Convergence of spectral sequences of cohomological type - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:46:10Z http://mathoverflow.net/feeds/question/18429 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18429/convergence-of-spectral-sequences-of-cohomological-type Convergence of spectral sequences of cohomological type ferret 2010-03-16T21:38:53Z 2010-03-18T13:44:01Z <p>Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the <strong>homological situation</strong>.</p> <p>An <em>exact couple</em> consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that</p> <p><code>$$\begin{array}{rcl}A&amp;\xrightarrow{i}&amp;A\newline {\scriptsize k}\nwarrow&amp;&amp;\swarrow{\scriptsize j}\newline&amp;E&amp;\end{array}$$</code></p> <p>is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=j\circ k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $a\in \mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees</p> <p><code>$$\begin{array}{rcl}A^{(r)}&amp;\xrightarrow{(1,-1)}&amp;A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&amp;&amp;\swarrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&amp;E^{(r)}&amp;\end{array}$$</code></p> <p>induces a spectral sequence $E_{p,q}^{r+a}=E_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}\circ k^{(r)}$. Here ${\scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.</p> <p>Here is the canonical example. Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them: <code>$$\begin{array}{lcccccr} \to H_{p+q}(C_p,C_{p-1})&amp;\xrightarrow{k}&amp;H_{p+q-1}(C_{p-1})&amp;\xrightarrow{i}&amp;H_{p+q-1}(C_p)&amp;\xrightarrow{j}&amp;H_{p+q-1}(C_p,C_{p-1})\to\newline \to H_{p+q}(C_{p-1},C_{p-2})&amp;\xrightarrow{k}&amp;H_{p+q-1}(C_{p-2})&amp;\xrightarrow{i}&amp;H_{p+q-1}(C_{p-1})&amp;\xrightarrow{j}&amp;H_{p+q-1}(C_{p-1},C_{p-2})\to\newline \to H_{p+q}(C_{p-2},C_{p-3})&amp;\xrightarrow{k}&amp;H_{p+q-1}(C_{p-3})&amp;\xrightarrow{i}&amp;H_{p+q-1}(C_{p-2})&amp;\xrightarrow{j}&amp;H_{p+q-1}(C_{p-2},C_{p-3})\to \end{array}$$</code></p> <p>There is an exact couple as above with $a=1$ and $E_{p,q}=H_{p+q}(C_p,C_{p-1})$ and $A_{p,q}=H_{p+q}(C_p)$. Please note, that all the bigrades are correct. To follow $d^1:E_{p,q}^1\to E_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.</p> <p>Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.</p> <blockquote> <p>The spectral sequence $E^1(C_\bullet)$ of homological type converges to $H_{p+q}(C)$ if it is bounded (=only finitely many non-zero entries on every fixed diagonal $p+q$). One has $$E^\infty_{p,q}=i(H_{p+q}(C_p))/i(H_{p+q}(C_{p-1})).$$ ($i$ denotes the image in the colimit $H_{p+q}(C)$.)</p> </blockquote> <p><em>Proof.</em> Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple</p> <p><code>$$\begin{array}{c} \to E^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}_{p+r,q-r}\newline \xrightarrow{j^{r}}E^{r}_{p,q}\xrightarrow{k^{r}}\newline A^{r}_{p-1,q}\xrightarrow{i^{r}}A^{r}_{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to. \end{array}$$</code></p> <p>The first and the last term are $0$ because of the bounding assumption. We have $A_{p,q}^{r}=i(A^1_{p-r,q+r})$ and since $C_n$ is zero for $n&lt;0$, the last three terms are $0$. Exactness implies the result.</p> <p>Now the <strong>cohomological situation</strong>.</p> <p>Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H_n(\hom(C,\mathbb{Z}))$, thehomology of the dualized complex as one does in topology.</p> <p>Again we have many long exact sequences: <code>$$\begin{array}{lcccccr} \to H^{p+q}(C_p,C_{p-1})&amp;\xrightarrow{k}&amp;H^{p+q}(C_p)&amp;\xrightarrow{i}&amp;H^{p+q}(C_{p-1})&amp;\xrightarrow{j}&amp;H^{p+q+1}(C_p,C_{p-1})\to\newline \to H^{p+q}(C_{p+1},C_p)&amp;\xrightarrow{k}&amp;H^{p+q}(C_{p+1})&amp;\xrightarrow{i}&amp;H^{p+q}(C_{p})&amp;\xrightarrow{j}&amp;H^{p+q+1}(C_{p+1},C_{p})\to\newline \to H^{p+q}(C_{p+2},C_{p+1})&amp;\xrightarrow{k}&amp;H^{p+q}(C_{p+2})&amp;\xrightarrow{i}&amp;H^{p+q}(C_{p+1})&amp;\xrightarrow{j}&amp;H^{p+q+1}(C_{p+2},C_{p+1})\to \end{array}$$</code></p> <p>One can again follow $d^1:E_{p,q}^1\to E_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C_p)$ and $E^{p,q}=H^{p+q}(C_p,C_{p-1})$ and $a=1$.</p> <p>An exact couple with bidegrees</p> <p><code>$$\begin{array}{rcl}A_{(r)}&amp;\xrightarrow{(-1,1)}&amp;A_{(r)}\newline {\scriptsize (0,0)}\nwarrow&amp;&amp;\swarrow{\scriptsize (a+r,-(a-1+r))}\newline&amp;E_{(r)}&amp;\end{array}$$</code></p> <p>induces a spectral sequence $E_{r+a}=E_{(r)}$ of cohomological type with differentials $d_{r+a}=j_{(r)}\circ k_{(r)}$. Here ${\scriptsize something}_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.</p> <p>Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:</p> <blockquote> <p>The spectral sequence $E_1(C_\bullet)$ of cohomological type converges to $H^{p+q}(C)$ if it is bounded. One has $$E_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C_{p})).$$</p> </blockquote> <p><em>Proof?</em> Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple</p> <p><code>$$\begin{array}{c} \to E_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline \xrightarrow{j_{r}}E_{r}^{p,q}\xrightarrow{k_{r}}\newline A_{r}^{p,q}\xrightarrow{i_{r}}A_{r}^{p-1,q+1}\xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}\to. \end{array}$$</code></p> <p>The last term is $0$ because of the bounding assumption. We have $A_{r}^{p,q}=i(A^{p+r,q-r}_1)$ and since $C_n$ is zero for $n&lt;0$, the second and the third term are $0$. Exactness implies that <code>$$E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))\to i(H^{p+q}(C_{p+r-1}))).$$</code> I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:</p> <blockquote> <p><em>How does Lemma 1.2. of Hatcher's text works in the cohomological case?</em></p> </blockquote> http://mathoverflow.net/questions/18429/convergence-of-spectral-sequences-of-cohomological-type/18590#18590 Answer by Allen Hatcher for Convergence of spectral sequences of cohomological type Allen Hatcher 2010-03-18T13:44:01Z 2010-03-18T13:44:01Z <p>The lemma you refer to has two halves. The first half covers the case of homology and the second half covers the case of cohomology. Proofs are given for both halves (though the last sentence of the proof in the cohomology case requires a moment's thought to convince oneself of). The way to derive the cohomology spectral sequence from the second half is explained on the page following the lemma. Perhaps there is something that seems unclear in what is written there? Please feel free to ask me about this, though this might be better done in email.</p>