# Algebraic structures on parallelizable manifolds

Sergey Grigorian
School of Mathematical & Statistical Sciences
University of Texas Rio Grande Valley
Edinburg, TX 78539
USA

###### Abstract

In this paper we explore algebraic and geometric structures that arise on
parallelizable manifolds. Given a parallelizable manifold $\mathbb{L}$, there
exists a global trivialization of the tangent bundle, which defines a map
$\rho_{p}:\mathfrak{l}\longrightarrow T_{p}\mathbb{L}$ for each point
$p\in\mathbb{L}$, where $\mathfrak{l}$ is some vector space. This allows us to
define a particular class of vector fields, known as fundamental vector
fields, that correspond to each element of $\mathfrak{l}$. Furthermore, flows
of these vector fields give rise to a product between elements of
$\mathfrak{l}$ and $\mathbb{L}$, which in turn induces a local loop structure
(i.e. a non-associative analog of a group). Furthermore, we also define a
generalization of a Lie algebra structure on $\mathfrak{l}$. We will describe
the properties and examples of these constructions.

###### Contents

  1. 1 Introduction
  2. 2 Parallelized manifolds
  3. 3 Automorphisms
  4. 4 Products of spheres
  5. A Appendix

##  1 Introduction

The tangent bundle is a fundamental aspect of smooth manifolds, and its global
trivialization, or absolute parallelism, has been a key area of study in
differential geometry since the field’s emergence in the early 20th century
[4, 5]. The study of parallelizable manifolds, as smooth manifolds with a
trivial tangent bundle are now generally referred to, has been continuing
since then in several directions. One approach, which is more topological in
nature, reframed the question in terms of _stably_ parallelizable manifolds,
i.e. manifolds for which the _stable_ tangent bundle is trivial. Given a real
smooth manifold $M$ with tangent bundle $TM$, and a trivial rank $1$ bundle
$\varepsilon$ over $M$ (i.e. $M\times\mathbb{R}$), the stable tangent bundle
is $TM\oplus\varepsilon$. The conditions for a manifold to be stably
parallelizable (also known as a $\pi$-manifold) in low dimensions can be
expressed in terms of vanishing of certain characteristic classes [7]. It also
follows from [2] that in dimensions $1,3,$ and $7,$ manifolds are
parallelizable if and only if they are stably parallelizable. Another
approach, which stems from the origins of Riemannian geometry, involves the
study of flat metric connections on Riemannian manifolds. Indeed, the
existence of a flat metric connection allows to parallel transport a frame
from a single point to the entire manifold, and thus obtain a trivialization
of the tangent bundle. The converse is also trivially true. Generally such
connections will admit torsion and then questions about classification of
parallelizable manifolds become related to the properties of torsion of flat
metric connections. In [4], Cartan and Schouten have shown initiated this
approach by showing that a Riemannian manifold with a flat metric connection
and totally skew-symmetric torsion is either a compact simple Lie group or the
$7$-sphere $S^{7},$ depending on whether the torsion is parallel or not. A
similar classification in the pseudo-Riemannian case has been obtained by Wolf
[16, 17], again under the assumption of totally skew-symmetric torsion. A more
modern treatment of these approaches has been given in [1]. However,
surprisingly, despite significant efforts from different points of view, there
is still no full classification of parallelizable manifolds and shows that
there is still a need to explore properties of parallelizable manifolds.

This paper builds upon these foundational concepts to explore new algebraic
structures on parallelizable manifold. These structures extend the traditional
notions of Lie algebras and Lie groups, providing a broader perspective on the
geometric and algebraic properties of these manifolds. Indeed, Lie groups form
one of the most well-known classes of parallelizable manifolds and are of
course characterized by a globally defined smooth associative product and a
Lie algebra structure on the tangent space at identity. The associativity
property may be relaxed to consider _smooth loops_[6]. The $7$-sphere from [4]
is an example of a smooth loop, when regarded as the set of unit octonions,
and is of course parallelizable. The smooth product allows to identify any
tangent space with the tangent space at identity, providing a global
trivialization of the tangent bundle. This also allows to define an bracket
algebra on the tangent space at identity, however, this is in general no
longer a Lie algebra. As shown in [6], on a smooth loop we may define a family
of brackets, defined for each point of the loop, leading to a _bracket
function_ defined on the loop. The non-constant nature of this function then
leads to a non-trivial right hand side in the Jacobi identity. On Lie group,
the bracket of left- or right-invariant vector fields is itself left- or
right-invariant, respectively, and hence only a unique bracket is defined.

In this paper, we expand some of the findings from [6], but for arbitrary
parallelizable manifolds. Indeed, several properties of Lie groups and smooth
loops only depend on the trivialization of the tangent bundle, which is a
weaker property than the existence of a global structure.

In Section 2, our exploration begins with a detailed examination of the global
trivialization of tangent bundles, a defining characteristic of parallelizable
manifolds. This global trivialization allows us to define fundamental vector
fields, which are parallel vector fields with respect to the trivial
connection. Integral curves of these vector lead a local loop structure, a
non-associative analog of a group structure. This structure enriches the
manifold’s algebraic framework and opens new avenues for studying its
geometric properties. Furthermore, we introduce a generalization of the Lie
algebra structure on parallelizable manifolds. This generalization provides a
new perspective on the algebraic underpinnings of these manifolds and offers
new insights into their properties. The interplay between the geometric and
algebraic aspects of parallelizable manifolds, enriched by these Lie-like
structures, is a key focus of this paper. It should be noted that in the
1960’s, Kikkawa [10, 11] has similarly defined local loop structures based on
geodesics of arbitrary affine connections. However, unlike in the
parallelizable case, such an approach does not allow for a richer algebra
structure.

The structure in Section 2 is defined by a parallelizable manifold together
with a fixed trivialization of the tangent bundle. We denote this by the
_parallelized manifold_ triple $(\mathbb{L},\mathfrak{l,\rho)}$, where
$\mathbb{L}$ is a smooth manifold, $\mathfrak{l}$ a vector space which is the
model fiber for the tangent bundle, and $\rho$ is a smooth family of linear
isomorphisms from $\mathfrak{l}\ $to each tangent space $T_{p}\mathbb{L}$. We
use the trivialization $\rho~{}$to define _fundamental vector fields_ on
$\mathbb{L}$, for each $s\in\mathbb{L}$ given by $\rho_{s}\left(\xi\right)\
$for some $\xi\in\mathfrak{l}.$ These are precisely the parallel vector fields
for the flat connection defined by the trivialization. We also use $\rho$
family of local loop structures $\circ_{s}$ on $\mathfrak{l}$ and
corresponding bracket algebra structures $b^{\left(s\right)}$, together with a
family of trilinear forms $a^{\left(s\right)}$ that are defined from
differentials of $b^{\left(s\right)}$ Theorem 2.24 gives an analogue of the
Maurer-Cartan structure equation, while Theorem 2.26 relates the brackets as
infinitesimal commutators of the products $\circ_{s}$. As discussed above for
the case of smooth loops, the bracket function is in general non-constant on
$\mathbb{L}$, and its differential is related to a skew-symmetrization of
infinitesimal associators of the local product on $\mathfrak{l}.$

In Section 3 we then define morphisms of parallelized manifolds and study how
the automorphisms of parallelized manifolds interact with the algebraic
operations defined in Section 2. In Theorem 3.8 we prove that the automorphism
group of $(\mathbb{L},\mathfrak{l,\rho)}$ is a finite-dimensional Lie subgroup
of the diffeomorphism group of $\mathbb{L}.$

As a particular example, in Section 4 we consider explicit trivializations of
products of spheres, as given in [3, 14, 15]. As shown by Kervaire in [9], any
product of spheres that contains an odd-dimensional sphere, is parallelizable.
However, we consider a simpler examples of $S^{m}\times S^{1}$ and
$S^{m}\times N$, where $m$ is arbitrary and $N$ is any parallelizable
manifold. The parallelizability of these spaces follows immediately from the
fact that any sphere is stably parallelizable. We find that in the case of
$\mathbb{L=}S^{m}\times S^{1}$, the standard trivialization yields a family of
bracket algebras $b^{\left(x\right)}$, for $x\in S^{m}$, each of which is
isomorphic to the semidirect sum Lie algebra
$\mathbb{R}^{n}\oplus_{S}\mathbb{R}$. Moreover, considering the triple product
$a^{\left(x\right)}$, the algebraic structure
$\left(\mathfrak{l},a^{\left(x\right)}\right)$ is that of a Lie triple system
[8]. We also explicitly compute the local product $\circ_{s}$ on
$\mathfrak{l}$ in certain special cases and find that the automorphism group
$\Psi\left(\mathbb{L}\right)\cong SO\left(n+1\right)\times U\left(1\right)$.

In the more general case of $\mathbb{L=}S^{m}\times N$, we compute the
brackets and the Jacobi identity, finding that for instance, in the case of
$N=S^{n}\times S^{1}$, for some values of $s\in\mathbb{L}$, then bracket
$b^{\left(s\right)}$ does not satisfy the Jacobi identity, so generally,
parallelizable manifolds do indeed induce an algebraic structure that is more
general than a Lie algebra or a Lie triple system.

Overall, we see that the framework presented in this paper allows to associate
certain algebraic structures to parallelized manifolds. This paves the way
towards using such algebraic invariants to study and classify parallelizable
manifolds. Moreover, this approach is also easily adaptable to the broader
class of stably parallelizable manifolds as well, as the example of
$S^{m}\times S^{1}$ shows.

### Acknowledgements

This work was supported by the National Science Foundation [DMS-1811754].

##  2 Parallelized manifolds

Suppose $\mathbb{L}$ is a connected smooth real $n$-dimensional manifold, with
a trivializable tangent bundle
$T\mathbb{L}\cong\mathbb{L}\times\mathbb{R}^{n}.$ Now given an $n$-dimensional
real vector space $\mathfrak{l},$ the global trivialization of $T\mathbb{L}$
induces a diffeomorphism $\rho:\mathbb{L}\times\mathfrak{l}\longrightarrow
T\mathbb{L},$ such that

$\mathbb{L}\times\mathfrak{l}\ni\left(p,\xi\right)\mapsto\rho_{p}\left(\xi\right)\in
T_{p}\mathbb{L},$ (2.1)

with $\rho_{p}:\mathfrak{l}\longrightarrow T_{p}\mathbb{L}$ being a linear
isomorphism for each $p\in\mathbb{L}$.

