Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 :).

share|improve this question
    
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
7  
$\mathbb{R}^8$ 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. $J = K$.

share|improve this answer

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.