# Monday, April 11

## Spectral sequences

:::{.proposition title="Leray spectral sequence"}
If $f\in \Top(X, Y)$ and $\mcf \in \Sh(X; \Ab\Grp)$, there is a spectral sequence
\[
E_2^{p, q} = H^p(X; \RR^{q} f_* \mcf) \abuts H^{p+q}(X; \mcf)
.\]
:::

:::{.example title="?"}
If $0 \to A\cocovers \cocomplex{J}$ is an injective resolution of a sheaf $A$, then $E_1^{p, q} = H^p(J^q) \abuts H^{p+q}(A)$.
More generally, for any functor $F \in \Cat(\cat A, \cat B)$,
\[
E_1^{p, q} = \RR^p F(J^q) \abuts \RR^{p+q} F(A)
.\]

So if $J^q$ are $F\dash$acyclic, then $\tau_{\geq 1}\cocomplex{\RR} F(J^q) = 0$ and thus $\RR^n F(A)$ is the homology of the complex $F\cocomplex{J}$.
:::

:::{.proposition title="Grothendieck"}
If 

- $\cat{A} \mapsvia{F} \cat{B} \mapsvia{G} \cat{C}$ are left-exact functors between abelian categories 
- $\cat A, \cat B$ have enough injectives, and 
- $F(I)$ for $I$ injective in $\cat{A}$ yields a $G\dash$acyclic object in $\cat{B}$, 

then there is a first-quadrant spectral sequence

\[
E_2^{p, q} = \RR^p G( \RR^q G(A)) \abuts \RR^{p+q}(F\circ G)(A)
.\]
:::

:::{.remark}
This recovers the Leray spectral sequence via $\Sh(X; \Ab\Grp) \mapsvia{f_*} \Sh(Y; \Ab\Grp) \mapsvia{\globsec{Y; \wait}} \Ab\Grp$, where the composition is $\globsec{X; \wait}$.
Note that injective sheaves are flasque, and pushforwards of flasque sheaves are again flasque.
Why flasque implies injective:

\begin{tikzcd}
	0 && {j_!\ro{ \OO_X }{U}} && {\OO_X} \\
	\\
	&& I
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[dashed, from=1-5, to=3-3]
	\arrow[from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwial8hXFxyb3sgXFxPT19YIH17VX0iXSxbNCwwLCJcXE9PX1giXSxbMiwyLCJJIl0sWzAsMV0sWzEsMl0sWzIsMywiIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzEsM11d)

:::

:::{.remark}
Recall that cohomology vanishes above the dimension of a Noetherian space. 
The analog for pushforward involves the relative dimension.
:::

:::{.remark}
General setup:

- $d_r: E_r^{p, q} \to E_r^{p+r, q-r+1}$ (down and to the right) moves between diagonals.
  - For a fixed $p, q$, all differentials out of $E_{p, q}$ land on the same diagonal.
- $E_{r+1} = H(E_r, d_r)$.
- Letting $E_n = \bigoplus_{p+q=n} E_\infty^{p, q}$, there is a descending filtration $\complex{\Fil} E_n$ such that $\gr_p \complex{\Fil} E_n \da \Fil_p E_n/ \Fil_{p+1} E_n = E_\infty^{p, n-p}$.
- Extension problem: $\ZZ \contains 2\ZZ \contains 0$ where $\gr_1 = C_2$ and $\gr_1 \cong \ZZ$, but another group and filtration may have the same associated graded, e.g. $\ZZ \oplus C_2 \contains \ZZ \contains 0$.
- Double complexes naturally arise by taking an injective resolution $A \cocovers \cocomplex{J}$ and individually resolving the pieces by $J^n \cocovers C^{n, \bullet}$.
  Writing $\Tot(C^{p, q})_n \da \oplus_{p+q=n} C^{p, q}$, there are maps $A\to C^{0,0} \to (C^{0,1} \oplus C^{1, 0}) \to \cdots$ by summing horizontal and vertical differentials.
  Using the sign trick makes this a differential (multiply the vertical differentials in every even column by $-1$).

- There are spectral sequences 
\[
E_1{p, q} &= H^p(C^{\bullet, q}, d_h) \abuts H^{p+q}( \Tot( C^{\bullet, \bullet} ) ) \\
E_1{p, q} &= H^q(C^{p, \bullet}, d_v) \abuts H^{p+q}( \Tot( C^{\bullet, \bullet} ) )
.\]
  - Why this is useful: resolve $A$ by $J$ which are not necessarily injective, and resolve each $J^n$ by injectives, then $\Tot$ is now an injective resolution.
- 
:::