###### Definition 2.1

The _fundamental vector field for_ $\xi\in\mathfrak{l}$ of $T\mathbb{L}$ is
defined to be the smooth vector field
$\rho\left(\xi\right)\in\Gamma\left(T\mathbb{L}\right)$.

In particular, a choice of basis on $\mathfrak{l}$ induces a global frame of
$T\mathbb{L},$ or equivalently a section of the corresponding frame bundle
$F\mathbb{L}.$ A change of trivialization is equivalent to a
$GL\left(\mathbb{R},n\right)$ gauge transformation of $F\mathbb{L},$ while a
change of basis of $\mathfrak{l}$ corresponds to the right action of
$GL\left(\mathbb{R},n\right)$ on the principal bundle $F\mathbb{L}.$

More explicitly, suppose $\left\\{X_{i}\right\\}_{i=1}^{n}$ is a frame of
$T\mathbb{L}.$ Then, assuming we have a basis
$\left\\{e_{i}\right\\}_{i=1}^{n}$ on $\mathfrak{l},$ define the corresponding
trivialization $\rho$ so that $\rho\left(e_{i}\right)=X_{i}$ and hence

$\mathbb{L}\times\mathfrak{l}\ni\left(p,\xi\right)\mapsto\xi^{i}\left.X_{i}\right|_{p}\in
T_{p}\mathbb{L}.$

###### Definition 2.2

The triple $\left(\mathbb{L},\mathfrak{l},\rho\right)$, as above, will be
called a _parallelized manifold_ , i.e. a parallelizable manifold with a fixed
trivialization.

###### Remark 2.3

The above definition is almost equivalent to the notion of _absolute
parallelism_. Absolute parallelism is defined as a smooth family of linear
isomorphisms $\phi_{pq}:T_{q}\mathbb{L}\longrightarrow T_{p}\mathbb{L}$ for
each pair of points $p,q\in\mathbb{L}.$ Given an absolute parallelism and
fixing one of the tangent spaces as th vector space $\mathfrak{l}$ gives the
parallelized manifold definition. Conversely, given the linear maps $\rho_{p}$
from (2.1), an absolute parallelism is defined as
$\phi_{pq}=\rho_{p}\rho_{q}^{-1}.$

###### Example 2.4

Any Lie group can be parallelized via left translation or right translation.
In that case, $\mathfrak{l}$ corresponds to the Lie algebra, and $\rho$ is
either the left or right translation map.

###### Example 2.5

Any smooth loop (i.e. a smooth quasigroup with identity and smooth products
and quotients) is parallelizable, in the same way a Lie group [6]. In that
case, $\mathfrak{l}$ corresponds to the tangent algebra at identity. A
specific example is $S^{7}$, regarded as the loop of unit octonions.

###### Example 2.6

Any orientable $3$-dimensional manifold is parallelizable. For example,
$S^{2}\times S^{1}$ and $3$-dimensional lens spaces $L\left(p;q\right)$, as
well as $S^{3},$ but this is of course also a Lie group.

###### Example 2.7

A $4$-manifold is parallelizable if and only if the Stiefel-Whitney classes
$w_{1},$ $w_{2}$ vanish, the Euler characteristic vanishes, and the first
Pontryagin class vanishes.

###### Example 2.8

Products of parallelizable manifolds are parallelizable.

###### Example 2.9

As shown by Kervaire in [9], a product of spheres, where at least one of the
spheres is odd-dimensional, is parallelizable. Explicit parallelizations of
products of spheres have been demonstrated in [14, 15].

###### Remark 2.10

In general, parallelizability of a manifold is a property of the smooth
structure, rather than the underlying topological space. Indeed, as shown by
Milnor [12, Corollary 1], one may have two smooth manifolds that are
homeomorphic, but one is parallelizable, while the other one is not. However,
in certain situations, the parallelizability does reduce to topological
considerations. A manifold is known as _stably parallelizable_ if its stable
tangent bundle $TM\oplus\varepsilon$, where $\varepsilon$ is a trivial rank 1
bundle, is trivial. Such manifolds are also known as _framed manifolds_ or
$\pi$-manifolds. Stable parallelizability is in some cases equivalent to the
vanishing of certain characteristic classes [7]. As shown in [2], an
$n$-dimensional stably parallelizable manifold is parallelizable if and only
if $S^{n}$ is parallelizable. This of course happens for $n=1,3,7$. Hence in
these dimensions, parallelizability is equivalent to stable parallelizability.

To investigate properties of parallelizable manifolds using approaches from
Riemannian geometry, we may use $\rho$ to define a metric and a connection on
$\mathbb{L}$.

###### Definition 2.11

Given the triple $\left(\mathbb{L},\mathfrak{l},\rho\right)$ and any inner
product $\left\langle{}\right\rangle$ on $\mathfrak{l},$ define the metric $g$
on $\mathbb{L}$ as the pullback metric with respect to $\rho^{-1},$ i.e. for
$s\in\mathbb{L},$ and $X_{s},Y_{s}\in T_{s}\mathbb{L},$

$g_{s}\left(X_{s},Y_{s}\right)=\left\langle\rho_{s}^{-1}\left(X_{p}\right),\rho_{s}^{-1}\left(Y_{p}\right)\right\rangle.$
(2.2)

###### Definition 2.12

Given the triple $\left(\mathbb{L},\mathfrak{l},\rho\right),$ define the flat
connection $\nabla=\left(\rho^{-1}\right)^{\ast}d,$ so that

$\nabla=\rho\circ d\circ\rho^{-1}.$ (2.3)

###### Lemma 2.13

The metric and the flat connection satisfy the following properties.

  1. 1.

$\nabla$ is a metric connection with respect to $g.$

  2. 2.

Every fundamental vector field has constant norm.

Proof. Both of these properties follow immediately from the definitions.

  1. 1.

Suppose $X$ and $Y$ are two vector fields, then

$\displaystyle g\left(\nabla X,Y\right)+g\left(X,\nabla Y\right)$
$\displaystyle=$ $\displaystyle
g\left(\rho\left(d\left(\rho^{-1}X\right)\right),Y\right)$
$\displaystyle+g\left(X,\rho\left(d\left(\rho^{-1}Y\right)\right)\right)$
$\displaystyle=$ $\displaystyle\left\langle
d\left(\rho^{-1}X\right),\rho^{-1}Y\right\rangle+\left\langle\rho^{-1}X,d\left(\rho^{-1}Y\right)\right\rangle$
$\displaystyle=$ $\displaystyle
d\left(\left\langle\rho^{-1}X,\rho^{-1}Y\right\rangle\right)$ $\displaystyle=$
$\displaystyle d\left(g\left(X,Y\right)\right),$

hence $\nabla$ is metric.

  2. 2.

Suppose $X=\rho\left(\xi\right)$ is a fundamental vector field, then it is
nowhere zero, and

$d\left(\left\|X\right\|^{2}\right)=d\left(g\left(\rho\left(\xi\right),\rho\left(\xi\right)\right)\right)=d\left(\left\langle\xi,\xi\right\rangle\right)=0,$

hence $\left\|X\right\|$ is constant.

###### Remark 2.14

Given a vector field $X\in$ $\Gamma\left(T\mathbb{L}\right)$ and
$s\in\mathbb{L},$ consider the maximal integral curve $\gamma_{X,s}$ on $L$ of
the vector field $X$ through a point $s\in\mathbb{L}.$ If $\mathbb{L}$ is
compact, then every vector field on $\mathbb{L}$ is complete, and hence
$\gamma_{X,s}\left(t\right)$ is defined for all $t.$ If $\mathbb{L}$ is non-
compact, consider the global frame $\left\\{X_{i}\right\\}$ on $\mathbb{L}.$
Each $X_{i}$ is nowhere-vanishing, and given an arbitrary complete Riemannian
metric on $\mathbb{L}$ (which always exists if $\mathbb{L}$ is connected [13])
we can rescale each of these vector fields to be of unit norm everywhere.
Then, as it is well-known, unit norm vector fields on a complete Riemannian
manifold are complete. Hence, by normalizing the global frame, we may assume
that the fundamental vector fields are complete. Thus, for non-compact
manifolds, without loss of generality, we’ll consider only _complete_
trivializations, i.e. those for which the fundamental vector fields are
complete. Then, if $X$ is fundamental, then $\gamma_{X,s}\left(t\right)$ is
defined for all $t.$ We will assume that $\mathbb{L}$ is connected or compact.

Let $\xi\in\mathfrak{l}$ and define the flow diffeomorphisms
$\Phi_{\xi,t}:\mathbb{L}\longrightarrow\mathbb{L}$ of $\xi$ via

$\Phi_{\xi,t}\left(s\right)=\gamma_{\rho\left(\xi\right),s}\left(t\right),$
(2.4)

with $\gamma_{\rho\left(\xi\right),s}\left(0\right)=s.$ By definition of
integral curves, $\Phi_{\xi,t}\left(s\right)$ is the solution of the following
initial value problem:

$\left\\{\begin{array}[]{c}\frac{dp\left(t\right)}{dt}=\left.\rho\left(\xi\right)\right|_{p\left(t\right)}\\\
p\left(0\right)=s\end{array}\right..$ (2.5)

Since $\rho\left(\xi\right)$ is assumed to be complete, the maps
$\Phi_{\xi,t}$ are defined for all $t$, and in particular, for each
$a\in\mathbb{R}$, $\Phi_{a\xi,t}=\Phi_{\xi,at}$, and this motivates the
following definition.

###### Definition 2.15

The product operation $\mu$ on $\mathfrak{l}$ and $\mathbb{L}$ is the map
$\mu:\mathfrak{l\times}\mathbb{L}\longrightarrow\mathbb{L}$ defined by

$\mu\left(\xi,s\right)=\Phi_{\xi,1}\left(s\right),$ (2.6)

where $\xi\in\mathfrak{l},$ $s\in L.$ In particular, we will denote this
product by $\xi\cdot s.$ Define the left and right product maps:

  1. 1.

