Every complex manifold is almost complex

Question

Recall that $M$ is a complex manifold if $M$ is a topological manifold and there is an atlas $\{ \varphi_i : U_i \to \mathbb C^n\}$ such that the transition functions $\varphi_j \circ \varphi_i^{-1}$ are holomorphic.

Why is that every complex manifold is almost complex? That is, how to construct the $(1,1)$ tensor $J$ so that $J^{2}=-id$?

Answer 1

If $M$ is a complex manifold, let $(z_1,...,z_n)$ be a local complex coordinates at the point $p$ of $M$. If $z_i=x_i+\sqrt{-1}y_i$ for $i=1,2,...,n$, then as a smooth manifold, $M$ has local coordinates $(x_1,...,x_n,y_1,...,y_n)$ and the tangent space at $p$ is spanned by $\displaystyle\frac{\partial}{\partial x_i}$ and $\displaystyle\frac{\partial}{\partial y_i}$ for $i=1,2,...,n.$ Define $J:T_pM\to T_pM$ by $$J(\frac{\partial}{\partial x_i})=\frac{\partial}{\partial y_i}\mbox{ and } J(\frac{\partial}{\partial y_i})=-\frac{\partial}{\partial x_i}\mbox{ for all }i=1,2,...,n.$$ Then $J^2=-id$ which shows that $J$ is an almost complex structure.

Answer 2

The purpose of this answer is to point out that the almost complex structure constructed in the other answer is well-defined, independent of the complex coordinates chosen.

If $(u_1, \cdots, u_n, v_1, \cdots, v_n)$ is another local complex coordinates of $M$ which intersects with the $(x_1, \cdots, x_n, y_1, \cdots, y_n)$ coordinates. Assume that $\tilde J$ is constructed so that

$$ \tilde J \left( \frac{\partial}{\partial u_i}\right) = \frac{\partial}{\partial v_i}, \ \ \ \tilde J \left( \frac{\partial}{\partial v_i}\right) =- \frac{\partial}{\partial u_i}$$

then by the first Cauchy Riemann equation $\frac{\partial u_j}{\partial x_i} = \frac{\partial v_j}{\partial y_i}$,

\begin{align} \tilde J \left( \frac{\partial}{\partial x_i}\right) &= \tilde J \left( \sum_j \frac{\partial u_j}{\partial x_i} \frac{\partial}{\partial u_j}\right) \\ &=\sum_j \frac{\partial u_j}{\partial x_i} \tilde J \left( \frac{\partial}{\partial u_j}\right) \\ &=\sum_j \frac{\partial v_j}{\partial y_i} \frac{\partial}{\partial v_j} \\ &= \frac{\partial}{\partial y_i}. \end{align}

and similarly $\tilde J \left( \frac{\partial }{\partial y_i}\right) =- \frac{\partial }{\partial x_i}$. Thus $\tilde J=J$ and so $J$ is well-defined.