---
date: 2018-01-13T23:41
---

Tags:
#spectral-sequences #homotopy/fibrations

# Filtrations and the Serre spectral sequence

## References

- **Rational Homotopy Theory and Differential Forms** by Griffiths and Morgan
- **Differential Forms in Algebraic Topology** by Bott and Tu
- **Differential Topology** by Hirsch
- **Comprehensive Introduction to Differential Geometry** by Spivak
- **Topology from the Differentiable Viewpoint** by Milnor
- **Topology and Geometry** by Bredon
- **User's Guide to Spectral Sequences** by Mcleary
  - View [Here](http://www.math.hcmuns.edu.vn/~nvdong/DoiDongDieuNhom/McCleary%20J.%20User%20s%20guide%20to%20spectral%20sequences%20(2ed.,%20CUP,%202001)(575s).pdf)
  - Lots of technical details


# Filtration from a SES

The standard Serre fibration: $\Omega X \to PX \mapsvia{f} X$ where $\Omega X$ is the loop space, $PX$ is the path space, and $f$ is the "evaluation at the endpoint" map. Note that $PX$ is contractible!

Consider a SES $0 \to A \to B \to C \to C$, then look at it as a 2-step filtration of $B$ so 

- $F^0B = B$, 
- $F^1B = A$,
- $F^2B = 0$. 

The sections are $G_0 = C, G_1 = A$. Can use this to obtain LES from SS.

Homology in the ring-theoretic setting: If $R$ is a Noetherian ring and $I \subseteq R$, then if $I$ can be generated by $n$ elements then $H_I^i(M) = 0$ for any $R$-module $M$ and $i > n$. Thus to prove $I$ can *not* be generated by $n$ elements, it suffices to find a module $M$ where $H_I^{n+1} \neq 0$.


## Serre Spectral Sequence

Lots of good examples of computations [here](https://en.wikipedia.org/wiki/Serre_spectral_sequence).

Some fibrations

- Hopf: $S^1 \to S^3 \to S^2$
- $S^1 \to S^{2n+1} \to \CP^n$
- Path space: $\Omega S^n \to PS^n \to S^n$

Serre Spectral Sequence Example:
For the fibration $S^1 \to S^3 \to S^2$, the $E_2$ page:

[Link to Diagram](https://q.uiver.app/?q=WzAsMTYsWzIsMSwiSF4wKFNeMjsgXFxaWikiXSxbMiwyLCJIXjAoU14yOyBcXFpaKSJdLFs0LDEsIkheMShTXjI7IFxcWlopIl0sWzQsMiwiSF4xKFNeMjsgXFxaWikiXSxbNiwxLCJIXjIoU14yOyBcXFpaKSJdLFs2LDIsIkheMihTXjI7IFxcWlopIl0sWzEsM10sWzAsMywiXFxidWxsZXQiXSxbNywzLCJcXGJ1bGxldCJdLFsxLDAsIlxcYnVsbGV0Il0sWzEsNCwiXFxidWxsZXQiXSxbMCwyLCIwIl0sWzIsNCwiMCJdLFswLDEsIjEiXSxbNCw0LCIxIl0sWzYsNCwiMiJdLFs3LDgsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbOSwxMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==)

\begin{tikzcd}
	& \bullet \\
	1 && {H^0(S^2; \ZZ)} && {H^1(S^2; \ZZ)} && {H^2(S^2; \ZZ)} \\
	0 && {H^0(S^2; \ZZ)} && {H^1(S^2; \ZZ)} && {H^2(S^2; \ZZ)} \\
	\bullet & {} &&&&&& \bullet \\
	& \bullet & 0 && 1 && 2
	\arrow[no head, from=4-1, to=4-8]
	\arrow[no head, from=1-2, to=5-2]
\end{tikzcd}

Which is equal to
[Link to Diagram](https://q.uiver.app/?q=WzAsMTYsWzIsMSwiSF4wKFNeMjsgXFxaWikiXSxbMiwyLCJIXjAoU14yOyBcXFpaKSJdLFs0LDEsIjAiXSxbNCwyLCIwIl0sWzYsMSwiSF4yKFNeMjsgXFxaWikiXSxbNiwyLCJIXjIoU14yOyBcXFpaKSJdLFsxLDNdLFswLDMsIlxcYnVsbGV0Il0sWzcsMywiXFxidWxsZXQiXSxbMSwwLCJcXGJ1bGxldCJdLFsxLDQsIlxcYnVsbGV0Il0sWzAsMiwiMCJdLFsyLDQsIjAiXSxbMCwxLCIxIl0sWzQsNCwiMSJdLFs2LDQsIjIiXSxbNyw4LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzksMTAsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)

\begin{tikzcd}
	& \bullet \\
	1 && {H^0(S^2; \ZZ)} && 0 && {H^2(S^2; \ZZ)} \\
	0 && {H^0(S^2; \ZZ)} && 0 && {H^2(S^2; \ZZ)} \\
	\bullet & {} &&&&&& \bullet \\
	& \bullet & 0 && 1 && 2
	\arrow[no head, from=4-1, to=4-8]
	\arrow[no head, from=1-2, to=5-2]
\end{tikzcd}


And $E_3 = E_\infty$, so $d_2^{0,1}$ is an isomorphism.

*Note: Probably a good starting point for basic calculations? Fill out the missing details for this table.*



:::{.exercise title="?"}
Prove
\[
\pi_4(S^2) = \frac{\ZZ}{2\ZZ}
.\]
:::