For each $\xi\in\mathfrak{l},$ $L_{\xi}:\mathbb{L}\longrightarrow\mathbb{L}$

  2. 2.

For each $s\in L,$ $R_{s}:\mathfrak{l}\longrightarrow\mathbb{L}.$

###### Example 2.16

Suppose $\mathbb{L}$ is a Lie group and $\rho$ is the right-invariant
trivialization of the tangent bundle. Then, $\mathfrak{l}$ is just the tangent
space at identity, i.e. the corresponding Lie algebra, and $\xi\cdot
s=\exp\left(\xi\right)s,$ where $\exp$ is the standard Lie algebra exponential
map.

###### Lemma 2.17

Let $\xi,\eta\in\mathfrak{l}$. The product operation has the following
properties:

  1. 1.

For $t_{1},t_{2}\in\mathbb{R},$
$\left(t_{1}\xi\right)\cdot\left(\left(t_{2}\xi\right)\cdot
s\right)=\left(\left(t_{1}+t_{2}\right)\xi\right)\cdot s.$

  2. 2.

For any $s\in L,$ $0\cdot s=s.$

  3. 3.

$L_{\xi}$ is a diffeomorphism with $L_{\xi}^{-1}=L_{-\xi}.$

  4. 4.

$\left.dR_{s}\right|_{0}=\rho_{s}$.

  5. 5.

For each $s\in L,$ $R_{s}$ is a local diffeomorphism, and hence the right
quotient $R_{s}^{-1}$ is defined in some neighborhood of $s.$

  6. 6.

Suppose $p\left(t\right)$ is a curve in $\mathbb{L}$ with $p\left(0\right)=p$.
Assuming that $p/s$ is defined, we have

$\left.\frac{d}{dt}R_{t\xi\cdot
s}^{-1}\left(p\left(t\right)\right)\right|_{t=0}=\left.\frac{d}{dt}p\left(t\right)/s\right|_{t=0}-\left.\frac{d}{dt}\left(p/s\right)\circ_{s}\left(t\xi\right).\right|_{t=0}$
(2.7)

Proof. The first three properties immediately follow from the definition and
the properties of the flow diffeomorphism $\Phi_{\xi,t}$. For the fourth item,
let $\xi\in\mathfrak{l}$ and consider the straight line from the origin
$t\xi.$ Then,

$\displaystyle\left.dR_{s}\right|_{0}\left(\xi\right)$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}R_{s}\left(t\xi\right)\right|_{t=0}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt}\Phi_{\xi,t}\left(s\right)\right|_{t=0}$
$\displaystyle=$ $\displaystyle\rho_{s}\left(\xi\right),$

by definition of $R_{s}.$

This shows that $\left.dR_{s}\right|_{0}=\rho_{s},$ hence is an isomorphism,
and thus by the Inverse Function Theorem is a local diffeomorphism.

Now for sufficiently small $t$, consider

$p\left(t\right)=\left(R_{t\xi\cdot
s}^{-1}\left(p\left(t\right)\right)\right)\cdot\left(t\xi\cdot s\right)$

Then, differentiating both sides, and noting that $\left.R_{t\xi\cdot
s}^{-1}\left(p\left(t\right)\right)\right|_{t=0}=p/s$, we have

$\displaystyle\left.\frac{d}{dt}p\left(t\right)\right|_{t=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}\left(R_{t\xi\cdot
s}^{-1}\left(p\left(t\right)\right)\right)\cdot\left(t\xi\cdot
s\right)\right|_{t=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}\left(R_{t\xi\cdot
s}^{-1}\left(p\left(t\right)\right)\right)\cdot
s\right|_{t=0}+\left.\frac{d}{dt}\left(\left(p/s\right)\cdot\left(t\xi\cdot
s\right)\right)\right|_{t=0}$

Applying $\rho_{s}^{-1}$ to both sides, and noting that
$\left.dR_{s}\right|_{0}=\rho_{s}$, we find

$\left.\frac{d}{dt}\left(R_{t\xi\cdot
s}^{-1}\left(p\left(t\right)\right)\right)\right|_{t=0}=\left.\frac{d}{dt}p\left(t\right)/s\right|_{t=0}-\left.\frac{d}{dt}\left(p/s\right)\circ_{s}\left(t\xi\right).\right|_{t=0}$

###### Remark 2.18

Lemma 2.17 shows that for each $s\in\mathbb{L},$ we may define a _local loop
structure_ on __ $\mathfrak{l}$. Indeed, suppose $\xi,\eta\in\mathfrak{l}$ are
in a sufficiently small neighborhood of $0\in\mathfrak{l}.$ Then, define

$\eta\circ_{s}\xi=\left(\eta\cdot\left(\xi\cdot
s\right)\right)/s\in\mathfrak{l.}$ (2.8)

With respect to this product, we see that $0\in\mathfrak{l}$ is a two-sided
identity element. The right quotient with respect to $\circ_{s}$ is defined by

$\xi/_{s}\eta=\left(\xi\cdot s\right)/\left(\eta\cdot s\right),$ (2.9)

while the left quotient is given by

$\xi\backslash_{s}\eta=\left(\xi\backslash\left(\eta\cdot s\right)\right)/s.$
(2.10)

Equivalently, we may define a product on a neighborhood of $s$. Given $p,q$ in
a sufficiently small neighborhood of $s\in\mathbb{L}$, define

$p\circ_{s}q=\left(p/s\right)\cdot q.$ (2.11)

Note that if $p=\xi\cdot s$ and $q=\eta\cdot s$, then
$p\circ_{s}q=\left(\xi\circ_{s}\eta\right)\cdot s.$ Similarly, the quotients
are also defined.

###### Lemma 2.19

The product $\circ_{s}$ is power-associative. In particular, for any
$\xi\in\mathfrak{l},$ the powers $\xi^{n}=n\xi$ are defined unambiguously and
independently of $s$.

Proof. Let $\xi\in\mathfrak{l.}$ Firstly, for real numbers $t_{1}$ and $t_{2}$
consider

$\displaystyle\left(t_{1}\xi\right)\circ_{s}\left(t_{2}\xi\right)$
$\displaystyle=$
$\displaystyle\left(\left(t_{1}\xi\right)\cdot\left(\left(t_{2}\xi\right)\cdot
s\right)\right)/s$ $\displaystyle=$
$\displaystyle\left(t_{1}+t_{2}\right)\xi.$

Hence $\xi\circ_{s}\xi=2\xi.$ Now, for third powers

$\left(\xi\circ_{s}\xi\right)\circ_{s}\xi=\left(2\xi\right)\circ_{s}\xi=3\xi=\xi\circ_{s}\left(\xi\circ_{s}\xi\right).$

By induction we can see that for any integer power $n$, $\xi^{n}=n\xi.$

###### Theorem 2.20

Suppose $\mathbb{L}$ is connected and $s\in\mathbb{L},$ then any element
$p\in\mathbb{L}$ can be written as
$p=\xi_{1}\cdot\left(\xi_{2}\cdot...\left(\xi_{k}\cdot s\right)\right),$ for
some finite sequence $\left\\{\xi_{1},...,\xi_{k}\right\\}$ in $\mathfrak{l}.$

Proof. Let $W\subset\mathbb{L}$ be the subset of elements of $\mathbb{L}$ that
can be represented as a product of $s$ by some finite sequence of elements of
$\mathfrak{l}$ on the left. We will show that $W$ is both closed and open, and
hence $\mathbb{L=}W.$

First note that since for each $p\in\mathbb{L}$, $R_{p}$ is a local
diffeomorphism of a neighborhood of $0\in\mathfrak{l}$ to a neighborhood of
$p\in\mathbb{L},$ we see that there exists an open neighborhood $V_{p}$ of
$p,$ where each $q\in V_{p}$ is given by $\eta\cdot p$ for some
$\eta\in\mathbb{L}.$

Now let $w\in W,$ then by the above, there exists an open neighborhood $V_{w}$
of $w,$ where each $q\in V_{w}$ is given by $\eta\cdot w.$ In particular,
$V_{w}\subset W,$ and hence $W$ is open.

To show that $W$ is closed, consider a converging sequence
$p_{i}\longrightarrow p$ with each $p_{i}\in W.$ Then, for $N$ large enough,
we will have $p_{N}\in V_{p}.$ In other words, we will have $p_{N}=\eta\cdot
p,$ for some $\eta\in\mathfrak{l}.$ This implies that
$p=\left(-\eta\right)\cdot p_{N}\in W,$ and thus $W$ is closed.

###### Remark 2.21

The above result shows that given a fixed $s\in\mathbb{L},$ any other element
of $\mathbb{L}$ may be reached in a finite number of “steps”, i.e. left
multiplications by elements of $\mathfrak{l}.$ Thus, we may define a discrete
“distance” function $d\left(p,q\right)$ for points $p,q\in\mathbb{L}$ given by
the minimum number of steps needed to reach $q$ from $p.$

Given a trivialization $\rho$ and fixing a point $s\in\mathbb{L},$ we induce a
bracket on $\mathfrak{l}.$ Let $\xi,\eta\in\mathfrak{l}.$ Then,

$\left[\xi,\eta\right]^{\left(\rho,s\right)}=-\rho_{s}^{-1}\left(\left.\left[\rho\left(\xi\right),\rho\left(\eta\right)\right]\right|_{s}\right).$
(2.12)

Note that here we have the negative sign to be compatible with [6] and also to
follow the same convention as right-invariant vector fields on Lie groups. As
before, if $\left\\{X_{i}\right\\}_{i=1}^{n}$ is a frame of $T\mathbb{L},$
suppose the brackets of the frame elements are given by

$\left[X_{i},X_{j}\right]=c_{\ ij}^{k\ }X_{k},$ (2.13)

where $c_{\ ij}^{k}$ are smooth $\mathbb{F}$-valued functions on $M.$ Then,

$\displaystyle\left[\xi,\eta\right]^{\left(\rho,s\right)}$ $\displaystyle=$
$\displaystyle-\rho_{s}^{-1}\left(\left.\left[\xi^{i}X_{i},\eta^{j}X_{j}\right]\right|_{s}\right).$
(2.14) $\displaystyle=$ $\displaystyle-\rho_{s}^{-1}\left(\xi^{i}\eta^{j}c_{\
ij}^{k}\left(s\right)\left.X_{k}\right|_{s}\right),$ (2.15)

and therefore

$\left[\xi,\eta\right]^{\left(\rho,s\right)}=-\xi^{i}\eta^{j}c_{\
ij}^{k}\left(s\right)e_{k}$ (2.16)

