Canonical almost complex structure on symplectic manifold I'm trying to prove the following (well-known) theorem in symplectic geometry.
Theorem . For $(M, \omega)$ a symplectic manifold with Riemannian metric $g, \exists$ a canonical almost complex structure $J$ compatible with $ω$.
(source)
Could someone give me a sketch and I can fill in the details. Thank you.
 A: This is well-explained in a number of standard references such as da Silva or McDuff-Salamon's Introduction to Symplectic Topology.  I'll tour the exposition in da Silva (ch. 12).

We start with some basic symplectic linear algebra.  Let $V$ be a vector space with nondegenerate symplectic form $\Omega$ and Riemannian form $G$.


*

*There is a unique linear map $A: V \to V$ such that $\Omega(u,v) = G(Au,v)$ for all $u, v \in V$.  (Use nondegeneracy.)

*Let $A^*$ be the adjoint of $A$.  The map $AA^*$ is symmetric and positive-definite.  (Use the Riemannian property.)
Hence $AA^*$ diagonalizes with positive eigenvalues by the spectral theorem, and thus has a square root (defined eigenspace-by-eigenspace).  


*Let $S^2 = AA^*$, and write $J = S^{-1}A$.  This is the so-called polar decomposition.  Then $J$ commutes with $S$ and $A$.  Furthermore, $J$ is orthogonal ($JJ^* = I$), skew-adjoint ($J^* = -J$), and thus gives a complex structure on $V$.  Also, this complex structure is compatible with $\Omega$ and $G$, in the sense that $\Omega(Ju,Jv) = G(u,v)$ and $\Omega(u,Ju) > 0$.  (It's all linear algebra here; check da Silva if you get stuck.)



You can think about the above as being a pointwise version of the result you want.  Recall that an almost complex structure on a manifold is a smoothly varying family of complex structures on the tangent spaces, so all you have to do is check that $J$ depends smoothly on $G$ and $\Omega$ (this is not complicated - if you are working hard something is going wrong).
All this is "canonical" in a certain sense because we've made no choices along the way.  $A$ is unique and depends only on the maps $G$ and $\Omega$, and $J$ depends only on $A$.  We've made no reference to bases anywhere.
