R page of the spectral sequence of a filtered complex

The $r$-page of a filtered complex is the family $E_{p,q}^r=Z_{p,q}^r\mathrm{/}{B}_{p,q}^r(r\ge 0),$ with the conventions of R cycles and r boundaries of an increasingly filtered complex. The quotient is licensed by R boundaries embed in r cycles. In particular $E_{p,q}^0=F_p{C}_{p+q}\mathrm{/}{F}_{p-1}{C}_{p+q}$.

For $r\ge 1$ this is canonically ${\bar{Z}}_{p,q}^r\mathrm{/}{\bar{B}}_{p,q}^r$: the kernel of $A_{p,n}^r\to {\mathrm{gr}}_p{C}_n$ is $A_{p,n}^r\cap F_{p-1}{C}_n=A_{p-1,n}^{r-1},$ since $(p-1)-(r-1)=p-r$. The image of the remaining denominator summand is exactly ${\bar{B}}^r$; nested quotients in Spectral sequence subquotient and local lifting calculus give the comparison.

The formula uses only a finite filtration window at each position. More precisely, fix $k\ge 0$ and $0\le r\le k$. The filtered complex $F_{p+k}C\mathrm{/}{F}_{p-k-1}C$, with the induced filtration, gives canonically the same $E_{p,q}^r$. Indeed the numerator test uses levels $p,p-r\ge p-k$, and the denominator uses levels $p-1,p+r-1\le p+k$. The removed subcomplex has level $p-k-1$ below every test level, and its contribution in degree $n$ lies in $A_{p-1,n}^{r-1}$ when $r\ge 1$, because its differential stays at level $p-k-1\le p-r$. Thus inverse-image tests and the final quotient are unchanged; for $r=0$ this is directly the nested quotient $F_p\mathrm{/}{F}_{p-1}$.

Weibel, Chapter 5, Construction 5.4.6 and Lemma 5.4.7, pp.133–134; Sharifi, Theorem 4.2.3, pp.91–92. Increasing homological indices are used here.