Wild Spectral Sequences Ep. 3: Cohomological Dimension

Last time in Wild Spectral Squences 2, we saw how to prove the five lemma using a spectral sequence. Today, we'll see a very simple application of spectral sequences to the concept of cohomological dimension in group cohomology.

We will define the well-known concept of cohomological dimension of a group $ G$, and then show how the dimension of $ G$ relates to the dimension of $ G/N$ and $ N$ for a normal subgroup $ N$. We do this with a spectral sequence. Although this application will appear to be very simple, it might be a good exercise for those just learning about spectral sequences.

The Category

For the sake of conreteness, let us work in the category of $ G$-modules where $ G$ is a profinite group. The $ G$-modules are the $ \mathbb{Z}G$-modules $ A$ with a continuous action of $ G$ where $ A$ is given the discrete topology, but we could be working with any group $ G$ with suitable, minor modifications, or Lie algebras, etc.

Like in all homology theories, there is the notion of cohomological dimension: a profinite group $ G$ has cohomological dimension $ n\in \mathbb{N}$ if for every $ r > n$ and every torsion $ G$-module $ A$, the group $ H^r(G,A)$ is trivial. If no such $ n$ exists, we say that $ G$ has cohomological dimension $ \infty$. The usual arithmetic rules in working with $ \infty$ apply.

(If we drop "torsion", we get the notion of strict cohomolgical dimension, so the terminology here is unfortunate, but it has stuck through history.)

We denote the cohomological dimension of $ G$ by $ \mathrm{cd}(G)$.

Let $ N$ be a normal subgroup of $ G$. Of course, we would like to compare the three numbers $ \mathrm{cd}(G)$, $ \mathrm{cd}(G/N)$ and $ \mathrm{cd}(N)$. In general, they will not be equal, but at the least we would like $ \mathrm{cd}(G) \leq \mathrm{cd}(G/N) + \mathrm{cd}(N)$. Under certain conditions we will get equality, but the purpose of today is just to prove the inequality, and this is something we can do with a spectral sequence.

The Spectral Sequence

We use the Hochschild-Serre spectral sequence for the group $ G$, the normal subgroup $ N\leq G$, and $ G$-module $ A$:

$ E_2^{p,q} = H^p(G/N,H^q(N,A)) \Rightarrow H^{p+q}(G,A)$.

This is a special case of the Grothendieck spectral sequence for the composition of functors $ -^G = ((-)^N)^{G/N}$. Now let $ m = \mathrm{cd}(G/N)$ and $ n = \mathrm{cd}(N)$. Notice that if $ p > m$ then $ E_2^{p,q} = 0$ and if $ q > n$ then $ E_2^{p,q} = 0$ as well! However, even if $ p > m$, this does not imply that $ H^p(G,A) = 0$, since there for a fixed $ r$ there are many ways to write $ r = p + q$.

We recall that the $ H^{r}(G,A)$ piece will have a canonical filtration, whose subquotients are isomoprhic to the $ E_2^{p,q}$ terms for all $ p + q = r$. Hence in order to conclude $ H^r(G,A) = 0$, we need $ E_2$ to vanish for all possible $ p$ and $ q$ such that $ r = p + q$.

In other words, for such an $ r$ and every $ p$ and $ q$ such that $ r = p + q$, either $ p > m$ or $ q > n$. If $ r > m + n$, then certainly at least one of $ p > m$ or $ q > n$ must be true for every $ r = p + q$. If $ r \leq m + n$ then we cannot guarantee this without further information.

Thus the best we can do without any further information is that $ r > m + n$. In other words, we known that $ H^r(G,A) = 0$ for every torsion $ A$ as long as $ r > m + n$, which by definition shows that

$ \mathrm{cd}(G)\leq \mathrm{cd}(N) + \mathrm{cd}(G/N)$.

Exercise. Provide a proof of this without using spectral sequences.

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