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The Dirac equation on metrics of Eguchi–Hanson type

  • Zhuohua Cai , 1 ,
  • Xiao Zhang , 2, 3, 4,
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  • 1School of Mathematics and Information Science, Guangxi University, Nanning, Guangxi 530004, China
  • 2Guangxi Center for Mathematical Research, Guangxi University, Nanning, Guangxi 530004, China
  • 3Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China
  • 4School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China

Author to whom any correspondence should be addressed.

Received date: 2022-12-19

  Revised date: 2023-03-28

  Accepted date: 2023-03-28

  Online published: 2023-05-10

Copyright

© 2023 Institute of Theoretical Physics CAS, Chinese Physical Society and IOP Publishing

Abstract

We investigate parallel spinors on the Eguchi–Hanson metrics and find the space of complex parallel spinors is complex two-dimensional. For the metrics of Eguchi–Hanson type with the zero scalar curvature, we separate variables for the harmonic spinors and obtain the solutions explicitly.

Cite this article

Zhuohua Cai , Xiao Zhang . The Dirac equation on metrics of Eguchi–Hanson type[J]. Communications in Theoretical Physics, 2023 , 75(5) : 055002 . DOI: 10.1088/1572-9494/acc7f3

1. Introduction

Eguchi–Hanson metrics were constructed in [1, 2], which are complete four-dimensional Ricci flat, anti-self-dual ALE Riemannian metrics on R≥0 × P3. They are one type of gravitational instanton, and play important roles in the Euclidean approach of gravitational quantization [3]. Since then, gravitational instantons including the Eguchi–Hanson metrics attract much attention both in geometry and in general relativity.
In order to investigate the physical behaviour of spin-$\tfrac{1}{2}$ particles in general relativity, Chandrasekhar solved the Dirac equation in Kerr spacetimes by separating variables [4]. We refer to [5, 6] and references therein for further developments of the Dirac equation in various spacetimes. In [7], Sucu and Ünal solved the Dirac equation on Eguchi–Hanson and Bianchi VII0 gravitational instanton metrics. In particular, the solutions are obtained as the product of two hypergeometric functions for the Eguchi–Hanson gravitational instanton metric.
As it actually relates to manifold's scalar curvature, it is interesting to study the Dirac equation on much more general metrics of Eguchi–Hanson type with the zero scalar curvature. This type of metric was constructed by LeBrun using the method of algebraic geometry [8] and by the second author solving an ordinary differential equation [9] on R≥0 × S3/Zd (d > 2), and they provide the counter-examples of Hawking and Pope's generalized positive action conjecture.
In this note we first introduce the Eguchi–Hanson metrics, the metrics of Eguchi–Hanson type with the zero scalar curvature as well as the spin connections on them. Then we investigate the parallel spinors and the harmonic spinors. As the existence of a nontrivial parallel spinor on a Riemannian spin manifold implies that the manifold is Ricci-flat, parallel spinors only exist on the Eguchi–Hanson metrics, and we find the space of complex parallel spinors is complex two-dimensional. For the metrics of Eguchi–Hanson type with the zero scalar curvature, we separate variables for the harmonic spinors and obtain the solutions explicitly.

