7
$\begingroup$

As is well-known (see Friedrich's book for example) every Kähler manifold is spin (or at least spin$^c$) and the Dirac is given (up to a twist) by $\partial + \partial^*$. What happens in the quaternionic-Kaehler and hyper-Kähler cases? Are they spin$^c$, does the Dirac admit a nice description? I'm particularly interested in the case of Wolf spaces (Quaternion-Kähler symmetric space).

$\endgroup$

2 Answers 2

6
$\begingroup$

A very natural Dirac operator for Wolf spaces is discussed in Köhler, K, Weingart, G., Quaternionic analytic torsion, Adv. Math. 178 (2003), 375–395. It acts on subcomplexes of the de Rham complex, in analogy with the Kähler situation.

EDIT: There is a hierarchy of groups $$Spin(n)\hookrightarrow Spin^c(n)=Spin(n)\cdot U(1)\hookrightarrow Spin^h(n)=Spin(n)\cdot Sp(1)\twoheadrightarrow SO(n)\;.$$ Here "$\cdot$" means "product divided by the diagonal $\mathbb Z/2$ action". If there is a principal bundle $P$ with structure group $Spin^x$ together with an equivariant map to the $SO(n)$ frame bundle of an oriented Riemannian manifold, then $P$ is called a spin$^x$ structure ($x$ being either void, or $c$, or $h$). Associated to $P$ is a real, complex, or quaternionic spinor bundle, respectively.

Every QK manifold comes with a natural spin$^h$ structure. For the Wolf spaces, these are automatically equivariant, see this answer. One of the Dirac operators in [op. cit.] acts on the corresponding quaternionic spinor bundles. Some Wolf spaces are spin, e.g. $\mathbb HP^k$, $G_2(\mathbb C^n)$ or $G_4(\mathbb R^n)$, the last two for $n$ even. Some are spin$^c$ but not spin, e.g. $G_2(\mathbb C^n)$ , and some are not even spin$^c$, e.g. $G_4(\mathbb R^n)$ for odd $n$, if I am not mistaken.

$\endgroup$
1
  • 1
    $\begingroup$ A reference for the spin$^h$ structure on (almost) quaternionic manifolds is Lemma 3.9 of Elliptic Symbols by Bar. $\endgroup$ Mar 4, 2021 at 23:10
2
$\begingroup$

It is a result of Salamon that $8n$-dimensional quaternion-Kähler manifolds are spin.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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