11
$\begingroup$

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).

$\endgroup$
3
  • 1
    $\begingroup$ Sorry, I'm confused by your definition of $\omega$, did you mean $\omega = \sum_{i} dx_i \wedge dy_i$ ? $\endgroup$ Commented Feb 27, 2011 at 13:18
  • $\begingroup$ 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. $\endgroup$ Commented Feb 27, 2011 at 16:01
  • $\begingroup$ @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. $\endgroup$ Commented Feb 27, 2011 at 16:26

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

The first glance observation is correct (in some manner, maybe not the one asked here) - the standard symplectic form$$\omega=\sum dx^i\wedge dy^i$$can indeed be written as $$\omega=\frac{\sqrt{-1}}{2}\sum dz^id\overline{z}^i.$$These are just two ways to represent the same form, using different frames of the complexified cotangent bundle.

However, the (real!!) bilinear form on the holomorphic tangent bundle induced by $\omega$ is something else. Let $u=u^i\partial_{z^i},v=v^i\partial_{z^i},$ be two holomorphic tangent vectors. Under the specified (real!) isomorphism of the two bundles, $u$ and $v$ are mapped to$$u_\mathbb{R}=\mathrm{Re }u^i\partial_{x^i}+\mathrm{Im}u^i\partial_{y^i},\quad v_\mathbb{R}=\mathrm{Re}v^i\partial_{x^i}+\mathrm{Im}v^i\partial_{y^i}.$$Evaluating $\omega$ on these two vectors,$$\omega(u_\mathbb{R},v_\mathbb{R})=\sum\mathrm{Re}u^i\mathrm{Im}v^i-\mathrm{Im}u^i\mathrm{Re}v^i=\mathrm{Im}\sum\overline{u^i}v^i.$$Hence, the desired form is given by$$\omega_\mathbb{C}(u,v)=\mathrm{Im}\sum\overline{u^i}v^i.$$Once again: This form is not complex-bilinear, only real-bilinear (since it originates from a form on a real bundle).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .