I am currently self-studying Bott and Tu. In chapter 2 the Cech-de Rham cohomology is introduced and I thought I had understood it well enough. However when I got to chapter 3 on spectral sequences I could not follow; some lines of reasoning used by the authors prompted to me think that maybe I have misunderstood something in the Cech-de Rham cohomology. So I would like some help on clarifying some concepts:

1) The Cech-de Rham complex $C^{p,q}$ is bi-graded, but when we take the D-cohomology (some call it the total cohomology) the groups $H^n_D$ are only 'single-graded', with index $n=p+q$ right? If yes then what does $H^{p,q}_D\{C^*(\mathfrak{U},\Omega^*)\}$ mean in P.167 (14.16.1)?

2) (If the answer to (1) is yes) Is there a natural way to grade $H^n_D$, ie. something like $H^n_D=\bigoplus_{p+q=n} H^{p,q}_D$? In particular can we talk about $H^n_D$ restricted to $K^{p,q}$? My intuition says this is problematic since the differential operators $D$ between $\bigoplus_{p+q=n}K^{p,q}$ 'blends' different $(p,q)$ together. The reason I am asking this is that in P.164 it is reasoned that 'the bidegrees $(p,q)$ of the double complex $K$ persist in the spectral sequence $E_r$', this I understand but what comes after is very puzzling for me: the filtration on $H^n(K)=F^n_0 \supset F^n_1 \supset F^n_2 \supset ...$ (I am using the notation from Theorem 14.6 for convenience here) has quotients $E^{0,n}_\infty,E^{1,n-1}_\infty,...$ repectively, that is, $F^n_0/F^n_1=E^{0,n}_\infty$ etc. I see no reason why this should hold, and have no idea how to prove it especially if the bigrading on $E_r$ does not carry over to the complex $A_\infty$.

It would be very appreciated if someone can help clarify point (1) for me, and point out how the statement (14.13) I have mentioned in P.164 holds (feel free to ignore my concerns and write a fresh proof, as I may have gone off in the wrong direction in the first place)


1 Answer 1

  1. You are quite right that $H_D^n$ is single-graded. However, there is a filtration$$H_D^n = F_0 \supset F_1 \supset F_2 \supset \ldots \supset F_n \supset 0$$on $H_D^n$, and $H_D^{p, n - p}$ is defined to be the quotient$$H_D^{p, n - p} = F_p/F_{p + 1}.$$Note that $H_D^{p, n - p}$ is not a subgroup of $H_D^n$. Rather, it is a subquotient of $H_D^n$, a quotient of a subgroup. Thus, there is an equality$$GH_D^n = \bigoplus_{p = 0} F_p/F_{p + 1} = \bigoplus_{p + q = n} H_D^{p, q},$$where $GH_D^n$ is the associated graded complex of the filtered complex $H_D^n$.

    In general, $GH_D^n$ is not isomorphic to $H_D^n$. For example, if $H_D^n$ is $\mathbb{Z}_4$, the cyclic group of order $4$ and the filtration is$$\underbrace{\mathbb{Z}_4}_{= \,F_0} \supset \underbrace{\mathbb{Z}_2}_{= \,F_1} \supset 0,$$then the associated graded is$$G\mathbb{Z}_4 = {{F_0}\over{F_1}} \oplus F_1 = \mathbb{Z}_2 \oplus \mathbb{Z}_2,$$which is not isomorphic to $\mathbb{Z}_4$.

    However, if we are working in the category of vector spaces, then since the dimension is the only invariant of a finite-dimensional vector space, $H_D^n$ is isomorphic to its associated graded complex.

    Can you show that $H_D^n$ and $GH_D^n$ have the same dimension? It is because$$\dim(F_p/F_{p + 1}) = \dim F_p - \dim F_{p + 1}.$$The Čech-de Rham complex is of course a vector space and so (14.16.1) is to be interpreted as$$H_{\text{dR}}^k(M) \simeq GH_{\text{dR}}^k(M) = \bigoplus_{p + q = k} H_D^{p, q}\{C^*(\mathfrak{U}, \Omega^*)\},$$where$$H_D^{p, q}\{C^*(\mathfrak{U}, \Omega^*)\} = F_p/F_{p + 1}$$for the induced filtration $\{F_p\}$ on $H_D^k\{C^*(\mathfrak{U}, \Omega^*)\}$.

    As with many texts in algebraic topology, "$=$" is often used to mean "isomorphism." This may have caused some confusion.

  2. Although in the category of vector spaces, one can write$$H_D^n \simeq \bigoplus_{p + q = n} H_D^{p, q},$$this isomorphism is not canonical, i.e. there are no natural ways of identifying certain subspaces of $H_D^n$ as $H_D^{p, q}$, because, as I pointed out earlier, $H_D^{p, q}$ is a subquotient of $H_D^n$, not a subspace.

    For a smooth complex projective variety, there is a canonical decomposition$$H^n \simeq \bigoplus_{p + q = n} H^{p, q},$$the famous Hodge decomposition, but that is an altogether different story.

    As for the equality $F_0^n/F_1^1 = E_\infty^{0, n}$, that is simply the definition of $E_\infty^{0, n}$. There is nothing to prove there.

    Thank you for the detailed reply, I understand point (1) completely now! However for point (2) I thought $E_\infty^{0, n}$ was defined to be the stable value of $E_1^{0, n}$, $E_2^{0, n}$, $\ldots$ , where $E_r^{0, n}$ is the $(0, n)$th entry in the $r$th page $E_r$. One definition or another, how can we show these two are same?

    You are quite right that $E_\infty^{p, q}$ is the stationary value of $E_r^{p, q}$ as $r \to \infty$. Now each $E_r$ is the cohomology of the preceding page $E_{r - 1}$, and on $E_r$ there is a filtration$$E_r = F_0E_r \supset F_1E_r \supset F_2E_r \supset \ldots.$$enter image description here

    We define $E_r^{p, n - p}$ to be the quotient $F_pE_r^n/F_{p + 1}E_r^n$, where $E_r^n$ is the antidiagonal of degree $n$.

    enter image description here

    For a fixed $n$, as we turn the pages $E_r$, each page $E_r$ is the cohomology of the preceding page $E_{r - 1}$. After finitely many steps, the pages $E_r$ become stationary. There is an $r \in \mathbb{Z}^+$ such that$$E_r = E_{r + 1} = \ldots = E_\infty.$$The filtration $\{F_pE_r\}$ also becomes stationary as $r$ increases. Thus,$$E_\infty^{p, n - p} = E_r^{p, n - p} = {{F_pE_r^n}\over{F_{p+1}E_r^n}} \simeq {{F_pE_\infty^n}\over{F_{p + 1}E_\infty^n}}.$$

  • $\begingroup$ Thank you for the detailed reply, I understand point (1) completely now! However for point (2) I thought $E^{0.n}_\infty$ was defined to be the stable value of $E^{0.n}_1, E^{0.n}_2, ...$, where $E^{0.n}_r$ is the $(0,n)$-th entry in the $r$-th page $E_r$. One definition or another, how can we show these two are same? $\endgroup$ Jul 29, 2016 at 1:34
  • $\begingroup$ I think I got it now, thanks a lot! $\endgroup$ Jul 31, 2016 at 2:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .