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In this popular 2019 MO question, user მამუკა ჯიბლაძე asked:

The spaces $\operatorname{S}^2(k^n)$ and $\Lambda^2(k^{n+1})$ from the title have equal dimensions. Is there a natural isomorphism between them?

The answer was affirmative: let $E=k^2$, then the Wronskian isomorphism gives an identification $\bigwedge^m(S^{m+r-1}(E)) \rightarrow S_m(S^r(E))$ as $\operatorname{SL}(E)$-modules.

(Here, $S_m(V)=(V^{\otimes m})^{S_m}$ and $S^m(V)$ is the symmetric power, as a quotient of $V^{\otimes m}$.)

The recent paper Modular plethystic isomorphisms for two-dimensional linear groups of McDowell–Wildon contains the proof of this over any field.

My question is in a similar vein. Let $\nabla^\lambda$ be the Schur functor of the partition $\lambda$, and consider $\lambda_0=(2,1)$. Then, the $\operatorname{SL}(E)$-character of $\nabla^{\lambda_0}(S^m(E))$ is simply $$ \newcommand\qbinom{\genfrac[]0{}}[2]\qbinom{m+2}3, $$where the brackets indicate that these are quantum numbers and binomials, so for instance $[2]=q+q^{-1}$.

Notice that this holds for any $m$. This suggests the possibility that there exists a natural isomorphism $\nabla^{\lambda_0}(S^m(E))\cong E\otimes V_m$, for some "uniformly defined" $\mathrm{SL}(E)$-module $V_m$ of dimension $\qbinom{m+2}3$.

Is there such a natural isomorphism?

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    $\begingroup$ I think there is a typo in your expression for the character. Should it be $\newcommand\qbinom{\genfrac[]0{}}[2]\qbinom{m+2}3$? $\endgroup$ Commented Nov 17, 2023 at 3:17
  • $\begingroup$ @AntoineLabelle that's right! Got confused with the labeling. Thanks $\endgroup$ Commented Nov 17, 2023 at 6:31
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    $\begingroup$ $\binom{m+2}{3}_q$ is the character of $S^3\left(V\right)$, right? $\endgroup$ Commented Nov 18, 2023 at 19:07
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    $\begingroup$ Perhaps confusingly, not in this case (hence my original typo). The character of $V=S^m(E)$ is $(m+1)_q$, so $\binom{m+1}{3}_q$ is the character of $\Lambda^3(V)$, and $\binom{m+3}{3}_q$ is the character of $S^3(V)$. I’m looking for the middle guy $\binom{m+2}{3}_q$. $\endgroup$ Commented Nov 18, 2023 at 19:24

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This is not a complete answer, but it strongly suggests where to look. Since $\binom{m+2}{3}_q$ is the $q$-character of $\mathrm{Sym}^3\mathrm{Sym}^{m-1}(E)$, working over the complex numbers we have

$$\nabla^{(2,1)} \bigl(\mathrm{Sym}^m E\bigr) \cong E \otimes \mathrm{Sym}^3 \mathrm{Sym}^{m-1}E $$

as representations of $\mathrm{SL}(E)$. Tensoring on the right-hand side with the determinant extends this to an isomorphism of representations of $\mathrm{GL}(E)$ of common polynomial degree $3m$.

In the spirit of my joint paper with McDowell, I now tried introducing suitable dualities into each side. After some experimentation with computer algebra, the only candidate that remains for a uniformly defined isomorphism is

$$\nabla^{(2,1)}\bigl(\mathrm{Sym}_mE\bigr) \cong E \otimes \mathrm{Sym}^3 \mathrm{Sym}_{m-1}E. $$

This isomorphism exists working with modules for $\mathrm{SL}_2(\mathbb{F}_q)$ when $q=16$, $q=27$ and $q=125$ for $m \le 10$. When $q=27$ and $m=10$, of the $8$ possible modules on either side defined by different combinations of the symmetric power functors, only the proposed pair are isomorphic. By the dual version of the modular Wronskian isomorphism proved in my joint paper, the right-hand side could also be written as $E \otimes \bigwedge^3 \mathrm{Sym}_mE$.

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  • $\begingroup$ This is a fantastic answer, thank you so much! $\endgroup$ Commented Nov 19, 2023 at 0:10
  • $\begingroup$ @Mark_Wildon by the way, what software do you use for such computations? I'm interested in finding variants of this isomorphism, as well as the explicit map. $\endgroup$ Commented Nov 19, 2023 at 0:29
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    $\begingroup$ I'm using the commercial computer algebra package Magma. I think it's possible to use it for free online for short runs. Please send me an email if you'd like the code. $\endgroup$ Commented Nov 19, 2023 at 9:28
  • $\begingroup$ Thank you, I emailed you. $\endgroup$ Commented Nov 20, 2023 at 0:14

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