0
$\begingroup$

Schur's lemma says that if $M,N$ are two irreducible representations of a group $G$, then either $Hom_G(M,N)=0$ if $M,N$ are not isomorphic, or every $\varphi\in Hom_G(M,N) $ is invertible if they are isomorphic.

If we look at the case $G=S_n$ and $M=V_\lambda,\ N=V_\mu$ for partitions $\lambda,\mu$ of $n$, we must have $Hom_{S_n}(V_\lambda,V_\mu)=0$ if $\lambda\neq \mu$. But $V_\lambda$ is self dual, so $Hom_{S_n}(V_\lambda,V_\mu)\cong V^*_\lambda\otimes V_\mu\cong V_\lambda\otimes V_\mu=0$.

But exercise 4.51 in Fulton-Harris implies that the tensor product $V_\lambda \otimes V_\mu$ decomposes as a direct sum of copies of $V_v$ for partitions $v$ of $n$. In particular, $V_{(n)}\otimes V_\lambda=V_\lambda$ and $V_{(1,...,1)}\otimes V_\lambda=V_{\lambda^\prime}$, where $\lambda^\prime$ is the conjugate partition.

What went wrong here?

$\endgroup$
3
$\begingroup$

You're confusing the internal and external hom here. $V_{\lambda}^{\ast} \otimes V_{\mu}$ is isomorphic to the internal hom $[V_{\lambda}, V_{\mu}]$, which is the vector space of all linear maps $V_{\lambda} \to V_{\mu}$ with the appropriate $S_n$-action. The external hom $\text{Hom}_{S_n}(V_{\lambda}, V_{\mu})$ is the $S_n$-invariant subspace of the internal hom.

This implies that $V_{\lambda} \otimes V_{\mu}$ has at most one trivial summand (it has one if $\lambda = \mu$ and zero otherwise), but it can have lots of other nontrivial summands.

$\endgroup$
  • 1
    $\begingroup$ He is also confusing the internal and the external tensor product, which I found to be the more obvious issue :) $\endgroup$ – darij grinberg Dec 13 '15 at 20:13
  • 1
    $\begingroup$ I guess, but if the OP knew that there even was such a thing as the external tensor product he would hopefully be denoting it by $\otimes_{S_n}$. $\endgroup$ – Qiaochu Yuan Dec 13 '15 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.