# Conjugation of Dolbeault cohomology and cup product.

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?

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You probably mean compact Kaehler manifold. In that case the comparison iso. between Dolbeault cohomology and singular cohomology is compatible with the cup product. But the cup product is defined over singular cohomology with real coefficients and therefore commutes with complex conjugation. So the answer to your question is positive. –  Damian Rössler Feb 2 '13 at 23:08