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# Complete quaternionic Kähler manifolds with finite volume ends

V. Cortés  Department of Mathematics
University of Hamburg
Bundesstraße 55, D-20146 Hamburg, Germany  M. Röser  Department of Mathematics
University of Hamburg
Bundesstraße 55, D-20146 Hamburg, Germany  D. Thung  Department of Mathematics
University of Hamburg
Bundesstraße 55, D-20146 Hamburg, Germany

###### Abstract

We construct examples of complete quaternionic Kähler manifolds with an end of
finite volume, which are not locally homogeneous. The manifolds are aspherical
with fundamental group which is up to an infinite cyclic extension a semi-
direct product of a lattice in a semi-simple group with a lattice in a
Heisenberg group. Their universal covering is a cohomogeneity one deformation
of a symmetric space of non-compact type.

_Keywords: quaternionic Kähler manifolds, $c$-map, one-loop deformation,
isometry groups, cohomogeneity one_

_MSC classification: 53C26._

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 The (one-loop deformed) $c$-map
    2. 2.2 Symmetries and the $c$-map
  3. 3 The $c$-map applied to complex hyperbolic spaces
    1. 3.1 The infinitesimal action
    2. 3.2 $\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}$ as an algebra of Killing fields
    3. 3.3 The action of $\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$
  4. 4 Quotients and ends of finite volume
    1. 4.1 Quotients from arithmetic lattices
    2. 4.2 The volume of fiberwise quotients by lattices

##  1 Introduction

One of the milestones of twentieth century differential geometry is Berger’s
classification of irreducible holonomy groups of non-locally symmetric
Riemannian manifolds [Ber55]. The problem to construct non-locally symmetric
Riemannian manifolds (or to show their non-existence) for each of the groups
in Berger’s list has been a major driving force for research in the theory of
Einstein manifolds, involving a combination of Lie theory [Ale68, Ale75],
differential geometry [Bry87, BS89, LeB89, LeB91], geometric analysis [Yau78,
Joy96a, Joy96b] and algebraic geometry [Bea83]. As a result, we have compact
non-locally symmetric examples of Riemannian manifolds for all of the groups
in Berger’s list with exception of the group $\mathrm{Sp}(n)\mathrm{Sp}(1)$.

Riemannian manifolds with holonomy group a subgroup of
$\mathrm{Sp}(n)\mathrm{Sp}(1)$ for $n\geq 2$ are known as quaternionic Kähler
manifolds. These manifolds are Einstein and, in fact, generalize half-
conformally flat Einstein four-manifolds, which are by definition the
quaternionic Kähler manifolds of dimension four.

All the known compact examples of non-zero scalar curvature are locally
symmetric. They fall into two classes: symmetric quaternionic Kähler manifolds
of compact type, known as Wolf spaces [Wol65], and locally symmetric spaces
obtained as smooth quotients of the non-compact duals of the Wolf spaces by
lattices of the isometry group [Bor63]. The manifolds in both classes are
rigid [LS94, Hor96] which shows that further examples cannot be obtained from
deformation theory. Moreover, in the case of positive scalar curvature it is
even conjectured that the Wolf spaces exhaust all complete (and hence compact,
by Bonnet-Myers) examples [LS94].

In this article we show that complete quaternionic Kähler manifolds of
negative scalar curvature can have ends of finite volume without being locally
homogeneous. Our constructions are based on the recently proven fact that the
known homogeneous quaternionic Kähler manifolds of negative scalar curvature
and higher rank can be deformed into complete quaternionic Kähler manifolds
with an isometric cohomogeneity one action [CST21, Cor. 3.19].

In this paper we focus on the above deformation $(\bar{N}_{n},g^{c})$, $c\geq
0$, for the symmetric spaces

$(\bar{N}_{n},g^{0})=\left(\frac{\mathrm{SU}(2,n)}{\mathrm{S}(\mathrm{U}(2)\times\mathrm{U}(n))},g_{\mathrm{can}}\right).$

The one-parameter family $(\bar{N}_{n},g^{c})$, $c\geq 0$, is obtained from an
indefinite hyper-Kähler manifold by the HK/QK correspondence. This implies
that the metrics $g^{c}$ are quaternionic Kähler [ACDM15, Cor. 1]. The
completeness of $(\bar{N}_{n},g^{c})$ for $c\geq 0$ was shown in [CDS17, Cor.
15]. The manifolds $(\bar{N}_{n},g^{c})$ are locally inhomogeneous for $c>0$
[CST21, Thm. 4.8].

We show in dimensions $4$ and $8$ (i.e. for $n=1$ and $2$) that the isometry
group of $(\bar{N}_{n},g^{c})$, $c\geq 0$, contains discrete subgroups which
give rise to smooth quotient manifolds $X=X_{\Gamma}=\bar{N}/\Gamma$ with
exactly two ends, see 4.2. The metric $g^{c}$ on $\bar{N}_{n}$ induces a
complete quaternionic Kähler metric on $X_{\Gamma}$, which we denote by the
same symbol. With respect to this metric one of the two ends is of finite
volume and the other is of infinite volume, see 4.5.

The fundamental groups $\Gamma$ of the eight-dimensional complete quaternionic
Kähler manifolds $(X_{\Gamma},g^{c})$, $c\geq 0$, have the structure of a
cyclic quotient $(\Gamma_{1}\ltimes\Gamma_{2})/\mathbb{Z}$ of a semi-direct
product $\Gamma_{1}\ltimes\Gamma_{2}$, where
$\Gamma_{1}\subset\widetilde{\mathrm{SU}}(1,1)$ is the preimage of an
arithmetic subgroup $\bar{\Gamma}_{1}\subset\mathrm{SU}(1,1)$ in the universal
covering and $\Gamma_{2}$ is a lattice in the five-dimensional Heisenberg
group. The subgroup $\mathbb{Z}\subset\Gamma_{1}\ltimes\Gamma_{2}$ is diagonal
for $c>0$, in the sense that it has trivial intersection with each of the
factors. We consider also $\bar{\Gamma}=\bar{\Gamma}_{1}\ltimes\Gamma_{2}$,
which is a lattice in $\mathrm{SU}(1,1)\ltimes\operatorname{Heis}_{5}$. In the
undeformed case $c=0$, we have $\mathbb{Z}\subset\Gamma_{1}$ and the above
construction reduces to a semi-direct product
$\Gamma=\bar{\Gamma}=\bar{\Gamma}_{1}\ltimes\Gamma_{2}$, of an arithmetic
group $\bar{\Gamma}_{1}\subset\mathrm{SU}(1,1)$ with a lattice
$\Gamma_{2}\subset\operatorname{Heis}_{5}$. In four dimensions, $\Gamma$
reduces to a lattice in the three-dimensional Heisenberg group and
$X_{\Gamma}$ is diffeomorphic to
$(\operatorname{Heis}_{3}/\Gamma)\times\mathbb{R}$.

Our results are developed in such a way that they immediately extend to higher
dimensions given lattices
$\bar{\Gamma}\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$,
which do always exist. In particular, the isometric action of
$\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$ on a cyclic quotient of
$\bar{N}_{n}$ is proven in all dimensions, see 3.16. The arithmetic part of
the construction consists in finding lattices $\bar{\Gamma}_{1}$ in
$\mathrm{SU}(1,n-1)$ which normalize a lattice in
$\operatorname{Heis}_{2n+1}$. For $n=2$ we were able to construct co-compact
lattices of this type using quaternion algebras. The lattices which we
obtained can be described as follows.

Let $a,b\in\mathbb{Z}_{>0}$ be such that $b$ is prime and $a$ is a quadratic
non-residue mod $b$, i.e. $a$ does not have a square root in the field
$\mathbb{Z}_{b}$. Denote by
$\mathcal{O}_{a,b}\subset\mathfrak{gl}(2,\mathbb{C})$ the $\mathbb{Z}$-span of
the four matrices

$\mathds{1}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix},\quad
I=\sqrt{a}i\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},\quad
J=\sqrt{b}\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},\quad
K=\sqrt{ab}i\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

Then
$\bar{\Gamma}_{1,a,b}=\mathcal{O}_{a,b}\cap\mathrm{SU}(1,1)=\\{A\in\mathcal{O}_{a,b}:\det
A=1\\}$ is a co-compact Fuchsian group, which preserves the lattice
$\Gamma_{2}$ in $\operatorname{Heis}_{5}\cong\mathbb{C}^{2}\times\mathbb{R}$
generated by $\mathcal{O}_{a,b}\cdot\begin{pmatrix}1\\\
0\end{pmatrix}\subset\mathbb{C}^{2}$. Up to passing to a suitable finite index
normal subgroup, $\bar{\Gamma}=\bar{\Gamma}_{1,a,b}\ltimes\Gamma_{2}$ gives a
lattice of the desired type, as long as the one-loop parameter $c$ is chosen
to be a rational multiple of $\sqrt{ab}$.

The following theorem summarizes the main results about quaternionic Kähler
manifolds of dimension $\leq 8$ obtained in this paper, compare 4.2 and 4.5.

######  Theorem 1.1.

There exist complete locally inhomogeneous quaternionic Kähler manifolds of
dimension $4$ and $8$ which are diffeomorphic to a product $X=\mathbb{R}\times
K$, where $K$ is a compact aspherical manifold described more in detail below
and the volume of one of the two ends of $X$ is finite, the other infinite.

  * •

For $\dim X=4$ the fiber $K$ is a nilmanifold.

  * •

For $\dim X=8$ it is a manifold with fundamental group of the form
$(\Gamma_{1}\ltimes\Gamma_{2})/\mathbb{Z}$, where $\Gamma_{1}$ is a lattice in
$\widetilde{\mathrm{SU}}(1,1)$ covering a finite index normal subgroup of any
of the above Fuchsian groups $\bar{\Gamma}_{1,a,b}\subset\mathrm{SU}(1,1)$,
$\Gamma_{2}$ is a lattice in $\mathrm{Heis}_{5}$ and $\mathbb{Z}$ is a
diagonally embedded central subgroup.

While co-compact lattices in $\mathrm{SU}(1,n-1)$ are known from complex
hyperbolic geometry, see for instance [Eps87], there is no such lattice
normalizing a lattice in $\mathbb{C}^{n}$ for $n\geq 3$
compatible111Compatibility is defined on page 4.2. with the Hermitian
structure (and a fortiori no such lattice normalizing a compatible log-lattice
in $\operatorname{Heis}_{2n+1}$). Indeed, such a lattice in
$\mathrm{SU}(1,n-1)$, $n\geq 3$, would give rise to an anisotropic Hermitian
sesquilinear form over an imaginary number field. The real part thereof would
be an anisotropic non-degenerate rational symmetric bilinear form in more than
$4$ variables that is indefinite over $\mathbb{R}$, contradicting Meyer’s
theorem (see [Ser73, Cor. 2 on p. 43]).

However, lattices of finite co-volume in
$\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$ can be easily obtained
by considering normalizers of certain lattices in
$\operatorname{Heis}_{2n+1}$. Using suitable lattices

$\bar{\Gamma}\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$ (1)

we construct quaternionic Kähler manifolds of dimension $4n$ fibering over
$\mathbb{R}$ with locally homogeneous fibers of finite volume. The results can
be summarized as follows, see 4.4 and 4.5.

######  Theorem 1.2.

In all dimensions $4n\geq 4$, there exist complete locally inhomogeneous
quaternionic Kähler manifolds diffeomorphic to $X=\mathbb{R}\times K$, where
the fibers $\\{t\\}\times K$ are aspherical of finite volume with fundamental
group an infinite cyclic extension of a group $\bar{\Gamma}$ of the form (1).
Infinitely many examples of such groups $\bar{\Gamma}$ do occur in every
dimension. The domains $\\{t>t_{0}\\}\subset X$ are of finite volume for all
$t_{0}\in\mathbb{R}$ while the domains $\\{t<t_{0}\\}$ are of infinite volume.

After this summary of the main results, we would like to briefly describe the
structure of our paper. Our constructions are based on an explicit
construction of quaternionic Kähler metrics of negative scalar curvature known
as the one-loop deformed $c$-map. This is reviewed in Section 2.

In Section 3 we specialize the construction to the complete quaternionic
Kähler $4n$-manifolds $\bar{N}=(\bar{N}_{n},g^{c})$, $c>0$, considered above.
We prove that the simply connected group
$\widetilde{\mathrm{SU}}(1,n-1)\ltimes\mathrm{Heis}_{2n+1}$ acts on $\bar{N}$
by isometries and characterize its image in the isometry group.

In Section 4 we study lattices in that semi-direct product, which act with co-
finite volume on the leaves of a certain codimension one foliation of
$\bar{N}$. When the action on the leaves is co-compact, we prove that the
corresponding quotients of $\bar{N}$ have an end of finite volume, while the
other end has infinite volume.

Acknowledgements

This research was partially funded by the Deutsche Forschungsgemeinschaft
(DFG, German Research Foundation) under Germany’s Excellence Strategy – EXC
2121 Quantum Universe – 390833306. V.C. thanks Oliver Baues, Florin Belgun,
and Vincent Koziarz for helpful discussions and Christoph Böhm for useful
comments on the manuscript. We are moreover thankful to Oliver Baues and
Vincent Koziarz for pointing out the references [Eps87] and [Ser73],
respectively. We thank the anonymous referee for the detailed report and for
making a number of useful suggestions to improve the manuscript.

##  2 Preliminaries

###  2.1 The (one-loop deformed) $c$-map

The one-loop deformed $c$-map is a string theory construction [RSV06] which,
to a given projective special Kähler manifold $\bar{M}$, associates a one-
parameter family of quaternionic Kähler manifolds. A geometric proof of the
quaternionic Kähler property was given in [ACDM15, Theorem 5] based on an
extension [ACM13] of Haydys’ HK/QK correspondence [Hay08]. These constructions
were elegantly recovered in [MS15] based on Swann’s twist approach [Swa10], as
we will now describe. For more information see [CST21, ACDM15, MS15] and
references therein.

We begin by recalling [Fre99, ACD02] the notion of a projective special Kähler
manifold, which is defined in terms of a corresponding conical affine special
Kähler manifold.

######  Definition 2.1.

A pseudo-Kähler manifold $(M,g,J,\omega)$ endowed with a flat, torsion-free
connection $\nabla$ which satisfies $\nabla\omega=0$ and
$\mathrm{d}^{\nabla}J=0$, where $J$ is viewed as a $TM$-valued one-form, is
called an affine special Kähler (ASK) manifold.

An ASK manifold $(M,g,J,\nabla)$ endowed with a vector field $\xi$ is called
conical, or a CASK manifold, if

  1. (i)

$\\{\xi,J\xi\\}$ generate a principal (that is, free and proper)
$\mathbb{C}^{*}$-action,

  2. (ii)

$g|_{\langle\xi,J\xi\rangle}$ is negative-definite while
$g|_{\langle\xi,J\xi\rangle^{\perp}}$ is positive-definite, and

  3. (iii)

$\nabla\xi=\nabla^{\mathrm{LC}}\xi=\operatorname{id}_{TM}$, where
$\nabla^{\mathrm{LC}}$ is the Levi-Civita connection of $g$.

The vector field $\xi$ is called the Euler field.

The $\mathbb{C}^{*}$-action on a CASK manifold casts it as the total space of
a $\mathbb{C}^{*}$-principal bundle. The base of this bundle is what we call a
projective special Kähler manifold.

######  Definition 2.2.

Let $(M,g,J,\nabla,\xi)$ be a CASK manifold. Then $\bar{M}\coloneqq
M/\mathbb{C}^{*}$ is called a projective special Kähler (PSK) manifold.

That every PSK manifold is indeed Kähler follows from the fact that we can
construct $\bar{M}$ as a Kähler quotient. Given a CASK manifold
$(M,g,J,\nabla,\xi)$, $-J\xi$ generates an $S^{1}$-action which preserves the
(pseudo-)Kähler structure. Its Hamiltonian function is
$\mu=-\frac{1}{2}g(\xi,\xi)$, whose level sets intersect the orbits of the
$\mathbb{R}_{+}$-action generated by the Euler field exactly once. Choosing
the level $\frac{1}{2}$, we then find
$M\mathbin{/\mkern-6.0mu/}S^{1}=\mu^{-1}\big{(}\frac{1}{2}\big{)}/S^{1}\cong
M/\mathbb{C}^{*}=\bar{M}$. Since $g|_{\langle\xi,J\xi\rangle^{\perp}}$ is
positive-definite, so is the induced Kähler metric on $\bar{M}$.

By definition, $\bar{M}$ arises from a CASK manifold
$(M,g_{M},J_{M},\omega_{M},\nabla,\xi)$. The ASK structure on $M$ gives rise
to a pseudo-hyper-Kähler structure on its cotangent bundle.

######  Theorem 2.3 ([CFG89, Fre99, ACD02]).

Let $(M,g_{M},J_{M},\omega_{M},\nabla)$ be an ASK manifold. Then $N\coloneqq
T^{*}M$ carries a pseudo-hyper-Kähler structure. With respect to the splitting
$TN\cong\pi^{*}TM\oplus\pi^{*}T^{*}M$ induced by $\nabla$, the hyper-Kähler
metric $g$ and complex structures $I_{k}$, $k=1,2,3$, are given by the
expressions

$g=\begin{pmatrix}g_{M}&0\\\ 0&g^{-1}_{M}\end{pmatrix}\hskip
35.56593ptI_{1}=\begin{pmatrix}J_{M}&0\\\ 0&J^{*}_{M}\end{pmatrix}\hskip
35.56593ptI_{2}=\begin{pmatrix}0&-\omega_{M}^{-1}\\\
\omega_{M}&0\end{pmatrix}\hskip 35.56593ptI_{3}=I_{1}I_{2},$

where $g^{-1}_{M}$ and $J_{M}^{*}$ denote the natural induced structures on
$T^{*}M$, $\omega_{M}$ is regarded as an isomorphism $TM\to T^{*}M$, and we
have omitted pullbacks throughout to simplify notation. ∎

This pseudo-hyper-Kähler structure, which is known as the rigid $c$-map
structure, is a natural generalization of the standard pseudo-hyper-Kähler
structure on
$\mathbb{H}^{n,1}\cong\mathbb{C}^{n,1}\oplus(\mathbb{C}^{n,1})^{*}$.

In the conical case, we additionally have a natural circle action, generated
by the vector field $-J_{M}\xi$.

######  Definition 2.4.

We call a vector field $Z\in\mathfrak{X}(N)$ on a (pseudo-)hyper-Kähler
manifold $(N,g,\omega_{k})$, $k=1,2,3$, a rotating Killing field if it
satisfies $L_{Z}g=0$, $L_{Z}\omega_{1}=0$, $L_{Z}\omega_{2}=\omega_{3}$ and
$L_{Z}\omega_{3}=-\omega_{2}$.

Using coordinates adapted to the CASK structure [ACD02], a straightforward
computation shows:

######  Proposition 2.5 ([ACM13]).

Let $(M,g_{M},J_{M},\nabla,\xi)$ be a CASK manifold, and endow $N=T^{*}M$ with
the rigid $c$-map structure. Then $Z=-\widetilde{J_{M}\xi}\in\mathfrak{X}(N)$,
where the tilde denotes the $\nabla$-horizontal lift, is a rotating Killing
field. Moreover, it is $\omega_{1}$-Hamiltonian, with Hamiltonian function
$f_{Z}=-\frac{1}{2}g(Z,Z)$. ∎

In summary, the CASK structure on $M$ induces a pseudo-hyper-Kähler structure
equipped with a free circle action, generated by an $\omega_{1}$-Hamiltonian
rotating Killing field, on its cotangent bundle. This is precisely the input
required for the so-called HK/QK correspondence [Hay08, ACM13, ACDM15], which
produces a one-parameter family of quaternionic Kähler manifolds
$(\bar{N},g^{c})$, $c\geq 0$, endowed with a nowhere-vanishing Killing field,
out of it. The fact that, for the data described in Proposition 2.5, the
resulting metric is precisely the one-loop deformed $c$-map metric of [RSV06]
was proven in [ACDM15].

The twist construction [Swa10] provides a duality between manifolds endowed
with a distinguished, closed two-form and a vector field which is Hamiltonian
with respect to it. In the case at hand, the relevant two-form is
$\omega_{\mathrm{H}}\coloneqq\omega_{1}+\mathrm{d}\iota_{Z}g$, and the vector
field is the rotating Killing field $Z$ introduced above. For any
$c\in\mathbb{R}$, the function
$f_{\mathrm{H}}=f_{Z}+g(Z,Z)=\frac{1}{2}g(Z,Z)-\frac{1}{2}c$ is a Hamiltonian
for $Z$ with respect to $\omega_{\mathrm{H}}$, i.e. satisfies
$\mathrm{d}f_{\mathrm{H}}=-\iota_{Z}\omega_{\mathrm{H}}$. The triple
$(Z,\omega_{\mathrm{H}},f_{\mathrm{H}})$ makes up the so-called twist data for
the HK/QK correspondence. In the following, we will assume that
$\omega_{\mathrm{H}}$ is an integral two-form, i.e. has integer periods.

