# Holomorphic (2,0)-form on HyperKähler manifolds

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 $(M,J)$.

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• $\sigma_I$ 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.