I heard that $Ext(M,N)$ is naturally isomorphic to $Ext(M^\*\otimes N,1)$ where 1 is the trivial representation and $M,N$ some representations of a group $G$.
Can anyone explain why?
Is there an explicit construction of a map from one to the other or does it just follow from some general considerations about derived functors?


Thanks.



Question about Ext
Perhaps the best way to think about this is as follows: pick your favorite injective resolution for N and favorite projective resolution of M. Then $\mathrm{Ext}(M,N)$ is given by taking Hom between these complexes (NOT chain maps, just all maps of representations between the underlying modules), and putting a differential on those in the usual way.


Now, use the usual identification of $\mathrm{Hom}(A,B)\cong \mathrm{Hom}(A\otimes B^\*,1)$ on this complex. So you see, it's the same as if we had tensored the projective resolution of $M$ with the dual of the injective resolution of $N$, which is a projective resolution of $N^\*$, and then taken Hom to 1. Of course, the tensor product of two projective resolutions is a projective resolution of the tensor product, so we see this complex also computes $\mathrm{Hom}(N^\*\otimes M,1)$.`


It also follows by abstract nonsense in one line: isomorphic functors have isomorphic derived functors.


