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A Riemannian manifold is hyperkähler, if there are three complex structures $I,J,K$, which are all compatible with the Riemannian metric (i.e., $(v,Iw)$ defines a symplectic form and similarly for $J$ and $K$). Furthermore, we also need the complex structures to satisfy the quaternionic relations $I^2=J^2=K^2=-1$ and $IJ+JI=0=IK+KI=JK+KJ$.

I was able to find only very few examples of closed (compact without boundary) hyperkähler manifolds, basically complex tori and K3-surfaces. In higher dimensions than complex dimension 2, there are also generalized Kummer varieties of tori. Based on K3-surfaces, there are Hilbert schemes and resolutions of singularities in some moduli spaces.

I am searching for an example of a slightly more general setting: Let $X$ be a closed Riemannian manifold and $J_1,..., J_r$ be complex structures, all compatible with the Riemannian metric. Furthermore, assume that the $J_l$ to satisfy the relations $J_l^2=-1$ and $J_lJ_k+J_kJ_l=0$ for $k\neq l$.

If we take $X$ to be a vector space of dimension $2^{4a+b}c$ with $c$ odd, I know the maximal number for $r$ is $8a+2^b-1$, i.e. the maximal number is always odd. This carries over to quotients of vector spaces, i.e. for tori and quotients of tori. These examples are flat.

This yields the question: Is there a closed, non-flat manifold admitting more than three such complex structures?

If $X$ admits more than three complex structures, it also admits three such structures. Therefore every such manifold has to be a hyperkähler manifold, i.e. one of the above mentioned examples or a product of them. Are there different examples and does one of them admit more complex structures than the 3 for hyperkähler?

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I don't know the answer, but hyperkähler manifolds arise naturally from the classification of Riemannian manifolds in terms of their holonomy. What restriction does the existence of your $r$ "orthogonal" complex structures compatible with the metric impose on the holonomy group? –  Victor Protsak Mar 9 '11 at 1:51
Thanks for you reply. I came across the holomony definition of hyperkähler manifolds myself. But I don't know anything about the holonomy conditions arising from more complex structures. But I am looking for the case $r>3$, so in particular, these manifolds are hyperkähler. That already fixes the holonomy group to be Sp(n) for a 4n-dim manifold. So the existence of more structures probably does not restrict anything more. –  Doris Hein Mar 9 '11 at 4:03
The structures you're looking for are typically called "Clifford Structures". See, e.g., the recent preprint "Clifford Structures on Riemannian Manifolds", by Moroianu and Semelmann. That preprint focuses on local almost complex structures, while you seem interested in global complex structures, but the preprint should give some ideas on what's possible, and what restrictions are out there. –  Marty Mar 9 '11 at 7:10
On another note - I'd change the title of the question. Hyperkahler manifolds have an infinite number of complex structures, so the title doesn't really reflect the otherwise interesting question. –  Marty Mar 9 '11 at 7:12
It looks like the paper you want to read is: Manifolds with Many Complex Structures by Dominic Joyce, in Quart. J. Math. Oxford, vol. 46 (1995), 169-184. It's available here: eprints.maths.ox.ac.uk/61/1/complex.pdf In it, he constructs examples of manifolds with a prescribed number of complex structures with the relations (and subsequent restrictions) as you desire. –  Brendan Foreman Mar 9 '11 at 11:37

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A manifold admitting a triple of complex structures satisfying quaternionic relations also admits a torsion-free connection (called "Obata connection") preserving the quaternionic structure. Such a connection is unique. For a hyperkaehler manifold, the Obata connection coinsides with the Levi-Civita. Therefore, the manifolds with this structure have Levi-Civita connection which preserves the complex structures $J_i$. From Berger's classification of holonomy it follows that they are products of hyperkaehler manifolds, and actually flat if $i>3$ (because $J_4$ must exchange the tangent bundles to the factors of the product).

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P. S. Of course, if you weaken your assumption by removing the Kaehler metric, the answer would be different - there are some examples of manifolds with big number of anticommuting complex structures, even compact ones, e.g. nilmanifolds. –  Misha Verbitsky Mar 9 '11 at 21:47

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