Using a similar argument in singular (co)homology, calculate the cohomology ring of the closed orientable surface \(\Sigma_g\) of genus \(g\).
Solution
1. Choose the standard symplectic basis\[ a_1,b_1,\ldots,a_g,b_g\in H_1(\Sigma_g;\mathbb Z) \]and let\[ \alpha_1,\beta_1,\ldots,\alpha_g,\beta_g\in H^1(\Sigma_g;\mathbb Z) \]be the Poincaré-dual basis. Let \(\omega\in H^2(\Sigma_g;\mathbb Z)\) be the orientation class.
A closed orientable genus-\(g\) surface has \[ H_0\cong H_2\cong\mathbb Z,\qquad H_1\cong\mathbb Z^{2g}, \] and the standard handle curves form a symplectic basis for the intersection pairing.
2. The only nonzero products of degree-one basis classes are\[ \alpha_i\smile\beta_j=\delta_{ij}\omega, \qquad \beta_i\smile\alpha_j=-\delta_{ij}\omega. \]All products \(\alpha_i\smile\alpha_j\) and \(\beta_i\smile\beta_j\) vanish.
Under Poincaré duality, evaluation of the cup product on the fundamental class equals algebraic intersection number: \[ \langle u\smile v,[\Sigma_g]\rangle=\operatorname{PD}(u)\cdot\operatorname{PD}(v). \] The standard curves satisfy \(a_i\cdot b_j=\delta_{ij}\), \(b_i\cdot a_j=-\delta_{ij}\), and all \(a_i\cdot a_j\), \(b_i\cdot b_j\) vanish.
3. Hence \(H^*(\Sigma_g;\mathbb Z)\) is the graded ring generated by\[ 1,\alpha_1,\beta_1,\ldots,\alpha_g,\beta_g,\omega \]with \(|\alpha_i|=|\beta_i|=1\), \(|\omega|=2\), relations from 2, and every product involving \(\omega\) and a positive-degree class equal to zero.
There is no cohomology above degree \(2\), so all products of total degree greater than \(2\) vanish. The additive groups and all remaining degree-two products are determined by <1>1–<1>2.