I am a little muddled and am hoping I can get some clarification about forms in a complex manifold. Since I am only concerned with local issues, consider $M = \mathbb C^n$ as a complex manifold. So I have complex coordinates $z_1,\ldots,z_n$ and corresponding real coordinates $x_1, y_1,\ldots, x_n, y_n$ with $z_j = x_j + i y_j$. Now I have a canonical complex structure $J$ on $M$ given by $J \partial_{x_j} = \partial_{y_j}, J \partial_{y_j} = -\partial_{x_j}$. Then I have an isomorphism of the complex bundles between $TM$ and the holomorphic tangent bundle $T_{(1,0)} M = span_{\mathbb C} \{\partial_{z_j}\}$. This isomorphism is given by
$$
(a+ib) \partial_{z_j} \mapsto a\partial_{x_j} + b\partial_{y_j}.
$$
Now let $\omega = \sum_i dx_i \wedge dy_i$ be the standard symplectic form on $M$. What does this correspond to under the above isomorphism. At first glance it is
$$
\tilde \omega = \frac{i}{2}\sum_j dz_j \wedge d\bar z_j
$$
where $dz_j = dx_j + i dy_j, d\bar{z_j} = dx_j – i dy_j$. At second glance this can't be correct since this is zero on the holomorphic tangent bundle. On third glance, everything seems to work out if I think of
$$
d\bar{z_j}(c\partial_{z_j}) = \bar c.
$$
But this is unsettling to me. Is this the right way to think about it though? Can I get in any trouble by thinking of it this way?
The problem seems to be that the differentials of the real coordinates give real dual vectors, but the covectors $dz_j$, $d\bar z_j$ are inherently complex (their real span is not the real dual to $T_{(1,0)} M$ unless I interpret $d\bar z_j$ as mentioned above).
I believe what is usually done is $\tilde\omega$ is considered to be a form on $TM \otimes \mathbb C$ and is a real symplectic form on
$$
\mathbb Rspan\{\partial_{x_j}, \partial_{z_j}\} = \mathbb Rspan\{\partial_{z_j} + \partial_{\bar z_j}, i(\partial_{z_j} – \partial_{\bar z_j})\}
$$
but I'd prefer to not complexify if I don't have to (since I am just concerned with the symplectic geometry and thus the real bundle, but I would like the convenience of the complex notation).
Best Answer
Your bundle isomorphism isn't the tangent map of a diffeomorphism of the base. This makes all of the constructions come out wrong. The symplectic form under this identification is $$ \sum e_i \wedge f_i,$$ where $e_i( c \partial_z ) = c$ and $f_i( c \partial_z ) = \bar c$. If you understand $d \bar z$ as actually $\overline {dz}$, then your formula is fine.
I think it is somewhat easier to take the identification of the tangent space $\mathbb R^{2}$ with $\mathbb C$ by $(a+ib)\partial_x \mapsto a \partial_x + b \partial_y$. (i.e. use $\partial_x$ instead of $\partial_z$)
The other option, as you suggest, is to take the complexified tangent space, and work there. I think this latter is the most convenient for calculations... you just then need to be sure that you are only working with real objects. This is not a big deal, and imo, is a cleaner way of doing computations.