2
$\begingroup$

Let $k$ be an algebraically closed field of positive characteristic, let $G=SL_2(k)$ acting naturally on $V=k^2$, and let $S^r V$ be the $r$-th symmetric power of the standard representation $V$.

Is it true that $Hom^G(S^p V, S^q V) \neq \{0\}$ if and only if $p=q$ (in which case it is $\{ \lambda Id; \lambda \in k\}$)? The problem is that $S^p V$ and $S^q V$ can be reducible in positive characteristic and so strange things might happen.

If yes, could someone please indicate a reference for this fact? This is easily checked in characteristic zero, but I don't see a straighfoward argument in positive characteristic.

Thank you in advance!

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ It's not true. You can check for yourself if p=2 then S^2(V) is indecomposable with a 2 dimensional socle with 1 dimensional (trivial) quotient. (Look at the action of the root groups---which generate G---on it.) So take p=2, q=0 and you can have a non-zero homomorphism. In fact this generalises to 'p'=p, 'q'=p-2 where p=char k. $\endgroup$
    – David Stewart
    Apr 28, 2017 at 12:49
  • $\begingroup$ Thanks for the answer David! And do you know if it is possible for $S^p V$ to contain a 1-dimensional subrepresentation when $p \geq 2$? $\endgroup$
    – Dupont
    Apr 28, 2017 at 13:06
  • 1
    $\begingroup$ A 1-dimensional subrepresentation of $S^dV$ would be an invariant on $V^*$ of degree $d$. All invariants have degree $0$, though, since $G$ has a dense orbit in $V^*$. $\endgroup$
    – Friedrich Knop
    Apr 28, 2017 at 16:09
  • $\begingroup$ @Dupont: There is a lot of scattered literature, ranging from dissertations to conference reports to research papers, dealing with the rank 1 case. These symmetric powers are the same as dual Weyl modules, so for example in the definining characteristic $p$ there is always a unique simple submodule determined by the highest weight occurring. (This is the trivial module only for highest weight 0.) Have you looked at Jantzen's 2003 AMS book? See also the references in my 2006 LMS Lecture Notes 326. Or try ams.org/mathscinet-getitem?mr=0399281 $\endgroup$
    – Jim Humphreys
    Apr 28, 2017 at 21:55
  • $\begingroup$ @Dupont: As Jim says, $S^r(V)$ only has one simple submodule, namely that of high weight $r$, so you would have to have $r=0$. You can get a trivial module as a /quotient/ of $S^{2p-2}(V)$ if that is of any interest. $\endgroup$
    – David Stewart
    Apr 30, 2017 at 10:51

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.