Can a quaternionic Kähler manifold be NOT Kähler?

Question

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?

Answer 1

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.