2. Metrics of Eguchi–Hanson type

Metrics of Eguchi–Hanson type are given by
$\begin{eqnarray}g={f}^{-2}{\rm{d}}{r}^{2}+{r}^{2}\left({\sigma }_{1}^{2}+{\sigma }_{2}^{2}+{f}^{2}{\sigma }_{3}^{2}\right),\end{eqnarray}$
where f is a real function of r, σ1, σ2 and σ3 are the Cartan–Maurer forms for SU(2) ≅ S3
$\begin{eqnarray*}\begin{array}{rcl}{\sigma }_{1} & = & \displaystyle \frac{1}{2}\left(\sin \psi {\rm{d}}\theta -\sin \theta \cos \psi {\rm{d}}\phi \right),\\ {\sigma }_{2} & = & \displaystyle \frac{1}{2}\left(-\cos \psi {\rm{d}}\theta -\sin \theta \sin \psi {\rm{d}}\phi \right),\\ {\sigma }_{3} & = & \displaystyle \frac{1}{2}\left({\rm{d}}\psi +\cos \theta {\rm{d}}\phi \right)\end{array}\end{eqnarray*}$
with three Euler angles θ, φ, ψ. The frame and coframe are
$\begin{eqnarray}{e}^{1}={f}^{-1}{\rm{d}}r,\quad {e}^{2}=r{\sigma }_{1},\quad {e}^{3}=r{\sigma }_{2},\quad {e}^{4}={rf}{\sigma }_{3},\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{rcl}{e}_{1} & = & f{\partial }_{r},\quad {e}_{2}=\displaystyle \frac{2}{r}\left(\sin \psi {\partial }_{\theta }-\displaystyle \frac{\cos \psi }{\sin \theta }{\partial }_{\phi }\right.\\ & & \left.+\displaystyle \frac{\cos \theta \cos \psi }{\sin \theta }{\partial }_{\psi }\right),\\ {e}_{3} & = & \displaystyle \frac{2}{r}\left(-\cos \psi {\partial }_{\theta }-\displaystyle \frac{\sin \psi }{\sin \theta }{\partial }_{\phi }+\displaystyle \frac{\cos \theta \sin \psi }{\sin \theta }{\partial }_{\psi }\right),\\ {e}_{4} & = & \displaystyle \frac{2}{{rf}}{\partial }_{\psi }\end{array}\end{eqnarray}$
respectively.
Let $\{{\omega }_{j}^{i}\}$ be the connection 1-forms given by ${\rm{d}}{e}^{i}=-{\omega }_{j}^{i}\wedge {e}^{j}$. Then [9]
$\begin{eqnarray}\begin{array}{rcl}{\omega }_{1}^{2} & = & \displaystyle \frac{f}{r}{e}^{2},\quad {\omega }_{1}^{3}=\displaystyle \frac{f}{r}{e}^{3},\quad {\omega }_{1}^{4}=\left(\displaystyle \frac{f}{r}+{f}^{{\prime} }\right){e}^{4},\\ {\omega }_{4}^{3} & = & \displaystyle \frac{f}{r}{e}^{2},\quad {\omega }_{2}^{4}=\displaystyle \frac{f}{r}{e}^{3},\quad {\omega }_{3}^{2}=\left(\displaystyle \frac{2}{{rf}}-\displaystyle \frac{f}{r}\right){e}^{4}.\end{array}\end{eqnarray}$
Let B > 0 be certain positive constant. Taking
$\begin{eqnarray*}f=\sqrt{1-\displaystyle \frac{B}{{r}^{4}}},\end{eqnarray*}$
and
$\begin{eqnarray*}r\geqslant \sqrt[4]{B},\quad 0\leqslant \theta \leqslant \pi ,\quad 0\leqslant \phi \leqslant 2\pi ,\quad 0\leqslant \psi \leqslant 2\pi ,\end{eqnarray*}$
metric (2.1) is the Eguchi–Hanson metric [1, 2]
$\begin{eqnarray}g={\left(1-\displaystyle \frac{B}{{r}^{4}}\right)}^{-1}{\rm{d}}{r}^{2}+{r}^{2}\left({\sigma }_{1}^{2}+{\sigma }_{2}^{2}+\left(1-\displaystyle \frac{B}{{r}^{4}}\right){\sigma }_{3}^{2}\right),\end{eqnarray}$
which is Ricci-flat and geodesically complete. Topologically, it is
$\begin{eqnarray*}{R}_{\geqslant 0}\times {SU}(2)/{Z}_{2}\cong {R}_{\geqslant 0}\times {SO}(3)\cong {R}_{\geqslant 0}\times {P}_{3}.\end{eqnarray*}$
Let d > 2 be a positive integer. Taking
$\begin{eqnarray*}\begin{array}{rcl}f & = & \sqrt{1-\displaystyle \frac{2A}{{r}^{2}}-\displaystyle \frac{B}{{r}^{4}}},\quad A=-\displaystyle \frac{d-2}{2}\sqrt{\displaystyle \frac{B}{d-1}},\\ {r}_{0} & = & \sqrt[4]{\displaystyle \frac{B}{d-1}}\end{array}\end{eqnarray*}$
and
$\begin{eqnarray*}r\geqslant {r}_{0},\quad 0\leqslant \theta \leqslant \pi ,\quad 0\leqslant \phi \leqslant 2\pi ,\quad 0\leqslant \psi \leqslant \displaystyle \frac{4\pi }{d},\end{eqnarray*}$
metric (2.1) is
$\begin{eqnarray}\begin{array}{rcl}g & = & {\left(1-\displaystyle \frac{2A}{{r}^{2}}-\displaystyle \frac{B}{{r}^{4}}\right)}^{-1}{\rm{d}}{r}^{2}\\ & & +{r}^{2}\left({\sigma }_{1}^{2}+{\sigma }_{2}^{2}+\left(1-\displaystyle \frac{2A}{{r}^{2}}-\displaystyle \frac{B}{{r}^{4}}\right){\sigma }_{3}^{2}\right),\end{array}\end{eqnarray}$
which is scalar-flat and geodesically complete, constructed in [8, 9]. Topologically, it is
$\begin{eqnarray*}{R}_{\geqslant 0}\times {SU}(2)/{Z}_{d}\cong {R}_{\geqslant 0}\times {S}^{3}/{Z}_{d}.\end{eqnarray*}$

