I am wondering if the following is true:

Let $(M,\omega)$ be a compact symplectic manifold which is also monotone, i.e. $c_1(TM)=\lambda [\omega]$. Moreover assume that it admits a Hamiltonian circle action with isolated fixed points. This forces $\lambda$ to be positive, hence we can rescale the symplectic form to satisfy $$ c_1(TM)=[\omega].$$

Is there any result already in the literature that says that in this case $(M,\omega)$ admits a Kähler structure (whose integrable $J$ is compatible with $\omega$)? This structure is not required to be invariant under the action.


Recall that a compact symplectic manifold $(M, \omega)$ is called Fano if $c_1(M)=[\omega]\in H^2(M, \mathbb{R})$. As is stated in the question body, compact monotone symplectic manifold admitting a Hamiltonian $S^1$-action with isolated fixed points provide examples of symplectic Fano manifolds (possibly after rescaling).

In real dimension 4, it's known that a compact symplectic Fano manifold is diffeomorphic to a del Pezzo surface (Ohta&Ono). Since any two cohomologous symplectic forms on a del Pezzo surface are equivalent and symplectic cone of a del Pezzo surface coincides with its Kaehler cone, we see that any compact symplectic Fano manifold $(M, \omega)$ of dimension 4 admits an integrable a.c.s compatible with $\omega$ (just pullback the a.c.s. from the corresponding del Pezzo surface). Therefore, the answer to your question in real dimension 4 is yes.

In real dimension 6, there is a conjecture stating that a compact symplectic Fano manifold admitting a Hamiltonian $S^1$-action is diffeomorphic to a Fano 3-fold. It has been actually proved that such a manifold is necessarily simply-connected and satisfies $c_1c_2=24$ (for symplectic Chern classes). The conjecture, as far as I know, is still open in full generality.

The picture in higher dimensions is probably even more complicated (for instance, starting from real dimension 12 there are examples of not simply-connected symplectic Fano manifolds; of course, such symplectic manifolds can not admit compatible integrable a.c.s. since they would have a trivial fundamental group then).


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.