Exploiting the integrality, we pass to a principal $S^{1}$-bundle
$\pi_{N}:P\to N$ endowed with a principal $S^{1}$-connection $\eta$ with
curvature $\omega_{\mathrm{H}}$. The next step is to lift
$Z\in\mathfrak{X}(N)$ to a vector field on $P$ which preserves $\eta$; such a
lift is provided by $Z_{P}\coloneqq\tilde{Z}+\pi^{*}_{N}f_{\mathrm{H}}X_{P}$,
where $\tilde{Z}\in\mathfrak{X}(P)$ is the $\eta$-horizontal lift of $Z$ and
$X_{P}\in\mathfrak{X}(P)$ is generator of the principal circle action on $P$.
It is known that, after imposing an integrality condition on the additive
constant $c\in\mathbb{R}$ which appears in the definition of $f_{\mathrm{H}}$,
one may assume that $Z_{P}$ generates a circle action [Swa10]. If the circle
action on $N$ is free, as in the case at hand, we may moreover assume that the
lifted action is free as well. The quotient of $P$ by this lifted circle
action is then a smooth manifold; this quotient realizes $P$ as the total
space of a second circle bundle $\pi_{\bar{N}}:P\to\bar{N}$. Since the two
principal circle actions on $P$ commute, $\bar{N}$ inherits a twisted circle
action, generated by $(\pi_{\bar{N}})_{*}X_{P}$.

The power of the twist construction lies in the fact that it enables one to
use geometric structures on $N$ to induce twisted structures on $\bar{N}$,
assuming that they are invariant under the initial circle action. This is done
by exploiting the double fibration structure (see (2)), which induces
pointwise identifications of the tangent spaces to both $N$ and $\bar{N}$ with
the $\eta$-horizontal subspaces of the tangent spaces to $P$.

$N$twist$P$$\pi_{N}$$\pi_{\bar{N}}$$\bar{N}$ (2)

In particular, the quaternionic structure bundle of the hyper-Kähler manifold
$N$, which is preserved by the rotating Killing field $Z$, endows $\bar{N}$
with an almost quaternionic structure. However, the hyper-Kähler metric $g$
does not twist to a quaternionic Kähler metric on $\bar{N}$. In order to
obtain a quaternionic Kähler structure, we must first perform a so-called
elementary deformation [MS15].

A (pseudo-)Riemannian metric $h$ on $N$ is called an elementary deformation if
it takes the following form:

$h=ag+b\sum_{\mu=0}^{3}\alpha_{\mu}\otimes\alpha_{\mu}$

where, after setting, $\omega_{0}\coloneqq g$, we use the notation
$\alpha_{\mu}=\iota_{Z}\omega_{\mu}$, and $a,b\in C^{\infty}(N)$ are nowhere-
vanishing functions. Assuming that $Z$ is nowhere-vanishing, the second term
is proportional to the restriction of $g$ to the quaternionic span
$\mathbb{H}Z=\langle Z,I_{1}Z,I_{2}Z,I_{3}Z\rangle$ of $Z$. Thus, we can think
of an elementary deformation as arising from the composition of a conformal
scaling with an independent conformal scaling along $\mathbb{H}Z$.

Building on earlier work of several authors, Macia and Swann showed that there
is an essentially unique combination of elementary deformation and twist data
which yield a (pseudo-)quaternionic Kähler metric:

######  Theorem 2.6 ([Hay08, ACM13, ACDM15, MS15]).

Let $(N,g,\omega_{k},Z,f_{Z})$ be a (pseudo-)hyper-Kähler manifold endowed
with an $\omega_{1}$-Hamiltonian, rotating Killing field. Assume that its
Hamiltonian function $f_{Z}$ is nowhere-vanishing and let
$k_{1},k_{2}\in\mathbb{R}\setminus\\{0\\}$. Then, with respect to the twist
data $(Z,\omega_{\mathrm{H}},f_{\mathrm{H}})$, where

$\omega_{\mathrm{H}}=k_{1}(\omega_{1}+\mathrm{d}\iota_{Z}g)\qquad\qquad\qquad
f_{\mathrm{H}}=k_{1}(f_{Z}+g(Z,Z))$

the elementary deformation

$g_{\mathrm{H}}\coloneqq
k_{2}\bigg{(}\frac{1}{f_{Z}}g+\frac{1}{f_{Z}^{2}}\sum_{\mu}\alpha_{\mu}\otimes\alpha_{\mu}\bigg{)}$

twists to a (pseudo-)quaternionic Kähler metric on $\bar{N}$. These are the
only choices of elementary deformation and twist data which lead to a
(pseudo-)quaternionic Kähler structure. Moreover, when the initial
(pseudo-)hyper-Kähler manifold arises from a CASK manifold in the manner
described above, the resulting quaternionic Kähler metrics are positive-
definite, and of negative scalar curvature. ∎

The constants $k_{1},k_{2}$ do not influence the local geometry of the
resulting quaternionic Kähler manifold, and will be set to $1$ in the
following. However, the additive constant $c$ parametrizing the freedom in the
choice of the Hamiltonian function $f_{Z}$ does play an important role as can
be seen from the explicit appearance of $f_{Z}$ in the formula for
$g_{\mathrm{H}}$.

In our setup, where $N$ is the cotangent bundle of a CASK manifold, we have
$f_{Z}=-\frac{1}{2}(g(Z,Z)+c)$. Denoting the resulting metrics by $g^{c}$, we
have in summary obtained a one-parameter family $(\bar{N},g^{c})$, $c\geq 0$,
of quaternionic Kähler manifolds out of a PSK manifold $\bar{M}$. The metric
$g^{0}$ is known as the undeformed $c$-map metric (or Ferrara–Sabharwal
metric); the metrics $g^{c}$, $c>0$, are called (one-loop) deformed $c$-map
metrics.

We will rely on the following completeness results:

######  Theorem 2.7 ([CHM12, CDS17]).

Let $(\bar{N},g^{c})$, $c\geq 0$, be the image of the PSK manifold $\bar{M}$
under the deformed $c$-map. Then $(\bar{N},g^{0})$ is complete if and only if
$\bar{M}$ is complete. If, additionally, $\bar{M}$ arises from the
(supergravity) $r$-map, or has regular boundary behavior, then
$(\bar{N},g^{c})$ is complete for all $c\geq 0$. ∎

Another fact to keep in mind is:

######  Proposition 2.8 ([CDS17]).

Let $(\bar{N},g^{c})$, $c\geq 0$, be the image of the PSK manifold $\bar{M}$
under the deformed $c$-map. Then, for arbitrary $c_{1},c_{2}>0$,
$(\bar{N},g_{c_{1}})$ is locally isometric to $(\bar{N},g_{c_{2}})$. ∎

The undeformed $c$-map metric, however, is generally distinct, as we will soon
demonstrate in explicit examples.

###  2.2 Symmetries and the $c$-map

In this section we review what is known about the symmetries of the
quaternionic Kähler metrics constructed via the $c$-map and its one-loop
deformation. In particular, we outline how one may obtain Killing vector
fields from infinitesimal automorphisms of the initial PSK manifold.

It is well-known that the $4n$-dimensional quaternionic Kähler manifold
$(\bar{N},g^{c})$ that arises from the $c$-map applied to a PSK manifold
$\bar{M}$ of complex dimension $n-1$ admits an isometric action of a
$2n+1$-dimensional Heisenberg group. In the examples considered in this paper,
where $\bar{M}$ is a so-called PSK domain, we can think of $(\bar{N},g^{c})$
as a bundle over $\bar{M}$ such that the Heisenberg group preserves fibers and
acts on them freely. In particular, the orbits are submanifolds of the fibers
of codimension one. For $c=0$, this can be improved: A one-dimensional
solvable extension of the Heisenberg group acts isometrically, freely and
transitively on the fibers, casting $(\bar{N},g^{0})$ as a bundle of Lie
groups. A detailed description, including an explicit formula for the action
with respect to a standard coordinate system on the fibers, is given in
[CST21, Lem. 2.20].

Given this large group of fiber-preserving isometries, it is natural to ask
whether (deformed) $c$-map metrics admit additional isometries that cover non-
trivial diffeomorphisms of the initial PSK manifold $\bar{M}$. In particular,
if the PSK structure of $\bar{M}$ admits non-trivial automorphisms, we may ask
if these can be lifted to symmetries of its image under the $c$-map.

In order to provide a precise answer, we must formalize what an automorphism
of a PSK manifold is. Keeping in mind the extrinsic definition of this class
of manifolds, the following is natural:

######  Definition 2.9.

Let $(M,g,J,\nabla,\xi)$ be a CASK manifold, and $\bar{M}$ the corresponding
PSK manifold. An automorphism of the CASK structure on $M$ is a diffeomorphism
of $M$ which preserves the pseudo-Kähler structure and the flat connection
$\nabla$, and commutes with the $\mathbb{C}^{*}$-action induced by
$\\{\xi,J\xi\\}$. An automorphism of the PSK manifold $\bar{M}$ is a
diffeomorphism of $\bar{M}$ which is induced by a CASK automorphism of the
corresponding CASK manifold $M$.

The matter of lifting PSK automorphisms to the image $(\bar{N},g^{c})$ under
the $c$-map was taken up in [CST21]. By definition, every one-parameter group
of PSK automorphisms lifts to a one-parameter group of CASK automorphisms of
the corresponding CASK manifold $M$. These CASK automorphisms can in turn be
lifted canonically to $N=T^{*}M$, using the pullback on one-forms. It turns
out that the canonical lift of a CASK automorphism to $N$ preserves all data
on $N$ that we have discussed, i.e. the pseudo-hyper-Kähler structure and
twist data $(Z,\omega_{\mathrm{H}},f_{\mathrm{H}})$. Moreover, every one-
parameter family of canonically lifted CASK automorphisms is
$\omega_{\mathrm{H}}$-Hamiltonian, and there is a canonical choice of
Hamiltonian function. In fact, there exists an equivariant moment map for the
lifted action of $\operatorname{Aut}M$:

######  Proposition 2.10.

Let $(M,g,\omega,\nabla,\xi)$ be an CASK manifold. Then an equivariant moment
map for the (canonically lifted) action of $\operatorname{Aut}M$ on $N$, with
respect to $\omega_{\mathrm{H}}$, is given by $\mu:N\to\mathfrak{g}^{*}$ where
$\langle\mu,X\rangle\eqqcolon\mu_{X}=\frac{1}{2}\big{(}g(Z,X)+\nu_{X}\big{)}$.
Here, $\nu_{X}(\alpha)=(\omega^{-1})_{\pi(\alpha)}(\alpha\circ\nabla
V^{X},\alpha)$, $\alpha\in T^{*}M$, with $V^{X}$ the fundamental vector field
corresponding to $X$.

###### Proof.

The expression given here for the Hamiltonian function corresponding to
$X\in\mathfrak{aut}M$ agrees (though the notation differs) with the results of
[CST21], so we will not verify (again) that it indeed provides a Hamiltonian.
One then checks the equivariance, i.e. that
$\mu_{X}(g\cdot\alpha)=\mu_{\operatorname{Ad}_{g^{-1}}X}(\alpha)$ for all
$g\in\operatorname{Aut}M$ and $\alpha\in T^{*}M$, by working through the
definitions of $\mu_{X}$ and $\nu_{X}$. ∎

######  Theorem 2.11 ([CST21]).

Let $\mathfrak{aut}(\bar{M})$ denote the algebra of infinitesimal
automorphisms of the PSK manifold $\bar{M}$, and
$\mathfrak{isom}(\bar{N},g^{c})$ the algebra of Killing fields of its image
$(\bar{N},g^{c})$ under the (deformed) $c$-map. Then there exists an
injective, linear map
$\mathfrak{aut}(\bar{M})\to\mathfrak{isom}(\bar{N},g^{c})$ for every $c\geq
0$.

###### Sketch of Proof.

After lifting an infinitesimal PSK automorphism to the corresponding CASK
manifold, and then to its cotangent bundle $N$, let us denote the generator of
the lifted one-parameter family by $X$, and its
$\omega_{\mathrm{H}}$-Hamiltonian by $f_{X}$. Then we certainly have
$L_{X}g_{\mathrm{H}}=0$, where $g_{\mathrm{H}}$ is the elementary deformation
from 2.6. However, since the twist of a Killing field is generally not a
Killing field with respect to the twisted metric, we must modify $X$ to obtain
a Killing field of the quaternionic Kähler metric $g^{c}$. This aim is
achieved by $X-\frac{f_{X}}{f_{\mathrm{H}}}Z$, whose twist $Y$ indeed
satisfies $L_{Y}g^{c}=0$. Note that $Y$ implicitly depends on the deformation
parameter $c$ through the appearance of $f_{\mathrm{H}}$. ∎

######  Remark 2.12.

  1. (i)

The proof shows that the entire procedure can be applied just as well to an
infinitesimal CASK automorphism of $M$ that does not arise as a lift of an
infinitesimal PSK automorphism of $\bar{M}$. Such infinitesimal automorphisms
can only exist in case the CASK manifold is flat (cf. [CST21]), in which case
the generator of the distinguished circle action is an infinitesimal
automorphism. The series of examples discussed in this paper is of this type.

  2. (ii)

The principal circle action on $P$ pushes down to a distinguished, isometric
circle action on $(\bar{N},g^{c})$ which commutes with the Killing fields
constructed by means of 2.11.

The map $\mathfrak{aut}(\bar{M})\to\mathfrak{isom}(\bar{N},g^{c})$ of 2.11 is
in general not a Lie algebra homomorphism. Upon twisting, the commutator of
vector fields (generically) picks up an additional term which is proportional
to the vector field generating the distinguished circle action on $\bar{N}$,
yielding a one-dimensional central extension:

######  Theorem 2.13 ([CST21]).

The Lie algebra $\mathfrak{aut}(\bar{M})$ induces a Lie algebra of Killing
fields on $(\bar{N},g^{c})$ which is isomorphic to a one-dimensional central
extension of $\mathfrak{aut}(\bar{M})$. ∎

######  Remark 2.14.

If the central extension is trivial, then we may arrange matters such that we
do obtain a subalgebra of $\mathfrak{isom}(\bar{N},g^{c})$ isomorphic to
$\mathfrak{aut}(\bar{M})$. This is the case in the series of examples on which
we focus in this work.

The Killing fields constructed via 2.11 are linearly independent of the
generators of the action of the $2n+1$-dimensional Heisenberg group. Putting
them together and taking into account 2.12, we have constructed an algebra of
Killing fields of $(\bar{N},g^{c})$ whose dimension is
$\dim\mathfrak{aut}(M)+2n+1$, where $\dim_{\mathbb{C}}M=n$. For $c=0$, this is
improved to $\dim\mathfrak{aut}(M)+2n+2$. Under the assumptions of 2.7, we
know that $(\bar{N},g^{c})$ is complete for all $c\geq 0$ and can integrate
these Killing fields to obtain a group of isometries of the (deformed) $c$-map
metric.

##  3 The $c$-map applied to complex hyperbolic spaces

###  3.1 The infinitesimal action

We will now focus on an important series of examples. Consider
$\mathbb{C}^{n}$, equipped with the flat pseudo-Kähler structure induced by
the indefinite Hermitian form
$h=-\mathrm{d}z^{0}\otimes\mathrm{d}\bar{z}^{0}+\sum_{a=1}^{n-1}\mathrm{d}z^{a}\otimes\mathrm{d}\bar{z}^{a}$.
Since the Levi-Civita connection is flat, we may regard it as endowing
$\mathbb{C}^{n}$ with the structure of an ASK manifold. The standard
$\mathbb{C}^{*}$-action by scalar multiplication endows the
$\mathbb{C}^{*}$-invariant domain $M_{n}=\\{z\in\mathbb{C}^{n}\mid-\lvert
z^{0}\rvert^{2}+\sum_{a=1}^{n-1}\lvert z^{a}\rvert^{2}<0\\}$ with the
structure of a CASK manifold. The Kähler quotient by
$\mathrm{U}(1)\subset\mathbb{C}^{*}$ is nothing but complex hyperbolic space
$\mathbb{C}\mathrm{H}^{n-1}$ equipped with the (symmetric) Bergman metric

$g_{\mathbb{C}\mathrm{H}^{n-1}}=\frac{1}{1-\lVert
X\rVert^{2}}\Bigg{(}\sum_{a=1}^{n-1}\lvert\mathrm{d}X^{a}\rvert^{2}+\frac{1}{1-\lVert
X\rVert^{2}}\bigg{|}\sum_{a=1}^{n-1}\bar{X}^{a}\mathrm{d}X^{a}\bigg{|}^{2}\Bigg{)}$
(3)

The indefinite unitary group $\mathrm{U}(1,n-1)$ acts on $M_{n}$ by
automorphisms of the CASK structure and projects down to yield a transitive
action on $\mathbb{C}\mathrm{H}^{n-1}$, which is therefore a homogeneous PSK
manifold.

It follows directly from the general theory developed in Section 2.2 that the
image $(\bar{N}_{n},g^{0})$ of $\mathbb{C}\mathrm{H}^{n-1}$ under the
undeformed $c$-map is a homogeneous quaternionic Kähler manifold (of negative
scalar curvature), and that the deformed metrics $(\bar{N}_{n},g^{c})$, $c>0$,
are complete and of cohomogeneity at most one. An explicit expression for the
metrics $g^{c}$, with respect to a global coordinate system
$(X^{a},w^{k},\tilde{\phi},\rho)\in\mathbb{C}^{n-1}\times\mathbb{C}^{n}\times\mathbb{R}\times\mathbb{R}_{>0}$,
where $a=1,\dots,n-1$ and $\lVert X\rVert^{2}=\sum_{a}\lvert
X^{a}\rvert^{2}<1$, and $k=0,\dots,n-1$, was given in [CDS17]:

$\displaystyle g^{c}$
$\displaystyle=\frac{\rho+c}{\rho}g_{\mathbb{C}\mathrm{H}^{n-1}}+\frac{1}{4\rho^{2}}\frac{\rho+2c}{\rho+c}\mathrm{d}\rho^{2}$
$\displaystyle\quad+\frac{1}{4\rho^{2}}\frac{\rho+c}{\rho+2c}\bigg{(}\mathrm{d}\tilde{\phi}-4\operatorname{Im}\bigg{(}\bar{w}^{0}\mathrm{d}w^{0}-\sum_{a=1}^{n-1}\bar{w}^{a}\mathrm{d}w^{a}\bigg{)}+\frac{2c}{1-\lVert
X\rVert^{2}}\operatorname{Im}\bigg{(}\sum_{a=1}^{n-1}\bar{X}^{a}\mathrm{d}X^{a}\bigg{)}\bigg{)}^{2}$
$\displaystyle\quad-\frac{2}{\rho}\bigg{(}\mathrm{d}w^{0}\mathrm{d}\bar{w}^{0}-\sum_{a=1}^{n-1}\mathrm{d}w^{a}\mathrm{d}\bar{w}^{a}\bigg{)}+\frac{\rho+c}{\rho^{2}}\frac{4}{1-\lVert
X\rVert^{2}}\Big{|}\mathrm{d}w^{0}+\sum_{a=1}^{n-1}X^{a}\mathrm{d}w^{a}\Big{|}^{2}$
(4)

Note that the case $n=1$, where we apply the $c$-map to a single point, is
exceptional. With the convention $\sum_{a=1}^{0}=0$, the expression (4)
remains valid and greatly simplifies. The resulting quaternionic Kähler four-
manifold is known as the universal hypermultiplet in the physics literature.
Despite the qualitative difference with the case $n\geq 2$, our results in
this paper remain valid if $n=1$, unless explicitly indicated.

As a smooth manifold, $\bar{N}_{n}$ is just a copy of $\mathbb{R}^{4n}$.
However, its Riemannian structure is interesting. In fact, it is known that
the undeformed $c$-map metric casts $\bar{N}_{n}$ as the non-compact symmetric
space $\frac{\mathrm{SU}(n,2)}{\mathrm{S}(\mathrm{U}(n)\times\mathrm{U}(2))}$.
Moreover, it was proven in [CST21] that $(\bar{N}_{n},g^{c})$ is of
cohomogeneity one. In summary, the family $(\bar{N}_{n},g^{c})$, $c\geq 0$ is
a one-parameter deformation of a symmetric space through cohomogeneity one
complete quaternionic Kähler metrics.

To develop a more concrete understanding of these manifolds, we will now show
in detail how the general theory works out in these examples. Our main goal is
to derive explicit formulas for the Killing fields obtained by means of 2.11.
The starting point is the (standard) action of $\mathrm{U}(1,n-1)$ on $M_{n}$
by CASK automorphisms, or rather the infinitesimal action of its Lie algebra
$\mathfrak{u}(1,n-1)$. Passing to
$N_{n}=T^{*}M_{n}=M_{n}\times\mathbb{C}^{n}\subset\mathbb{C}^{n}\times\mathbb{C}^{n}$,
the canonically lifted vector field $\mathcal{X}_{A}$ corresponding to
$A\in\mathfrak{u}(1,n-1)$ is given by
$\mathcal{X}_{A}(z,w)=(Az+\overline{Az},-A^{\top}w-\overline{A^{\top}w})$,
where we identify the tangent spaces with the underlying vector space.

