Spectral sequence of a two step filtration

Goal

We want to explicitly consider all of the objects, maps, and differentials in a particular spectral sequence arising from a space that admits a filtration that terminates in two steps. There are several concrete examples that should fit into this framework:

\(0 \hookrightarrow S^k \hookrightarrow S^n\) for any \(k < n\)
\(0 \hookrightarrow S^n \hookrightarrow{\mathbf{CP}}^n\)
\(0 \hookrightarrow{\mathbf{RP}}^n \hookrightarrow S^n\)
- Using \(S^n\) as a double cover of \({\mathbf{RP}}^n\)

Setup: Space and Filtration

Let \(X\) be a space and let \(A\subset X\) be a subspace, inducing the inclusion \(A\xrightarrow{i} X\), so we have the following inclusions of spaces:

\begin{align*}0 \hookrightarrow A \hookrightarrow B\end{align*}

Then consider applying the “chain functor” \(C_*(\cdot): \textbf{Top} \to\textbf{Ab}\) that sends a space \(X\) to a singular chain complex \begin{align*}C_*(X) \coloneqq\cdots \xrightarrow{{\partial}_{i-1}} C_i(X) \xrightarrow{{\partial}_i} C_{i+1}(X) \xrightarrow{{\partial}_{i+1}} \cdots\end{align*}

Applying this functor to the above inclusion induces an inclusion of chain complexes:

\(0 \hookrightarrow C_*(A) \hookrightarrow C_*(X)\)

We regard this as a two step filtration on \(C^*(X)\) by making the following identifications:

\(F_0C_*(X) \coloneqq C_*(X)\)
\(F_1C_*(X) \coloneqq C_*(A)\)
\(F_2C_*(X) \coloneqq 0\)

And we obtain the primary object of interest for this spectral sequence:

\(0 = F_2C_*(X) \hookrightarrow F_1C_*(X) \hookrightarrow F_0C_*(X) = C_*(X)\)

This process is roughly summarized in the following diagram: \begin{align*} \begin{CD} 0 @>>\hookrightarrow> A @>i>\hookrightarrow> X \\ @VVV @VVC_*(\cdot)V @VVV\\ 0 @>>\hookrightarrow> C_*(A) @>i_*>\hookrightarrow> C_*(X)\\ @| @| @|\\ F_2C_*(X) @>>\hookrightarrow> F_1C_*(X) @>i_*>\hookrightarrow> F_0C_*(X) \end{CD} \end{align*}

Setup: Spectral Sequence

A few definitions to recall:

\(G_iC_*(X) \coloneqq\frac{F_iC_*(X)}{F_{i+1}C_*(X)}\)

\(E_0^{p,q} = G_pC_{p+q}(X)\)

\(E_1^{p,q} = H(E_0^{p,q}, d_0)\)

Computation of Pages

\(E_{-1}\)

Not standard usage, here I consider the “\(E_{-1}\) page” to be simply a presentation of the double complex itself. The formula works out to be something like \(E_{-1}^{p,q} = F_pC_q(X)\)

\begin{align*} \begin{array}{l:r|cccc} q= n &\hspace{4em} 0 & F_0C_n(X) & F_1C_n(X) & F_2C_n(X) \\ \vdots & \vdots & \vdots & \vdots \\ q=3 &0 & F_0C_3(X) & F_1C_3(X) & F_2C_3(X) \\ q=2 &0 & F_0C_2(X) & F_1C_2(X) & F_2C_2(X) \\ q=1 &0 & F_0C_1(X) & F_1C_1(X) & F_2C_1(X) \\ q=0 &0 & F_0C_0(X) & F_1C_0(X) & F_2C_0(X) \\ \hline \\ q=-1 &0 & 0 & 0 & 0 \\ q-2 &0 & 0 & 0 & 0 \\ \\\hdashline\\ p = -2 & p=-1& p=0 & p=1 & p=2 \\ \end{array} \end{align*}

For clarity, we unpack definitions here to show how the actual original chain complexes sit inside of this page: \begin{align*} \begin{array}{l:r|cccc} q= n &\hspace{4em} 0 & C_n(X) & C_n(A) & 0 \\ \vdots & \vdots & \vdots & \vdots \\ q=3 &0 & C_3(X) & C_3(A) & 0 \\ q=2 &0 & C_2(X) & C_2(A) & 0 \\ q=1 &0 & C_1(X) & C_1(A) & 0 \\ q=0 &0 & C_0(X) & C_0(A) & 0 \\ \hline \\ q=-1 &0 & 0 & 0 & 0 \\ q-2 &0 & 0 & 0 & 0 \\ \\\hdashline\\ p = -2 & p=-1& p=0 & p=1 & p=2 \\ \end{array} \end{align*}

Focusing on the area \(p,q >= -1\), we use the fact that the chain complexes come with natural boundary maps to define the differentials \(d_{-1}\coloneqq{\partial}_n: C_n(X) \to C_{n-1}(X)\). \begin{align*} \begin{CD} 0 @<<< 0 @<<< 0 @<<< 0 \\ @VVV @VVV @VVV @VVV \\ 0 @<<< C_n(X) @<

\(E_0\)

Here we use the following formulas/facts:

