12
$\begingroup$
  1. Is there an analogue of the Lyndon–Hochschild–Serre spectral sequence for a non-normal subgroup?
  2. What can you say about it? Can you describe $E^{p, q}_1$ ? What is about $E^{p, q}_2$?
  3. What is the best technique to get the spectral sequence? For me the Grothendieck spectral sequence us much better than the spectral sequence of a filtered complex.

There is a parallel question which is likely easier.

  1. Is there an analogue of the Hochschild–Serre spectral sequence for a Lie subalgebra which is not an ideal?

2 and 3 remain the same.

I already asked a version of this question on MathSE but got no responses.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ The main difficulty is that, if $H < G$, there is not really a functor that will take the $H$-fixed points and produce the $G$-fixed points unless you include extra data. There are methods to get the cohomology of $G$, but almost all of them will require as extra input the cohomology of intersections of conjugates of $H$. $\endgroup$
    – Tyler Lawson
    Commented Feb 1, 2015 at 1:22
  • $\begingroup$ It is very interesting. Can you provide references? $\endgroup$
    – quinque
    Commented Feb 1, 2015 at 7:56
  • 4
    $\begingroup$ The shortest way to say it is the following. For a group $G$, the category of $G$-modules is equivalent to the category of quasicoherent (etale) sheaves on the classifying stack $BG$, and the global section functor is the fixed-point functor. There is a faithfully flat cover $BH \to BG$, and so there is a Cech-to-derived / descent spectral sequence. But to compute effectively with it, you need to know the iterated fiber products of $BH$ over $BG$, which correspond to $G$-orbits in $(G/H)^k$. $\endgroup$
    – Tyler Lawson
    Commented Feb 1, 2015 at 17:24
  • 1
    $\begingroup$ From the point of view of filtered complexes, this is a little easier to do because there is a nonprojective resolution of $\Bbb Z$ by a complex with terms $\Bbb Z[(G/H)^k]$, with the same boundary operator as on homogeneous chains. You can apply $\Bbb RHom_G(-,M)$ to this resolution and get a filtered complex, and this gives you the spectral sequence too. $\endgroup$
    – Tyler Lawson
    Commented Feb 1, 2015 at 17:27
  • $\begingroup$ (In answer to your explicit question, no, I do not know an immediate reference for this. I do know that this technique, from the topological point of view, appears in calculations of homological stability (I think it's used in Quillen's calculations for number fields). The chain complex I just described is the simplicial chain complex of a "classifying space for the family of subgroups of $H$".) $\endgroup$
    – Tyler Lawson
    Commented Feb 1, 2015 at 17:30

3 Answers 3

Reset to default
8
$\begingroup$

Sorry for reviving an old question, but it seems that the Kropholler spectral sequence exactly answers the first 3 questions:

Kropholler, P.H., A generalization of the Lyndon-Hochschild-Serre spectral sequence with applications to group cohomology and decompositions of groups., J. Group Theory 9, No. 1, 1-25 (2006). ZBL1115.20042.

Improve this answer
$\endgroup$
4
$\begingroup$

I don't think so. The LHS spectral sequence can be thought of as the Serre spectral sequence associated to the fiber sequence

$$BN \to BG \to B(G/N)$$

where $G$ is a group and $N$ is a normal subgroup of it. If $N$ is not required to be normal then the third term in this fiber sequence no longer exists, so it's unclear to me in what sense we can have a reasonable analogue of the LHS spectral sequence here.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ What you said is just that this approach does not work. But I mean something different. I even edited my question, wrote "analog of Lyndon–Hochschild–Serre spectral sequence". $\endgroup$
    – quinque
    Commented Jan 31, 2015 at 21:53
  • $\begingroup$ math.ru.nl/~solleveld/scrip.pdf Here you can find a definition of relative Lie algebra cohomology. It is done by means of explicit complex but it is still a way to make sense of such kind of objects. $\endgroup$
    – quinque
    Commented Jan 31, 2015 at 21:57
  • $\begingroup$ @quinque: that notion of relative Lie algebra cohomology is a little different. If you think of the cohomology of Lie algebras as an algebraic model of the de Rham cohomology of compact Lie groups $G$, then relative Lie algebra cohomology should be an algebraic model of the de Rham cohomology of homogeneous spaces $G/H$. Of course these can make sense if $H$ is not normal, but for group cohomology we want to compute the cohomology of the delooping of these spaces, and we just can't deloop homogeneous spaces in general. $\endgroup$
    – Qiaochu Yuan
    Commented Feb 1, 2015 at 0:50
  • $\begingroup$ Yes, relative cohomology are cohomology of homogenious space! And moreover there is a spectral sequence of a bundle. But I want to get this sequence for arbitrary Lie algebra. Here have to be purely algebraic approach for this. $\endgroup$
    – quinque
    Commented Feb 1, 2015 at 8:21
1
$\begingroup$

For Lie algebras $\mathfrak h \subseteq \mathfrak g$ and a $\mathfrak g$-module $M$ there is a spectral sequence arising from the following filtration $F_i C^q(\mathfrak g, M)$ on the Chevalley-Eilenberg complex $C^q(\mathfrak g,M)$. The filtration is defined by

$$ F_iC^q(\mathfrak g, M) = \{ \gamma \mid i_{h_1}i_{h_2}\cdots i_{h_{q+1-i}}\gamma = 0 \text{ for all }h_i \in \mathfrak h\},$$ where $i_{h}$ means contraction by $h$ (putting $h$ into one argument).

The $E_1$-page may be identified as $$ E_1^{j,i} = H^i(\mathfrak h, Hom(\wedge^j \mathfrak g/\mathfrak h, M)).$$ When $\mathfrak h$ is an ideal, all of this coincides with the Grothendieck spectral sequence for the composition of $(-)^{\mathfrak h}$ and $(-)^{\mathfrak g/\mathfrak h}$.

Reference: Hochschild, G., & Serre, J. P. (1953). Cohomology of Lie algebras. Annals of Mathematics, 591-603.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .