Spectral Sequences III

July 9, 2018 by aram Leave a comment

Bigraded Algebrae

McCleary introduces the concept of a differential graded algebra in section 1.3 (Definition 1.6, p. 11). These are algebras (over a field $k$ ), which tend to be $\N$ -graded, and importantly carry with them a map $d$ called a differential which is $k$ -linear, shifts the degree of elements (in the grading) up by one, and satisfies a “Leibniz rule:”

$d(a\cdot a')=d(a)\cdot a'+(-1)^{\deg(a)} a\cdot d(a')$

for $a, a'$ in our algebra $A^*$ . This is a twisted version of what is usually called Leibniz’ rule in calculus (which is basically just product rule), which coincides with how the differential works in the algebra of differential forms.

This idea is easily extended to the notion of a differential bigraded algebra $(E^{*,*}, d)$ , where now the elements are $\N^2$ graded (for the time being, later we’ll have $\Z^2$ ), but $d$ remains a total-degree 1 mapping. That is,

$d: \bigoplus_{p+q=n} E^{p,q} \lra \bigoplus_{r+s=n+1}E^{r,s},$

and $d$ still satisfies the Leibniz rule

(1) $\begin{equation*} d(e\cdot e')=d(e)\cdot e'+(-1)^{p+q}e\cdot d(e') \end{equation*}$

where $e\in E^{p,q}$ .

A standard construction is to form a bigraded algebra by tensoring two graded algebras together. This would work with just component-wise multiplication, but to get a working differential that satisfies our version of the Leibniz rule 1 as well, we introduce an extra sign: we mean, supposing $(A^*,d)$ and $(B^*, d')$ are differential graded algebras, then we can assign $E^{p,q}:=A^p\ox B^q$ , and furthermore

(2) $\begin{equation*} (a_1\ox b_1 )\cdot (a_2\ox b_2):= (-1)^{(\deg a_2)(\deg b_1)}a_1a_2\ox b_1 b_2.\end{equation*}$

Then if we define a differential $d_\ox$ on $E^{*,*}$ by

(3) $\begin{equation*} d_\ox(a\ox b)=d(a)\ox b + (-1)^{\deg a} a \ox d'(b), \end{equation*}$

then $d_\ox$ satisfies the Leibniz rule 1. It is clarifying to check this, so we’ll record it here. Switching notation a bit, we will write $(-1)^{\ab{a}}$ instead of $(-1)^{\deg a}$ . To satisfy 1 we need

$\begin{align*}d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2))& = d_\ox (a_1\ox b_1)\cdot (a_2 \ox b_2) + \\ &(-1)^{\ab{a_1}+\ab{b_1}} (a_1\ox b_1) \cdot d_\ox(a_2\ox b_2) \end{align*}$

we then apply 3 to the individual terms on the right side above to get

$\begin{align*} d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2)) = [d(a_1)\ox b_1+((-1)^{\ab{a_1}} a_1 \ox d'(b_1)]\cdot (a_2\ox b_2)+\\ (-1)^{\ab{a_1}+\ab{b_1}} (a_1\ox b_1) \cdot [d(a_2)\ox b_2+(-1)^{\ab{a_2}} a_2\ox d'(b_2)]. \end{align*}$

Now applying the multiplication rule 2 and distributing, we find

(4) $\begin{align*} d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2)) = (-1)^{\ab{a_2}\ab{b_1}}d(a_1)a_2\ox b_1b_2 +(-1)^{\ab{a_1}+\\\ab{d'(b_1)}\ab{a_2}}a_1a_2\ox d'(b_1)b_2 + (-1)^{\ab{a_1}+\ab{b_1}+\ab{d(a_2)}\ab{b_1}} a_1 d(a_2)\ox b_1b_2 +\\ (-1)^{\ab{a_1}+\ab{a_2}+\ab{b_1} +\ab{b_1}\ab{a_2}} a_1a_2 \ox b_1d'(b_2). \end{align*}$

To check the rule holds, we perform this computation by instead multiplying first and then applying the differential. That calculation looks like

$\begin{align*} d_\ox((a_1\ox b_1)\cdot (a_2 \ox b_2)) &= d_\ox((-1)^{\ab{a_2}\ab{b_1}} a_1 a_2 \ox b_1b_2) \\ &=(-1)^{\ab{a_2}\ab{b_1}}[d(a_1a_2)\ox b_1b_2+ (-1)^{\ab{a_1}+\ab{a_2}}a_1a_2\ox d'(b_1b_2)] \\ &=(-1)^{\ab{a_2}\ab{b_1}}[(d(a_1)a_2+(-1)^{\ab{a_1}}a_1d(a_2)\ox b_1b_2 +\\ &(-1)^{\ab{a_1}+\ab{a_2}}a_1a_2\ox(d'(b_1)b_2+(-1)^{\ab{b_1}}b_1d'(b_2))] \qquad \\ &=(-1)^{\ab{a_2}\ab{b_1}}d(a_1)a_2\ox b_1b_2+(-1)^{\ab{a_1}+\ab{a_2}\ab{b_1}} a_1d(a_2)\ox b_1b_2+ \\ &(-1)^{\ab{a_1}+\ab{a_2}+\ab{a_2}\ab{b_1}} a_1 a_2\ox d'(b_1) b_2 +\qquad\\ & (-1)^{\ab{a_1}+\ab{a_2}+\ab{b_1} +\ab{a_2}\ab{b_1}}a_1a_2\ox b_1d'(b_2). \end{align*}$

Finally, remarking that $\ab{d'(b_1)}=\ab{b_1}+1$ and $\ab{d(a_2)}+1=\ab{a_2}$ shows that terms of the last line above match with those of 4, so everything checks out and $(A^*\ox B^*, d_\ox)$ becomes a differential bigraded algebra.

