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$\DeclareMathOperator\GL{GL}$In this question, representations means finite-dimensional complex representations.

Fix some $n,m \geq 2$ and some prime $p$. I'm interested in representations $V$ of $\GL(n,\mathbb{F}_p)$ and $W$ of $\GL(m,\mathbb{F}_p)$ such that the action of $\GL(n,\mathbb{F}_p) \times \GL(m,\mathbb{F}_p)$ on $V \oplus W$ can be extended to an action of $\GL(n+m,\mathbb{F}_p)$. This is trivially possible if both $V$ and $W$ are vector spaces equipped with trivial actions. Are there any other examples? My guess is that the answer is no, aside from maybe some sporadic examples for small $n$ and $m$ and $p$.

Note that it is important that we are restricting to complex representations. If we allowed characteristic $p$ representations, then the standard representation of $\GL(n+m,\mathbb{F}_p)$ on $\mathbb{F}_p^{n+m}$ would give another example.

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  • $\begingroup$ "Are there any other examples" (than the trivial modules): Obviously yes: when $V,W$ are the standard $n$- and $m$-dimensional modules. $\endgroup$
    – YCor
    Commented Nov 28, 2023 at 6:59
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    $\begingroup$ @YCor: The standard representations are characteristic $p$ representations, but the question is looking for complex representations. $\endgroup$
    – Andy Putman
    Commented Nov 28, 2023 at 10:03
  • $\begingroup$ @AndyPutman ah, sure, thanks! $\endgroup$
    – YCor
    Commented Nov 28, 2023 at 11:03
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    $\begingroup$ I'll note though that the example YCor pointed out is the only other example in characteristic p -- this can be seen by looking at weight spaces. Similarly, the standard $n+m-1$ dimensional representation of $S_{n+m}$ is the only non-trivial irreducible representation of $S_{n+m}$ with a decomposition like this when we restrict to $S_n \times S_m$. So any example would be very constrained in terms of both the composition factors in its reduction to characteristic p and its restriction to $S_{n+m}$. I'm pretty sure no other examples exist, but I don't see an argument yet. $\endgroup$
    – Nate
    Commented Nov 28, 2023 at 15:26
  • $\begingroup$ The question makes sense even when $m$ or $n$ is $1$, and might be easier then, and also help into having an intuition for the general case. $\endgroup$
    – YCor
    Commented Nov 28, 2023 at 19:35

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