Define the bracket function

$b^{\left(\rho\right)}:\mathbb{L}\longrightarrow\Lambda^{2}\mathfrak{l}^{\ast}\otimes\mathfrak{l}$
(2.17)

given by

$b^{\left(\rho\right)}\left(s\right)\left(\xi,\eta\right)=\left[\xi,\eta\right]^{\left(\rho,s\right)}$

for any $s\in\mathbb{L}$ and $\xi,\eta\in\mathfrak{l}.$ We can also define the
$\mathop{\mathrm{a}d}$ map:

$\mathop{\mathrm{a}d}_{\xi}^{\left(\rho,s\right)}:\mathfrak{l}\longrightarrow\mathfrak{l}$

by

$\mathop{\mathrm{a}d}_{\xi}^{\left(\rho,s\right)}\left(\eta\right)=\left[\xi,\eta\right]^{\left(\rho,s\right)}.$

###### Definition 2.22

Let the _Maurer-Cartan form_
$\theta^{\left(\rho\right)}\in\Omega^{1}\left(\mathbb{L},\mathfrak{l}\right)$
be an $\mathfrak{l}$-valued $1$-form, defined by

$\left.\theta^{\left(\rho\right)}\right|_{s}\left(X\right)=\rho_{s}^{-1}\left(X\right)$
(2.18)

for any $X\in T_{s}M,$ where $s\in M.$ In particular, for any
$\eta\in\mathfrak{l},$ given a fundamental vector field
$\rho\left(\eta\right),$

$\theta^{\left(\rho\right)}\left(\rho\left(\eta\right)\right)=\eta.$ (2.19)

In particular, we can now rewrite (2.12) as

$b^{\left(\rho\right)}\left(\theta^{\left(\rho\right)}\left(X\right),\theta^{\left(\rho\right)}\left(Y\right)\right)=-\theta^{\left(\rho\right)}\left(\left[X,Y\right]\right).$
(2.20)

###### Remark 2.23

To simplify notation, we will drop $\left(\rho\right)$ from $b$ and $\theta.$
We still need to remember that these objects depend on the trivialization
$\rho,$ but once the trivialization is fixed, it is not necessary to keep
writing it.

######  Theorem 2.24 (Structural Equation)

The $1$-form $\theta$ satisfies the following relation.

$d\theta-\frac{1}{2}b\left(\theta,\theta\right)=0,$ (2.21)

where $b\left(\theta,\theta\right)$ is the bracket of $\mathfrak{l}$-valued
$1$-forms such that for any $X,Y\in T_{p}\mathbb{L}$,
$\frac{1}{2}\left.b\left(\theta,\theta\right)\right|_{s}\left(X,Y\right)=\left[\theta\left(X\right),\theta\left(Y\right)\right]^{\left(s\right)}.$

Proof. It is sufficient to check this for fundamental sections of
$T\mathbb{L}.$ Let $X=\rho\left(\xi\right)$ and $Y=\rho\left(\eta\right)$ for
$\xi,\eta\in\mathfrak{l}.$ Then from (2.18),

$\displaystyle\left(d\theta\right)\left(X,Y\right)$ $\displaystyle=$
$\displaystyle
X\left(\theta\left(Y\right)\right)-Y\left(\theta\left(X\right)\right)-\theta\left(\left[X,Y\right]\right)$
$\displaystyle=$ $\displaystyle
X\left(\eta\right)-Y\left(\xi\right)-\theta\left(\left[X,Y\right]\right)$
$\displaystyle=$ $\displaystyle
b\left(\theta\left(X\right),\theta\left(Y\right)\right),$

where we have used (2.20) and the fact that $\eta$ and $\xi$ are constant.

Applying the exterior derivative to the structural equation, we find the
generalization of the Jacobi identity:

######  Theorem 2.25 (Generalized Jacobi identity)

The bracket $b$ and the Maurer-Cartan form $\theta$ satisfy:

$b\left(\theta,b\left(\theta,\theta\right)\right)=\left(db\right)\left(\theta,\theta\right)\text{,}$
(2.22)

where exterior derivatives are implied.

Define the _skew-associator_ map
$a:\mathbb{L}\longrightarrow\Lambda^{2}\mathfrak{l}^{\ast}\otimes\mathfrak{l}^{\ast}\otimes\mathfrak{l}$
such that for $s\in\mathbb{L}$ and $\xi,\eta,\gamma\in\mathfrak{l,}$

$a^{\left(s\right)}\left(\xi,\eta,\gamma\right)=\left.d_{\rho\left(\gamma\right)}b\left(\xi,\eta\right)\right|_{p}.$
(2.23)

Suppose $\xi,\eta,\gamma\in\mathfrak{l}.$ Then, substituting
$\rho\left(\xi\right),\rho\left(\eta\right),\rho\left(\gamma\right)\in\Gamma\left(T\mathbb{L}\right)$
into (2.22) and evaluating at a point $s\in\mathbb{L}$ gives

$\displaystyle\left[\xi,\left[\eta,\gamma\right]^{\left(s\right)}\right]^{\left(s\right)}+\left[\eta,\left[\gamma,\xi\right]^{\left(s\right)}\right]^{\left(s\right)}+\left[\gamma,\left[\xi,\eta\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle
a^{\left(s\right)}\left(\xi,\eta,\gamma\right)$
$\displaystyle+a^{\left(s\right)}\left(\eta,\gamma,\xi\right)$
$\displaystyle+a^{\left(s\right)}\left(\gamma,\xi,\eta\right).$

We thus see that the obstruction to satisfying the Jacobi identity is
precisely the cyclic permutation of $a$ or equivalently $db.$ In particular,
if $b$ is constant, then $\mathfrak{l}$ is a Lie algebra. In this case, the
bracket is independent of $p,$ and $\mathfrak{l}$ has a unique bracket that
satisfies the Jacobi identity.

Both the bracket and the skew-associator maps can be related to the product
$\circ_{s}$on $\mathfrak{l}.$ In particular, as we see below, the bracket is
the infinitesimal commutator of $\circ_{s}$ and $a^{\left(s\right)}$ is a
skew-symmetrization of the infinitesimal associator of $\circ_{s}$.

###### Theorem 2.26

Let $\xi,\eta,\gamma\in\mathfrak{l},$ and $s\in\mathbb{L}$, then

$\displaystyle\left[\xi,\eta\right]^{\left(s\right)}$ $\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{dt_{1}dt_{2}}\left(\left(t_{1}\xi\right)\circ_{s}\left(t_{2}\eta\right)-\left(t_{2}\eta\right)\circ_{s}\left(t_{1}\xi\right)\right)\right|_{t_{1}=t_{2}=0}$
(2.25a) $\displaystyle a^{\left(s\right)}\left(\xi,\eta,\gamma\right)$
$\displaystyle=$
$\displaystyle\left[\xi,\eta,\gamma\right]^{\left(s\right)}-\left[\eta,\xi,\gamma\right]^{\left(s\right)}\text{,}$
(2.25b)

where

$\left[\xi,\eta,\gamma\right]^{\left(s\right)}=\left.\frac{d^{3}}{dt_{1}dt_{2}dt_{3}}\left(\left(t_{1}\xi\right)\circ_{s}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{3}\gamma\right)\right)-\left(\left(t_{1}\xi\right)\circ_{s}\left(t_{2}\eta\right)\right)\circ_{s}\left(t_{3}\gamma\right)\right)\right|_{t_{1}=t_{2}=t_{3}=0}.$

Proof. By definition (2.12), we have

$\left[\xi,\eta\right]^{\left(s\right)}=-\rho_{s}^{-1}\left(\left.\left[X,Y\right]\right|_{s}\right).$
(2.26)

where $X=\rho\left(\xi\right)$ and $Y=\rho\left(\eta\right)$ are vector fields
on $\mathbb{L}.$ However, the bracket of two vector fields $X$ and $Y$ is
given by

$\left[X,Y\right]_{s}=\left.\frac{d}{dt}\left(\left(\Phi_{\xi,t}^{-1}\right)_{\ast}\left(Y_{\Phi_{\xi,t}\left(s\right)}\right)\right)\right|_{t=0},$
(2.27)

where $\Phi_{\xi,t}$ is the flow generated by $X,$ and hence,
$\Phi_{\xi,t}\left(s\right)=\left(t\xi\right)\cdot s.$ Therefore

$Y_{\Phi_{\xi,t}\left(s\right)}=\rho_{\left(t\xi\cdot
s\right)}\left(\eta\right).$

Hence

$\left(\Phi_{\xi,t}^{-1}\right)_{\ast}\left(Y_{\Phi_{\xi,t}\left(s\right)}\right)=\left(L_{t\xi}^{-1}\right)_{\ast}\rho_{\left(t\xi\cdot
s\right)}\left(\eta\right),$ (2.28)

and from Lemma 2.17, $L_{t\xi}^{-1}=L_{-t\xi}$. Then, differentiating, we find
that

$\left.\frac{d}{dt}\left(\left(L_{-t\xi}\right)_{\ast}\rho_{\left(t\xi\cdot
s\right)}\left(\eta\right)\right)\right|_{t=0}=\left.\frac{d}{dt}\left(\rho_{\left(t\xi\cdot
s\right)}\left(\eta\right)\right)\right|_{t=0}-\left.\frac{d}{dt}\left(\left(L_{t\xi}\right)_{\ast}\rho_{\left(s\right)}\left(\eta\right)\right)\right|_{t=0}.$

Also from Lemma 2.17, we know that
$\rho_{\left(s\right)}=\left.\left(R_{s}\right)_{\ast}\right|_{0}.$ Hence,
since for some parameter $t_{2}$, $t_{2}\eta$ is a path in $\mathfrak{l}$
through $0$ with tangent vector $\eta$ at $0$, we can rewrite

$\displaystyle\rho_{\left(t\xi\cdot s\right)}\left(\eta\right)$
$\displaystyle=$ $\displaystyle\left.\left(R_{t\xi\cdot
s}\right)_{\ast}\right|_{0}\left(\eta\right)$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt_{2}}\left(\left(t_{2}\eta\right)\cdot\left(t\xi\cdot
s\right)\right)\right|_{t_{2}=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt_{2}}R_{s}\left(\left(t_{2}\eta\right)\circ_{s}\left(t\xi\right)\right)\right|_{t_{2}=0}$

and similarly,

$\displaystyle\left(L_{t\xi}\right)_{\ast}\rho_{\left(s\right)}\left(\eta\right)$
$\displaystyle=$
$\displaystyle\left.\left(L_{t\xi}\right)_{\ast}\right|_{s}\left.\left(R_{s}\right)_{\ast}\right|_{0}\left(\eta\right)$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt_{2}}\left(t\xi\right)\cdot\left(\left(t_{2}\eta\right)\cdot
s\right)\right|_{t_{2}=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt_{2}}R_{s}\left(\left(t\xi\right)\circ_{s}\left(t_{2}\eta\right)\right)\right|_{t_{2}=0}$

Overall,

$\left.\left[X,Y\right]\right|_{s}=\left(R_{s}\right)_{\ast}\left.\frac{d^{2}}{dt_{1}dt_{2}}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{1}\xi\right)-\left(t_{1}\xi\right)\circ_{s}\left(t_{2}\eta\right)\right)\right|_{t_{1}=t_{2}=0},$

and therefore, we indeed get (2.25a).

