Let $H^p(X,\Omega_X^q)$ denote the (p,q) Dolbeault cohomology group of a Kähler manifold X. Conjugation of forms induces an isomorphism $$H^p(X,\Omega_X^q) \simeq \overline{H^q(X,\Omega_X^p)}$$
Let $v\in H^1(X,\mathcal{O}_X)$ and denote by $\bullet \cup v$ the cup product with this class, and consider the following diagram
$\begin{array}{ccc} H^p(X,\Omega_X^q) & \stackrel{\cup v}{\longrightarrow} & H^{p+1}(X, \Omega_X^q) \\ \downarrow & & \downarrow \\ H^q(X,\Omega_X^p) & \longrightarrow &H^q(X,\Omega_X^{p+1}) \end{array}$
where the vertical maps are isomorphisms induced by conjugation. The bottom horizontal map looks like wedge product with the conjugate class $\bar{v}\in H^0(X,\Omega_X^1)$. Is this true?
