Editing Spectral sequence (section)

== Examples of degeneration ==

=== The spectral sequence of a filtered complex, continued ===

Notice that we have a chain of inclusions:

:<math>Z_0^{p,q} \supe Z_1^{p,q} \supe Z_2^{p,q}\supe\cdots\supe B_2^{p,q} \supe B_1^{p,q} \supe B_0^{p,q}</math>

We can ask what happens if we define

:<math>Z_\infty^{p,q} = \bigcap_{r=0}^\infty Z_r^{p,q},</math>
:<math>B_\infty^{p,q} = \bigcup_{r=0}^\infty B_r^{p,q},</math>
:<math>E_\infty^{p,q} = \frac{Z_\infty^{p,q}}{B_\infty^{p,q}+Z_\infty^{p+1,q-1}}.</math>

<math>E_\infty^{p,q}</math> is a natural candidate for the abutment of this spectral sequence. Convergence is not automatic, but happens in many cases. In particular, if the filtration is finite and consists of exactly ''r'' nontrivial steps, then the spectral sequence degenerates after the ''r''th sheet. Convergence also occurs if the complex and the filtration are both bounded below or both bounded above.

To describe the abutment of our spectral sequence in more detail, notice that we have the formulas:

:<math>Z_\infty^{p,q} = \bigcap_{r=0}^\infty Z_r^{p,q} = \bigcap_{r=0}^\infty \ker(F^p C^{p+q} \rightarrow C^{p+q+1}/F^{p+r} C^{p+q+1})</math>
:<math>B_\infty^{p,q} = \bigcup_{r=0}^\infty B_r^{p,q} = \bigcup_{r=0}^\infty (\mbox{im } d^{p,q-r} : F^{p-r} C^{p+q-1} \rightarrow C^{p+q}) \cap F^p C^{p+q}</math>

To see what this implies for <math>Z_\infty^{p,q}</math> recall that we assumed that the filtration was separated. This implies that as ''r'' increases, the kernels shrink, until we are left with <math>Z_\infty^{p,q} = \ker(F^p C^{p+q} \rightarrow C^{p+q+1})</math>.  For <math>B_\infty^{p,q}</math>, recall that we assumed that the filtration was exhaustive. This implies that as ''r'' increases, the images grow until we reach <math>B_\infty^{p,q} = \text{im }(C^{p+q-1} \rightarrow C^{p+q}) \cap F^p C^{p+q}</math>. We conclude

:<math>E_\infty^{p,q} = \mbox{gr}_p H^{p+q}(C^\bull)</math>,

that is, the abutment of the spectral sequence is the ''p''th graded part of the ''(p+q)''th homology of ''C''. If our spectral sequence converges, then we conclude that:

:<math>E_r^{p,q} \Rightarrow_p H^{p+q}(C^\bull)</math>

==== Long exact sequences ====

Using the spectral sequence of a filtered complex, we can derive the existence of [[long exact sequence]]s. Choose a short exact sequence of cochain complexes 0 → {{var|A}}{{i sup|•}} → {{var|B}}{{i sup|•}} → {{var|C}}{{i sup|•}} → 0, and call the first map {{var|f}}{{i sup|•}} : {{var|A}}{{i sup|•}} → {{var|B}}{{i sup|•}}. We get natural maps of homology objects ''H<sup>n</sup>''({{var|A}}{{i sup|•}}) → ''H<sup>n</sup>''({{var|B}}{{i sup|•}}) → ''H<sup>n</sup>''({{var|C}}{{i sup|•}}), and we know that this is exact in the middle. We will use the spectral sequence of a filtered complex to find the connecting homomorphism and to prove that the resulting sequence is exact.To start, we filter {{mvar|B}}{{i sup|•}}:

:<math>F^0 B^n = B^n</math>
:<math>F^1 B^n = A^n</math>
:<math>F^2 B^n = 0</math>

This gives:

:<math>E^{p,q}_0
= \frac{F^p B^{p+q}}{F^{p+1} B^{p+q}} = \begin{cases}
0 & \text{if } p < 0 \text{ or } p > 1 \\
C^q & \text{if } p = 0 \\
A^{q+1} & \text{if } p = 1 \end{cases}</math>
:<math>E^{p,q}_1
= \begin{cases}
0 & \text{if } p < 0 \text{ or } p > 1 \\
H^q(C^\bull) & \text{if } p = 0 \\
H^{q+1}(A^\bull) & \text{if } p = 1 \end{cases}</math>

The differential has bidegree (1, 0), so ''d''<sub>0,''q''</sub> : ''H<sup>q</sup>''({{var|C}}{{i sup|•}}) → ''H''<sup>''q''+1</sup>({{var|A}}{{i sup|•}}). These are the connecting homomorphisms from the [[snake lemma]], and together with the maps {{mvar|A}}{{i sup|•}} → {{mvar|B}}{{i sup|•}} → {{mvar|C}}{{i sup|•}}, they give a sequence:

