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!)

The Chase?

Doesn’t everyone love diagram chasing? Proving that certain sequences are exact always involves some kind of diagram chasing, doesn’t it? We can certainly prove the snake lemma using diagram chasing. All the maps in the 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$

are the obvious ones, except for $ \ker(h)\to\mathrm{coker}(f)$, which just requires a little trick to produce. The reader is encouraged to try proving the exactness of this sequence using a diagram chase.

Well, much diagram chasing is encoded in the formalism of spectral sequences. Hence, sometimes we can reuse all that diagram chasing we had to do in learning about spectral sequences to avoid chasing elements. So let’s try it! Of course, since typing diagrams is a bit tricky without the great xy-pic, I shall just explain the gist of the proof, and the reader should be able to fill in the unwritten details, all of which are straightforward calulculations.

Proof of Snake Lemma. To our diagram $ D$, we have two spectral sequences

$ ~^IE_{pq}^2 = H_p^h(H_q^v(D)) \Rightarrow H_{p+q}(\mathrm{Tot},(D))$
$ ~^{II}E_{pq}^2 = H_p^v(H_q^h(D)) \Rightarrow H_{p+q}(\mathrm{Tot}(D))$.

The notation $ H_p^h(H_q^v(D))$ means taking the vertical homology and then the horizontal homology to get the $ E^2$ terms. Now, since the rows are exact, the horizontal homology vanishes, and hence the $ E^2$-terms in the second spectral sequence are all zero, so that the homology of the total complex also vanishes everywhere.

This means that the $ E^2$ terms in the first spectral sequence converge to zero, so that the $ E^\infty$ terms all vanish. This gives us the two exact sequences:

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

All we have to do is patch them up somehow. Of course, as in the diagram-chase proof, this part is the tricky part, but it really is actually quite straightforward. Like we said, the $ E^\infty$ terms in the first spectral sequence all vanish. Thus from the $ E^2$ differential, we get an acyclic complex (exact sequence)

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

where the quotients are given via the maps in the diagram giving the exact sequences. Thus the nontrivial map is an isomorphism. If we look at the inverse, then we get that $ \ker(h)$ maps onto the kernel of the map $ \mathrm{coker}(f)\to\mathrm{coker}(g)$, and its kernel is $ \ker(g)$, which is exactly what we need to paste the two exact sequences together! QED

Do you still want to see a proof of the snake lemma the old-fashioned way? Check out the It’s My Turn movie clip on YouTube!

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