Posted by Jason Polak on 21. April 2016 · 1 comment · Categories: math

I was looking through the past episodes of Wild Spectral Sequences, where we did the snake lemma, five lemma, cohomological dimension formula, Schanuel’s lemma (my favourite), and an application of the LHS sequence to calculating H^1 of tori. However, I’m surprised we never did the 3×3 lemma:

Theorem.If we have a 3×3 diagram:


where all the rows are exact, and two out of the three columns are exact, then all three columns are exact.

Even though the proof in of the 3×3 lemma is not much different than the other proofs in this series, there are at least three cool things about it:

  1. The spectral sequence argument easily allows you to remember the 3×3 lemma.
  2. The spectral sequence argument shows that there’s nothing special about the number 3 in the 3×3 lemma: the argument gives you the corresponding nxn lemma.
  3. If you’re learning spectral sequences, it’s good to have repitition!

Double Complexes

Recall that a double complex is a two dimensional array of objects $C_{p,q}$ together with two maps: $d^h:C_{p,q}\to C_{p-1,q}$ and $d^v:C_{p,q}\to C_{p,q-1}$ that satisfy $d^hd^h = 0$ and $d^vd^v = 0$, and are required to anticommute: $d^hd^v + d^vd^h = 0$. The reason for this latter anticommutativity condition is so that the corresponding total sum complex, whose $n$th degree term is $\oplus_{p+q = n} C_{p,q}$ with the obvious differentials, is actually a chain complex. Having anticommutativity in the squares is a lot easier to do than having commutativity and then describing the differential.

If one has a double complex $C$, then one can take in spot $p,q$ the horizontal homology or vertical homology. Applying vertical and the horizontal gives the spectral sequence $E^2_{p,q} = H_p^hH_q^v(C)\Rightarrow H_{p+q}({\rm Tot}(C))$. The other way around gives the spectral sequence $E^2_{p,q} = H_p^vH_q^h({\rm Tot}(C))$. One thing to be careful here is that in the first spectral sequence, the $E^2_{p,q}$ term is the vertical-then-horizontal homology at the $p,q$-spot. However, in the second spectral sequence, the $E^2_{p,q}$ is the horizontal-then-vertical homology in the $q,p$-spot.

The logic for the notation is that to derive these spectral sequences, we are using the spectral sequence of a filtration, and if we use the index $p$ for the $p$th filtrant, then if we filter by rows first, $p$ becomes the second variable corresponding to the “y-axis”, since changing the “y-variable” changes the row.

Back to the 3×3 Lemma

Now, the careful reader should note that we’re trying to prove something about the 3×3 diagram, which is actually not a double complex. Of course, we’re just trying to prove something about the exactness of certain columns; therefore, it’s harmless to replace the 3×3 diagram with a double complex by multiplying the odd-numbered row differentials by $-1$, which won’t affect exactness. So let’s just assume we’ve done this.

So, in our 3×3 lemma, we know that one of the assumptions is that all rows are exact. Hence, in the spectral sequence where we take $H^h$ first, we know that all the $E^2_{p,q}$ terms, and hence all the $E^\infty_{p,q}$ terms, are zero. Thus, the total complex is acyclic.

Now, we look at the other spectral sequence, the one that starts out by taking vertical homology first. If we do this, it’s clear that an assumption like “all the columns but one are exact”; in this case, taking horizontal homology after taking vertical won’t make any difference, because all the other columns have zero vertical homology! Hence, under this additional assumption, the $E^2_*$ terms in this spectral sequence are just obtained by taking vertical homology.

So we’ve assumed that all but one “special” column has zero homology. In the corresponding column of the spectral sequence, these $E^2$ terms are the $E^\infty$ terms, all of which correspond to a filtration of the homology of the total complex, which is zero! Therefore, all the $E^2$-terms in this “special” column are zero, and thus the vertical homology of the special column is also zero: in other words, all columns are exact!

You’ll notice that we didn’t use anything special about the size of the diagram. In other words, we have:

Theorem. Suppose we have an $n\times n$ diagram, all of whose rows are acyclic. If all but one of the columns are acyclic, then all the columns are acyclic.

Although I’ve taken a long time to explain the proof, once you’re more familiar with spectral sequences, the proof should be encapsulated into: the assumption on horizontal rows shows that the total complex is exact; every column but one being acyclic implies the homology groups of the remaining column are sucessive quotients in a necessarily zero filtration of the homology of the total complex; hence the vertical homology of the “special” row is also acyclic.

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