# Möbius fluid dynamics on the unitary groups

Daniela Emmanuele1, Marcos Salvai2 and Francisco Vittone3

###### Abstract.

We study the nonrigid dynamics induced by the standard birational actions of
the split unitary groups $G=O_{o}\left(n,n\right)$, $SU\left(n,n\right)$ and
$Sp\left(n,n\right)$ on the compact classical Lie groups $M=SO_{n}$, $U_{n}$
and $Sp_{n}$, respectively. More precisely, we study the geometry of $G$
endowed with the kinetic energy metric associated with the action of $G$ on
$M,$ assuming that $M$ carries its canonical bi-invariant Riemannian metric
and has initially a homogeneous distribution of mass. By the least action
principle, force free motions (thought of as curves in $G$) correspond to
geodesics of $G$. The geodesic equation may be understood as an inviscid
Burgers equation with Möbius constraints. We prove that the kinetic energy
metric on $G$ is not complete and in particular not invariant, find symmetries
and totally geodesic submanifolds of $G$ and address the question under which
conditions geodesics of rigid motions are geodesics of $G$. Besides, we study
equivalences with the dynamics of conformal and projective motions of the
sphere in low dimensions.

Keywords: force free motion, kinetic energy metric, nonrigid dynamics, unitary
group, split unitary group, Möbius action, maximal isotropic subspace,
inviscid Burgers equation

##  1\. Introduction and statements of the results

††footnotetext: 1<EMAIL_ADDRESS>Universidad Nacional de Rosario,
Pellegrini 250, 2000, Rosario, Argentina. ††footnotetext: 2
<EMAIL_ADDRESS>FaMAF, Universidad Nacional de Córdoba - CIEM,
CONICET, Ciudad Universitaria, (5000) Córdoba, Argentina. ††footnotetext: 3
<EMAIL_ADDRESS>Universidad Nacional de Rosario - CONICET,
Pellegrini 250, 2000, Rosario, Argentina.

Geometric mechanics, introduced by Arnold in [1], describes the force free
motions of a system as geodesics of a Riemannian metric on the configuration
space, given by the kinetic energy. Among the first examples we find invariant
metrics on $SO_{3}$ and on the diffeomorphism group
Diff${}_{\mu}\left(M\right)$ of a compact Riemannian manifold $M$ preserving
the volume form $\mu$, in order to describe, respectively, force free motions
of the rigid body and the dynamics of an ideal flow on $M$.

This approach had an even greater impact when many classical differential
equations were recognized within this theory –that is, as geodesic equations–,
among others, Korteweg-de Vries [24] (see more examples for instance in [32,
3]).

In this paper we study the dynamics of certain finite dimensional Lie groups
endowed with the kinetic energy metric induced by the action of them on
Riemannian manifolds, which do not preserve the volume. We begin by putting
the situation in a broader context. Ebin and Marsden [11] extended Arnold’s
ideas for the whole diffeomorphism group Diff$\left(M\right)$, studying in
this way compressible flows. In this case, the metric is no longer invariant.
The force free motions $t\mapsto\gamma\left(t\right)$, or equivalently, the
geodesics in Diff$\left(M\right)$, are rendered in a simple manner:
$t\mapsto\gamma\left(t\right)\left(q\right)$ turn out to be geodesics in $M$
for all $q\in M$. This description is no longer valid if the flow is
constrained, for instance in the case of Diff${}_{\mu}\left(M\right)$, which,
with the metric inherited from that of Diff$\left(M\right)$, is not totally
geodesic.

Geodesics of the induced metric have also been considered for groups of
diffeomorphisms preserving other structures on $M$, for example for the group
of symplectomorphisms [9, 18] and contactomorphisms [10]. For these groups,
and in general for subgroups of Diff${}_{\mu}\left(M\right)$, the kinetic
energy metric is invariant and has been studied extensively. In the finite
dimensional case, invariant metrics appear in the rigid context, and force
free motions, that is, geodesics, have been much explored and relatively well
understood.

Concerning non-invariant metrics, they arise for compressible flows (the basic
example being Diff$\left(M\right)$, which we mentioned above) and have been
less studied. Bauer and Modin have recovered the interest in this subject in
[4].

By non-rigid dynamics we understand the study of a dynamical system given by a
kinetic energy metric on a Lie group, induced by the action of it on a
Riemannian manifold, which is not by isometries. Among the reasons why the
nonrigid dynamics in finite dimension has been addressed to a lesser extent,
we can mention that the most significant actions of noncompact Lie groups on
compact Riemannian manifolds are in general scarce, for instance, for the
important family of compact symmetric spaces: the celebrated Nagano’s Theorem
[22] states that if a Lie group $G$ acts on a compact symmetric space $M$ and
contains the identity connected component of its isometry group, then $M$ is
essentially a symmetric $R$-space (or equivalently, it is extrinsically
symmetric) and $G$ is the so called big group, or group of basic
transformations in Takeuchi’s terminology in [30]. The best known examples are
the conformal and projective transformations of the sphere $S^{n}$, the
biholomorphic maps of the complex quadrics and the diffeomorphisms on
Grassmannians which are induced by linear isomorphisms. More generally, the
$R$-spaces, homogeneous models of the important parabolic geometries [8, 27],
admit actions of finite dimensional Lie groups containing the isometries.

The nonrigid (and non-invariant) dynamics of the simplest among the symmetric
$R$-spaces, the sphere, is studied in [25] and [19], for its two big groups,
associated with conformal and projective transformations. Also in the nonrigid
finite dimensional setting, we can cite [7] (force free affine motions in
Euclidean space) and the recent article [28], where a non-invariant metric on
$SL\left(3,\mathbb{R}\right)$ appears in relation to affine motions of 3D
ideal fluids surrounded by vacuum.

A further fundamental problem in this area, apart from existence and
regularity of the geodesic flow, is its completeness, related with the onset
of shocks. The first two issues (existence and regularity) are of interest
only in infinite dimension. The third one (completeness), also in finite
dimension (of course, only when the configuration space is not Riemannian
homogeneous), and arises naturally, as we have pointed out, when the group
acts without preserving the volume.

In finite dimension, in the invariant case, the kinetic energy metric (as any
invariant metric) is complete. Since for infinite dimension the Hopf-Rinow
theorem is no longer valid, not even for invariant metrics, completeness
depends on regularity. Bruveris and Vialard deal with this topic in depth in
[6]. In [4] sufficient conditions for completeness are found in the semi-
invariant case.

In this article we study the dynamics of Möbius motions of the compact
classical Lie groups, mainly drawing on the symmetries of the systems, in the
interest of advancing the understanding of non-rigid motions in finite
dimension. In this context, non-invariant metrics on Lie groups arise
naturally.

We consider the unitary groups $SO_{n}$, $U_{n}$ and $Sp_{n}$, over the real,
complex and quaternionic numbers (that is, the identity components of the
automorphism groups of the positive definite Hermitian forms over these scalar
fields, see Subsection 2.1).

We recall the standard birational action $\ast$ of the split unitary group
$G=O_{o}\left(n,n\right)$, $SU\left(n,n\right)$ or $Sp\left(n,n\right)$ on the
compact classical Lie group $M=SO_{n}$, $U_{n}$ or $Sp_{n}$, respectively:

(1) $\ast:G\times M\rightarrow M\text{, \ \ \ \ \ \ \
}\left(\begin{array}[]{cc}A&B\\\ C&D\end{array}\right)\ast
U=(AU+B)(CU+D)^{-1}\text{,}$

which we call the Möbius action of $G$ on $M$. This is one of the actions
studied in [22], which we mentioned above (see also [31]). Geometrically, it
comes from the identification of $M$ with the Grassmannian of maximal
isotropic subspaces, assigning to each operator $U$ in $M$ its graph from the
second to the first factor (see Subsection 2.2 below) and the natural action
of $G$ on this Grassmannian: if

$\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\left(\begin{array}[]{c}Ux\\\
x\end{array}\right)=\left(\begin{array}[]{c}AUx+B\\\
CUx+D\end{array}\right)=\left(\begin{array}[]{c}U^{\prime}y\\\
y\end{array}\right)$

for $x,y=y\left(x\right)\in\mathbb{F}^{n}$, then $U^{\prime}$ is the right
hand side of (1).

Up to coverings and connected components, $G$ is the big group of $M$, which
is characterized in [26] as the group of diffeomorphisms of $M$ sending
circles in circles (certain distinguished curves in $M$ determined by three
generic points).

Now, we consider the Riemannian structure on $M$ induced by the inclusion in
the space $\mathbb{F}^{n\times n}$ of $\left(n\times n\right)$-matrices with
entries in $\mathbb{F}={\mathbb{R}},\,{\mathbb{C}}$ or the quaternions
${\mathbb{H}}$, endowed with the real inner product given by $\left\langle
X,Y\right\rangle=\operatorname{Re}\left(\operatorname{tr}\left(\overline{X}^{T}Y\right)\right)$,
where the superscript $T$ denotes transposition (beware of the peculiar
relationship of the multiplication of quaternionic matrices with transposition
and conjugation; see Theorem 4.1 in [33]). It turns out that the metric on $M$
is bi-invariant and also that Möbius maps are not necessarily isometries.

We are concerned with the nonrigid dynamics induced by this non-isometric
action. We take $G$ as the configuration space. This means that, if $M$ has
initially a homogeneous mass distribution of constant density $1$, the points
of $M$ are allowed to move only in such a way that two configurations differ
in an element of $G$. We will see that the action of $G$ on $M$ is almost
effective (only a finite number of elements act as the identity), except for
$G=O_{o}\left(2,2\right)$. For this reason, unless otherwise stated, from now
on we assume that $n>2$ when $M=SO_{n}$. Nevertheless, for the sake of
completeness, we describe the Möbius action of $O_{o}\left(2,2\right)$ on
$SO_{2}$ in Proposition 3.

A curve $\gamma$ in $G$ may be thought of as a motion of $M$. The Riemannian
metric on $M$ allows us to consider the kinetic energy of a motion and this
induces a Riemannian metric on $G$, introduced originally by V.I. Arnold in
[1], called the kinetic energy metric on $G$ associated with the action on
$M$, with the property that force free motions (thought of as curves in $G$)
correspond to geodesics of $G$, via the least action principle.

For each $g\in G$ and $X\in T_{g}G$, the square norm of $X$ for the kinetic
energy metric turns out to be

$\left\|X\right\|^{2}=\int_{M}\left|\left.\frac{d}{dt}\right|_{0}\left(ge^{tX}\ast
q\right)\right|^{2}d\mu\left(q\right)\text{,}$

where $\left|v\right|$ denotes the norm of a tangent vector to $M$ (see
Subsection 2.3).

The kinetic energy metric on $G$ is the restriction to $G$ of the usual weak
Riemannian metric on $\operatorname{Diff}\left(M\right)$ and the equation for
geodesics may be understood as the inviscid Burgers equation with Möbius
constraints (see § 4.1 in [17]).

Next we state our first result on the geometry of $G$, its incompleteness. We
denote by $I_{n}$ and $0_{n}$ the identity and the zero $\left(n\times
n\right)$-matrices, respectively.

