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A Family of Cohomological Complex Projective Spaces A family of cohomological complex projective spaces MustafaKalafat DersimKaya October 23, 2018 Abstract We exhibit a family of complex manifolds, which has a member at each odd complex dimension and which has the same cohomology groups as the complex projective space at that dimension, but not homotopy equivalent to it. We also analyse the even dimensional analogue. 1 Introduction This paper is about various examples of complex manifolds. Conventionally we define the complex degree 2 hypersurface in the n-dimensional complex projective space by V 2 2 CP 2 := {[Z0 : ··· : Zn] | Z0 + ··· + Zn = 0}⊂ n. Our main result is as follows. arXiv:1810.09029v1 [math.AG] 21 Oct 2018 V CP Theorem 1.1. The algebraic variety 2 ⊂ 2k has the same cohomology groups as CP 2k−1 but it is not homotopy equivalent to it for k > 1. The idea of the proof is to work on the underlying smooth structure of this hy- persurface. One can show that [Sal89] this complex manifold is the set of totally isotropic complex 2-dimensional subspaces of Cn+1 and indeed diffeomorphic +Rn+1 to the oriented Grassmannian G2 , the space of oriented 2-planes in real 1 V +R3 n+1-space. In the lower dimension for k=1, 2 is the projective line and G2 is a topological sphere, so all these spaces are the same. However if one goes further in the upper dimensions the spaces start diverging from each other. To be able o distinguish, furthermore using Serre spectral sequence techniques one can compute the cohomology ring of the Grassmannian as follows. Theorem 3.1. The cohomology ring of the Grassmann manifold is the following trun- cated polynomial ring for which degxm = m. ∗ +R2k+1 Z Z k 2 H (G2 ; ) ≈ [x2, x2k]/hx2 − 2x2k, x2ki. One can compare the cohomology rings to distinguish this space with the pro- jective space or one can use homotopy groups. We give the details in the follow- ing. To give a complete discussion we also give the cohomology ring in the even case as follows. Theorem 4.2. The cohomology ring of the Grassmann manifold is the following trun- cated polynomial ring for which degxm = m. k k−1 2 2k−2 x2k−2 + y2k−2 + (−1) x2 , 2x2k−2 − x2 ∗ +R2k Z Z k k 2 2k−1 H (G2 ; ) ≈ [x2, x2k−2, y2k−2, x2k]/ 2x2x2k−2 + (−1) x2 , 2y2k−2 − x2 . * 2 2k−1 + x2k−2y2k−2 , x2k , x2 There has been a growing interest in Grassmann manifolds because of their role in analysing submanifolds of smooth manifolds or Riemannian manifolds with special or exceptional holonomy. This is another reason that makes our work interesting. Interested reader may consult to [AK16] or [Kal18] for more information. This paper is organized as follows. In section §2 we deal with the case n=7, in section §3 we deal with the odd and in section §4 with the even dimensional analogues. Acknowledgements. The first author would like to thank to his father and fam- ily for their support during the preparation of this paper. This work is partially supported by the grant ♯114F320 of Tubitak¨ 1. 1Turkish science and research council. 2 2 Typical Case In order to work on the Grassmannian we need the setting of a cohomological Serre spectral sequence of a fiber bundle. The reader may consult to [Hat04] or [Sat99] for the fundamental tools in spectral sequences. We will be working on the following fiber bundle S1 R7 +R7 → V2 −→ G2 (1) obtained by sending an ordered orthonormal 2-frame of R7 to the oriented 2- plane spanned by it. One could work with any ordered 2-frames as well, which 7 constitutes a space that deformation retracts to our V2R . Having a simply connected cell complex as a base is crucial as one can show in (8). Since we want to understand the cup product structures we are interested in cohomology. In this case the Serre spectral sequence is defined as p,q p q S1 Z E2 := H (G; H ( ; )) which is guaranteed to converge to the following limit p,q p,q p+1,q−1 E∞ ≈ F /F where Fp,q is a filtration of abelian groups satisfying, Hn(V; Z)= F0,n ⊃ F1,n−1 ⊃···⊃ Fn+1,−1 = 0. p,q p′,q′ p+p′,q+q′ We have a product En ⊗ En → En on the pages of the spectral sequence satisfying a Leibniz rule, ′ ′ p+q p,q p ,q dn(ab) = (dna)b + (−1) adnb for a ∈ En , b ∈ En . (2) On the second page this product, ′ ′ ′ ′ p,q p q Z p ,q p′ q′ Z p+p ,q+q p+p′ q+q′ Z E2 (≈ H (B; H (F; ))) ⊗ E2 (≈ H (B; H (F; ))) → E2 (≈ H (B; H (F; ))) ′ equals (−1)qp times the cup product of the base space B, whose coefficient product is the cup product of the fiber F. The product for En+1 is derived from the product for En, and the product for E∞ is derived from the cup product for E. 3 The homology groups of various Grassmann manifolds are computed in [AK16] using the Serre spectral sequence. We can reinterpret one of the re- sults in there through Poincare´ duality and write the table for the cohomology +R7 of the Grassmann manifold G2 of oriented 2-planes in real 7-space as follows, ∗ +R7 Z Z Z Z Z Z Z H (G2 ; ) = ( ,0, ,0, ,0, ,0, ,0, ). (3) Next, we will set up the cohomological Serre spectral sequence of the fiber bun- dle (1). Imposing the definition and using (3) we obtain the second page as illustrated in the Table 1. +R7 Table 1: The second page of the cohomological Serre spectral sequence for G2 . 1 Za ❍0 Zax2 0 Zax4 0 Zax6 0 Zax8 0 Zax10 ❍❍❥ E2 0 Z1 0 Zx2 0 Zx4 0 Zx6 0 Zx8 0 Zx10 012345678910 To be able to figure out the cup product structure, we have to find the limit of this spectral sequence. For this reason, we need information on the total space of 7 the fiber bundle. The cohomology of the 11-dimensional Stiefel manifold V2R is well known (see [Hat02] for a reference) and is given by, ∗ R7 Z Z 2 Z 2 H (V2 ; ) ≈ [x11]/hx11i⊕ 2[x6]/hx6i where deg xk = k. Since the total space at the levels 1,2,3,4,5 and 7,8,9,10 has vanishing cohomology, the limiting diagonals at these levels are totally zero. Let us figure out the 5,1 remaining diagonals. The term E∞ = 0 automatically since it is the limit of 5,1 5 Z E2 = H (G; )= 0. Since, 6 0,6 1,5 5,1 6,0 Z2 = H (V; Z)= F ⊃ F ⊃···⊃ F ⊃ F ⊃ 0 0,6 4,2 1,5 6,0 and E∞ = ··· = E∞ = 0 from the second page yields Z2 = F = ··· = F . 6,0 6,0 7,−1 Then we can compute the limit E∞ = F /F = Z2. 0,11 Next, in a similar fashion starting with Z = H11(V; Z)= F0,11 and using E∞ = 9,2 ··· = E∞ = 0 from the second page we get Z = F1,10 = ··· = F10,1. Again, 11,0 from the second page we have E∞ = 0. Finally, we can compute the limit 4 +R7 Table 2: The limiting page of cohomological Serre spectral sequence for G2 . 1 0 0 0 0 0 0 0 0 0 0 Z E∞ 0 Z 0 0 0 0 0 Z2 0 0 0 0 0 012345 6 78910 10,1 E∞ = F10,1/F11,0 = Z. Combining these results we obtain Table 2 as the limit of the spectral sequence. Now we are ready to compute the ring structure of the 0,1 cohomology. Let the symbols a and xk denote generators of the groups E2 and k,0 Z E2 all of which are isomorphic copies of as shown in Table 1. The generators on the upper row are a times the generators of the lower row since the product 0,q r,s r,s+q E2 × E2 → E2 is nothing but multiplication of coefficients. Realize that the 0,1 2,1 6,1 8,1 differentials d2 , d2 and d2 , d2 are isomorphisms. So if we rearrange the sign of a or x2 to make the relation d2a = x2 hold, we can use the Leibniz rule and rearrange the sign of x4 to make the following relation hold, 2 x4 = d2(ax2) = (d2a)x2 ± a(d2 x2) = (d2a)x2 = x2 (4) 6,0 Z 4,1 since d2x2 is out of the table. Since the limit E∞ = 2, the differential d2 has to double the generators. So that again applying the Leibniz rule and rearranging we impose, 3 2x6 = d2(ax4) = (d2a)x4 = x2x4 = x2. (5) The remaining differentials are also isomorphisms so we get, x8 = d2(ax6) = (d2a)x6 = x2x6 (6) 2 x10 = d2(ax8) = (d2a)x8 = x2x8 = x2x6 (7) So, combining the relations (4,5,6,7) we can conclude that the cohomology ring of the Grassmann manifold is the following truncated polynomial ring for which degxk = k. ∗ +R7 Z Z 3 2 H (G2 ; ) ≈ [x2, x6]/hx2 − 2x6, x6i. One can also compute the ring structure of CP5 in a similar fashion using Serre spectral sequence as in [BT82] to get ∗ CP Z Z 6 H ( 5; )= [x2]/hx2i 5 2 CP Z k 2k CP Z where x2 generates H ( 5; ) so that x2 generates H ( 5; ) additively. Once we compare these two ring structures, we decide that they are different, +R7 CP and hence these two spaces G2 and 5 are not homotopy equivalent.
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