For the purposes of this question, a hyper-Kähler manifold will be a complete connected Riemannian manifold $(\mathcal{M},g)$ whose holonomy representation is isomorphic to the natural representation of the compact symplectic group $Sp_n$.

There is a $3$-dimensional real representation naturally associated to such a space. This can be given by the action of the isometry group permuting the holonomy subbundles within their common $Sp_nSp_1$-enlargement, or equivalently by rotating the $2$-sphere of parallel complex structures. This is summarized in the exact sequence \begin{equation*} Tri(\mathcal{M},g) \to Isom(\mathcal{M},g) \stackrel{\phi}\to SO(Im \mathbb{H}) \end{equation*} where the middle entry is the isometry group and on the left are the isometries that preserve all complex structures (tri-holomorphic). I would like to know about the representation $\phi$.

A Killing vector field on a compact Ricci-flat space is parallel. Therefore, provided the dimension is larger than one, the isometry group of a locally-irreducible compact Ricci-flat space is discrete. It is furthermore finite by compactness.

The finite subgroups of $SO_3$ are known to be of various forms: cyclic groups, dihedral groups, and the rotation groups of a regular tetrahedron, octahedron/cube and icosahedron/dodecahedron.

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(Fundamental domains of $C_4$, $D_4$ and the three polyhedral subgroups, images from Wikipedia.)

Q) Are examples known of compact hyper-Kähler manifolds that realise all of these types of subgroup through $\phi$?

I would appreciate any related remarks, as I have not been able to deduce much from the examples of compact hyper-Kähler manifolds that I have seen.

  • $\begingroup$ I don't really understand what you are asking. When $\mathcal{M}$ is compact and $g$ is a hyperKähler metric on $\mathcal{M}$, the isometry group is finite (not merely discrete) since it must be compact in the compact open topology and have the structure of a Lie group of dimension $0$, so the image of $\phi$ must be finite as well. $\endgroup$ – Robert Bryant Nov 9 '14 at 10:35
  • $\begingroup$ @RobertBryant, yes of course, I should have noticed that. I have edited the question to clarify. $\endgroup$ – Paul Reynolds Nov 9 '14 at 12:01
  • $\begingroup$ "It is a well-known fact that the isometry group of a compact Ricci-flat space is finite": this is false, take for example a compact flat torus. $\endgroup$ – YangMills Nov 17 '14 at 4:38
  • $\begingroup$ @YangMills, I ought to be more careful with my well-known 'facts'. I'll fix it now. $\endgroup$ – Paul Reynolds Nov 17 '14 at 20:29
  • $\begingroup$ I don't understand $\phi$. You take an isometry $F$ of a compact manifold $M$ and you map to a quaternionic matrix $A=\phi(F)$, apparently. What is this $\phi$? $\endgroup$ – Ben McKay Nov 17 '14 at 21:50

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