:<math>\cdots\rightarrow H^q(B^\bull) \rightarrow H^q(C^\bull) \rightarrow H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) \rightarrow\cdots</math>

It remains to show that this sequence is exact at the ''A'' and ''C'' spots.  Notice that this spectral sequence degenerates at the ''E''<sub>2</sub> term because the differentials have bidegree (2, &minus;1). Consequently, the ''E''<sub>2</sub> term is the same as the ''E''<sub>∞</sub> term:

:<math>E^{p,q}_2
\cong \text{gr}_p H^{p+q}(B^\bull)
= \begin{cases}
0 & \text{if } p < 0 \text{ or } p > 1 \\
H^q(B^\bull)/H^q(A^\bull) & \text{if } p = 0 \\
\text{im } H^{q+1}f^\bull : H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) &\text{if } p = 1 \end{cases}</math>

But we also have a direct description of the ''E''<sub>2</sub> term as the homology of the ''E''<sub>1</sub> term. These two descriptions must be isomorphic:

:<math> H^q(B^\bull)/H^q(A^\bull) \cong \ker d^1_{0,q} : H^q(C^\bull) \rightarrow H^{q+1}(A^\bull)</math>
:<math> \text{im } H^{q+1}f^\bull : H^{q+1}(A^\bull) \rightarrow H^{q+1}(B^\bull) \cong H^{q+1}(A^\bull) / (\mbox{im } d^1_{0,q} : H^q(C^\bull) \rightarrow H^{q+1}(A^\bull))</math>

The former gives exactness at the ''C'' spot, and the latter gives exactness at the ''A'' spot.

=== The spectral sequence of a double complex, continued ===

Using the abutment for a filtered complex, we find that:

:<math>H^\textrm{I}_p(H^\textrm{II}_q(C_{\bull,\bull})) \Rightarrow_p H^{p+q}(T(C_{\bull,\bull}))</math>
:<math>H^\textrm{II}_q(H^\textrm{I}_p(C_{\bull,\bull})) \Rightarrow_q H^{p+q}(T(C_{\bull,\bull}))</math>

In general, ''the two gradings on {{tmath|H^{p+q}(T(C_{\bull,\bull}))}} are distinct''.  Despite this, it is still possible to gain useful information from these two spectral sequences.

==== Commutativity of Tor ====

Let ''R'' be a ring, let ''M'' be a right ''R''-module and ''N'' a left ''R''-module. Recall that the derived functors of the tensor product are denoted [[Tor functor|Tor]]. Tor is defined using a [[projective resolution]] of its first argument. However, it turns out that <math>\operatorname{Tor}_i(M,N) =\operatorname{Tor}_i(N,M)</math>. While this can be verified without a spectral sequence, it is very easy with spectral sequences.

Choose projective resolutions <math>P_\bull</math> and <math>Q_\bull</math> of ''M'' and ''N'', respectively. Consider these as complexes which vanish in negative degree having differentials ''d'' and ''e'', respectively. We can construct a double complex whose terms are <math>C_{i,j} = P_i \otimes Q_j</math> and whose differentials are <math>d \otimes 1</math> and <math>(-1)^\textrm{I}(1 \otimes e)</math>. (The factor of &minus;1 is so that the differentials anticommute.) Since projective modules are flat, taking the tensor product with a projective module commutes with taking homology, so we get:

:<math>H^\textrm{I}_p(H^\textrm{II}_q(P_\bull \otimes Q_\bull)) = H^\textrm{I}_p(P_\bull \otimes H^\textrm{II}_q(Q_\bull))</math>
:<math>H^\textrm{II}_q(H^\textrm{I}_p(P_\bull \otimes Q_\bull)) = H^\textrm{II}_q(H^\textrm{I}_p(P_\bull) \otimes Q_\bull)</math>

Since the two complexes are resolutions, their homology vanishes outside of degree zero. In degree zero, we are left with

:<math>H^\textrm{I}_p(P_\bull \otimes N) = \operatorname{Tor}_p(M,N)</math>
:<math>H^\textrm{II}_q(M \otimes Q_\bull) = \operatorname{Tor}_q(N,M)</math>

In particular, the <math>E^2_{p,q}</math> terms vanish except along the lines ''q'' = 0 (for the {{rn|I}} spectral sequence) and ''p'' = 0 (for the {{rn|II}} spectral sequence). This implies that the spectral sequence degenerates at the second sheet, so the ''E''<sup>∞</sup> terms are isomorphic to the ''E''<sup>2</sup> terms:

:<math>\operatorname{Tor}_p(M,N) \cong E^\infty_p = H_p(T(C_{\bull,\bull}))</math>
:<math>\operatorname{Tor}_q(N,M) \cong E^\infty_q = H_q(T(C_{\bull,\bull}))</math>

Finally, when ''p'' and ''q'' are equal, the two right-hand sides are equal, and the commutativity of Tor follows.