The Lyndon-Hochschild(-Serre) spectral sequence applies to group extensions in a manner analogous to the Serre-Leray spectral sequence applied to a fibration.

Does anyone know of a good description (or reference) of the transgression maps in the Lyndon-Hochschild spectral sequence? MacLane describes them in terms of an additive relation, but I don't find this helpful in computing them.

More generally, I don't know how to calculate the differentials in this spectral sequence. In the Serre spectral sequence, I can see how an exact couple arises and the differentials are straightforward to see if not easy to calculate. But the LHSS arises from a double complex and I'm not sure how to get an exact couple from this.


If you only need the first couple of transgressions, there is a nice description of them in the paper

MR0641328 (83a:18021) Huebschmann, Johannes Automorphisms of group extensions and differentials in the Lyndon\mhy Hochschild\mhy Serre spectral sequence. J. Algebra 72 (1981), no. 2, 296--334.

For concrete calculations, this description is sometimes easier than the abstract nonsense description from the double complex.


A double complex is actually one of the nicest situations to describe differentials explicitly.

Say C{p,q} is a double complex (I'll take the convention that this means that the vertical differential commutes with the horizontal differential) and you have an element x on the r'th page of the spectral sequence in bidegree (p,q).

That actually means that there exists a lift, given by a sequence of elements (x0,x1,x2,...,x{r-1}) with xi in C{p-i,q+i} with the following properties:

x0 is a lift your given element x to the double complex, the vertical differential applied to x0 gives zero, and the horizontal differential applied to any xi coincides with the vertical differential applied to x{i+1}.

Then the d_r differential of x is the equivalence class of the horizontal differential applied to the last class x{r-1}. You can show that this is independent of the choice of lift.

In terms of the exact couple, this comes from filtering the double complex C{p,q} by the subcomplexes just consisting of the first few columns.

edit: I'm using homological conventions rather than cohomological conventions; for cohomology everything goes in the opposite direction and you filter by quotient complexes rather than subcomplexes.


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