Editing Spectral sequence (section)

== Worked-out examples ==
=== First-quadrant sheet ===
Consider a spectral sequence where <math>E_r^{p,q}</math> vanishes for all <math> p </math> less than some <math> p_0 </math> and for all <math> q </math> less than some <math> q_0 </math>.  If <math> p_0 </math> and <math> q_0 </math> can be chosen to be zero, this is called a '''first-quadrant spectral sequence'''.
The sequence abuts because <math> E_{r+i}^{p,q} = E_r^{p,q} </math> holds for all <math> i\geq 0 </math> if <math> r>p </math> and <math> r>q+1 </math>. To see this, note that either the domain or the codomain of the differential is zero for the considered cases. In visual terms, the sheets stabilize in a growing rectangle (see picture above). The spectral sequence need not degenerate, however, because the differential maps might not all be zero at once. Similarly, the spectral sequence also converges if <math>E_r^{p,q}</math> vanishes for all <math> p </math> greater than some <math> p_0 </math> and for all <math> q </math> greater than some <math> q_0 </math>.

=== 2 non-zero adjacent columns ===
Let <math>E^r_{p, q}</math> be a homological spectral sequence such that <math>E^2_{p, q} = 0</math> for all ''p'' other than 0, 1. Visually, this is the spectral sequence with <math>E^2</math>-page
:<math>\begin{matrix}
& \vdots & \vdots & \vdots & \vdots & \\
\cdots & 0 & E^2_{0,2} & E^2_{1,2} & 0 & \cdots \\
\cdots & 0 & E^2_{0,1} & E^2_{1,1} & 0 & \cdots \\
\cdots & 0 & E^2_{0,0} & E^2_{1,0} & 0 & \cdots \\
\cdots & 0 & E^2_{0,-1} & E^2_{1,-1} & 0 & \cdots \\
& \vdots & \vdots & \vdots & \vdots & 
\end{matrix}</math>
The differentials on the second page have degree (-2, 1), so they are of the form
:<math>d^2_{p,q}:E^2_{p,q} \to E^2_{p-2,q+1}</math>
These maps are all zero since they are
:<math>d^2_{0,q}:E^2_{0,q} \to 0</math>, <math>d^2_{1,q}:E^2_{1,q} \to 0</math>
hence the spectral sequence degenerates: <math>E^{\infty} = E^2</math>. Say, it converges to <math>H_*</math> with a filtration
:<math>0 = F_{-1} H_n \subset F_0 H_n \subset \dots \subset F_n H_n = H_n</math>
such that <math>E^{\infty}_{p, q} = F_p H_{p+q}/F_{p-1} H_{p+q}</math>. Then <math>F_0 H_n = E^2_{0, n}</math>, <math>F_1 H_n / F_0 H_n = E^2_{1, n -1}</math>, <math>F_2 H_n / F_1 H_n = 0</math>, <math>F_3 H_n / F_2 H_n = 0</math>, etc. Thus, there is the exact sequence:<ref>{{harvnb|Weibel|1994|loc=Exercise 5.2.1.}}; there are typos in the exact sequence, at least in the 1994 edition.</ref>
:<math>0 \to E^2_{0, n} \to H_n \to E^2_{1, n - 1} \to 0</math>.
Next, let <math>E^r_{p, q}</math> be a spectral sequence whose second page consists only of two lines ''q'' = 0, 1. This need not degenerate at the second page but it still degenerates at the third page as the differentials there have degree (-3, 2). Note <math>E^3_{p, 0} = \operatorname{ker} (d: E^2_{p, 0} \to E^2_{p - 2, 1})</math>, as the denominator is zero. Similarly, <math>E^3_{p, 1} = \operatorname{coker}(d: E^2_{p+2, 0} \to E^2_{p, 1})</math>. Thus,
:<math>0 \to E^{\infty}_{p, 0} \to E^2_{p, 0} \overset{d}\to E^2_{p-2, 1} \to E^{\infty}_{p-2, 1} \to 0</math>.
Now, say, the spectral sequence converges to ''H'' with a filtration ''F'' as in the previous example. Since <math>F_{p-2} H_{p} / F_{p-3} H_{p} = E^{\infty}_{p-2, 2} = 0</math>, <math>F_{p-3} H_p / F_{p-4} H_p = 0</math>, etc., we have: <math>0 \to E^{\infty}_{p - 1, 1} \to H_p \to E^{\infty}_{p, 0} \to 0</math>. Putting everything together, one gets:<ref>{{harvnb|Weibel|1994|loc=Exercise 5.2.2.}}</ref>
:<math>\cdots \to H_{p+1} \to E^2_{p + 1, 0} \overset{d}\to E^2_{p - 1, 1} \to H_p \to E^2_{p, 0} \overset{d}\to E^2_{p - 2, 1} \to H_{p-1} \to \dots.</math>

=== Wang sequence ===
The computation in the previous section generalizes in a straightforward way. Consider a [[fibration]] over a sphere:
:<math>F \overset{i}\to E \overset{p}\to S^n</math>
with ''n'' at least 2. There is the [[Serre spectral sequence]]:
:<math>E^2_{p, q} = H_p(S^n; H_q(F)) \Rightarrow H_{p+q}(E)</math>;
that is to say, <math>E^{\infty}_{p, q} = F_p H_{p+q}(E)/F_{p-1} H_{p+q}(E)</math> with some filtration <math>F_\bullet</math>.

