# Friday, April 15

## Filtrations and Gradings

:::{.remark}
Given $\Fil A$ a descending filtration, define $\gr_i A \da \Fil_i A/ \Fil_{i+1} A$.
Convention: everywhere we'll set $p+q\da n, p = n-q$, etc.

This results in a collection of short exact sequences:
\[
0 \to \Fil_{i+1} A\to \Fil_i A \to \gr_i A \to 0
.\]
:::

:::{.remark}
Our main example: a double complex $\cobicomplex C$ with $\cocomplex A \da \cocomplex{\Tot} \cobicomplex C$ with $A^n \da \oplus _{p+q=n} C^{p ,q}$ and differentials $\bd = (d_v, d_h)$ producing skew-commuting squares.
The main question is computing $H^*(A)$.

Each $A^n$ is a filtration $\Fil A^n$ where $\bd \Fil^i A^n \subseteq \Fil^{i+1} A^n$.
The filtration is defined by 
\[
\Fil^{p_0} A^n = \bigoplus _{p+q=n, p\geq p_0} C^{p, q}
,\]
taking everything to the right of column $p_0$.
The claim is that this induces a filtrations on $Z^n(A), B^n(A), H^n(A)$ (cycles, boundaries, and homology).
One can restrict the differential on $\cocomplex{A}$ to $\Fil\cocomplex{A}$; note that cycles $Z_n\mapsto 0$ and boundaries are the image and we're taking cycles mod boundaries.
Writing $\Fil^p Z^n \da \Fil^p A^n \intersect Z_n$ and similarly for $B^n, H^n$, one gets a filtration $\Fil H(\Fil^p A)$ on $H(\Fil^p A)$.
This yields
\[
E_\infty^{p, q} = \gr_p H^n = \Fil^p H^n / \Fil^{p+1} H^n
.\]
If all of the SESs split, then $H^n = \bigoplus _{p+q=n} E^{p, q}_\infty$.
:::

:::{.remark}
Set

- $E_0^{p, q} \da C^{p, q}$
- $E_1^{p, q} = H^n( C^{p, \bullet}, d_v )$.
- $E_2^{p, q} = H^n( E_1^{p, q}, d_v) = H^*(\cdots\to H^{n-1}C^{p, \bullet} \to H^n(C^{p, \bullet}) \to H^{n+1} C^{p+1,\bullet}\to \cdots )$.

What are the cycles in $E_0$?
To map to zero under the total differential $\del$, things emanating from column $p$ must go to zero, and for the columns $p+k$, images under $d_h^{p+k, \ell}$ must cancel with images under $d_h^{p+k+1, \ell-1}$.
Define the *approximate homology*
\[
\Fil^p H^{\approx}_{p \pm r} = {\del\inv (\Fil^{p+r} A^{n+1} ) \over \del(\Fil^{p-r+1} A^{n-1} ) }
.\]
Note that this *increases* the number of allowed cycles and *decreases* the number of allowed boundaries.
Then $E_r^{p, q} = \gr_p H^n_{p \pm r}$.
:::

:::{.remark}
Note that the statement is not the $E_r$ is computed as $H^*(E_{r-1})$; instead there is a formula for $E_r^{p, q}$ for all $r,p,q$ a priori, and it is a property that taking homology of pages computes this.
:::

:::{.remark}
Claim: $\gr_p H^n_{p\pm 0} = C^{p, q}$.
Check that
\[
\Fil^{p_0} H^n_{p\pm 0} = {\bigoplus _{p+q=n, p\geq p_0} C^{p, q} \over d\qty{ \bigoplus _{p+q=n-1, p\geq p_0 + 1} C^{p, q} } }
.\]
:::