# Why can't hyper-kahler manifolds have a connection with torsion?

I have often seen the statement that Hyper-Kahler (HK) manifolds have torsion-free connections. In general relativity, however, one is usually taught that the connection is something that you can "choose". You might choose to use the Levi-Civita connection or not. Is it not possible for a HK manifold to choose a connection with torsion?

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I suppose the question is what the putative connection with torsion is supposed to satisfy. Presumably you would like it to be metric and preserving the hyperkähler structure? –  José Figueroa-O'Farrill Jul 4 '13 at 22:56

• 3 complex structures $I$, $J$, $K$ satisfying the quaternion relations,
• a Riemannian metric $g$ Hermitian with respect to $I$, $J$ and $K$,
• a connection $\nabla$ relative to which $g$, $I$, $J$, $K$ are parallel,
such that: the tensor $g(\cdot,T(\cdot,\cdot))$ defined by the torsion $T$ of $\nabla$ is totally antisymmetric (a 3-form).
The irony, observed in the very review of their introduction, is that unless $T=0$, these manifolds are not Hyper-Kähler in the classical sense of having their (Levi-Civita) holonomy contained in $\operatorname{Sp}(n)$.