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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?


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The key is that the tensor product of a projective module and any module is projective. Then it follows from the definitions. – Bruce Westbury Mar 13 '10 at 19:16
@Bruce: is it really as automatic as you say? I mean, the second Ext group in general doesn't even make sense if we're working in the module category of an arbitrary algebra. Your comment also isn't true for arbitrary module categories (I think). – Yemon Choi Mar 13 '10 at 19:57
Yemon- remember, we're dealing with tensor products in the sense of Hopf algebras here, so tensoring with an object is exact as a functor. – Ben Webster Mar 13 '10 at 20:16
Better yet, it has exact adjoint, since the adjoint is just tensoring with the dual representation. A functor between representations of finite dimensional algebras sends projectives to projectives if its adjoint is exact. – Ben Webster Mar 13 '10 at 20:19
OK, I was misinterpreting the tensor product of two modules as being a bimodule. If we're only taking the module action on, say, the first factor, then yes I agree that projectives are preserved. (I notice that Ben has immediately gone to finite-dimensional algebras; which is where I may have been talking at cross-purposes to everyone else.) – Yemon Choi Mar 13 '10 at 20:22
up vote 11 down vote accepted

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.

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Thanks! It was informative. – Adam Gal Mar 13 '10 at 21:26
Surely, Ext(M,N) is the homology of the Hom-complex between the resolutions, right? – Mikael Vejdemo-Johansson Mar 13 '10 at 22:49
That depends on your indexing conventions, doesn't it? – Ben Webster Mar 14 '10 at 0:52

This is probably typographic, but the original question writes

Ext(M,N) is naturally isomorphic to Ext(M*⊗N,1)

when what is true (see other answers) is that Ext(M,N) is isom. to Ext(M $\otimes$ N*,1).

By the way, it is often useful to view Ext(M,N) as Ext(1,M* $\otimes$ N) instead, since this later ext-group coincides with the group cohomology H(G,M* $\otimes$ N); here H*(G,-) = derived functor of "G-fixed-points".

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