3. Spin connection and parallel spinors

We refer to [10] for an introductory knowledge of spin geometry. It is clear that the underline manifolds of both metric (2.5) and metric (2.6) are spin and their complex spinor bundles are complex four-dimensional equipped with spin connection
$\begin{eqnarray*}{{\rm{\nabla }}}_{{e}_{k}}{\rm{\Phi }}={e}_{k}{\rm{\Phi }}+\displaystyle \frac{1}{4}\displaystyle \sum _{i,j=1}^{4}g\left({{\rm{\nabla }}}_{{e}_{k}}{e}_{i},{e}_{j}\right){e}_{i}\cdot {e}_{j}\cdot {\rm{\Phi }}\end{eqnarray*}$
for spinor ${\rm{\Phi }}={\left({{\rm{\Phi }}}_{1},{{\rm{\Phi }}}_{2},{{\rm{\Phi }}}_{3},{{\rm{\Phi }}}_{4}\right)}^{t}$. Using (2.4), we obtain
$\begin{eqnarray}\begin{array}{rcl}{{\rm{\nabla }}}_{{e}_{1}}{\rm{\Phi }} & = & {e}_{1}{\rm{\Phi }},\\ {{\rm{\nabla }}}_{{e}_{2}}{\rm{\Phi }} & = & {e}_{2}{\rm{\Phi }}+\displaystyle \frac{f}{2r}\left({e}_{1}\cdot {e}_{2}-{e}_{3}\cdot {e}_{4}\right)\cdot {\rm{\Phi }},\\ {{\rm{\nabla }}}_{{e}_{3}}{\rm{\Phi }} & = & {e}_{3}{\rm{\Phi }}+\displaystyle \frac{f}{2r}\left({e}_{1}\cdot {e}_{3}+{e}_{2}\cdot {e}_{4}\right)\cdot {\rm{\Phi }},\\ {{\rm{\nabla }}}_{{e}_{4}}{\rm{\Phi }} & = & {e}_{4}{\rm{\Phi }}+\displaystyle \frac{1}{2}\left(\left(\displaystyle \frac{f}{r}+{f}^{{\prime} }\right){e}_{1}\cdot {e}_{4}\right.\\ & & \left.+\left(\displaystyle \frac{f}{r}-\displaystyle \frac{2}{{rf}}\right){e}_{2}\cdot {e}_{3}\right)\cdot {\rm{\Phi }}.\end{array}\end{eqnarray}$
Throughout the paper, we fix the following representations for ei in Spin(4).
$\begin{eqnarray}\begin{array}{l}{e}_{1}\mapsto \left(\begin{array}{cccc}0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\end{array}\right),\qquad {e}_{2}\mapsto \left(\begin{array}{cccc}0 & 0 & 0 & {\rm{i}}\\ 0 & 0 & {\rm{i}} & 0\\ 0 & {\rm{i}} & 0 & 0\\ {\rm{i}} & 0 & 0 & 0\end{array}\right),\\ {e}_{3}\mapsto \left(\begin{array}{cccc}0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\end{array}\right),\qquad {e}_{4}\mapsto \left(\begin{array}{cccc}0 & 0 & {\rm{i}} & 0\\ 0 & 0 & 0 & -{\rm{i}}\\ {\rm{i}} & 0 & 0 & 0\\ 0 & -{\rm{i}} & 0 & 0\end{array}\right).\end{array}\end{eqnarray}$
Spinor Φ is parallel if ∇XΦ = 0 for any tangent vector X. Any Riemannian metric carried a parallel spinor must be Ricci-flat, see, e.g. [10]. Thus it is interesting to investigate the parallel spinors for the Eguchi–Hanson metric.

Parallel spinors for the Eguchi–Hanson metric (2.5) are complex constant spinors which form a complex two-dimensional space.

