Define $V_{p,q}=\underset{p}{\underbrace{V\otimes\cdots\otimes V}}\otimes\underset{q}{\underbrace{V^{*}\otimes\cdots\otimes V^{*}}}$.
In a previous question here I was shown that $\hom_{k}\left(V_{1,0},V_{1,0}\right)=\mbox{End}V\simeq V\otimes V^{*}=V_{1,1}$. Now I've heard that this could be generalized to the form: $\hom_{k}\left(V_{p,q},V_{r,s}\right)\simeq V_{q+r,p+s}$ for all $p,q,r,s\in\mathbb{N}$. How do you prove this?
Thanks!
[Assuming that everything is finite-dimensional.] You could combine other well-known isomorphisms. Using the ideas from NKS's answer, show that there is a natural isomorphism $$ \operatorname{Hom}(V, W) \approx V^* \otimes W. $$ You also have a nice identification $$ V^* \otimes W^* \to (V \otimes W)^*, $$ which you can then extend to the case of finitely many (but possibly more than two) factors. Just send $f \otimes g$ to the functional on $V \otimes W$ mapping $v \otimes w \mapsto f(v)g(w)$. And of course $(V^*)^* \approx V$. I think these are all of the ingredients that you need: apply the first isomorphism to $V_{p, q}$ and $V_{r, s}$, apply the second to show that $V_{p, q}^* \approx V_{q, p}$, and you're almost home.
