MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In one of their papers (before Theorem 7.2), Benson and Carlson state that the transfer map is Tate-dual to the restriction homomorphisms (also see Remark 1.3 of this recent paper).

More precisely: If $H \le G$ are finite groups and $k$ a field those characteristic divides the order of $H$, then the there should be a commutative diagramm $$\begin{array}{ccc} \hat{H}^{-s-1}(G,k) & \cong & \text{Hom}_k(\hat{H}^sS(G,k),k) \newline res^G_H \downarrow & & \downarrow (tr^G_H)^\ast \newline \hat{H}^{-s-1}(H,k) & \cong & \text{Hom}_k(\hat{H}^sS(H,k),k) \end{array}$$ where the horizontal isomorphisms are Tate duality.

Does anyone know a reference with a proof or can provide a proof of this statement ? Thanks in advance.

share|cite|improve this question
If you pass to a broader context then you can use degree-shifting. Namely, consider the category of all finite-dimensional $k$-linear representations $V$ of $G$, so the cup product pairing in Tate cohomology ${\rm{H}}^s(G,V) \times {\rm{H}}^{-s-1}(G,V^{\ast}) \rightarrow {\rm{H}}^{-1}(G,k) = k$ is a perfect duality (if I remember correctly), and likewise for cohomology of $H$. Then by the two-sided "erasability" of Tate cohomology and variation in $V$ you should be able to degree-shift the problem to the case $s = 0$, which can be analyzed by inspection. – user29720 Dec 10 '12 at 5:41
To clarify, at the end of my preceding comment I meant "the case $s = 0$ with coefficients in $V$ and $V^{\ast}$" (not just $s = 0$ with coefficients in $k$, which is too restrictive). – user29720 Dec 10 '12 at 5:43
Do you mean $\hat{H}$ instead of $H$ in your diagramm ? – Ralph Dec 11 '12 at 15:31
up vote 3 down vote accepted

I don't know of a reference, but the duality in question can be proved by results from Brown's book on group cohomology. I'll show the case $s\ge 0$. First note that for each integer $j$ there is an isomorphism

$$\psi: \hat{H}^j(G,k) \xrightarrow{\sim} \text{Hom}_k(\hat{H}_j(G,k),k)$$

(Brown, VI.7.2) and for $s\ge 0$ there is an isomorphism $$\varphi: \hat{H}_{-s-1}(G,k) \xrightarrow{\sim} H^s(G,k)$$

(Brown, VI.4). Denote the $k$-dual of a vector space or of a homomorphism by $(-)^\ast$. Tate duality is then the composition $$t = (\varphi^{-1})^\ast\circ \psi: \hat{H}^{-s-1}(G,k) \xrightarrow{\sim} \hat{H}_{-s-1}(G,k)^\ast \xrightarrow{\sim} H^s(G,k)^\ast.$$ Hence we have to show the commutativity of the diagramm

$$\begin{array}{ccccc} \hat{H}^{-s-1}(G,k) & \xrightarrow{\psi} & \hat{H}_{-s-1}(G,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(G,k)^\ast \newline {\scriptstyle \widehat{res}} \downarrow & & \downarrow \scriptstyle res^\ast & & \downarrow \scriptstyle tr^\ast\newline \hat{H}^{-s-1}(H,k) & \xrightarrow{\psi} & \hat{H}_t(H,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(H,k)^\ast \end{array}$$

($t$ stands for $-s-1$ which the editor doesn't accept!?) The commutativity of the left hand square follows right from the definition of the maps and the right hand square commutes if we can show the commutativity of the following square: $$\begin{array}{ccc} \hat{H}_{-s-1}(G,k) & \xrightarrow{\varphi} & H^s(G,k)\newline {\scriptstyle \widehat{res}} \uparrow & & \uparrow \scriptstyle tr^G_H \newline \hat{H}_{-s-1}(H,k) & \xrightarrow{\varphi} & H^s(H,k) \end{array}\tag{1}$$ In order to describe $\varphi$ on chain level, let $P \to k$ be a projective resolution over $kG$ and let $F$ be a complete resolution such that $F_i=P_i$ and $F_{-i-1} = P_i^\ast$ for $i \ge 0$. Then $\varphi$ is induced by the composition $$\varphi: F_{-i-1}\otimes_{kG}k=P_i^\ast \otimes_{kG}k \xrightarrow{\alpha\otimes id} \text{Hom}_{kG}(P_i,kG) \otimes_{kG} k\xrightarrow{\beta}\text{Hom}_{kG}(P_i,k)$$ where $\alpha(f)(x)=\sum_{g \in G}f(g^{-1}x)g$ (Brown, VI.3.4) and $\beta(f \otimes a)(x)=f(x)a$ (Brown, I.8.3). Hence $$\varphi(f \otimes a)(x)=\sum_{g \in G}f(g^{-1}x)(ga)=\sum_{g \in G}f(g^{-1}x)a=tr^G_E(f)(x)a\tag{2}$$ where $f \in P_i^\ast, a \in k, x \in P_i$ and $E=\{1\}$.

On chain level $(1)$ is given by the diagramm $$\begin{array}{ccc} P_i^\ast \otimes_{kG} k & \xrightarrow{\varphi_G} & \text{Hom}_{kG}(P_i,k) \newline {\scriptstyle \kappa} \uparrow & & \uparrow \scriptstyle tr^G_H \newline P_i^\ast \otimes_{kH} k & \xrightarrow[\varphi_H]{} & \text{Hom}_{kH}(P_i,k) \newline \end{array}\tag{3}$$

where $\kappa(f \otimes_H a)=f \otimes_G a$. With $f,a,x$ as above, we obtain $$(tr^G_H \circ \varphi_H)(f \otimes_H a)(x)=\sum_{g \in G/H}\varphi_H(f\otimes_H a)(g^{-1}x) \overset{(2)}{=}\sum_{g \in G/H}tr^H_E(f)(g^{-1}x)a$$ $$\qquad=tr^G_H(tr^H_E(f))(x)a=tr^G_E(f)(x)a$$ $$\qquad\qquad=\varphi_G(f \otimes_G a)(x)=(\varphi_G \circ res)(f \otimes_H a)(x)$$ Thus the commutativity of $(3)$ is shown. QED

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.