# Monday, March 22 ## 5.4: Spectral Sequence of a Filtration :::{.remark} We have an increasing filtration $F_p C \subseteq F_{p+1}C$, where we defined \[ E_{p, q}^0 = { F_p C_{p+q} \over F_{p-1} C_{p+1} } && E_{p,q}^1 = H_{p+q} E_{p, *}^0 .\] 1. We have a map \[ \eta_p: F_p C \surjects {F_p C \over F_{p-1}C } = E_p^0 ,\] where we've dropped the $q$ from notation. 2. \[ A_{p, q}^r = \ts{ c \in C_p C \st dc \in F_{p-1} C } ,\] the eventual cycles. We defined $Z_p^r = \eta_p A_p^r$ and $B_p^r = \eta_p dA_{p+r-1}^{r-1}$, and wrote $A_p^r \intersect F_{p-1}C = A_{p-1}^{r-1}$. 3. We had the chain of inclusions \[ 0 = B_p^r \subseteq \cdots \subseteq B_p^{\infty} \subset Z_p^{\infty } \subset \cdots \subseteq Z_p^1 = E_p^) .\] 4. We also have $E_p^r = Z_p^r/B_p^r = A_p^r / dA_{p+r-1}^{r-1} + A_{p-1}^{r-1}$ 5. $Z_{p}^{r+1}/B_pr \cong {A_p^{r+1} +A_{p-1}^{r-1} \over dA_{p+r-1} ^{r-1} + A_{p-1}^{r-1}}$. 6. $dA_p^r \intersect F_{p-r-1} C = dA_P^{r+1}$. \todo[inline]{See video for missed spoken details!} Obviously we have \[ d: A_p^r &\to A_{p-r}^{r} \\ d: A_{p-1}^r &\to dA_{p-1}^{r-1} ,\] so $d$ induces a well-defined map $d_p^r: E_p^r \mapsvia{} E_{p-r}^r$, which of course squares to zero, which goes $r$ columns to the left and decreases the total degree $n$ by 1 since the original $d$ did on $C_n$. This is what we need to set up a spectral sequence, since we now have pages and differentials, and it just remains to show that $E^{r+1} \cong H_*(E^r, d^r)$. ::: :::{.lemma title="?"} $d$ determines isomorphisms $Z_{p}^r/Z_p^{r+1} \mapsvia{\sim} B_{p-r}^{r+1} / B_{p-r}^r$. ::: :::{.proof title="?"} Unwind definitions! Note that we have $B_{p-r}^{r+1} = \eta_{p-r} dA_p^r$, using that the lower index on $B$ and upper index on $A$ should sum to the lower index on $A$. This is equal to $dA_p^r / dA_p^r \intersect F_{p-r-1} C$, where the latter term is $\ker\eta_{p-r}$ and $B_{p-r}^r = \eta_{p-r} dA_{p-1}^{r-1}$. This yields \[ {B^{r+1}_{p-r} \over B_{p-r}^r} \cong { dA_{p}^r \over dA_{p-1}^{r-1} + (dA_p^r \intersect F_{p-r-1} C) } .\] Similarly, \[ {Z_p^r \over Z_p^{r+1}} \da { \eta_p A_p^r \over \eta_p A_p^{r+1} } \cong {A_p^r \over A_p^{r+1} + (A_p^r \intersect F_{p-1} C )} \congbecause{(3)} {A_p^r \over A_p^{r+1} + A_{p-1}^{r-1} } .\] Now applying the map induced by $d: A_p^r \to F_{p-r}C$ to this quotient, we have $\ker \ro{d}{A_p^r} \subseteq A_p^{r+1}$. These go down $r$ steps, but everything in the kernel goes down as far as you'd like! So $d$ kills one of the denominator terms, and thus induces an injective map on the quotient. Thus \({Z_p^r \over Z_p^{r+1}} \mapsvia{\sim} {dA_p^r \over dA_p^{r+1} + dA_{p-1}^{r-1} } \), which is exactly the previous expression with the order switched, so this is isomorphic to $B_{p-r}^{r+1} / B_{p-r}^r$. ::: :::{.proposition title="The $r+1$st page is the homology of the $r$th page"} \[ { \ker d_p^r \over \im d_{p+r}^r } \cong E_p^{r+1} \da {Z_p^{r+1} \over B_p^{r+1} } .