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.

Thank you in advance.


1 Answer 1


By Ascola-Arzelà theorem, the group of isometries of a compact metric space is itself compact and then a compact connected Lie Group. The condition that the isometries preserve the complex structure is a closed condition, so your subgroup is compact.

  • $\begingroup$ I asked for Hamiltonian isometries, are you saying that the requirement of being hamiltonian is equivalent to being a complex isometry? $\endgroup$
    – student
    Mar 9, 2015 at 15:12
  • $\begingroup$ Sorry, I was thinking of symplectomorphisms. Anyhow, the group of Hamiltonian diffeos is closed as well (more than that, it's an infinite dimensional Lie group), so your group is compact. $\endgroup$ Mar 9, 2015 at 15:49
  • $\begingroup$ 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. $\endgroup$
    – student
    Mar 9, 2015 at 16:46
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 9, 2015 at 17:13
  • 1
    $\begingroup$ @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 $J$. $\endgroup$ Mar 11, 2015 at 16:07

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