######  Theorem 1.

Let $G$ be endowed with the kinetic energy metric. Then the curve
$\gamma:\mathbb{R}\rightarrow G$ defined by

$\gamma(t)=\exp\left(t\left(\begin{array}[]{cc}0_{n}&I_{n}\\\
I_{n}&0_{n}\end{array}\right)\right)=\left(\begin{array}[]{cc}\cosh
t~{}I_{n}&\sinh t~{}I_{n}\\\ \sinh t~{}I_{n}&\cosh
t~{}I_{n}\end{array}\right)$

is the reparametrization of an inextendible geodesic in $G$ of finite length.
In particular, $G$ is not complete and so the kinetic energy metric on $G$ is
neither left nor right invariant.

Incompleteness, in a classical context, is related to the existence of shocks
(collapses in finite time).

We have mentioned above the work of Bruveris and Vialard [6] about
completeness in the invariant case. For the semi-invariant case, in the recent
paper [4], Bauer and Modin give _sufficient_ conditions for completeness.
Notice that the metric on $G$ in this article is semi-invariant in the sense
of [4], as we will see in Proposition 2.

The following proposition shows the behavior of the mass when
$\lim_{t\rightarrow\pm\infty}\gamma\left(t\right)$.

######  Proposition 1.

If $M\neq SO_{n}$ with odd $n$, under the action of $\gamma$, the mass in $M$
concentrates on $\pm I_{n}$ as $t\rightarrow\pm\infty$. More precisely, for
$\varepsilon=\pm 1$,

$\lim_{t\rightarrow\varepsilon\infty}\gamma\left(t\right)\ast q=\varepsilon
I_{n}$

for all $q\in\left\\{p\in M\mid p\text{ has no eigenvalue
}-\varepsilon\right\\}$, which is an open and dense set in $M$. If $M=SO_{n}$
with odd $n$, then the same property holds for $\varepsilon=1$.

Next we present some totally geodesic submanifols $N$ of $G$ endowed with the
kinetic energy metric. In our dynamical setting, this means that if a force
free motion begins tangent to $N$, then it remains in $N$.

######  Theorem 2.

The following inclusions determine totally geodesic submanifolds _(_ in any
case the ambient group carries the kinetic energy metric and subgroup, the
induced metric _):_

a) $O_{o}\left(n,n\right)\subset SU\left(n,n\right)$, |  b) $U\left(n,n\right)\subset Sp\left(n,n\right)$, |  c) $O_{o}\left(n,n\right)\subset Sp\left(n,n\right)$,
---|---|---
d) $U(n,n)\subset O_{o}(2n,2n)$, |  e) $Sp\left(n,n\right)\subset O_{o}(4n,4n)$. |

Totally geodesic subgroups when the kinetic energy metric is invariant are
addressed in [14, 21].

Now we consider the maximal compact subgroups $K=SO_{n}\times SO_{n}$,
$S(U_{n}\times U_{n})$ and $Sp_{n}\times Sp_{n}$ of $G$ (for the three
different possibilities for $G$, namely $O_{o}\left(n,n\right)$,
$SU\left(n,n\right)$ and $Sp\left(n,n\right)$). Up to coverings and connected
components, $K$ is the isometry group of $M$, or, in our dynamical context,
the group of rigid transformation of $M$.

We comment, without trying to be precise, that $M$ appears, on the one hand,
in a passive manner, as the manifold containing the fluid whose motion is
subject to Möbius constraints, and also, on the other hand, in an active way,
acting on itself on the right and on the left. Notice that $K$ is equal to
$M\times M$ or is contained in it and that the Möbius action (1) restricted to
$K$ coincides essentially with the canonical action
$\operatorname{diag}\left(A,D\right)\ast U=AUD^{-1}$ of $M\times M$ on $M$ by
isometries (here $\operatorname{diag}(A,D)$ denotes the matrix with diagonal
blocks $A$ and $D$).

Although the kinetic energy metric on $G$ is not invariant, it has a certain
symmetry given by the action of $K\times K$, as the following proposition
states, placing the topic in the framework of [4].

######  Proposition 2.

_a)_ The action of $K\times K$ on $G$ given by
$\left(k_{1},k_{2}\right)\left(g\right)=k_{1}gk_{2}^{-1}$ is by isometries of
the kinetic energy metric of $G$.

_b)_ The Riemannian metric on $K$ induced by the kinetic energy metric of $G$
is bi-invariant; moreover, it is the restriction to $K$ of a product metric on
$M\times M$ such that the map $\rho:K\rightarrow K$ given by
$\rho\left(g,h\right)=\left(h,g\right)$ is an isometry.

Since the metric on $K$ is bi-invariant, the geodesics $\sigma$ in $K$ through
the identity have the form $\sigma\left(t\right)=\exp\left(tX\right)$ with
$X\in\mathfrak{k}$, where $\mathfrak{k}$ denotes the Lie algebra of $K$. The
following theorem answers partially the question whether they are also
geodesics in $G$. We denote by $\mathfrak{m}$ the Lie algebra of $M$.

######  Theorem 3.

Let $G$ be endowed with the kinetic energy metric and suppose that
$Z\in\mathfrak{k}$ equals $\operatorname{diag}(X,0_{n})$ or
$\operatorname{diag}(0_{n},X)$, where $X\in\mathfrak{m}$. Then the geodesic
$\gamma(t)=\exp(tZ)$ of $K$ is also a geodesic of $G$.

In [25] and [19] similar problems as in this article have been studied for the
conformal and the projective maps of spheres. We recall the actions

$\ast_{p}:SL_{m+1}\left(\mathbb{R}\right)\times S^{m}\rightarrow S^{m}\text{,\
\ \ \ \ \ \ \ }\ast_{c}:O_{o}\left(1,m+1\right)\times S^{m}\rightarrow
S^{m}\text{,}$

on $S^{m}$ by projective and conformal transformations, respectively, which
are given by

(2) $A\ast_{p}z=Az/\left|Az\right|\text{ \ \ \ \ \ \ and \ \ \ \ \ \
}A\ast_{c}z=z^{\prime}\text{,}$

where $z^{\prime}$ is the unique point in $S^{m}$ such that

(3) $A\left(1,z\right)^{T}\in\mathbb{R}\left(1,z^{\prime}\right)^{T}\text{.}$

It is well known that in low dimensions there are isomorphisms (up to
coverings) between $SL_{n}\left(\mathbb{R}\right)$ or $O_{o}\left(1,n\right)$
and some of the groups $G$ that have been considered in this paper (see for
instance [16], pp 517–528). In Section 7 we explore whether the corresponding
actions on spheres or unitary groups are equivalent (the task turned out to be
more demanding than expected). This will have implications in the dynamics of
Möbius motions of the compact classical Lie groups: Translating the results of
[25] and [19], Corollary 1 provides a refinement of Theorem 3 in some low
dimensional cases.

##  2\. Preliminaries and basic facts

###  2.1. Unitary groups

In this section we recall some basic facts about the classical compact Lie
groups. For more details on the subject we refer to [15] and [31]. Let
$\mathbb{F}=\mathbb{R}$, ${\mathbb{C}}$ or ${\mathbb{H}}$ (the skew field of
quaternions) and let $N\in\mathbb{N}$. Then $\mathbb{F}^{N}$ is an
$\mathbb{F}$-vector space, where in the case of $\mathbb{F}={\mathbb{H}}$, we
choose it to be a right ${\mathbb{H}}$-vector space. As usual, we denote by
$GL(N,\mathbb{F})$ the group of invertible $\mathbb{F}$-linear endomorphisms
of $\mathbb{F}^{N}$.

For $n\in\mathbb{N}$ and $m\in\mathbb{N}\cup\left\\{0\right\\}$ set
$b_{n,m}^{\mathbb{F}}:\mathbb{F}^{n+m}\times\mathbb{F}^{n+m}\rightarrow\mathbb{F}$
the sesquilinear form

(4)
$b_{n,m}^{\mathbb{F}}((x,y),(\xi,\eta))=\overline{x}^{T}\,\xi-\overline{y}^{T}\,\eta\text{,}$

where $x,\,\xi\in\mathbb{F}^{n}$, and $y,\,\eta\in\mathbb{F}^{m}$ are column
vectors, the bar denotes conjugation and the superscript $T$ indicates
transposition, as before.

The form $b_{n,m}^{\mathbb{F}}$ is called the $\mathbb{F}$_-Hermitian
symmetric inner product of signature_ $(n,m)$ on $\mathbb{F}^{n+m}$ and the
space $\mathbb{F}^{n+m}$ endowed with this form is called $\mathbb{F}^{n,m}$.
Let $U(n,m,\mathbb{F})$ be the Lie group

$U(n,m,\mathbb{F})=\\{A\in GL(n+m,\mathbb{F})\mid
b_{n,m}^{\mathbb{F}}(Ax,Ay)=b_{n,m}^{\mathbb{F}}(x,y)\ \ \text{for all
}\,x,y\in\mathbb{F}^{n+m}\\}\text{.}$

We shall denote by $U_{o}(n,m,\mathbb{F})$ the connected component of the
identity of $U(n,m,\mathbb{F})$. For $\mathbb{F}={\mathbb{R}}$ or
${\mathbb{C}}$, $SU(n,m,\mathbb{F})=\left\\{A\in
U(n,m,\mathbb{F})\mid{\mathrm{det}}A=1\right\\}$.

We are interested in the cases where $m=n$ or $m=0$. For brevity, we call
$U\left(n,\mathbb{F}\right)=U(n,0,\mathbb{F})$. Then, for each
$n\in\mathbb{N}$ we have:

\- the orthogonal, unitary and symplectic groups $U(n,{\mathbb{R}})=O_{n}$,
$U(n,\mathbb{C})=U_{n}$ and $U(n,\mathbb{H})=Sp_{n}$ (also called
$HU\left(n\right)$),

\- $U(n,n,{\mathbb{R}})=O(n,n)$, $U(n,n,{\mathbb{C}})=U(n,n)$ and
$U(n,n,{\mathbb{H}})=Sp\left(n,n\right)$ (also called $HU\left(n,n\right)$).

###  2.2. The action of $U(n,n,\mathbb{F})$ on $U(n,\mathbb{F})$

An $\mathbb{F}$-subspace $V$ of $\mathbb{F}^{n,n}$ is said to be totally
isotropic if $\left.b_{n,n}^{\mathbb{F}}\right|_{V\times V}\equiv 0$. A
maximal totally isotropic subspace must have dimension $n$ (see [15]). Denote
by Grass${}_{0}(n,\mathbb{F})$ the Grassmannian of maximal totally isotropic
subspaces of $\mathbb{F}^{n,n}$. Then, it is not difficult to see that the map

(5) $f:U(n,\mathbb{F})\rightarrow\text{Grass}_{0}(n,\mathbb{F}),\ \ \ \ \ \ \
f(T)=\operatorname{rgraph}(T)=\\{(Tx,x))\mid x\in\mathbb{F}^{n}\\}$

is a diffeomorphism, where $\operatorname{rgraph}(T)$ is the graph of $T$
reflected with respect to the diagonal, or equivalently, the graph of
$T^{-1}$, or the graph of $T$ as a map from the second factor into the first.

