Why do hyperkaehler manifolds have complex structure valued in a 2-sphere?

85 Views Asked by At

I understand that hyperkaehler manifolds have almost quaternionic structure, whereby there are three complex structures $I,J$ and $K$ which satisfy $$ I^2=J^2=K^2=IJK=-1. $$ It is also said that $aI+bJ+cK$ is also a complex structure, so long as $$ a^2+b^2+c^2=1, $$ where $a,b,c\in \mathbb{R}$. This last requirement implies that the complex structure of a hyperkaehler manifold is valued in a 2-sphere. But what is the reason we have this last requirement on the parameters $a,b$ and $c$?

2

There are 2 best solutions below

1
On BEST ANSWER

Hint Consider a general endomorphism of the form $L := a I + b J + c K$. Then, if $L$ is a complex structure, we must have $$-1 = L^2 = (a I + b J + c K)^2 = a^2 I^2 + ab IJ + ac IK + ab JI + \cdots .$$ We can deduce from the defining equation that $I, J, K$ all anticommute, i.e., satisfy $IJ = -JI$, etc., so we can see above that the contribution of the second and fourth terms together is $$ab IJ + ab JI = ab IJ + ab (-IJ) = 0 ,$$ and by symmetry the terms in $J, K$ and $K, I$ similarly vanish. On the other hand, since, e.g., $I$ is a complex structure, we have $I^2 = -1$, and likewise for $J, K$. Putting this all together and clearing signs gives the condition.

0
On

Note that

\begin{align*} & (aI + bJ + cK)(aI + bJ + cK)\\ =&\ a^2I^2 + abIJ + acIK + abJI + b^2J^2 + bcJK + acKI + bcKJ + c^2K^2\\ =&\ -a^2\operatorname{id} + abIJ + acIK - abIJ -b^2\operatorname{id} + bcJK -acIK - bcJK - c^2\operatorname{id}\\ =&\ -(a^2 + b^2 + c^2)\operatorname{id}. \end{align*}

So $aI + bJ + cK$ is an almost complex structure if and only if $a^2 + b^2 + c^2 = 1$.