Antisymmetric tensors of a tensor product: $\Lambda^k(V \otimes W)$ Given two vector spaces $V, W$ over $\mathbb{R}$, it's true that
$\Lambda^2 (V \otimes W) \cong \left(\Lambda^2 V \otimes S^2 W \right) \oplus \left( S^2 V \otimes \Lambda^2 W \right)$. If I'm seeing it right, this should be the decomposition into irreducible submodules as a module over the symmetric group on two elements.
Is there an easy way to see how this generalizes to something like $\Lambda^k(V \otimes W)$ for $k \geq 3$, using Young symmetrizers or something like that?
For example when $k = 3$, again, the "trivial" summands $S^3V \otimes \Lambda^3 W$ and $\Lambda^3V \otimes S^3 W$ can be found, but by playing around with the dimensions (in the finite-dimensional case), it seems that there should be more. My guess is that they should come from a clever combination of the spaces $V^{(2,1)}, W^{(2,1)}$, by which I mean the image of the Young symmetrizers belonging to the $(2,1)$-Young tableau. However, I'm not very familiar with these latter objects, and the calculations I have attempted to understand this low-dimensional case just make my head spin. If there's a good piece of literature that'd help me there, I'd welcome a recommendation.
 A: In Fulton's "Young Tableaux with applications to representation theory and geometry", §8.3, after Corollary 3, he derives an expression of $\Lambda^k (V \otimes W)$ and $S^k (V \otimes W)$ in terms of Young symmetrizers. The proof isn't explicitly spelled out, but he argues it arises from considering representations of $GL(U) \times GL(V)$.
Regardless, what we end up with is the following: if $p \geq 0$ is an integer, consider the set of all partitions $\lambda = (\lambda_1,\dots,\lambda_m)$ with $\lambda_1 + \dots + \lambda_m = p$ and $\lambda_1 \geq \dots \geq \lambda_m > 0$. We write $\lambda \vdash p$ for such partitions.
Such a partition gives rise to a young diagram with $\lambda_i$ boxes in the $i$-th row. We identify such a Young diagram with the partition $\lambda$.
Also, with the Young symmetrizer, these partitions give rise to irreducible submodules $U^\lambda \subset U^{\otimes p}, V^\lambda \subset V^{\otimes p}$, with respect to the action of the $p$-th symmetric group.
Recall now also that every Young diagram $\lambda$ has a transpose $\tilde{\lambda}$ resulting from flipping the Young diagram along its diagonal. Then, we have
$$\Lambda^p(U \otimes V) = \bigoplus_{\lambda \vdash p} U^\lambda \otimes V^{\tilde{\lambda}}, \quad 
S^p(U \otimes V) = \bigoplus_{\lambda \vdash p} U^\lambda \otimes V^{\lambda}.$$
When $p = 2$, the only partitions of $p$ are $2 = 1 + 1$, resulting in the Young diagram of two boxes in a single column, and $2 = 2$, two boxes in a single row, and both of them are obviously transpose to one another. The former corresponds to $U^\lambda = S^2 U$, the latter corresponds to $U^\lambda = \Lambda^2 U$, from which follow then
$$S^2(U \otimes V) = (S^2 U \otimes S^2 V) \oplus (\Lambda^2 U \otimes \Lambda^2 V), 
\quad 
\Lambda^2(U \otimes V) = (S^2 U \otimes \Lambda^2 V) \oplus (\Lambda^2 U \otimes S^2 V).$$
The more complicated $p = 3$ case again has a single column, a single row as Young diagrams, but now also the $3 = 2 + 1$ diagram, which is its own transpose. Hence:
$$\Lambda^3(U \otimes V) = (S^3 U \otimes \Lambda^3 V) \oplus (U^{(2,1)} \otimes V^{(2,1)} )\oplus (\Lambda^3 U\otimes S^3 V)$$
and similarly for symmetric tensors.
While not necessarily easy to perform any explicit calculations with that, I'll say this is at least quite pretty :)
