Computing with differential forms on the tangent space of a complex manifold

Question

Consider $\mathbb{C}^n$ as a holomorphic manifold. For $z=(z_1,\cdots z_n)\in\mathbb{C}^n$ fixed, consider the linear subspace of $T_z\mathbb{C}^n$ spanned by $v_1=\sum_{i=1}^n z_i\frac{\partial}{\partial z_i}$, where $\{\frac{\partial}{\partial z_i}\}$ is a basis of $T_z\mathbb{C}^n$. Consider the (1,1)-form $\omega=\sum_{j=1}^n dz_j\wedge d\overline{z}_j$. If I evaluate $\omega(v_1,v_2)$, where $v_2$ is just an arbitrary element of $T_z\mathbb{C}^n$, is the following expression correct? $$ \omega_z(v_1,v_2)=\sum_{j=1}^n (dz_j)(v_1)(d\overline{z}_j)(v_2)-(dz_j)(v_2)d(\overline{z}_j)(v_1)= \sum_{j=1}^n z_j(d\overline{z}_j)(v_2)-\overline{z}_j(dz_j)(v_2)? $$ I.e., this question boils down to knowing whether $(dz_j)(z_j\frac{\partial}{\partial{z_j}})=z_j$ and $(d\overline{z}_j)(z_j\frac{\partial}{\partial z_j})=\overline{z}_j,$ or do we have that $d\overline{z}_j(\frac{\partial}{\partial z_j})=0,$ I seem to find both, but which is correct to use in this particular case? The last one seems wrong, tho, since it would imply that $d\overline{z}_j=0$ which is nonsense.

Answer 1

This is a matter of authors' preference. If you think of forms as operating on the complexified tangent space, then $dz^j$ is dual to $\partial/\partial z^j$ and $d\bar z^j$ is dual to $\partial/\partial\bar z^j$. However, if you're working on the tangent space itself (e.g., thinking of the Kähler metric as giving a hermitian form on the tangent space), then for $v$ a holomorphic tangent vector you interpret $d\bar z^j(v) = \overline{dz^j(v)} = \overline{v^j}$. This is what you are referring to in your question, insofar as you're working with the (holomorphic) tangent space, not its complexification.

Comment: Either way, forms are linear over complex-valued functions, so the coefficients just pull out.