Now, there is a canonical action of $U(n,n,\mathbb{F})$ on
Grass${}_{0}(n,\mathbb{F})$. The identification $f$ induces an action of
$U(n,n,\mathbb{F})$ on $U\left(n,\mathbb{F}\right)$, as follows: Given $U\in
U(n,\mathbb{F})$, let $V=\operatorname{rgraph}(U)$. So, if $T\in
U(n,n,\mathbb{F})$ has the matrix form

$T=\left(\begin{array}[]{cc}A&B\\\ C&D\end{array}\right)\text{,}$

then $T(V)=\\{(AUx+Bx,CUx+Dx)\mid\,x\in\mathbb{F}^{n}\\}$. On the other hand,
there exists $U^{\prime}\in U(n,\mathbb{F})$ such that
$T(V)=\operatorname{rgraph}(U^{\prime})=\\{(U^{\prime}y,y)\mid\,y\in\mathbb{F}^{n}\\}$
and so we must have $y=(CU+D)x$ and consequently the action reads, as in (1),

(6) $T\ast U=U^{\prime}=(AU+B)(CU+D)^{-1}\text{.}$

By Witt’s Theorem, the canonical action of $U(n,n,\mathbb{F})$ on
Grass${}_{0}(n,\mathbb{F})$ is transitive, and then so is the action
restricted to $U_{o}(n,n,\mathbb{F})$. Through $f$ we obtain a transitive
action of $U_{o}(n,n,\mathbb{F})$ on $U(n,\mathbb{F})$.

At this point we should make some remarks:

\- As it is well known, $O(n,n)$ is not connected (see [23]). However, since
it acts transitively on $O_{n}$, then $O_{o}(n,n)$ acts transitively on
$(O_{n})_{o}=SO_{n}$.

\- The action of $U(n,n,{\mathbb{C}})=U(n,n)$ on $U_{n}$ is not effective. It
is not difficult to see that actually, for each $u\in
S^{1}\subset{\mathbb{C}}$ and $T\in U\left(n,n\right)$, we have $(uT)\ast
U=T\ast U$. In order to obtain an effective action, we can take the group
$PU(n,n)=U(n,n)/S^{1}$. For convenience, we will consider the action of the
matrix group $SU(n,n)$ on $U_{n}$, which is almost effective since for $T$ in
this group, $T\ast U=U$ for all $U\in U_{n}$ if and only if $\ T=uI_{2n}$ with
$u^{2n}=1$.

We comment that the group $G=U\left(n,n,\mathbb{F}\right)$ acts canonically on
the Grassmannian of maximal space-like subspaces of $\mathbb{F}^{n,n}$, from
which $M=U\left(n,\mathbb{F}\right)$, via the identification (5), is a border
inheriting the action.

###  2.3. The kinetic energy metric

In this section we review the restriction to a finite dimensional subgroup of
the usual $L^{2}$ weak Riemannian metric on the space of all diffeomorphisms
of a compact Riemannian manifold (see for instance Theorem 9.1 (i) in [11]),
setting in a broader context some of the concepts presented in [25] and [19].

Let $G$ be a connected Lie group acting smoothly and almost effectively on a
compact oriented Riemannian manifold $M$. Suppose that $M$ has initially a
homogeneous distribution of mass of constant density $1$ and that the points
are allowed to move only in such a way that two configurations differ in an
element of $G$. Then the configuration space may be naturally identified with
$G$ (aside from the fact that a finite number of elements act by the identity)
and a curve $\gamma$ in $G$ may be thought of as a motion of $M$.

The total kinetic energy $E(\gamma)$ of the motion $\gamma$ at the instant $t$
is given by

(7) $E(t)=\frac{1}{2}\int_{M}\rho_{t}(q)|v_{t}(q)|^{2}d\mu,$

where $d\mu$ is the volume form on $M$ and if $q=\gamma(t)(p)$ for some $p\in
M$, then

$v_{t}(q):=\left.\frac{d}{ds}\right|_{t}\gamma(s)(p)\in T_{q}M,\ \ \ \ \ \ \
\rho_{t}(q)=1/{\mathrm{det}}(d\gamma(t)_{p})$

are the velocity of the point $q$ and the density at $q$ at the instant $t$,
respectively. This is the smooth analogue of
$\sum_{i}\frac{1}{2}m_{i}v_{i}^{2}$, once one recognizes $\rho d\mu$ as the
mass element.

Changing variables in (7), one obtains

(8)
$E(t)=\frac{1}{2}\int_{M}\left|\left.\frac{d}{ds}\right|_{t}\gamma(s)(p)\right|^{2}d\mu.$

Based on the principle of least action, we say that a smooth curve $\gamma$ in
$G$, thought of as a motion of $M$, is force free if it is a critical point of
the kinetic energy functional.

In order to introduce a Riemannian metric on $G$ whose geodesics correspond to
force free motions of $G$, for $g\in G$ and $X\in T_{g}G$, we define the map
$\widetilde{X}:M\rightarrow TM$ by

$\widetilde{X}(q)=\left.\frac{d}{dt}\right|_{0}\gamma(t)(q)\in T_{g(q)}M,$

where $\gamma$ is any smooth curve in $G$ with $\gamma(0)=g$ and
$\gamma^{\prime}(0)=X$. Standard arguments show that the map $\widetilde{X}$
is well-defined and smooth, and that $\widetilde{X}$ is a vector field on $M$
if and only if $X\in T_{e}G$. Following [25, Prop. 1], one has that

(9) $\langle
X,Y\rangle=\int_{M}\langle\widetilde{X}(q)\,,\,\widetilde{Y}(q)\rangle
d\mu\text{,}$

for $X,Y\in T_{g}G$ and $g\in G$, defines a Riemannian metric on $G$ such that
a curve $\gamma$ in $G$ is a geodesic if and only if it is force free (on the
basis of the principle of least action). This metric is called the kinetic
energy metric of $G$ induced by the action on $M$.

##  3\. Symmetries of the kinetic energy metric

From now on, we will consider $G=O_{o}(n,n),\,SU(n,n)$ or
$Sp\left(n,n\right)$, their maximal compact subgroups $K=SO_{n}\times SO_{n}$,
$S(U_{n}\times U_{n})$ or $Sp_{n}\times Sp_{n}$, and $M=SO_{n},\,U_{n}$ or
$Sp_{n}$, respectively.

Proof of Proposition 2.  a) For $k_{1},k_{2}\in K$, set

(10) $F_{k_{1},k_{2}}:G\rightarrow G,\ \ \ \ \ g\mapsto
F_{k_{1},k_{2}}(g)=k_{1}gk_{2}{}^{-1}\text{.}$

We have to verify that $F_{k_{1},k_{2}}$ is an isometry of the kinetic energy
metric on $G$.

Let $X\in T_{g}G$ and let $\gamma$ be a curve in $G$ such that $\gamma(0)=g$,
$\gamma^{\prime}(0)=X$. Then
$dF_{k_{1},k_{2}}(X)=\left.\frac{d}{dt}\right|_{0}k_{1}\gamma(t)k_{2}^{-1}$
and so for each $q\in M$ we have that

$\widetilde{dF_{k_{1},k_{2}}(X)}(q)=\left.\frac{d}{dt}\right|_{0}(k_{1}\gamma(t))\ast(k_{2}^{-1}\ast
q)\text{.}$

Observe that since $k_{i}\in K$, then $\displaystyle
k_{i}=\operatorname{diag}\left(A_{i},B_{i}\right)$ with $A_{i},B_{i}\in M$,
and so $k_{i}\ast q=A_{i}qB_{i}^{-1}=R_{B_{i}^{-1}}L_{A_{i}}(q)$. Define

$f:M\rightarrow\mathbb{R}\text{,\ \ \ \ \ \ \
}f(q)=\left|\left.\frac{d}{dt}\right|_{0}(k_{1}\gamma(t))\ast
q\right|^{2}\text{,}$

and $h:M\rightarrow M$ by $h=R_{B_{2}}L_{A_{2}^{-1}}$. Then,

(11) $\left\|dF_{k_{1},k_{2}}X\right\|^{2}=\int_{M}\left(f\circ
h\right)\,d\mu=\int_{M}f(h^{\ast}d\mu)=\int_{M}fd\mu,$

since $d\mu$ is bi-invariant by the action of $M$.

On the other hand,

$f(q)=\left|dR_{B_{1}^{-1}}dL_{A_{1}}\left(\left.\frac{d}{dt}\right|_{0}\gamma(t)\ast
q\right)\right|^{2}=\left|\left.\frac{d}{dt}\right|_{0}\gamma(t)\ast
q\right|^{2}\text{,}$

and so

$\left\|dF_{k_{1},k_{2}}X\right\|^{2}=\int_{M}\left|\left.\frac{d}{dt}\right|_{0}\gamma(t)\ast
q\right|^{2}d\mu(q)=\left\|X\right\|^{2}\text{.}$

Then $F_{k_{1},k_{2}}$ is an isometry.

b) We start with the least simple case $\mathbb{F}=\mathbb{C}$, when $K$ is
strictly contained in $M\times M$, that is, $K=S\left(U_{n}\times
U_{n}\right)$ and $M=U_{n}$. The cases $M=SO_{n}$ (with $n>2$) and $Sp_{n}$,
which are semisimple Lie groups, are dealt with in the same way, but in a more
direct manner. We have to show that the kinetic energy metric restricted to
$K=S\left(U_{n}\times U_{n}\right)$ is the restriction of a product metric on
$M\times M=U_{n}\times U_{n}$ such that $\rho$ is an isometry.

By the first part of the proof, the kinetic energy metric restricted to $K$ is
bi-invariant, in particular, the inner product on
$\mathfrak{k}=T_{\left(I_{n},I_{n}\right)}K$ is Ad$\left(K\right)$-invariant.
We write

$\mathfrak{k}=s\left(u_{n}\oplus u_{n}\right)=su_{n}\oplus
su_{n}\oplus\mathbb{R}iI_{2n}^{-}\text{,}$

where $I_{2n}^{-}=\operatorname{diag}\left(I_{n},-I_{n}\right)$. Since
$su_{n}$ is simple and the last direct summand has dimension 1, the summands
are orthogonal.

Now we check that $d\rho_{\left(I_{n},I_{n}\right)}$ restricted to
$su_{n}\oplus su_{n}$ is a linear isometry, or equivalently, that
$\left\|\left(X,0\right)\right\|=\left\|\left(0,X\right)\right\|$ for all
$\left(X,0\right)\in\mathfrak{k}$ (using again that $\mathfrak{k}$ carries the
restriction of a product metric). We compute

$\left|\left.\frac{d}{dt}\right|_{0}e^{t\left(X,0\right)}\ast
q\right|^{2}=\left|\left.\frac{d}{dt}\right|_{0}e^{tX}q\right|^{2}=\left|Xq\right|^{2}=\left|X\right|^{2}\text{,}$

for all $q\in M$, since the inner product on $\mathbb{F}^{m\times m}$ is bi-
invariant by $M.$ Integrating over $M$ yields
$\left\|\left(X,0\right)\right\|^{2}=\operatorname{vol}\left(M\right)\left|X\right|^{2}$.
One obtains similarly that
$\left\|\left(0,X\right)\right\|^{2}=\operatorname{vol}\left(M\right)\left|X\right|^{2}$.

