Spectral Sequences I

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

    \[E_\infty^{2,0}=E_3^{2,0} = E_2^{2,0}/\im d_2.\]

To splice all this together, we recall that we can connect

    \[\dots\oset{s}{\lra} L\overset{\alpha}{\lra} M\lra 0 \quad \text{and} \quad 0 \lra M \overset{\beta}{\lra} N \oset{t}{\lra} \dots \]

as

    \[\dots \oset{s}{\lra} L \overset{\gamma}{\lra}  N \oset{t}{\lra} \dots \]

where \gamma=\beta\circ\alpha. We maintain exactness since \ker \gamma=\ker \alpha=\im s and \ker t = \im \beta = \im \gamma.

Performing this surgery on sequences 2 and 3 yields the main exact sequence claimed by the example, namely

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

Stay tuned for more clarifications from Chapter 1.

Leave a Reply

Your email address will not be published.