\(G_iC_*(X) \coloneqq\frac{F_iC_*(X)}{F_{i+1}C_*(X)}\)
\(E_0^{p,q} \coloneqq G_pC_{p+q}(X)\)
\(C_n(X, A) \coloneqq\frac{C_n(X)}{C_n(A)}\)
- This can be done because there is a SES \(0 \to C_*(A) \to C_*(X) \to\frac{C_*(X)}{C_*(A)} \to 0\) Then since \({\partial}_n : C_n(X) \to C_{n-1}(X)\) has the property that \({\partial}_n(C_*(A)) = C_*(A)\), it factors through the quotient \(\frac{C_*(X)}{C_*(A)}\) to yield a map \(\widehat{{\partial}}_n: \frac{C_n(X)}{C_n(A)} \to\frac{C_{n-1}(X)}{C_{n-1}(A)}\). Shorten notation by calling \(\frac{C_*(X)}{C_*(A)} \coloneqq C_*(X, A)\) the relative chain complex; this yields relative homology with respect to \(\widehat{{\partial}}\), i.e. \(H_n(X,A) \coloneqq\frac{\ker {\partial}_n}{\operatorname{im}{\partial}_{n+1}} \subset C_n(X, A)\).

which explicitly yields \begin{align*} G_0C_*(X) = \frac{F_0C_*(X)}{F_1C_*(X)} = \frac{C_*(X)}{C_*(A)} \coloneqq C_*(X, A) \\ G_1C_*(X) = \frac{F_1C_*(X)}{F_2C_*(X)} = \frac{C_*(A)}{ 0 } = C_*(A) \\ G_2C_*(X) = \frac{0}{0} = 0 \end{align*} \(E_0^{p,q} \coloneqq G_pC_q(X)\)

\(C_n(X, A) \coloneqq\frac{C_n(X)}{C_n(A)}\)

\begin{align*} \begin{array}{l:r|cccc} q= n &\hspace{4em} 0 & G_0C_n(X) & G_1C_{n+1}(X) & 0 \\ \vdots & \vdots & \vdots & \vdots \\ q=3 &0 & G_0C_3(X) & G_1C_4(X) & 0 \\ q=2 &0 & G_0C_2(X) & G_1C_3(X) & 0 \\ q=1 &0 & G_0C_1(X) & G_1C_2(X) & 0 \\ q=0 &0 & G_0C_0(X) & G_1C_1(X) & 0 \\ \hline \\ q=-1 &0 & 0 & G_1C_0(X) & 0 \\ q-2 &0 & 0 & 0 & 0 \\ \\\hdashline\\ p = -2 & p=-1& p=0 & p=1 & p=2 \\ \end{array} \end{align*}

Which unpacks as \begin{align*} \begin{array}{l:r|cccc} q= n &\hspace{4em} 0 & \frac{F_0C_n(X)}{F_1C_n(X)} & \frac{F_1C_{n+1}(X)}{F_2C_{n+1}(X)} & 0 \\ \vdots & \vdots & \vdots & \vdots \\ q=3 &0 & \frac{F_0C_3(X)}{F_1C_3(X)} & \frac{F_1C_4(X)}{F_2C_4(X)} & 0 \\ q=2 &0 & \frac{F_0C_2(X)}{F_1C_2(X)} & \frac{F_1C_3(X)}{F_2C_3(X)} & 0 \\ q=1 &0 &\frac{F_0C_1(X)}{F_1C_1(X)} & \frac{F_1C_2(X)}{F_2C_2(X)} & 0 \\ q=0 &0 & \frac{F_0C_0(X)}{F_1C_0(X)} & \frac{F_1C_1(X)}{F_2C_1(X)} & 0 \\ \hline \\ q=-1 &0 & 0 & \frac{F_1C_0(X)}{F_2C_0(X)} & 0 \\ q-2 &0 & 0 & 0 & 0 \\ \\\hdashline\\ p = -2 & p=-1& p=0 & p=1 & p=2 \\ \end{array} \end{align*}

Which further unpacks as \begin{align*} \begin{array}{l:r|cccc} q= n &\hspace{4em} 0 & \frac{C_n(X)}{C_n(A)} & \frac{C_{n+1}(A)}{0} & 0 \\ \vdots & \vdots & \vdots & \vdots \\ q=3 &0 & \frac{C_3(X)}{C_3(A)} &\frac{C_4(A)}{0} & 0 \\ q=2 &0 & \frac{C_2(X)}{C_2(A)} & \frac{C_3(A)}{0} & 0 \\ q=1 &0 &\frac{C_1(X)}{C_1(A)} &\frac{C_2(A)}{0} & 0 \\ q=0 &0 & \frac{C_0(X)}{C_0(A)} & \frac{C_1(A)}{0} & 0 \\ \hline \\ q=-1 &0 & 0 & \frac{C_0(A)}{0} & 0 \\ q-2 &0 & 0 & 0 & 0 \\ \\\hdashline\\ p = -2 & p=-1& p=0 & p=1 & p=2 \\ \end{array} \end{align*}

Which by definition is \begin{align*} \begin{array}{l:r|cccc} q= n &\hspace{4em} 0 & C_n(X, A) & C_{n+1}(A) & 0 \\ \vdots & \vdots & \vdots & \vdots \\ q=3 &0 & C_3(X, A) & C_4(A) & 0 \\ q=2 &0 & C_2(X, A) & C_3(A)& 0 \\ q=1 &0 &C_1(X, A) &C_2(A) & 0 \\ q=0 &0 &C_0(X, A) & C_1(A) & 0 \\ \hline \\ q=-1 &0 & 0 & C_0(A) & 0 \\ q-2 &0 & 0 & 0 & 0 \\ \\\hdashline\\ p = -2 & p=-1& p=0 & p=1 & p=2 \\ \end{array} \end{align*}

For any pair \((X, A)\), there is a long exact sequence

\(\cdots H_n(A) \xrightarrow{} H_n(X) \xrightarrow{} H_n(X, A) \xrightarrow{\delta_n} H_{n-1}(A) \cdots\) \begin{align*} \begin{CD} 0 @<<< 0 @<<< 0 @<<< 0 \\ @VVV @VVV @VVV @VVV \\ 0 @<<< C_n(X, A) @<