Editing Spectral sequence (section)

==== 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.