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I'm trying to solve exercise 3.3.26 in Hatcher's Algebraic Topology:

Compute the cup product structure in $H^{*}((S^{2}\times S^{8})\#(S^{4}\times S^{6});\mathbb{Z})$, and in particular show that the only nontrivial cup products are those dictated by Poincaré duality.

What I know: I can use Künneth formula to compute the cohomology ring and individual groups of each product space. I also know that the cohomology group of the connected sum is isomorphic to the direct sum of cohomology groups at the same dimension for $0 < i < 10$. This was an earlier exercise I solved.

What I'm struggling with is inferring the cup product structure. Since "connecting" is happening at dimension $10$, I can guess that the cup product at lower dimensions is unaffected. In other words it happens in each component of direct sums individually. I can also guess that the connected sum identifies the $10$-cell of both products into one.

I have no idea how to show this rigorously. Could somebody please show me how to do this? Also how is Poincaré duality helpful here as hinted by the exercise?

Thanks

PeterM
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1 Answers1

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Let M denote $(S^2 \times S^8) \sharp (S^4 \times S^6)$

  1. To calculate the cohomology of $M$, I think you can use cellular cohomology. There is one $0, 2, 8, 10$-cell in $S^2 \times S^8$; and there is one $0, 4, 6, 10$-cell in $S^4 \times S^6$. Overall in $M$, there is one $0, 2, 4, 6, 8, 10$ cell. So the cohomology of $M$ is $\mathbb{Z}$ in dimensions $0, 2, 4, 6, 8 , 10$ and zero otherwise.

  2. The cup product between the $2-$cell and the $8-$cell, as well as the cup product between $4-$cell and $6-$cell are nontrivial, and is trivial for the other pairings. More precisely, let $\alpha_i$ be a nonzero generator of $H^i(M)$, there is a nonzero generator $\beta_{10-i}$ of $H^{n - i}(M)$ such that $\alpha_i \cup \beta_{n-i}$ is a generator of $H^{10}(M) \cong \mathbb{Z}$. This agrees with the statement of Corollary 3.39 in Hatcher (the Poincare Duality section). In this case, the cohomology classes that do not pair to zero are like 'Poincare duals'.

Qiao
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  • How can you show that the pairing of the 2 with the 6 cell is trivial? – Dimitris Sep 19 '14 at 04:38
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    This is because the $2$-cell is in $S^2 \times S^8$ while the $6$-cell is in $S^4 \times S^6$. In $M$ these two cycles do not intersect transversely, so the cup product in cohomology is zero. I think this is explained somewhere in Hatcher's book. – Qiao Sep 20 '14 at 17:11