# First cohomology group on a Riemann surface with all wedge products equal to zero

Sorry for the strong edit, but I realized my question had a easier formulation:

Can there be a Riemann surface $X$ with the property $\sigma\wedge \tau=0$ for every $\sigma,\tau\in H^1(X,\mathbb{C})$?

Here I mean $\sigma\wedge\tau =0\in H^2(X,\mathbb{C})$, i.e. $\sigma\wedge\tau$ is an exact 1-form.

My guess is no but I can't find a contradiction...

• $\sigma \wedge \sigma$ is always zero. – user98602 Jun 29 '16 at 14:09
• yes, but here I mean the wedge product between every two different forms.. – user0182 Jun 29 '16 at 14:14
• Oh, I see. Sorry for my misreading. This is possible on noncompact Riemann surfaces (eg a cylinder, or connected sum of two cylinders, etc), but not for closed surfaces, by what's called Poincare duality. – user98602 Jun 29 '16 at 14:15

Following Mike Miller's suggestion, consider the cylinder $X =S^1 \times \mathbb{R}$ (as a Riemann surface, you may view it as either $\mathbb{C} \setminus \{0\}$ or $\mathbb{C}/\mathbb{Z}$). As this deformation retracts onto the base circle and homology is a homotopy invariant, we know that $H_2(X;\mathbb{C}) \cong H_2(S^1;\mathbb{C}) =0$. As $\mathbb{C}$ is a field, cohomology is dual to homology, and so $H^2(X;\mathbb{C}) \cong (H_2(X;\mathbb{C}))^* = 0$.
Alternatively, if you want to work exclusively with (complex) de Rham cohomology, you may simply observe that $H_{dR}^2(X) \cong H_{dR}^2(S^1) = 0$ because there are no nontrivial $2$-forms on $S^1$.
Or, using the Künneth formula, \begin{align}H_{dR}^2(X) &= \left(H^2_{dR}(S^1) \otimes H^0_{dR}(\mathbb{R})\right) \oplus \left(H^1_{dR}(S^1) \otimes H^1_{dR}(\mathbb{R})\right) \oplus \left(H^0_{dR}(S^1) \otimes H^2_{dR}(\mathbb{R})\right)\\ &= (0 \otimes \mathbb{C}) \oplus (\mathbb{C}\otimes 0) \oplus (\mathbb{C} \otimes 0) = 0.\end{align}