2
$\begingroup$

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 $\mathcal{J}_{\tau}(M,\omega)$?

$\endgroup$
1
$\begingroup$

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. $\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.