I am looking for a reference for the following. Say we have a $G$-space $X$ whose homology groups (in field coefficients $k$) are non-zero only in dimension zero and for a fixed $n>0$. Let $M$ denote the top homology group of $X$. Then in the cohomology spectral sequence for the fibration $X\rightarrow (EG\times X)/G\rightarrow BG$ the transgression is multiplication by the $k$-invariant of $M$. The $k$-invariant lives in $Ext^{n+1}_{kG}(k,M)$. I believe the product looks like $Ext^{0}_{kG}(k,Hom(M,k))\times Ext^{n+1}_{kG}(k,M)\rightarrow Ext^{n+1}(k,k)$ induced by the pairing $Hom(M,k)\times M\rightarrow k$.
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1$\begingroup$ Is $G$ a discrete group? What is $M$? Do you mean $k$-invariant as in Postnikov towers? If so then I'm surprised to see $k$ appearing in the Ext expression. I've voted to close the question until these points are clarified. $\endgroup$– Mark GrantCommented Nov 16, 2015 at 8:48
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$\begingroup$ I'm also confused, especially by what "multiplication by" means, as this is a map from H^n(X) to H^(n+1) (BG). But I'm very interested if there is a general description of this transgression map. I suggest that M is meant to be the top cohomology group of X. Also perhaps multiplication means composition product, but then shouldn't it be Ext(M,k) rather than (k,M)? $\endgroup$– David SprehnCommented Nov 16, 2015 at 16:10
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$\begingroup$ I think you are misusing the term $k$-invariant (especially as $k$ is also being used to denote the ground field). It seems to me that the class you are talking about is more like a generalized Euler class (since $X$ is a generalized homology sphere and so you have something like a spherical fibration). $\endgroup$– Mark GrantCommented Nov 17, 2015 at 11:25
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1$\begingroup$ This is explained in Section 7 of "Zur Homotopietheorie der Kenenkomplexe" of Dold but in German unfortunately. $\endgroup$– NickCommented Nov 18, 2015 at 5:05
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2$\begingroup$ k-invariants are by definition integral, which is apparently confusing people. In this case, the k-invariant can perhaps be thought of as an element in $H^{n+1}(BG, H_n(X,M)) = \mathrm{Ext}^{n+1}_{\mathbb{Z}G}(\mathbb{Z}, M)$, where $M=H_n(X,\mathbb{Z})$. This projects under $\mathbb{Z}\to k$ to an element of $\mathrm{Ext}^{n+1}_{kG}(k,M\otimes k)$; it is this image that appears in your proposed formula for the transgression. $\endgroup$– Charles RezkCommented Nov 18, 2015 at 20:31
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I think the difficulty is that you are assuming that $X$ only has homology in two degrees, but are then looking at the cohomology spectral sequence. (To get sensible answers I seem to have to take cohomology to be negatively graded.)
Suppose instead that $H^0(X;k)=k$ and $H^{-n}(X;k)=N$ are the only two non-trivial cohomology groups. The inclusion of the constant $0$-cochains gives a map $k \to C^*(X;k)$ whose mapping cone has homology $N[-n]$, so there is an exact triangle $$N[-n-1] \overset{\Sigma^{-n-1} \kappa}\to k \to C^*(X;k)$$ in the derived category of complexes of $kG$-modules, defining the $k$-invariant $\kappa \in Ext_{kG}^{n+1}(N,k)$.
Then $C^*((EG \times X)/G;k) \cong (C^*(EG;k) \otimes C^*(X;k))^G$, and $C^*(EG;k)$ is a complex of finitely generated free $kG$-modules so $(C^*(EG;k) \otimes -)^G$ is exact, hence there is an exact triangle $$(C^*(EG;k) \otimes N[-n-1])^G \to C^*(G;k) \to C^*((EG \times X)/G;k).$$ in the derived category of $k$-modules. The associated long exact sequence $$H^*(G;n) \to H^*((EG \times X)/G;k) \to H^{*+n}(G;N) \overset{d}\to H^{*-1}(G;k)$$ is then the cohomology spectral sequence, and the connecting map $d$ is now manifestly given by product with $\kappa$.
answered Nov 19, 2015 at 0:08
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$\begingroup$ Thank you for the answer. I think also we have $Ext^{n+1}_{kG}(N,k)\cong Ext^{n+1}_{kG}(k,N^*\otimes k)\cong Ext^{n+1}_{kG}(k,M)$ where $M$ as in the question. $\endgroup$– NickCommented Nov 19, 2015 at 19:47
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$\begingroup$ You're right: I had overlooked that $k$ was a field. $\endgroup$ Commented Nov 19, 2015 at 23:32