# Schur-Weyl Duality (Example for k = 3)

I am an undergraduate student interested in representation theory. I know that you can decompose the iterated tensor product $(C^n)^{\otimes k}$ into the direct sum of irreducible $S_k \times GL(V)$ representations $S^\lambda \otimes U^\lambda$, where the length of $\lambda$ is smaller equal n. For the partitions $(k)$ and $(1^k)$ these are just the symmetric / skewsymmetric tensors.

However, I am interested in the third part (example for k = 3) $S^{(21)} \otimes U^{(21)}$. I tried to compute both of them seperatly, but I'm stuck somewhere: I think $U^{(21)}$ is the kernel of the canonical map $Alt^2(V) \otimes V \rightarrow Alt^3(V)$. How does $S^{(21)} \otimes U^{(21)}$ look like though?

And how does this decompositions helps us in any way? I assume the general goal is to compute the multiplicities of irreducible representations?

Let $\tau:V^{\otimes 3}\to V^{\otimes 3}$ be defined by $\tau(u\otimes v\otimes w)=v\otimes w\otimes u$. Then the $S^{(21)}$ part of $V^{\otimes 3}$ is the kernel of $\textrm{id}+\tau+\tau^2$.
• Thanks for your quick answer! If I understand you correctly, those are tensors of the form $\sum_{\sigma \in A_k} v_{\sigma(1)}...v_{\sigma(3)}$. How does that fit/interact with $U^{(21)}$ though? Shouldn't I get just one subspace, let's say $H^{(21)}$? – user2103480 Jul 31 '18 at 15:45
• No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+\tau+\tau^2$ is a complement to that. @Amjad – Lord Shark the Unknown Jul 31 '18 at 15:51
• Okay, got it. And how does $S^{(21)} \otimes U^{(21)}$ look like? Isn't $S^{(21)}$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing \$U^{(21)}? Sorry that I've got so many questions. – user2103480 Jul 31 '18 at 17:21