Finally, we verify that the metric on $K$ is the restriction to $K$ of a
product metric on $M\times M$. We observe that $u_{n}\oplus
u_{n}=\mathfrak{k}\oplus\mathbb{R}iI_{2n}$. Let
$\epsilon_{1}=\operatorname{diag}\left(iI_{n},0_{n}\right)$ and
$\epsilon_{2}=\operatorname{diag}\left(0_{n},iI_{n}\right)$, which form a
basis of $\mathbb{R}iI_{2n}^{-}\oplus\mathbb{R}iI_{2n}$. We have that

$u_{n}\oplus
u_{n}=\left(su_{n}\oplus\mathbb{R}\epsilon_{1}\right)\oplus\left(su_{n}\oplus\mathbb{R}\epsilon_{2}\right)\equiv\left(su_{n}\oplus
su_{n}\right)\oplus\mathbb{R}\epsilon_{1}\oplus\mathbb{R}\epsilon_{2}\text{.}$

Suppose that $\left\|iI_{2n}^{-}\right\|=\ell$. We consider an inner product
such that the three summands in the last expression are orthogonal,
$su_{n}\oplus su_{n}$ has the metric induced from the kinetic energy and
$\left\|\epsilon_{1}\right\|=\left\|\epsilon_{2}\right\|=\ell/\sqrt{2}$. Since
$iI_{2n}^{-}=\epsilon_{1}-\epsilon_{2}$, this inner product on $u_{n}\oplus
u_{n}$ satisfies the required conditions.∎

##  4\. Some totally geodesic subgroups of $G$

Before proving Theorem 2 we comment on the canonical inclusions. First, notice
that

(12) $U\left(n,n\right)=\\{A\in Sp(n,n)\mid iA=Ai\\}\text{, \
}O\left(n,n\right)=\\{A\in Sp(n,n)\mid iA=Ai\text{ y }jA=Aj\\}\text{,}$

since for each entry $q=a+bi+cj+dk$ of a matrix in $Sp\left(n,n\right)$,
$iq=qi$ holds if and only if $c=d=0$, and also $jq=qj$ if and only if $b=d=0$.

Now we realize the inclusions of $U(n,n)$ and $Sp\left(n,n\right)$ as
subgroups of $O(2n,2n)$ and $O(4n,4n)$, respectively. We follow Chapter 2 of
[15]. We recall first that under a suitable identification of
$\mathbb{C}^{2n}$ with $\mathbb{R}^{4n}$, it is possible to describe

(13) $U(n,n)=\\{A\in O_{o}(2n,2n)\mid JA=AJ\\},$

where $J:\mathbb{R}^{4n}\rightarrow\mathbb{R}^{4n}$ is the matrix with $2n$
diagonal blocks of the form $J_{0}=\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right)$.

Consider now the right vector space ${\mathbb{H}}^{2n}$ over $\mathbb{H}$.
Now, any $q\in{\mathbb{H}}$ has the form $q=a+bi+cj+dk$, with
$a,b,c,d\in{\mathbb{R}}$ and so ${\mathbb{H}}^{2n}$ is naturally isomorphic to
${\mathbb{R}}^{8n}$. Via this identification,

(14) $Sp\left(n,n\right)=\\{A\in O_{o}(4n,4n)\mid R_{i}A=AR_{i}\text{ and
}R_{j}A=AR_{j}\\}\text{,}$

where $R_{p}$ denotes right multiplication by $p\in{\mathbb{H}}$. The
($8n\times 8n$)-matrices of $R_{i}$ and $R_{j}$ with respect to the canonical
basis consist, respectively, of $2n$ diagonal blocks of the forms

$\left(\begin{array}[]{cc}J_{0}&0\\\ 0&-J_{0}\end{array}\right)\text{\ \ \ \ \
\ and\ \ \ \ \ \ }\left(\begin{array}[]{cc}0&-I_{2}\\\
I_{2}&0\end{array}\right)\text{.}$

The following lemma will be useful to prove that $O_{o}\left(n,n\right)$ is
totally geodesic in $SU\left(n,n\right)$. Observe that the arguments in the
proof are not valid for $Sp(n,n)$ (see [33]).

######  Lemma.

If $G=SU\left(n,n\right)$ is endowed with the kinetic energy metric, then
conjugation $C:G\rightarrow G$, $C\left(q\right)=\overline{q},$ is an
isometry.

######  Proof.

Let $X\in T_{p}G$ and let $\gamma$ be a curve such that $\gamma(0)=p$ and
$\gamma^{\prime}(0)=X$. We have

$\displaystyle\|dC_{p}(X)\|^{2}$ $\displaystyle=$
$\displaystyle\int_{U_{n}}\left|\left.\frac{d}{dt}\right|_{0}\left(\overline{\gamma(t)}\ast
q\right)\right|^{2}d\mu=\int_{U_{n}}\left|\overline{\left.\frac{d}{dt}\right|_{0}(\gamma(t)\ast\overline{q})}\right|^{2}d\mu$
$\displaystyle=$
$\displaystyle\int_{U_{n}}\left|\left.\frac{d}{dt}\right|_{0}(\gamma(t)\ast\overline{q})\right|^{2}d\mu=\|X\|^{2}\text{.}$

The last equality holds by the change of variables theorem, since conjugation
in $U_{n}$ is an isometry of $U_{n}$. ∎

Proof of Theorem 2. We use repeatedly that each connected component of the set
of fixed points of an isometry is a totally geodesic submanifold (see for
instance Proposition 10.3.6 in [5]). This and the previous lemma yield that
the the first inclusion is totally geodesic, since $O_{o}(n,n)$ is a connected
component of the set of fixed points of the conjugation.

For $k\in K$, let $F_{k}:G\rightarrow G$ be the conjugation by $k$, that is,
$F_{k}\left(g\right)=kgk^{-1}$. Notice that $F_{k}=F_{k,k}$ with the notation
of (10). By Proposition 2 (a), it suffices to show in each case the existence
of $k\in K$ such that the corresponding subgroup is the identity component of
the set of fixed points of the isometry $F_{k}$. Actually, in the cases (c)
and (e) we need to consider the intersection of such sets for different
elements of $K$. It is not difficult to see that the following assertion is
true: Let $N_{1}$ and $N_{2}$ be two totally geodesic submanifolds. If a
connected component of $N_{1}\cap N_{2}$ is a submanifold, then it is totally
geodesic.

Now, $U\left(n,n\right)$ is a totally geodesic subgroup of
$Sp\left(n,n\right)$ since it is connected and by (12) is the set of fixed
points of $F_{iI_{2n}}$, which is an isometry of $Sp\left(n,n\right)$, as
$iI_{n}\in Sp_{n}$. Analogously, $O_{o}\left(n,n\right)$ is a totally geodesic
subgroup of $Sp\left(n,n\right)$, given that (12) is the identity component of
the intersection of the sets of fixed points of $F_{iI_{2n}}$ and
$F_{jI_{2n}}$, which are isometries of $Sp\left(n,n\right)$.

By (13), the group $U\left(n,n\right)$ is the set of fixed points of the map
$F_{J}:O_{o}(2n,2n)\rightarrow O_{o}(2n,2n)$ given by
$F_{J}(A)=JAJ^{-1}=-JAJ$. Now, by the description of $J$ above, $J\in
SO_{2n}\times SO_{2n}$. Then, $F_{J}$ is an isometry of $O_{o}(2n,2n)$ endowed
with the kinetic energy metric. Since $U(n,n)$ is connected, it is a totally
geodesic submanifold of $O_{o}(2n,2n)$.

By (14), the group $Sp\left(n,n\right)$ is the intersection of the sets of
fixed points of the maps $F_{R_{i}},F_{R_{j}}:O_{o}(4n,4n)\rightarrow
O_{o}(4n,4n)$ given by $F_{R_{i}}\left(A\right)=R_{i}AR_{i}^{-1}$, and
similarly for $F_{R_{j}}$. From the form of the matrices of $R_{i}$ and
$R_{j}$ (see above) one deduces that they belong to $SO_{4n}\times SO_{4n}$.
Thus, $F_{R_{i}}$ and $F_{R_{j}}$ are isometries of $O_{o}(4n,4n)$ endowed
with the kinetic energy metric. Thus, $Sp\left(n,n\right)$, which is
connected, is totally geodesic in $O_{o}(4n,4n)$. ∎

##  5\. The incompleteness of the kinetic energy metric

In this section we prove Theorem 1. Let
$\delta=\operatorname{diag}(I_{n},-I_{n})$. Then $b_{n,n}^{\mathbb{F}}$ as in
(4) is given by $b_{n,n}^{\mathbb{F}}(x,y)=\overline{x}^{T}\delta y$, for
$x,y\in\mathbb{F}^{2n}$. Thus, $U\in U(n,n,\mathbb{F})$ if and only if
$\overline{U}^{T}\delta U=\delta$. From this expression one gets that
$u\in\mathfrak{u}(n,n,\mathbb{F})$, the Lie algebra of $U(n,n,\mathbb{F})$, if
and only if $\overline{u}^{T}\delta+\delta u=0$. Hence,

(15) $\mathfrak{g}=\left\\{\left(\begin{array}[]{cc}a&c\\\
\overline{c}^{T}&b\end{array}\right)\mid c\in\mathbb{F}^{n\times n}\text{,
}\overline{a}^{T}=-a\text{, }\overline{b}^{T}=-b\right\\}$

is the Lie algebra of $G$ for $\mathbb{F}=\mathbb{R}$ and $\mathbb{H}$,
whereas the elements of $su\left(n,n\right)$ have the same form, with the
extra condition $\operatorname{tr}(a+b)=0$.

######  Lemma.

For $i=2,\dots,n$, let $T_{i}$ be the $\mathbb{F}$-linear transformation on
$\mathbb{F}^{n}$ such that $T_{i}\left(e_{1}\right)=e_{i}$,
$T_{i}\left(e_{i}\right)=-e_{1}$ and $T_{i}\left(e_{s}\right)=e_{s}$ for
$1<s\leq n$, $s\neq i$. Then an $\mathbb{F}$-linear transformation $T$ on
$\mathbb{F}^{n}$ commutes with $T_{i}$ for all $i=2,\dots,n$ if and only if
$T=\lambda I_{n}$, for some $\lambda\in\mathbb{F}$.

######  Proof.

The matrix of $T_{i}$ with respect to the canonical basis is $A^{i}$, where

$A_{1,i}^{i}=1\text{,\ \ \ }A_{i,1}^{i}=-1\text{, \ }A_{j,j}^{i}=1\text{ \ \
for }1<j\neq i$

and $A_{s,t}^{i}=0$ otherwise. First observe that for any matrix $X$, $A^{i}X$
is the matrix whose rows coincide with those of $X$, except that its first row
is the $i$-th row of $X$ and its $i$-th row is minus the first row of $X$. On
the other hand, $XA^{i}$ is obtained by a similar procedure, but interchanging
columns of $X$ instead of rows. So we obtain that

(16)
$X_{ii}=\left(A^{i}X\right)_{1,i}=\left(XA^{i}\right)_{1,i}=X_{11}\text{.}$

for all $1<i\leq n$. Also, for $t\neq i$,

(17)
$X_{ti}=\left(A^{i}X\right)_{ti}=\left(XA^{i}\right)_{ti}=X_{t1}=\left(A^{i}X\right)_{t1}=\left(XA^{i}\right)_{t1}=-X_{ti}$

Combining conditions (16) and (17), we get $X_{ii}=X_{11}$ for every $i$ and
$X_{ti}=0$ for $t\neq i$. Therefore $X=\lambda I_{n}$, for some
$\lambda\in\mathbb{F}$. ∎

_Proof of Theorem 1_. The proof will consist of two parts.

Part 1. We prove that the image of $\gamma$ is a totally geodesic submanifold
of $G$ and so $\gamma$ is the reparametrization of a geodesic in $G$, which
will be inextendible since $\lim_{t\rightarrow\pm\infty}\gamma\left(t\right)$
does not exist.

