# a Kähler manifold is symplectic

Let $\omega(X,Y)=g(JX,Y)$ on a Kähler manifold. We will prove that $\omega$ is a symplectic form.

• $\omega$ is anti-symmetric

$\omega(X,Y)=g(JX,Y)=g(Y,JX)=g(JY,J^{2}X)=g(JY,-X)=-g(JY,X)=-\omega(Y,X)$. Here we used the fact that $g$ is an Hermitian tensor on a Kähler manifold ($g(X,Y)=g(JX,JY)$)

• $\omega$ is linear

Due to anti-symmetry, we just need to check linearity on the second slot. Since $g(JX,\cdot)$ is by definition linear, $\omega$ will also be linear.

• $\omega$ is non degenerate

On a given point on the manifold, pick a non null vector $X$, $\alpha_{X}(\cdot)=\omega(X,\cdot)=g(JX,\cdot)$. Since $g$ is non-degenerate11no vector but the null vector is orthogonal to every other vector, $\alpha$ is also non-degenerate (for all $X$). $\omega$ is thus non degenerate.

• $\omega$ is closed

First note that

 $\displaystyle X(\omega(Y,Z))$ $\displaystyle=$ $\displaystyle\nabla_{X}(\omega(Y,Z))$ $\displaystyle=$ $\displaystyle\nabla_{X}(g(JY,Z))$ $\displaystyle=$ $\displaystyle g(\nabla_{X}(JY),Z)+g(JY,\nabla_{X}Z)$ $\displaystyle=$ $\displaystyle g(J\nabla_{X}Y,Z)+g(JY,\nabla_{X}Z)$ $\displaystyle=$ $\displaystyle\omega(\nabla_{X}Y,Z)+\omega(Y,\nabla_{X}Z)$

Here we used the fact that both $g$ and $J$ are covariantly constant ($\nabla g=0$ and $\nabla J=0$)

We aim to prove that $d\omega=0$ which is equivalent to proving $(d\omega)(X,Y,Z)=0$ for all vector fields $X,Y,Z$.

Since this is a tensorial identity, WLOG we can assume that at a specific point $p$ in the Kähler manifold $[X,Y]_{p}=[Y,Z]_{p}=[Z,X]_{p}=0$ and prove the indentity for these vector fields22in particular this works for the canonical base of $T_{p}M$ associated with a local coordinate system.

Consider $X,Y,Z$ with the previous commutation relations at $p$, using the formulas for differential forms of small valence:

 $\displaystyle(d\omega)(X,Y,Z)$ $\displaystyle=$ $\displaystyle X(\omega(Y,Z))+Y(\omega(Z,X)+Z(\omega(X,Y)))$ $\displaystyle=$ $\displaystyle\omega(\nabla_{X}Y,Z)+\omega(Y,\nabla_{X}Z)+$ $\displaystyle\omega(\nabla_{Y}Z,X)+\omega(Z,\nabla_{Y}X)+$ $\displaystyle\omega(\nabla_{Z}X,Y)+\omega(X,\nabla_{Z}Y)$ $\displaystyle=$ $\displaystyle\omega(\nabla_{X}Y-\nabla_{Y}X,Z)+\omega(\nabla_{Y}Z-\nabla_{Z}Y,% X)+\omega(\nabla_{Z}X-\nabla_{X}Z,Y)$

The Levi-Civita connection is torsion-free, $\nabla_{X}Y-\nabla_{Y}X=[X,Y]$ thus:

 $(d\omega)(X,Y,Z)=\omega([X,Y],Z)+\omega([Y,Z],X)+\omega([Z,X],Y)$

And since all the commutators are null at $p$ (by assumption) we get that:

 $(d\omega)(X,Y,Z)=0$

$\omega$ is therefore closed.

Title a Kähler manifold is symplectic AKahlerManifoldIsSymplectic 2013-03-22 16:07:54 2013-03-22 16:07:54 cvalente (11260) cvalente (11260) 15 cvalente (11260) Result msc 53D99 KahlerManifold