Since <math>H_p(S^n)</math> is nonzero only when ''p'' is zero or ''n'' and equal to '''Z''' in that case, we see <math>E^2_{p, q}</math> consists of only two lines <math>p = 0,n</math>, hence the <math>E^2</math>-page is given by
:<math>\begin{matrix}
& \vdots & \vdots & \vdots &  & \vdots  & \vdots & \vdots & \\
\cdots & 0 & E^2_{0,2} & 0 & \cdots & 0 & E^2_{n,2} & 0 & \cdots \\
\cdots & 0 & E^2_{0,1} & 0 & \cdots & 0  & E^2_{n,1} & 0 & \cdots \\
\cdots & 0 & E^2_{0,0} & 0 & \cdots & 0  & E^2_{n,0} & 0 & \cdots \\
\end{matrix}</math>
Moreover, since
:<math>E^2_{p, q} = H_p(S^n;H_q(F)) = H_q(F)</math>
for <math>p = 0,n</math> by the [[universal coefficient theorem]], the <math>E^2</math> page looks like
:<math>\begin{matrix}
& \vdots & \vdots & \vdots &  & \vdots  & \vdots & \vdots & \\
\cdots & 0 & H_2(F) & 0 & \cdots & 0 & H_2(F) & 0 & \cdots \\
\cdots & 0 & H_1(F) & 0 & \cdots & 0  & H_1(F) & 0 & \cdots \\
\cdots & 0 & H_0(F) & 0 & \cdots & 0  & H_0(F) & 0 & \cdots \\
\end{matrix}</math>
Since the only non-zero differentials are on the <math>E^n</math>-page, given by
:<math>d^n_{n,q}:E^n_{n,q} \to E^n_{0,q+n-1}</math>
which is
:<math>d^n_{n,q}:H_q(F) \to H_{q+n-1}(F)</math>
the spectral sequence converges on <math>E^{n+1} = E^{\infty}</math>. By computing <math>E^{n+1}</math> we get an exact sequence
:<math>0 \to E^{\infty}_{n, q-n} \to E^n_{n, q-n} \overset{d}\to E^n_{0, q-1} \to E^{\infty}_{0, q-1} \to 0.</math>
and written out using the homology groups, this is
:<math>0 \to E^{\infty}_{n, q-n} \to H_{q-n}(F) \overset{d}\to H_{q-1}(F) \to E^{\infty}_{0, q-1} \to 0.</math>
To establish what the two <math>E^\infty</math>-terms are, write <math>H = H(E)</math>, and since <math>F_1 H_q/F_0 H_q = E^{\infty}_{1, q - 1} = 0</math>, etc., we have: <math>E^{\infty}_{n, q-n} = F_n H_q / F_0 H_q</math> and thus, since <math>F_n H_q = H_q</math>,
:<math>0 \to E^{\infty}_{0, q} \to H_q \to E^{\infty}_{n, q - n} \to 0.</math>
This is the exact sequence
:<math>0 \to H_q(F) \to H_q(E) \to H_{q-n}(F)\to 0.</math>
Putting all calculations together, one gets:<ref>{{harvnb|Weibel|1994|loc=Application 5.3.5.}}</ref>
:<math>\dots \to H_q(F) \overset{i_*}\to H_q(E) \to H_{q-n}(F) \overset{d}\to H_{q-1}(F) \overset{i_*}\to H_{q-1}(E) \to H_{q-n -1}(F) \to \dots</math>
(The [[Gysin sequence]] is obtained in a similar way.)

=== Low-degree terms ===
With an obvious notational change, the type of the computations in the previous examples can also be carried out for cohomological spectral sequence. Let <math>E_r^{p, q}</math> be a first-quadrant spectral sequence converging to ''H'' with the decreasing filtration
:<math>0 = F^{n+1} H^n \subset F^n H^n \subset \dots \subset F^0 H^n = H^n</math>
so that <math>E_{\infty}^{p,q} = F^p H^{p+q}/F^{p+1} H^{p+q}.</math>
Since <math>E_2^{p, q}</math> is zero if ''p'' or ''q'' is negative, we have:
:<math>0 \to E^{0, 1}_{\infty} \to E^{0, 1}_2 \overset{d}\to E^{2, 0}_2 \to E^{2, 0}_{\infty} \to 0.</math>
Since <math>E_{\infty}^{1, 0} = E_2^{1, 0}</math> for the same reason and since <math>F^2 H^1 = 0,</math>
:<math>0 \to E_2^{1, 0} \to H^1 \to E^{0, 1}_{\infty} \to 0</math>.
Since <math>F^3 H^2 = 0</math>, <math>E^{2, 0}_{\infty} \subset H^2</math>. Stacking the sequences together, we get the so-called [[five-term exact sequence]]:
:<math>0 \to E^{1, 0}_2 \to H^1 \to E^{0, 1}_2 \overset{d}\to E^{2, 0}_2 \to H^2.</math>