My notes ask me to confirm the result:

$\mathrm{Hom}(V \otimes W, U) \cong \mathrm{Hom}(V, \mathrm{Hom}(U,W))$ as $\mathbb C$-representations of a finite group $G$.

But it seems to be incorrect. $\chi_{\mathrm{Hom}(V \otimes W, U)} = \overline{\chi_{V \otimes W}} \chi_U = \overline{\chi_V}\ \overline{\chi _ W} \chi_U$, whilst $\chi_{\mathrm{Hom}(V, \mathrm{Hom}(U,W))} = \overline{\chi_V} \chi_{\mathrm{Hom}(U,W)} = \overline{\chi_V} \ \overline{\chi_U} \chi_W$. So it looks like $U$ and $W$ should be switched in the RHS of the isomorphism.

Am I right?


  • 2
    $\begingroup$ Yes, you are right. $\endgroup$ – Qiaochu Yuan Mar 22 '12 at 20:17

Yes, you're right. This is a version of hom-tensor adjunction. Instead of equality of characters you could also write down an ismorphism of vector spaces: in one direction it should send $f: V\otimes W \to U$ to the map that sends $v\in V$ to $w \mapsto f(v\otimes w) \in \hom(W,U)$.


An alternative proof (for all fields is) ${\rm Hom}(X\otimes Y,Z)\cong (X\otimes Y)^*\otimes Z\cong (X^* \otimes Y^*)\otimes Z\cong X^* \otimes (Y^*\otimes Z)\cong {\rm Hom}(X,{\rm Hom}(Y,Z)).$


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.