The twisting two-form $\omega_{\mathrm{H}}$ is

$\omega_{\mathrm{H}}=\frac{i}{2}\bigg{(}\mathrm{d}z^{0}\wedge\mathrm{d}\bar{z}^{0}-\mathrm{d}w^{0}\wedge\mathrm{d}\bar{w}^{0}-\sum_{a=1}^{n-1}\big{(}\mathrm{d}z^{a}\wedge\mathrm{d}\bar{z}^{a}-\mathrm{d}w^{a}\wedge\mathrm{d}\bar{w}^{a}\big{)}\bigg{)},$

so that the Hamiltonian corresponding to $\mathcal{X}_{A}$ is given by
$f_{\mathcal{X}_{A}}(z,w)=-\frac{1}{2}\operatorname{Im}\big{(}\langle
Az,z\rangle+\langle Aw,w\rangle\big{)}$, where $\langle
z_{1},z_{2}\rangle=-z_{1}^{0}\bar{z}_{2}^{0}+\sum_{a=1}^{n-1}z_{1}^{a}\bar{z}_{2}^{a}$.
The rotating Killing field $Z$, which is the horizontal lift of the (negative)
generator of the standard $\mathrm{U}(1)$-action on the first factor, is given
by $Z(z,w)=(-iz,0)$. Its Hamiltonian is

$f_{\mathrm{H}}=f_{Z}+g(Z,Z)=\frac{1}{2}(g(Z,Z)-c)=\frac{1}{2}(\langle
z,z\rangle-c).$

(Recall that for the particular vector field $Z$ we denoted by
$f_{Z}=-\frac{1}{2}(g(Z,Z)+c)$ its Hamiltonian with respect to $\omega_{1}$.
That is why we are using the notation $f_{H}$ for the $\omega_{H}$
Hamiltonian.) Following the proof of 2.11, we now know that the twist of
$\mathcal{X}_{A}^{\mathrm{H}}\coloneqq\mathcal{X}_{A}-\frac{f_{\mathcal{X}_{A}}}{f_{\mathrm{H}}}Z$
will be a Killing field on $(\bar{N}_{n},g^{c})$.

In order to derive explicit expressions for these twisted vector fields, we
first construct a circle bundle $P$, equipped with a connection $\eta$ with
curvature $\omega_{\mathrm{H}}$. Since $\omega_{\mathrm{H}}$ is exact, $P$ is
trivial and we may choose
$\eta=\mathrm{d}s+\frac{1}{2}\iota_{\Xi}\omega_{\mathrm{H}}$, where $s$ is the
periodic coordinate parametrizing $S^{1}=\mathbb{R}/2\pi\mathbb{Z}$ and $\Xi$
is the position vector field

$\Xi=\sum_{k=0}^{n-1}\bigg{(}z^{k}\frac{\partial}{\partial
z^{k}}+\bar{z}^{k}\frac{\partial}{\partial\bar{z}^{k}}+w^{k}\frac{\partial}{\partial
w^{k}}+\bar{w}^{k}\frac{\partial}{\partial\bar{w}^{k}}\bigg{)}.$

Explicitly,

$\eta=\mathrm{d}s+\frac{i}{4}\Big{(}z^{0}\mathrm{d}\bar{z}^{0}-\bar{z}^{0}\mathrm{d}z^{0}-w^{0}\mathrm{d}\bar{w}^{0}+\bar{w}^{0}\mathrm{d}w^{0}-\sum_{a=1}^{n-1}\big{(}z^{a}\mathrm{d}\bar{z}^{a}-\bar{z}^{a}\mathrm{d}z^{a}-w^{a}\mathrm{d}\bar{w}^{a}+\bar{w}^{a}\mathrm{d}w^{a}\big{)}\Big{)}.$

We lift $Z$ to the vector field $Z_{P}$, which generates the circle action
defining the principal circle bundle $P\to\bar{N}_{n}$. Recall that
$Z_{P}=\tilde{Z}+f_{\mathrm{H}}\frac{\partial}{\partial s}$, where the tilde
denotes the $\eta$-horizontal lift and we regard $f_{\mathrm{H}}$ as a
function on $P=N_{n}\times S^{1}$. Regarding $Z$ as a vector field on $P$, a
short computation shows that

$Z_{P}=Z-\frac{c}{2}\frac{\partial}{\partial
s}=-i\sum_{k=0}^{n-1}\bigg{(}z^{k}\frac{\partial}{\partial
z^{k}}-\bar{z}^{k}\frac{\partial}{\partial\bar{z}^{k}}\bigg{)}-\frac{c}{2}\frac{\partial}{\partial
s}.$

Similarly, for $A\in\mathfrak{u}(1,n-1)$, one may check that
$\eta(\mathcal{X}_{A}^{\mathrm{H}})=-\frac{f_{\mathcal{X}_{A}}}{f_{\mathrm{H}}}\frac{c}{2}$,
so that

$\tilde{\mathcal{X}}_{A}^{\mathrm{H}}=\mathcal{X}_{A}^{\mathrm{H}}+\frac{f_{\mathcal{X}_{A}}}{f_{\mathrm{H}}}\frac{c}{2}\frac{\partial}{\partial
s}=\mathcal{X}_{A}-\frac{f_{\mathcal{X}_{A}}}{f_{\mathrm{H}}}Z_{P}.$

We now want to push this vector field down to $\bar{N}_{n}$. To this end, we
realize $\bar{N}_{n}$ as a submanifold of $P$, using a global slice for the
action obtained by integrating $Z_{P}$, which is given by
$(z,w,s)\mapsto(e^{-it}z,w,s-\frac{c}{2}t)$, $t\in\mathbb{R}$. The flow is not
periodic for arbitrary $c\geq 0$, but for $c\in 2\mathbb{Z}$ we have a circle
action covering the circle action on $N_{n}$ (this is the previously mentioned
quantization condition). Since the local geometry of the metrics $g^{c}$,
$c>0$, is independent of $c$ (cf. Proposition 2.8) we will only consider the
values $c=0$ and $c=2$. The submanifold $\bar{N}_{n}\coloneqq\\{(z,w,s)\in
N_{n}\times S^{1}\mid\arg z^{0}=0\\}$ then provides a global slice for the
action, and can therefore be identified as the twist manifold.

The final ingredient required is an identification of the horizontal tangent
spaces to $P$ with the tangent spaces of $\bar{N}_{n}$, given by an
$S^{1}$-principal connection with respect to the circle action generated by
$Z_{P}$, namely $\theta=\frac{1}{f_{\mathrm{H}}}\eta$. Now we have to write
$\tilde{\mathcal{X}}_{A}^{\mathrm{H}}$ in the form
$\tilde{\mathcal{X}}_{A}^{\mathrm{H}}=Y_{A}-\theta(Y_{A})Z_{P}$, where $Y_{A}$
is tangent to $\bar{N}_{n}\subset P$ (at every point in $\bar{N}_{n}$).
$Y_{A}$ is then be the twist of $\mathcal{X}_{A}^{\mathrm{H}}$, whose
$\theta$-horizontal lift indeed agrees with $\eta$. Since $\theta(Z_{P})=1$
and $\theta(\tilde{\mathcal{X}}_{A}^{\mathrm{H}})=0$ by construction, we
already have

$\tilde{\mathcal{X}}_{A}^{\mathrm{H}}=\mathcal{X}_{A}-\theta(\mathcal{X}_{A})Z_{P},$

but $\mathcal{X}_{A}$ will not be tangent to $\bar{N}_{n}$ in general. To
remedy this, we have to consider a vector field of the form
$\mathcal{X}_{A}+\varphi Z_{P}$, where $\varphi$ is a function. This vector
field restricts to $\bar{N}_{n}$ if and only if it satisfies $\mathrm{d}(\arg
z^{0})(\mathcal{X}_{A}+\varphi Z_{P})=\mathrm{d}(\arg
z^{0})(\mathcal{X}_{A})-\varphi=0$, where we used that $\mathrm{d}(\arg
z^{0})(Z_{P})=-1$, showing that we must have $\varphi=\mathrm{d}(\arg
z^{0})(\mathcal{X}_{A})$. The twisted vector field $Y_{A}$ on $\bar{N}_{n}$ is
therefore given by

$Y_{A}\coloneqq\mathcal{X}_{A}+\mathrm{d}(\arg z^{0})(\mathcal{X}_{A})Z_{P}.$
(5)

Now we turn to the algebra $\mathfrak{u}(1,n-1)$ of infinitesimal CASK
automorphisms of $M_{n}$, which we consider as a real form of
$\mathfrak{gl}(n,\mathbb{C})$. Writing $A\in\mathfrak{gl}(n,\mathbb{C})$ in
block form with respect to the splitting $\mathbb{C}^{n}=\mathbb{C}e_{0}\oplus
e_{0}^{\perp}$ (where $\\{e_{k}\\}_{k=0,\dots,n-1}$ is an orthonormal basis
with respect to the indefinite Hermitian form), i.e.

$A=\begin{pmatrix}\lambda&v^{\top}\\\ w&B\end{pmatrix},$ (6)

with $\lambda\in\mathbb{C},v,w\in\mathbb{C}^{n-1}$ and
$B\in\mathfrak{gl}(n-1,\mathbb{C})$, we will think of $\mathfrak{u}(1,n-1)$ as
the fixed-point set of the anti-linear involutive automorphism

$\sigma:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{gl}(n,\mathbb{C}),\qquad
A^{\sigma}\coloneqq\sigma(A)\coloneqq-I\bar{A}^{\top}I,$

where $I=\begin{pmatrix}-1&0\\\ 0&\mathds{1}_{n-1}\end{pmatrix}$. We may now
write any $A\in\mathfrak{gl}(n,\mathbb{C})$ uniquely in the form

$A=\operatorname{Re}_{\sigma}(A)+i\operatorname{Im}_{\sigma}(A),$

where

$\operatorname{Re}_{\sigma}(A)=\frac{1}{2}(A+A^{\sigma}),\qquad\operatorname{Im}_{\sigma}(A)=\frac{1}{2i}(A-A^{\sigma})$

are both elements of $\mathfrak{u}(1,n-1)$. Note that
$A^{\sigma}=\operatorname{Re}_{\sigma}(A)-i\operatorname{Im}_{\sigma}(A)$.

The Lie algebra $\mathfrak{sl}(n,\mathbb{C})$ is generated by the matrices

$U_{a}=\begin{pmatrix}0&e_{a}^{\top}\\\ 0&0\end{pmatrix},\qquad
U_{a}^{\sigma}\coloneqq\sigma(U_{a})=\begin{pmatrix}0&0\\\
e_{a}&0\end{pmatrix},\qquad a=1,\dots,n-1.$ (7)

We observe that

$[U_{a},U_{b}]=0,\qquad[U_{a}^{\sigma},U_{b}^{\sigma}]=0,\qquad[U_{a},U_{b}^{\sigma}]=\begin{pmatrix}\delta_{ab}&0\\\
0&-e_{b}e_{a}^{\top}\end{pmatrix}.$ (8)

Thus,

$\\{U_{a},U_{a}^{\sigma},[U_{a},U_{b}^{\sigma}]:a,b=1,\dots,n-1\\}$

is a basis of $\mathfrak{sl}(n,\mathbb{C})$. From this, together with the
central element $C=i\mathds{1}_{n}$, they form a basis for
$\mathfrak{gl}(n,\mathbb{C})$, which is therefore generated (as a complex Lie
algebra) by $\\{C,U_{a},U_{a}^{\sigma}\colon a=1,\dots,n-1\\}$.

The infinitesimal action of $\mathfrak{u}(1,n-1)$ on
$M_{n}\subset\mathbb{C}^{n}$ by CASK automorphisms extends complex-linearly to
an infinitesimal action
$\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{X}^{\mathbb{C}}(M_{n})$ by complex
vector fields. Since both canonically lifting and the twisting procedure
extend to complex vector fields without problems, we can complexify the above
discussion of $\mathfrak{u}(1,n-1)$ to obtain a linear map
$\alpha^{\mathbb{C}}:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{X}^{\mathbb{C}}(\bar{N}_{n})$.

######  Proposition 3.1.

The linear map
$\alpha^{\mathbb{C}}:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{isom}(\bar{N}_{n},g^{c})^{\mathbb{C}}$,
$A\mapsto Y_{A}$ defines an infinitesimal action of
$\mathfrak{gl}(n,\mathbb{C})$ by complex Killing fields on
$(\bar{N}_{n},g^{c})$. It maps $\mathfrak{u}(1,n-1)$ into
$\mathfrak{isom}(\bar{N}_{n},g^{c})$ and is given on generators by

$\displaystyle Y_{C}$
$\displaystyle\coloneqq\alpha^{\mathbb{C}}(C)=-i\sum_{k=0}^{n-1}\bigg{(}w^{k}\frac{\partial}{\partial
w^{k}}-\bar{w}^{k}\frac{\partial}{\partial\bar{w}^{k}}\bigg{)}-2c\frac{\partial}{\partial\tilde{\phi}},$
(9) $\displaystyle Y_{a}$
$\displaystyle\coloneqq\alpha^{\mathbb{C}}(U_{a})=\frac{\partial}{\partial\bar{X}^{a}}-X^{a}\sum_{b=1}^{n-1}X^{b}\frac{\partial}{\partial
X^{b}}-w^{0}\frac{\partial}{\partial
w^{a}}-\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{0}}+icX^{a}\frac{\partial}{\partial\tilde{\phi}},$
$\displaystyle\bar{Y}_{a}$
$\displaystyle=\alpha^{\mathbb{C}}(U_{a}^{\sigma}),$

where $(X^{a}\coloneqq\frac{z^{a}}{z^{0}},w^{k},\tilde{\phi}\coloneqq
4s,\rho\coloneqq 2f_{Z}=\lvert z^{0}\rvert^{2}-\sum_{a=1}^{n-1}\lvert
z^{a}\rvert^{2}-c)$, $a=1,\dots,n-1$, $k=0,\dots,n-1$, are global coordinates
on $\bar{N}_{n}\cong B^{n-1}\times\mathbb{R}^{2n+1}\times\mathbb{R}_{>0}$.
Here $B^{n-1}\subset\mathbb{C}^{n-1}$ denotes the unit ball.

######  Corollary 3.2.

The real vector fields
$Y_{C},\operatorname{Re}(Y_{a}),\operatorname{Im}(Y_{a})\in\mathfrak{X}(\bar{N}_{n})$
generate a subalgebra of $\mathfrak{isom}(\bar{N}_{n},g^{c})$ isomorphic to
$\mathfrak{u}(1,n-1)$. ∎

###### Proof of Proposition 3.1.

First observe that it follows immediately from the fact that
$\alpha^{\mathbb{C}}$ satisfies
$\alpha^{\mathbb{C}}(A^{\sigma})=\overline{\alpha^{\mathbb{C}}(A)}$ for every
$A\in\mathfrak{gl}(n,\mathbb{C})$ that
$\alpha^{\mathbb{C}}(\mathfrak{u}(1,n-1))\subset\mathfrak{isom}(\bar{N}_{n},g^{c})$.

Our next aim is to prove (9), for which we use (5). Denoting the canonical
lift of the fundamental vector field (on $M_{n}$) corresponding to $U_{a}$ by
$\mathcal{X}_{a}$, we obtain

$\mathcal{X}_{a}=z^{a}\frac{\partial}{\partial
z^{0}}+\bar{z}^{0}\frac{\partial}{\partial\bar{z}^{a}}-w^{0}\frac{\partial}{\partial
w^{a}}-\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{0}}.$

Since $\mathrm{d}(\arg
z^{0})=-\frac{i}{2}\big{(}\frac{\mathrm{d}z^{0}}{z^{0}}-\frac{\mathrm{d}\bar{z}^{0}}{\bar{z}^{0}}\big{)}$,
$\mathrm{d}(\arg z^{0})(\mathcal{X}_{a})=-\frac{i}{2}\frac{z^{a}}{z^{0}}$ and
the twist of $\mathcal{X}_{a}$ is

$\displaystyle Y_{a}$
$\displaystyle=\mathcal{X}_{a}-\frac{i}{2}\frac{z^{a}}{z^{0}}Z_{P}$
$\displaystyle=z^{a}\frac{\partial}{\partial
z^{0}}+\bar{z}^{0}\frac{\partial}{\partial\bar{z}^{a}}-w^{0}\frac{\partial}{\partial
w^{a}}-\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{0}}-\frac{1}{2}\frac{z^{a}}{z^{0}}\sum_{k=0}^{n-1}\bigg{(}z^{k}\frac{\partial}{\partial
z^{k}}-\bar{z}^{k}\frac{\partial}{\partial\bar{z}^{k}}\bigg{)}+\frac{ic}{4}\frac{z^{a}}{z^{0}}\frac{\partial}{\partial
s}.$

Expressing this vector field in terms of the conventional (global) coordinate
functions $(X^{a}\coloneqq\frac{z^{a}}{z^{0}},w^{k},\tilde{\phi}\coloneqq
4s,\rho\coloneqq 2f_{Z}=\lvert z^{0}\rvert^{2}-\sum_{a=1}^{n-1}\lvert
z^{a}\rvert^{2}-c)$ on $\bar{N}_{n}$, we obtain $Y_{a}$ as in (9). Applying
the same procedure to the central element $C=i\mathds{1}_{n}$, we obtain
$Y_{C}$ as in (9). Finally, the third line of (9) follows from
$\alpha^{\mathbb{C}}(U_{a}^{\sigma})=\overline{\alpha^{\mathbb{C}}(U_{a})}$.

Next, we explicitly compute the commutators of the complex Killing fields
$Y_{C}$, $Y_{a}$, $\bar{Y}_{a}$ and compare them with the structure constants
of $\mathfrak{gl}(n,\mathbb{C})$ with respect to the basis
$\\{C,U_{a},U_{a}^{\sigma},[U_{a},U_{b}^{\sigma}]\colon a,b=1,\dots,n-1\\}$.

Since $C=i\mathds{1}_{n}$ lies in the center of $\mathfrak{gl}(n,\mathbb{C})$,
we start by verifying that $Y_{C}$ lies in the center of
$\operatorname{Im}\alpha^{\mathbb{C}}\subset\mathfrak{X}^{\mathbb{C}}(\bar{N}_{n})$.
Since $C\in\mathfrak{u}(1,n-1)$, $Y_{C}$ is real and

$\displaystyle[Y_{C},Y_{a}]$
$\displaystyle=-i\Bigg{[}\sum_{k=0}^{n-1}\bigg{(}w^{k}\frac{\partial}{\partial
w^{k}}-\bar{w}^{k}\frac{\partial}{\partial
w^{k}}\bigg{)}+2c\frac{\partial}{\partial\tilde{\phi}}\,,\,\frac{\partial}{\partial\bar{X}^{a}}-\sum_{j}X^{a}X^{j}\frac{\partial}{\partial
X^{j}}+icX^{a}\frac{\partial}{\partial\tilde{\phi}}\Bigg{]}$
$\displaystyle\quad\
+i\Bigg{[}\sum_{k=0}^{n-1}\bigg{(}w^{k}\frac{\partial}{\partial
w^{k}}-\bar{w}^{k}\frac{\partial}{\partial
w^{k}}\bigg{)}\,,\,w^{0}\frac{\partial}{\partial
w^{a}}+\bar{w}^{a}\frac{\partial}{\partial w^{0}}\Bigg{]}$
$\displaystyle=i\Bigg{[}\sum_{k=0}^{n-1}\bigg{(}w^{k}\frac{\partial}{\partial
w^{k}}-\bar{w}^{k}\frac{\partial}{\partial
w^{k}}\bigg{)},w^{0}\frac{\partial}{\partial
w^{a}}+\bar{w}^{a}\frac{\partial}{\partial w^{0}}\Bigg{]}=0.$

It follows that $Y_{C}$ commutes with $\bar{Y}_{a}$ and thus also with the
commutators $[Y_{a},\bar{Y}_{b}]$ for all $a,b$. This means that $Y_{C}$
commutes with every vector field in the image of $\alpha^{\mathbb{C}}$, as
desired.

Next, since $[U_{a},U_{b}]=0$ we have to check that $[Y_{a},Y_{b}]=0$ for all
$a,b$:

$\displaystyle[Y_{a},Y_{b}]$
$\displaystyle=\Bigg{[}\frac{\partial}{\partial\bar{X}^{a}}-\sum_{j}X^{a}X^{j}\frac{\partial}{\partial
X^{j}}+icX^{a}\frac{\partial}{\partial\tilde{\phi}}\,,\,\frac{\partial}{\partial\bar{X}^{b}}-\sum_{k}X^{b}X^{k}\frac{\partial}{\partial
X^{k}}+icX^{b}\frac{\partial}{\partial\tilde{\phi}}\Bigg{]}$
$\displaystyle\quad\ +\bigg{[}w^{0}\frac{\partial}{\partial
w^{a}}+\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{0}}\,,\,w^{0}\frac{\partial}{\partial
w^{b}}+\bar{w}^{b}\frac{\partial}{\partial\bar{w}^{0}}\bigg{]}$
$\displaystyle=\Bigg{[}-\sum_{j}X^{a}X^{j}\frac{\partial}{\partial
X^{j}}+icX^{a}\frac{\partial}{\partial\tilde{\phi}}\,,\,-\sum_{k}X^{b}X^{k}\frac{\partial}{\partial
X^{k}}+icX^{b}\frac{\partial}{\partial\tilde{\phi}}\Bigg{]}$
$\displaystyle=\sum_{j,k}\Big{[}X^{a}X^{j}\frac{\partial}{\partial
X^{j}}\,,\,X^{b}X^{k}\frac{\partial}{\partial
X^{k}}\Big{]}-ic\bigg{(}\sum_{j}X^{a}X^{j}\delta_{jb}-\sum_{k}X^{b}X^{k}\delta_{ka}\bigg{)}\frac{\partial}{\partial\tilde{\phi}}$
$\displaystyle=0.$

This relation immediately implies $[\bar{Y}_{a},\bar{Y}_{b}]=0$.

