# Hyper-Complex and quaternionic Kahler Geometry

What is the exact relationship hyper-complex and quaternionic Kahler manifolds? From Wikipedia we get that hyper-Kahler manifolds are both hyper-complex and quaternionic Kahler. Thus, the two families have a non-empty intersection. But this is all I can conclude.

Moreover, quaternionic projective space is quaternionic Kahler, but is it also hyper-Kahler? Is it even hyper-complex?

-

By a recent result of Gauduchon, Moroianu and Semmelmann if a positive quaternion Kahler manifold admits any almost complex structure then it must be the complex Grassmannians $Gr_2(\mathbb C^{n+2})$. The latter is not hypercomlex so there are no hypercomplex manifolds which are also positive Quaternion Kahler. Note that it's even easier if you assume that the hypercomplex structure is compatible with the quaternion Kahler structure which the above argument does not assume but I think the original question did assume. That implies that the twistor bundle is trivial which is well-known not to be possible for positive quaternion Kahler manifolds. I don't know what happens in the negative quaternion Kahler case but I suspect there are no examples there either at least among compact ones. So I would guess that any compact hypercomplex manifold with holonomy in $Sp(n)\cdot Sp(1)$ must be hyper-Kahler.

-
The link has just been useful, thanks @Vitali. – Paul Reynolds Nov 19 '13 at 21:39

[First paragraph has been edited, after Vitali's comments below.] According to one convention, hyperKahler manifolds are not actually quaternionic-Kahler. This is the case if you define hyperKahler as having holonomy exactly $\mathrm{Sp}(m)$, and quaternionic-Kahler as having holonomy exactly $\mathrm{Sp}(n) \cdot \mathrm{Sp}{1} = (\mathrm{Sp}(n) \times \mathrm{Sp}(1)) / \lbrace \pm 1 \rbrace$, where in both cases the real dimension is $4m$.

In fact, these "strictly" quaternionic-Kahler manifolds are NOT even Kahler, so they are poorly named, but unfortunately, the name has stuck. HyperKahler manifolds are always Ricci-flat, but quaternionic-Kahler manifolds are always Einstein with nonzero Einstein constant.

By contrast, hypercomplex manifolds need not have a Riemannian metric at all. Being hypercomplex means that one has a triple $I, J, K$ of almost complex structures which satisfy the quaternion multiplication relations. I can't remember if one requires them to all be integrable or if this is always true. On a hyperKahler manifold, we certainly have such a structure, so it is hypercomplex. A quaternionic-Kahler manifold does not admit such complex structures globally, only locally, but the $4$-form $\Omega = \omega_I^2 + \omega_J^2 + \omega_K^2$ is actually globally defined, where $\omega_I = g(I \cdot, \cdot)$ is the associated Kahler form to $I$.

Added later: Quaternionic projective space $\mathbb H \mathbb P^n$ is not hyperKahler. For example, $\mathbb H \mathbb P^1 \cong S^4$, which cannot even be Kahler, since $b_2(S^4) = 0$.

-
I once asked Calabi why the terminology "quaternionic-Kähler" was chosen since these manifolds are, after all, not Kähler, and he said that they wanted the quaternionic projective spaces and Grassmannians to be "quaternionic" (which means a torsion-free $GL(n,{\mathbb{H}})Sp(1)$ structure) and the natural quaternionic analog of a metric reduction would then be "quaternionic-Kahler", in analogy with the notion of the natural metric reduction of a complex manifold would be "Kähler". – Robert Bryant Dec 14 '11 at 21:17
This depends on your definitions but I would personally disagree with your point of view that hyper-Kahler manifolds are not quaternionic Kahler. $Sp(n)$ is a subgroup of $Sp(n)\cdot Sp(1)$ and I would argue that anything which has holonomy $Sp(n)$ is a special case of holonomy in $Sp(n)\cdot Sp(1)$. Likewise, I would never say that Calabi-Yau manifolds are not Kahler. According to your logic they would have to be because they have holonomy $SU(n)$ and not full $U(n)$. – Vitali Kapovitch Dec 14 '11 at 21:21
@Vitali: your logic is completely correct, of course. I am just stating what (I believe) is the more widely held opinion on the terminology. – Spiro Karigiannis Dec 15 '11 at 2:35
I edited my answer to make it clear that whether hyperKahler manifolds are quaternionic-Kahler is a matter of convention. It is unambiguous, however, that $\mathbb H \mathbb P^n$ is not hyperKahler. – Spiro Karigiannis Dec 15 '11 at 15:54