**Situation**


Let $G$ be a finite group and provide $G\text{-mod} := {\mathbb Z}G\text{-mod}$ with the Frobenius structure of ${\mathbb Z}$-split short exact sequences. Denote by $\underline{G\text{-mod}}$ the associated stable category with loop functor $\Omega$.


For any Frobenius category $({\mathcal A},{\mathcal E})$ and a complete projective-injective resolution $P\_{\bullet}$ of some $X\in{\mathcal A}$, we have for any $Y\in{\mathcal A}$ a canonical isomorphism of abelian groups 


$H^n(\text{Hom}\_{\mathcal A}(P\_{\bullet},Y))\cong [\Omega^n X,Y]$,


where $[-,-] := \text{Hom}\_{\underline{{\mathcal A}}}(-,-)$.


Applying this to $G\text{-mod}$ yields an isomorphism


$\widehat{H}^k(G;M)\cong [\Omega^k{\mathbb Z},M]$,


where $\widehat{H}^k(G;M)$ denotes the Tate-Cohomology of $G$ with values in $M$.


If I didn't mix things up, in this language Tate-Duality should mean that the canonical map


$[{\mathbb Z},\Omega^k{\mathbb Z}]\otimes\_{\mathbb Z}[\Omega^k{\mathbb Z},{\mathbb Z}]\to[{\mathbb Z},{\mathbb Z}]\cong{\mathbb Z}/|G|{\mathbb Z}$


is a duality.


**Question**


I'd like to know sources which introduce and treat Tate cohomology in the way described above, i.e. using the language of Frobenius categories and its associated stable categories. *In particular*, I would be interested in a proof of Tate Duality using this more abstract language instead of resolutions.


Does anybody know such sources?


**Remark**


It seems to be more difficult to work over the integers instead of some field, for in this case, the exact sequences in the Frobenius structure $G\text{-mod}$ are required to be ${\mathbb Z}$-split, which is not automatic. As a consequence, there may be projective/injective objects in $(G\text{-mod},{\mathcal E}^{G}\_{\{e\}})$ which are not projective/injective as ${\mathbb Z}G$-modules. Further, the long exact cohomology sequence exists only for ${\mathbb Z}$-split exact sequences of $G$-modules (not good, because Brown uses the exact sequence $0\to {\mathbb Z}\to{\mathbb Q}\to{\mathbb Q}/{\mathbb Z}\to 0$ in his proof of Tate duality); of course, one can choose particular complete resolutions of ${\mathbb Z}$ consisting of ${\mathbb Z}G$-projective modules, and such a resolution yields a long exact cohomology sequence for any short exact sequence of coefficient modules, but this seems somewhat unnatural and doesn't fit into the picture right now.


**Partial Results**


**(1)** For any subgroup $H\leq G$ there are restriction and corestriction morphisms 


$[\Omega^k {\mathbb Z},-]^{\underline{G}}=\widehat{H}^\*(G;-)\leftrightarrows\widehat{H}^\*(H;-)=[\Omega^k{\mathbb Z},-]^{\underline{H}}$


defined as follows: for any $G$-module $M$, the abelian group $[{\mathbb Z},M]^{\underline{G}}$ is in canonical bijection with $M^G / |G| M^G$, and there are restriction and transfer maps


$\text{res}: M^G / |G| M^G\longrightarrow M^H / |H| M^H,\quad [m]\mapsto [m]$,


$\text{tr}: M^H / |H| M^H\longrightarrow M^G / |G| M^G\quad [m]\mapsto\left[\sum\limits\_{g\in G/H} g.m\right]$,


respectively. Now


$[\Omega^k{\mathbb Z},M]^{\underline{G}}\cong [{\mathbb Z},\Omega^{-k}M]^{\underline{G}}\stackrel{\text{res}}{\longrightarrow} [{\mathbb Z},\Omega^{-k}M]^{\underline{H}}\cong[\Omega^k{\mathbb Z},M]^{\underline{H}}$


