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.

  • 1
    $\begingroup$ What does canonical mean? And where is the source? $\endgroup$ – user99914 Jan 5 '17 at 7:05
  • $\begingroup$ @JohnMa Just added source $\endgroup$ – Ashley Jan 5 '17 at 16:41
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    $\begingroup$ @Ashley: You will find all this done in Proposition 1 at page 2 in lecture 7 of the MIT OCW lecture notes on the geometry of manifolds by Denis Auroux. $\endgroup$ – Alex M. Jan 5 '17 at 18:03
  • $\begingroup$ Thanks, hadn't seen those before! $\endgroup$ – Ashley Jan 5 '17 at 22:16

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 $G$.

  1. 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.)

  2. 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).

  1. 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.

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    $\begingroup$ The Riemannian form $G$ is in general NOT compatible with $J$ an $\Omega$. $\endgroup$ – Vít Tuček Nov 28 '18 at 8:29
  • $\begingroup$ @VítTuček Yes, but we've built a specific $J$ (smoothly in $G$ and $\Omega$) so that they are compatible. Or did you have another objection that I'm not understanding? $\endgroup$ – Thurmond Apr 2 at 2:58
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    $\begingroup$ Nope, the constructed almost complex structure $J$ is not compatible with $G$ but with other scalar product. In other words, you have constructed $J$ such that $\Omega(u,Jv)$ is a scalar product which can be different from $G$. $\endgroup$ – Vít Tuček Apr 19 at 16:33

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