Let $(M, g, J)$ be a closed Kaehler manifold. Is there some more-or-less non-tautological condition ensuring that its group of orientation-preserving isometries is connected? Also the same question, but for holomorphic isometries.

In the case of curves, holomorphic isometry is the same thing as orientation-preserving isometry. For sphere and torus, I believe, the group under consideration is connected, but for curves of genus 2 it is not (being a group of order at least 2 and at most 48).

For projective spaces, I believe, all orientation-preserving isometries are Hamiltonian, so both groups coincide and are known to be connected.

The underlying smooth manifold of a K3 surface, being a closed $4k$-dimensional manifold with non-zero signature, does not admit orientation-reversing diffeomorphisms. However, there do exist K3 surfaces with anti-holomorphic isometries: this implies that the group of isometries can not be connected. Namely, for any closed Kaehler manifold, every Killing vector field is holomorphic, and since the isometry group is a compact Lie group (in compact-open topology), Lie group of holomorphic isometries is a union of some connected components of the group of isometries; existence of a non-holomorphic isometry implies non-connectedness of the group of isometries.

  • 2
    $\begingroup$ For a general Riemann surface of genus $\geq 3$ the automorphism group is trivial, hence connected, while it is disconnected for some special surfaces. I think that for holomorphic isometries this will be the standard situation. $\endgroup$ – abx Feb 25 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.