For $i=2,\dots,n$, let $k_{i}=\operatorname{diag}(A^{i},A^{i})\in K$, where
$A^{i}$ is the matrix introduced in Lemma Lemma (notice that $A^{i}\in
SO_{n}$), and let $F_{i}:G\rightarrow G,$
$F_{i}\left(g\right)=k_{i}gk_{i}^{-1}$. Let $L$ be the identity component of
the Lie group

$\left(\cap_{i=2}^{n}\,\operatorname{Fix}_{0}\left(F_{i}\right)\right)\cap
O_{o}\left(n,n\right)\text{.}$

We will prove that $L$ coincides with the image of $\gamma$. Thus, this will
be totally geodesic in $G$, since Fix${}_{0}\left(F_{i}\right)$ and
$O_{o}\left(n,n\right)$ are totally geodesic in $G$ by Proposition 2 (a) and
Theorem 2 (a) and (c), respectively (as we mentioned in the proof of the
latter, if a connected component of the intersection of a finite number of
totally geodesic submanifolds is a submanifold, then it is totally geodesic).

The Lie algebra of $L$ is $\mathfrak{f}\cap o\left(n,n\right)$, where

$\mathfrak{f}=\cap_{i=2}^{n}\,\operatorname{Fix}\left(\operatorname{Ad}\left(k_{i}\right)\right)\text{,}$

since the Lie algebra of the identity component of the intersection of a
finite number of Lie subgroups coincides with the intersection of its Lie
algebras.

For each $i=2,\dots,n$ we apply $\operatorname{Ad}(k_{i})$ to a generic
element of $\mathfrak{g}$ as in the beginning of the section. One easily
verifies that

$\operatorname{Ad}(k_{i})\left(\begin{array}[]{cc}a&\overline{c}^{T}\\\
c&b\end{array}\right)=\left(\begin{array}[]{cc}A_{i}aA_{i}^{-1}&A_{i}\overline{c}^{T}A_{i}^{-1}\\\
A_{i}cA_{i}^{-1}&A_{i}bA_{i}^{-1}\end{array}\right)\text{.}$

Since
$\operatorname{Fix}\left(\operatorname{Ad}\left(k_{i}\right)\right)=\left\\{X\in\mathfrak{g}\mid
X\text{ conmuta con }k_{i}\right\\}$, by the conditions on $a$ and $b$ in
(15), using Lemma Lemma we have that

$\mathfrak{f}=\left\\{\left(\begin{array}[]{cc}\alpha
I_{n}&\overline{\lambda}I_{n}\\\ \lambda I_{n}&\beta
I_{n}\end{array}\right)\mid\alpha,\lambda\in\mathbb{F}\text{,
}\overline{\alpha}=-\alpha\text{, }\overline{\beta}=-\beta\right\\}$

(requiring additionally that $\alpha+\beta=0$ in the complex case). Now, since
$o\left(n,n\right)$ consists of matrices whose entries are real numbers, we
have that

$\mathfrak{f}\cap
o\left(n,n\right)=\mathbb{R}\left(\begin{array}[]{cc}0&I_{n}\\\
I_{n}&0\end{array}\right)\text{.}$

Therefore, the Lie algebras of $L$ and $\gamma$ coincide. Consequently,
$L=\gamma\left(\mathbb{R}\right)$, given that both are connected.

Part 2. We prove that $\gamma$ has finite length. It will be convenient to
consider the reparametrization $\sigma:\left(-1,1\right)\rightarrow G$ of
$\gamma$ given by

$\sigma(s)=\gamma(\tanh^{-1}(s))=\frac{1}{\sqrt{1-s^{2}}}\left(\begin{array}[]{cc}I_{n}&sI_{n}\\\
sI_{n}&I_{n}\end{array}\right)\text{.}$

Then, by (9),

(18)
$\left\|\sigma^{\prime}\left(t\right)\right\|^{2}=\int_{M}\left|\widetilde{\sigma^{\prime}(t)}(A)\right|^{2}d\mu\left(A\right)=\int_{M}\left|\left.\frac{d}{ds}\right|_{t}\sigma(s)\ast
A\right|^{2}d\mu\left(A\right).$

Let
$T=\left\\{D(\theta_{1},\cdots,\theta_{m})\mid\theta_{j}\in\mathbb{R}\right\\}$
be a maximal torus of $M$, where $D(\theta_{1},\cdots,\theta_{m})$ equals

\- $\operatorname{diag}(R_{\theta_{1}},\cdots,R_{\theta_{m}})$ with
$R_{\theta}=\left(\begin{array}[]{cc}\cos\theta&-\sin\theta\\\
\sin\theta&\cos\theta\end{array}\right)$, if $M=SO_{2m}$.

\- $\operatorname{diag}(R_{\theta_{1}},\cdots,R_{\theta_{m}},1)$, if
$M=SO_{2m+1}$.

\- $\operatorname{diag}(e^{i\theta_{1}},\cdots,e^{i\theta_{m}})$, if $M=U_{m}$
or $M=Sp_{m}$.

We recall the general form of the Weyl integration formula on $M$. Provided
that $f$ is a class function, that is,
$f\left(BAB^{-1}\right)=f\left(A\right)$ for all $A,B\in M$, we have

$\int_{M}f\left(A\right)d\mu\left(A\right)=\int_{T}f\left(D(\theta_{1},\cdots,\theta_{m})\right)F\left(\theta_{1},\cdots,\theta_{m}\right)d\theta_{1}\cdots
d\theta_{m}\text{,}$

where $F$ is some bounded function. The precise formula for $U_{n}$ can be
found in Theorem 3.1 in [20], and for the remaining groups, in Theorems
IX.9.3, IX.9.4, IX.9.5 in [29].

It is not difficult to see that for every $A,B\in M$,
$\sigma(s)\ast(BAB^{-1})=B(\sigma(s)\ast A)B^{-1}$ and this implies that the
integrand in the right hand side of (18) is a class function of $A$. Then,

(19) $\int_{M}\left|\left.\frac{d}{ds}\right|_{t}\sigma(s)\ast
A\right|^{2}d\mu\left(A\right)\leq
C\int_{T}\left|\left.\frac{d}{ds}\right|_{t}\sigma(s)\ast
D(\theta_{1},\cdots,\theta_{m})\right|^{2}d\theta_{1}\cdots d\theta_{m}$

for some constant $C>0$.

Consider first the case $M=SO_{n}$ with $n=2m$. Since

$\sigma(s)\ast
D(\theta_{1},\cdots,\theta_{m})=\operatorname{diag}(\sigma_{2}(s)\ast
R_{\theta_{1}},\cdots,\sigma_{2}(s)\ast R_{\theta_{m}}),$

where
$\sigma_{2}(s)=\frac{1}{\sqrt{1-s^{2}}}\left(\begin{array}[]{cc}I_{2}&sI_{2}\\\
sI_{2}&I_{2}\end{array}\right)\in O_{o}\left(2,2\right)$, then

$\left|\left.\frac{d}{ds}\right|_{t}\sigma(s)\ast
D(\theta_{1},\cdots,\theta_{m})\right|^{2}=\sum_{j=1}^{m}\left|\left.\frac{d}{ds}\right|_{t}\sigma_{2}(s)\ast
R_{\theta_{j}}\right|^{2}\text{.}$

We compute

(24) $\displaystyle\sigma_{2}\left(s\right)\ast R_{\theta}$ $\displaystyle=$
$\displaystyle\left(R_{\theta}+sI_{2}\right)\left(sR_{\theta}+I_{2}\right)^{-1}$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}s+\cos\theta&-\sin\theta\\\
\sin\theta&s+\cos\theta\end{array}\right)\left(\begin{array}[]{cc}1+s\cos\theta&-s\sin\theta\\\
s\sin\theta&1+s\cos\theta\end{array}\right)^{-1}$ (27) $\displaystyle=$
$\displaystyle\frac{1}{1+2s\cos\theta+s^{2}}\left(\begin{array}[]{cc}2s+\left(1+s^{2}\right)\cos\theta&(s^{2}-1)\sin\theta_{j}\\\
(1-s^{2})\sin\theta&2s+\left(1+s^{2}\right)\cos\theta\end{array}\right)\text{.}$

Now we identify $SO_{2}$ with $S^{1}$ in the usual way, that is, $R_{\alpha}$
is identified with $e^{i\alpha}$. Then (27) reads

$\sigma_{2}\left(s\right)\ast
e^{i\theta}=\frac{e^{i\theta}+s}{1+se^{i\theta}}\text{,}$

since the real and imaginary parts of this complex number are equal to the
first and second entries of the first column of the matrix (27), respectively.
Replacing in (19) we obtain

$\left\|\sigma^{\prime}\left(t\right)\right\|^{2}=\int_{M}\left|\left.\frac{d}{ds}\right|_{t}\sigma(s)\ast
A\right|^{2}d\mu\left(A\right)\leq
C\sum_{j=1}^{m}\int_{S^{1}}\left|\left.\frac{d}{ds}\right|_{t}\frac{e^{i\theta_{j}}+s}{1+se^{i\theta_{j}}}\right|^{2}d\theta_{j}\text{.}$

Setting $z=e^{i\theta_{j}}$, the integrals in the right hand side can be
written as integrals over the circle of a meromorphic function on the two-
punctured disc. They can be computed by residues and shown to be equal to
$4\pi/\left(1-t^{2}\right)$. This is done in [12], where the force free Möbius
motions of the circle are studied (the $j$-th integrand equals
$\left|\widetilde{Y}\left(e^{i\theta_{j}}\right)\right|^{2}$, with
$\widetilde{Y}$ as in the proof of Theorem 3 of that article). Therefore,

$\text{length}\left(\left.\sigma\right|_{\left[0,1\right)}\right)=\lim_{T\rightarrow
1^{-}}\int_{0}^{T}\left\|\sigma^{\prime}\left(t\right)\right\|~{}dt\leq
C^{\prime}\int_{0}^{1}\frac{2}{\sqrt{1-t^{2}}}\,dt=C^{\prime}\pi$

(with $C^{\prime}=\sqrt{Cm\pi}$) and so the length of
$\left.\sigma\right|_{\left[0,1\right)}$ is finite, as desired.

