most recent 30 from http://mathoverflow.net 2013-05-24T15:01:07Z http://mathoverflow.net/feeds/question/57900 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57900/existence-of-closed-manifolds-with-more-than-3-linearly-independent-complex-struc Doris Hein 2011-03-09T01:46:56Z 2011-03-09T13:44:48Z

<p>A Riemannian manifold is <em>hyperkähler</em>, 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 $IJ+JI=0=IK+KI=JK+KJ$.</p> <p>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.</p> <hr> <p>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 $k\neq l$.</p> <p>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.</p> <p>This yields the question: Is there a closed, non-flat manifold admitting more than three such complex structures?</p> <p>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?</p>

http://mathoverflow.net/questions/57900/existence-of-closed-manifolds-with-more-than-3-linearly-independent-complex-struc/57932#57932 Misha Verbitsky 2011-03-09T08:27:50Z 2011-03-09T08:27:50Z

<p>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).</p>