# How to Compute Transgressions in a Serre Spectral Sequence?

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?

• Is this the Lyndon-Hochschild-Serre spectral sequence of the extension $A\to B\to C$ ? Apr 18, 2012 at 19:23
• Have you looked at McCleary's User's Guide?
– Josh
Apr 18, 2012 at 19:26
• @Ralph, yes, it is. However, the name "Lyndon-Hochschild-Serre spectral sequence" was not explicitly mentioned in Hatcher's book. Apr 18, 2012 at 19:32
• @Josh, it seems to be quite a big book. May I know which page/section should I read for this question? Apr 18, 2012 at 19:38
• 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.
– Josh
Apr 19, 2012 at 1:40

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.

• 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. Apr 18, 2012 at 22:39
• @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. Apr 18, 2012 at 23:39
• Nice answer, +1. Small notational quibble: should $f^\ast$ be $f_\ast$, since it's covariant? Apr 19, 2012 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). Apr 19, 2012 at 10:18