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I'm learning about spectral sequences in Ravi Vakil's notes, and can't quite figure out how to define the map ($d_2$) on the bottom of page 59 (he describes it as a worthwhile exercise). It should be a morphism of the cohomology complexes on page $E_1$ of the sequence. I tried using short exact sequences to get a connecting homomorphism, but I haven't been able to do it.

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1 Answer 1

I was puzzled at first as well, turns out this is just an exercise on drawing and reading.

Introduction

Let me recall/introduce the notation I will be using for completeness (it is a slight variation of the one in Vakil's notes):

We have a double graded complex $E^{p,q}$, with differentials $\partial^{p,q}$ going left-to-right ($E^{p,q}\rightarrow E^{p+1,q}$), and differentials $d^{p,q}$ going bottom-up ($E^{p,q}\rightarrow E^{p,q+1}$). Let's say the complex has commuting squares.

The $0$-th sheet consists just from the objects o the double complex (i.e. $E_0^{p,q}=E^{p,q}$) with left-to-right differentials (i.e. $d_{0}^{p,q}=\partial^{p,q}$). The first sheet then consists of homologies of these, i.e. $E_{1}^{p,q}=\mathrm{Ker}\, \partial_{p,q}/\mathrm{Im}\; \partial_{p-1,q}$ and the induced maps by the original bottom-up differentials, i.e. $d_1^{p,q}=\bar{d}^{p,q}: [x] \mapsto [d^{p,q}(x)]$, where $x \in \mathrm{Ker}\,\partial^{p,q}$ and the brackets denote factorization modulo $\mathrm{Im}\,\partial$ in the respective places.

Key point

Now, from all this data, we want to define an arrow going two steps up and one to the left on the homologies. So let us first think of how to get there. I think this is the key point:

Take an element $[x] \in \mathrm{Ker}\,\bar{d}^{p,q}.$ That means that

  • $x \in \mathrm{Ker}\,\partial^{p,q}$,
  • $\bar{d}^{p,q}([x])=0,$ i.e. $d^{p,q}(x) \in \mathrm{Im}\,\partial^{p-1,q+1}.$

Now, this tells you exactly what you should do, if you remember the diagram chasing for the snake lemma/long exact sequence of homologies:

Push $x$ to $d^{p,q}(x).$ This is in $\mathrm{Im}\,\partial^{p-1,q+1},$ so take some preimage $x' \in E^{p-1,q}$, i.e. element such that $\partial^{p-1,q}(x')=d^{p,q}(x)$. Now push this once further up to get $x''=d^{p-1,q+1}(x')$.

Some verifications:

Fist, let us convince ourselves that the map taking such $x$ to the class $[[x'']]$ (where double brackets represent the succesive factorizations, i.e modulo $\mathrm{Im}\,\partial$ and then modulo $\mathrm{Im}\, \bar{d}$), is well-defined.

1) $x'' \in \mathrm{Ker}\, d^{p-1,q+2}$ :

Obvious, since $$d^{p-1,q+2}(x'')=d^{p-1,q+2}(d^{p-1,q+1}(x'))=0.$$

2) $x'' \in \mathrm{Ker}\, \partial^{p-1,q+2}$ :

Easy: $$\partial^{p-1,q+2}(x'')=\partial^{p-1,q+2}(d^{p-1,q+1}(x'))=\partial^{p-1,q+2}(d^{p-1,q+1}(x'))=d^{p,q+1}(\partial^{p-1,q+1}(x'))=$$ $$=d^{p,q+1}(d^{p,q}(x))=0.$$

These two points showed that the symbol $[[x'']]$ (explained above) makes sense (i.e. is an element of $E_2^{p-1,q+2}$).

3) "Independence of the choice of $x'$:"

Assume that $x_1', x_2'$ are two preimages of $d^{p,q}(x)$, i.e. $\partial^{p-1,q+1}(x_i')=d^{p,q}(x), \; i=1, 2$. This means that $$\partial^{p-1,q+1}(x_1'-x_2')=0, \;\; \text{i.e. }x_1'-x_2' \in \mathrm{Ker}\,\partial^{p-1,q+1}.$$

So the class $[x_1'-x_2']\in E_1^{p-1,q+1}$ is well defined. In particular, since $\bar{d}^{p-1,q+1}([x_1'-x_2'])$ is tautologically in the image of $\bar{d}^{p-1,q+1}$, we have

$$0=[\bar{d}^{p-1,q+1}([x_1'-x_2'])]=[[d^{p-1,q+1}(x_1'-x_2')]]=[[d^{p-1,q+1}(x_1')-d^{p-1,q+1}(x_2')]]=[[x_1''-x_2'']].$$

In other words, we have $[[x_1'']]=[[x_2'']]$, so the choice between $x_1'$ and $x_2'$ really does not matter.

4) The map $x \mapsto [[x'']]$ is a well-defined group (module) hom:

This is straghtforward. The point is that given $x,y$ at the start, one can choose now $(x+y)'$ as $x'+y'$, since we establish that the choice of preimage does not matter. The same goes for multiplication by scalars if one needs that.

5) $\mathrm{\mathrm{Im}\, \partial^{p-1,q}} \subseteq \mathrm{Ker}\,(x \mapsto [[x'']])$:

If one starts with $x = \partial^{p-1,q}(y)$, it follows that one can choose $x'=d^{p-1,q}(y)$. This works and the choice of $x'$ does not matter. Then $$x''=d^{p-1,q+1}(x')=d^{p-1,q+1}(d^{p-1,q}(y))=0.$$ In particular, $[[x'']]=0.$

So, now we have a map $[x]\mapsto [[x'']]$.

6) $\mathrm{\mathrm{Im}\, \bar{d}^{p,q-1}} \subseteq \mathrm{Ker}\,([x] \mapsto [[x'']])$:

Suppose $[x]=\bar{d}^{p,q-1}([y])$. Then $[[x'']]=[[({d}^{p,q-1}(y))'']]$ by the previous point. However, we have that ${d}^{p,q}({d}^{p,q-1}(y))=0$, so obviously by the construction of "$x \mapsto x''$", we have $$[[({d}^{p,q-1}(y))'']]=0.$$

Now we are finished.

We have defined a map $d_2^{p,q}:[[x]]\mapsto[[x'']]$. What remains is to check that $d_2\circ d_2=0$. I'll leave that to you, since this post is long enough as it is.

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