A Chain Rule

Given the length and detail of section 1.3, surprisingly we find no glaring errors in this section, but the use of the differential becomes somewhat muddled in calculation in section 1.4. Again, perhaps as an undesirable side effect of the fact that we remain at the “informal stage,” it’s always difficult to keep track of what assumptions we’re working with in each example. Case in point, example 1.H, p. 20. The paragraph preceding definition 1.11 seems to indicate that all graded algebras are assumed to be graded commutative — at least for the rest of the section, one guesses, though the language is vague. Let’s try this here with a bit more force.

Assumption: All graded algebras are graded commutative for the rest of the post. This is to say, for all $x,y$ in any $A^*$ , we have $x\cdot y =(-1)^{\ab{x}\ab{y}} y\cdot x$ . Now let’s have a look at the example. We suppose a spectral sequence of algebras $(E_r^{*,*}, d_r)$ with $E_2^{*,*}\cong V^*\ox W^*$ , converging to the graded algebra $H^*$ which is $\Q$ in degree 0 and $\{0\}$ in all others. The example asserts that if $V^*$ is a graded commutative polynomial algebra in one generator/variable, then $W^*$ is a graded commutative exterior algebra in one generator, and vice versa.

The first confusion appears in a restatement of the Leibniz rule near the bottom of page 20, except this time there are tensors involved. This appears to be a mixed use/abuse of notation, which was slightly different in the first edition of the book, but not more consistent. The idea is as follows. $V*$ and $W^*$ embed into $V^*\ox W^*$ under the maps $v \mapsto v \ox 1$ and $w \mapsto 1\ox w$ . Then one can also write an element $w\ox z \in V^*\ox W^*$ (mind the inexplicable inconsistent choice of letters) as

(5) $\begin{equation*} w \ox z = (-1)^0 (w\ox 1)\cdot (1 \ox z)= (w\ox 1) \cdot (1 \ox z) \end{equation*}$

since the degree of 1 is zero in each graded algebra. Note that this also allows us to regard $V^*\ox W^*$ as graded commutative with the tensor product as multiplication between pure $V^*$ and pure $W^*$ elements, writing

$z\ox w:=(1\ox z) \cdot (w \ox 1)=(-1)^{\ab{z}\ab{w}} (w\ox z).$

One can apply Leibniz rule to the product in 5 so that if $V^*\ox W^*$ comes with a differential $d$ , we get

$d(w\ox z) = d((w\ox 1) (1\ox z)) =d(w\ox 1)(1\ox z)+(-1)^{\ab{w}}(w\ox 1) d(1\ox z).$

The thing is we really need not write the tensor product $w\ox 1$ ; it is just as correct to write $w$ on it’s own, as we often do with polynomial algebras and so on. Then the above can be written instead as

$d(w\ox z) = d(w)\ox z +(-1)^{\ab{w}} w \ox d(z)$

as McCleary does near the bottom of page 20. What makes this confusing is that up to this point we had only seen differentials acting on tensors by defining the bigraded differential from tensoring two differential graded algebras together, seen above. In this context, the differential of the bigraded algebra must act on an element of the algebra coming from $V^* \ox W$ , it cannot act on just one side of the tensor. What’s different here is that the tensor product is actually the multiplication operation on each page of the spectral sequence. Thus, the restatement of the familiar rule with new notation.

Nevertheless, the next equality is also a bit confounding at first, partly because McCleary, goes back to writing the extra $1$ in the tensor, suggesting that we need to pay attention to its effect. He says that if $d_i(1\ox u) =\sum_j v_j\ox w_j$ , then

(6) $\begin{equation*} d_i(1\ox u^k) = k \lt(\sum_j v_j \ox ( w_ju^{k-1})\rt) \end{equation*}$

which looks sort of reasonable as it resembles something like a chain rule, $d(u^k)=k u^{k-1} d(u)$ . It is presented as if it should follow immediately from the Leibniz rule stated before. But this seems weird when the degree of $u$ is odd. To be totally transparent about this, let’s illustrate the case where $k=2$ , suppressing the subscript on the differential again, but maintaining the tensorial notation.

$\begin{align*} d(1\ox u^2) & =d((1\ox u)(1\ox u)) \\ & = d(1\ox u) (1\ox u) +(-1)^{\ab{u}} (1\ox u) d (1\ox u) \\ &= \sum_j (v_j \ox w_j)(1\ox u) + (-1)^{\ab{u}} \sum_j (1\ox u) (v_j \ox w_j) \\ &=\sum_j v_j\ox w_j u + (-1)^{\ab{u}} \sum_j (-1)^{\ab{u}\ab{v_j}} v_j \ox u w_j \\ &=\sum_j v_j\ox w_j u + (-1)^{\ab{u}} \sum_j (-1)^{\ab{u}\ab{v_j}+\ab{u} \ab{w_j}} v_j \ox w_j u \\ &=\sum_j v_j\ox w_j u + (-1)^{\ab{u}} \sum_j v_j \ox w_j u \\ \end{align*}$

where the last line follows since $v_j\ox w_j$ has total degree $\ab{u}+1$ , so the sign inside the sum there has exponent $\ab{u}(\ab{u}+1)$ which is even. We see that if $u$ has odd degree then, these terms cancel and we get 0. So you say “wait a minute, that’s not right, I wan’t my chain rule looking thing” until you eventually realize that if $u$ has odd degree, since it’s sitting in a graded commutative algebra, $u^2$ is actually zero! And the same goes for all higher powers of $u$ . Then, $d(u^2)=d(0)=0$ makes complete sense. Meanwhile, if $u$ has even degree, the terms will pile up with positive sign and we get the chain rule looking thing that was claimed. So the statement 6 is in fact true, though it really breaks down into two distinct cases.

Going forward in the example, McCleary only really seems to use the chain rule (liberally mixing in the described sort of abuse of notation) on terms of even degree, so it’s tempting to think that it only applies there, but it is sort of “vacuously true” in odd degree as well. Oh well. Onwards.

Spectral Sequences II

June 25, 2018 by aram Leave a comment

Two Stripes

The next thing to address in McCleary is an apparent mistake on p. 9 of section 1.2. Here we again assume a first quadrant spectral sequence converging to a graded vector space $H^*$ . This is mentioned at the beginning of the section, but it’s easy to forget that when a bold-faced and titled example (1.D) seems to be presenting a reset of assumptions, rather than building upon prior discussion. Furthermore, in this example, McCleary seems to be working again with the assumption from example 1.A that $F^p+kH^p=0$ for $k>0$ . On the other hand, this can be seen as a consequence of the fact that our spectral sequence is limited to the first quadrant, provided the filtration is finite in the sense of Weibel’s Homological Algebra, p. 123 ( $F^mH^s=0$ for some $m$ ). But then it would be unclear why McCleary took this as an additional assumption rather than as a consequence of prior assumptions in the first case. : /

The new part of this example is the assumption that $E^{p,q}_2=0$ unless $q=0$ or $q=n$ , so all terms of the spectral sequence are to be found just in two horizontal stripes. In particular $E_\infty^{p,q}$ is only possibly non-zero in these stripes, and since these correspond to filtration quotients, the filtration takes a special form.

First, we might look at the filtration on $H^s$ where $0\leq s \leq n-1$ . Note that the spectral sequence terms that give information about $H^s$ are those along the diagonal line where $p+q=s$ . Since $s\leq n-1$ , the only place where anything interesting might happen is when this line crosses the $p$ -axis, i. e. when $q=0$ . This forces $p=s$ , so the only possible nonzero filtration quotient is

$E_\infty^{s,0}=F^sH^s/F^{s+1}H^s=F^sH^s=F^0H^s=H^s$

working with the assumption that $F^{s+1}H^s=0$ . So on the one hand, we get no interesting filtration of $H^s$ for $s<n$ , but on the other hand we can see exactly what it is from the spectral sequence limit.

Now we treat the case of $H^{n+p}$ , where $p\geq 0$ . I find this awkward notation again, preferring to reserve $p$ for a pure arbitrary spectral sequence index, but since we are trying to address the mistake in this notation, we should keep it for now. The filtration of this vector space/cohomology is interesting when $q=n$ and $q=0$ , where the quotients are given by

$E^{p,n}_\infty=F^pH^{p+n}/F^{p+1}H^{p+n}\quad\text{and}\quad E^{p+n,0}_\infty=F^{p+n}H^{p+n}/0.$

Every where else, successive quotients are 0, meaning the filtration looks like…

$0\sus F^{n+p}H^{n+p}= \dots =F^{p+1}H^{n+p} \sus F^pH^{n+p}=\dots=F^1H^{n+p}\sus H^0{n+p}$

In the filtration on page 9, McCleary puts one of the (possibly) non-trivial quotients at $F^nH^{n+p}$ instead of at $F^pH^{n+p}$ where it should be. That’s all I’m saying.

This situation is modeled on a spectral sequence for sphere bundles i.e. bundles where the fibers are spheres of a given dimension. The stripes coincide with the fact that a sphere $\mathbb{S}^n$ has nontrivial cohomology only at $H^n$ and $H^0$ . This sort of computation is famous enough that it has a name: the Thom-Gysin sequence (or just Gysin sequence).

As a final remark on section 1.2, McCleary says that the sequence in example 1.C is the Gysin sequence. Example 1.C doesn’t exist, we mean example 1.D : )

Spectral Sequences I

June 21, 2018 by aram Leave a comment

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