The case $M=SO_{2m+1}$ is analogous. If $M=U_{n}$ or $Sp_{n}$, we get that

$\displaystyle\sigma(s)\ast D(\theta_{1},\cdots,\theta_{n})$ $\displaystyle=$
$\displaystyle(D(\theta_{1},\cdots,\theta_{n})+sI_{n})(sD(\theta_{1},\cdots,\theta_{n})+I_{n})^{-1}$
$\displaystyle=$
$\displaystyle\operatorname{diag}\left(\frac{e^{i\theta_{1}}+s}{1+se^{i\theta_{1}}},\cdots,\frac{e^{i\theta_{n}}+s}{1+se^{i\theta_{n}}}\right)$

and the proof follows from similar arguments. ∎

_Proof of Proposition 1_. For $q\in M$ we compute

$\gamma\left(t\right)\ast q=\left(\cosh t\ q+\sinh t\ I_{n}\right)\left(\sinh
t\ q+\cosh t\ I_{n}\right)^{-1}=\left(q+\tanh t\ I_{n}\right)\left(\tanh t\
q+I_{n}\right)^{-1}\text{. }$

For $\varepsilon=\pm 1$, if $-\varepsilon$ is not an eigenvalue of $q$, we
have that

$\lim_{t\rightarrow\varepsilon\infty}\gamma\left(t\right)\ast
q=\left(q+\varepsilon I_{n}\right)\left(\varepsilon
q+I_{n}\right)^{-1}=\varepsilon I_{n}\text{.}$

Suppose that $M=SO_{n}$ with $n$ even. Given $q\in M$ with eigenvalue
$-\varepsilon,$ we verify that any neighborhood of $q$ contains elements with
no eigenvalue $-\varepsilon.$ Let $R_{t}$ denote the rotation in
$\mathbb{R}^{2}$ through the angle $t$. We have

$q=O\operatorname{diag}\left(R_{\alpha_{1}},\dots,R_{\alpha_{k}},-\varepsilon
I_{\ell}\right)O^{-1}$

for some $O\in SO_{n}$, where $\ell$ is even and $R_{\alpha_{i}}$ has no
eigenvalue $-\varepsilon$. Now, the curve

$t\mapsto
q_{t}:=O\operatorname{diag}\left(R_{\alpha_{1}},\dots,R_{\alpha_{k}},-\varepsilon
R_{t},\dots,-\varepsilon R_{t}\right)O^{-1}$

has initial point $q$ and $q_{t}$ has no eigenvalue $-\varepsilon$ for $t>0$
near $0$. Similar arguments hold for $SO_{n}$ with $n$ odd if $\varepsilon=1$.

The cases $M=U_{n}$ or $Sp_{n}$ can be dealt with analogously, with
$e^{i\alpha_{j}}$ and and $e^{it}$ instead of $R_{\alpha_{j}}$ and $R_{t}$,
respectively. ∎

##  6\. Force free rigid motions

We recall that the maximal compact subgroup $K$ of $G$ is a product
$SO_{n}\times SO_{n}$ or $Sp_{n}\times Sp_{n}$ or is contained as the
hypersurface $S\left(U_{n}\times U_{n}\right)$ in the product $U_{n}\times
U_{n}$. Theorem 3 asserts that geodesics in $K$ contained in one of the
factors are also geodesics in the ambient Lie group $G$ endowed with the
kinetic energy metric. Next we prove the theorem.

Proof of Theorem 3. _Let $\widehat{Z}$ be the right invariant vector field of
$G$ such that $\widehat{Z}_{I_{2n}}=Z$. Then its flow is
$t\mapsto\varphi_{t}=L_{\exp(tZ)}$. Now, for all $t$, $\varphi_{t}$ is an
isometry of $G$ by Proposition 2 (a). Therefore, $\widehat{Z}$ is a Killing
vector field of $G$ and it is standard to check that $\gamma$ is an integral
curve of $\widehat{Z}$ through the identity. So, in order to see that $\gamma$
is a geodesic of $G$, we need to show that
$(\nabla_{\widehat{Z}}\widehat{Z})_{\gamma(t)}=0$, where $\nabla$ is the Levi-
Civita connection of $G$ endowed with the kinetic energy metric. Since $K$
acts on $G$ by isometries and
$dR_{\exp\left(tZ\right)}(\widehat{Z})=\widehat{Z}\circ
R_{\exp\left(tZ\right)}$, it suffices to see that
$(\nabla_{\widehat{Z}}\widehat{Z})_{I_{2n}}=0$, or equivalently, since
$\widehat{Z}$ is Killing, that_

(28)
$-2\langle\nabla_{Z}\widehat{Z},Y\rangle=2\langle\nabla_{Y}\widehat{Z},Z\rangle=Y(\|\widehat{Z}\|^{2})=\left.\frac{d}{dt}\right|_{0}\|\widehat{Z}(\exp(tY))\|^{2}$

_vanishes for every $Y\in\mathfrak{g}$ (the second equality holds by the
Koszul formula). We have that_

$\widehat{Z}(\exp(tY))=\left.\frac{d}{ds}\right|_{0}e^{sZ}e^{tY}$

_and so, calling $q_{t}=e^{tY}\ast q\in M$, by (9),_

(29)
$\|\widehat{Z}(\exp(tY))\|^{2}=\int_{M}\left|\left.\frac{d}{ds}\right|_{0}\left(e^{sZ}e^{tY}\right)\ast
q\right|^{2}d\mu\left(q\right)=\int_{M}\left|\left.\frac{d}{ds}\right|_{0}e^{sZ}\ast
q_{t}\right|^{2}d\mu\left(q\right)\text{.}$

_Now suppose that $Z=\operatorname{diag}\left(X,Y\right)$ with
$X,Y\in\mathfrak{m}$ and $X=0$ or $Y=0$. Hence, by (1),_

$e^{sZ}\ast q_{t}=\operatorname{diag}\left(e^{sX},e^{sY}\right)\ast
q_{t}=e^{sX}q_{t}e^{-sY}$

_and so the integrand in ( 29) is $\left|Xq_{t}-q_{t}Y\right|^{2}$, which
equals $\left|Xq_{t}\right|^{2}=\left|X\right|^{2}$ or
$\left|q_{t}Y\right|^{2}=\left|Y\right|^{2}$ if $Y=0$ or $X=0$, respectively
(notice that the inner product on $\mathbb{F}^{n\times n}$ is bi-invariant).
In both cases we have independence on $t$. Therefore, the expression (28)
vanishes, as desired._ ∎

##  7\. Equivalences with conformal and projective motions of the sphere in
low dimensions

As we mentioned in the introduction, in this section we explore, on the one
hand, whether the projective and conformal actions on spheres (2) are
equivalent to the Möbius actions on unitary groups in low dimensions, and on
the other hand, the implications of this in the dynamics of Möbius motions of
the compact classical Lie groups.

The simplest case is the Möbius action $SU\left(1,1\right)\times
U_{1}\rightarrow U_{1}$. The equivalences up to coverings with the usual
actions of $SL_{2}\left(\mathbb{R}\right)$ and $O_{o}\left(1,2\right)$ on the
circle (Möbius or projective and conformal, respectively) is well known. In
[12] the first two authors gave a complete description of the geometry of the
kinetic energy metric in this simple case. See also [13], where the
relationship with the article [2] is addressed, providing a dynamical
presentation of the metric on the disc where everything is better than nice.
The kinetic energy metric projects to a non-complete negatively curved disc
whose geodesics are hypocycloids.

The next simplest case is the Möbius action $O_{o}\left(2,2\right)\times
SO_{2}\rightarrow SO_{2}$, which is the only one of those actions which is not
even almost effective. For the sake of completeness we comment on its
relationship with projective transformations of the circle.

We recall from [15] the split quaternions $\mathbb{M}$, which is the skew
algebra $M\left(2,\mathbb{R}\right)=\mathbb{R}^{2\times 2}$ with the usual
matrix multiplication and the inner product whose square norm is given by
$\left\langle X,X\right\rangle={\mathrm{det}}X$. In particular,
$SL\left(2,\mathbb{R}\right)$ is the unit sphere in $\mathbb{M}$. The matrices

$\mathbf{1}=\left(\begin{array}[]{cc}1&0\\\ 0&1\end{array}\right)\text{,\ \ \
\ }\mathbf{i}=\left(\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right)\text{,\ \
\ \ }\mathbf{j}=\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right)\text{,\
\ \ \ }\mathbf{k}=\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right)$

form a basis of $\mathbb{M}$ and the following relations hold:

$-\mathbf{i}^{2}=\mathbf{j}^{2}=\mathbf{k}^{2}=\mathbf{1}\text{,\ \ \ \
}\mathbf{ij}=-\mathbf{ji}=\mathbf{k}\text{,\ \ \ \
}\mathbf{jk}=-\mathbf{kj}=-\mathbf{i}\text{ \ \ \ and\ \ \ \
}\mathbf{ki}=-\mathbf{ik}=\mathbf{j}\text{.}$

The square norm is given by

$\left\langle
a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k},a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}\right\rangle=a^{2}+b^{2}-c^{2}-d^{2}$

and hence the inner product has signature $\left(2,2\right)$. Observe that
$p\overline{p}=\left\langle p,p\right\rangle$ for all $p\in\mathbb{M}$ and
$p^{-1}=\overline{p}/\left\langle p,p\right\rangle$ if $\left\langle
p,p\right\rangle\neq 0$. Let $L_{p}$ and $R_{p}$ denote left and right
multiplication by $p$, respectively, which are linear isometries of
$\mathbb{M}$ if $\left\langle p,p\right\rangle=1$. One has that

$SL\left(2,\mathbb{R}\right)\times SL\left(2,\mathbb{R}\right)\longrightarrow
O_{o}\left(\mathbb{M}\right)\text{,}\ \ \ \ \ \ \ \left(p,q\right)\mapsto
L_{p}\circ R_{\overline{q}}$

is a double covering morphism with kernel
$\left\\{\left(\mathbf{1},\mathbf{1}\right),\left(-\mathbf{1},-\mathbf{1}\right)\right\\}$.
In particular,

$O_{o}\left(\mathbb{M}\right)=\left\\{L_{p}\circ R_{\overline{q}}\mid p,q\in
SL_{2}\left(\mathbb{R}\right)\right\\}\text{.}$

We set $\mathbb{R}\mathbf{1}=\mathbb{R}$,
$\mathbb{C}=\mathbb{R}+\mathbf{i}\mathbb{R}$,
$\mathbb{M}=\mathbb{C}+\mathbf{j}\mathbb{C}$ and write the operators in
$O_{o}\left(\mathbb{M}\right)$ in blocks with respect to this decomposition.
The ordered bases $\left\\{\mathbf{1},\mathbf{i}\right\\}$ and
$\left\\{\mathbf{j},\mathbf{ji}=-\mathbf{k}\right\\}$ of $\mathbb{C}$ and
$\mathbf{j}\mathbb{C}$ determine an isomorphism from $SO_{2}\cong S^{1}$ onto
$SO\left(\mathbf{j}\mathbb{C},\mathbb{C}\right)$, assigning to $u\in S^{1}$
the map $\mathbf{j}\mathbb{C}\ni\mathbf{j}z\mapsto uz$. The juxtaposition of
these bases provides the identification $O_{o}\left(\mathbb{M}\right)\cong
O_{o}\left(2,2\right)$.

Now we consider the Möbius action $\ast:O_{o}\left(2,2\right)\times
SO_{2}\rightarrow SO_{2}$, using the presentation in (6), identifying $u\in
SO_{2}\equiv S^{1}$ with the set $\left\\{uz+\mathbf{j}z\mid
z\in\mathbb{C}\right\\}$, according to (5). Thus,

(30) $L_{p}R_{\overline{q}}\ast u=v\left(p,q,u\right)\text{,}$

where $v\left(p,q,u\right)$ is the only point of $S{{}^{1}}$ that satisfies
that for all $z\in\mathbb{C}$ exists $w\left(z\right)\in\mathbb{C}$ such that

$p\left(uz+\mathbf{j}z\right)\overline{q}=v\left(p,q,u\right)w\left(z\right)+\mathbf{j}w\left(z\right)\text{.}$

######  Proposition 3.

_a)_ The Möbius action $O_{o}\left(2,2\right)\times SO_{2}\rightarrow SO_{2}$
is not almost effective. More precisely, $\left(L_{p}\circ
R_{\overline{q}}\right)\ast u=L_{p}\ast u$ for all $p,q\in
SL_{2}\left(\mathbb{R}\right)$ and $u\in S^{1}$.

