Differentials in the Lyndon-Hochschild spectral sequence

Question

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.

Answer 1

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.

Answer 2

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 $(x_0,x_1,x_2, \cdots ,x_{r-1})$ with $x_i$ in $C^{p-i,q+i}$ with the following properties:

$x_0$ is a lift your given element $x$ to the double complex, the vertical differential applied to $x_0$ gives zero, and the horizontal differential applied to any $x_i$ 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.

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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.