Finally, we show that
$\alpha^{\mathbb{C}}([U_{a},U_{b}^{\sigma}])=-[Y_{a},\bar{Y}_{b}]$; note the
sign, which arises due to the fact that the map sending a Lie algebra element
to its fundamental vector field is a Lie algebra anti-homomorphism in our
setting. We start by computing the commutator $[Y_{a},\bar{Y}_{b}]$:

$\displaystyle[Y_{a},\bar{Y}_{b}]$
$\displaystyle=\Bigg{[}\frac{\partial}{\partial\bar{X}^{a}}-\sum_{j}X^{a}X^{j}\frac{\partial}{\partial
X^{j}}+icX^{a}\frac{\partial}{\partial\tilde{\phi}}\,,\,\frac{\partial}{\partial
X^{b}}-\sum_{j}\bar{X}^{b}\bar{X}^{j}\frac{\partial}{\partial\bar{X}^{j}}-ic\bar{X}^{b}\frac{\partial}{\partial\tilde{\phi}}\Bigg{]}$
$\displaystyle\quad\ +\bigg{[}w^{0}\frac{\partial}{\partial
w^{a}}+\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{0}}\,,\,\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{b}}+w^{b}\frac{\partial}{\partial
w^{0}}\bigg{]}$
$\displaystyle=-\sum_{j}\bigg{(}(\delta_{ab}\bar{X}^{j}+\delta_{aj}\bar{X}^{b})\frac{\partial}{\partial\bar{X}^{j}}-(\delta_{ab}X^{j}+\delta_{bj}X^{a})\frac{\partial}{\partial
X^{j}}\bigg{)}-2ic\delta_{ab}\frac{\partial}{\partial\tilde{\phi}}$
$\displaystyle\quad\ +w^{0}\delta_{ab}\frac{\partial}{\partial
w^{0}}-w^{b}\frac{\partial}{\partial
w^{a}}+\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{b}}-\delta_{ab}\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{0}}$
$\displaystyle=\delta_{ab}\Bigg{(}\sum_{j}\bigg{(}X^{j}\frac{\partial}{\partial
X^{j}}-\bar{X}^{j}\frac{\partial}{\partial\bar{X}^{j}}\bigg{)}+w^{0}\frac{\partial}{\partial
w^{0}}-\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{0}}-2ic\frac{\partial}{\partial\tilde{\phi}}\Bigg{)}$
$\displaystyle\quad\ +X^{a}\frac{\partial}{\partial
X^{b}}-\bar{X}^{b}\frac{\partial}{\partial\bar{X}^{a}}+\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{b}}-w^{b}\frac{\partial}{\partial
w^{a}}.$

On the other hand, the canonical lift of the fundamental vector field
corresponding to $[U_{a},U_{b}^{\sigma}]$ is

$\mathcal{X}_{[U_{a},U_{b}^{\sigma}]}=\delta_{ab}\bigg{(}z_{0}\frac{\partial}{\partial
z_{0}}-\bar{z}_{0}\frac{\partial}{\partial\bar{z}_{0}}-w_{0}\frac{\partial}{\partial
w_{0}}+\bar{w}_{0}\frac{\partial}{\partial\bar{w}_{0}}\bigg{)}-z_{a}\frac{\partial}{\partial
z_{b}}+\bar{z}_{b}\frac{\partial}{\partial\bar{z}_{a}}+w_{b}\frac{\partial}{\partial
w_{a}}-\bar{w}_{a}\frac{\partial}{\partial\bar{w}_{b}}.$

Upon twisting, we indeed obtain

$\displaystyle\alpha^{\mathbb{C}}([U_{a},U_{b}^{\sigma}])$
$\displaystyle=-\delta_{ab}\Bigg{(}\sum_{j}\bigg{(}X^{j}\frac{\partial}{\partial
X^{j}}-\bar{X}^{j}\frac{\partial}{\partial\bar{X}^{j}}\bigg{)}+w^{0}\frac{\partial}{\partial
w^{0}}-\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{0}}-2ic\frac{\partial}{\partial\tilde{\phi}}\Bigg{)}$
$\displaystyle\quad\ -X^{a}\frac{\partial}{\partial
X^{b}}+\bar{X}^{b}\frac{\partial}{\partial\bar{X}^{a}}-\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{b}}+w^{b}\frac{\partial}{\partial
w^{a}}.$

This completes the proof. ∎

The Proposition 3.1 provides an (infinitesimal) action of the unitary Lie
algebra $\mathfrak{u}(1,n-1)$ on $\bar{N}_{n}$. As explained in Section 2.2,
there is an additional action of a $(2n+1)$-dimensional Heisenberg group. As
we will now show, this action also has a natural interpretation in terms of
the twist construction of $\bar{N}_{n}$.

On the hyper-Kähler side of the twist correspondence, translations of the
$w$-coordinates define an action of $\mathbb{R}^{2n}$ on $N_{n}$. The action
preserves the pseudo-hyper-Kähler structure and twist data
$(Z,\omega_{\mathrm{H}},f_{\mathrm{H}})$, and is
$\omega_{\mathrm{H}}$-Hamiltonian. Thus, we may apply the procedure outlined
in the proof of 2.11 to turn its infinitesimal generators into Killing fields
on $\bar{N}_{n}$.

We will once again work with the complexified generators
$\frac{\partial}{\partial w^{k}}$, $k=0,1\dots,n-1$, and their conjugates. A
natural choice of $\omega_{\mathrm{H}}$-Hamiltonian function for
$\frac{\partial}{\partial w^{k}}$ is $f_{w^{k}}=\pm\frac{i}{2}\bar{w}^{k}$,
where the sign is positive for $k=0$ and negative otherwise. Carrying out the
exact same procedure as outlined for the vector fields arising from the
underlying CASK manifold, we now obtain Killing fields on $\bar{N}_{n}$ by
twisting the modified vector fields $\mathcal{U}_{k}=\frac{\partial}{\partial
w^{k}}-\frac{f_{w^{k}}}{f_{\mathrm{H}}}Z$. Horizontally lifting, we find

$\displaystyle\tilde{\mathcal{U}}_{k}$
$\displaystyle=\mathcal{U}_{k}-\eta(\mathcal{U}_{k})\frac{\partial}{\partial
s}=\mathcal{U}_{k}-\frac{1}{2}\big{(}\Xi(f_{w^{k}})-\frac{f_{w^{k}}}{f_{\mathrm{H}}}\Xi(f_{\mathrm{H}})\big{)}\frac{\partial}{\partial
s}$
$\displaystyle=\mathcal{U}_{k}+\frac{f_{w^{k}}}{2f_{\mathrm{H}}}(f_{\mathrm{H}}+c)\frac{\partial}{\partial
s}=\frac{\partial}{\partial w^{k}}+\frac{f_{w^{k}}}{2}\frac{\partial}{\partial
s}-\frac{f_{w^{k}}}{f_{\mathrm{H}}}Z_{P}$

where we used that $\Xi(f_{w^{k}})=f_{w^{k}}$ and
$\Xi(f_{\mathrm{H}})=2f_{\mathrm{H}}+c$ in passing to the second line. We note
that the first two terms determine a vector field $V_{k}$ on $P$ which is
tangent to $\bar{N}_{n}$ (at every point in $\bar{N}_{n}$) and satisfies
$\tilde{\mathcal{U}}_{k}=V_{k}-\theta(V_{k})Z_{P}$, which means that $V_{k}$
restricts to the sought-after Killing field on $\bar{N}_{n}$. After
substituting $\tilde{\phi}=4s$, we have:

$V_{a}=\frac{\partial}{\partial
w^{a}}-i\bar{w}^{a}\frac{\partial}{\partial\tilde{\phi}},\qquad
V_{0}=\frac{\partial}{\partial
w^{0}}+i\bar{w}^{0}\frac{\partial}{\partial\tilde{\phi}}.$ (10)

where $a=1,\dots,n-1$. The complex conjugate vector fields $\bar{V}_{k}$,
$k=0,\dots,n-1$, naturally arise in the same way from
$\frac{\partial}{\partial\bar{w}^{k}}$.

Now let $e_{0},f_{0},e_{a},f_{a}$, $a=1,\dots,n-1$, be the standard basis of
$\mathbb{R}^{2n}$. We define the one-dimensional central extension
$\mathfrak{heis}_{2n+1}$ of $\mathbb{R}^{2n}$ by setting

$[e_{k},e_{l}]=0,\quad[f_{k},f_{l}]=0,\quad[e_{k},f_{l}]=\bigg{(}\delta_{k0}\delta_{l0}-\sum_{a=1}^{n-1}\delta_{ka}\delta_{la}\bigg{)}T$
(11)

for every $k,l=0,1,\dots,n-1$, where $T$ denotes the generator of the center.
Complexifying and extending the Lie bracket complex-bilinearly, we obtain
$\mathfrak{heis}_{2n+1}^{\mathbb{C}}$.

######  Proposition 3.3.

The vector fields $V_{k}$ and $\bar{V}_{k}$, $k=0,\dots,n-1$, together with
$\frac{\partial}{\partial\tilde{\phi}}$, generate an infinitesimal action of a
complexified Heisenberg algebra $\mathfrak{heis}_{2n+1}^{\mathbb{C}}$ by
complex Killing fields on $(\bar{N}_{n},g^{c})$ which sends
$\mathfrak{heis}_{2n+1}$ into $\mathfrak{isom}(\bar{N}_{n},g^{c})$.

###### Proof.

We define a (complex) infinitesimal action, i.e. a Lie algebra anti-
homomorphism
$\beta^{\mathbb{C}}:\mathfrak{heis}_{2n+1}^{\mathbb{C}}\to\mathfrak{X}^{\mathbb{C}}(\bar{N})$,
by setting $\beta^{\mathbb{C}}(E_{k})=V_{k}$ and
$\beta^{\mathbb{C}}(\bar{E}_{k})=\bar{V}_{k}$, where $k=0,1,\dots,n-1$ and
$E_{k}\coloneqq e_{k}-if_{k}$, as well as
$\beta(T)=\frac{\partial}{\partial\tilde{\phi}}$. The fact that
$\beta^{\mathbb{C}}$ defines an infinitesimal action follows from

$[V_{k},\bar{V}_{l}]=-2i\bigg{(}\delta_{k0}\delta_{l0}-\sum_{a=1}^{n-1}\delta_{ka}\delta_{la}\bigg{)}\frac{\partial}{\partial\tilde{\phi}}$

and the vanishing of all other Lie brackets, which is easily deduced from
(10). The vector fields $V_{k}$, $\bar{V}_{k}$ are complex Killing fields by
construction. Their real and imaginary parts, which generate
$\mathbb{R}^{2n}\subset\mathfrak{heis}_{2n+1}$, are real Killing fields on
$\bar{N}_{n}$. The general theory of the twist construction also dictates that
$\frac{\partial}{\partial\tilde{\phi}}$, which is the restriction of the
generator of the principal circle action of the bundle $P\to N_{n}$, is
Killing. Indeed, the quaternionic Kähler metric $g^{c}$ arises by pushing down
the pullback of the elementary deformation $g_{\mathrm{H}}$, which is
certainly invariant under the principal circle action. ∎

The next logical step is to study how the generators of the infinitesimal
(complex) $\mathfrak{gl}(n,\mathbb{C})$\- and
$\mathfrak{heis}_{2n+1}^{\mathbb{C}}$-actions interact with one another. To
determine this interaction we recall that, on the hyper-Kähler side of the
HK/QK correspondence, the vector spaces spanned by the $w$-coordinates are the
fibers of $N_{n}=T^{*}M_{n}$. Therefore, $\mathfrak{u}(1,n-1)$ naturally acts
on them via the dual of the defining representation.

Passing to the quaternionic Kähler side, these same $w$-coordinates span
$\mathbb{R}^{2n}\subset\mathfrak{heis}_{2n+1}$. We therefore expect that
$\mathfrak{u}(1,n-1)$ acts on this subspace via the dual representation. We
extend this to an action on all of $\mathfrak{heis}_{2n+1}$ by stipulating
that $\mathfrak{u}(1,n-1)$ acts trivially on the center. This defines a
representation of $\mathfrak{u}(1,n-1)$ by derivations of
$\mathfrak{heis}_{2n+1}$ and therefore yields a semi-direct product structure
$\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}$. Complexifying and
extending the Lie bracket complex-bilinearly, we obtain a complex Lie algebra
$\mathfrak{gl}(n,\mathbb{C})\ltimes\mathfrak{heis}_{2n+1}^{\mathbb{C}}$.

Recall that $\\{U_{a},U_{a}^{\sigma},[U_{a},U_{b}^{\sigma}]\colon
a,b=1,\dots,n-1\\}$, together with $C$, form a basis of
$\mathfrak{gl}(n,\mathbb{C})$. We complete to a basis of
$\mathfrak{gl}(n,\mathbb{C})\ltimes\mathfrak{heis}_{2n+1}^{\mathbb{C}}$ by
adding the basis elements $T$, $E_{k}$, $\bar{E}_{k}$, $k=0,1,\dots,n-1$ of
$\mathfrak{heis}_{2n+1}^{\mathbb{C}}$ (as in the proof of Proposition 3.3). We
now extend the infinitesimal action
$\alpha^{\mathbb{C}}:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{isom}(\bar{N}_{n},g^{c})^{\mathbb{C}}$
to an infinitesimal action
$\alpha^{\mathbb{C}}:\mathfrak{gl}(n,\mathbb{C})\ltimes\mathfrak{heis}_{2n+1}^{\mathbb{C}}\to\mathfrak{isom}(\bar{N}_{n},g^{c})^{\mathbb{C}}$
in the manner prescribed in the proof of Proposition 3.3, i.e. by setting

$\alpha^{\mathbb{C}}(E_{k})=V_{k},\quad\alpha^{\mathbb{C}}(\bar{E}_{k})=\bar{V}_{k},\quad\alpha^{\mathbb{C}}(T)=\frac{\partial}{\partial\tilde{\phi}}.$
(12)

where $E_{k}=e_{k}-if_{k}$.

######  Proposition 3.4.

The linear map
$\alpha^{\mathbb{C}}\colon\mathfrak{gl}(n,\mathbb{C})\ltimes\mathfrak{heis}_{2n+1}^{\mathbb{C}}\to\mathfrak{isom}(\bar{N}_{n},g^{c})^{\mathbb{C}}$
as defined by Proposition 3.1 and Equation 12 is an infinitesimal action on
$(\bar{N},g^{c})$ by complex Killing fields which sends
$\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}$ into
$\mathfrak{isom}(\bar{N}_{n},g^{c})$.

###### Proof.

It is clear from the preceding discussion that $\alpha^{\mathbb{C}}$ indeed
maps into $\mathfrak{isom}(\bar{N}_{n},g^{c})^{\mathbb{C}}$, and that the
image of $\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}$ consists of real
Killing fields of $(\bar{N}_{n},g^{c})$. All that remains is to verify that
$\alpha^{\mathbb{C}}$ defines a Lie algebra anti-homomorphism.

By Propositions 3.1 and 3.3, we need only check those Lie brackets that
involve one element of $\mathfrak{gl}(n,\mathbb{C})$ and one element of
$\mathfrak{heis}_{2n+1}^{\mathbb{C}}$. In
$\mathfrak{gl}(n,\mathbb{C})\ltimes\mathfrak{heis}^{\mathbb{C}}_{2n+1}$, the
following brackets can be calculated from (7) and the fact that, by definition
of the semi-direct product structure, the bracket of
$\mathfrak{gl}(n,\mathbb{C})\ltimes\mathfrak{heis}_{2n+1}^{\mathbb{C}}$
evaluated on $A\in\mathfrak{u}(1,n-1)$ and $v\in\mathbb{R}^{2n}$ is just
$[A,v]=-A^{\top}v$ (where $A^{\top}$ is identified with a real $2n\times
2n$-matrix). This prescription yields

$\displaystyle[C,E_{k}]=-iE_{k},$
$\displaystyle\qquad[C,\bar{E}_{k}]=i\bar{E}_{k},$
$\displaystyle[U_{a},E_{k}]=-\delta_{k0}E_{a},$
$\displaystyle\qquad[U^{\sigma}_{a},E_{k}]=-\delta_{ka}E_{0},$
$\displaystyle[U_{a},\bar{E}_{k}]=-\delta_{ka}\bar{E}_{0},$
$\displaystyle\qquad[U^{\sigma}_{a},\bar{E}_{k}]=-\delta_{k0}\bar{E}_{a}$

The remaining brackets involving one element of $\mathfrak{gl}(n,\mathbb{C})$
and one of $\mathfrak{heis}_{2n+1}^{\mathbb{C}}$ can be deduced from these via
the Jacobi identity. We now compare to the brackets of the corresponding
vector fields on the quaternionic Kähler manifold, using the formulas (9) and
(10). They yield the expected results, namely

$\displaystyle[Y_{C},V_{k}]=iV_{k},$
$\displaystyle\qquad[Y_{C},\bar{V}_{k}]=-i\bar{V}_{k},$
$\displaystyle[Y_{a},V_{k}]=\delta_{k0}V_{a},$
$\displaystyle\qquad[\bar{Y}_{a},V_{k}]=\delta_{ka}V_{0},$
$\displaystyle[Y_{a},\bar{V}_{k}]=\delta_{ka}\bar{V}_{0},$
$\displaystyle\qquad[\bar{Y}_{a},\bar{V}_{k}]=\delta_{k0}\bar{V}_{a}.$

This confirms that $\alpha^{\mathbb{C}}$ defines an infinitesimal action. ∎

This shows that, despite the $c$-dependence of the Killing fields constructed
via the HK/QK correspondence, the isomorphism type of the Lie algebra of
Killing fields is independent of $c$. The $c$-dependence arises only once we
consider the ($c$-dependent) infinitesimal action
$\alpha:\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}\to\mathfrak{isom}(\bar{N}_{n},g^{c})$
obtained by restricting $\alpha^{\mathbb{C}}$.

###  3.2 $\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}$ as an algebra of
Killing fields

The goal of this section is to provide a characterization of
$\mathfrak{g}\coloneqq\operatorname{Im}\alpha$ as a subalgebra of the Killing
algebra of $\mathfrak{isom}(\bar{N}_{n},g^{c})$, see Proposition 3.10. We
start by determining the stabilizer of a distinguished point:

######  Lemma 3.5.

Consider for fixed $\rho_{0}>0$ the point
$n_{0}=(0,0,0,\rho_{0})\in\bar{N}_{n}$. Then the stabilizer of $n_{0}$ in
$\mathfrak{g}$ is given by

$\mathfrak{g}_{n_{0}}=\operatorname{span}\bigg{\\{}Y_{C}+2c\frac{\partial}{\partial\tilde{\phi}},\operatorname{Re}([Y_{a},\bar{Y}_{b}]),\operatorname{Im}([Y_{a},\bar{Y}_{b}])+2c\delta_{ab}\frac{\partial}{\partial\tilde{\phi}}\colon
a,b=1,\dots,n-1\bigg{\\}}.$

In particular, $\mathfrak{g}_{n_{0}}$ is isomorphic to
$\mathfrak{u}(1)\times\mathfrak{u}(n-1)$.

###### Proof.

Recall from the proof of Proposition 3.1 the formula

$\displaystyle=\delta_{ab}\Bigg{(}\sum_{d=1}^{n-1}\bigg{(}X^{d}\frac{\partial}{\partial
X^{d}}-\bar{X}^{d}\frac{\partial}{\partial\bar{X}^{d}}\bigg{)}+w^{0}\frac{\partial}{\partial
w^{0}}-\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{0}}-2ic\frac{\partial}{\partial\tilde{\phi}}\Bigg{)}$
(13) $\displaystyle\quad\ +X^{a}\frac{\partial}{\partial
X^{b}}-\bar{X}^{b}\frac{\partial}{\partial\bar{X}^{a}}+\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{b}}-w^{b}\frac{\partial}{\partial
w^{a}}.$

From this we read off

$[Y_{a},\bar{Y}_{b}](n_{0})=-2ic\delta_{ab}\frac{\partial}{\partial\tilde{\phi}}.$

Similarly, $Y_{C}(n_{0})=-2c\frac{\partial}{\partial\tilde{\phi}}$; the
inclusion of the right-hand side in the left-hand side now follows. The other
inclusion can be seen directly from (9), (10) and (12).

To see that this subalgebra is isomorphic to
$\mathfrak{u}(1)\times\mathfrak{u}(n-1)$, consider its preimage under the
(injective!) infinitesimal action $\alpha$, which is

$\mathfrak{g}^{\prime}=\operatorname{span}\big{\\{}C+2cT,\operatorname{Re}([U_{a},U_{b}^{\sigma}]),\operatorname{Im}([U_{a},U_{b}^{\sigma}])+2c\delta_{ab}T\colon
a,b=1,\dots,n-1\big{\\}}$ (14)

Since
$\mathfrak{heis}_{2n+1}\subset\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}$
is an ideal, the natural projection map
$\pi_{1}:\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}\to\mathfrak{u}(1,n-1)$
is a homomorphism of Lie algebras. Its restriction to $\mathfrak{g}^{\prime}$
is injective since
$\alpha(\mathfrak{heis}_{2n+1})\cap\mathfrak{g}_{n_{0}}=\\{0\\}$, and from
(14) one reads off that

$\pi_{1}(\mathfrak{g}^{\prime})=\operatorname{span}\big{\\{}C,\operatorname{Re}([U_{a},U_{b}^{\sigma}]),\operatorname{Im}([U_{a},U_{b}^{\sigma}])\colon
a,b=1,\dots,n-1\big{\\}},$

which is nothing but the standard embedding
$\mathfrak{u}(1)\times\mathfrak{u}(n-1)\subset\mathfrak{u}(1,n-1)$. ∎

Using this, we can determine the subalgebra consisting of Killing fields
tangent to the submanifold
$H\coloneqq\\{X=0,\rho=\rho_{0}\\}\subset\bar{N}_{n}$, which is acted on
simply transitively and isometrically by $\operatorname{Heis}_{2n+1}$. Since
the isometry group of the symmetric space $(\bar{N}_{n},g^{0})$ is well-
understood, we focus on $(\bar{N}_{n},g^{c})$, $c>0$.

