$\mathbb{C}$ can be interpreted as $\mathbb{R}^2$ with a complex structure $J$ .

Parametrize $\mathbb{R}^2$ via the usual coordinates $(x,y)$ .

A point $z$ in the complex plane can thus be written $z=x+iy$ .

The tangent space at each point is generated by the $\mathrm{span}_{\mathbb{R}} \left\{ \frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right\}$ and the complex structure $J$ is defined by ¹

	(1)
	(2)

So lets verify all the points in the definition.

• $\mathbb{C}$ is a Riemannian Manifold
• $g$ is Hermitian. $$ g\left(J\frac{\partial}{\partial x}, J\frac{\partial}{\partial y}\right) = g\left(\frac{\partial}{\partial y},-\frac{\partial}{\partial x} \right)=0=g\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right) \\ $$ $$ g\left(J\frac{\partial}{\partial x}, J\frac{\partial}{\partial x}\right) = g\left(\frac{\partial}{\partial y}, \frac{\partial}{\partial y}\right)=1 = g\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial x}\right) $$

  $$ g\left(J\frac{\partial}{\partial y}, J\frac{\partial}{\partial y}\right) = g\left(-\frac{\partial}{\partial x}, -\frac{\partial}{\partial x}\right)=1 = g\left(\frac{\partial}{\partial y}, \frac{\partial}{\partial y}\right) $$

• $J$ is covariantly constant because its components in the $(x,y)$ coordinates are constant and as previously stated, the covariant derivatives are just partial derivatives in this example.

$\mathbb{C}$ is therefore a Kähler manifold.

The symplectic form for this example is

$$\omega = dx \wedge dy $$

This is the simplest example of a Kähler manifold and can be seen as a template for other less trivial examples. Those are generalizations of this example just as Riemannian manifolds are generalizations of $\mathbb{R}^n$ seen as a metric space.

Footnotes

... by¹

notice $J$ acts as a counterclockwise rotation by $\frac{\pi}{2}$ , just as expected

... coordinates²

the Christoffel symbols on these coordinates vanish