# How does Tate cohomology fit into a derived categories framework?

I've read through one class field theory text after another, but there's something very non-intuitive for me about cohomology that makes it hard for me to understand why Tate cohomology was invented.

In order to make up for my lack of intuition regarding anything cohomological, I have gone through a few texts introducing the derived category framework. This approach has been very helpful for me, and has helped me gain intuition about cohomological methods in Algebraic Topology and Algebraic Geometry. My hope is that, with your help, it could also shed light on "Tate cohomology", which, at the moment, seems to me like a completely arbitrary definition that helps in mysterious and miraculous ways.

Is there a definition of Tate cohomology via derived categories that sheds light on why they are a natural thing to consider? What would be a reference of such a treatment? (If you think that I am looking in the wrong place for intuition and you have an alternative suggestion, I would be very happy to hear that as well!)

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It's the 2-sided derived functor of its degree-0 part (via the evident generalization of the usual "erasable $\delta$-functor" formalism to the case of 2-sided $\delta$-functors). – Marguax Oct 1 '13 at 21:53

To any ring $R$, we can associate the bounded derived category $D^b(R)$. The full subcategory $D^{perf}(R)$ is spanned by bounded complexes of projectives. If $R$ is self-injective, e.g. $R=k[G]$ the group-algebra of a finite group $G$ over a field $k$, then the Verdier quotient $D^b(R)/D^{perf}(R)$ is equivalent to the stable module category $\underline{mod}(R)$. If $R=k[G]$, and $M$ is an $R$-module, the Tate cohomology groups are

$$H^n(G,M)=\underline{mod}(R)(k,M[n]),\quad n\in\mathbb Z,$$ where $k$ is, as usual, the trivial $R$-module. The identification of the quotient with the stable module category is the non-trivial part of the story, and is due to Rickard ("Derived categories and stable equivalence").

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whistle in awe... I need to look up some of these terms, but this answer sounds promising. – Tavish Jordan Oct 1 '13 at 20:48
So, in a few words, can you explain why from this viewpoint Tate cohomology makes things more simple than plain group cohomology? (Or in general, what the impetus for inventing it is.) – Tavish Jordan Oct 1 '13 at 20:50
I don't know how Tate cohomology originally arises, so I cannot answer why it was invented. I don't think that Tate cohomology is simpler than group cohomology in any way, since the latter is part of the former. – Fernando Muro Oct 1 '13 at 21:00
It is essential in practice that one considers $G$-modules which are not over a field, and $\mathbf{Z}[G]$ is not self-injective. – Marguax Oct 1 '13 at 21:53
@Fernando: That isn't true. In fact, the historical reason Tate invented Tate cohomology in the first place was for applications to class field theory, for which the interesting examples are not over fields. The Tate cohomology functors most definitely are defined on the category of $\mathbf{Z}[G]$-modules for any finite group $G$ (see number theory books of Cassels-Frohlich, Lang, etc.). – Marguax Oct 1 '13 at 22:21

More of a comment to Fernando Muro's answer: Alexander Beilinson has nice handwritten lecture notes on class field theory which can be found here: http://www.math.lsa.umich.edu/~mityab/beilinson/. An expansion on the point of view explained by Fernando Muro can be found in Lecture 2 with many illuminating explanations, examples and applications.

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Ah, great. I've come across these notes before, but I've forgotten they existed. I am probably mathematically mature enough now to understand them better than I did the first time around. – Tavish Jordan Oct 1 '13 at 20:59