homalg – mathlab

A goose chase through homological algebra etc. has led us to start reading McCleary’s A User’s Guide to Spectral Sequences. The book seems like a nice introduction for those that know their way around graduate topology and geometry, but haven’t yet encountered cause to pull out this extra machinery to compute (co)homology. First published by Spivak’s Publish or Perish Press in the 80’s, a second edition was released by Cambridge University Press in 2000. Though it sounds like the second edition is rather improved, there seem to be a number of mistakes remaining which may frustrate those trying to learn a notoriously complicated subject for the first time. Pending an official list of errata, we may as well collect some of them here.

Section 1.1 – Notation

The first comment worth making is regarding some confusing notation, largely an overuse of the letter $E$ . The first use comes on p. 4 (Section 1.1), given a graded vector space $H^*$ and a filtration $F^*$ of $H^*$ , by defining

$E_0^p(H^*) := F^p(H^*)/F^{p+1}(H^*).$

Here, the symbol $E_0^p$ seems to designate an endofunctor on (graded) vector spaces — it eats one and gives back another; transporting morphisms through the filtration and quotient shouldn’t be a problem either. It isn’t really clear what the $0$ subscript is supposed to indicate at this point, but the reader sits tight expecting the truth to be revealed.

However, on the very same page, McCleary twists things by making the assignment

(1) $\begin{equation*} E^{p,q}_0 := F^pH^{p+q} / F^{p+1} H^{p+q} , \end{equation*}$

where $F^pH^r : = F^pH^*\cap H^r$ , the $r\ts{th}$ graded piece of the $p\ts{th}$ filtration. Now, with the extra index $q$ , $E^{p,q}_0$ is a vector space on it’s own. The notation doesn’t indicate reference to $H^*$ , though in this case it really depends on $H^*$ . For instance, McCleary indicates that we should write something like

$E^p_0(H^*)=\bigoplus_q E^{p,q}_0.$

The definition immediately afterwards (Definition 1.1) indicates $E^{p,q}_r$ is to be used to designate a vector space in a spectral sequence which is irrespective of any $H^*$ for all $r\geq 1$ . The typical way to relate a spectral sequence $\{E^{*,*}_r, d_r\}$ , to a graded vector space $H^*$ is the situation of convergence (Definition 1.2, p. 5) where instead

$E_\infty^{p,q} \cong E_0^{p,q}(H^*).$

The right hand side above has nothing to do with the spectral sequence $E^{*,*}_r$ (since we take $r\geq 1$ in our definition), it is just an instance of the definition from equation 1… but with distinct use of notation… oh. So on the one hand, $E^{p,q}_0$ should be a standalone vector space, like the other $E_r^{p,q}$ ‘s, but also it needs to come from an $H^*$ so one should really write $E_0^{p,q}(H^*)$ as in Definition 1.2. Wha? Shoot. Couldn’t we have used like an $A$ instead or something?

Perhaps there is good reasoning for all of this to be discovered once we get further in. Also, it seems so far that initial terms are usually $E_2^{*,*}$ . Why not $E_1^{*,*}$ ? And why don’t we allow $0$ -pages? In these cases the differentials would be vertical and horizontal (resp.) instead of diagonal, which feels less interesting somehow, though this doesn’t seem like it would be totally frivolous… TBD.

Splicing Short Exact Sequences

Finishing out the first section, we address what seems to be a typo in example 1.A (p. 6). McCleary’s expository style consists of many statements which are not obvious, though usually not difficult to work out. This is perhaps for the best, as the community seems to indicate that the only real way to learn spectral sequences (make that: all math?) is by working them out. Nevertheless, it is a bit discouraging to find yourself at odds with the author at the first example…

We have assumed a first quadrant spectral sequence with initial term $E_2^{*,*}$ converging to $H^*$ with a filtration satisfying $F^{p+k}H^p=\{0\}$ for all $k> 0$ . Then we have a filtration on $H^1$ in particular, given by

$\{0\}\sus F^1H^1\sus H^1,$

since, by the assumption, $F^{2}H^1=\{0\}$ etc., and $F^0H^1=H^1$ by definition. By convergence, then,

$E_\infty^{1,0}=F^1H^1/F^2H^2=F^1H^1,$

so $E_\infty^{1,0}$ is a submodule of $H^1$ . But also because $E_r^{1,0}$ lies on the $p$ -axis (depicted as what is usually the $x$ -axis) and our spectral sequence has only first quadrant terms, $d_r(E_r^{1,0})$ must be the zero map for all $r$ . Furthermore, $E_r^{1,0}$ is too close to the $q$ -axis to get hit by any differential $d_r$ , thus $E_2^{1,0}$ survives as the kernel of every $d_r$ , mod the image of a $d_r$ from a zero vector space in the second quadrant. This is all to say

$E_2^{1,0}=E_\infty^{1,0}\cong F^1H^1.$

We then have part of the short exact sequence McCleary gives for $H^1$ as

$0 \lra E_2^{1,0}\lra H^1 \lra H^1/F^1H^1 \lra 0 .$

How can we describe the third term using the spectral sequence? Well, from our definitions, $H^1/F^1H^1=F^0H^1/F^1H^1\cong E_\infty^{0,1}$ . The book seems to be indicating that $E_\infty^{0,1}=E_2^{0,1}$ but this is not necessarily the case! It also doesn’t make sense with how the short exact sequences are spliced later on.

Let’s address the first claim first. Because $E_r^{0,1}$ lies on the $q$ -axis, and the differentials point “southeast” towards the empty fourth quadrant, $d_r(E_r^{0,1})$ is the zero map for any $r\geq 3$ , but it can’t be hit by anything so we have now

$E_\infty^{0,1}=E_3^{0,1} = \frac{\ker d_2 : E_2^{0,1}\to E_2^{2,0} }{\im d_2: E_2^{-2,0}\to E_2^{0,1}}.$

The denominator is the image of a map from a zero vector space, so it is zero, and thus $E_\infty^{1,0}$ is a subspace of $E_2^{0,1}$ , but this latter space can be larger! This is all to say, the short exact sequence for $H^1$ is misprinted, and should go

(2) $\begin{equation*} 0 \lra E_2^{1,0}\lra H^1 \lra E_\infty^{0,1} \lra 0 . \end{equation*}$

One can confirm this by examining the SES given just below, where we see $E_\infty^{0,1}$ injecting into $E_2^{0,1}$ :

(3) $\begin{equation*} 0 \lra E_\infty^{0,1}\lra E_2^{0,1}\overset{d_2}{\lra} E_2^{2,0}\lra E_\infty^{2,0}\lra 0. \end{equation*}$

This is a standard decomposition of the map $d_2$ in the middle: for any morphism $\phi:A\to B$ (in an abelian category at least, we suppose) there is a SES

$0\lra\ker\phi\lra A \overset{\phi}{\lra} B\lra B/\im \phi \lra 0 .$

It remains to see that $E_\infty^{2,0}\cong E_2^{2,0}/\im d_2$ . Because of where $E_r^{2,0}$ sits on the $p$ -axis, it is again the kernel of $d_r$ for all $r$ . Further, it can only possibly be hit by $d_2$ , so in fact $E_3^{2,0}$ survives through all further terms to give the desired equality