Under (3.2), we have

$\begin{eqnarray}{{\rm{\nabla }}}_{{e}_{1}}{\rm{\Phi }}={e}_{1}\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\\ {{\rm{\Phi }}}_{3}\\ {{\rm{\Phi }}}_{4}\end{array}\right)=0,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}_{{e}_{2}}{\rm{\Phi }}={e}_{2}\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\\ {{\rm{\Phi }}}_{3}\\ {{\rm{\Phi }}}_{4}\end{array}\right)+\displaystyle \frac{f}{2r}\left(\begin{array}{c}0\\ 0\\ -2{\rm{i}}{{\rm{\Phi }}}_{4}\\ -2{\rm{i}}{{\rm{\Phi }}}_{3}\end{array}\right)=0,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}_{{e}_{3}}{\rm{\Phi }}={e}_{3}\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\\ {{\rm{\Phi }}}_{3}\\ {{\rm{\Phi }}}_{4}\end{array}\right)+\displaystyle \frac{f}{2r}\left(\begin{array}{c}0\\ 0\\ 2{{\rm{\Phi }}}_{4}\\ -2{{\rm{\Phi }}}_{3}\end{array}\right)=0,\end{eqnarray}$
$\begin{eqnarray}{{\rm{\nabla }}}_{{e}_{4}}{\rm{\Phi }}={e}_{4}\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\\ {{\rm{\Phi }}}_{3}\\ {{\rm{\Phi }}}_{4}\end{array}\right)+\left(\displaystyle \frac{1}{{rf}}+\displaystyle \frac{{f}^{{\prime} }}{2}\right)\left(\begin{array}{c}0\\ 0\\ -{\rm{i}}{{\rm{\Phi }}}_{3}\\ {\rm{i}}{{\rm{\Phi }}}_{4}\end{array}\right)=0.\end{eqnarray}$
Firstly, (3.3) implies
$\begin{eqnarray*}{\partial }_{r}{\rm{\Phi }}=0.\end{eqnarray*}$
Secondly, (3.4) gives
$\begin{eqnarray*}\left(\sin \psi {\partial }_{\theta }-\displaystyle \frac{\cos \psi }{\sin \theta }{\partial }_{\phi }+\displaystyle \frac{\cos \theta \cos \psi }{\sin \theta }{\partial }_{\psi }\right)\left(\begin{array}{c}{{\rm{\Phi }}}_{3}\\ {{\rm{\Phi }}}_{4}\end{array}\right)=\displaystyle \frac{f}{r}\left(\begin{array}{c}{\rm{i}}{{\rm{\Phi }}}_{4}\\ {\rm{i}}{{\rm{\Phi }}}_{3}\end{array}\right).\end{eqnarray*}$
Since the left hand side is independent on r, but the right hand side contains function $f(r)$, it must be
$\begin{eqnarray*}{{\rm{\Phi }}}_{3}={{\rm{\Phi }}}_{4}=0.\end{eqnarray*}$
Now (3.6) gives
$\begin{eqnarray*}{\partial }_{\psi }{\rm{\Phi }}=0.\end{eqnarray*}$
Thus (3.4), (3.5) give
$\begin{eqnarray*}\begin{array}{l}\displaystyle \frac{2}{r}\left(\sin \psi {\partial }_{\theta }-\displaystyle \frac{\cos \psi }{\sin \theta }{\partial }_{\phi }\right)\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\end{array}\right)=0,\\ \displaystyle \frac{2}{r}\left(-\cos \psi {\partial }_{\theta }-\displaystyle \frac{\sin \psi }{\sin \theta }{\partial }_{\phi }\right)\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\end{array}\right)=0.\end{array}\end{eqnarray*}$
Therefore
$\begin{eqnarray*}{\partial }_{\theta }\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\end{array}\right)\,=\,{\partial }_{\phi }\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\end{array}\right)\,=\,0.\end{eqnarray*}$
So ${{\rm{\Phi }}}_{1}$ and ${{\rm{\Phi }}}_{2}$ are arbitrary complex constants and the proof of the theorem is complete.

4. Harmonic spinors

Harmonic spinors are the solutions of the Dirac equation
$\begin{eqnarray*}{ \mathcal D }=\displaystyle \sum _{k=1}^{4}{e}_{k}\cdot {{\rm{\nabla }}}_{{e}_{k}}=0.\end{eqnarray*}$
We shall find the explicit solutions of harmonic spinors on metric (2.6) of Eguchi–Hanson type with zero scalar curvature (d > 2). In the case d = 2, harmonic spinors on Eguchi–Hanson metric (2.5) were solved by Sucu and Ünal in [7], and the solutions can be obtained in terms of hypergeometric functions.

Let positive integer $d\gt 2$. There exists harmonic spinors

$\begin{eqnarray*}{\rm{\Phi }}={{\rm{e}}}^{{\rm{i}}\left(n+\displaystyle \frac{1}{2}\right)\phi }\left(\begin{array}{c}{C}_{1}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\psi }{u}_{1}(r){v}_{1}(\theta )\\ {C}_{2}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\psi }{u}_{2}(r){v}_{2}(\theta )\\ {C}_{3}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\psi }{u}_{3}(r){v}_{3}(\theta )\\ {C}_{4}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\psi }{u}_{4}(r){v}_{4}(\theta )\end{array}\right),\end{eqnarray*}$
on metric (2.6) of Eguchi–Hanson type with zero scalar curvature, where m, n are integers, Ci are complex constants
$\begin{eqnarray}\left\{\begin{array}{l}{u}_{1}(r)={\left({r}^{2}-{r}_{0}^{2}\right)}^{\tfrac{1-d-{dm}}{2d}}{\left({r}^{2}+(d-1){r}_{0}^{2}\right)}^{-\tfrac{1-d-{dm}}{2d}-\tfrac{d\left(1+2m\right)}{4}},\\ {u}_{2}(r)={\left({r}^{2}-{r}_{0}^{2}\right)}^{\tfrac{1-d+{dm}}{2d}}{\left({r}^{2}+(d-1){r}_{0}^{2}\right)}^{-\tfrac{1-d+{dm}}{2d}-\tfrac{d\left(1-2m\right)}{4}},\\ {u}_{3}(r)=\displaystyle \frac{1}{r}{\left({r}^{2}-{r}_{0}^{2}\right)}^{-\tfrac{1}{2}-\tfrac{1-d-{dm}}{2d}}{\left({r}^{2}+(d-1){r}_{0}^{2}\right)}^{-\tfrac{1}{2}+\tfrac{1-d-{dm}}{2d}+\tfrac{d\left(1+2m\right)}{4}},\\ {u}_{4}(r)=\displaystyle \frac{1}{r}{\left({r}^{2}-{r}_{0}^{2}\right)}^{-\tfrac{1}{2}-\tfrac{1-d+{dm}}{2d}}{\left({r}^{2}+(d-1){r}_{0}^{2}\right)}^{-\tfrac{1}{2}+\tfrac{1-d+{dm}}{2d}+\tfrac{d\left(1-2m\right)}{4}},\end{array}\right.\end{eqnarray}$
$\begin{eqnarray}\left\{\begin{array}{l}{v}_{1}(\theta )={v}_{3}(\theta )={\left(\sin \displaystyle \frac{\theta }{2}\right)}^{n+\tfrac{1}{2}-\tfrac{d}{2}\left(m+\tfrac{1}{2}\right)}{\left(\cos \displaystyle \frac{\theta }{2}\right)}^{-n-\tfrac{1}{2}-\tfrac{d}{2}\left(m+\tfrac{1}{2}\right)},\\ {v}_{2}(\theta )={v}_{4}(\theta )={\left(\sin \displaystyle \frac{\theta }{2}\right)}^{-n-\tfrac{1}{2}+\tfrac{d}{2}\left(m-\tfrac{1}{2}\right)}{\left(\cos \displaystyle \frac{\theta }{2}\right)}^{n+\tfrac{1}{2}+\tfrac{d}{2}\left(m-\tfrac{1}{2}\right)}.\end{array}\right.\end{eqnarray}$

