E 2 01 H 1 E (2) Formulate and prove an analogous statement for a first quadrant cohomological spectral sequence.
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1 Josh Swanson Math 583 Spring 014 Group Cohomology Homework 1 May nd, 014 Problem 1 (1) Let E pq H p+q be a first quadrant (homological) spectral sequence converging to H. Show that there is an exact sequence ( The five-term exact sequence ): H E 0 d E 01 H 1 E () Formulate and prove an analogous statement for a first quadrant cohomological spectral sequence. For (1), let δ = d 0 : E 0 E 01. It s easy to see that if we take homology at the second page, the third page is E 01/ im δ E 00 E 10 ker δ whereupon the four indicated terms stabilize. Since this is a first quadrant spectral sequence, it s also easy to see the relationship between these E pq terms and the filtration on H p+q is E 01 = F 0 H 1 E 00 = H 0 E 10 = H 1 /F 0 H 1 E 0 = H /F 1 H Hence we have a sequence H H /F 1 H = ker δ E 0 δ E 01 E 01/ im δ = F 0 H 1 H 1 H 1 /F 0 H 1 = E 10 0 One can quickly check exactness of the induced five-term sequence, so the result follows. For (), let δ = d 01 : E 01 E 0. The third page is ker δ E 00 E 10 E 0 / im δ which again collapses, and the E pq terms relate to the filtration on H via E 01 = H 1 /F 1 H 1 E 00 = H 0 E 10 = F 1 H 1 E 0 = F H 1
2 Hence we have a sequence 0 E 10 = F 1 H 1 H 1 H 1 /F 1 H 1 = ker δ E 01 δ E 0 E 0 / im δ = F H H which induces the exact sequence 0 E 10 H 1 E 01 E 0 H. Problem Let 0 A B C 0 be a short exact sequence of complexes. Using spectral sequences, show that there is an exact sequence in homology: H n+1 (C ) H n (A ) H n (B ) H n (C ) H n 1 (A ) Consider the double complex whose bottom two rows are 0 C 1 B 1 A C 0 B 0 A 0 0 (where as usual we ve toggled the sign on the B column s maps). Taking horizontal homology gives 0 s everywhere since the rows are exact, so II E r pq collapses to 0 at r = 1, forcing the abutment to be trivial. Hence I E r pq 0. Take vertical homology to get I E 1 pq as β 1 α 1 0 H 1 (C ) H 1 (B ) H 1 (A ) 0 β 0 α 0 0 H 0 (C ) H 0 (B ) H 0 (A ) 0 It s easy to see the B column stabilizes on the next page, so it must stabilize at 0, i.e. the above is exact at H n (B ). Take horizontal homology to get I E pq as H 1 (C )/ im β 1 0 ker α 1 H 0 (C )/ im β 0 0 ker α 0 It s easy to see the C and A columns stabilize on the next page, so the above is exact at ker α 0 and H 1 (C )/ im β 1, i.e. the connecting map is an isomorphism. Hence we have a sequence H 1 (A ) α1 H 1 (B ) β1 H 1 (C ) H 1 (C )/ im β 1 = ker α0 H 0 (A ) α0 which gives the desired long exact sequence.
