I'm starting to study hyperKähler manifolds and I have a question for you. I hope it's not too stupid...


I know that given a Kähler metric $g$ on an hyperKähler manifold $X=(M,I)$ ($M$ is the differential structure and $I$ the complex structure) there are $J$, $K$ complex structures on $M$ with quaternionic relations such that for every $(a,b,c)\in S^2$ the linear combination $aI+bJ+cK$ is a complex structure which makes $g$ Kähler.


The form $\sigma\_I := g(J-,-)+ig(K-,-)$ should be a $(2,0)$-form for $X$ which gives the isomorphism $K\_X\simeq \mathcal{O}\_X$.


But I have trouble proving it: let's take $u$ and $v$ eigenvectors for $I$ (seen as an almost complex structure) with eigenvalue $i$. Then
$$\sigma\_I(u,v)=g(Ju,v)+ig(Ku,v)=2g(Ju,v)$$
But $\sigma\_I$ should be a nowhere zero $(2,0)$-form, so it should be $2g(Ju,v)\neq 0$. Why is that? I just know that $g(J-,-)$ is a $(1,1)$-form for $(M,J)$.



Holomorphic (2,0)-form on HyperKähler manifolds
Not sure which of the following is giving you trouble, so here is both:


* $\sigma\_I$ is **nondegenerate** (i.e. $\sigma\_I(u,\cdot) = 0$ implies $u=0$) for the simple reason that its real and imaginary parts are nondegenerate, since $g$ is.
* $\sigma\_I$ is a **(2,0)-form** relative to the complex structure $I$ because, letting $z^a$ be complex coordinates for that structure and $(\partial/\partial z^a,\partial/\partial\bar z^a)$ the resulting basis of $TX\otimes\mathbf C$, we have $I\partial/\partial\bar z^a = -i\partial/\partial\bar z^a$ and hence
\begin{align}
\sigma\_I(\partial/\partial\bar z^a,\cdot)
&=g(J\partial/\partial\bar z^a,\cdot)+ig(K\partial/\partial\bar z^a,\cdot)\\
&=g(KI\partial/\partial\bar z^a,\cdot)+ig(K\partial/\partial\bar z^a,\cdot)\\
&=g(-iK\partial/\partial\bar z^a,\cdot)+ig(K\partial/\partial\bar z^a,\cdot)\\
&=0.
\end{align}
Thus we get $\sigma\_I=\sigma\_{ab}dz^a\wedge dz^b$ where $\sigma\_{ab}=\sigma\_I(\partial/\partial z^a,\partial/\partial z^b)$; i.e. no $d\bar z^a$ terms, as claimed.


