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Spectral sequences

Having computed $E_{pq}^2=\rH^h_q(\rH^v_{p,\bullet})$ of a bicomplex $C_{\bullet\bullet}$ it seems that we have used all data, that is vertical and horizontal differentials in the bicomplex. However, there is a piece of information which we can extract in addition to $E_{pq}^2$. We can define a homomorphism

\begin{displaymath}
d^2\: E_{pq}^2\to E_{p-2, q+1}^2
\end{displaymath}

as follows.

\begin{displaymath}
\xymatrix{
C_{p-2, q+1} & C_{p-1, q+1} \ar[l]_{d_h} \ar[d]_{d^v} & \\
& C_{p-1, q} & C_{pq} \ar[l]_{d^h}
}
\end{displaymath}

Using a horizontal cycle $x\in \rZ_p(C_{\bullet, q})$ we want to define an element in $C_{p-2, q-1}$ which represents an element in horizontal cycles of vertical homology complex, that is in $\rZ^h_p(\rH^v_q(C_{\bullet\bullet}))$. Our $x$ gives $[x]\in \rH_q^v(C_{\bullet\bullet})$. Using the induced map

\begin{displaymath}
d_*^h\:\rH^v_q(C_{p,\bullet}) \to \rH^v_q(C_{p-1, \bullet})
\end{displaymath}

we have $d_*^h([x])=0=[d^h(x)]$. Saying that the homology class is zero means that the cycle is in fact a boundary. Therefore there exists an $y\in C_{p-1, q+1}$ such that $d^v(y)=d^h(x)$. Now we define our cycle as $d^h(y)\in C_{p-2, q+1}$.

\begin{displaymath}
\xymatrix{
d^h(y) & y \ar[l]_{d_h} \ar[d]_{d^v} & \\
& d^v(y)=d^h(x) & x \ar[l]_{d^h}
}
\end{displaymath}

We claim that this element defines an element in $E^2_{p-2, q+1}$ which does not depend on the choice of $y$ nor on the choice of the representative of $[x]$. Thus we have defined

\begin{displaymath}
d^2\: E_{pq}^2\to E_{p-2, q+1}^2,\quad [x]\mapsto [d^h(y)].
\end{displaymath}

Furthermore $d^2\circ d^2=0$, so now we can take homology to obtain $E^3_{pq}$ and

\begin{displaymath}
d^3\: E_{pq}^2\to E_{p-3, q+2}^3.
\end{displaymath}

This procedure can be continued and as a result we get a sequence of arrays $E^r_{pq}$ for any $r\geq 2$ and maps

\begin{displaymath}
d^r\: E_{pq}^r\to E_{p-r, q+r-1}^r
\end{displaymath}

such that $E^r_{pq}$ is the homology of the complex $(E^{r-1}, d^{r-1})$ at the place $(p, q)$. Furthermore there are subspaces $B^r_{pq}$, $Z^r_{pq}$ of $C_{pq}$

\begin{displaymath}
B^2_{pq}\subseteq B^3_{pq} \subseteq \hdots \subseteq B^{\in...
...q \hdots \subseteq Z^2_{pq}
\subseteq Z^2_{pq}\subseteq C_{pq}
\end{displaymath}

such that $E^r_{pq}=Z^r_{pq}/B^r_{pq}$.

\begin{displaymath}
\xymatrix{
& \bullet & \bullet & \bullet & \bullet & \bullet...
...ullet & \bullet \\
\ar[uuuuuuuu]^q \ar[rrrrrrrr]_p &&&&&&&&
}
\end{displaymath}

When both differentials (leaving and entering) for $E_{pq}^r$ are zero, this component does not change furthermore and we have $E_{pq}^r=E_{pq}^{r+1}=\hdots$. We denote this stable component by $E_{pq}^{\infty}$.

There is a filtration on the total complex

\begin{displaymath}
\rF_p\Tot C_{\bullet\bullet} := \Tot \bigoplus_{k\leq p} C_{k\bullet},
\end{displaymath}


\begin{displaymath}
0\subseteq F_0\subseteq F_1\subseteq \hdots\subseteq F_{p-1}\subseteq F_p \subseteq \hdots\subseteq
\Tot C_{\bullet\bullet}.
\end{displaymath}

This filtration induces a filtration on $\rH_*(\Tot C_{\bullet\bullet})$

\begin{displaymath}
\rF_p:= \rF_p\rH_*(\Tot C_{\bullet\bullet}):=
\im(\rH_*(\rF_p\Tot C_{\bullet\bullet}) \to \rH_*(\Tot C_{\bullet\bullet}))
\end{displaymath}

Denote the quotient

\begin{displaymath}
F_p/F_{p-1} =: \Gr_p(\rH_{p+q}(\Tot C_{\bullet\bullet})).
\end{displaymath}

All data defined above, that is $\{E_{pq}^r, d^r\}_{p,q,r}$ and a filtration $\{F_p\}_p$ define a spectral sequence of a bicomplex $C_{\bullet\bullet}$. We say that the spectral sequence abuts to $\rH_*(\Tot C_{\bullet\bullet})$, which means that there is an isomorphism

\begin{displaymath}
E_{pq}^{\infty}\isom \Gr_p(\rH_{p+q}(\Tot C_{\bullet\bullet}))
\end{displaymath}

We write

\begin{displaymath}
E_{pq}^2=\rH_p^h(\rH^v_q(C_{\bullet\bullet}))\implies
\rH_{p+q}(\Tot C_{\bullet\bullet}).
\end{displaymath}

which is to read as: there is a spectral sequence starting at $E^2_{pq}$ and converging to $\rH_{p+q}(\Tot C_{\bullet\bullet})$
\begin{example}
The typical theorem using spectral sequences in algebraic topolo...
...tration. That is why these data are often omitted in the theorems.
\end{example}

\begin{example}
Let $X$\ be an $S^1$-space, $\E S^1$\ contractible space of path...
...ow{d^2} \rH_{n-2}(B)
\to \rH_{n-1}(E) \to \hdots.
\end{displaymath}\end{example}
Recall that for the bicomplex we took the vertical homology and then horizontal homology. We could have done it the other way. Any bicomplez gives a rise to two spectral sequences
\begin{align*}
E'^2_{pq} &= \rH_p^h(\rH^v_q(C_{\bullet\bullet})) \implies \rH_{p...
...{\bullet\bullet})) \implies \rH_{p+q}(\Tot(C_{\bullet\bullet})) \\
\end{align*}
But remark that the filtrations are different on $\Tot(C_{\bullet\bullet})$.
next up previous contents
Next: Cyclic homology Up: Cyclic category Previous: Bicomplexes   Contents
Pawel Witkowski 2006-11-07