Let $(M^{2n},\omega)$ be a symplectic manifold and suppose that $J$ is an almost complex structure on $M$ such that $\omega$ tames $J$, i.e. $\omega(X,JX)>0$ for all nonzero $X\in TM$.

The space $\mathcal{J}_{\tau}(M,\omega)$ of almost complex structures tamed by $\omega$ is path-connected. In particular, any two almost complex structures in $\mathcal{J}_{\tau}(M,\omega)$ are homotopic (actually isotopic).

I am curious about the converse. Let $J'$ be an almost complex structure on $M$ which is in the same homotopy class as $J$.

Question: Is $J'$ necessarily tamed by $\omega$, i.e. in $\mathcal{J}_{\tau}(M,\omega)$?


In dimension 2, any almost complex structure (a priori unrelated to the symplectic form $\omega$) which is homotopic (among almost complex structures) to a $\omega$-tamed almost complex structure also turns out to be $\omega$-tamed. On any symplectic manifold of dimension $4$ or higher, this is not the case.

Proof: Since almost complex structures are endomorphisms of the tangent bundle, it is sufficient to prove the statement for symplectic vector spaces using only linear algebra arguments.

Given a symplectic vector space $(V^{2n}, \omega)$, there exists a symplectic basis, so we can assume without lost of generality that we work with $(\mathbb{R}^{2n}, \omega_{0})$ with cartesian coordinates $(x_1, \dots, x_n, y_1, \dots, y_n)$ and standard symplectic form $\omega_0 = \sum_{i=1}^n dx_i \wedge dy_i$.

Dimension 2: Consider the basis element $e_1$ of the symplectic basis $(e_1, f_1)$, so that $\omega_0(e_1, f_1) = 1$. Let $L := \mathbb{R}\langle e_1 \rangle$ be the 1-dimensional subspace spawned by $e_1$ and set $S = \mathbb{R}^2 \setminus L$. Notice that $S$ has two (path-)connected components. I claim that the set of (almost) complex structures on $\mathbb{R}^2$ is in bijection with $S$ (in fact diffeomorphic to $S$) via the map $J \mapsto J(e_1)$; I leave the proof of this fact (i.e. that the map is well-defined, injective and surjective) as an exercise. Assuming this claim, it is easy to see that $J$ is $\omega_0$-tamed if and only if it is homotopic to the standard complex structure $J_0$ (which sends $e_1$ to $f_1$).

Higher dimensions: It suffices to prove the statement for dimension 4. Consider the symplectic basis $(e_1, e_2, f_1, f_2)$ and the standard symplectic form $\omega_0$ for that basis. Set (with respect to that basis)

$$ J_t = \left( \begin{array}{cc} \sin(\pi t) j_0 & - \cos(\pi t) I \\ \cos(\pi t) I & -\sin(\pi t) j_0 \end{array} \right) \quad \mbox{ where } t \in [0,1], \; I = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right), \, j_0 = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) \; .$$

Simple computations show that $J_t^2 = - Id$, that $J_0$ is $\omega_0$-tamed and that $J_1 = - J_0$ is not $\omega_0$-tamed. $\square$


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