Denote

$\begin{eqnarray*}{F}_{1}=\displaystyle \frac{f}{r}+\displaystyle \frac{{f}^{{\prime} }}{2}-\displaystyle \frac{1}{{rf}},\quad {F}_{2}=\displaystyle \frac{2f}{r}+\displaystyle \frac{{f}^{{\prime} }}{2}+\displaystyle \frac{1}{{rf}}.\end{eqnarray*}$
Under (3.2), harmonic spinors satisfy
$\begin{eqnarray}\begin{array}{l}\left(f{\partial }_{r}-\displaystyle \frac{2{\rm{i}}}{{rf}}{\partial }_{\psi }+{F}_{1}\right){{\rm{\Phi }}}_{1}\\ \quad -\displaystyle \frac{2}{r}{{\rm{e}}}^{{\rm{i}}\psi }\left({\partial }_{\theta }-{\rm{i}}\,\csc \,\theta {\partial }_{\phi }+{\rm{i}}\,\cot \,\theta {\partial }_{\psi }\right){{\rm{\Phi }}}_{2}=0,\\ \left(f{\partial }_{r}+\displaystyle \frac{2{\rm{i}}}{{rf}}{\partial }_{\psi }+{F}_{1}\right){{\rm{\Phi }}}_{2}\\ \quad +\displaystyle \frac{2}{r}{{\rm{e}}}^{-{\rm{i}}\psi }\left({\partial }_{\theta }+{\rm{i}}\,\csc \,\theta {\partial }_{\phi }-{\rm{i}}\,\cot \,\theta {\partial }_{\psi }\right){{\rm{\Phi }}}_{1}=0,\\ \left(f{\partial }_{r}+\displaystyle \frac{2{\rm{i}}}{{rf}}{\partial }_{\psi }+{F}_{2}\right){{\rm{\Phi }}}_{3}\\ \quad +\displaystyle \frac{2}{r}{{\rm{e}}}^{{\rm{i}}\psi }\left({\partial }_{\theta }-{\rm{i}}\,\csc \,\theta {\partial }_{\phi }+{\rm{i}}\,\cot \,\theta {\partial }_{\psi }\right){{\rm{\Phi }}}_{4}=0,\\ \left(f{\partial }_{r}-\displaystyle \frac{2{\rm{i}}}{{rf}}{\partial }_{\psi }+{F}_{2}\right){{\rm{\Phi }}}_{4}\\ \quad -\displaystyle \frac{2}{r}{{\rm{e}}}^{-{\rm{i}}\psi }\left({\partial }_{\theta }+{\rm{i}}\,\csc \,\theta {\partial }_{\phi }-{\rm{i}}\,\cot \,\theta {\partial }_{\psi }\right){{\rm{\Phi }}}_{3}=0.\end{array}\end{eqnarray}$
Separating variables
$\begin{eqnarray}\left(\begin{array}{c}{{\rm{\Phi }}}_{1}\\ {{\rm{\Phi }}}_{2}\\ {{\rm{\Phi }}}_{3}\\ {{\rm{\Phi }}}_{4}\end{array}\right)={{\rm{e}}}^{{\rm{i}}\left(n+\displaystyle \frac{1}{2}\right)\phi }\left(\begin{array}{c}{{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\psi }{h}_{1}(r,\theta )\\ {{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\psi }{h}_{2}(r,\theta )\\ {{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\psi }{h}_{3}(r,\theta )\\ {{\rm{e}}}^{{\rm{i}}\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\psi }{h}_{4}(r,\theta )\end{array}\right),\end{eqnarray}$
where m, n are integers, the Dirac equation gives
$\begin{eqnarray}{{\rm{e}}}^{{\rm{i}}\left(\frac{d}{2}-1\right)\psi }\vec{L}(r,\theta )=\vec{R}(r,\theta ),\end{eqnarray}$
where
$\begin{eqnarray*}\begin{array}{rcl}\vec{L}(r,\theta ) & = & \left(\begin{array}{c}\displaystyle \frac{{rf}}{2}\left({\partial }_{r}+\displaystyle \frac{d\left(2m+1\right)}{2{{rf}}^{2}}+\displaystyle \frac{1}{r}+\displaystyle \frac{{f}^{{\prime} }}{2f}-\displaystyle \frac{1}{{{rf}}^{2}}\right){h}_{1}\\ \left({\partial }_{\theta }-\left(n+\displaystyle \frac{1}{2}\right)\,\csc \,\theta +\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\,\cot \,\theta \right){h}_{1}\\ -\displaystyle \frac{{rf}}{2}\left({\partial }_{r}-\displaystyle \frac{d\left(2m+1\right)}{2{{rf}}^{2}}+\displaystyle \frac{2}{r}+\displaystyle \frac{{f}^{{\prime} }}{2f}+\displaystyle \frac{1}{{{rf}}^{2}}\right){h}_{3}\\ \left({\partial }_{\theta }-\left(n+\displaystyle \frac{1}{2}\right)\,\csc \,\theta +\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\,\cot \,\theta \right){h}_{3}\end{array}\right),\\ \vec{R}(r,\theta ) & = & \left(\begin{array}{c}\left({\partial }_{\theta }+\left(n+\displaystyle \frac{1}{2}\right)\,\csc \,\theta -\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\,\cot \,\theta \right){h}_{2}\\ -\displaystyle \frac{{rf}}{2}\left({\partial }_{r}-\displaystyle \frac{d\left(2m-1\right)}{2{{rf}}^{2}}+\displaystyle \frac{1}{r}+\displaystyle \frac{{f}^{{\prime} }}{2f}-\displaystyle \frac{1}{{{rf}}^{2}}\right){h}_{2}\\ \left({\partial }_{\theta }+\left(n+\displaystyle \frac{1}{2}\right)\,\csc \,\theta -\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\,\cot \,\theta \right){h}_{4}\\ \displaystyle \frac{{rf}}{2}\left({\partial }_{r}+\displaystyle \frac{d\left(2m-1\right)}{2{{rf}}^{2}}+\displaystyle \frac{2}{r}+\displaystyle \frac{{f}^{{\prime} }}{2f}+\displaystyle \frac{1}{{{rf}}^{2}}\right){h}_{4}\end{array}\right).\end{array}\end{eqnarray*}$
Note that $\vec{L}(r,\theta )$ and $\vec{R}(r,\theta )$ are independent with ψ, and $d\geqslant 3$, we obtain
$\begin{eqnarray}\vec{L}(r,\theta )=\vec{R}(r,\theta )=0.\end{eqnarray}$
Thus, for $i=1,3$, j = 2, 4, we have
$\begin{eqnarray*}\begin{array}{rcl}{\partial }_{\theta }{h}_{i} & = & \left(\left(n+\displaystyle \frac{1}{2}\right)\,\csc \,\theta -\displaystyle \frac{d}{2}\left(m+\displaystyle \frac{1}{2}\right)\,\cot \,\theta \right){h}_{i},\\ {\partial }_{\theta }{h}_{j} & = & \left(-\left(n+\displaystyle \frac{1}{2}\right)\,\csc \,\theta +\displaystyle \frac{d}{2}\left(m-\displaystyle \frac{1}{2}\right)\,\cot \,\theta \right){h}_{j}.\end{array}\end{eqnarray*}$
Solving these ordinary differential equations, we obtain
$\begin{eqnarray*}\begin{array}{rcl}{h}_{i} & = & {u}_{i}(r){\left(\sin \displaystyle \frac{\theta }{2}\right)}^{n+\tfrac{1}{2}-\tfrac{d}{2}\left(m+\tfrac{1}{2}\right)}{\left(\cos \displaystyle \frac{\theta }{2}\right)}^{-n-\tfrac{1}{2}-\tfrac{d}{2}\left(m+\tfrac{1}{2}\right)},\\ {h}_{j} & = & {u}_{j}(r){\left(\sin \displaystyle \frac{\theta }{2}\right)}^{-n-\tfrac{1}{2}+\tfrac{d}{2}\left(m-\tfrac{1}{2}\right)}{\left(\cos \displaystyle \frac{\theta }{2}\right)}^{n+\tfrac{1}{2}+\tfrac{d}{2}\left(m-\tfrac{1}{2}\right)}.\end{array}\end{eqnarray*}$
Substituting them into (4.6), we have
$\begin{eqnarray*}\begin{array}{rcl}{u}_{1}^{{\prime} } & = & \displaystyle \frac{1}{{r}^{4}-2{{Ar}}^{2}-B}\left({Ar}-d\left(m+\displaystyle \frac{1}{2}\right){r}^{3}\right){u}_{1},\\ {u}_{2}^{{\prime} } & = & \displaystyle \frac{1}{{r}^{4}-2{{Ar}}^{2}-B}\left({Ar}+d\left(m-\displaystyle \frac{1}{2}\right){r}^{3}\right){u}_{2},\\ {u}_{3}^{{\prime} } & = & \displaystyle \frac{1}{r\left({r}^{4}-2{{Ar}}^{2}-B\right)}\left(3{{Ar}}^{2}+B\right.\\ & & \left.+\left(d\left(m+\displaystyle \frac{1}{2}\right)-3\right){r}^{4}\right){u}_{3},\\ {u}_{4}^{{\prime} } & = & \displaystyle \frac{1}{r\left({r}^{4}-2{{Ar}}^{2}-B\right)}\left(3{{Ar}}^{2}+B\right.\\ & & \left.-\left(d\left(m-\displaystyle \frac{1}{2}\right)+3\right){r}^{4}\right){u}_{4}.\end{array}\end{eqnarray*}$
Solving these ordinary differential equations, we obtain (4.1) and (4.2). This gives the proof of theorem.