3 Problem 3 Prove a subtler version of the 5-lemma: namely, what are the minimal conditions you need to put on the following commutative diagram with exact rows to conclude that γ is injective? What about surjective? A B C D E α β γ δ ɛ A B C D E Consider the diagram as a double complex by flipping it horizontally and toggling the signs of the second and fourth vertical arrows without loss of generality We find II E 1 pq is where represents the kernel or cokernel of the appropriate map. The remaining pages do not change the n = and n = 3 antidiagonals, hence the filtration on these pieces of the abutment H n are trivial, so H = H 3 = 0. In particular I E pq = 0 for p + q = n =, 3. Now compute I E 1 pq: ker ɛ ker δ ker γ ker β ker α coker ɛ coker δ coker γ coker β coker α If ker δ = ker β = coker α = 0, taking homology at ker γ does nothing at this page or the next, so ker γ = I E 1 = 0. Likewise if coker δ = coker β = ker ɛ, it follows that coker γ = 0. So, we have γ is injective if δ, β are injective and α is surjective γ is surjective if δ, β are surjective and ɛ is injective This seems to essentially be the two four lemmas ; I m not sure if this is a minimal set of conditions in any reasonable sense. They seem to be the most obvious conditions, if that s worth anything. Problem 4 Let f : (A, d A ) (B, d B ) be a map of complexes. The mapping cone Cone(f) is the total f complex of the double complex A B. It can be described explicitly as follows: d A 0 f d B Cone(f) n := A n 1 B n, d n : A n 1 B n A n B n 1. Show that there is a long exact sequence H n+1 (Cone(f) ) H n (A ) H n (B ) H n (Cone(f) ) H n 1 (A ). 3
4 f Minor note: I assume the double complex A B is anticommutative, whereas a map of complexes in my experience has commutative squares. Consider the double complex π A 0 A n 1 A n 1 B n B n 0 d A d n d B π A 0 A n A n B n 1 B n 1 0 One can check this has exact rows and columns. Hence we have an exact sequence of chain complexes 0 B Cone(f) A[ 1] 0, which from Problem gives rise to a long exact sequence H n+1 (Cone(f) ) H n A H n (B ) H n (Cone(f) ) H n 1 (A ). Problem 5 Establish the Künneth spectral sequence for complexes (it s ok to use the classical Künneth formula as in [Weibel, 3.6.3] if you feel that you need to): Let R be a (commutative) ring and C, D complexes of R-modules bounded below. Assume the C n are flat for all n. Show that there is a convergent spectral sequence Epq = Tor R p (H s (C ), H t (D )) H p+q (C R D ) where H p+q (C R D ) stands for the homology of the total complex. Let P D be a Cartan-Eilenberg resolution. Consider the double chain complex E 0 pq = Tot q (C P p ) where the vertical maps are the usual total complex maps, and the horizontal maps are induced by the horizontal maps from P D. Take horizontal homology to get ( ( II Epq) 1 T = H p (Tot q (C R P )) = H p C s R P t ) = H p (C s R P t ) = = δ p0 Tot q (C R D ), C s R H p (P t ) = C s R δ p0 D t where we ve used the following facts: is finite; homology commutes with finite sums; homology commutes with C s R since C s is flat; P t D t is a projective resolution, so taking homology gives zeros except at the very bottom when it gives D t. That is, we re left with just Tot q (C R D ) in the p = 0 column. Thus we ll be able to recover the abutment exactly, not just the associated graded object. Take vertical homology to get ( II E pq) T = δ p0 H q (Tot (C R D )). The sequence stabilizes here, so the nth piece of the abutment is H n (Tot (C R D )) (since p = 0 gives the only non-zero term and n = p + q). 4
5 On the other hand, take vertical homology of E 0 pq to get and then take horizontal homology to get I E 1 pq = H q (Tot (C R P p )) I E pq = H p (H q (Tot (C R P ))) = H p ( = H p (H s (C ) R H t (P )) = H s (C ) R H t (P ) ) Tor R p (H s (C ), H t (D )), where we ve used the following facts: the Künneth formula quoted below; is finite; homology commutes with finite sums; H t (P ) H t (D ) is a projective resolution; Tor is computed in the usual way. This application of the Künneth formula uses the fact that P is projective, hence flat, and B(P ) is projective, hence flat, which was proved in class; it also uses the fact that H t (P ) is projective, hence flat, so the Tor term vanishes, and we have an isomorphism. In all we have a (convergent) spectral sequence Tor R p (H s (C ), H t (D )) H p+q (C R D ). 1 Theorem (Künneth formula for complexes) Let P and Q be right and left R-module complexes, respectively. If P n and d(p n ) are flat for each n, then there is an exact sequence 0 H p (P ) H q (Q) H n (P R Q) Tor R 1 (H p (P ), H q (Q)) 0. p+q=n p+q=n 1 5
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