######  Lemma 3.6.

Let $c>0$ and let $K$ be a Killing field on $(\bar{N}_{n},g^{c})$ which is
tangent to $H$ at every point contained in $H$. Then there exists an element
of the Lie subalgebra
$\alpha((\mathfrak{u}(1)\times\mathfrak{u}(n-1))\ltimes\mathfrak{heis}_{2n+1})\subset\mathfrak{g}$
which coincides with $K$ along $H$.

###### Proof.

Clearly, $K|_{H}$ is a Killing field on $(H,g_{H}^{c})$ where we wrote
$g_{H}^{c}$ for the metric on $H$ induced by $g^{c}$. From (4), we find

$\displaystyle g^{c}_{H}$
$\displaystyle=\frac{1}{2\rho_{0}}\sum_{a=1}^{n-1}\big{|}\mathrm{d}w^{a}\big{|}^{2}+\bigg{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\bigg{)}\big{|}\mathrm{d}w^{0}\big{|}^{2}$
$\displaystyle\quad+\frac{1}{4\rho_{0}^{2}}\frac{\rho_{0}+c}{\rho_{0}+2c}\bigg{(}\mathrm{d}\tilde{\phi}-2i\bigg{(}\bar{w}^{0}\mathrm{d}w^{0}-w^{0}\mathrm{d}\bar{w}^{0}-\sum_{a=1}^{n-1}(\bar{w}^{a}\mathrm{d}w^{a}-w^{a}\mathrm{d}\bar{w}^{a})\bigg{)}\bigg{)}^{2},$

which is a left-invariant metric on the Heisenberg group
$\operatorname{Heis}_{2n+1}\cong H$ by construction. A result of Wilson
[Wil82] asserts that the Killing algebra of $(H,g_{H}^{c})$ can be constructed
as a semi-direct product $\mathfrak{s}\ltimes\mathfrak{heis}_{2n+1}$, where
$\mathfrak{s}$ is the Lie algebra of skew-symmetric derivations of
$\mathfrak{heis}_{2n+1}=T_{(0,0,0)}H$.

Note that

$g^{c}_{H}(0,0,0)=\frac{1}{2\rho_{0}}\sum_{a=1}^{n-1}\lvert\mathrm{d}w^{a}\rvert^{2}+\bigg{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\bigg{)}\lvert\mathrm{d}w^{0}\rvert^{2}+\frac{1}{4\rho_{0}^{2}}\frac{\rho_{0}+c}{\rho_{0}+2c}\mathrm{d}\tilde{\phi}^{2}$

is diagonal, and recall, see (11), that the bracket on
$\mathfrak{heis}_{2n+1}$ is given by
$[v_{1},v_{2}]=\omega(v_{1},v_{2})\frac{\partial}{\partial\tilde{\phi}}$,
where
$\omega=\frac{i}{2}\big{(}\mathrm{d}w^{0}\wedge\mathrm{d}\bar{w}^{0}-\sum_{a=1}^{n-1}\mathrm{d}w^{a}\wedge\mathrm{d}\bar{w}^{a}\big{)}$.

Now let $A\in\operatorname{End}(\mathfrak{heis}_{2n+1})$ be a skew-symmetric
derivation, so that in particular

$\omega(v_{1},v_{2})A\bigg{(}\frac{\partial}{\partial\tilde{\phi}}\bigg{)}=\big{(}[Av_{1},v_{2}]+[v_{1},Av_{2}]\big{)}\frac{\partial}{\partial\tilde{\phi}},\quad\mbox{for
all}\quad v_{1},v_{2}\in\mathbb{R}^{2n}.$ (15)

For $v\in\mathbb{R}^{2n}$ we write
$Av=\tilde{A}v+l_{A}(v)\frac{\partial}{\partial\tilde{\phi}}$ and
$A\big{(}\frac{\partial}{\partial\tilde{\phi}}\big{)}=\lambda_{A}\frac{\partial}{\partial\tilde{\phi}}$
(every derivation preserves the center) with
$l_{A}\in(\mathbb{R}^{2n})^{*},\lambda_{A}\in\mathbb{R}$. Since $g^{c}_{H}$ is
diagonal at $(0,0,0)$, skew-symmetry implies that $\lambda_{A}=0$, $l_{A}=0$.
Then equation (15) simplifies to

$0=\omega(\tilde{A}v_{1},v_{2})+\omega(v_{1},\tilde{A}v_{2}),$

that is $\tilde{A}\in\mathfrak{sp}(2n,\mathbb{R})$.

Introducing real coordinates $\\{\tilde{\zeta}_{k},\zeta^{k}\\}$ by writing
$w^{0}=\frac{1}{2}(\tilde{\zeta}_{0}+i\zeta^{0})$ and
$w^{a}=\frac{1}{2}(\tilde{\zeta}_{a}-i\zeta^{a})$, we have
$\omega=\frac{1}{2}\sum_{k=0}^{n-1}\mathrm{d}\zeta^{k}\wedge\mathrm{d}\tilde{\zeta}_{k}$
and

$g^{c}_{H}=\frac{1}{8\rho_{0}}\bigg{(}\sum_{a=1}^{n-1}(\mathrm{d}\zeta^{a})^{2}+(\mathrm{d}\tilde{\zeta}^{a})^{2}\bigg{)}+\frac{1}{4}\bigg{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\bigg{)}(\mathrm{d}\zeta^{0})^{2}+(\mathrm{d}\tilde{\zeta}^{0})^{2})+\frac{1}{4\rho_{0}^{2}}\frac{\rho_{0}+c}{\rho_{0}+2c}\mathrm{d}\tilde{\phi}^{2}.$

Write $\tilde{A}$ in block form with respect to the splitting
$\mathbb{R}^{2n}=\mathbb{R}^{2}\oplus\mathbb{R}^{2n-2}$.

$\tilde{A}=\begin{pmatrix}A_{1}&A_{2}\\\ A_{3}&A_{4}\end{pmatrix}$

The Gram matrices of $g^{c}_{H}$ (restricted to
$\mathbb{R}^{2n}=\operatorname{span}\\{\tilde{\zeta}_{k},\zeta^{k}\\}$) and
$\omega$ are

$G=\frac{1}{4}\begin{pmatrix}\big{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\big{)}\mathds{1}_{2}&0\\\
0&\frac{1}{2\rho_{0}}\mathds{1}_{2n-2}\end{pmatrix},\qquad\Omega=\frac{1}{2}\begin{pmatrix}J_{2}&0\\\
0&J_{2n-2}\end{pmatrix},$

where $J_{2}=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$ and $J_{2n-2}$ consists
of $(n-1)$ copies of $J_{2}$ on its diagonal. Skew-symmetry of $A$ implies
$\tilde{A}^{\top}G+G\tilde{A}=0$ and
$\tilde{A}\in\mathfrak{sp}(2n,\mathbb{R})$ implies that
$\tilde{A}^{\top}\Omega+\Omega\tilde{A}=0$. This means that

$\displaystyle\begin{pmatrix}A_{1}^{\top}&A_{3}^{\top}\\\
A_{2}^{\top}&A_{4}^{\top}\end{pmatrix}\\!\begin{pmatrix}\big{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\big{)}\mathds{1}_{2}&0\\\
0&\frac{1}{2\rho}\mathds{1}_{2n-2}\end{pmatrix}\\!\\!+\\!\\!\begin{pmatrix}\big{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\big{)}\mathds{1}_{2}&0\\\
0&\frac{1}{2\rho_{0}}\mathds{1}_{2n-2}\end{pmatrix}\\!\begin{pmatrix}A_{1}&A_{2}\\\
A_{3}&A_{4}\end{pmatrix}\\!$ $\displaystyle=0,$
$\displaystyle\begin{pmatrix}A_{1}^{\top}&A_{3}^{\top}\\\
A_{2}^{\top}&A_{4}^{\top}\end{pmatrix}\begin{pmatrix}J_{2}&0\\\
0&J_{2n-2}\end{pmatrix}+\begin{pmatrix}J_{2}&0\\\
0&J_{2n-2}\end{pmatrix}\begin{pmatrix}A_{1}&A_{2}\\\
A_{3}&A_{4}\end{pmatrix}\\!$ $\displaystyle=0.$

This implies

$\displaystyle A_{1}+A_{1}^{\top}$ $\displaystyle=0$ $\displaystyle
A_{1}^{\top}J_{2}+J_{2}A_{1}$ $\displaystyle=0$ $\displaystyle
A_{4}+A_{4}^{\top}$ $\displaystyle=0$ $\displaystyle
A_{4}^{\top}J_{2}+J_{2}A_{4}$ $\displaystyle=0$
$\displaystyle\bigg{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\bigg{)}A_{2}^{\top}+\frac{1}{2\rho_{0}}A_{3}$
$\displaystyle=0$ $\displaystyle A_{2}^{\top}J_{2}+J_{2n-2}A_{3}$
$\displaystyle=0$

The last two lines imply $A_{2}=0$ and $A_{3}=0$, since the parameter $c>0$.
Thus, $\tilde{A}$ is of the form

$\tilde{A}=\begin{pmatrix}A_{1}&0\\\ 0&A_{4}\end{pmatrix}$

with $A_{1}$ and $A_{4}$ skew-symmetric and complex linear. Put differently,
we have shown that $A\in\mathfrak{u}(1)\times\mathfrak{u}(n-1)$, and thereby
that $\mathfrak{s}\cong\mathfrak{u}(1)\times\mathfrak{u}(n-1)$.

The stabilizer (in $\mathfrak{g}=\operatorname{Im}\alpha$) of the point
$(0,0,0,\rho_{0})$, determined in Lemma 3.5, consists of Killing fields that
are tangent to $H$ and span a subalgebra isomorphic to
$\mathfrak{u}(1)\times\mathfrak{u}(n-1)$. Combining this with
$\alpha(\mathfrak{heis}_{2n+1})$, we see that
$\alpha((\mathfrak{u}(1)\times\mathfrak{u}(n-1))\ltimes\mathfrak{heis}_{2n+1})\cong\mathfrak{s}\ltimes\mathfrak{heis}_{2n+1}$,
so the full algebra of Killing fields of $H$ arises from $\alpha$ as claimed.
∎

The image of $\alpha^{\mathbb{C}}$, which we denote by
$\mathfrak{g}^{\mathbb{C}}$, contains the subalgebra

$\mathfrak{h}=\operatorname{span}_{\mathbb{C}}\Big{\\{}Y_{a},V_{a},V_{0},\frac{\partial}{\partial\tilde{\phi}}:a=1,\dots,n-1\Big{\\}}$

This is a $2$-step nilpotent Lie algebra of complex dimension $2n$, whose
center is

$\mathfrak{z}(\mathfrak{h})=\operatorname{span}_{\mathbb{C}}\Big{\\{}V_{a},\frac{\partial}{\partial\tilde{\phi}}:a=1,\dots,n-1\Big{\\}}.$

The only non-trivial brackets are $[Y_{a},V_{0}]=V_{a}$, for $a=1,\dots,n-1$.
Analogous statements hold for $\overline{\mathfrak{h}}$. Note that
$\mathfrak{h}\cap\overline{\mathfrak{h}}=\mathbb{C}\frac{\partial}{\partial\tilde{\phi}}$,
and that $\mathfrak{h}+\bar{\mathfrak{h}}$ generates the Lie algebra
$\mathfrak{g}^{\mathbb{C}}$. Note that the elements of
$\mathfrak{g}^{\mathbb{C}}$ preserve the coordinate function $\rho$ and are
therefore tangent to its level sets, which we shall denote by
$\bar{N}_{\rho}$.

######  Lemma 3.7.

There is a natural isomorphism
$T^{\mathbb{C}}\bar{N}_{\rho}\cong\bar{N}_{\rho}\times(\mathfrak{h}+\bar{\mathfrak{h}})$.

###### Proof.

We show that the $4n-1$ vector fields
$Y_{a},V_{a},V_{0},\bar{Y}_{a},\bar{V}_{a},\bar{V}_{0},\frac{\partial}{\partial\tilde{\phi}}$
give a global frame for $T^{\mathbb{C}}\bar{N}_{\rho}$. Comparing dimensions,
it is enough to check that these complex vector fields are pointwise linearly
independent. We have the decomposition

$T^{\mathbb{C}}\bar{N}_{\rho}\cong(\mathrm{pr}_{X}^{*}TB^{n-1})^{\mathbb{C}}\oplus(\mathrm{pr}_{w}^{*}T\mathbb{C}^{n})^{\mathbb{C}}\oplus\mathbb{C}\frac{\partial}{\partial\tilde{\phi}},$

where $B^{n-1}\subset\mathbb{C}^{n-1}$ denotes the open unit ball, and we used
the natural projections
$\mathrm{pr}_{X}:\bar{N}_{\rho}=B^{n-1}\times\mathbb{C}^{n}\times\mathbb{R}\to
B^{n-1}$,
$\mathrm{pr}_{w}:\bar{N}_{\rho}=B^{n-1}\times\mathbb{C}^{n}\times\mathbb{R}\to\mathbb{C}^{n}$.
Now consider a linear combination

$0=\sum_{a=1}^{n-1}\lambda_{a}Y_{a}+\mu_{a}\bar{Y}_{a}+\sum_{k=0}^{n-1}\delta_{k}V_{k}+\epsilon_{k}\bar{V}_{k}+\gamma\frac{\partial}{\partial\tilde{\phi}},$

with coefficients
$\lambda_{a},\mu_{a},\delta_{k},\epsilon_{k},\gamma\in\mathbb{C}$. Projecting
onto the $(TB^{n-1})^{\mathbb{C}}=T^{1,0}B^{n-1}\oplus T^{0,1}B^{n-1}$
component and considering $(1,0)$ and $(0,1)$ parts separately yields the
relations

$\mu_{a}=X^{a}\sum_{j=1}^{n-1}X^{j}\lambda_{j},\qquad\lambda_{a}=\bar{X}^{a}\sum_{j=1}^{n-1}\bar{X}^{j}\mu_{j}.$

Using vector notation
$\lambda=(\lambda_{1},\dots,\lambda_{n-1})\in\mathbb{C}^{n-1}$ and similarly
for $\mu$ and $X$, we can use the standard Hermitian inner product on
$\mathbb{C}^{n-1}$ to rewrite these relations as
$\mu=\langle\lambda,\bar{X}\rangle X$ and
$\lambda=\langle\mu,X\rangle\bar{X}$. The Cauchy–Schwarz inequality now yields

$\lVert\lambda\rVert\leq\lVert\mu\rVert\lVert
X\rVert^{2},\qquad\lVert\mu\rVert\leq\lVert\lambda\rVert\lVert X\rVert^{2},$

which implies $(1-\lVert X\rVert^{4})\lVert\lambda\rVert=0$ since $\lVert
X\rVert<1$ (cf. the discussion preceding (4)). Since $\lVert X\rVert<1$, this
means that $\lambda=0$, hence also $\mu=0$. Projecting onto the
$T\mathbb{C}^{n-1}$-factor and considering $(1,0)$ and $(0,1)$ parts, we see
immediately from (10) that $\delta_{k}=0=\epsilon_{k}$ for all
$k=0,\dots,n-1$, which in turn implies $\gamma=0$, completing the proof. ∎

Consider once again the submanifold $H=\\{X=0\\}\subset\bar{N}_{\rho_{0}}$ and
note that, at $p=(0,w,\tilde{\phi},\rho_{0})\in H$, we have

$Y_{a}(p)=\frac{\partial}{\partial\bar{X}^{a}}-w^{0}\frac{\partial}{\partial
w^{a}}-\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{0}},$

and

$V_{a}(p)=\frac{\partial}{\partial
w^{a}}-i\bar{w}^{a}\frac{\partial}{\partial\tilde{\phi}},\qquad
V_{0}(p)=\frac{\partial}{\partial
w^{0}}+i\bar{w}^{0}\frac{\partial}{\partial\tilde{\phi}}.$

The metric (given in (4)) also simplifies on $T_{p}\bar{N}_{\rho_{0}}$:

$\displaystyle g^{c}$
$\displaystyle=\frac{\rho_{0}+c}{\rho_{0}}\sum_{a=1}^{n-1}\big{|}\mathrm{d}X^{a}\big{|}^{2}+\frac{1}{2\rho_{0}}\sum_{a=1}^{n-1}\big{|}\mathrm{d}w^{a}\big{|}^{2}+\bigg{(}\frac{\rho_{0}+c}{\rho_{0}^{2}}-\frac{1}{2\rho_{0}}\bigg{)}\big{|}\mathrm{d}w^{0}\big{|}^{2}$
(16)
$\displaystyle\quad+\frac{1}{4\rho_{0}^{2}}\frac{\rho_{0}+c}{\rho_{0}+2c}\bigg{(}\mathrm{d}\tilde{\phi}-2i\bigg{(}\bar{w}^{0}\mathrm{d}w^{0}-w^{0}\mathrm{d}\bar{w}^{0}-\sum_{a=1}^{n-1}(\bar{w}^{a}\mathrm{d}w^{a}-w^{a}\mathrm{d}\bar{w}^{a})\bigg{)}\bigg{)}^{2}.$

######  Lemma 3.8.

The vector fields $\frac{\partial}{\partial
X^{a}},\frac{\partial}{\partial\bar{X}^{a}}$, $a=1,\dots,n-1$ form a parallel
frame for the complexified normal bundle of the submanifold
$H\subset\bar{N}_{\rho_{0}}$ with respect to the connection induced by the
Levi-Civita connection of $(\bar{N}_{\rho_{0}},g^{c})$.

###### Proof.

By looking at the form of the metric $g^{c}$ at $p\in H$ given in (16) we see
that the vector fields $\frac{\partial}{\partial
X^{a}},\frac{\partial}{\partial\bar{X}^{a}}$, $a=1,\dots,n-1$ are orthogonal
to $T^{\mathbb{C}}_{p}H$. Moreover, the complex bilinear extension of the
metric satisfies

$\displaystyle g^{c}\bigg{(}\frac{\partial}{\partial
X^{a}},\frac{\partial}{\partial X^{b}}\bigg{)}$ $\displaystyle=0$
$\displaystyle
g^{c}\bigg{(}\frac{\partial}{\partial\bar{X}^{a}},\frac{\partial}{\partial\bar{X}^{b}}\bigg{)}$
$\displaystyle=0$ $\displaystyle g^{c}\bigg{(}\frac{\partial}{\partial
X^{a}},\frac{\partial}{\partial\bar{X}^{b}}\bigg{)}$
$\displaystyle=\frac{\rho_{0}+c}{\rho_{0}}\delta_{ab}.$

Now it follows from the Koszul formula that for any vector field
$W\in\mathfrak{X}^{\mathbb{C}}(H)$ the covariant derivatives
$\nabla_{W}\frac{\partial}{\partial X^{b}}$ and
$\nabla_{W}\frac{\partial}{\partial\bar{X}^{b}}$ must be tangent to $H$. ∎

Given a vector field $K$ on $\bar{N}_{\rho_{0}}$, we shall denote the normal
component of the restriction of $K$ to the submanifold $H$ by $K_{H}^{\perp}$.

######  Lemma 3.9.

Denote by
$\mathfrak{k}=\\{K\in\mathfrak{X}^{\mathbb{C}}(\bar{N}_{\rho_{0}}):\nabla^{\perp}(K_{H}^{\perp})=0\\}$,
the space of complex vector fields on $\bar{N}_{\rho_{0}}$ whose normal
component along the submanifold $H$ is a parallel section of the normal bundle
of $H$. Then $\mathfrak{g}\subset\mathfrak{k}$.

###### Proof.

We can write any complex vector field in the form

$K=\sum_{a}f_{a}\frac{\partial}{\partial
X^{a}}+g_{a}\frac{\partial}{\partial\bar{X}^{a}}+\sum_{k}\phi_{k}\frac{\partial}{\partial
w^{k}}+\psi_{k}\frac{\partial}{\partial\bar{w}^{k}}+h\frac{\partial}{\partial\tilde{\phi}}$

with $f_{a},g_{a},\phi_{a},\psi_{k},h\in C^{\infty}(\bar{N}_{\rho_{0}})$. By
Lemma 3.8, we see that $K\in\mathfrak{k}$ if and only if the functions
$f_{a},g_{a}$ are constant when restricted to the submanifold $H$. Clearly,
this is satisfied by the vector fields
$Y_{C},Y_{a},\bar{Y}_{a},V_{k},\bar{V}_{k},\frac{\partial}{\partial\tilde{\phi}}$.
Moreover, at $p=(0,w,\tilde{\phi},{\rho_{0}})$ we have

$[Y_{a},\bar{Y}_{b}](p)=\delta_{ab}\Bigg{(}w^{0}\frac{\partial}{\partial
w^{0}}-\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{0}}-2ic\frac{\partial}{\partial\tilde{\phi}}\Bigg{)}+\bar{w}^{a}\frac{\partial}{\partial\bar{w}^{b}}-w^{b}\frac{\partial}{\partial
w^{a}},$

which is also tangent to $H$ and thus is an element of $\mathfrak{k}$. This
means that we have verified our claim for the basis
$\\{Y_{C},Y_{a},\bar{Y_{a}},[Y_{a},\bar{Y}_{b}],V_{k},\bar{V_{k}},\frac{\partial}{\partial\tilde{\phi}}\\}$
of $\mathfrak{g}^{\mathbb{C}}$. ∎

######  Proposition 3.10.

For every $c>0$, the elements of the algebra $\mathfrak{g}$ can be
characterized, modulo Killing fields that are zero at each point of $H$, as
those Killing fields on $(\bar{N}_{\rho_{0}},g^{c})$ whose normal component
along $H$ is parallel.

