53
Cohomology of two points in projective spaces
term now takes the form !1" {0} !2" !2" !3" {2} !4" !4" !5" {4} !6" !6" !7" · · ·
5 4 3
Z
!2" !1" !3" !2" !4" !3" !5" !4" !6" !5" !7" !6" · · · !1" !1" !2" !2" !3" !3" !4" !4" !5" !5" !6" !6" !7" · · ·
2 1 0
Z 0
1
!2" !1" {2} !2" !4" !3" {4} !4" !6" !5" {6} !6" · · · 2
3
4
5
6
7
8
9
10
11
12
13
where again only dimensions at most 13 are shown. At this point it is convenient to observe that the last statement in the paragraph following (41) fails under the current hypothesis. Indeed, the generator of E30,3 is twice the generator of E20,3 , breaking up the vertical symmetry of d3 -differentials holding under (40)—of course, the groups in the current E3 -term already lack the vertical symmetry we had in the case of (40). In order to deal with such an asymmetric situation we need to make a differential-wise measurement of all the groups involved in the current E3 -term (we will simultaneously analyze the possibilities for the two horizontal families of d3 -differentials). To begin with, note that the arguments dealing, in the case of (40), with the two differentials E31,2 → E34,0 and E32,2 → E35,0 apply without change under the current hypothesis to yield that these two differentials are injective, the former with cokernel E44,0 = Z2 ⊕ Z4 (i.e., both yield injective maps after tensoring with Z2 ). Note that any other group not appearing as the domain or codomain of these two differentials must be eventually wiped out in the spectral sequence, either because H i (B(P3 , 2)) = 0 for i ≥ 6, or else because the already observed E45,0 = Z2 accounts for all there is in H 5 (B(P3 , 2)) in view of Corollary 3.2. This observation is the key in the analysis of further differentials, which uses repeatedly the following three-step argument (the reader is advised to keep handy the previous chart in order to follow the details): Step 1. The groups E3p,q not yet considered and having smallest p+q are E33,2 and E32,3 . Both are isomorphic to !2"; none can be hit a differential. Since E36,0 = !4", we must have injective differentials d3 : E33,2 → E36,0 and d4 : E42,3 → E46,0 , clearing the E∞ -term at nodes (2, 3), (3, 2), and