# For genus $g$ Riemann surface, $\operatorname{H}^1(X)$ has complex dimension $2g$?

Let $X$ be a compact Riemann surface of genus $g$. Let $\operatorname{H}^{0,1}(X)=\Omega ^{0,1}(X)/ \bar{\partial}( \Omega^0(X) )$ and $\operatorname{H}^{1,0}(X)$ be space of holomorphic 1-forms on X. Then \begin{align*} \operatorname{H}^{1,0}(X) \oplus \operatorname{H}^{0,1}(X) \simeq \operatorname{H}^{1}(X) \end{align*} where $\operatorname{H}^{1}(X)$ is the first de-Rham cohomology of $X$. \begin{align*} \operatorname{dim}_{\mathbb{C}} \operatorname{H}^{0,1}(X) = g, \ \operatorname{dim}_{\mathbb{C}} \operatorname{H}^{1,0}(X) = g \end{align*} Thus $\operatorname{H}^{1}(X)$ has dimension $2g$ as a vector space over $\mathbb{C}$?

$\operatorname{H}^{1}(X)$ has dimension $2g$ as a vector space over $\mathbb{R}$ so shouldn't it be $g$ dimensional over $\mathbb{C}$?

Does $\operatorname{H}^{1}(X)$ have two different meanings, one of which has comples dimension $2g$ and the other has real dimension $2g$?

• Yes, you can take singular cohomology with coefficients in either the real or the complex numbers. The Hodge decomposition is about singular cohomology with complex coefficients. – Qiaochu Yuan Jun 2 '18 at 8:10