## FANDOM

162 Pages Attribution: This question is taken from this Maths Overflow question by "Xin Nie" with it's answer given by "Robert Bryant".

## Question Edit

Every expository article on hyperkähler manifolds that I have read states without detailed proof the following fact:

It follows from Yau's theorem (i.e. a compact Kähler manifold $M$ with $c_1(M)=0$ admits a Ricci-flat Kähler metric) that if a compact Kähler manifold $M$ has a complex symplectic form $\omega_\mathbb{C}$ then there exists a Kähler form $\omega$ on $M$ such that $\omega$ and $\omega_\mathbb{C}$ constitute a hyperkähler structrue.

I have trouble trying to see this. Indeed, for the metric to be hyperkähler we require the holonomy group to be in $Sp(n)$ (where $\mathrm{dim}_\mathbb{R}M=4n$), so a least the holonomy should be in $Sp(2n,\mathbb{C})$, which means that the Levi-Civita covariant derivation of $\omega_\mathbb{C}$ is $0$. This is a condition stronger than Ricci-flatness at first glance and I can't see how to realize this only knowing the existence of Ricci-flat metric.

## Answers Edit

### Answer 1 Edit

The point is that a Bochner formula shows that, if the Kähler metric is Ricci flat and the manifold is compact, then every global holomorphic $p$-form must be parallel.

In particular, if you have a complex, compact symplectic manifold that is Kähler, then you have a nonvanishing holomorphic volume form (the top power of the holomorphic symplectic $2$-form). You can then use Yau's theorem to find a Kähler metric whose volume form is a constant multiple of this holomorphic volume form wedged with its conjugate and this metric will have vanishing Ricci curvature. Then, by the Bochner formula, the symplectic form has to be parallel with respect to the Levi-Civita connection of the Kähler metric. In particular, this implies that the holonomy preserves this form and hence is a subgroup of $\mathrm{Sp}(n,\mathbb{C})\cap\mathrm{U}(2n)=\mathrm{Sp}(n)$. If you know, for some reason, that the manifold is not a product, then Berger's classification of holonomy coupled with the deRham holonomy theorem implies that the metric must have restricted holonomy equal to $\mathrm{Sp}(n)$.

#### Comments Edit

Community content is available under CC-BY-SA unless otherwise noted.