# Symplectic form on a complex manifold

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).

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Sorry, I'm confused by your definition of $\omega$, did you mean $\omega = \sum_{i} dx_i \wedge dy_i$ ? –  Tim van Beek Feb 27 '11 at 13:18
Thanks thats what i meant. I just edited it. –  Eric O. Korman Feb 27 '11 at 15:47
I'm a little confused as to what the actual question is. If all you care about is the the real symplectic form in complex notation, then your "at first glance" is correct---why does it matter that it vanishes on the (1,0)-part of the complexitied tangent bundle? Note that there is a difference between a real symplectic form on a complex manifold vs a holomorphic symplectic form: the latter are the things like $dz_1 \wedge dz_2$ on $\mathbb C^2$ for example, so are the correct analogs of sympletic forms in the complex world. –  Santiago Canez Feb 27 '11 at 16:01
@Santiago: I want to identify $T_{(1,0)} M$ with $TM$ in the way I mentioned. Then I want to know what the corresponding symplectic form looks like on $T_{(1,0)}$. It can only be $\frac{i}{2}\sum_j dz_j \wedge d\bar z_j$ if I interpret $d\bar z_j$ differently. So my basic question is if that is the right way to think about it. –  Eric O. Korman Feb 27 '11 at 16:26

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 $\partial_z$)