# Showing that filtered complexes induce a spectral sequence

Let $(F_p C_\bullet)_p$ be a filtered complex of $R$-modules, with chain maps $$\cdots \xrightarrow{}C_n\xrightarrow{d_n}C_{n-1}\xrightarrow{}\cdots$$ such that $$\cdots \hookrightarrow F_pC_n\hookrightarrow F_{p+1}C_n\hookrightarrow\cdots\hookrightarrow C_n$$ and $d_n(F_pC_n)\subseteq F_p C_{n-1}$ for all $p$ and $n$. Then we define the spectral sequence $\{E^r_{p,q}\}_{p,q,r}$ as follows:

$$E_{p,q}^r = \frac{ \{x\in F_pC_{p+q}\,\vert\,d_{p+q}x\in F_{p-r}C_{p+q-1}\} }{ F_{p-1}C_{p+q} + d_{p+q+1}(F_{p+r-1}C_{p+q+1})}$$ where we quotient by the intersection of the denominator with the numerator, and we let the map $$d_{p,q}^r:E_{p,q}^r\to E_{p-r,q+r-1}^r$$ be induced by $d_{p+q}:C_{p+q}\to C_{p+q-1}$, such that $d_{p-r,q+r-1}^r\circ d_{p,q}^r=0$. Then the last thing we need to check for this to be a spectral sequence is that $$E_{p,q}^{r+1} = \frac{\mathrm{ker\,} d_{p,q}^r}{\mathrm{im\,} d_{p+r,q-r+1}^r}$$ and this is where I'm having difficulties. Supposing $[x]\in\ker d_{p,q}^r$, then why should it be that $d_{p+q}x\in F_{p-r-1}C_{p+q-1}$? I can see that we must have that $$d_{p+q}x=y+d_{p+q}z$$ for some $y\in F_{p-r-1}C_{p+q-1}$ and $z\in F_{p-1}C_{p+q}$, but I don't see why we must have that $$d_{p+q}z\in F_{p-r-1}C_{p+q-1}.$$ Am I missing something?

• One of the first things one needs to learn is to live without subindices. =) – Pedro Tamaroff Jan 3 '17 at 3:32
• (This is explained in Weibel's book, by the way. What you need is to use isomorphism theorems.) – Pedro Tamaroff Jan 3 '17 at 3:47

Perhaps you want to send the class $[x]$ not to (the class of) $x$ but to $x-z$. Note that $x-z \in F_pC_{p+q}$ and $d_{p+q}(x-z)=y \in F_{p-r-1}C_{p+q-1}$. Also since $z \in F_{p-1}C_{p+q}$, $[x] = [x-z]$ in $E^r_{p,q}$.
• But is $[x-z]\in E_{p,q}^{r+1}$ independent of the choice of $z$? – Monstrous Moonshine Jan 3 '17 at 22:35
• @MonstrousMoonshine yes. If z' were another choice, then the difference between x-z and x-z' is z-z' which lies in $F_{p-1}C_{p+q}$ (since both z,z' lie in there) and this is quotiented out in the definition of $E^{r+1}_{p,q}$ – usr0192 Jan 4 '17 at 0:14