Finally we study the singular sets of harmonic spinors constructed in theorem 4.1. The singularity occurs when at least one of the power indices of rr0 is negative in (4.1) or at least one of the power indices of $\sin \displaystyle \frac{\theta }{2}$ or $\cos \displaystyle \frac{\theta }{2}$ is negative in (4.2). The relations between polar coordinates {r, θ, ψ, φ} and Cartesian coordinates {x0, x1, x2, x3} of the metric (2.6) are [9]
$\begin{eqnarray*}\begin{array}{rcl}{x}_{1} & = & r\cos \displaystyle \frac{\theta }{2}\cos \displaystyle \frac{\psi +\phi }{2},\qquad {x}_{2}=r\cos \displaystyle \frac{\theta }{2}\sin \displaystyle \frac{\psi +\phi }{2},\\ {x}_{3} & = & r\sin \displaystyle \frac{\theta }{2}\cos \displaystyle \frac{\psi -\phi }{2},\qquad {x}_{0}=r\sin \displaystyle \frac{\theta }{2}\sin \displaystyle \frac{\psi -\phi }{2}.\end{array}\end{eqnarray*}$
Since rr0, we obtain
$\begin{eqnarray}\sin \displaystyle \frac{\theta }{2}=0\,\Longleftrightarrow \,{x}_{3}={x}_{0}=0,\quad \cos \displaystyle \frac{\theta }{2}=0\,\Longleftrightarrow \,{x}_{1}={x}_{2}=0.\end{eqnarray}$
Now we denote
$\begin{eqnarray*}\begin{array}{rcl}{a}_{m}^{\pm } & = & \displaystyle \frac{1-d\pm {dm}}{2d},\\ {a}_{{mn}}^{\pm } & = & n+\displaystyle \frac{1}{2}+\displaystyle \frac{d}{2}\left(-m\pm \displaystyle \frac{1}{2}\right),\\ {b}_{{mn}}^{\pm } & = & n+\displaystyle \frac{1}{2}+\displaystyle \frac{d}{2}\left(m\pm \displaystyle \frac{1}{2}\right).\end{array}\end{eqnarray*}$
It is clearly
$\begin{eqnarray}{a}_{{mn}}^{+}-{a}_{{mn}}^{-}=\displaystyle \frac{d}{2},\quad {b}_{{mn}}^{+}-{b}_{{mn}}^{-}=\displaystyle \frac{d}{2}.\end{eqnarray}$

The harmonic spinor Φ, constructed in theorem 4.1, as well as its norm $| {\rm{\Phi }}| $ are singular at $r={r}_{0}$. Moreover, it holds that

(1) If $-\tfrac{d}{2}\lt {a}_{{mn}}^{-}\lt 0$, $-\tfrac{d}{2}\lt {b}_{{mn}}^{-}\lt 0$, v1, v2, v3, v4 and $| {\rm{\Phi }}| $ are singular either at ${x}_{1}{x}_{2}$-plane, or at ${x}_{3}{x}_{0}$-plane;

(2) If $-\tfrac{d}{2}\lt {a}_{{mn}}^{-}\lt 0$, ${b}_{{mn}}^{-}\geqslant 0$, either v1, v2, v3, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{1}{x}_{2}$-plane, or v1, v3, $| {\rm{\Phi }}| $ are singular at ${x}_{3}{x}_{0}$-plane;

(3) If $-\tfrac{d}{2}\lt {a}_{{mn}}^{-}\lt 0$, ${b}_{{mn}}^{-}\leqslant -\tfrac{d}{2}$, either v1, v2, v3, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{1}{x}_{2}$- plane, or v2, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{3}{x}_{0}$-plane;

(4) If ${a}_{{mn}}^{-}\geqslant 0$, $-\tfrac{d}{2}\lt {b}_{{mn}}^{-}\lt 0$, either v2, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{1}{x}_{2}$-plane, or v1, v2, v3, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{3}{x}_{0}$-plane;

