For the closed orientable surface \(M\) of genus \(g \geq 1\), show that for each nonzero \(\alpha \in H^1(M; \mathbb{Z})\) there exists \(\beta \in H^1(M; \mathbb{Z})\) with \(\alpha \beta \neq 0\). Deduce that \(M\) is not homotopy equivalent to a wedge sum \(X \vee Y\) of CW complexes with nontrivial reduced homology. Do the same for closed nonorientable surfaces using cohomology with \(\mathbb{Z}_2\) coefficients.
Solution
1. \(H^1(M;\ZZ) \cong \ZZ^{2g}\), and the cup product pairing \(H^1 \times H^1 \to H^2 \cong \ZZ\) is a nondegenerate symplectic form.
standard computation of the cohomology ring of a closed orientable surface of genus \(g\); the cup product is the intersection form, which is symplectic (nondegenerate alternating).
2. For each nonzero \(\alpha \in H^1(M;\ZZ)\) there is \(\beta \in H^1(M;\ZZ)\) with \(\alpha \smile \beta \neq 0\).
nondegeneracy of the symplectic form in 1: if \(\alpha \smile \beta = 0\) for all \(\beta\), then \(\alpha = 0\).
3. \(M\) is not homotopy equivalent to a wedge \(X \vee Y\) with \(\widetilde{H}_*(X) \neq 0\) and \(\widetilde{H}_*(Y) \neq 0\).
3.1. For a wedge \(X \vee Y\), the cup product of any class in \(H^1(X)\) with any class in \(H^1(Y)\) is zero.
the cup product \(H^1(X \vee Y) \times H^1(X \vee Y) \to H^2(X \vee Y)\) is zero, since \(H^1(X \vee Y) \cong H^1(X) \oplus H^1(Y)\) and cross terms vanish (the two summands are “disjoint” in the wedge).
3.2. But \(M\) has a nonzero cup product \(H^1 \times H^1 \to H^2\).
2.
3.3. Hence \(M\) is not homotopy equivalent to any such wedge.
the cup product structure is a homotopy invariant, and 1 contradicts 2.
4. Nonorientable case: for a closed nonorientable surface \(N\), \(H^1(N;\ZZ_2) \cong \ZZ_2^{g}\) (where \(g\) is the genus), and the cup product pairing \(H^1(N;\ZZ_2) \times H^1(N;\ZZ_2) \to H^2(N;\ZZ_2) \cong \ZZ_2\) is nondegenerate.
standard computation of the \(\ZZ_2\)-cohomology ring of a nonorientable surface.
5. Hence for each nonzero \(\alpha \in H^1(N;\ZZ_2)\) there is \(\beta\) with \(\alpha \smile \beta \neq 0\), and \(N\) is not homotopy equivalent to a wedge \(X \vee Y\) with nontrivial reduced homology.
the same argument as 2 and 3, with \(\ZZ_2\) coefficients.
6. Q.E.D.
2, 3, and 5.