**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?

`$\widehat{H}^k(G;{\mathbb Z})\otimes_{\mathbb Z}\widehat{H}^{-k}(G;{\mathbb Z})\to\widehat{H}^0(G;{\mathbb Z})\cong{\mathbb Z}/|G|{\mathbb Z}$`

is a duality between the finite`$|G|$`

-torsion groups`$\widehat{H}^k(G;{\mathbb Z})$`

and`$\widehat{H}^{-k}(G;{\mathbb Z})$`

I thought I had seen it called "Tate Duality", but I'm not sure. $\endgroup$4more comments