_b)_ The following diagram commutes _:_

$\begin{array}[]{ccc}SL_{2}\left(\mathbb{R}\right)\times
S^{1}&\overset{\cdot}{\longrightarrow}&S^{1}\\\
\downarrow\left(F,\rho\right)&&\downarrow\rho\\\ O_{o}\left(2,2\right)\times
SO_{2}&\overset{\ast}{\longrightarrow}&SO_{2}\text{,}\end{array}$

where $F\left(p\right)=L_{p}$, $\rho\left(u\right)\left(z\right)=uz$ and
$\cdot$ denotes the usual action by conformal transformations of the plane
preserving the circle, that is,

(31) $p\cdot u=\left(\alpha u+\overline{\beta}\right)\left(\beta
u+\overline{\alpha}\right)^{-1}$

if $p=\alpha+\mathbf{j}\beta$, with $\alpha,\beta\in\mathbb{C}$,
$\left|\alpha\right|^{2}-\left|\beta\right|^{2}=1$.

######  Proof.

a) Let $q=\alpha+\mathbf{j}\beta$ with $\alpha,\beta\in\mathbb{C}$. Given
$u\in S^{1}$, we compute

$R_{\overline{\alpha}-\mathbf{j}\beta}\left(uz+\mathbf{j}z\right)=\left(uz+\mathbf{j}z\right)\left(\overline{\alpha}-\mathbf{j}\beta\right)=uz\overline{\alpha}-\overline{z}\beta+\mathbf{j}\left(z\overline{\alpha}-\overline{u}\,\overline{z}\beta\right)$

(we have used that $w\mathbf{j}=\mathbf{j}\overline{w}$ for all
$w\in\mathbb{C}$). Since
$uz\overline{\alpha}-\overline{z}\beta=u\left(z\overline{\alpha}-\overline{u}\,\overline{z}\beta\right)$
for all $z\in\mathbb{C}$, we have by (30) that $v\left(1,q,u\right)=u$, and
so, $R_{\overline{q}}\ast u=u$.

b) Now let $p=\alpha+\mathbf{j}\beta\in SL_{2}\left(\mathbb{R}\right)$ with
$\alpha,\beta\in\mathbb{C}$. We compute

$L_{\alpha+\mathbf{j}\beta}\left(uz+\mathbf{j}z\right)=\left(\alpha+\mathbf{j}\beta\right)(uz+\mathbf{j}z)=\left(\alpha
u+\overline{\beta}\right)z+\mathbf{j}\left(\beta
u+\overline{\alpha}\right)z\text{.}$

Calling $w=\left(\beta u+\overline{\alpha}\right)z$, we have $z=\left(\beta
u+\overline{\alpha}\right)^{-1}w$, and so $v\left(p,1,u\right)$ as in (30)
equals the right hand side of (31), as desired. ∎

###  7.1. Relationship with conformal motions of the three sphere

######  Lemma.

Let $f:sp\left(1,1\right)\rightarrow o\left(1,4\right)$ be defined by

$f\left(\begin{array}[]{cc}\xi&\overline{\alpha}\\\
\alpha&\eta\end{array}\right)=\left(\begin{array}[]{cc}0&2\alpha^{T}\\\
2\alpha&L_{\eta}-R_{\xi}\end{array}\right),$

where $\xi,\eta\in\operatorname{Im}(\mathbb{H})$ and $\alpha\in\mathbb{H}$.
Then $f$ is a Lie algebra isomorphism which induces a morphism
$F:Sp\left(1,1\right)\rightarrow O_{o}\left(1,4\right)$.

######  Proof.

Standard arguments show that $f$ is a morphism, using that

$4\left(\alpha\beta^{T}-\beta\alpha^{T}\right)=L_{\alpha\overline{\beta}-\beta\overline{\alpha}}-R_{\overline{\alpha}\beta-\overline{\beta}\alpha}$

holds for all $\alpha,\beta\in\mathbb{H}$ (in the left hand side, quaternions
are thought of as column vectors, under the canonical identification of
$\mathbb{H}$ with $\mathbb{R}^{4}$). Now we sketch how to verify this
identity. Since each side defines an skew symmetric $\mathbb{R}$-bilinear map

$\left(\alpha,\beta\right)\in\mathbb{H}\times\mathbb{H}\rightarrow\left\\{\text{skew-
symmetric }\mathbb{R}\text{-linear maps of }\mathbb{H}\right\\}\text{,}$

it suffices to consider only the case when $\left\\{\alpha,\beta\right\\}$ is
orthonormal. By linearity, it is enough to evaluate both sides at $q=\alpha$,
$q=\beta$ and $q\in\left\\{\alpha,\beta\right\\}^{\bot}$.

The last assertion follows, since $Sp\left(1,1\right)$ is simply connected (by
the Cartan decomposition, it is diffeomorphic to $\left(S^{3}\times
S^{3}\right)\times\mathbb{R}^{6}$). ∎

######  Proposition 4.

The following diagram is commutative _:_

$\begin{array}[]{ccc}Sp\left(1,1\right)\times
Sp_{1}&\overset{\ast}{\longrightarrow}&Sp_{1}\\\
\downarrow\left(F,\rho\right)&&\downarrow\rho\\\ O_{o}\left(1,4\right)\times
S^{3}&\overset{\ast_{c}}{\longrightarrow}&S^{3}\text{,}\end{array}$

where $\rho\left(u\right)=\overline{u}$, $\ast$ denotes the Möbius action and
$\ast_{c}$ the conformal action.

######  Proof.

First, we evaluate $F$ at the elements of the maximal compact subgroup
$K=Sp_{1}\times Sp_{1}$ of $Sp\left(1,1\right)$. For all $p,q,u\in S^{3}$,

$F\left(\operatorname{diag}\left(p,q\right)\right)\ast_{c}\rho\left(u\right)=L_{q}R_{\overline{p}}\ast_{c}\overline{u}=q\,\overline{u}\,\overline{p}=\rho\left(\operatorname{diag}\left(p,q\right)\ast
u\right)$

holds, putting $q=e^{t\eta}$ for $\eta\in$ $\operatorname{Im}(\mathbb{H})$,
since $\exp\left(tL_{\eta}\right)=L_{e^{t\eta}}$ (and similarly for
$R_{\overline{p}}$).

Given that $F$ is a morphism, by the $KAK$ decomposition of
$Sp\left(1,1\right)$, where $A=\exp\left(\mathbb{R}X\right)$ with
$X=\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right)$, it only remains to
check that

$F\left(\exp
tX\right)\ast_{c}\rho\left(u\right)=\rho\left(\exp\left(tX\right)\ast
u\right)$

holds for all $t\in\mathbb{R}$. We call

$H_{t}=\exp\left(tX\right)=\left(\begin{array}[]{cc}\cosh t&\sinh t\\\ \sinh
t&\cosh t\end{array}\right)\in O_{o}\left(1,1\right)\subset
Sp\left(1,1\right)\text{.}$

By definition of the Möbius action we have

(32) $\rho\left(\exp\left(tX\right)\ast u\right)=\rho\left(H_{t}\ast
u\right)=\left(\overline{u}\sinh t+\cosh t\right)^{-1}\left(\overline{u}\cosh
t+\sinh t\right)\text{.}$

On the other hand,

$F\left(\exp
tX\right)=\exp\left(tf\left(X\right)\right)=\exp\left(t\operatorname{diag}\left(2X,0_{3}\right)\right)=\operatorname{diag}\left(H_{2t},I_{3}\right)\text{.}$

Now, putting $u=a+w\in S^{3}$, with $a\in\mathbb{R}$,
$w\in\operatorname{Im}\left(\mathbb{H}\right)$, we compute

$\left(\begin{array}[]{ccc}\cosh 2t&\sinh 2t&0\\\ \sinh 2t&\cosh 2t&0\\\
0&0&I_{3}\end{array}\right)\left(\begin{array}[]{c}1\\\ a\\\
-w\end{array}\right)=\left(\begin{array}[]{c}\cosh\left(2t\right)+\sinh\left(2t\right)a\\\
\sinh\left(2t\right)+\cosh\left(2t\right)a\\\ -w\end{array}\right)\text{.}$

So, by (3),

(33) $F\left(\exp
tX\right)\ast_{c}\overline{u}=\operatorname{diag}\left(H_{2t},I_{3}\right)\ast_{c}\overline{u}=\frac{\sinh\left(2t\right)+a\cosh\left(2t\right)-w}{\cosh(2t)+a\sinh(2t)}\text{.}$

Using that $\left|u\right|^{2}=a^{2}+\left|w\right|^{2}=1$ and
$1=u\overline{u}=\left(a+w\right)\left(a-w\right)=a^{2}-w^{2}$, a lengthy but
straightforward computation yields that expressions (32) and (33) coincide, as
desired. ∎

###  7.2. Relationship with projective motions of the three sphere

Identifying $\mathbb{R}^{4}=\mathbb{H}$ and
$\mathbb{R}^{3}=\operatorname{Im}(\mathbb{H})$, the orthogonal groups for
dimensions 3 and 4 can be described as

$SO_{3}=\left\\{L_{p}\circ R_{\overline{p}}\mid p\in S^{3}\right\\}\text{ \ \
\ \ \ \ and \ \ \ \ \ \ \ }SO_{4}=\left\\{L_{p}\circ R_{\overline{q}}\mid
p,q\in S^{3}\right\\}\text{.}$

Next we recall the well known double covering morphism from
$SL_{4}(\mathbb{R})$ to $O_{o}\left(3,3\right)$. Let $o_{4}=\left\\{\text{skew
symmetric }4\times 4\text{ real matrices}\right\\}$. There is an isomorphism
$\omega:o_{4}\rightarrow\Lambda^{2}\left(\mathbb{R}^{4}\right)^{\ast}$,
$Z\mapsto\omega_{Z}$, where $\omega_{Z}\left(x,y\right)=x^{T}Zy$. We consider
on $o_{4}$ the inner product $g$ of signature $\left(3,3\right)$ induced via
$\omega$ from the inner product $h$ on
$\Lambda^{2}\left(\mathbb{R}^{4}\right)^{\ast}$ given by
$h\left(\sigma,\tau\right){\mathrm{det}}=\frac{1}{2}\,\sigma\wedge\tau$.

We also have the decomposition

$o_{4}=\mathcal{L}\oplus\mathcal{R}\cong\operatorname{Im}(\mathbb{H})\times(\mathbb{H})\text{,}$

where
$\mathcal{L}=\left\\{L_{\xi}\mid\xi\in\operatorname{Im}(\mathbb{H})\right\\}$
and
$\mathcal{R}=\left\\{R_{\eta}\mid\eta\in\operatorname{Im}(\mathbb{H})\right\\}$.
The set
$\mathcal{B}=\left\\{L_{i},L_{j},L_{k}\right\\}\cup\left\\{R_{i},R_{j},R_{k}\right\\}$
is a $\left(3,3\right)$-orthonormal basis of $o_{4}$ with respect to $g$,
where the first elements are space-like and the last ones are time-like. Now,
the group $SL_{4}\left(\mathbb{R}\right)$ acts on
$\Lambda^{2}\left(\mathbb{R}^{4}\right)^{\ast}$ by linear isometries with
respect to $h$ via $A.\sigma=\left(A^{T}\right)^{\ast}\sigma$. This action,
through the isomorphism $\omega,$ induces the morphism

$\Phi:SL_{4}(\mathbb{R})\rightarrow O_{o}\left(o_{4},g\right)\text{,\ \ \ \ \
\ \ \ \ \ \ }\Phi\left(A\right)\left(Z\right)=AZA^{T}\text{.}$

By means of the basis $\mathcal{B}$, we identify $o_{4}=\mathbb{R}^{3,3}$ and
so $\Phi$ induces a morphism $F:SL_{4}(\mathbb{R})\rightarrow
O_{o}\left(3,3\right)$. We also define $\rho:S^{3}\rightarrow SO_{3}$ by
$\rho\left(q\right)=L_{q}R_{\overline{q}}$.

