6
$\begingroup$

Let $k$ be an algebraically closed field of positive characteristic, and let $G$ be a reductive algebraic group over $k$ (for instance a classical group).

Let $V$ be a (rational) $G$-module. We say that $V$ admits a good filtration [see for instance Jantzen, Representations of Algebraic Groups, §4.19] if there exists an ascending chain of $G$-modules $0=V_0 \subset V_1 \subset V_2 \subset ... \subset V$ such that $V=\cup V_i$ and, for each $i \geq 1$, the quotient $V_i/V_{i-1}$ is isomorphic to some $H^0(\lambda) \otimes N_i$, where $N_i$ is a (non necessary finite dimensional) vector space on which $G$ acts trivially, and $H^0(\lambda)$ is the usual induced module $ind_B^G(k_{-\lambda})$, $B \subset G$ being a Borel subgroup, $\lambda$ a dominant weight, and $k_{-\lambda}$ the one dimensional representation of $B$ corresponding to $-\lambda$.

My questions are the following:

1) If $M_1$ and $M_2$ are two $G$-submodules of $M$ such that $M_1,M_2$, and $M$ admit a good filtration, does $M_1 \cap M_2$ also admit a good filtration?

2) If $M$ is a $G$-module with a good filtration and if $N \subset M$ is any $G$-submodule, does there exist a unique minimal $G$-submodule $N'$ such that $N \subset N' \subset M$ and $N'$ admits a good filtration?

$\endgroup$
  • $\begingroup$ Note that Jantzen and van der Kallen start with slightly different definitions of "good filtration" in infinite dimension. In any case, your question and Jeremy's answer already work for finite dimensional modules if you take advantage of Jantzen's construction of injective objects in "bounded" categories (which works at least when the prime is not too small). $\endgroup$ – Jim Humphreys Jun 18 '14 at 23:41
  • $\begingroup$ One may also observe that the two definitions are equivalent for submodules of the injective hull of a finite dimensional module. The definition of Jantzen gets unpleasant only when multiplicities of weight spaces of $U$-fixed points become infinite. $\endgroup$ – Wilberd van der Kallen Jun 19 '14 at 7:00
7
$\begingroup$

The answer to both questions is no.

For 1) let $V=H^0(\lambda)$ with $\lambda$ dominant, $U\leq V$ any submodule that doesn't have a good filtration and $\Delta(U)=\{(u,u): u\in U\}\leq V\oplus V$. Take $M_1=V\oplus0\leq (V\oplus V)/\Delta(U)$, $M_2=0\oplus V\leq(V\oplus V)/\Delta(U)$, and $M$ an injective hull of $(V\oplus V)/\Delta(U)$. Then $M_1\cap M_2\cong U$ doesn't have a good filtration.

For 2) take the same example, with $N=M_1\cap M_2$. Then $M_1$ and $M_2$ are both minimal submodules of $M$ with good filtrations that have $N$ as a submodule.

$\endgroup$
  • $\begingroup$ Thank you very much, this is exactly the kind of counter-example I was looking for! $\endgroup$ – sabrebooth Jun 18 '14 at 21:58
  • $\begingroup$ @Jeremy: Maybe say something more about the "minimal" condition in 2)? (For both questions it might be simpler if you start with a specific example for a low-rank group involving few composition factors.) $\endgroup$ – Jim Humphreys Jun 18 '14 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.