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It is well known that given a short exact sequence $1\to H \to G \to G/H \to 1$ the transgression map $$ H^{p-1}(G/H, H^1(H,A)) \to H^{p+1}(G/H,A^H) $$ in the inflation-restriction sequence is in fact the cup product with the opposite of the class of the extension $1 \to H^{ab} \to G/H' \to G/H \to 1$ in $H^2(G/H, H^{ab})$, given that the action of $H$ on the module $A$ is trivial.

What is known about the transgression map in case $A$ is a non-trivial module? Is it possible then to describe transgression in similar terms?

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Is the upper $G/H$ in $H^{p-1}(...)^{G/H}$ correct ? –  Ralph May 28 '13 at 21:57
    
yes, you are right, shouldn't be there (I started writing this down for p=1 and then forgot to erase $G/H$) –  Dima Sustretov May 29 '13 at 7:32
    
What does $H^*(G/H,A)$ mean if $H$ doesn't act trivially on $A$? –  Matthew Towers May 29 '13 at 16:34
    
oops, it should be $A^H$ (which obviously is $A$ when the action is trivial) –  Dima Sustretov May 29 '13 at 17:36

1 Answer 1

up vote 2 down vote accepted

I can offer the following generalization for non-trivial coefficients $A$: Write $Q = G/H$. The cap product $$H_1(H;\mathbb{Z}) \otimes H^1(H;A) \to \mathbb{Z}\otimes_H A = A_H$$ is $G$-linear with trivial $H$-action and induces a cup product $$\cup: H^p(Q;H_1(H;\mathbb{Z})) \otimes H^q(Q;H^1(H;A))\xrightarrow{} H^{p+q}(Q;A_H).$$ Let $u \in H^2(Q;H_1(H;\mathbb{Z}))$ be the class that corresponds to the extension $$1 \to H^{ab} \to G/H' \to Q \to 1$$ and let $\kappa: A^H \to A_H$ be the canonical homomorphism.

Theorem: The compostion $$H^p(Q;H^1(H;A)) \xrightarrow{d_2}H^{p+2}(Q;A^H)\xrightarrow{\kappa^\ast}H^{p+2}(Q;A_H)$$ is (up to sign) cup product with $u$, i.e. $(\kappa^\ast \circ d_2^{p,1})(x) = - u \cup x$.

Proof: By abuse of notation let $\kappa: A \to A_H$. Since $\kappa$ is $G$-linear, it induces a map of spectral sequences $\kappa_r^{pq}: E_r^{pq}(A) \to E_r^{pq}(A_H)$. In particular, $\kappa_2^{p+2,0}\circ d_2^{p,1}(A)=d_2^{p,1}(A_H)\circ \kappa_2^{p,1}$. Clearly, $\kappa_2^{p+2,0}$ is the map $\kappa^\ast$ in the theorem and since $H$ acts trivially on $A_H$, we obtain $(\kappa^\ast \circ d_2^{p,1}(A))(x) = d_2^{p,1}(A_H)(\kappa_2^{p,1}x)=-u \cup \kappa_2^{p,1}x$. Hence, it remains to show $u \cup x = u \cup \kappa_2^{p,1}x$ (note the cup products are w.r.t. different pairings). But this follows immediately by applying $H^\ast(Q;-)$ to the following commutative diagram of pairings: $$\begin{array}{ccc} H_1(H;\mathbb{Z}) \otimes H^1(H;A) \;\; & \xrightarrow{\cap} & A_H \newline \scriptstyle id\otimes \kappa^\ast \displaystyle\downarrow\qquad\quad & & \downarrow\scriptstyle id \newline H_1(H;\mathbb{Z}) \otimes H^1(H;A_H) & \xrightarrow{\cap} & A_H. \end{array}$$

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@Ralph: I fixed a typo by changing an $H$ to a $Q$, I hope it's OK. –  Mark Grant Jun 1 '13 at 12:08
    
Yes, it's OK. Thank you Mark. –  Ralph Jun 3 '13 at 9:49
    
Thanks, Ralph. This seems to be the closest to what I have been hoping for. It's a pity that there is no natural pairing $H^{ab} \times H^1(H, A) \to A$, but alas, this how the things are. –  Dima Sustretov Jun 3 '13 at 10:46

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