5
$\begingroup$

I am interested in quaternionic-Kahler metrics that are "as inhomogeneous as possible."

Every complete quaternionic-Kahler manifold $X$ I can remember hearing of is a discrete quotient of some $Y$, such that $Isom(Y)$ contains a nontrivial connected Lie group. Are there any known examples of complete quaternionic-Kahler $X$ that don't arise in this way?

(By "quaternionic-Kahler manifold" I mean one with holonomy contained in $Sp(n)Sp(1)$ but not in $Sp(n)$ -- in other words, I want to exclude the hyperkahler case.)

$\endgroup$

1 Answer 1

6
$\begingroup$

I'm not familiar with the nonpositive case (negative since you exclude hyperkahler case) but there are no such examples known for postive quaternion Kahler manifolds (i.e. those with positive scalar curvature). They are all conjectured to be symmetric spaces (conjecture of LeBrun and Salamon) and this conjecture has been verified in dimensions 4 (Hitchin) and and 8 (Poon and Salamon). Certainly, only symmetric examples are known such as $\mathbb HP^n, Gr_2(\mathbb C^{n+2})$, $\widetilde {Gr_4}(\mathbb R^{n+4})$ (the Grassmanian of oriented real $4$-planes) and a few exceptional spaces such as $G_2/SO(4)$.

I looked around for what's known in negative case and in this paper LeBrun constructs an infinite dimensional family of negative quaternion Kahler metrics on $\mathbb R^{4n}$. I suspect most of these have no symmetries. But apparently, only locally symmetric compact examples are known though.

$\endgroup$
3
  • 1
    $\begingroup$ It has also been verified in dimension 12 by Herrera and Herrera: ams.org/mathscinet-getitem?mr=1979364 $\endgroup$ Commented Oct 21, 2011 at 23:46
  • 1
    $\begingroup$ there is a known gap in that paper and the 12-dimensional case is still open. $\endgroup$ Commented Oct 21, 2011 at 23:54
  • $\begingroup$ @Vitali: I didn't know about this. I heard about this result at a conference in 2006 in honour of Nigel Hitchin and at the time no doubt was expressed. I'll keep this in mind! Thank you. $\endgroup$ Commented Oct 22, 2011 at 0:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .