This is the little brother of [question 68071](https://mathoverflow.net/questions/68071): elementary, simple-looking and probably much easier to answer. Of course, it is just a small part of question 68071, as anybody with $\lambda$-rings experience will see.


Let $n$ and $m$ be positive integers. Let $k$ be a field of characteristic $0$. Let $U$ be the canonical $n$-dimensional representation of $\mathrm{GL}\_n\left(k\right)$. Let $V$ be the canonical $m$-dimensional representation of $\mathrm{GL}\_m\left(k\right)$. A folk result I cannot prove tells me that


$\wedge^2\left(U\otimes V\right)\oplus \left(\wedge^2\left(U\right)\otimes\wedge^2\left(V\right)\right)^{\oplus 2} \cong U^{\otimes 2}\otimes\wedge^2 V \oplus \wedge^2 U\otimes V^{\otimes 2}$


(don't confuse direct powers with tensor powers in this equation; also, the usual precedence rules apply where $\oplus$ is seen as addition and $\otimes$ as multiplication) as representations of $\mathrm{GL}\_n\left(k\right)\times \mathrm{GL}\_m\left(k\right)$.


Since the representation theory of $\mathrm{GL}\_n\left(k\right)\times \mathrm{GL}\_m\left(k\right)$ is semisimple (or at least I hope so?), this isomorphism must somehow "factor" into isomorphisms of parts. Or not? I am far from feeling safe here.


Anyway, is there a nice explicit description of the above isomorphism, that shows us what it maps stuff to, like the Clebsch-Gordan formulae for $\mathrm{SL}\_2\left(k\right)$ ?



An isomorphism of 2-Schur modules
As long as 2 is invertible, we can use the isomorphism $W^{\otimes 2} \cong S^2 W \oplus \wedge^2 W$ to split $(U \otimes V)^{\otimes 2}$ in two different ways:


1. $(S^2U \otimes S^2V) \oplus (\wedge^2 U \otimes \wedge^2 V)$ is the symmetric summand, hence isomorphic to $S^2(U \otimes V)$.
2. $(S^2U \otimes \wedge^2 V) \oplus (\wedge^2 U \otimes S^2V)$ is the alternating summand, hence isomorphic to $\wedge^2(U \otimes V)$.


We use the decomposition of $\wedge^2(U \otimes V)$ to build up the right side of your equation:


1. $(S^2U \otimes \wedge^2 V) \oplus (\wedge^2U \otimes \wedge^2 V) \cong U^{\otimes 2} \otimes \wedge^2 V$
2. $(\wedge^2 U \otimes S^2V) \oplus (\wedge^2U \otimes \wedge^2 V) \cong \wedge^2 U \otimes V^{\otimes 2}$.


If you are only concerned with algebraic representations, then $GL\_n \times GL\_m$ is linearly reductive, so you get complete reducibility. I don't know what happens for representations of groups of field-valued points in general.


