How to Compute Transgressions in a Serre Spectral Sequence? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:39:27Z http://mathoverflow.net/feeds/question/94424 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94424/how-to-compute-transgressions-in-a-serre-spectral-sequence How to Compute Transgressions in a Serre Spectral Sequence? Zuriel 2012-04-18T16:23:32Z 2012-04-19T10:19:24Z <p>For a short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$ there is an associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, which can be constructed by realizing the homomorphism $B\rightarrow C$ by a map $K(B,1)\rightarrow K(C,1)$ and the convert it into a fibration. The fiber is $K(A,1)$ (from the associated long exact sequence of homotopy groups). </p> <p>For a fibration $F\rightarrow X\rightarrow B$, the differential $d_n\colon E_{n,0}^n\to E_{0,n-1}^n$ in the Serre spectral sequence was shown to be equal to the transgression in Hatcher's book on Spectral Sequences (Proposition 1.13). The transgression was defined using (relative) homology groups.</p> <p>My questions is: From the short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$, is there any method to directly compute the transgression of the associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, at least for the case $n=2$, without constructing $K(G,1)$'s and considering their homologies?</p> http://mathoverflow.net/questions/94424/how-to-compute-transgressions-in-a-serre-spectral-sequence/94463#94463 Answer by Ralph for How to Compute Transgressions in a Serre Spectral Sequence? Ralph 2012-04-18T21:11:17Z 2012-04-19T10:19:24Z <p>I can give a description in case of cohomology: Let $$1 \to H \to G \to G/H \to 1$$ be an extension of groups. Then we obtain an extension with abelian kernel $$1 \to H_{ab} \to G/H' \to G/H \to 1$$ Let $\varepsilon \in H^2(G/H;H_{ab})$ be its extension class. If $M$ is a trivial $G$-module, then the differential (which equals the transgression) $$d_2^{0,1}: E_2^{0,1}=H^1(H;M)^{G/H} \to H^2(G/H;M) = E_2^{2,0}$$ is given as follows: Let $f \in H^1(H;M)^{G/H} \le Hom(H,M)=Hom(H_{ab},M)$. Since $f$ is $G/H$-invariant, we have a hom. of $G/H$-modules $f:H_{ab}\to M$ and an induced hom. $f_\ast: H^2(G/H;H_{ab}) \to H^2(G/H;M)$. Then: </p> <blockquote> <p>$\hspace{120pt}d_2^{0,1}(f) = f_\ast(\varepsilon)$ </p> </blockquote> <p>A good reference for this is Theorem 2.1.8 in Neukirch et. al.: Cohomology of Number Fields. </p> <p>In case of $M=\mathbb{F}_p$, Kudo's transgression theorem may also be of relevance. </p> <p>In case of homology you can try to dualize the result above. But I personally would always prefer to use cohomology, since here the cup product is available that is very helpful in computing spectral sequences. </p>