$[\Omega^k{\mathbb Z},M]^{\underline{H}}\cong [{\mathbb Z},\Omega^{-k}M]^{\underline{H}}\stackrel{\text{tr}}{\longrightarrow} [{\mathbb Z},\Omega^{-k}M]^{\underline{G}}\cong[\Omega^k{\mathbb Z},M]^{\underline{G}}$


seems to be the natural thing to define restriction and transfer. (This is very similar to the usual method of giving a morphism of $\delta$-functors only in degree $0$ and extend it by dimension shifting, though a bit more elegant in my opinion)


Note that it was implicitly used that $\Omega^k$ commutes with the forgetful functor $G\text{-mod}\to H\text{-mod}$


**(2)** For any subgroup $H\leq H$, $g\in G$ and a $G$-module $M$ there is a map 


$g\_\*:\ \widehat{H}^\*(H;-)\to\widehat{H}^\*(gHg^{-1};M)$


extending the canonical map


$M^H/|H|M^H\longrightarrow M^{gHg^{-1}}/|H|M^{gHg^{-1}},\quad [m]\mapsto [g.m]$.


(1) and (2) fit together in the usual way; there is a transfer formula and a lifting criterion for elements of Sylow-subgroups.


**(3)** The cup product on $\widehat{H}^\*(G;{\mathbb Z})$ is given simply by composition of maps:


$[\Omega^p{\mathbb Z},{\mathbb Z}]\otimes\_{\mathbb Z}[\Omega^q{\mathbb Z},{\mathbb Z}]\stackrel{\Omega^q\otimes\text{id}}{\longrightarrow}[\Omega^{p+q}{\mathbb Z},\Omega^q{\mathbb Z}]\otimes\_{\mathbb Z}[\Omega^q{\mathbb Z},{\mathbb Z}]\longrightarrow [\Omega^{p+q}{\mathbb Z},{\mathbb Z}]$


Does anybody see why this product is graded-commutative?



Tate Cohomology via stable categories
To address Hanno's question about checking that composition gives a graded-commutative ring structure on $End^{\*}(\mathbb{Z}) = \oplus\_i [\mathbb{Z}, \Omega^{-i} \mathbb{Z}]$ suppose first that  

$a \stackrel{f}{\to} b \stackrel{g}{\to} c \stackrel{h}{\to} \Omega^{-1} a$  

is a distinguished triangle in the stable category. Then we can produce from this two isomorphic triangles: one by rotation namely  

$a \stackrel{-\Omega^{-1}f}{\to} b \stackrel{-\Omega^{-1}g}{\to} c \stackrel{-\Omega^{-1}h}{\to} \Omega^{-1} a$  

and one by applying $\Omega^{-1}\mathbb{Z} \otimes\_\mathbb{Z}$  

$\Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} a \stackrel{1\_{\Omega^{-1}\mathbb{Z}}\otimes f}{\to} \Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} b \stackrel{\Omega^{-1}\mathbb{Z}\otimes g}{\to}\Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} c \stackrel{\Omega^{-1}\mathbb{Z}\otimes h}{\to} \Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z} \Omega^{-1}a$  

The point of this is that the natural isomorphism commuting the loops functor across introduces a sign change, which is precisely the one you pick up by changing the order of composition since changing the composition order is equivalent to applying symmetry isomorphisms to the tensor product which is equivalent to commuting loops across.


To be completely explicit about this there are two functors naturally isomorphic to $\Omega^{-1}$ namely $\Omega^{-1}\mathbb{Z}\otimes\_\mathbb{Z}(-)$ and $\mathbb{Z}\otimes \Omega^{-1}(-)$ since $\otimes\_\mathbb{Z}$ is biexact there are natural transformations commuting $\Omega^{-1}$ with $\otimes\_\mathbb{Z}$ namely  

$\Omega^{-1}(-)\otimes\_\mathbb{Z} (-) \stackrel{\sim}{\rightarrow} \Omega^{-1}((-)\otimes\_\mathbb{Z}(-)) \stackrel{\sim}{\leftarrow} (-)\otimes\_\mathbb{Z} \Omega^{-1}(-)$  

Our example triangles above which can be obtained from one another by first moving the loops to the right and then applying the unit transformation show that there must be a sign attached to this map for this to give an isomorphism of these triangles. In particular for $\otimes\_\mathbb{Z}$ to be compatible with the triangulated structure the two natural isomorphisms moving $\Omega$ about must have different signs. These natural isomorphisms are the precise cause of the sign change.


