An isomorphism of 2-Schur modules This is the little brother of question 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 $\mathrm{SL}_2\left(k\right)$ ?
 A: 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:


*

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

*$(S^2U \otimes \wedge^2 V) \oplus (\wedge^2 U \otimes S^2V)$ 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:


*

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

*$(\wedge^2 U \otimes S^2V) \oplus (\wedge^2U \otimes \wedge^2 V) \cong \wedge^2 U \otimes V^{\otimes 2}$.
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.
A: As I'm sure you know, this is an easy computation with symmetric polynomials/characters over the field $\mathbb{C}$. I will give a proof which shows how to leverage that fact and show that, for any commutative $\mathbb{Q}$-algebra $R$, and $U$ and $V$ any free $R$-modules, we have the required isomorphism. At the end, I'll try to make a general statement.

Lemma 1: Let $U$ a finite dimensional $\mathbb{C}$ vector space. For $\sigma$ a permutation in $S_d$, let $\sigma : U^{\otimes d} \to U^{\otimes d}$ act by permuting the factors according to $\sigma$. So we get a map
$$\mathbb{C}[S_d] \to \mathrm{End}_{GL(U)}(U^{\otimes d}).$$
This map is surjective. 
Proof: This is part of the statement of Schur-Weyl duality. 
Corollary 2: Let $U$ and $V$ be finite dimensional $\mathbb{C}$ vector spaces. We have similar surjections:
$$\mathbb{C}[S_d \times S_e] \to \mathrm{End}_{GL(U) \times GL(V)}(U^{\otimes d} \otimes V^{\otimes e})$$
and 
$$\mathrm{Mat}_{r \times s}( \mathbb{C}[S_e \times S_d]) \to \mathrm{Hom}_{GL(U) \times GL(V)} \left( \left(U^{\otimes d} \otimes V^{\otimes e}\right)^{\oplus r},  \left(U^{\otimes d} \otimes V^{\otimes e}\right)^{\oplus s}  \right).$$
Now, let $U$ and $V$ be finite dimensional $\mathbb{Q}$-vector spaces. Observe that 
$$\bigwedge\nolimits^{2}\left(U\otimes V\right)\oplus \left(\bigwedge\nolimits^2\left(U\right)\otimes\bigwedge\nolimits^2\left(V\right)\right)^{\oplus 2} \quad (\ast)$$
is naturally a quotient of $(U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 3}$. Even better, there is an idempotent $\pi_1$ in $\mathrm{Mat}_{3 \times 3}(\mathbb{Q}[S_2 \times S_2]$ such that $(\ast)$ is the image of $\pi_1$.
Similarly, we can find $\pi_2$, an idempotent in $\mathrm{Mat}_{2 \times 2}(\mathbb{Q}[S_d \times S_e])$ such that the image of $\pi_2$ acting on $(U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 2}$ is 
$$U^{\otimes 2}\otimes\bigwedge\nolimits^2 V \oplus \bigwedge\nolimits^2 U\otimes  V^{\otimes 2} \quad (\ast\ast)$$
We have a commutative diagram
$$\begin{matrix}
\mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_2 \times S_2]) & \rightarrow & \mathrm{Mat}_{2 \times 3}(\mathbb{C}[S_2 \times S_2]) \\
\downarrow & & \downarrow \\
\mathrm{Hom}_{GL(U) \times GL(V)} \left( (U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 3},  (U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 2} \right) & \rightarrow & \mathrm{Hom}_{GL(U_{\mathbb{C}}) \times GL(V_{\mathbb{C}})} \left( (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 3},  (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 2} \right)
\end{matrix}$$
where for brevity I use a subscript $\mathbb{C}$ for tensor with $\mathbb{C}$. Call the vertical maps $\alpha$. 
Let $A \subset \mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_d])$ be those maps $\phi$ such that 
$$\alpha(\phi) = \alpha(\phi \pi_1) = \alpha(\pi_2 \phi). \quad (\dagger)$$
Let $A_{\mathbb{C}}$ be the analogous subspace of $\mathrm{Mat}_{2 \times 3}(\mathbb{C}[S_d])$. Since $A$ and $A_{\mathbb{C}}$ are cut out by the same equations, we have $A_{\mathbb{C}} = A \otimes \mathbb{C}$.
The images of $\pi_1$ and $\pi_2$ are isomorphic over $\mathbb{C}$. Composing that isomorphism with the projection $(U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 3} \to (\ast)_{\mathbb{C}}$ and the injection $(\ast \ast)_{\mathbb{C}} \to (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 2}$, we get a map $\overline{\phi}$ in the bottom right hand corner of the diagram, obeying $\overline{\phi} = \overline{\phi} \alpha(\pi_1) = \alpha(\pi_1) \overline{\phi}$. Moreover, this map has rank $\dim (\ast) = \dim (\ast \ast)$.
Using Corollary 2, we can find $\phi$ with $\alpha(\phi) = \overline{\phi}$. So $\phi \in A_{\mathbb{C}}$ and $\alpha(\phi)$ has rank $\dim (\ast)$.
Crucial paragraph The subspace of $A_{\mathbb{C}}$ consisting of maps $\phi$ such that $\alpha(\phi)$ has rank $\dim (\ast)$ is Zariski open. And $A$ is Zariski dense in $A_{\mathbb{C}}$. (Any polynomial which vanishes at all rational inputs is zero.) So there is some $\phi \in A$ obeying $(\dagger)$. This $\phi$ must induce an isomorphism between $(\ast)$ and $(\ast \ast)$.
So we have shown that $(\ast) \cong (\ast \ast)$, and this isomorphism is induced by a matrix in $\mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_d])$. For any $\mathbb{Q}$ algebra $R$, this gives a corresponding isomorphism between the analogous representations of $GL(U \otimes R) \times GL(V \otimes R)$. 

