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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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  • $\begingroup$ 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? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jul 4, 2013 at 22:56

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You may be understandably confused by a terminological inconsistency. People study so-called "Hyper-Kähler manifolds with torsion" (a.k.a. HKT manifolds), which by definition have

  • 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 $\operatorname{Sp}(n)$.

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