In the compact Hermitian manifold,we can define the $H^{p,q}(X) = H^{q}(X,\Omega ^p)$ which has Dolbeault complex as resolution, therefore the dimension of the Dolbeault cohomology can be defined to be the Hodge number, on the compact Hermitian manifold Dolbeault cohomology isomorphic to Harmonic space for $\bar{\partial}$, therefore it reduce to the problem on the harmonic space.
Secondly, since the harmonic space for $d$ and $\bar{\partial}$ are equal on the Kahler manifold:
$$\mathscr{H}_d^{p,q}(X) = \mathscr{H}_{\bar{\partial}}^{p,q}(X)$$
While the conjugation commute with $\Delta_d$ and take $(p,q)$ form to $(q,p)$ form , we have the additional symmetric $h^{p,q} = h^{q,p}$.
However I was wondering if this symmetric holds true on general Hermitian manifold which is non Kahler?
Best Answer
There is a easier example take the $X=\Bbb{C}\setminus \{0\}$ with $H^{1,0}(X) = \{fdz \mid \bar{\partial}f = 0\} \ne 0$ (since there are many holomorphic function on $X$) and $H^{0,1}(X) =0$. (since it's complex 1 dimension the only basis is $dz$.)