Now consider

$\left.\frac{d^{3}}{dt_{1}dt_{2}dt_{3}}\left(t_{1}\xi\right)\circ_{\left(t_{3}\gamma\cdot
s\right)}\left(t_{2}\eta\right)\right|_{t_{1}=t_{2}=t_{3}=0}.$

Expanding, we get

$\displaystyle\left(t_{1}\xi\right)\circ_{\left(t_{3}\gamma\cdot
s\right)}\left(t_{2}\eta\right)$ $\displaystyle=$
$\displaystyle\left(R_{t_{3}\gamma\cdot
s}\right)^{-1}\left(\left(t_{1}\xi\right)\cdot\left(\left(t_{2}\eta\right)\cdot\left(t_{3}\gamma\cdot
s\right)\right)\right)$ $\displaystyle=$
$\displaystyle\left(R_{t_{3}\gamma\cdot
s}\right)^{-1}\left(\left(\left(t_{1}\xi\right)\circ_{s}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{3}\gamma\right)\right)\right)\cdot
s\right).$

Using (2.7) with
$p\left(t_{3}\right)=\left(\left(t_{1}\xi\right)\circ_{s}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{3}\gamma\right)\right)\right)\cdot
s$, we find

$\displaystyle\left.\frac{d}{dt_{3}}\left(R_{t_{3}\gamma\cdot
s}\right)^{-1}\left(\left(\left(t_{1}\xi\right)\circ_{s}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{3}\gamma\right)\right)\right)\cdot
s\right)\right|_{t_{3}=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt_{3}}\left(t_{1}\xi\right)\circ_{s}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{3}\gamma\right)\right)\right|_{t_{3}=0}$
$\displaystyle-\left.\frac{d}{dt_{3}}\left(\left(t_{1}\xi\right)\circ_{s}\left(t_{2}\eta\right)\right)\circ_{s}\left(t_{3}\gamma\right)\right|_{t_{3}=0}.$

Hence,

$\left.\frac{d^{3}}{dt_{1}dt_{2}dt_{3}}\left(t_{1}\xi\right)\circ_{\left(t_{3}\gamma\cdot
s\right)}\left(t_{2}\eta\right)\right|_{t_{1}=t_{2}=t_{3}=0}=\left[\xi,\eta,\gamma\right]^{\left(s\right)}.$

Consider now

$\left.d_{\rho\left(\gamma\right)}b\left(\xi,\eta\right)\right|_{s}=a^{\left(s\right)}\left(\xi,\eta,\gamma\right),$

where we use the definition of $a^{\left(s\right)}$ (2.23). However, from
(2.25a),

$\displaystyle\left.d_{\rho\left(\gamma\right)}b\left(\xi,\eta\right)\right|_{s}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt_{3}}\left[\xi,\eta\right]^{t_{3}\gamma\cdot
s}\right|_{t_{3}=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d^{3}}{dt_{1}dt_{2}dt_{3}}\left(\left(t_{1}\xi\right)\circ_{\left(t_{3}\gamma\cdot
s\right)}\left(t_{2}\eta\right)-\left(t_{2}\eta\right)\circ_{\left(t_{3}\gamma\cdot
s\right)}\left(t_{2}\xi\right)\right)\right|_{t_{1}=t_{2}=t_{3}=0}$
$\displaystyle=$
$\displaystyle\left[\xi,\eta,\gamma\right]^{\left(s\right)}-\left[\eta,\xi,\gamma\right]^{\left(s\right)}.$

So indeed,

$a^{\left(s\right)}\left(\xi,\eta,\gamma\right)=\left[\xi,\eta,\gamma\right]^{\left(s\right)}-\left[\eta,\xi,\gamma\right]^{\left(s\right)}.$

Consider now the connection $\nabla$ from (2.3). Its torsion
$T\left(X,Y\right)$ for vector fields $X$ and $Y$ is given by

$T\left(X,Y\right)=\nabla_{X}Y-\nabla_{Y}X-\left[X,Y\right],$

so in particular for $X=\rho\left(\xi\right)$ and $Y=\rho\left(\eta\right),$
for $\xi,\eta\in\mathfrak{l}$, we have

$T\left(\rho\left(\xi\right),\rho\left(\eta\right)\right)=-\left[\rho\left(\xi\right),\rho\left(\eta\right)\right].$

Hence, using (2.12), we find that at a point $s\in\mathbb{L}$, we have

$T\left(\rho\left(\xi\right),\rho\left(\eta\right)\right)_{s}=\rho_{s}\left(\left[\xi,\eta\right]^{\left(\rho,s\right)}\right).$
(2.29)

Now let $Z=\rho\left(\gamma\right)$ for $\gamma\in\mathfrak{l,}$ hence

$g_{s}\left(T\left(\rho\left(\xi\right),\rho\left(\eta\right)\right)_{s},\rho_{s}\left(\gamma\right)\right)=\left\langle\left[\xi,\eta\right]^{\left(\rho,s\right)},\gamma\right\rangle.$
(2.30)

Recall that a torsion of a metric connection is said to be totally skew-
symmetric if for any vector fields $X,Y,Z,$

$g\left(T\left(X,Y\right),Z\right)=-g\left(Y,T\left(X,Z\right)\right).$ (2.31)

From (2.30), we see that the torsion of the flat connection $\nabla$ is
totally skew-symmetric if and only if for any $s\in\mathbb{L}$,

$\left\langle\left[\xi,\eta\right]^{\left(\rho,s\right)},\gamma\right\rangle=-\left\langle\eta,\left[\xi,\gamma\right]^{\left(\rho,s\right)}\right\rangle.$
(2.32)

Equivalently, for any $\xi\in\mathfrak{l}$ and any $s\in\mathbb{L},$ the map
$\mathop{\mathrm{a}d}_{\xi}^{\left(\rho,s\right)}$ is skew-adjoint with
respect to the inner product on $\mathfrak{l}.$

Moreover, consider $\nabla T.$ Then,

$\displaystyle\left(\nabla_{\rho\left(\gamma\right)}T\right)\left(\rho\left(\xi\right),\rho\left(\eta\right)\right)$
$\displaystyle=$
$\displaystyle\nabla_{\rho\left(\gamma\right)}\left(T\left(\rho\left(\xi\right),\rho\left(\eta\right)\right)\right)$
$\displaystyle=$
$\displaystyle\nabla_{\rho\left(\gamma\right)}\left(\rho\left(b\left(\xi,\eta\right)\right)\right)$
$\displaystyle=$
$\displaystyle\rho\left(d_{\rho\left(\gamma\right)}\left(b\left(\xi,\eta\right)\right)\right)$
$\displaystyle=$
$\displaystyle\rho\left(a\left(\xi,\eta,\gamma\right)\right).$

A classical theorem by Cartan and Schouten [4] (and [1] for a more modern
treatment) states the following.

######  Theorem 2.27 ([4])

Let $\left(M,g\right)$ be a simply-connected, complete irreducible Riemannian
manifold with a flat metric connection $\nabla$ with a non-zero totally skew-
symmetric torsion $T.$ Then

  1. 1.

If $\nabla T=0,$ then $M$ is isometric to a compact simple Lie groups.

  2. 2.

If $\nabla T\neq 0$, then $M$ is isometric to $S^{7}.$

In our formulation, we have the following corollary.