###### Proof.

Let $K$ be such a Killing field. Expand it in the frame
$T^{\mathbb{C}}\bar{N}_{\rho_{0}}\cong\bar{N}_{\rho_{0}}\times(\mathfrak{h}+\bar{\mathfrak{h}})$
constructed in Lemma 3.7. Then the coefficient functions in front of
$Y_{a},\bar{Y}_{a}$ are constant along $H$. By subtracting a linear
combination of $Y_{a},\bar{Y}_{a}$ with constant coefficients, we can thus
obtain a Killing field that is tangent to $H$. By Lemma 3.6, this must then
coincide with an element of
$\alpha((\mathfrak{u}(1)\times\mathfrak{u}(n-1))\ltimes\mathfrak{heis}_{2n+1})$
along $H$. This shows that, up to addition of a Killing field vanishing along
$H$, the vector field $K$ lies in $\mathfrak{g}$. ∎

###  3.3 The action of
$\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$

In the previous section we have constructed an injective ($c$-dependent)
infinitesimal action
$\alpha:\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}\to\mathfrak{g}\subset\mathfrak{isom}(\bar{N}_{n},g^{c})$.
Since $(\bar{N},g^{c})$ is a complete Riemannian manifold, any Killing field
on it is necessarily complete. We can therefore integrate the infinitesimal
action
$\alpha:\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}\to\mathfrak{isom}(\bar{N}_{n},g^{c})$
to an isometric action of the corresponding simply connected group
$\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$. In this way
we obtain a group homomorphism
$\beta:\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}\to\operatorname{Isom}(\bar{N}_{n},g^{c})$.
Let us collect some useful pieces of notation, which we will repeatedly make
use of in the following.

######  Notation.

We will abbreviate
$\tilde{\mathcal{G}}\coloneqq\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$
and denote its image $\beta(\tilde{\mathcal{G}})$ by $\mathcal{G}$. The
subgroups $\beta(\widetilde{\mathrm{U}}(1,n-1))$ and
$\beta(\operatorname{Heis}_{2n+1})$ will be called $\mathcal{U}$ and
$\mathcal{H}$, respectively. We will write $\mathcal{F}$ for the intersection
$\mathcal{U}\cap\mathcal{H}$. We will also consider
$\tilde{\mathcal{G}}^{\prime}\coloneqq\widetilde{\mathrm{SU}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$;
the restriction of $\beta$ to $\tilde{\mathcal{G}^{\prime}}$ will be denoted
by $\beta^{\prime}$. Its image is
$\mathcal{G}^{\prime}\coloneqq\beta^{\prime}(\tilde{\mathcal{G}}^{\prime})$,
and we abbreviate $\beta^{\prime}(\widetilde{\mathrm{SU}}(1,n-1))$ to
$\mathcal{U}^{\prime}$ (note that
$\beta^{\prime}(\operatorname{Heis}_{2n+1})=\mathcal{H}$). Finally, we set
$\mathcal{F}^{\prime}\coloneqq\mathcal{U}^{\prime}\cap\mathcal{H}$.

We are interested in $\mathcal{G}$, which acts effectively on
$\operatorname{Isom}(\bar{N}_{n},g^{c})$. In order to understand
$\mathcal{G}$, we have to determine the kernel of $\beta$.

Since the infinitesimal action
$\alpha:\mathfrak{u}(1,n-1)\ltimes\mathfrak{heis}_{2n+1}\to\mathfrak{isom}(\bar{N}_{n},g^{c})$
is injective, $\ker\beta$ is discrete and normal, hence central. The elements
in $\ker\beta$ stabilize every point of $\bar{N}_{n}$, and in particular the
point $n_{0}=(0,0,0,\rho_{0})$, for any fixed $\rho_{0}>0$. Thus, $\ker\beta$
must be a discrete central subgroup of the stabilizer
$\tilde{\mathcal{G}}_{n_{0}}$. We computed its Lie algebra
$\mathfrak{g}_{n_{0}}$ in Lemma 3.5, and it turns out that we can integrate
the vector fields in the center of $\mathfrak{g}_{n_{0}}$ explicitly. This
allows us to determine $\ker\beta$, as we will now show.

######  Lemma 3.11.

For $n\geq 2$, the kernel of $\beta$ is the subgroup of
$Z(\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1})\cong\mathbb{R}\times
2\pi\mathbb{Z}\times\mathbb{R}$ generated by $(2\pi,0,4\pi c)$ and
$\big{(}-\frac{2\pi}{n},-2\pi,4\pi\frac{n-2}{n}c\big{)}$. In particular,
$\ker\beta\cong\mathbb{Z}^{2}$.

###### Proof.

As shown in Lemma 3.5, the Lie algebra $\mathfrak{g}_{n_{0}}$ of the
stabilizer of $n_{0}$ is isomorphic to
$\mathfrak{u}(1)\times\mathfrak{u}(n-1)$. Its two-dimensional center is
spanned by the vector fields

$\displaystyle\mathcal{C}_{1}$ $\displaystyle\coloneqq
Y_{C}+2c\frac{\partial}{\partial\tilde{\phi}},$ (17)
$\displaystyle\mathcal{C}_{2}$
$\displaystyle\coloneqq\sum_{a=1}^{n-1}\operatorname{Im}([Y_{a},\bar{Y}_{a}])+2(n-1)c\frac{\partial}{\partial\tilde{\phi}}-\mathcal{C}_{1},$

as can be easily checked using (8). Let us reproduce the explicit coordinate
expressions, obtained from (9) and (13), for convenience:

$\displaystyle\mathcal{C}_{1}$
$\displaystyle=-i\sum_{k=0}^{n-1}\bigg{(}w^{k}\frac{\partial}{\partial
w^{k}}-\bar{w}^{k}\frac{\partial}{\partial\bar{w}^{k}}\bigg{)},$
$\displaystyle\mathcal{C}_{2}$
$\displaystyle=-in\sum_{a=1}^{n-1}\bigg{(}X^{a}\frac{\partial}{\partial
X^{a}}-\bar{X}^{a}\frac{\partial}{\partial\bar{X}^{a}}+w^{0}\frac{\partial}{\partial
w^{0}}-\bar{w}^{0}\frac{\partial}{\partial\bar{w}^{0}}\bigg{)}.$

Integrating these vector fields, we see that the corresponding one-parameter
groups of diffeomorphisms act as follows:

$\displaystyle\Phi_{t}^{\mathcal{C}_{1}}(X^{a},w^{0},w^{a},\tilde{\phi},\rho)$
$\displaystyle=(X^{a},e^{-it}w^{0},e^{-it}w^{a},\tilde{\phi},\rho)$
$\displaystyle\Phi_{t}^{\mathcal{C}_{2}}(X^{a},w^{0},w^{a},\tilde{\phi},\rho)$
$\displaystyle=(e^{-int}X^{a},e^{-int}w^{0},w^{a},\tilde{\phi},\rho)$

It is clear that the flows are periodic with periods $2\pi$ and
$\frac{2\pi}{n}$, respectively. Thus, the $\mathbb{Z}^{2}$-subgroup generated
by $\exp(2\pi\mathcal{C}_{1})$ and $\exp(\frac{2\pi}{n}\mathcal{C}_{2})$ acts
trivially on all of $\bar{N}_{n}$ and is therefore contained in $\ker\beta$.
In fact, it is equal to $\ker\beta$, since we know that $\ker\beta$ is
certainly contained in the preimage of the stabilizer.

The center of $\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$
is

$Z(\widetilde{\mathrm{U}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1})=\exp(\mathbb{R}\cdot
C)\times\exp(2\pi\mathbb{Z}\cdot C^{\prime})\times\exp(\mathbb{R}\cdot
T)\cong\mathbb{R}\times 2\pi\mathbb{Z}\times\mathbb{R},$

where
$C^{\prime}\coloneqq\frac{i}{n}\operatorname{diag}(1-n,1,\dots,1)\in\mathfrak{su}(1,n-1)$.

Using (17) and combining the fact that
$\sum_{a}\operatorname{Im}_{\sigma}[U_{a},U_{a}^{\sigma}]=nC^{\prime}$ and
that $\alpha$ is an anti-homomorphism, we can directly read off that the
kernel of $\beta$ is the subgroup generated by $(2\pi,0,4\pi c)$ and
$\big{(}-\frac{2\pi}{n},-2\pi,4\pi\frac{n-2}{n}c\big{)}$. ∎

######  Remark 3.12.

We observe that, for all $c>0$, the subgroup $\ker\beta\subset
Z(\tilde{\mathcal{G}})$ has a non-trivial projection onto the subgroup
$Z(\operatorname{Heis}_{2n+1})$.

We will now consider the subgroups $\mathcal{U}$ and $\mathcal{H}$ of
$\mathcal{G}$.

######  Lemma 3.13.

For every $n\geq 2$, the following hold:

  1. (i)

The restricted map $\beta:\operatorname{Heis}_{2n+1}\to\mathcal{H}$ is an
isomorphism.

  2. (ii)

The kernel $\ker\beta$ is contained in the subgroup
$\widetilde{\mathrm{U}}(1,n-1)$ if $c=0$. If $c>0$, then the kernel of the
restricted map $\beta:\widetilde{\mathrm{U}}(1,n-1)\to\mathcal{U}$ is equal to
the infinite cyclic group
$2\pi\mathbb{Z}\cdot(n-1,n)\subset\mathbb{R}\times\mathbb{Z}=Z(\widetilde{\mathrm{U}}(1,n-1))$
if $n$ is odd and $\pi\mathbb{Z}\cdot(n-1,n)$ if $n$ is even.

  3. (iii)

The intersection $\mathcal{F}=\mathcal{U}\cap\mathcal{H}$ is given by
$\langle\beta(0,0,\frac{8\pi c}{n})\rangle$ if $n$ is even and
$\langle\beta(0,0,\frac{4\pi c}{n})\rangle$ if $n$ is odd. In particular, it
is trivial if $c=0$ and infinite cyclic if $c>0$.

###### Proof.

  1. (i)

By Lemma 3.11, the kernel of $\beta$ is the subgroup of
$Z(\tilde{\mathcal{G}})\cong\mathbb{R}\times 2\pi\mathbb{Z}\times\mathbb{R}$
generated by $(2\pi,0,4\pi c)$ and
$\big{(}-\frac{2\pi}{n},-2\pi,4\pi\frac{n-2}{n}c\big{)}$. But this group has
trivial intersection with $\operatorname{Heis}_{2n+1}$ for any value of $c$,
so the restriction of $\beta$ is injective, hence an isomorphism.

  2. (ii)

We compute $\ker\beta\cap Z(\widetilde{\mathrm{U}}(1,n-1))$, using the
generators provided by Lemma 3.11. For $c=0$, our claim is obvious. Now assume
that $c>0$ and let $x,y\in\mathbb{Z}$. The third component of $x(2\pi,0,4\pi
c)+y\big{(}-\frac{2\pi}{n},-2\pi,4\pi\frac{n-2}{n}c\big{)}$ is $4\pi
c(x+\frac{n-2}{n}y)$, which vanishes if and only if $nx+(n-2)y=0$, which in
turn holds if and only if $(x,y)$ is an integral multiple of $(2-n,n)$ if $n$
is odd and $\frac{1}{2}(2-n,n)$ if $n$ is even. Thus, if $n$ is odd, then
$\ker\beta$ is generated by the element $(n-2)\cdot(2\pi,0,4\pi
c)-n\cdot\big{(}-\frac{2\pi}{n},-2\pi,4\pi\frac{n-2}{n}c\big{)}=2\pi\cdot(n-1,n,0)$.
If $n$ is even we obtain $\pi\cdot(n-1,n,0)$ as a generator instead.

  3. (iii)

Since $\dim\mathcal{G}=\dim\mathcal{U}+\dim\mathcal{H}$, $\mathcal{F}$ is
discrete and normal in $\mathcal{U}$. Thus $\mathcal{U}\subset
Z(\mathcal{U})$. Moreover, since left-translation by $\mathcal{U}$ fixes the
origin in $\mathbb{R}^{2n}\times\\{0\\}\subset\mathcal{H}$, we must have
$\mathcal{F}\subset Z(\mathcal{H})$. It follows that
$\mathcal{F}=Z(\mathcal{U})\cap Z(\mathcal{H})$. The preimage
$\beta^{-1}(\mathcal{F})=\beta^{-1}(Z(\mathcal{U}))\cap\beta^{-1}(Z(\mathcal{H}))$
is therefore a discrete subgroup of the center $\mathbb{R}\times
2\pi\mathbb{Z}\times\mathbb{R}$ of $\tilde{\mathcal{G}}$.

By part (i), $\beta$ restricts to an isomorphism
$\operatorname{Heis}_{2n+1}\to\mathcal{H}$. Consequently, we have
$\beta^{-1}(Z(\mathcal{H}))=(\\{(0,0)\\}\times\mathbb{R})+\ker\beta$. On the
other hand $\beta^{-1}(Z(\mathcal{U}))=(\mathbb{R}\times
2\pi\mathbb{Z}\times\\{0\\})+\ker\beta$. Now, a point $2\pi\cdot(0,0,\lambda)$
with $\lambda\in\mathbb{R}$ lies in $\beta^{-1}(\mathcal{F})$ if and only if
we can find some $\kappa\in\ker\beta$, $a\in\mathbb{R}$ and $b\in\mathbb{Z}$
such that $2\pi\cdot(0,0,\lambda)+\kappa=2\pi\cdot(a,b,0)$. For any
$x,y\in\mathbb{Z}$ we have

$\displaystyle(0,0,2\pi\lambda)\\!+x(2\pi,0,4\pi
c)\\!-y(\tfrac{2\pi}{n},2\pi,-4\pi\tfrac{n-2}{n}c)=2\pi\big{(}x\\!-\\!\tfrac{y}{n},-y,\lambda\\!+\\!2c(x+\tfrac{n-2}{n}y)\big{)}.$

To make the third entry zero, we must have
$\lambda=-2c\big{(}x+\frac{n-2}{n}y\big{)}\in
2c\cdot\operatorname{span}_{\mathbb{Z}}\\{1,\frac{n-2}{n}\\}$, which equals
$\frac{4c}{n}\mathbb{Z}$ if $n$ is even and $\frac{2c}{n}\mathbb{Z}$ if $n$ is
odd. Thus, $\beta^{-1}(\mathcal{F})=\mathbb{Z}\cdot(0,0,\frac{8\pi
c}{n})+\ker\beta$ if $n$ is even and
$\beta^{-1}(\mathcal{F})=\mathbb{Z}\cdot(0,0,\frac{4\pi c}{n})+\ker\beta$ if
$n$ is odd, which means that $\mathcal{F}=\langle\beta(0,0,\frac{8\pi
c}{n})\rangle$ in the former case and $\mathcal{F}=\langle\beta(0,0,\frac{4\pi
c}{n})\rangle$ in the latter, just as claimed.∎

Next, we study $\mathcal{G}^{\prime}$, which is obtained by restricting
$\beta$ to
$\tilde{\mathcal{G}}^{\prime}=\widetilde{\mathrm{SU}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$.

######  Lemma 3.14.

For every $n\geq 2$, the following hold:

  1. (i)

If $c=0$, then the kernel of
$\beta^{\prime}:\widetilde{\mathrm{SU}}(1,n-1)\to\mathcal{U}^{\prime}$ is
equal to $2\pi n\mathbb{Z}\subset
2\pi\mathbb{Z}=Z(\widetilde{\mathrm{SU}}(1,n-1))$. Thus,
$\mathcal{U}^{\prime}\cong\mathrm{SU}(1,n-1)$.

  2. (ii)

If $c>0$ then
$\beta^{\prime}:\widetilde{\mathrm{SU}}(1,n-1)\to\mathcal{U}^{\prime}$ is an
isomorphism.

  3. (iii)

The subgroup $\mathcal{F}^{\prime}=\mathcal{U}^{\prime}\cap\mathcal{H}$ is
given by $\langle\beta^{\prime}(0,0,4\pi c(n-1))\rangle$, hence trivial if
$c=0$ and infinite cyclic if $c>0$.

###### Proof.

  1. (i)

Since $c=0$, Lemma 3.11 implies that the kernel of $\beta$ intersects
$Z(\widetilde{\mathrm{SU}}(1,n-1))\cong 2\pi\mathbb{Z}$ in the cyclic subgroup
generated by $2\pi n$. The assertion follows.

  2. (ii)

We have
$\widetilde{\mathrm{U}}(1,n-1)\cong\mathbb{R}\times\widetilde{\mathrm{SU}}(1,n-1)$
and $\mathcal{U}^{\prime}=\beta(\\{0\\}\times\widetilde{\mathrm{SU}}(1,n-1))$.
By part (ii) of Lemma 3.13, the intersection of $\ker\beta$ and
$\\{0\\}\times\widetilde{\mathrm{SU}}(1,n-1)$ is trivial.

  3. (iii)

By the same arguments as in the proof of part (iii) of Lemma 3.13,
$\mathcal{F}^{\prime}=Z(\mathcal{U}^{\prime})\cap Z(\mathcal{H})$. The center
of
$\tilde{\mathcal{G}}^{\prime}=(\\{0\\}\times\widetilde{\mathrm{SU}}(1,n-1))\ltimes\operatorname{Heis}_{2n+1}$
is $\\{0\\}\times 2\pi\mathbb{Z}\times\mathbb{R}$, which intersects the kernel
of $\beta$ in the cyclic subgroup $\mathbb{Z}\cdot(0,-2\pi n,4\pi c(n-1))$.
This observation together with a calculation analogous to the proof of part
(iii) of Lemma 3.13 shows that
$\mathcal{F}^{\prime}=\langle\beta^{\prime}(0,0,4\pi c(n-1))\rangle$.∎

The following Proposition is now an immediate application of the results
proven in Lemmata 3.11, 3.13 and 3.14.

######  Proposition 3.15.

Let $n\geq 2$ and consider
$\mathcal{G}^{\prime}\subset\operatorname{Isom}(\bar{N}_{n},g^{c})$.

  1. (i)

If $c=0$, then
$\mathcal{G}^{\prime}\cong\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$.

  2. (ii)

If $c>0$, then
$\mathcal{G}^{\prime}\cong(\widetilde{\mathrm{SU}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1})/\mathcal{F}^{\prime}$,
where $\mathcal{F}^{\prime}$ is the infinite cyclic subgroup
$\langle\beta^{\prime}(0,0,4\pi c(n-1))\rangle$. ∎

For $c>0$ and $n\geq 2$ we now consider the cyclic quotient
$\hat{N}_{n}^{c}\coloneqq\bar{N}_{n}/\mathcal{F}^{\prime}$, endowed with the
induced quaternionic Kähler metric which we continue to denote by $g^{c}$.
This amounts to considering the coordinate $\tilde{\phi}$ periodic with period
$4\pi c(n-1)$. We extend this family to $c=0$ by setting
$\hat{N}_{n}^{0}=\bar{N}_{n}$.

Since $\mathcal{F}^{\prime}\subset\tilde{\mathcal{G}}$ is central, the action
of $\tilde{\mathcal{G}}$ on $\bar{N}_{n}$ descends to $\hat{N}_{n}^{c}$. We
extend our notation to the resulting action in the obvious way:

######  Notation.

We denote by
$\hat{\beta}:\tilde{\mathcal{G}}\to\operatorname{Isom}(\hat{N}_{n}^{c},g^{c})$
and
$\hat{\beta}^{\prime}:\tilde{\mathcal{G}}^{\prime}\to\operatorname{Isom}(\hat{N}_{n}^{c},g^{c})$
the corresponding group homomorphisms with images $\hat{\mathcal{G}}$ and
$\hat{\mathcal{G}}^{\prime}$, respectively. We also define
$\hat{\mathcal{H}}\coloneqq\hat{\beta}(\operatorname{Heis}_{2n+1})$ and
$\hat{\mathcal{U}}=\hat{\beta}(\widetilde{\mathrm{U}}(1,n-1))$, as well as
$\hat{\mathcal{U}}^{\prime}=\hat{\beta}^{\prime}(\widetilde{\mathrm{SU}}(1,n-1))$.

We can now state the main result of this section.

######  Theorem 3.16.

For $n\geq 2$ and any value of $c\geq 0$,
$\hat{\mathcal{G}}^{\prime}\subset\operatorname{Isom}(\hat{N}_{n}^{c},g^{c})$
is isomorphic to
$\mathrm{SU}(1,n-1)\ltimes(\operatorname{Heis}_{2n+1}/\mathcal{F}^{\prime})$.
In particular, the group
$\mathrm{SU}(1,n-1)\ltimes(\operatorname{Heis}_{2n+1}/\mathcal{F}^{\prime})$
acts effectively and isometrically on $(\hat{N}_{n}^{c},g^{c})$.

