Is the group $Ham(M,\omega)\cap Iso_{0}(M,g)$ compact?

let $(M,J,g,\omega)$ be a compact K\"ahler manifold of complex dimension at least $2$. As usual $J$ is the complex structure, $\omega$ is the symplectic form, $g$ is the Riemannian metric and $$\omega(\cdot,\cdot)=g(J\cdot, \cdot)\,.$$

I denote with $Ham(M,\omega)$ the group of Hamiltonian symplectomorphisms w.r.t. the symplectic form $\omega$ and with $Iso_{0}(M,g)$ the identity component of the group of isometries w.r.t. the metric $g$.

Is the group $Ham(M,\omega)\cap Iso_{0}(M,g)$ compact? Maybe it is an easy question but i haven't figured out the answer yet. Any suggestion or hint is welcome and if the question is not well suited for this site i'll move it to MSE.

• The group of isometries that are also symplectomorphisms is closed, ok. But, a priori, this group is bigger than the one of hamiltonian isometries. How do you see that $Ham(M,\omega)$ is closed? Moreover, which topology you put on it to say that? The only topology i can think about is the metric one induced by the Hofer norm and still i cannot prove that the group of hamiltonian isometries is closed. Mar 9, 2015 at 16:46
• You should give a look at Chapter 10 of the McDuff-Salamon book: it is almost an exercise to prove that $Ham(M,\omega)$ is normal in $Symp(M,\omega)$. More interestingly, it corresponds to the kernel of a very interesting (continuous) morphism. At the end of the chapter there is a proof that $Ham(M,\omega)$ is a Lie group, whose Lie algebra is the algebra of Hamiltonian vector fields. Mar 9, 2015 at 17:13
• @student The Hofer topology on $Ham(M, \omega)$ is not particularly well-behaved. A better topology is the smooth compact-open topology (or sometimes called $C^{\infty}$-topology). Moreover, for Kähler manifolds a diffeomorphim which preserves two of the structures $g, \omega, J$ also preserves the third one -- for example an isometry of $g$ which preserves $\omega$ also preserves $J$. Mar 11, 2015 at 16:07