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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).

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.

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?

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Is this the Lyndon-Hochschild-Serre spectral sequence of the extension $A\to B\to C$ ? –  Ralph Apr 18 '12 at 19:23
    
Have you looked at McCleary's User's Guide? –  jd.r Apr 18 '12 at 19:26
    
@Ralph, yes, it is. However, the name "Lyndon-Hochschild-Serre spectral sequence" was not explicitly mentioned in Hatcher's book. –  Zuriel Apr 18 '12 at 19:32
    
@Josh, it seems to be quite a big book. May I know which page/section should I read for this question? –  Zuriel Apr 18 '12 at 19:38
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The Lyndon-Hochschild-Serre SS is associated to a group extension $A /to B \to C$. The Serre-Leray (or Serre) SS is associated to any fibration $F \to E \to B$. @Zuriel Try Section 6.2 in McCleary's book for the trangression of the SLSS and chapter 8b for the LHSSS. Also, you might find the answers to mathoverflow.net/questions/590/… useful. –  jd.r Apr 19 '12 at 1:40
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1 Answer 1

up vote 8 down vote accepted

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:

$\hspace{120pt}d_2^{0,1}(f) = f_\ast(\varepsilon)$

A good reference for this is Theorem 2.1.8 in Neukirch et. al.: Cohomology of Number Fields.

In case of $M=\mathbb{F}_p$, Kudo's transgression theorem may also be of relevance.

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.

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That's interesting. Probably that's still the case if the action of $G$ on $M$ factors through the abelianization? For general $M$, though, I'm hoping someone is about to describe $d^{01}_2$ using the bar complex. –  Graham Denham Apr 18 '12 at 22:39
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@Graham: Such a description is given in Prop. 1.6.5 of the Neukirch book. But I don't believe it's of much help in practice. –  Ralph Apr 18 '12 at 23:39
    
Nice answer, +1. Small notational quibble: should $f^\ast$ be $f_\ast$, since it's covariant? –  Mark Grant Apr 19 '12 at 7:22
    
@Mark: Yes, it seems that $f_\ast$ is more in use, for example in MacLane's homology book. (I tend to use basically superscripts for induced maps in cohomology (either in 1st or 2nd argument) and subscripts for homology). –  Ralph Apr 19 '12 at 10:18
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