Decompose $\Lambda^3(V \otimes W)$ [closed]

Question

Let $V, W$ be two vector spaces. We have $\Lambda^2(V \otimes W) \cong (\Lambda^2 V \otimes S^2 W) \oplus (S^2 V \otimes \Lambda^2 W)$. I am trying to find similar results for $\Lambda^3(V \otimes W)$. I think that $\Lambda^3 V \otimes S^3 W$ and $S^3 V \otimes \Lambda^3 W$ are subspaces of $\Lambda^3(V \otimes W)$. It seems that $\Lambda^3(V \otimes W)$ has other subspaces since if the dimensions of $V$ and $W$ are $n, m$ respectively, then the dimension of $\Lambda^3(V \otimes W)$ is $\frac{m\, n\, \left(m\, n - 1\right)\, \left(m\, n - 2\right)}{6}$ and the sum of dimensions of $\Lambda^3 V \otimes S^3 W$ and $S^3 V \otimes \Lambda^3 W$ is $\frac{m\, n\, \left(m^2\, n^2 + 2\, m^2 - 9\, m\, n + 2\, n^2 + 4\right)}{18}$. The difference of these two numbers is $\frac{m\, n\, \left(n - 1\right)\, \left(n + 1\right)\, \left(m - 1\right)\, \left(m + 1\right)}{9}$. Therefore we have $$ \Lambda^3(V \otimes W) \cong (\Lambda^3 V \otimes S^3 W) \oplus (S^3 V \otimes \Lambda^3 W) \oplus W_1. $$ How to compute $W_1$ explicitly? Any help will be greatly appreciated!

Answer 1

EDIT: The group $GL(V)\times GL(W)$ acts naturally on $\wedge^n(V\otimes W)$ for any $n\in \mathbb{N}$ (in your case $n=3$). One can describe decomposition of this space into irreducible components. For example for $n=2$ the decomposition you described coincides with that one.

Any irreducible representation of $GL(V)\times GL(W)$ has the form $\pi\otimes \rho$ where $\pi$ and $\rho$ are irreducible representation of $GL(V)$ and $GL(W)$ respectively. In our case relevant irreducible representations $\pi$ and $\rho$ of $GL(V)$ and $GL(W)$ are parameterized by Young diagrams with $n$ cells. Moreover for every irreducible component $\pi\otimes \rho$ the corresponding to $\pi$ and $\rho$ Young diagrams are transposed to each other (and have $n$ cells). Any such pair of diagrams appears in the decomposition at most once, and in fact exactly once provided the dimensions of $V$ and $W$ are large enough. Thus for $n=3$ there are three such diagrams, hence 3 irreducible components. The first two summands you wrote in this case are the two irreducible $GL(V)\times GL(W)$-components. The last summand is the tensor product of two Young diagrams both equal to $(2,1)$, i.e. in the first row there are two cells, and in the second one only one cell. Actually is was described in the comment by Robert Bryant as the kernel of the symmetrization map $V\otimes S^2(V)\to S^3(V)$. Similarly for $W$.

In representation theoretical literature there is a lot of information about the properties of representations with given Young diagram, e.g. how to compute the dimension. See e.g. the paper "Remarks on classical invariant theory" by R. Howe, p. 560. The space can be described explicitly using e.g. Young symmetrizers.