# Exterior and symmetric powers of external tensor products of representations

Let us assume that $\pi: G\to Aut(V)$ and $\rho : K\to Aut(W)$ are two finite-dimensional representations of two Lie groups $G$ and $K$, and consider the representation
$\pi\hat{\otimes}\rho : G\times K\to Aut(V\otimes W)$, the so called external tensor product of $\pi$ and $\rho$, given by

$$(\pi\hat{\otimes}\rho)(g, k)(v\otimes w):= \pi(g)v\otimes \rho(k)w,$$

for any $g\in G$, $k\in K$, $v\in V$ and $w\in W$. For the second exterior power of the representation $\pi\hat{\otimes}\rho$ is known the following isomorphism:

$$\Lambda^{2}(\pi\hat{\otimes}\rho)=(\Lambda^{2}\pi \ \hat{\otimes}\ Sym^{2}\rho) \ \oplus (Sym^{2}\pi \ \hat{\otimes} \ \Lambda^{2}\rho).$$ Similarly, for the second symmetric power it holds that

$$Sym^{2}(\pi\hat{\otimes}\rho)=(Sym^{2}\pi \ \hat{\otimes}\ Sym^{2}\rho) \ \oplus (\Lambda^{2}\pi \ \hat{\otimes} \ \Lambda^{2}\rho).$$

I would like to understand how these formulas can be generalized for exterior and symmetric powers of bigger degree. For example, what we can say about

$$\Lambda^{3}(\pi\hat{\otimes}\rho), \ \Lambda^{4}(\pi\hat{\otimes}\rho), \ Sym^{3}(\pi\hat{\otimes}\rho), \ Sym^{4}(\pi\hat{\otimes}\rho), \ \dots \ ?$$

For the isomorphisms above, you can see for example the link

An isomorphism of 2-Schur modules

Thank you!

• I don't think you can say much. Determining these relationships boils down to determining some symmetric function identities (by looking at characters), and I don't expect there to be nice corresponding identities in general. – Qiaochu Yuan Apr 2 '13 at 2:01
• why is this in CW mode? – Marc Palm Apr 2 '13 at 7:29

One has $$Sym^k(\pi\otimes\rho) = \bigoplus_{|\alpha|=k} \Sigma^\alpha(\pi)\otimes\Sigma^\alpha(\rho),$$ the sum is over all Young diagrams with $k$ boxes, $\Sigma^\alpha$ is the Schur functor. Similarly, $$\Lambda^k(\pi\otimes\rho) = \bigoplus_{|\alpha|=k} \Sigma^\alpha(\pi)\otimes\Sigma^{\alpha^T}(\rho),$$ where $\alpha^T$ is the transposed diagram.
• Apparently, if we don't insist on direct sums but are OK with a filtration, then it also works in finite characteristic as well (except that the second factor in the formula for $\Lambda$ is "the other $\Sigma$ which generalizes divided powers). This is explained in Theorem 2.3.2 of Weyman's book. – Vladimir Baranovsky Apr 11 '13 at 17:40