oplus V(\lambda)$. $\endgroup$ – Jeremy Rickard Mar 12 '18 at 14:38
  • $\begingroup$ @JeremyRickard Hmmm, you might be right. I think I want the $\lambda_i$ ordered such that $\lambda_i < \lambda_j$ implies $i < j$ but I am not sure whether that is possible. $\endgroup$ – Michiel Van Couwenberghe Mar 12 '18 at 15:33
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    $\begingroup$ @Michiel: The ordering of Weyl modules in a Weyl filtrattion suggested in your edit isn't always possible, as shown in my rank one example. The main point of introducing these filtrations is to study injective modules, or tilting modules, for $G$ in cases where the modules are indecomposable. $\endgroup$ – Jim Humphreys Mar 12 '18 at 20:15
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    $\begingroup$ Basic example: conjugation action of SL2 on two by two matrices in characteristic two. Find the filtration. $\endgroup$ – Wilberd van der Kallen Mar 13 '18 at 8:47

Your conclusion about direct sums is false. It's helpful here to have some examples in mind, such as a typical projective/injective module for the Lie algebra of $G:=\mathrm{SL}_2$, lifted to $G$. This is indecomposable but might have a Weyl filtration with two quotients: for example, the trivial 1-dimensional module $V(0) =L(0)$ at the top, but the linked Weyl module $V(2p-2)$ at the bottom. (Here I am referring to a non-negative integral multiple of the single fundamental weight just by the integer coefficient, while $L$ as in Jantzen denotes a simple module.)

In any case, you seem to be referring to the expanded second edition of Jantzen's 1987 Academic Press book Representations of Algebraic Groups (AMS, 2003), where the results you want are in Part II. For example, he refers to 4.19 as II.4.19 if the part is unclear from the context.

Note too that Jantzen's treatment is for a connected semisimple algebraic group, though adaptations to a reductive group with nontrivial (semisimple) derived subgroup can be deduced easily. And the field of definition doesn't play a role here.

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    $\begingroup$ @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 $T(2)$). $\endgroup$ – Tobias Kildetoft Mar 14 '18 at 8:41
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    $\begingroup$ @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. $\endgroup$ – Jim Humphreys Mar 14 '18 at 14:30
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    $\begingroup$ @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. $\endgroup$ – Tobias Kildetoft Mar 14 '18 at 20:36
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    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Mar 14 '18 at 20:38
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    $\begingroup$ @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 $\mathbb{Z}$. $\endgroup$ – Tobias Kildetoft Mar 15 '18 at 10:04

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