###### Proof.

By construction, and the proof of part (iii) of Lemma 3.14, the kernel of
$\hat{\beta}^{\prime}$ is generated by $(0,2\pi n,0)$ and $(0,0,4\pi c(n-1))$.
This means that
$\operatorname{im}\hat{\beta}^{\prime}\cong(\widetilde{\mathrm{SU}}(1,n-1)\ltimes\operatorname{Heis}_{2n+1})/\ker\hat{\beta}^{\prime}\cong\mathrm{SU}(1,n-1)\ltimes(\operatorname{Heis}_{2n+1}/\mathcal{F}^{\prime})$,
as claimed. ∎

With the notation introduced above, we may rephrase this result as follows.
Firstly, $\hat{\mathcal{U}}^{\prime}\cong\mathrm{SU}(1,n-1)$ and
$\hat{\mathcal{H}}\cong\operatorname{Heis}_{2n+1}/\mathcal{F}^{\prime}$.
Secondly
$\hat{\mathcal{U}}^{\prime}\cap\hat{\mathcal{H}}=\\{\operatorname{id}\\}$ so
that
$\hat{\mathcal{G}}^{\prime}\cong\hat{\mathcal{U}}^{\prime}\ltimes\hat{\mathcal{H}}$.

The above results have been formulated for the case $n\geq 2$ only. The
arguments in the case $n=1$ are very similar and the differences are mostly
notational. We state them separately in the next Proposition.

######  Proposition 3.17.

Let $n=1$ and $c\geq 0$ arbitrary. Then

  1. (i)

$\ker\beta=\mathbb{Z}\cdot(2\pi,4\pi c)\subset\mathbb{R}\times\mathbb{R}\cong
Z(\widetilde{\mathrm{U}(1)}\ltimes\operatorname{Heis}_{3})$.

  2. (ii)

The restricted map $\beta:\operatorname{Heis}_{3}\to\mathcal{H}$ is an
isomorphism.

  3. (iii)

The kernel of the restricted map
$\beta:\widetilde{\mathrm{U}(1)}\to\mathcal{U}$ is
$2\pi\mathbb{Z}\subset\mathbb{R}=\widetilde{\mathrm{U}(1)}$ if $c=0$ and
trivial if $c>0$.

  4. (iv)

The intersection $\mathcal{F}=\mathcal{U}\cap\mathcal{H}$ is given by
$\langle\beta(0,4\pi c)\rangle$. In particular, it is trivial if $c=0$ and
infinite cyclic if $c>0$.

  5. (v)

$\mathcal{G}=(\widetilde{\mathrm{U}(1)}\ltimes\operatorname{Heis}_{3})/\mathcal{F}$.

  6. (vi)

$\mathcal{G}^{\prime}=\operatorname{Heis}_{3}$. ∎

In the following, it will be convenient to set
$\hat{N}_{1}^{c}\coloneqq\bar{N}_{1}/\mathcal{F}$ for $c>0$ and
$\hat{N}_{1}^{0}\coloneqq\bar{N}_{1}$ so that we can speak of
$\hat{N}_{n}^{c}$ for all $c\geq 0$ and $n\in\mathbb{N}$.

Our next aim is to prove that $\mathcal{G}$ and $\hat{\mathcal{G}}$ are closed
subgroups of $\operatorname{Isom}(\bar{N}_{n},g^{c})$ and
$\operatorname{Isom}(\hat{N}_{n}^{c},\hat{g}^{c})$, respectively. We first
consider the subgroup
$\hat{\mathcal{U}}=\hat{\beta}(\widetilde{\mathrm{U}}(1,n-1))\subset\hat{\mathcal{G}}$.

######  Lemma 3.18.

For all $n\in\mathbb{N}$ and any $c\geq 0$,
$\hat{\mathcal{U}}\subset\operatorname{Isom}(\hat{N}_{n}^{c},g^{c})$ has
compact center.

###### Proof.

The $\mathfrak{u}(1)$-factor of
$\operatorname{Lie}(\hat{\mathcal{U}})\cong\mathfrak{u}(1)\oplus\mathfrak{su}(1,n-1)$
is generated by the vector field $Y_{C}$. If $c=0$, it is clear that $Y_{C}$
generates a $\mathrm{U}(1)$-subgroup. In the case $c>0$, the periodicity of
the coordinate $\tilde{\phi}$ ensures that this assertion remains valid, so
$\hat{\mathcal{U}}=\mathrm{U}(1)\cdot\hat{\mathcal{U}}^{\prime}$, where
$\hat{\mathcal{U}}^{\prime}$ is the group of isometries generated by the
subalgebra $\mathfrak{su}(1,n-1)$. Since the $\mathrm{U}(1)$-factor is
central, we have $Z(\hat{\mathcal{U}})=\mathrm{U}(1)\cdot
Z(\hat{\mathcal{U}}^{\prime})$, and it therefore suffices to prove that
$Z(\hat{\mathcal{U}}^{\prime})$ is compact. In the case $n=1$,
$\hat{\mathcal{U}}^{\prime}$ is trivial so there is nothing to prove. For
$n\geq 2$, 3.16 shows that
$\hat{\mathcal{U}}^{\prime}\cong\mathrm{SU}(1,n-1)$, so its center is cyclic
of order $n$ and in particular finite. ∎

For all $c>0$, the periodicity of the coordinate $\tilde{\phi}$ implies that
the center of $\hat{\mathcal{H}}=\hat{\beta}(\operatorname{Heis}_{2n+1})$ is
compact as well. By [Mac93, Prop. 4.2] the compactness of the centers of
$\hat{\mathcal{U}}$ and $\hat{\mathcal{H}}$ implies that they are closed
subgroups of $\operatorname{Isom}(\hat{N}_{n}^{c},g^{c})$.

######  Proposition 3.19.

For all $n\in\mathbb{N}$ and $c\geq 0$, $\hat{\mathcal{G}}$ is a closed
subgroup of $\operatorname{Isom}(\hat{N}_{n}^{c},g^{c})$ and $\mathcal{G}$ is
a closed subgroup of $\operatorname{Isom}(\bar{N}_{n},g^{c})$.

###### Proof.

We start by proving the result in the case $c>0$. The idea is to exploit the
closedness of the subgroups $\hat{\mathcal{U}}$ and $\hat{\mathcal{H}}$ of
$\hat{\mathcal{G}}$ to prove the first claim, and then apply a lifting
argument to establish the second claim.

Note that $\hat{\mathcal{H}}\subset\hat{\mathcal{G}}$ is normal, i.e.
$\hat{\mathcal{G}}$ is contained in the normalizer $N(\hat{\mathcal{H}})$,
which is a closed subgroup, since it coincides with the normalizer of the Lie
algebra of $\hat{\mathcal{H}}$. Thus, it suffices to prove that
$\hat{\mathcal{G}}$ is closed as a subgroup of $N(\hat{\mathcal{H}})$. Let us
therefore consider a sequence $\\{x_{j}\\}$ in $\hat{\mathcal{G}}$ which
converges in $N(\hat{\mathcal{H}})$. Under the projection
$N(\hat{\mathcal{H}})\to N(\hat{\mathcal{H}})/\hat{\mathcal{H}}$, which is a
submersion because $\hat{\mathcal{H}}$ is closed, this projects to a sequence
in $\hat{\mathcal{U}}/(\hat{\mathcal{U}}\cap\hat{\mathcal{H}})\subset
N(\hat{\mathcal{H}})/\hat{\mathcal{H}}$.

It follows from the description of $\mathcal{U}\cap\mathcal{H}$ given in Lemma
3.13 and Proposition 3.17, and the definition of $\hat{N}_{n}^{c}$, that
$\hat{\mathcal{U}}\cap\hat{\mathcal{H}}$ is a finite cyclic subgroup, so the
map
$\hat{\mathcal{U}}\to\hat{\mathcal{U}}/(\hat{\mathcal{U}}\cap\hat{\mathcal{H}})$
is a finite cyclic covering. The center of
$\hat{\mathcal{U}}/(\hat{\mathcal{U}}\cap\hat{\mathcal{H}})$ is compact
(because the center of $\hat{\mathcal{U}}$ is) and we may appeal once again to
[Mac93, Prop. 4.2] to deduce that it is a closed subgroup of
$N(\hat{\mathcal{H}})/\hat{\mathcal{H}}$. The projected sequence therefore
converges in $\hat{\mathcal{U}}/(\hat{\mathcal{U}}\cap\hat{\mathcal{H}})$, and
lifts to the covering space, producing a convergent sequence $\\{g_{j}\\}$ in
$\hat{\mathcal{U}}$. We may now write $x_{j}=g_{j}h_{j}$, and it follows from
convergence of $\\{x_{j}\\}$ and $\\{g_{j}\\}$ that $\\{h_{j}\\}$ converges in
$N(\hat{\mathcal{H}})$. In fact, since $\hat{\mathcal{H}}$ is closed in
$N(\hat{\mathcal{H}})$ it converges in $\hat{\mathcal{H}}$, and we conclude
that $\\{x_{j}\\}$ converges in
$\hat{\mathcal{G}}=\hat{\mathcal{U}}\cdot\hat{\mathcal{H}}$. This proves that
$\hat{\mathcal{G}}\subset\operatorname{Isom}(\hat{N}_{n}^{c},\hat{g}^{c})$ is
closed.

By definition, $\hat{\mathcal{G}}=\mathcal{G}/\mathcal{F}^{\prime}$ where
$\mathcal{F}^{\prime}\cong\mathbb{Z}$ is a discrete subgroup of the center
$\mathcal{Z}$ of $\mathcal{H}=\beta(\operatorname{Heis}_{2n+1})$. Let
$C(\mathcal{Z})\subset\operatorname{Isom}(\bar{N}_{n},g^{c})$ denote its
centralizer. Then the covering map $\bar{N}_{n}\to\hat{N}_{n}^{c}$ induces a
covering $C(\mathcal{Z})\to C(\mathcal{Z}/\mathcal{F}^{\prime})$. Under this
covering map, $\mathcal{G}$ is the preimage of the closed subgroup
$\hat{\mathcal{G}}\subset C(\mathcal{Z}/\mathcal{F}^{\prime})$, and therefore
$\mathcal{G}$ is itself closed in $C(\mathcal{Z})$, hence in
$\operatorname{Isom}(\bar{N}_{n},g^{c})$ as well.

In the case $c=0$ we proceed analogously, by passing to the quotient
$(\bar{N}_{n}/\mathbb{Z},g^{0})$ obtained by making $\tilde{\phi}$ periodic
with period $2\pi$, showing that $\mathcal{G}/\mathbb{Z}$ is a closed subgroup
of $\operatorname{Isom}(\bar{N}_{n}/\mathbb{Z},g^{0})$ and then performing the
same lifting argument. ∎

We note the following consequence, which we will use later.

######  Corollary 3.20.

Let $n\in\mathbb{N}$ and $c\geq 0$ and consider arbitrary points
$p\in\bar{N}_{n}$ and $\hat{p}\in\hat{N}_{n}^{c}$. Then
$\mathcal{G}_{0}\coloneqq\\{g\in\mathcal{G}\mid g\cdot p=p\\}$ and
$\hat{\mathcal{G}}_{0}\coloneqq\\{\hat{g}\in\hat{\mathcal{G}}\mid\hat{g}\cdot\hat{p}=\hat{p}\\}$
are compact. ∎

##  4 Quotients and ends of finite volume

###  4.1 Quotients from arithmetic lattices

We will now use the group actions discussed in the previous section to
construct interesting quotients of $(\bar{N}_{n},g^{c})$ by dividing out
appropriate discrete subgroups of isometries. The resulting manifolds will be
complete and quaternionic Kähler (since the isometries automatically preserve
the quaternionic structure), with non-trivial fundamental group.

For $c=0$, the metric is symmetric and much is known about its quotients by
discrete subgroups. In particular, it is known that
$(\bar{N}_{n},g^{0})=\frac{\mathrm{SU}(n,2)}{\mathrm{S}(\mathrm{U}(n)\times\mathrm{U}(2))}$
admits a compact quotient [Bor63]. Since any isometry of
$(\bar{N}_{n},g^{c})$, $c>0$, preserves222This follows from the fact that
$\rho$ is a curvature invariant for $c>0$ [CST22]. For $c=0$, this is of
course not true, as witnessed by the fact that $(\bar{N}_{n},g^{c})$ is the
symmetric space
$\frac{\mathrm{SU}(n,2)}{\mathrm{S}(\mathrm{U}(n)\times\mathrm{U}(2))}$. the
level sets of the global coordinate function $\rho$, we cannot expect to
obtain compact quotients in this case. The best one can hope for is to obtain
manifolds that have the structure of a fiber bundle over $\mathbb{R}_{>0}$,
which is parametrized by $\rho$, with locally homogeneous and compact fibers.

If $n\leq 2$ we shall explicitly construct (infinitely many) discrete
subgroups of $\operatorname{Isom}(\bar{N}_{n},g^{c})$, $c>0$, which yield
complete quaternionic Kähler manifolds of this type. We shall see that if one
does not insist on compactness of the fibers and only requires them to be of
finite volume, suitable discrete subgroups can be found for all values of
$n\in\mathbb{N}$.

For the purpose of studying quotients of $(\bar{N}_{n},g^{c})$, we may
(recalling the notation of Section 3.3) restrict our attention to the subgroup
$\mathcal{G}^{\prime}\subset\mathcal{G}$, whose orbits are the same as those
of the full group. This is most easily seen by noting that
$\mathfrak{g}=\operatorname{Lie}(\mathcal{G})=\mathfrak{u}(1)\oplus\operatorname{Lie}(\mathcal{G}^{\prime})$,
where the first factor is generated by the vector field $Y_{C}$, which is
tangent to the fibers of the projection
$(\bar{N}_{n},g^{c})\to\mathbb{C}\mathrm{H}^{n}$ and preserves the coordinate
function $\rho$. But the level sets of $\rho$ in these fibers are already
acted upon transitively by $\mathcal{H}\cong\operatorname{Heis}_{2n+1}$. As we
will see shortly, restricting to $\mathcal{G}^{\prime}$ affords us certain
technical advantages.

Let us now turn to the construction of discrete subgroups. If $c=0$, then
$\mathcal{G}^{\prime}$ has the structure of a semi-direct product (cf.
Proposition 3.15). This no longer holds true for $c>0$, but can be remedied by
passing to the cyclic quotient $\hat{N}_{n}^{c}$: by 3.16, the induced group
of isometries $\hat{\mathcal{G}}^{\prime}$ is always a semi-direct product of
the form $\mathrm{SU}(1,n-1)\ltimes\hat{\mathcal{H}}$, where
$\hat{\mathcal{H}}=\operatorname{Heis}_{2n+1}/\mathcal{F}^{\prime}$, with
$\mathcal{F}^{\prime}\cong\mathbb{Z}$ a central, cyclic subgroup of
$\operatorname{Heis}_{2n+1}$.

Thinking of $\hat{\mathcal{G}}^{\prime}$ as determined by an action of
$\mathrm{SU}(1,n-1)$ on $\hat{\mathcal{H}}$ by automorphisms suggests the
following strategy. First, we construct a discrete subgroup $\bar{\Gamma}_{1}$
of $\mathrm{SU}(1,n-1)$. Next, we construct a discrete subgroup
$\hat{\Gamma}_{2}\subset\hat{\mathcal{H}}$ which is invariant under the action
of $\bar{\Gamma}_{1}$. Then
$\hat{\Gamma}\coloneqq\bar{\Gamma}_{1}\ltimes\hat{\Gamma}_{2}$ will be a
discrete subgroup of the semi-direct product group.

The first step, thus, is to construct a suitable discrete subgroup of
$\mathrm{SU}(1,n-1)$. Recall that a lattice in a unimodular Lie group $G$ is a
discrete subgroup $\Gamma\subset G$ such that $\operatorname{vol}(G/\Gamma)$
is finite. Here $\operatorname{vol}(G/\Gamma)=\operatorname{vol}(F)$, where
$F\subset G$ is a fundamental domain and $\operatorname{vol}(F)$ is computed
with respect to the Haar measure on $G$. A lattice is called co-compact if
$G/\Gamma$ is compact. The study of lattices in semi-simple Lie groups is a
well-developed area of research (see e.g. [Mor15]), and we will now show how
to apply some of the known constructions to the problem at hand.

In the case $n=1$ the group $\mathrm{SU}(1,n-1)$, and thus the first step in
our construction, is trivial. We can take $\hat{\Gamma}=\hat{\Gamma}_{2}$
where $\hat{\Gamma}_{2}$ is a discrete subgroup in
$\hat{\mathcal{H}}=\operatorname{Heis}_{3}/\mathcal{F}^{\prime}$. If
$\hat{\Gamma}$ is a co-compact lattice in $\hat{\mathcal{H}}$, the resulting
quotient $\hat{N}_{1}/\hat{\Gamma}$ is topologically
$\mathbb{R}_{>0}\times\hat{{\mathcal{H}}}/\hat{\Gamma}$ and hence of the
desired type.

The next case, $n=2$, is much less simple. The first step is to construct co-
compact lattices in $\mathrm{SU}(1,1)$. It is well-known that
$\mathrm{SU}(1,1)$ is isomorphic to $\mathrm{SL}(2,\mathbb{R})$, so discrete
subgroups of $\mathrm{SU}(1,1)$ are the same thing as Fuchsian groups.
Therefore, we are looking for co-compact Fuchsian groups which preserve a
lattice in $\operatorname{Heis}_{5}$. One way to construct Fuchsian groups is
via quaternion algebras, and this is what we recall next. Let $F$ be a field,
and let $a,b\in F^{*}=F\setminus\\{0\\}$. Then the quaternion algebra over $F$
associated to the pair $(a,b)$ is the unital algebra over $F$ with generators
$I,J,K$ and relations

$I^{2}=a,\quad J^{2}=b,IJ=K=-JI$

Note that this implies $K^{2}=-IJJI=-ab$. We will denote this algebra by
$\big{(}\frac{a,b}{F}\big{)}$. The case relevant for us is $F=\mathbb{Q}$. It
is known that if $a,b$ are positive integers such that $b$ is prime and $a$ is
a quadratic non-residue mod $b$, then $\big{(}\frac{a,b}{\mathbb{Q}}\big{)}$
is a division algebra (see, for instance, [Kat92, Ch. 5])

Note that we may realize $A\coloneqq\big{(}\frac{a,b}{\mathbb{Q}}\big{)}$ as a
$\mathbb{Q}$-subalgebra of $\mathrm{Mat}_{2}(\mathbb{C})$ via

$\mathds{1}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix},\quad
I=\sqrt{a}i\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},\quad
J=\sqrt{b}\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},\quad
K=\sqrt{ab}i\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$ (18)

The reduced norm of an element $Q=q_{0}\mathds{1}+q_{1}I+q_{2}J+q_{3}K\in
A\subset\mathrm{Mat}_{2}(\mathbb{C})$ is given by $\det(Q)$. Explicitly, we
have

$Q=\begin{pmatrix}q_{0}+\sqrt{ab}iq_{3}&\sqrt{a}iq_{1}+\sqrt{b}q_{2}\\\
-\sqrt{a}iq_{1}+\sqrt{b}q_{2}&q_{0}-\sqrt{ab}iq_{3}\end{pmatrix},$

so

$\det(Q)=q_{0}^{2}-aq_{1}^{2}-bq_{2}^{2}+abq_{3}^{2}.$

Note that if $a$ and $b$ are positive and $\det(Q)=1$, then
$Q\in\mathrm{SU}(1,1)$.

The _standard order_ in $A$ is given by

$\mathcal{O}=\\{Q=q_{0}\mathds{1}+q_{1}I+q_{2}J+q_{3}K:q_{i}\in\mathbb{Z}\\}$

We can now define the Fuchsian group associated with the standard order in
$A$. It is given by the elements of $\mathcal{O}$ of unit reduced norm:

$\Gamma(A,\mathcal{O})\coloneqq\\{Q\in\mathcal{O}:\det(Q)=1\\}\subset\mathrm{SU}(1,1).$

It is known that $\Gamma(A,\mathcal{O})$ constructed as above is a Fuchsian
group. Moreover, the fact that $A$ is a division algebra guarantees that the
quotient space $\mathbb{C}\mathrm{H}^{1}/\Gamma(A,\mathcal{O})$ is compact
[Kat92, Ch. 5]. In particular, $\Gamma(A,\mathcal{O})$ is a co-compact
Fuchsian group.

We shall see next that $\Gamma(A,\mathcal{O})$ preserves a lattice in
$\operatorname{Heis}_{5}$, as desired. In what follows, we shall use the
following realization of $\operatorname{Heis}_{2n+1}$ for any
$n\in\mathbb{N}$. Consider $\mathbb{C}^{n}$ with the Hermitian structure
$(h,\bar{\cdot})$ of signature $(1,n-1)$ and associated symplectic form
$\omega=\mathrm{Im}(h)$. Then $\operatorname{Heis}_{2n+1}$ is the set
$\mathbb{C}^{n}\times\mathbb{R}$ with multiplication law

$(v,t)\cdot(v^{\prime},t^{\prime})=(v+v^{\prime},t+t^{\prime}+\tfrac{1}{2}\omega(v,w)),$
(19)

for any $v,v^{\prime}\in\mathbb{C}^{n},t,t^{\prime}\in\mathbb{R}$

######  Proposition 4.1.

Let $b$ be a prime number and $a\in\mathbb{N}$ a quadratic non-residue modulo
$b$. Equip $A=\big{(}\frac{a,b}{\mathbb{Q}}\big{)}$ with the standard order
$\mathcal{O}$. Then there exists a $\Gamma(A,\mathcal{O})$-invariant and co-
compact lattice $\Gamma_{2}$ in the five-dimensional Heisenberg group. Thus,
$\bar{\Gamma}_{a,b}\coloneqq\Gamma(A,\mathcal{O})\ltimes\Gamma_{2}$ is a co-
compact lattice in $\mathrm{SU}(1,1)\ltimes\operatorname{Heis}_{5}$.