Wow, that took a lot longer than I expected it to. I dread the notation that would be needed to make this precise, but a theorem like the following should follow by similar arguments: Let $\mathcal{S}$ and $\mathcal{T}$ be two formulas made up from atoms $V_1$, $V_2$ ..., $V_r$ joined together direct sums, tensor products and Schur functors. For any commutative ring $R$ and sequence of positive integers $(n_1, \ldots, n_r)$, we can "plug in" $R^{n_i}$ for $V_i$ and get a $\prod GL_{n_i}(R)$ representation.
"Theorem" If $\mathcal{S}(\mathbb{C}, n_1, \ldots, n_r) \cong \mathcal{T}(\mathbb{C}, n_1, \ldots, n_r)$, then $\mathcal{S}(R, n_1, \ldots, n_r) \cong \mathcal{T}(R, n_1, \ldots, n_r)$ holds for any $\mathbb{Q}$-algebra $R$ and there is a "universal formula" for the isomorphism as a $\mathbb{Q}$-linear combination of reordering various tensor products.
When I set out to prove this, I expected to only need that all positive integers $\leq \max(n_1, \ldots, n_r)$ were invertible in $R$, not that $R$ was a $\mathbb{Q}$ algebra. But I couldn't get the Zariski density argument in the Crucial Paragraph to work without a ground field. I'm curious whether this was an artifact of the proof method, or a genuine obstacle.
A: This is not an answer, but a comment to S. Carnahan's answer:
Let me write down the isomorphism $\left(S^2 U \otimes \wedge^2 V\right) \oplus \left(\wedge^2 U \otimes S^2 V\right) \to \wedge^2\left(U\otimes V\right)$ explicitly: It sends $\left(uu^{\prime}\otimes v\wedge v^{\prime}, p\wedge p^{\prime} \otimes qq^{\prime}\right)$ to $\dfrac12\left(\left(u\otimes v\right)\wedge\left(u^{\prime}\otimes v^{\prime}\right) - \left(u\otimes v^{\prime}\right)\wedge\left(u^{\prime}\otimes v\right) + \left(p\otimes q\right)\wedge\left(p^{\prime}\otimes q^{\prime}\right) + \left(p\otimes q^{\prime}\right)\wedge\left(p^{\prime}\otimes q\right)\right)$ (where both $\wedge$ and multiplication operators bind more strongly than $\otimes$). The inverse isomorphism sends $\left(u\otimes v\right)\wedge\left(u^{\prime}\otimes v^{\prime}\right)$ to $\left(uu^{\prime}\otimes v\wedge v^{\prime}, u\wedge u^{\prime}\otimes vv^{\prime}\right)$.
Note that the inverse isomorphism is fraction-free. Hence, the isomorphism $\wedge^2\left(U\otimes V\right)\oplus \left(\wedge^2\left(U\right)\otimes\wedge^2\left(V\right)\right)^{\oplus 2} \to U^{\otimes 2}\otimes\wedge^2 V \oplus \wedge^2 U\otimes  V^{\otimes 2}$ is fraction-free as well (but its inverse probably not).
As it comes to generalizing to $\wedge^n$, I assume the inverse isomorphism is the better bet.
