Any Hyperkähler manifold has 3 complex structures $I\_{1}, I\_{2}, I\_{3}$. Assume that there is an additional complex structure $J$. Can this be written as $J = aI\_{1} + bI\_{2} + cI\_{3}$, where $(a,b,c) \in S^{2} \subset \mathbb{R}^{3}$? Hope the question is not too trivial :).



Rotation in Hyperkähler manifolds
Assume $J$ is compatible with $g$ so that $J^\dagger = -J$. Then $J$ is a linear combination of $I\_i$ as above if and only if $J I\_i + I\_i J = -2 a\_i \mathbf{1}$, where $\mathbf{1}$ is the identity endomorphism and $a\_i$ are real numbers satisfying $a\_1^2 + a\_2^2 + a\_3^2 = 1$.


**Proof**: Necessity is obvious, since if $J = a\_1 I\_1 + a\_2 I\_2 + a\_3 I\_3$ then $I\_i J + J I\_i = -2a\_i \mathbf{1}$ directly from the quaternion relations. Conversely, suppose that $J I\_i + I\_i J = -2 a\_i \mathbf{1}$, with $\mathbf{a} \in S^2$. Set $K = a\_1 I\_1 + a\_2 I\_2 + a\_3 I\_3$. Then $K^2 = -\mathbf{1}$ and:
\begin{align}
(J-K)^2 &= (J - \sum a\_i I\_i)(J - \sum a\_i I\_i) \\\
&= J^2 - \sum a\_i (I\_i J + J I\_i) + K^2 \\\
&= -\mathbf{1} + 2 \sum a\_i^2 \mathbf{1} - \mathbf{1} \\\
&= -\mathbf{1} + 2 \mathbf{1} - \mathbf{1} = 0.
\end{align}
So $J-K$ is nilpotent. On the other hand, $J-K$ is skew adjoint, so in fact it must be identically zero, i.e. $J = K$.


