Spectral Sequences

Now Let $K$ be a filtered module and $d : K \to K$ be $A$-linear s.t. $d^2 = 0$. We call $(K, d)$ a defferential filtered module if $d(K_p) \subset K_p \,\,\forall\, p$. The theory of spectral sequences essentially is meant to construct successive approximation to $H(K)$ using the filtration. Use the convension $K_i = K$ for $i \le 0$.

Let $r \in \ensuremath{\mathbb{N}}$. Put

$$\ensuremath{{Z}^{p}_{r}} = \{ x \in K_p : dx \in K_{p+r}\} = d^{-1} (K_{p+r}) \cap K_p$$

Evidently $\ensuremath{{Z}^{p}_{r}}= K_p$ for $r \le 0$. Define

$$\ensuremath{{B}^{p}_{r}} = d \ensuremath{{Z}^{p-r+1}_{r-1}} = d K_{p-r+1} \cap K_p$$

$$\ensuremath{{E}^{p}_{r}} = \ensuremath{{Z}^{p}_{r}}/ (\ensuremath{{B}^{p}_{r}}+ \ensuremath{{Z}^{p+1}_{r-1}})$$

$$E_r = \oplus_p \ensuremath{{E}^{p}_{r}}$$

Note that if $x \in K_p$ is a cycle (ie. $dx = 0$) then $x$ defines an element of $\forall\, r$ and if $x$ is a boundary then its image in is zero for large $r$.

Theorem 3   For every $r$ there exists a derivation of $E_r$ of degre $r$ such that the module $H(E_r)$ is canonically isomorphic to $E_{r+1}$.

Proof:The map $d$ restricted to gives

$$d: \ensuremath{{Z}^{p}_{r}}\to \ensuremath{{Z}^{p+r}_{r}}$$

and $d(\ensuremath{{B}^{p}_{r}}) = 0$ and $d\ensuremath{{Z}^{p+1}_{r-1}} \subset \ensuremath{{B}^{p+r}_{r}}$. Hence we have a map

$$d^p_r : \ensuremath{{E}^{p}_{r}}\to \ensuremath{{E}^{p+r}_{r}}$$

if $d^p_r(\bar{x}) = 0$ then $d (x) \in \ensuremath{{B}^{p+r}_{r}} + \ensuremath{{Z}^{p+r+1}_{r-1}}$, ie. $d(x) = d(y) + z,\,\, y \in \ensuremath{{Z}^{p+1}_{r-1}},\, z \in \ensuremath{{Z}^{p+r+1}_{r-1}}$. Put $u = x - y$, then $u \in K_p$ and $d(u) = z \in K_{p+r+1}$, ie. $u \in \ensuremath{{Z}^{p}_{r+1}}$. Hence $ker\, (d^p_r) = (\ensuremath{{Z}^{p}_{r+1}} + \ensuremath{{Z}^{p+1}_{r-1}}) / (\ensuremath{{B}^{p}_{r}}+ \ensuremath{{Z}^{p+1}_{r-1}}$).

On the other hand, $im (d^{p-r}_r) = (d\ensuremath{{Z}^{p-r}_{r}} + \ensuremath{{Z}^{p+1}_{r-1}}) / (\ensuremath{{B}^{p}_{r}}+ \ensuremath{{Z}^{p+1}_{r-1}})$.

$$% latex2html id marker 1129 \therefore ker\,(d^p_r) / im (d^... ...(d\ensuremath{{Z}^{p-r}_{r}} + \ensuremath{{Z}^{p+1}_{r-1}})).$$

Now, $d\ensuremath{{Z}^{p-r}_{r}} \subset \ensuremath{{Z}^{p}_{r+1}} \mbox{ and } \en... ...{{Z}^{p+1}_{r-1}} \cap \ensuremath{{Z}^{p}_{r+1}} = \ensuremath{{Z}^{p+1}_{r}}$. Hence,

$$H(E^p_r) \cong \ensuremath{{Z}^{p}_{r+1}} / (d\ensuremath{{Z}... ...r}} + \ensuremath{{Z}^{p+1}_{r}}) = \ensuremath{{E}^{p}_{r+1}}.$$

we are done.

Notations:

$$K_\infty = 0$$

$$K_{-\infty} = K$$

$$\ensuremath{{Z}^{p}_{\infty}} = \{ x \in K_p : dx = 0 \}$$

$$\ensuremath{{B}^{p}_{\infty}} = \{ dx \in K_p : x \in K \}$$

$$\ensuremath{{E}^{p}_{\infty}} = \ensuremath{{Z}^{p}_{\infty}} /(\ensuremath{{B}^{p}_{\infty}} + \ensuremath{{Z}^{p+1}_{\infty}})$$

$$\ensuremath{{E}^{}_{\infty}} = \amalg\, \ensuremath{{E}^{p}_{\infty}}$$

Theorem 4   Let $H(K)_p$ be the image of $H(K_p)$ in $H(K)$. Then $\{H(K)_p\}$ is a filtration of $H(K)$, and

$$E_\infty \cong G(H(K))$$

Proof:We have a commutative diagram:

$$\xymatrix{ \ensuremath{{Z}^{p}_{\infty}} \ar[r] & H (K_p) \a... ...} \ar[r]\ar[u] & H (K_{p+1}) \ar[r]\ar[u] & H(K)_{p+1}\ar[u] }$$

Hence a commutative diagram with exact rows:

$$\xymatrix{ \ensuremath{{Z}^{p}_{\infty}} \ar[rr]^{\ensuremat... ...{\ensuremath{\varphi_{p+1}}} && H(K)_{p+1}\ar[u] \ar[r] & 0 }$$

$\ensuremath{\varphi_{p}}(x) \in H(K)_{p+1} \iff \e... ...}.\,\,\therefore \ensuremath{{E}^{p}_{\infty}} \cong \frac{H(K)_p}{H(K)_{p+1}}$.