(5) If ${a}_{{mn}}^{-}\geqslant 0$, ${b}_{{mn}}^{-}\geqslant 0$, v2, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{1}{x}_{2}$-plane, or v1, v3, $| {\rm{\Phi }}| $ are singular at ${x}_{3}{x}_{0}$-plane,

(6) If ${a}_{{mn}}^{-}\geqslant 0$, ${b}_{{mn}}^{-}\leqslant -\tfrac{d}{2}$, v2, v4, $| {\rm{\Phi }}| $ are singular either at ${x}_{1}{x}_{2}$- plane or at ${x}_{3}{x}_{0}$-plane;

(7) If ${a}_{{mn}}^{-}\leqslant -\tfrac{d}{2}$, $-\tfrac{d}{2}\lt {b}_{{mn}}^{-}\lt 0$, either v1, v3, $| {\rm{\Phi }}| $ are singular at ${x}_{1}{x}_{2}$-plane, or v1, v2, v3, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{3}{x}_{0}$-plane;

(8) If ${a}_{{mn}}^{-}\leqslant -\tfrac{d}{2}$, ${b}_{{mn}}^{-}\geqslant 0$, v1, v3, $| {\rm{\Phi }}| $ are singular either at ${x}_{1}{x}_{2}$-plane, or at ${x}_{3}{x}_{0}$-plane;

(9) If ${a}_{{mn}}^{-}\leqslant -\tfrac{d}{2}$, ${b}_{{mn}}^{-}\leqslant -\tfrac{d}{2}$, either v1, v3, $| {\rm{\Phi }}| $ are singular at ${x}_{1}{x}_{2}$-plane, or v2, v4, $| {\rm{\Phi }}| $ are singular at ${x}_{3}{x}_{0}$-plane.

The norm of harmonic spinors in theorem 4.1 is

$\begin{eqnarray*}| {\rm{\Phi }}{| }^{2}=\displaystyle \sum _{i=1}^{4}| {{\rm{\Phi }}}_{i}{| }^{2}=\displaystyle \sum _{i=1}^{4}| {C}_{i}{| }^{2}{u}_{i}^{2}{v}_{i}^{2}.\end{eqnarray*}$
If any ui or vi goes to $\infty $ at some subset of the space, the harmonic spinors and their norm become singular there. The power orders of $r-{r}_{0}$ in ui are
$\begin{eqnarray*}{a}_{m}^{-},\quad {a}_{m}^{+},\quad -\displaystyle \frac{1}{2}-{a}_{m}^{-},\quad -\displaystyle \frac{1}{2}-{a}_{m}^{+}\end{eqnarray*}$
respectively. Note that m, d are integers and $d\gt 2$. Thus
$\begin{eqnarray*}\begin{array}{l}m=0\Longrightarrow {a}_{m}^{\pm }\lt 0,\quad -\displaystyle \frac{1}{2}-{a}_{m}^{\pm }\lt 0,\\ m\leqslant -1\Longrightarrow {a}_{m}^{-}\gt 0,\quad -\displaystyle \frac{1}{2}-{a}_{m}^{+}\gt 0,\\ {a}_{m}^{+}\lt 0,\quad -\displaystyle \frac{1}{2}-{a}_{m}^{-}\lt 0,\\ m\geqslant 1\Longrightarrow {a}_{m}^{-}\lt 0,\quad -\displaystyle \frac{1}{2}-{a}_{m}^{+}\lt 0,\\ {a}_{m}^{+}\gt 0,\quad -\displaystyle \frac{1}{2}-{a}_{m}^{-}\gt 0.\end{array}\end{eqnarray*}$
This indicates that r0 is the singular point of the harmonic spinor Φ as well as its norm $| {\rm{\Phi }}| $.

From (4.2), we obtain

$\begin{eqnarray*}\begin{array}{rcl}{v}_{1}(\theta ) & = & {v}_{3}(\theta )={\left(\sin \displaystyle \frac{\theta }{2}\right)}^{{a}_{{mn}}^{-}}{\left(\cos \displaystyle \frac{\theta }{2}\right)}^{-{b}_{{mn}}^{-}-\tfrac{d}{2}},\\ {v}_{2}(\theta ) & = & {v}_{4}(\theta )={\left(\sin \displaystyle \frac{\theta }{2}\right)}^{-{a}_{{mn}}^{-}-\tfrac{d}{2}}{\left(\cos \displaystyle \frac{\theta }{2}\right)}^{{b}_{{mn}}^{-}}.\end{array}\end{eqnarray*}$
Therefore the rest parts of the theorem follow from the above formulas, (4.7) and the fact that $\sin \tfrac{\theta }{2}$, $\cos \tfrac{\theta }{2}$ cannot be both zero.

The authors would like to thank the referees for valuable suggestions. This work is supported by Chinese NSF grants 11731001, the special foundations for Guangxi Ba Gui Scholars and Junwu Scholars of Guangxi University.

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