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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?

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Sorry to show my ignorance, but what do you mean by Tate Duality? I've heard this phrase used in the context of Galois cohomology of local and global fields but I don't think I've heard it in the context of Tate cohomology groups when $G$ is finite. –  Kevin Buzzard Jan 23 '10 at 22:30
    
I mean the statement that the map $\widehat{H}^k(G;{\mathbb Z})\otimes_{\mathbb Z}\widehat{H}^{-k}(G;{\mathbb Z})\to\widehat{H}^0(G;{\mathbb Z})\cong{\mathbb Z}/|G|{\mathbb Z}$ 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. –  Hanno Becker Jan 23 '10 at 22:43
    
I saw Beilinson develop Tate cohomology somewhat along these lines in a lecture at U of Chicago maybe 9 years ago, but I've not seen a written source. –  Emerton Jan 24 '10 at 4:13
    
There are references for developing analogues of Tate cohomology in various triangulated settings but I am not sure if these would be helpful. I have not seen this written down anywhere for groups either. Is there any reason you expect that going via the stable category would give an abstract proof of Tate duality? Really one is still working with resolutions and it isn't immediately obvious to me that such a statement should follow from general nonsense. –  Greg Stevenson Jan 24 '10 at 4:48
    
The cup product you have defined above is graded-commutative due to $\Omega$ itself being graded-commutative in some sense. One can see this from playing around with the rotation axiom for triangles (usually denoted [TR2]). You might also want to see mathoverflow.net/questions/5901/… for further details. –  Greg Stevenson Jan 24 '10 at 20:17

5 Answers 5

up vote 3 down vote accepted

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.

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Ahh, I finally got it :) ... the maps $\Omega^{-1}{\mathbb Z}\otimes_{\mathbb Z}\Omega^{-1}a\cong{\mathbb Z}\otimes_{\mathbb Z}\Omega^{-2}a\cong\Omega^{-2}a$ and $\Omega^{-1}{\mathbb Z}\otimes_{\mathbb Z}\Omega^{-1}a\cong\Omega^{-1}(\Omega^{-1}{\mathbb Z}\otimes_{\mathbb Z} a)\to\Omega^{-1}({\mathbb Z}\otimes_{\mathbb Z}\Omega^{-1}a)\cong\Omega^{-2}a$ agree only up to sign (and this is also what Mariano requires axiomatically in his paper). Now everything's clear! Thanks alot, Greg, for your help and your patience! –  Hanno Becker Jan 25 '10 at 22:37

The notes by Beilinson which Emerton is referring to (or at least a very similar course) have been hand written and are up online at http://www.math.uchicago.edu/~mitya/beilinson/ . They develop the theory via $G$-modulations.

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One way to show that your product is commutative is given in http://arxiv.org/abs/math/0209029.

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Thank you, Mariano! How do you prove that a tensored Frobenius Category is suspended in the sense of your paper, i.e. how do you get the anti-commutativity of the diagram in (1.4)? Is it a special case of (2.1) in some way? If not, one could apply the following: if we have an 3x3-array $(x_{ij})_{0\leq i,j\leq 2}$ of elements in a Frobenius category with arrows lowering the indices such that any column and any row is exact, then the two induced maps $\Omega^2 x_{22}\to\Omega x_{20}\to x_{00}$ and $\Omega^2 x_{22}\to\Omega x_{02}\to x_{00}$ agree up to the sign $-1$. –  Hanno Becker Jan 25 '10 at 16:46

Google gives the following paper: Greenlees, Tate cohomology in axiomatic stable homotopy theory. It gives a definition of the Tate construction using Bousfield localization and completion, and has some duality theorems, although I couldn't tell if any of them yield Tate duality as you state it.

I think if $M$ is a complex of abelian groups, the Tate construction $M^{TG}$ is the cofiber of the norm map $N: M_{hG} \to M^{hG}$ from homotopy orbits to homotopy fixed points. Lurie's lecture notes introduce the construction in the special case when $G \cong \mathbb{Z}/2 \mathbb{Z}$ and $M$ is a complex of $\mathbb{F}_2$-vector spaces, and give some properties.

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Maybe you want to check out the book 'Homological and Homotopical Aspects of Torsion Theories

In Chapter IX, they used the stable categories to develop this theory. Hope it is helpful

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