###### Corollary 2.28

Suppose $\left(\mathbb{L},\mathfrak{l},\rho\right)$ is a parallelized
manifold. If for $s\in\mathbb{L}$, there exists a smooth family of inner
products $\left\langle\cdot,\cdot\right\rangle_{s}$ on $\mathfrak{l}$ such
that the bracket $\left[\cdot,\cdot\right]^{\left(\rho,s\right)}$ is skew-
adjoint with respect to it, then there exists a parallelization $\tilde{\rho}$
of $\mathbb{L}$ and an inner product $\left\langle\cdot,\cdot\right\rangle$ on
$\mathfrak{l},$ such that the Riemannian manifold $\left(\mathbb{L},g\right)$
with $g=\left\langle\tilde{\rho}^{-1},\tilde{\rho}^{-1}\right\rangle$ is
isometric to a product of compact simple Lie groups and copies of $S^{7}.$

Proof. Suppose for each $s\in\mathbb{L},$
$\left\langle\cdot,\cdot\right\rangle_{s}$ is such that for any
$\xi,\eta,\gamma\in\mathfrak{l},$

$\left\langle\left[\xi,\eta\right]^{\left(\rho,s\right)},\gamma\right\rangle_{s}=-\left\langle\eta,\left[\xi,\gamma\right]^{\left(\rho,s\right)}\right\rangle_{s}.$

Then, choosing any inner product $\left\langle\cdot,\cdot\right\rangle$ on
$\mathfrak{l},$ we can write
$\left\langle\xi,\eta\right\rangle_{s}=\left\langle
Q_{s}^{-1}\xi,Q_{s}^{-1}\eta\right\rangle,$ for some map $Q_{s}\in
GL\left(\mathfrak{l}\right).$ Since the family of inner products is smooth, we
obtain a smooth family of maps $Q_{s}.$ Now define a new parallelization of
$\mathbb{L}$ given by $\tilde{\rho}_{s}=\rho_{s}\circ Q_{s}.$ Then,

$\displaystyle\left.g\left(X,Y\right)\right|_{s}$ $\displaystyle=$
$\displaystyle\left\langle\tilde{\rho}_{s}^{-1}X,\tilde{\rho}_{s}^{-1}Y\right\rangle$
$\displaystyle=$ $\displaystyle\left\langle
Q_{s}^{-1}\rho_{s}^{-1}X,Q_{s}^{-1}\rho_{s}^{-1}Y\right\rangle$
$\displaystyle=$
$\displaystyle\left\langle\rho_{s}^{-1}X,\rho_{s}^{-1}Y\right\rangle_{s}.$

In particular, note that by definition (2.12)

$\displaystyle\left[\xi,\eta\right]^{\left(\tilde{\rho},s\right)}$
$\displaystyle=$
$\displaystyle-\tilde{\rho}_{s}^{-1}\left(\left.\left[\tilde{\rho}\left(\xi\right),\tilde{\rho}\left(\eta\right)\right]\right|_{s}\right)$
$\displaystyle=$ $\displaystyle-
Q_{s}^{-1}\rho_{s}^{-1}\left(\left.\left[\rho\left(Q\left(\xi\right)\right),\rho\left(Q\left(\eta\right)\right)\right]\right|_{s}\right)$
$\displaystyle=$ $\displaystyle
Q_{s}^{-1}\left[Q_{s}\left(\xi\right),Q_{s}\left(\eta\right)\right]^{\left(\rho,s\right)}$

and hence

$\displaystyle\left\langle\left[\xi,\eta\right]^{\left(\tilde{\rho},s\right)},\gamma\right\rangle$
$\displaystyle=$ $\displaystyle\left\langle
Q_{s}^{-1}\left[Q_{s}\left(\xi\right),Q_{s}\left(\eta\right)\right]^{\left(\rho,s\right)},\gamma\right\rangle$
$\displaystyle=$
$\displaystyle\left\langle\left[Q_{s}\left(\xi\right),Q_{s}\left(\eta\right)\right]^{\left(\rho,s\right)},Q_{s}\left(\gamma\right)\right\rangle$
$\displaystyle=$ $\displaystyle-\left\langle
Q_{s}\left(\eta\right),\left[Q_{s}\left(\xi\right),Q_{s}\left(\gamma\right)\right]^{\left(\rho,s\right)}\right\rangle$
$\displaystyle=$
$\displaystyle-\left\langle\eta,\left[\xi,\gamma\right]^{\left(\tilde{\rho},s\right)}\right\rangle.$

In particular, we now see that with $g$ and $\nabla$ defined using
$\tilde{\rho},$ the torsion of $\nabla$ satisfies (2.31), and hence by Theorem
2.27, $\left(\mathbb{L},g\right)$ is isometric to a product of compact simple
Lie groups and copies of $S^{7}.$

##  3 Automorphisms

###### Definition 3.1

The category of (complete, connected) parallelized manifolds is defined in the
following way. An object in this category is the triple
$\left(\mathbb{L},\mathfrak{l},\rho\right),$ where $\mathbb{L}$ is an
$n$-dimensional parallelizable connected manifold, $\mathfrak{l}$ is an
$n$-dimensional vector space, and $\rho$ is the trivialization map, such that
the fundamental vector fields are complete.

Suppose $\left(\mathbb{L}_{1},\mathfrak{l}_{1},\rho_{1}\right)$ and
$\left(\mathbb{L}_{2},\mathfrak{l}_{2},\rho_{2}\right)$ are two parallelized
manifolds, then a morphism between them is a smooth map
$h:\mathbb{L}_{1}\longrightarrow\mathbb{L}_{2}\ $for which there exists a
linear map $h^{\prime}:\mathfrak{l}_{1}\longrightarrow\mathfrak{l}_{2}$ such
that the following diagram commutes

${T_{p}\mathbb{L}_{1}}$${T_{h(p)}\mathbb{L}_{2}}$${\mathfrak{l}_{1}}$${\mathfrak{l}_{2}}$$\scriptstyle{\left.h_{\ast}\right|_{p}}$$\scriptstyle{\left(\rho_{1}\right)_{p}}$$\scriptstyle{h^{\prime}}$$\scriptstyle{\left(\rho_{2}\right)_{h\left(p\right)}}$
(3.1)

We will refer to $\left(h,h^{\prime}\right)$ as a compatible pair.

###### Remark 3.2

the notation $h^{\prime}$ is appropriate since this map is closely related to
the derivative of $h.$ Any smooth $h$ will induce an invertible linear map
$h_{p}^{\prime}=\rho_{f\left(p\right)}^{-1}\circ\left.h_{\ast}\right|_{p}\circ\rho_{p}:\mathfrak{l}_{1}\longrightarrow\mathfrak{l}_{2},$
however in general it will depend on $p.$ The key point of the above
definition is that for a morphism of parallelized manifolds, $h_{p}^{\prime}$
is independent of $p$. A morphism between parallelized manifolds is a special
case of a vector bundle morphism. Indeed, consider the tangent bundles
$T\mathbb{L}_{1}$ and $T\mathbb{L}_{2}.$ Then, the bundle map
$\left(h,\rho_{2}\circ
h^{\prime}\circ\rho_{1}^{-1}\right):T\mathbb{L}_{1}\longrightarrow
T\mathbb{L}_{2}$ covers $h:\mathbb{L}_{1}\longrightarrow\mathbb{L}_{2}$ and
induces a linear map of each fiber.

###### Example 3.3

Suppose we have a parallelized manifold
$\left(\mathbb{L},\mathfrak{l},\rho_{1}\right)$, then any linear isomorphism
of the vector space $\mathfrak{l}$ induces an isomorphism of parallelized
manifolds. Indeed, suppose
$h^{\prime}:\mathfrak{l}\longrightarrow\mathfrak{l}$ is an invertible linear
map. Then, the compatible pair $\left(\mathop{\mathrm{i}d},h^{\prime}\right)$,
where $\mathop{\mathrm{i}d}$ is the identity map on $\mathbb{L}$, give an
isomorphism from $\left(\mathbb{L},\mathfrak{l},\rho_{1}\right)$ to
$\left(\mathbb{L},\mathfrak{l},\rho_{2}\right),$ where
$\rho_{2}=\rho_{1}\circ\left(h^{\prime}\right)^{-1}.$ This just corresponds to
a $GL\left(n\right)$ transformation of the global frame on $T\mathbb{L}.$

###### Lemma 3.4

Suppose $\left(\mathbb{L}_{1},\mathfrak{l}_{1},\rho_{1}\right)$ and
$\left(\mathbb{L}_{2},\mathfrak{l}_{2},\rho_{2}\right)$ are two parallelized
manifolds. A pair $\left(h,h^{\prime}\right)$ of a smooth map
$h:\mathbb{L}_{1}\longrightarrow\mathbb{L}_{2}$ and a linear map
$h^{\prime}:\mathfrak{l}_{1}\longrightarrow\mathfrak{l}_{2}$ satisfy the
definition of a morphism of parallelized manifolds if and only if for any
$s\in\mathbb{L}_{1}$ and $\xi\in\mathfrak{l}_{1}$,

$h\left(\xi\cdot s\right)=h^{\prime}\left(\xi\right)\cdot h\left(s\right).$
(3.2)

Proof. Let $\xi\in\mathfrak{l}_{1},$ and let $s\in\mathbb{L}.$ Recall that
$\xi\cdot s=\gamma_{\rho_{1}\left(\xi\right),s}\left(1\right)$ is the solution
$x\left(1\right)$ at $t=1$ of the initial value problem

$\left\\{\begin{array}[]{c}\frac{dx\left(t\right)}{dt}=\rho_{1}\left(\xi\right)_{x\left(t\right)}\\\
x\left(0\right)=s\end{array}\right..$

Then, applying $h_{\ast}$ to the equation, we get

$\displaystyle\left.h_{\ast}\right|_{x\left(t\right)}\left(\frac{dx\left(t\right)}{dt}\right)$
$\displaystyle=$
$\displaystyle\frac{d\left(h\left(x\left(t\right)\right)\right)}{dt}$
$\displaystyle=$
$\displaystyle\left.h_{\ast}\right|_{x\left(t\right)}\left(\rho_{1}\left(\xi\right)_{x\left(t\right)}\right)$

Suppose now the pair $\left(h,h^{\prime}\right)$ satisfies the definition of a
morphism of parallelized manifolds. Using the property of $h^{\prime}$ from
(3.1), we have

$\left.h_{\ast}\right|_{x\left(t\right)}\left(\rho_{1}\left(\xi\right)_{x\left(t\right)}\right)=\rho_{2}\left(h^{\prime}\left(\xi\right)\right)_{h\left(x\left(t\right)\right)}.$

Therefore, we see that $y=h\left(x\left(t\right)\right)$ satisfies the
equation

$\left\\{\begin{array}[]{c}\frac{dy\left(t\right)}{dt}=\rho_{2}\left(h^{\prime}\left(\xi\right)\right)_{y\left(t\right)}\\\
y\left(0\right)=h\left(s\right)\end{array}\right.$ (3.4)

and overall we see that
$y\left(t\right)=\gamma_{\rho_{2}\left(\xi\right),h\left(s\right)}\left(t\right)$
For $t=1,$ we hence have

$h\left(\xi\cdot s\right)=h^{\prime}\left(\xi\right)\cdot h\left(s\right).$

Conversely, suppose (3.2) holds for any $s\in\mathbb{L}_{1}$ and
$\xi\in\mathfrak{l}_{1}.$ Then, we have

$h\left(t\xi\cdot s\right)=\left(th^{\prime}\left(\xi\right)\right)\cdot
h\left(s\right).$ (3.5)

Differentiating both sides, we find

$\left.h_{\ast}\right|_{x\left(t\right)}\left(\rho_{1}\left(\xi\right)_{x\left(t\right)}\right)=\rho_{2}\left(h^{\prime}\left(\xi\right)\right)_{y\left(t\right)},$

where $x\left(t\right)=t\xi\cdot s$ and
$y\left(t\right)=\left(th^{\prime}\left(\xi\right)\right)\cdot
h\left(s\right)=h\left(x\left(t\right)\right)$. Setting $t=0$, we find that
the pair $\left(h,h^{\prime}\right)$ indeed satisfies the definition of a
morphism of parallelizable manifolds.