######  Proposition 5.

The following diagram is commutative _:_

(34) $\begin{array}[]{ccc}SL_{4}(\mathbb{R})\times
S^{3}&\overset{\ast_{p}}{\longrightarrow}&S^{3}\\\
\downarrow\left(F,\rho\right)&&\downarrow\rho\\\ O_{o}\left(3,3\right)\times
SO_{3}&\overset{\ast}{\longrightarrow}&SO_{3}\text{,}\end{array}$

where $\ast$ denotes the Möbius action and $\ast_{p}$ the projective action.

######  Proof.

First we verify the commutativity of the diagram for the elements of
$SO_{4}\subset SL_{4}\left(\mathbb{R}\right)$. We have that
$\left.F\right|_{SO_{4}}:SO_{4}\rightarrow SO_{3}\times SO_{3}\subset
O_{o}\left(3,3\right)$ (canonical immersion) is given by

$F\left(L_{p}\circ R_{\overline{q}}\right)=\left(L_{p}\circ
R_{\overline{p}},L_{q}\circ R_{\overline{q}}\right)\text{.}$

Indeed, calling $L_{p}\circ R_{\overline{q}}=A$, $A\ast_{p}u=pu\overline{q}$
holds for all $u\in S^{3}$, since $\left|pu\overline{q}\right|=1$. Now,
$F\left(A\right)\ast\rho\left(u\right)=\rho\left(A\ast_{p}u\right)$ if and
only if

$L_{p}R_{\overline{p}}L_{u}R_{\overline{u}}\left(L_{q}R_{\overline{q}}\right)^{-1}=L_{pu\overline{q}}\left(R_{pu\overline{q}}\right)^{-1}$

for all $p,q,u\in S^{3}$, and this can be easily checked.

Since $F$ is a morphism, by the $KAK$ decomposition of
$SL_{4}\left(\mathbb{R}\right)$, it remains to check that the diagram commutes
for any diagonal matrix $E$ in $SL_{4}\left(\mathbb{R}\right)$. Any such $E$
has the form $E=\exp\left(x_{1}D+x_{2}D_{2}+x_{3}D_{3}\right)$, where

$D=\operatorname{diag}\left(1,1,-1,-1\right),\ \
D_{2}=\operatorname{diag}\left(1,-1,1,-1\right)\text{\ \ \ \ and \ \ \
}D_{3}=\operatorname{diag}\left(1,-1,-1,1\right).$

Now, there exist $k_{2},k_{3}\in SO_{4}$ such that $D_{l}=k_{l}Dk_{l}^{-1}$
for $l=2,3.$ Hence,

$E=\exp\left(x_{1}D\right)k_{2}\exp\left(x_{2}D\right)k_{2}^{-1}k_{3}\exp\left(x_{1}D\right)k_{3}^{-1}.$

Therefore, it suffices to check the commutativity of the diagram for

$A_{t}:=\exp\left(tD\right)=\operatorname{diag}\left(e^{t},e^{t},e^{-t},e^{-t}\right).$

Let $B_{t}$ be the matrix of the transformation
$\Phi\left(A_{t}\right)\left(L_{i}\right)=A_{t}L_{i}A_{t}^{T}\in
o\left(\mathbb{H}\right)$ with respect to the basis
$\left\\{1,i,j,k\right\\}$. We compute

$\displaystyle B_{t}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}e^{t}I_{2}&0_{2}\\\
0_{2}&e^{-t}I_{2}\end{array}\right)\left(\begin{array}[]{cc}J&0\\\
0&J\end{array}\right)\left(\begin{array}[]{cc}e^{t}I_{2}&0_{2}\\\
0_{2}&e^{-t}I_{2}\end{array}\right)=\left(\begin{array}[]{cc}e^{2t}J&0_{2}\\\
0_{2}&e^{-2t}J\end{array}\right)$ $\displaystyle=$
$\displaystyle\cosh\left(2t\right)\left(\begin{array}[]{cc}J&0_{2}\\\
0_{2}&J\end{array}\right)+\sinh\left(2t\right)\left(\begin{array}[]{cc}J&0_{2}\\\
0_{2}&-J\end{array}\right)$

for all $t\in\mathbb{R}$, where $J=\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right)$. Similar computations yield

$\displaystyle A_{t}L_{ai+bj+ck}A_{t}^{T}$ $\displaystyle=$
$\displaystyle\cosh\left(2t\right)L_{ai}+\sinh\left(2t\right)R_{ai}+L_{bj+ck}$
$\displaystyle A_{t}R_{ai+bj+ck}A_{t}^{T}$ $\displaystyle=$
$\displaystyle\sinh\left(2t\right)L_{ai}+\cosh\left(2t\right)R_{ai}+R_{bj+ck}$

for all $a,b,c,t\in\mathbb{R}$. Consequently, calling

$C_{t}=\operatorname{diag}\left(\cosh 2t,I_{2}\right)\text{\ \ \ \ \ and\ \ \
\ \ }S_{t}=\operatorname{diag}\left(\sinh 2t,0_{2}\right)\text{,}$

both in $\mathbb{R}^{3\times 3}$, the matrix of the operator
$\Phi\left(A_{t}\right)$ with respect to the basis $\mathcal{B}$, which equals
the matrix of $F\left(A_{t}\right)$, is

$\left(\begin{array}[]{cc}C_{t}&S_{t}\\\ S_{t}&C_{t}\end{array}\right)\in
O_{o}\left(3,3\right)\text{.}$

Now, for $w\neq 0$ in $\mathbb{H}$, we call $I_{w}$ the matrix of the operator
$L_{w}\left(R_{w}\right)^{-1}$ on $\operatorname{Im}(\mathbb{H})$ with respect
to the basis $\left\\{i,j,k\right\\}$. If $w=a+bi+cj+dk$, we compute

$I_{w}=\frac{1}{\left|w\right|^{2}}\left(\begin{array}[]{ccc}a^{2}+b^{2}-d^{2}-c^{2}&-2ad+2bc&2bd+2ac\\\
2ad+2bc&a^{2}-b^{2}+c^{2}-d^{2}&-2ab+2cd\\\
-2ac+2bd&2ab+2cd&a^{2}-b^{2}+d^{2}-c^{2}\end{array}\right)\text{.}$

The commutativity of the diagram for $A_{t}$ amounts to the validity of

$\left(C_{t}I_{u}+S_{t}\right)\left(S_{t}I_{u}+C_{t}\right)^{-1}=L_{\left(A_{t}u\right)/\left|A_{t}u\right|}\left(R_{\left(A_{t}u\right)/\left|A_{t}u\right|}\right)^{-1}=L_{A_{t}u}\left(R_{A_{t}u}\right)^{-1}=I_{A_{t}u}$

for all $u\in S^{3}$ and all $t\in\mathbb{R}$, or equivalently,
$C_{t}I_{u}+S_{t}=I_{A_{t}u}\left(S_{t}I_{u}+C_{t}\right)$. This can be
checked with a lengthy but straightforward computation. ∎

###  7.3. Dynamical implications for Möbius motions in low dimensions

Let $Sp\left(1,1\right)$ and $O_{o}\left(3,3\right)$ be endowed with the
kinetic energy metrics. By Proposition 2, the induced Riemannian metrics on
the maximal compact subgroups $Sp_{1}\times Sp_{1}$ and $SO_{3}\times SO_{3}$
are bi-invariant. Theorem 3 asserts that each factor is totally geodesic. The
results in this section give more information on the matter. At the same time,
they provide an alternative statement (perhaps more illuminating) for Theorem
3 (b) in [19].

######  Corollary 1.

Let $Sp(1,1)$ and $O_{o}\left(3,3\right)$ be endowed with the kinetic energy
metrics. Then:

_a)_ The subgroup $Sp_{1}\times Sp_{1}$ is totally geodesic in
$Sp\left(1,1\right)$.

_b)_ A geodesic in $SO_{3}\times SO_{3}$ is a geodesic in
$O_{o}\left(3,3\right)$ if and only if it remains in one of the factors.

######  Proof.

By Propositions 4 and 5, the Möbius actions of $Sp\left(1,1\right)$ on
$Sp_{1}$ and of $O_{o}\left(3,3\right)$ on $SO_{3}$ are equivalent, up to
double coverings, to the conformal action of $O_{o}\left(1,4\right)$ on
$S^{3}$ and the projective action of $SL_{4}\left(\mathbb{R}\right)$ on
$S^{3}$, respectively. The morphisms preserve the corresponding kinetic energy
metrics.

By Theorem 2 (b) in [25], $SO_{4}$ is totally geodesic in
$O_{o}\left(1,4\right)$ endowed with the kinetic energy metric associated with
the conformal action. Since $Sp_{1}\times Sp_{1}$ corresponds to $SO_{4}$
under the corresponding morphism, part (a) of the corollary follows.

Also, Theorem 3 (b) in [19] asserts that for $Z\in o_{4}$, the geodesic
$\sigma(t)=\exp(tZ)$ of $SO_{4}$ through the identity is also a geodesic of
$SL_{4}\left(\mathbb{R}\right)$ if and only if $Z=\lambda X$ for some
$\lambda\in\mathbb{R}$ and $X\in o_{4}$ with $X^{2}=-I_{4}$. Now,
$Z=L_{\xi}+R_{\eta}$ for some $\xi,\eta\in\operatorname{Im}(\mathbb{H})$. We
compute

$-\lambda^{2}I_{4}=Z^{2}=\left(L_{\xi}+R_{\eta}\right)^{2}=L_{\xi}^{2}+R_{\eta}^{2}+2L_{\xi}R_{\eta}=-\left(\left|\xi\right|^{2}+\left|\eta\right|^{2}\right)I_{4}+2L_{\xi}R_{\eta}$

($L_{\xi}$ and $R_{\eta}$ commute). This implies that $L_{\xi}R_{\eta}=\mu
I_{4}$ for some $\mu\in\mathbb{R}$. If $\eta\neq 0$, we have
$L_{\xi}=R_{\mu\bar{\eta}/\left|\eta\right|^{2}}$. Now, $p,q\in\mathbb{H}$
satisfy $L_{p}=R_{q}$ only if $p,q$ are in the center of $\mathbb{H}$, that
is, $p,q\in\mathbb{R}$. Since $\xi\in\operatorname{Im}(\mathbb{H})$, then
$\xi=0$, as desired. ∎

Acknowledgements. This work was supported by Consejo Nacional de
Investigaciones Científicas y Técnicas and Secretarías de Ciencia y Técnica of
Universidad Nacional de Córdoba and Universidad Nacional de Rosario. We would
like to thank Carlos Olmos for useful comments and Simon Chiossi for a
valuable reference.

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