###### Proof.

We write $e_{1}=\begin{pmatrix}1\\\ 0\end{pmatrix}\in\mathbb{C}^{2}$. Looking
at (18), we see that

$\mathcal{O}\cdot
e_{1}\coloneqq\operatorname{span}_{\mathbb{Z}}\\{e_{1},Ie_{1},Je_{1},Ke_{1}\\}=\operatorname{span}_{\mathbb{Z}}\\{e_{1},-\sqrt{a}ie_{2},\sqrt{b}e_{2},\sqrt{ab}ie_{1}\\}$

is a lattice in the real vector space $\mathbb{C}^{2}\cong\mathbb{R}^{4}$. Now
we consider the subgroup of $\operatorname{Heis}_{5}$ generated by the lattice
$\mathcal{O}\cdot e_{1}$. To see that the result is a lattice in the
Heisenberg group, we need only check that the possible entries in the
additional $\mathbb{R}$-factor are discrete. These are determined by the
symplectic form corresponding to the Hermitian metric of signature $(1,1)$ on
$\mathbb{C}^{2}$. By (19), the values that occur in the subgroup generated by
a lattice $\Lambda\subset\mathbb{C}^{2}$ are half-integer multiples of the
evaluation of this symplectic form on its generators. Thus, it suffices to
determine these:

$\displaystyle\omega(Ie_{1},e_{1})=\omega(Je_{1},e_{1})=0=\omega(Ie_{1},Ke_{1})=\omega(Je_{1},Ke_{1})$
$\displaystyle\omega(e_{1},Ke_{1})=\sqrt{ab}=\omega(Ie_{1},Je_{1})$

Clearly, we only obtain multiples of $\sqrt{ab}$ by evaluation. This shows
that the subgroup generated by this lattice in $\mathbb{C}^{2}$ is a co-
compact lattice in $\operatorname{Heis}_{5}$. ∎

We have now constructed infinitely many co-compact lattices
$\bar{\Gamma}_{a,b}\subset\mathrm{SU}(1,1)\ltimes\operatorname{Heis}_{5}$,
labeled by pairs $(a,b)\in\mathbb{N}^{2}$, where $b$ is prime and $a$ a
quadratic non-residue modulo $b$.

######  Theorem 4.2.

Let $a,b\in\mathbb{N}$ be such that $b$ is prime and $a$ is a quadratic non-
residue modulo $b$. Choose $c\geq 0$ such that $\tfrac{1}{2}\sqrt{ab}$ and
$4\pi c$ are linearly dependent over $\mathbb{Q}$. Then $\bar{\Gamma}_{a,b}$
admits a co-compact sub-lattice $\tilde{\Gamma}_{a,b}$ such that
$(\hat{N}_{2}^{c},\hat{g}^{c})/\tilde{\Gamma}_{a,b}$ is a complete
quaternionic Kähler manifold diffeomorphic to $\mathbb{R}_{>0}\times K$, where
$K$ is a compact and locally homogeneous seven-dimensional manifold.

###### Proof.

By Proposition 3.15 and 3.16 and since $\tfrac{1}{2}\sqrt{ab}$ and $4\pi c$
are linearly dependent over $\mathbb{Q}$, we see that $\bar{\Gamma}_{a,b}\cap
Z(\operatorname{Heis}_{5})+\mathcal{F}^{\prime}$ is a discrete subgroup of
$Z(\operatorname{Heis}_{5})\cong\mathbb{R}$. It follows that
$\bar{\Gamma}_{a,b}$ descends to a lattice $\hat{\Gamma}_{a,b}$ in
$\mathrm{SU}(1,n-1)\ltimes(\operatorname{Heis}_{5}/\mathcal{F}^{\prime})$
which acts effectively on $\hat{N}_{2}^{c}$. If $\hat{\Gamma}_{a,b}$ acts
freely on $(\hat{N}_{2}^{c},\hat{g}^{c})$, we may directly take the quotient
and obtain a quaternionic Kähler manifold with the required properties. We may
think of
$\hat{N}_{2}^{c}\cong\mathbb{R}_{>0}\times\bar{N}_{\rho_{0}}/\mathcal{F}^{\prime}$,
where $\bar{N}_{\rho_{0}}=\\{\rho\equiv\rho_{0}>0\\}\subset\bar{N}_{2}$, as a
fiber bundle over $\mathbb{R}_{>0}$ with homogeneous fibers. The
$\rho$-coordinate is preserved by all isometries, so
$\hat{N}_{2}^{c}/\hat{\Gamma}_{a,b}$ is a fiber bundle over $\mathbb{R}_{>0}$
with locally homogeneous fibers which are moreover compact since
$\hat{\Gamma}_{a,b}$ is co-compact.

However, the constructed lattices $\hat{\Gamma}_{a,b}$ do not necessarily act
freely on $\hat{N}_{2}^{c}$. Nevertheless their intersection with the
stabilizer of a point, which is compact by 3.20, is finite. Since a finite-
index subgroup of a co-compact lattice is once again a co-compact lattice, it
now suffices to find a finite-index subgroup of $\hat{\Gamma}_{a,b}$ which
only intersects this finite group in the identity.

The existence of such a subgroup is guaranteed by Selberg’s lemma, which
asserts that every finitely generated subgroup of $\mathrm{GL}(n,\mathbb{C})$
admits a finite-index normal subgroup which is torsion-free. To apply this
result, it remains to check that $\bar{\Gamma}_{a,b}$ is finitely generated.
Observe that $\bar{\Gamma}_{a,b}=\Gamma(A,\mathcal{O})\ltimes\Gamma_{2}$,
where $\Gamma_{2}\subset\operatorname{Heis}_{5}$ is the lattice generated by
$\\{e_{1},Ie_{1},Je_{1},Ke_{1}\\}$, hence finitely generated. It therefore
suffices to check that for the Fuchsian group $\Gamma(A,\mathcal{O})$ is
finitely generated. But $\Gamma(A,\mathcal{O})$ is a lattice in a semi-simple
Lie group and all such lattices are even finitely presented [Mor15, Ch. 4], so
Selberg’s lemma applies. Thus, we obtain a co-compact lattice
$\tilde{\Gamma}_{a,b}$ which acts freely and isometrically on
$(\hat{N}_{2}^{c},\hat{g}^{c})$ so that the corresponding quotient space
possesses all the claimed properties. ∎

We now move on to describe a class of lattices
$\Gamma\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$ for
arbitrary $n\geq 2$. These are of the form
$\Gamma=\bar{\Gamma}_{1}\ltimes\Gamma_{2}$, where $\Gamma_{2}$ is a lattice in
$\operatorname{Heis}_{2n+1}$ with the property that its normalizer
$\bar{\Gamma}_{1}\subset\mathrm{SU}(1,n-1)$ is again a lattice. We consider
lattices $\Gamma_{2}\subset\operatorname{Heis}_{2n+1}$ that are generated by
lattices $\Lambda\subset\mathbb{C}^{n}$ satisfying a compatibility condition
with the Hermitian structure $(h,\overline{\cdot})$ on $\mathbb{C}^{n}$ of
signature $(1,n-1)$.

To state this condition, let $F$ be a totally imaginary quadratic number
field, i.e. $F=\mathbb{Q}[i\sqrt{d}]$, where $d\in\mathbb{N}$ is a square-free
integer. If we denote by $\mathcal{O}_{F}$ the ring of integers of $F$, then
we have $\mathcal{O}_{F}=\mathbb{Z}[i\sqrt{d}]$, if $d\equiv 1,2\mod 4$ and
$\mathcal{O}_{F}=\mathbb{Z}[\tfrac{1+i\sqrt{d}}{2}]$ if $d\equiv 3\mod 4$.

We call a lattice $\Lambda\subset\mathbb{C}^{n}$ that admits an action of the
ring $\mathcal{O}_{F}$ of integers in $F$ _compatible_ with $(h,\bar{\cdot})$
if $\bar{\Lambda}\subset\Lambda$, i.e. $\Lambda$ is invariant under complex
conjugation, and $h|_{\Lambda\times\Lambda}$ induces a sesquilinear form over
$\mathcal{O}_{F}$.

We remark that in the co-compact lattices
$\bar{\Gamma}_{a,b}=\Gamma(A,\mathcal{O})\ltimes\Gamma_{2}$ constructed above
for $n=2$, the lattice $\Gamma_{2}\subset\operatorname{Heis}_{5}$ is generated
by a lattice $\Lambda\subset\mathbb{C}^{2}$ satisfying this compatibility
condition with $d=ab$, provided $d$ is square-free and $d\equiv 1,2\mod 4$.

######  Proposition 4.3.

Let $n\in\mathbb{N}$ be arbitrary and denote the generator of
$Z(\operatorname{Heis}_{2n+1})\cong\mathbb{R}$ by $T$. Then, for any square-
free $d\in\mathbb{N}$ with $d\equiv 1,2\mod 4$ there exists a lattice
$\Gamma_{2}\subset\operatorname{Heis}_{2n+1}$ generated by a lattice
$\Lambda\subset\mathbb{C}^{n}$ compatible with $(h,\bar{\cdot})$ such that
$\Gamma_{2}\cap Z(\operatorname{Heis}_{2n+1})=\tfrac{1}{2}\sqrt{d}\mathbb{Z}T$
and the stabilizer $\bar{\Gamma}_{1}$ in $\mathrm{SU}(1,n-1)$ is also a
lattice. In particular,
$\bar{\Gamma}\coloneqq\bar{\Gamma}_{1}\ltimes\Gamma_{2}\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$
is a lattice.

If $n\geq 3$ then a lattice
$\bar{\Gamma}\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$
constructed in this way is never co-compact.

###### Proof.

The assertion is clear for $n=1$, so assume that $n\geq 2$. Let
$d\in\mathbb{N}$ be square-free with $d\equiv 1,2\mod 4$, so that
$F=\mathbb{Q}[i\sqrt{d}]$ is a totally imaginary quadratic number field with
ring of integers $\mathcal{O}_{F}=\mathbb{Z}[i\sqrt{d}]$.

We start by constructing a suitable lattice
$\Lambda_{d}\subset\mathbb{C}^{n}$, which is compatible with
$(h,\bar{\cdot})$. Let $\\{e_{j}:j=1,\dots,n\\}$ be the standard basis of
$\mathbb{C}^{n}$, which is orthonormal with respect to the Hermitian metric
$h$ of signature $(1,n-1)$ on $\mathbb{C}^{n}$. Setting $f_{j}=ie_{j}$, we may
consider the lattice $\Lambda_{d}\subset\mathbb{R}^{2n}\cong\mathbb{C}^{n}$
generated by $\\{e_{j},\sqrt{d}f_{j}:j=1,\dots,n\\}$. Since the symplectic
form associated to $h$ takes values in $\mathbb{Z}\cdot\sqrt{d}$ when
evaluated on this lattice in $\mathbb{C}^{n}$, the subgroup $L_{d}$ of
$\operatorname{Heis}_{2n+1}$ that this lattice generates is in fact a co-
compact lattice. By construction, the above lattice $\Lambda_{d}$ is preserved
by the natural action of $\mathcal{O}_{F}$. Moreover, $\Lambda_{d}$ is
invariant under complex conjugation and $h|_{\Lambda\times\Lambda}$ induces a
sesquilinear form over $\mathcal{O}_{F}$, so $\Lambda$ is compatible with
$(h,\bar{\cdot})$ in the sense defined above.

In this fashion, any square-free $d\in\mathbb{N}$ with $d\equiv 1,2\mod 4$
gives rise to a (co-compact) lattice $L_{d}\subset\operatorname{Heis}_{2n+1}$
which is generated by a lattice $\Lambda_{d}\subset\mathbb{C}^{n}$ compatible
with $(h,\bar{\cdot})$.

Recall that we view $\operatorname{Heis}_{2n+1}$ as the set
$\mathbb{C}^{n}\times\mathbb{R}$ with multiplication law given by

$(v,t)\cdot(v^{\prime},t^{\prime})=(v+v^{\prime},t+t^{\prime}+\tfrac{1}{2}\omega(v,w)),$

where $\omega=\mathrm{Im}(h)$. Note that the Lie algebra
$\mathfrak{heis}_{2n+1}$ is the (real) vector space
$\mathbb{C}^{n}\oplus\mathbb{R}$ with Lie bracket

$[(v,t),(v^{\prime},t^{\prime})]=\omega(v,v^{\prime}).$

In this setting the exponential map
$\exp:\mathfrak{heis}_{2n+1}\to\operatorname{Heis}_{2n+1}$ is just the
identity on $\mathbb{C}^{n}\times\mathbb{R}$.

In general, we have the following description of the lattice
$\Gamma_{2}\subset\operatorname{Heis}_{2n+1}$ generated by a lattice
$\Lambda\subset\mathbb{C}^{n}$ compatible with $(h,\bar{\cdot})$. Let $rT$ be
a generator for the image $\omega(\Lambda\times\Lambda)\subset\mathbb{R}$.
Then we have

$\Gamma_{2}=\\{(v,t)\in\mathbb{C}^{n}\times\mathbb{R}:v\in\Lambda,t\in\tfrac{1}{2}rT\mathbb{Z}\\}.$

This follows from the observation that if $v,v^{\prime}\in\Lambda$ satisfy
$\omega(v,v^{\prime})=rT$, then we have in $\operatorname{Heis}_{2n+1}$:

$(v,0)\cdot(v^{\prime},0)\cdot(-(v+v^{\prime}),0)=(0,\tfrac{1}{2}rT).$

In particular, for the lattice $L_{d}$ constructed above, we find $L_{d}\cap
Z(\operatorname{Heis}_{2n+1})=\tfrac{1}{2}\sqrt{d}\mathbb{Z}T$.

The set
$\log\Gamma_{2}\subset\mathfrak{heis}_{2n+1}=\mathbb{C}^{n}\oplus\mathbb{R}$
is closed under addition and equals the lattice
$\Lambda\oplus\frac{1}{2}rT\mathbb{Z}$. Thus, the lattice $\Gamma_{2}$
generated by a compatible lattice $\Lambda\subset\mathbb{C}^{n}$ is a log-
lattice. The lattice $\log\Gamma_{2}\subset\mathfrak{heis}_{2n+1}$ induces the
$\mathbb{Q}$-structure
$(\mathfrak{heis}_{2n+1})_{\mathbb{Q}}=\mathrm{span}_{\mathbb{Q}}(\log\Gamma_{2})$
and the compatibility of $\Lambda$ ensures that $\mathrm{SU}(1,n-1)$ is
defined over $\mathbb{Q}$ with respect to
$(\mathfrak{heis}_{2n+1})_{\mathbb{Q}}$. It then follows from [MM94, Thm. 2.2]
that the stabilizer $\bar{\Gamma}_{1}\subset\mathrm{SU}(1,n-1)$ of
$\Gamma_{2}$ is a lattice in $\mathrm{SU}(1,n-1)$. We now take
$\Gamma_{2}=L_{d}$ and the above discussion shows that
$\bar{\Gamma}_{1}=\mathrm{Stab}_{\mathrm{SU}(1,n-1)}(\Gamma_{2})$ is a
lattice. We may then form the semi-direct product to obtain a lattice
$\bar{\Gamma}=\bar{\Gamma}_{1}\ltimes\Gamma_{2}\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$.

If $n\geq 3$, and $\Lambda\subset\mathbb{C}^{n}$ is a lattice compatible with
$(h,\bar{\cdot})$ generating a lattice
$\Gamma_{2}\subset\operatorname{Heis}_{2n+1}$, then the lattice
$\bar{\Gamma}_{1}=\mathrm{Stab}_{\mathrm{SU}(1,n-1)}(\Gamma_{2})\subset\mathrm{SU}(1,n-1)$
contains a unipotent element and is therefore not co-compact by Godement’s
compactness criterion [Mor15, Prop. 5.3.1]. To find such a unipotent element,
we consider the indefinite rational quadratic form given by the real part of
$h$ on the $2n$-dimensional $\mathbb{Q}$-vector space
$\mathrm{span}_{\mathbb{Q}}(\Lambda)$. Since $n\geq 3$, i.e. $2n\geq 6$, we
can apply Meyer’s theorem [Ser73, Cor. 2 on p.43] to find a vector
$v\in\Lambda$ such that $h(v,v)=0$. Furthermore, we can choose $w\in\Lambda$
such that $v$ and $w$ are linearly independent and $h(v,w)=0$. Then
$A\in\operatorname{End}(\mathbb{C}^{n})$ given by $A=h(\cdot,v)w-h(\cdot,w)v$
is not zero, skew-hermitian with respect to $h$ and nilpotent, in particular
$A\in\mathfrak{su}(1,n-1)$. Thus, the unipotent element
$\exp(A)\in\mathrm{SU}(1,n-1)$ stabilizes $\Lambda$, and hence $\Gamma_{2}$,
i.e. $\exp(A)\in\Gamma_{1}$. For the lattices $L_{d}$ constructed above, we
can take $v=e_{1}+e_{2}$ and $w=\sqrt{d}(f_{1}+f_{2})$. This shows that in
fact also for $n=2$ the stabilizer of $L_{d}$ in $\mathrm{SU}(1,n-1)$ is not
co-compact. ∎

Proceeding as in the proof of 4.2, we now easily obtain:

######  Theorem 4.4.

Let $n\in\mathbb{N}$ be arbitrary and let $d\in\mathbb{N}$ be square-free such
that $d\equiv 1,2\mod 4$. Then there exists a lattice
$\bar{\Gamma}\subset\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$ such
that $\bar{\Gamma}\cap
Z(\operatorname{Heis}_{2n+1})=\mathbb{Z}\cdot\frac{1}{2}\sqrt{d}T$, where $T$
is the generator of $Z(\operatorname{Heis}_{2n+1})\cong\mathbb{R}$. Choose
$c\geq 0$ such that $4\pi c$ and $\tfrac{1}{2}\sqrt{d}$ are linearly dependent
over $\mathbb{Q}$. Then $\bar{\Gamma}$ contains a lattice $\tilde{\Gamma}$
such that $(\hat{N}_{n}^{c},g^{c})/\tilde{\Gamma}$ is a complete quaternionic
Kähler manifold diffeomorphic to $\mathbb{R}\times K$, where the fibers
$\\{t\\}\times K$ are locally homogeneous and of finite volume. ∎

###  4.2 The volume of fiberwise quotients by lattices

Given a lattice in $\mathrm{SU}(1,n-1)\ltimes\operatorname{Heis}_{2n+1}$, we
have outlined above how to obtain a quaternionic Kähler manifold
$\bar{N}_{n}/\Gamma$ which can be viewed as a fiber bundle with locally
homogeneous fibers of finite volume. In the previous section we have shown how
to obtain infinitely many lattices for any $n\in\mathbb{N}$, and constructed
infinitely many co-compact examples in the case $n\leq 2$. We are now
interested in the dependence of the volume of the fibers
$\bar{N}_{\rho}/\Gamma\subset\bar{N}_{n}/\Gamma$ on $\rho$.

Recall the explicit expression for the one-loop deformed $c$-map metrics on
$\bar{N}_{n}$ given in (3) and (4):

$\displaystyle g^{c}$
$\displaystyle=\frac{\rho+c}{\rho}g_{\mathbb{C}\mathrm{H}^{n-1}}+\frac{1}{4\rho^{2}}\frac{\rho+2c}{\rho+c}\mathrm{d}\rho^{2}$
$\displaystyle\quad+\frac{1}{4\rho^{2}}\frac{\rho+c}{\rho+2c}\bigg{(}\mathrm{d}\tilde{\phi}-4\operatorname{Im}\bigg{(}\bar{w}^{0}\mathrm{d}w^{0}-\sum_{a=1}^{n-1}\bar{w}^{a}\mathrm{d}w^{a}\bigg{)}+\frac{2c}{1-\lvert
X\rvert^{2}}\operatorname{Im}\bigg{(}\sum_{a=1}^{n-1}\bar{X}^{a}\mathrm{d}X^{a}\bigg{)}\bigg{)}^{2}$
$\displaystyle\quad-\frac{2}{\rho}\bigg{(}\mathrm{d}w^{0}\mathrm{d}\bar{w}^{0}-\sum_{a=1}^{n-1}\mathrm{d}w^{a}\mathrm{d}\bar{w}^{a}\bigg{)}+\frac{\rho+c}{\rho^{2}}\frac{4}{1-\lvert
X\rvert^{2}}\Big{|}\mathrm{d}w^{0}+\sum_{a=1}^{n-1}X^{a}\mathrm{d}w^{a}\Big{|}^{2},$

where

$g_{\mathbb{C}\mathrm{H}^{n-1}}=\frac{1}{1-\lvert
X\rvert^{2}}\Bigg{(}\sum_{a=1}^{n-1}\lvert\mathrm{d}X^{a}\rvert^{2}+\frac{1}{1-\lvert
X\rvert^{2}}\bigg{|}\sum_{a=1}^{n-1}\bar{X}^{a}\mathrm{d}X^{a}\bigg{|}^{2}\Bigg{)}.$

Ordering our coordinates as $(\rho,X,w,\tilde{\phi})$, where we wrote
$X=(X^{1},\dots,X^{n-1})$ and $w=(w^{0},\dots,w^{n-1})$, the Gram matrix