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Aug 2, 2017 at 11:57 vote accept Mtheorist
Aug 2, 2017 at 11:34 comment added Malkoun As another remark, the equation you are interested in is the same for quaternion-kähler manifolds too, but you need to be careful that, while the metric $g$ is globally defined, the almost complex structures $I$, $J$ and $K$ (or $J_u$ in your notation) are only locally defined, and there is no canonical choice for them, in the sense that an $SO(3)$-rotated choice of such local $I$, $J$ and $K$ is an equally valid choice.
Aug 2, 2017 at 11:25 comment added Malkoun In short, to answer your question directly for the hyperkähler case, it is enough to require that each $J_u$ is $g$-orthogonal, for each value of $u$ (between 1 and 3). Let us write $I = J_1$, $J = J_2$ and $K = J_3$, which is a notation close to that of the quaternions. Then for instance $g(Iv,Jw) = g(Iv,-IKw) = -g(v,Kw)$, since $I$ is $g$-orthogonal, so all other relations can be deduced using $g$-orthogonality, and the equation for the products of the $J_u$ that you wrote down. But please read my answer below as well.
Aug 2, 2017 at 10:35 answer added Malkoun timeline score: 2
Aug 2, 2017 at 10:35 history edited Mtheorist CC BY-SA 3.0
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Aug 2, 2017 at 10:33 comment added Mtheorist @Malkoun Thank you for pointing that out, I am mostly interested in hyperkahler manifolds, though don't mind knowing the distinction in the quaternionic-Kahler case. I have edited the question appropriately.
Aug 2, 2017 at 10:21 comment added Malkoun Look up hyperkähler geometry and quaternion-kähler geometry, and the distinction between the two. I did not want to give you straight the answer, because there are some subtleties involved, and you should be careful about them.
Aug 2, 2017 at 10:06 history asked Mtheorist CC BY-SA 3.0