Let $C$ be a compact Riemann surface without boundary. In particular, $C$ is a surface of a genus $g$. Then the vector space of holomorphic 1-forms $\mathrm{H}^0(C,\mathrm{K}_C)$, where $\mathrm{K}_C$ is the canonical line bundle, has a basis $\eta_1,\ldots, \eta_g$.
I call a holomorphic 1-form on $C^{\times 2}$ a bidifferential. I.e. a bidifferential is an element $\omega \in \mathrm{H}^0(C^{\times 2},\mathrm{K}_{C^{\times 2}})$. I call a bidifferential $\omega$ symmetric if $\omega(z_1,z_2) = \omega(z_2,z_1)$ for all $z_i \in C$ with $i=1,2$. Any two 1-forms $\omega_1,\omega_2 \in \mathrm{H}^0(C,\mathrm{K}_C)$ canonically induce a bidifferential which I denote by $\omega_1 \boxtimes \omega_2$.
My question is the following: can any symmetric bidifferential be written as $\sum_{i,j=1}^g A_{ij} \, \eta_i \boxtimes \eta_j$ where $A_{ij} \in \mathbb{C}$ satisfy $A_{ij}=A_{ji}$?
I know that the statement is true locally, but I couldn't find a counter-example that shows that it doesn't hold globally. If $C$ is not compact, then there are obvious counter-example, e.g. for $C=\mathbb{C}$ the bidifferential $\omega= \sin(z_1 z_2) \, \mathrm{d} z_1 \boxtimes \mathrm{d} z_2$ is not in a product form.
I found a solution: the statement is indeed correct and the symmetry is not needed. Assume that $\omega$ is a bidifferential. Let $p \in C$ and let $v_p \in \mathrm{T}^{1,0}_pC$ be a tangent vector. Inserting $v_p$ into the first component yields a holomorphic 1-form $\omega(v_p \boxtimes \cdot)$. Therefore, it can be decomposed as $$ \omega(v_p \boxtimes \cdot) = \sum_{i=1}^g \alpha_i(v_p) \eta_i $$ where $\alpha_i(v_p) \in \mathbb{C}$ is a constant that depends on $v_p$. We can view $\alpha_i$ as a 1-form, so that $\omega = \sum_{i=1}^g \alpha_i \boxtimes \eta_i$. Since $\omega$ is holomorphic, also $\alpha_i$ must be holomorphic and hence, it can be written as $\alpha_i = \sum_{j=1}^g A_{ij} \eta_j$. Thus, we have $$ \omega = \sum_{i,j=1}^g A_{ij} \, \eta_i \boxtimes \eta_j \,. $$ This completes the proof.