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$\DeclareMathOperator\SL{SL}$Let $K$ be an algebraically closed field. Let $V$ be a 2-dimensional vector space over $K$. The group $G=\SL_2(K)$ acts naturally on $V$ by left-multiplication on column vectors.

If $K=\mathbb{C}$ then we have an isomorphism of $KG$-modules

\begin{equation} S^d(S^m(V)^{\star}) \cong \Lambda^d(S^{m+d-1}(V)^{\star}) \tag{1}\label{469536_1} \end{equation}

for all $d>0$ and $m>0$. This is a consequence of basic character theory. The stars aren't actually needed because the modules $S^m(V)$ are self-dual for all $m$. Since symmetric and exterior powers commute with restriction, this also implies the same isomorphism holds when $G$ is replaced by a subgroup of $\SL_2(K)$, such as the additive group of $\mathbb{C}$ which is embedded in $G$ as the subgroup of upper unitriangular matrices, or its discrete subgroup $H$ isomorphic to $\mathbb{Z}$ generated by the matrix

$$\begin{pmatrix} 1&1 \\ 0 &1 \end{pmatrix}.$$

Now let $K$ be an algebraically closed field of characteristic $p$ and let $H$ be the subgroup of $G$ generated by the above matrix. So $H$ is finite group of order $p$. There is an isomorphism of $KH$-modules given by the formula \eqref{469536_1}. This is usually stated somewhat differently — there are exactly $p$ indecomposable modules for $H$, namely $V_1,V=V_2, \dotsc, V_p$, where the generator of $H$ acts on $V_n$ as multiplication by a $n \times n$ Jordan block. Each of these is self-dual, and $S^m(V_2) \cong V_{m+1}$ for $m<p$, so we usually write \eqref{469536_1} as $S^d(V_m) \cong \Lambda^d(V_{m+d})$ where $m+d \leq p$. You can't prove it by reducing the proof in the characteristic 0 case modulo $p$, it's considerably more involved than that.

I'd like to know whether an isomorphism of $KG$-modules \eqref{469536_1} exists when $K$ is an algebraically closed field of characteristic $p$. I've trawled around and found some work on tensor products for representations of $\SL_2(K)$ but nothing on symmetric and exterior powers.

For context: I'm really interested in proving \eqref{469536_1} for finite subgroups of the additive group of $K$, which are elementary abelian $p$-groups of order $q=p^n$, when $d+m<q$. I've found a plausible looking formula for an explicit isomorphism, which doesn't seems to have any dependence on $q$, so actually my proof should apply to the additive group of $K$. I thought this question for the additive group of $K$ was unlikely to have attracted much attention in itself, but that the analogue for $\SL_2(K)$ might already be known to the experts in representation theory of reductive linear algebraic groups, which would imply everything I want.

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    $\begingroup$ I think a previous discussion may be relevant <mathoverflow.net/questions/455163/…>. See in particular Abdelmalek Abdesselam's comment, which mentions an article of his with Chipalkatti on this isomorphism. I hope this helps. $\endgroup$
    – Malkoun
    Commented Apr 19 at 10:10
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    $\begingroup$ Thanks for the reference. If I've understood it correctly, this paper gives an explicit isomorphism when $K = \mathbb{C}$, and an isomorphism for $K$ a field of characteristic $p$ provided $d+m<p$ (the formula involves multiplying by $m!/(d+m)!$). But the case $d+m \geq p$ still seems to be open. $\endgroup$
    – Jon Elmer
    Commented Apr 19 at 15:24
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    $\begingroup$ The isomorphism was designed with char zero in mind. I am not too familiar with the details of it but there is also work about this Wronskian isomorphism and the related Hermite map for positive characteristic. See this article sciencedirect.com/science/article/abs/pii/S0021869322001259 by McDowell and Wildon, and references therein, especially the Inventiones article by Aprodu et al. See also link.springer.com/chapter/10.1007/978-3-030-89694-2_23 by Raicu and Sam. $\endgroup$ Commented Apr 19 at 17:58
  • $\begingroup$ That's exactly what I was looking for, thanks a lot! $\endgroup$
    – Jon Elmer
    Commented Apr 22 at 7:59

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If I remember correctly, one of the symmetric powers should become divided powers. A quick google search brought me to Section 3.4 of Aprodu, Farkas, Papadima, Raicu, and Weyman - Koszul modules and Green's conjecture which seems to contain the correct result.

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