The flat dimension of an $R$-module $M$ is the infimum over lengths of flat resolutions of $M$, and the weak dimension (or $\mathrm{Tor}$-dimension) of $R$ is the supremum over all possible flat dimensions of modules. Let's use $\mathrm{w.dim}(R)$ to denote the weak dimension of $R$. As with the global dimension, the weak dimension of $R$ can be computed as the supremum over the set of flat dimensions of the modules $R/I$ for $I$ running over the set of all left-ideals or right-ideals, either is fine!

So, if every ideal is flat, then $\mathrm{w.dim}(R) \leq 1$. What about the converse? If $\mathrm{w.dim}(R) \leq 1$, is it true that every ideal is flat? Let's make a side remark in that if we replace weak dimension with global dimension, and flat with projective, then the answer follows from Schanuel's lemma. However, as far as I know there is no Schanuel's lemma when 'projective' is replaced by 'flat'.

However, we can get away with using part of the proof of Schanuel's lemma. Before continuing, the reader may wish to check out the statement and proof of Schanuel's lemma using a double complex spectral sequence.
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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.
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Last time on Wild Spectral Sequences, we conquered the snake lemma using a spectral sequence argument. This time, we meet a new beast: the five lemma. The objective is the usual: prove the five lemma using spectral sequences.

Recall that the five lemma states that given a diagram


in an abelian category, if the rows are exact and $ a,b,d,e$ are isomorphisms, then so is $ c$. Actually, the hypotheses are too strong. It suffices to have $ b,d$ isomorphisms, $ a$ an epimorphism and $ e$ a monomorphism. One can deduce this via J. Leicht's "strong four lemma" (which we might try and prove via a spectral sequence too) or just by using the regular diagram-chasing proof of the five lemma.
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Posted by Jason Polak on 21. November 2012 · Write a comment · Categories: homological-algebra · Tags: ,

Welcome ladies and gentlemen to a new feature on AZC called Wild Spectral Sequences. This will be a regular feature on spectral sequences for as long as I can find new examples of using spectral sequences. Showing a small, neat, easily digestible example of using spectral sequences will be the aim of each episode.

Now, it is widely believed that spectral sequences are scary and dangerous. Of course this is not true, and in fact spectral sequences are 100% safe for children eight and older! The prerequisites of these posts will be essentially basic knowledge of spectral sequences, such as the kind that can be found in Weibel's book or in many other texts.

Today's feature is the following: prove the snake lemma using spectral sequences. Recall:

Snake Lemma. Given an diagram

in an Abelian category with exact rows, there is a long exact sequence

$ 0\to \ker(f)\to\ker(g)\to\ker(h)\to\mathrm{coker}(f)\to\mathrm{coker}(g)\to\mathrm{coker}(h)\to 0$.

Read on for the solution! (But try it first: it's fun!) More »