In reading Gauduchon's notes (cannot link them, anyway it is standard material) I ran into the following construction.


Let $(M, \omega\_0)$ be a compact symplectic manifold which is fixed.
An almost complex structure $J$ on $M$ is said to be compatible with $\omega\_0$ if $\omega\_0( \cdot, J \cdot)$ is a Riemannian metric.
Let $\mathcal J$ be the space of such $J$'s and let $\mathcal J\_\text{int}$ be the subspace of integrable ones.


Then a Riemannian metric and a complex structure are constructed on the infinite dimensional manifold $\mathcal J$ which turns out to be Kaehler and it is proved that $\mathcal J\_\text{int}$ is a Kaehler submanifold, that is its tangent space at $J$, which is constructed by deriving the equation $N\_J=0$, is invariant under the complex structure on $\mathcal J$.


My question is: can it happen that $\mathcal J\_\text{int}$ is made by isolated points? I can see no way to say that its tangent space has positive dimension.


Thank you


David



Integrable compatible complex structures
The space of integrable $J$ is acted on by the diffeomorphism group, making it infinite dimensional. Symplectomorphisms will preserve the compatibility, so the orbit of an integrable $J$ under the symplectomorphism group is still infinite dimensional. The quotient by the symplectomorphism group is not known to be even an orbifold, as far as I know.


