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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 :).

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I suppose it depends on how one interprets the question. The hyperkahler manifold comes equipped with a hyperkahler metric $g$. Do you want $J$ to be compatible with $g$ or not? –  Gunnar Þór Magnússon Nov 6 '12 at 9:29
Yes, I forgot to mention this? Is it then true? –  hapchiu Nov 6 '12 at 10:40
what is meant by compatibility with orientation ? –  hapchiu Nov 6 '12 at 10:56
$\mathbb{R}^8$ admits an $S^6$'s worth of compatible complex structures, given by multiplication by unit imaginary octonions. Do you want to add more restrictions? –  Paul Reynolds Nov 6 '12 at 16:05
It should be pointed out that the $4$-dimensional cases and the conformally flat cases are quite different from the general case (even the hyperKähler case). In a generic sense, a conformal $2n$-manifold that admits a compatible complex structure admits only one (up to sign). I suspect that the corresponding statement for hyperKähler manifolds is that, for a 'generic' hyperKähler manifold of dimension greater than $4$, its only compatible complex structures (even locally) are the ones belonging to the obvious $2$-sphere family mentioned above. –  Robert Bryant Jan 6 '13 at 14:01

2 Answers 2

up vote 4 down vote accepted

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. $J = K$.

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This is really just a comment but it doesn't quite fit so I'll make it an answer.

It seems worth commenting on the difference between the infinitesimal and local/global versions of this question.

Let's say we're in real dimension $4n$. Infinitesimally, i.e., on the tangent space at any point, orthogonal complex structures compatible with the orientation are parameterised by $SO(4n)/U(2n)$ (where the compatibility with orientation is to require the Pfaffian be positive). Since $\dim(SO(4n)/U(2n)) = 2n(2n-1) > 2$ for $n > 1$ it is clear that the property in the question fails infinitesimally for $n > 1$. The claim could thus only hold locally (or indeed globally) if the condition of being integrable somehow miraculously restricted to the hyperkahler $S^2 \subset SO(4n)/U(n)$ at each point, which is false. The easiest counter examples are any complex structure on $\mathbb{H}^n$ not in the standard hyperkahler family as in Paul Reynolds's very helpful comment.

In dimension $4$ things are slightly more interesting since $SO(4)/U(2) \simeq S^2$ so infinitesmally in 4 dimensions all the relevant complex structures are a linear combination of any hyperkahler triple. However the coefficients $a_i$ are not constant in general. To settle the matter we must therefore exhibit an integrable complex structure for which $a_i$ are non-constant. I confess I cannot think of a trivial example off the top of my head but at first glance this paper appears to discuss the matter, at least locally. (Certainly they exist.)

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