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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.

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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).

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