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? 



Why can't hyper-kahler manifolds have a connection with torsion?
You may be understandably confused by a terminological inconsistency. People study so-called ***"[Hyper-Kähler manifolds with torsion](http://www.scielo.org.ar/scielo.php?script=sci_arttext&pid=S0041-69322008000200012)"*** (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](http://www.ams.org/mathscinet-getitem?mr=1396267) 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 $\operatorname{Sp}(n)$.


