Use the Kunneth formula to compute \(H^*(S^2\cross S^2; \ZZ)\).
- Known to be \([\ZZ, 0, \ZZ^2, 0, \ZZ, 0, 0, \cdots]\).
Solution
1. \(H^*(S^2;\ZZ) = \ZZ\) in degree \(0\), \(\ZZ\) in degree \(2\), and \(0\) otherwise.
standard computation of the cohomology of \(S^2\).
2. By the Künneth formula (with \(\ZZ\) coefficients, and all the relevant groups free so the Tor terms vanish),\[H^n(S^2 \times S^2;\ZZ) \cong \bigoplus_{i+j=n} H^i(S^2;\ZZ) \otimes H^j(S^2;\ZZ).\]
Künneth theorem; since \(H^*(S^2;\ZZ)\) is free, \(\operatorname{Tor}\) vanishes.
3. The nonzero tensor products are:- \(n=0\): \(H^0 \otimes H^0 = \ZZ\);- \(n=2\): \(H^2 \otimes H^0 \oplus H^0 \otimes H^2 = \ZZ \oplus \ZZ = \ZZ^2\);- \(n=4\): \(H^2 \otimes H^2 = \ZZ\).
1 and 2.
4. Hence\[H^n(S^2 \times S^2;\ZZ) = \begin{cases} \ZZ & n = 0, 4 \\ \ZZ^2 & n = 2 \\ 0 & \text{otherwise}. \end{cases}\]
3.
5. Q.E.D.
4, matching the known answer \([\ZZ, 0, \ZZ^2, 0, \ZZ, 0, \ldots]\).