Loading web-font TeX/Math/Italic

The Stacks project

15.62 Spectral sequences for Tor

In this section we collect various spectral sequences that come up when considering the Tor functors.

Example 15.62.1. Let R be a ring. Let K_\bullet be a chain complex of R-modules with K_ n = 0 for n \ll 0. Let M be an R-module. Choose a resolution P_\bullet \to M of M by free R-modules. We obtain a double chain complex K_\bullet \otimes _ R P_\bullet . Applying the material in Homology, Section 12.25 (especially Homology, Lemma 12.25.3) translated into the language of chain complexes we find two spectral sequences converging to H_*(K_\bullet \otimes _ R^\mathbf {L} M). Namely, on the one hand a spectral sequence with E_2-page

(E_2)_{i, j} = \text{Tor}^ R_ j(H_ i(K_\bullet ), M) \Rightarrow H_{i + j}(K_\bullet \otimes ^{\mathbf{L}}_ R M)

and differential d_2 given by maps \text{Tor}^ R_ j(H_ i(K_\bullet ), M) \to \text{Tor}^ R_{j - 2}(H_{i + 1}(K_\bullet ), M). Another spectral sequence with E_1-page

(E_1)_{i, j} = \text{Tor}^ R_ j(K_ i, M) \Rightarrow H_{i + j}(K_\bullet \otimes ^{\mathbf{L}}_ R M)

with differential d_1 given by maps \text{Tor}^ R_ j(K_ i, M) \to \text{Tor}^ R_ j(K_{i - 1}, M) induced by K_ i \to K_{i - 1}.

Example 15.62.2. Let R \to S be a ring map. Let M be an R-module and let N be an S-module. Then there is a spectral sequence

\text{Tor}^ S_ n(\text{Tor}^ R_ m(M, S), N) \Rightarrow \text{Tor}^ R_{n + m}(M, N).

To construct it choose a R-free resolution P_\bullet of M. Then we have

M \otimes _ R^{\mathbf{L}} N = P^\bullet \otimes _ R N = (P^\bullet \otimes _ R S) \otimes _ S N

and then apply the first spectral sequence of Example 15.62.1.

Example 15.62.3. Consider a commutative diagram

\xymatrix{ B \ar[r] & B' = B \otimes _ A A' \\ A \ar[r] \ar[u] & A' \ar[u] }

and B-modules M, N. Set M' = M \otimes _ A A' = M \otimes _ B B' and N' = N \otimes _ A A' = N \otimes _ B B'. Assume that A \to B is flat and that M and N are A-flat. Then there is a spectral sequence

\text{Tor}^ A_ i(\text{Tor}_ j^ B(M, N), A') \Rightarrow \text{Tor}^{B'}_{i + j}(M', N')

The reason is as follows. Choose free resolution F_\bullet \to M as a B-module. As B and M are A-flat we see that F_\bullet \otimes _ A A' is a free B'-resolution of M'. Hence we see that the groups \text{Tor}^{B'}_ n(M', N') are computed by the complex

(F_\bullet \otimes _ A A') \otimes _{B'} N' = (F_\bullet \otimes _ B N) \otimes _ A A' = (F_\bullet \otimes _ B N) \otimes ^{\mathbf{L}}_ A A'

the last equality because F_\bullet \otimes _ B N is a complex of flat A-modules as N is flat over A. Hence we obtain the spectral sequence by applying the spectral sequence of Example 15.62.1.

Example 15.62.4. Let K^\bullet , L^\bullet be objects of D^{-}(R). Then there is a spectral sequence with

E_2^{p, q} = H^ p(K^\bullet \otimes _ R^{\mathbf{L}} H^ q(L^\bullet )) \Rightarrow H^{p + q}(K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet )

and another spectral sequence with

E_2^{p, q} = H^ p(H^ q(K^\bullet ) \otimes _ R^{\mathbf{L}} L^\bullet ) \Rightarrow H^{p + q}(K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet )

Both spectral sequences have d_2^{p, q} : E_2^{p, q} \to E_2^{p + 2, q - 1}. After replacing K^\bullet and L^\bullet by bounded above complexes of projectives, these spectral sequences are simply the two spectral sequences for computing the cohomology of \text{Tot}(K^\bullet \otimes L^\bullet ) discussed in Homology, Section 12.25.


Comments (2)

Comment #5790 by Brad Dirks on

Is there a typo in the indices for Example 061Z? Homological spectral sequences have with bidegree , so should go from to . This would be a map . It seems that the page in the same example does have the correct indices.

Comment #5805 by on

I think you are right about the bidegree of being but I think you swapped the roles of and . So I think the differential for the first spectral squence of the example on the -page is given by maps which is what it says.


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.