###### Example 3.5

In the case of Lie groups, the definition of a morphism of parallelized
manifolds is equivalent to taking the pair of a Lie group homomorphism and its
differential at identity.

###### Definition 3.6

Suppose $\left(\mathbb{L},\mathfrak{l},\rho\right)$ is a parallelized
manifold. Then, an automorphism of $\left(\mathbb{L},\mathfrak{l},\rho\right)$
is a diffeomorphism $h:\mathbb{L}\longrightarrow\mathbb{L}$ for which there
exists an invertible linear map
$h^{\prime}:\mathfrak{l}\longrightarrow\mathfrak{l}$, such that the following
diagram commutes

${T_{p}\mathbb{L}}$${T_{h(p)}\mathbb{L}}$${\mathfrak{l}}$${\mathfrak{l}}$$\scriptstyle{\left.h_{\ast}\right|_{p}}$$\scriptstyle{\left(\rho\right)_{p}}$$\scriptstyle{h^{\prime}}$$\scriptstyle{\left(\rho\right)_{h\left(p\right)}}$
(3.6)

###### Remark 3.7

Given a trivialization $\rho,$ we may define the trivial connection $\nabla$
on $T\mathbb{L}$ via

$\nabla=\rho\circ d\circ\rho^{-1},$ (3.7)

where $\rho$ is interpreted now as a map
$\rho:C^{\infty}\left(\mathbb{L},\mathfrak{l}\right)\longrightarrow\Gamma\left(T\mathbb{L}\right).$
Given a diffeomorphism $h$ of $\mathbb{L},$ the pullback connection
$h^{\ast}\nabla$ is defined via

$\displaystyle h^{\ast}\nabla$ $\displaystyle=$ $\displaystyle
h_{\ast}^{-1}\circ\nabla\circ h_{\ast}$ $\displaystyle=$ $\displaystyle
h_{\ast}^{-1}\circ\rho\circ d\circ\rho^{-1}\circ h_{\ast}.$

Setting $h^{\prime}=\rho^{-1}\circ h_{\ast}\circ\rho,$ we get

$h^{\ast}\nabla=\rho^{-1}\circ\left(h^{\prime}\right)^{-1}\circ d\circ
h\circ\rho.$

Hence we see that $h^{\ast}\nabla=\nabla$ if and only if $h^{\prime}$ is
constant and equivalently $h$ is an automorphism of
$\left(\mathbb{L},\mathfrak{l},\rho\right).$ In other words, $h$ is a
connection-preserving diffeomorphism, or as it is also known as, an _affine
diffeomorphism_ for the trivial connection $\nabla.$

The set of automorphisms of $\left(\mathbb{L},\mathfrak{l},\rho\right)$ is
clearly non-empty, since the identity map is always an automorphism, and it’s
clearly a Lie group, with the Lie group structure induced from
$\mathop{\mathrm{D}iff}\left(\mathbb{L}\right)$. Let us denote it by
$\Psi\left(\mathbb{L}\mathfrak{,l,\rho}\right),$ or just $\Psi$ if there is no
ambiguity. Each $h\in\Psi\left(\mathbb{L}\mathfrak{,l,\rho}\right)$ defines a
corresponding linear isomorphism $h^{\prime}$ of $\mathfrak{l},$ so denote
this subgroup of $GL\left(\mathfrak{l}\right)$ by
$\Psi^{\prime}\left(\mathbb{L},\mathfrak{l,\rho}\right),$ or just
$\Psi^{\prime}$ if there is no ambiguity. Under an automorphism, a fundamental
vector field $\rho\left(\xi\right)$ is pushed forward to
$\rho\left(h^{\prime}\left(\xi\right)\right).$ The map $h\mapsto h^{\prime}$
is a Lie group homomorphism, so $\Psi^{\prime}$ is a Lie group of dimension at
most $n^{2}$ where $n=\dim\mathbb{L}.$

Let $\pi:\Psi\longrightarrow\Psi^{\prime}$ be the homomorphism that takes $h$
to $h^{\prime}.$ Let $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ be the Lie
algebras of $\Psi$ and $\Psi^{\prime},$ respectively, then at least for small
$t$,
$\pi\left(\exp_{\mathfrak{p}}\left(t\eta\right)\right)=\exp_{\mathfrak{p}^{\prime}}\left(t\pi_{\ast}\eta\right)$
for $\eta\in\mathfrak{p}.$

###### Theorem 3.8

Suppose $\mathbb{L}$ is a connected parallelizable manifold of dimension $n$.
The automorphism group $\Psi=\Psi\left(\mathbb{L}\mathfrak{,l,\rho}\right)$ of
$\left(\mathbb{L},\mathfrak{l},\rho\right)$ is a finite-dimensional Lie group
of dimension at most $n^{2}+n.$

Proof. The Lie group structure of $\Psi$ is clearly induced from
$\mathop{\mathrm{D}iff}\left(\mathbb{L}\right),$ so it is sufficient to show
that $\Psi$ is a finite-dimensional manifold. Fix any $s\in\mathbb{L}$, and
define the smooth map $\Psi\longrightarrow\Psi^{\prime}\times\mathbb{L}$ given
for any $h\in\Psi$ by

$h\mapsto\left(h^{\prime},h\left(s\right)\right)$ (3.9)

where $h^{\prime}\in GL\left(\mathfrak{l}\right)$ is the corresponding element
of $\Psi^{\prime}.$ We will show that this map is injective. Indeed, suppose
$\tilde{h}\in\Psi$ also maps to $\left(h^{\prime},h\left(s\right)\right).$
Then, consider $k=h^{-1}\tilde{h}\mapsto\left(\mathop{\mathrm{i}d},s\right).$
Recall from Theorem 2.20 that any element $p\in\mathbb{L}$ can be written as a
finite product $p=\xi_{1}\cdot\left(\xi_{2}\cdot...\left(\xi_{m}\cdot
s\right)\right)$ for $\xi_{i}\in\mathfrak{l}.$ Then, applying $k$ to $p$ using
Lemma 3.4 gives

$k\left(p\right)=k^{\prime}\left(\xi_{1}\right)\cdot\left(k^{\prime}\left(\xi_{2}\right)\cdot...\left(k^{\prime}\left(\xi_{m}\right)\cdot
k\left(s\right)\right)\right)=p,$

since $k^{\prime}=\mathop{\mathrm{i}d}$ and $k\left(s\right)=s$. Hence,
$k=\mathop{\mathrm{i}d}$, and hence (3.9) is an injective map.

Suppose $h_{t}\in\Psi\subset\mathop{\mathrm{D}iff}\left(\mathbb{L}\right),$ is
a smooth $1$-parameter family with $h_{0}=\mathop{\mathrm{i}d}.$ Then, for a
fixed $s\in\mathbb{L},$ $h_{t}\left(s\right)$ is a smooth curve on
$\mathbb{L},$ with
$u_{s}=\left.\frac{d}{dt}h_{t}\left(s\right)\right|_{t=0}\in T_{s}\mathbb{L}.$
The vector field $u\in\Gamma\left(T\mathbb{L}\right)$ is then the tangent
vector to the curve $h_{t}\in\Psi$ at $t=0$.

Recall from Theorem 2.20 that any element $p\in\mathbb{L}$ can be written as a
finite product $p=\xi_{1}\cdot\left(\xi_{2}\cdot...\left(\xi_{m}\cdot
s\right)\right)$ for $\xi_{i}\in\mathfrak{l},$ then

$\displaystyle u_{p}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}h_{t}\left(p\right)\right|_{t=0}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt}h_{t}\left(\xi_{1}\cdot\left(\xi_{2}\cdot...\left(\xi_{m}\cdot
s\right)\right)\right)\right|_{t=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}h_{t}^{\prime}\left(\xi_{1}\right)\cdot\left(h_{t}^{\prime}\left(\xi_{2}\right)\cdot...\left(h_{t}^{\prime}\left(\xi_{m}\right)\cdot
h_{t}\left(s\right)\right)\right)\right|_{t=0}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}h_{t}^{\prime}\left(\xi_{1}\right)\cdot\left(\xi_{2}\cdot...\left(\xi_{m}\cdot
s\right)\right)\right|_{t=0}+...$
$\displaystyle+\left.\frac{d}{dt}\xi_{1}\cdot\left(\xi_{2}\cdot...\left(h_{t}^{\prime}\left(\xi_{m}\right)\cdot
s\right)\right)\right|_{t=0}$
$\displaystyle+\left.\frac{d}{dt}\xi_{1}\cdot\left(\xi_{2}\cdot...\left(\xi_{m}\cdot\left(h_{t}\left(s\right)/s\cdot
s\right)\right)\right)\right|_{t=0}$

In particular, consider the linear map

$u\mapsto\left(\left.\frac{d}{dt}h_{t}^{\prime}\right|_{t=0},\left.\frac{d}{dt}h_{t}\left(s\right)/s\right|_{t=0}\right)\in\mathfrak{p}^{\prime}\times\mathfrak{l.}$
(3.11)

This is the differential of the map (3.9) at identity. From (3) we see that if
$\left.\frac{d}{dt}h_{t}^{\prime}\right|_{t=0}=0$ and
$\left.\frac{d}{dt}h_{t}\left(s\right)/s\right|_{t=0}=0$, then $u=0$. Thus, we
see that the map (3.11) is injective. Similarly, by translation, the
differential of (3.9) will be injective at any $h\in\Psi.$ Therefore, (3.9) is
an injective smooth immersion, and thus a smooth embedding. Hence, $\Psi\ $is
smoothly embedded in $\Psi^{\prime}\times\mathbb{L},$ which is at most
$\left(n^{2}+n\right)$-dimensional.

