I have an explicit construction of the metric on the quaternionic Kähler manifold $$\mathcal M = \frac{Sp(1, 1)}{Sp(1) \times Sp(1)}.$$ Arranging the four real degrees of freedom into two complex ones - denoted by $z, \bar z, c, \bar c$ - I find that all components of the metric vanish, except $$ g_{z, \bar z} = g_{c, \bar c} = \frac{1}{(1 - z \bar z - c \bar c)^2}.$$ For details on the construction see this paper. Since the manifold is quaternionic Kähler, I figured there must be a Kähler potential $\mathcal K$ such that $$ \partial_i \partial_{\bar j} \mathcal K = g_{i, \bar j}.$$ However, directly integrating the above expressions for $g_{z, \bar z}$ and $g_{c, \bar c}$ does not work.

I tried to read up on quaternionic Kähler manifolds (and hyperkähler manifolds, since the two are closely related). Nowhere did I find an explicit statement that quaternionic Kähler manifolds are indeed Kähler manifolds!

Thus, my question is: Are quaternionic Kähler manifolds actually Kähler in the sense that the metric can be expressed as the double derivative of a real function?

  • 2
    $\begingroup$ Quaternionic Kahler is not Kahler in general. The former has its holonomy contained in $Sp(n)Sp(1)$. The latter in $U(2n)$. The former is not a subgroup of latter. $\endgroup$ – Gil Bor Jul 24 '14 at 18:18

This question was already answered by Gil Bor, but I just wanted to add some remarks.

Consider quaternionic projective space $\mathbb{H}P^n$ for instance. It is quaternion-Kahler, but cannot be Kahler for topological reasons, since it has $b_2 = 0$. On the other hand, a Kahler manifold has by definition a closed Kahler form, which is not exact, since its $m$'th power is a volume form, where $m$ is the complex dimension of the Kahler manifold. So a Kahler manifold must have $b_2 > 0$.

(As a remark, I believe that $\mathbb{H}P^n$ is not even almost complex for $n \geq 2$, but please check in the literature...)

Edit: $\mathbb{H}P^n$ is not almost complex for $n \geq 1$, see Michael Albanese's comment below.

  • 1
    $\begingroup$ This paper shows that $\mathbb{HP}^n$ does not admit an almost complex structure for any $n \geq 1$ (note that $n = 1$ is the case of the four-sphere). $\endgroup$ – Michael Albanese May 9 '17 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.