\] ::: :::{.proof title="?"} Recall that $d_p^r: E_p^r \to E_{p-r}^r$ and by (4), $E_p^r \cong {A_p^r \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1}}$. Substituting $p \mapsfrom p-r$, we have \[ \ker d_p^r = { \ts{ z\in A_p^r \st dz \in dA_{p-1}^{r-1} + A_{p-r-1}^{r-1} } \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1} } = { A_{p-1}^{r-1} + A_{p}^{r+1} \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1} } \congbecause{(5)} {Z_p^{r+1} \over B_p^r} && \text{which is } (6) .\] Here we've used that $x\in F_p C\implies dx \in F_{p-r-1} C \implies dx\in A^{?}_{p-r-1}$. What is the image of $d_p^r$ in general? Note that later we can replace $p\mapsfrom p+r$. By the 1st isomorphism theorem, we have \[ d_p^r: E_p^r = Z_p^r / B_p^r \mapsvia{\sim} {Z_p^r / B_p^r \over Z_p^{r+1} / B_p^r} \mapsvia{\sim} {Z_p^r \over Z_p^{r+1}} \mapsvia{d} {B_{p-r}^{r+1} \over B_{p-r}^r} \injects {Z_{p-r}^r \over B_{p-r}^r} = E_{p-r}^r ,\] where we've applied the lemma from last time, and we've used the fact that in the last map, all of the $B$ are contained in all of the $Z$, so we can choose any superscript we want. These are all isomorphisms up until the last part, so \[ \im d_p^r \cong B_{p-r}^{r+1} / B_{p-r}^{r+1} .\]. Replacing $p\mapsfrom p+r$, we get a 7th fact :::{.fact title="7"} \[ \im d_{p+r}^r \cong B_{p}^{r+1} / B_{p}^{r+1} .\] ::: Now combining (6) and (7), we have \[ {\ker d_p^r \over \im d_{p+r}^{r} } \mapsvia{\sim} {Z_p^{r+1} / B_p^r \over B_{p}^{r+1} / B_p^r } \cong {Z_p^{r+1} \over B_p^{r+1}} = E_p^{r+1} .\] ::: ## 5.5: Convergence of the Spectral Sequence of a Filtration :::{.remark} We'll restrict our attention to bounded complexes. ::: :::{.remark} A filtration $F$ on a chain complex $C$ induces a filtration on the homology $H_*C$, where $H_p H_n C = \im ( H_n F_p C \to H_n C)$: \begin{tikzcd} {F_p C_{n+1}} && {C_{n+1}} \\ \\ {F_p C_{n}} && {C_{n}} \\ \\ {F_p C_{n-1}} && {C_{n-1}} \arrow[hook, from=1-1, to=1-3] \arrow[hook, from=3-1, to=3-3] \arrow[hook, from=5-1, to=5-3] \arrow["d"{description}, from=1-1, to=3-1] \arrow["d"{description}, from=3-1, to=5-1] \arrow["d"{description}, from=1-3, to=3-3] \arrow["d"{description}, from=3-3, to=5-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJGX3AgQ197bisxfSJdLFsyLDAsIkNfe24rMX0iXSxbMiwyLCJDX3tufSJdLFsyLDQsIkNfe24tMX0iXSxbMCwyLCJGX3AgQ197bn0iXSxbMCw0LCJGX3AgQ197bi0xfSJdLFswLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzQsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDQsImQiLDFdLFs0LDUsImQiLDFdLFsxLDIsImQiLDFdLFsyLDMsImQiLDFdXQ==) \todo[inline]{See video for missed details.} These inclusions induce a map from the homology of the subcomplex to the homology of the total complex. ::: :::{.remark} If the filtration on $C$ is bounded, say $0 = F_s C_n \subseteq \cdots \subseteq F_t C_n = C_n$ for some $s