Suppose $s\in\mathbb{L}$ is fixed, then recall from (2.8) that the product
$\circ_{s}$on $\mathfrak{l}$ is defined by
$\eta\circ_{s}\xi=\left(\eta\cdot\left(\xi\cdot s\right)\right)/s$ for
sufficiently small $\eta,\xi\in\mathfrak{l}.$ Now for $h\in\Psi$ and the
corresponding $h^{\prime}\in\Psi^{\prime}$, consider

$\displaystyle h^{\prime}\left(\eta\circ_{s}\xi\right)$ $\displaystyle=$
$\displaystyle h\left(\eta\cdot\left(\xi\cdot s\right)\right)/h\left(s\right)$
(3.12) $\displaystyle=$ $\displaystyle
h^{\prime}\left(\eta\right)\cdot\left(h^{\prime}\left(\xi\right)\cdot
h\left(s\right)\right)/h\left(s\right)$ $\displaystyle=$ $\displaystyle
h^{\prime}\left(\eta\right)\circ_{h\left(s\right)}h^{\prime}\left(\xi\right).$

Equivalently, we can write

$h^{\prime}\left(\eta\circ_{s}\xi\right)\cdot
h\left(s\right)=h^{\prime}\left(\eta\right)\cdot\left(h^{\prime}\left(\xi\right)\cdot
h\left(s\right)\right),$ (3.13)

or

$h^{\prime}\left(\eta\circ_{s}\xi\right)\circ_{s}\left(h\left(s\right)/s\right)=h^{\prime}\left(\eta\right)\circ_{s}\left(h^{\prime}\left(\xi\right)\circ_{s}\left(h\left(s\right)/s\right)\right),$
(3.14)

###### Remark 3.9

In the theory of loops and quasigroups, a map $h^{\prime}$ that satisfies the
property
$h^{\prime}\left(xy\right)A=h^{\prime}\left(x\right)\left(h^{\prime}\left(y\right)A\right)$
is known as a _pseudoautomorphism_ with _companion_ $A$. Using this language,
we can say that $h^{\prime}$ has _companion_ $h\left(s\right)/s$ with respect
to $s.$ The group $\mathop{\mathrm{A}ut}\left(\mathfrak{l},\circ_{s}\right)$
of automorphisms of $\circ_{s}$ is then defined as the subset of elements of
$\Psi^{\prime}$ which have companion $0.$ This then corresponds to the
stabilizer $\mathop{\mathrm{S}tab}_{\Psi}\left(s\right)$ of $s.$

The group $\Psi$ also acts on the bracket
$\left[\cdot,\cdot\right]^{\left(s\right)}.$ As we noted above, on Lie groups,
the map $h^{\prime}$ is the differential of $h$ at identity, and we know this
is a Lie algebra isomorphism. However, in the more general case, the map
$h^{\prime}$ is an algebra isomorphism from
$\left(\mathfrak{l},\left[\cdot,\cdot\right]^{\left(s\right)}\right)$ to
$\left(\mathfrak{l},\left[\cdot,\cdot\right]^{\left(h\left(s\right)\right)}\right).$

###### Theorem 3.10

Suppose $s\in\mathbb{L}$ and $h\in\Psi$, then for any
$\xi,\eta\in\mathfrak{l},$

$h^{\prime}\left(\left[\xi,\eta\right]^{\left(s\right)}\right)=\left[h^{\prime}\left(\xi\right),h^{\prime}\left(\eta\right)\right]^{\left(h\left(s\right)\right)}.$
(3.15)

Proof. Indeed, from Theorem 2.26 and using (3.12), we obtain

$\displaystyle h^{\prime}\left(\left[\xi,\eta\right]^{\left(s\right)}\right)$
$\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{dt_{1}dt_{2}}\left(h^{\prime}\left(\left(t_{1}\xi\right)\circ_{s}\left(t_{2}\eta\right)\right)-h^{\prime}\left(\left(t_{2}\eta\right)\circ_{s}\left(t_{1}\xi\right)\right)\right)\right|_{t_{1}=t_{2}=0}$
$\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{dt_{1}dt_{2}}\left(\left(t_{1}h^{\prime}\left(\xi\right)\right)\circ_{h\left(s\right)}\left(t_{2}h^{\prime}\left(\eta\right)\right)-\left(\left(t_{2}h^{\prime}\left(\eta\right)\right)\circ_{h\left(s\right)}\left(t_{1}h^{\prime}\left(\xi\right)\right)\right)\right)\right|_{t_{1}=t_{2}=0}$
$\displaystyle=$
$\displaystyle\left[h^{\prime}\left(\xi\right),h^{\prime}\left(\eta\right)\right]^{h\left(s\right)}.$

##  4 Products of spheres

As mentioned in Example 2.9, a large family of parallelizable manifolds that
are not Lie groups or Moufang loops is given by products of spheres where one
factor is an odd-dimensional sphere. As it is well-known, the only
parallelizable spheres are $S^{1},S^{3},S^{7}$, which correspond to unit norm
sets of complex numbers, quaternions, and octonions. In particular, $S^{1}$
and $S^{3}$ are both Lie groups, while $S^{7}$ is the Moufang loop of unit
octonions. In [3, 14, 15] explicit parallelizations on products of spheres
have been computed and this will serve as an important example of the
framework developed in this paper.

In particular, let us consider in detail the case of $S^{m}\times S^{1}$ from
[14]. Consider $S^{m}$ as embedded in $\mathbb{R}^{m+1}$ in a standard way

$S^{m}=\left\\{\left(x_{1},...,x_{m+1}\right)\in\mathbb{R}^{m+1}:\left|x\right|^{2}=x_{1}^{2}+...+x_{m+1}^{2}=1\right\\}.$
(4.1)

Consider the standard coordinate frame $\left\\{\partial_{x_{i}}\right\\}$ on
$\mathbb{R}^{m+1}$. Then, the orthogonal projections $\left\\{M_{i}\right\\}$
of these vector fields onto $S^{m}$ are given by

$M_{i}=\partial_{x_{i}}-x_{i}M,$ (4.2)

where $M=\sum_{i=1}^{m+1}x_{i}\partial_{x_{i}}$ is the normal vector field to
$S^{m}\subset\mathbb{R}^{m+1}.$ Also consider $\phi$ as a coordinate on
$S^{1}$ with the associated vector field $\partial_{\phi}.$ Then, a
parallelization of $S^{m}\times S^{1}$ is given by the following.

###### Theorem 4.1

Consider $S^{m}\subset\mathbb{R}^{m+1}$ and for $i=1,...,n+1$, define

$f_{i}=M_{i}+x_{i}\partial_{\phi}.$ (4.3)

Then, the set $\left\\{f_{i}\right\\}$ is global frame on $S^{m}\times S^{1}.$
In particular, it is orthonormal with respect to the product metric and
satisfies

$\left[f_{i},f_{j}\right]=x_{i}f_{j}-x_{j}f_{i}.$ (4.4)

Now let $\mathbb{L}=S^{m}\times S^{1}$, so that each $s\in\mathbb{L}$ is given
by $s=\left(x,\phi\right)$ for $x\in S^{m}\subset\mathbb{R}^{m+1}$ and
$\phi\in S^{1}.$ Also suppose $\mathfrak{l}$ is an
$\left(m+1\right)$-dimensional real vector space with an inner product
$\left\langle\cdot,\cdot\right\rangle$ and an orthonormal basis
$\left\\{e_{i}\right\\}.$ Then, for each $s=\left(x,\phi\right)\in\mathbb{L},$
define

$\begin{array}[]{c}\rho_{s}:\mathfrak{l}\longrightarrow T_{s}\mathbb{L}\\\
e_{i}\mapsto\left.f_{i}\right|_{s}\end{array}.$ (4.5)

In particular, if $\xi=\xi^{i}e_{i}\in\mathfrak{l},$ then
$\rho\left(\xi\right)=\xi^{i}f_{i}.$

Consider the derivatives of the coordinate functions in the directions of the
basis vectors $\left\\{f_{i}\right\\}.$ We have

$\displaystyle f_{i}\left(x_{j}\right)$ $\displaystyle=$
$\displaystyle\left(\partial_{x_{i}}-x_{i}M\right)x_{j}$ $\displaystyle=$
$\displaystyle\delta_{ij}-x_{i}x_{j}$ $\displaystyle f_{i}\left(\phi\right)$
$\displaystyle=$ $\displaystyle x_{i}$

In particular, define $c_{ij}=f_{i}\left(x_{j}\right)=\delta_{ij}-x_{i}x_{j}.$

The bracket on $\mathfrak{l}$ is defined from (4.4) via (2.12). In particular,
for the basis $\left\\{e_{i}\right\\}$ on $\mathfrak{l}$ and
$s=\left(x,\phi\right)\in\mathbb{L}$, we have

$b_{ij}^{\left(s\right)}=:\left[e_{i},e_{j}\right]^{\left(s\right)}=e_{i}x_{j}-e_{j}x_{i}.$
(4.6)

Equivalently, given any $\xi,\eta\in\mathfrak{l}$, we have

$\left[\xi,\eta\right]^{\left(s\right)}=\xi\left\langle\eta,x\right\rangle-\eta\left\langle\xi,x\right\rangle.$
(4.7)

In particular, note that if both $\xi$ and $\eta$ are orthogonal to $x$, then
$\left[\xi,\eta\right]^{\left(s\right)}=0$ and for any $\xi\in\mathfrak{l},$
$\left[\xi,x\right]^{\left(s\right)}=\xi$.

###### Lemma 4.2

For each fixed $s$, $\mathfrak{l}$ with the bracket
$\left[\cdot,\cdot\right]^{\left(s\right)}$ defines a Lie algebra that is
isomorphic to a semidirect sum $\mathbb{R}^{n}\oplus_{S}\mathbb{R},$ with
bracket
$\left[\left(\xi,\lambda\right),\left(\eta,\mu\right)\right]=\left[\mu\xi-\lambda\eta,0\right].$

Proof. First let us show that $\left[\cdot,\cdot\right]^{\left(s\right)}$, as
given by (4.7) defines a Lie algebra on $\mathfrak{l}.$ We just need to check
the Jacobi identity. Indeed, for $\gamma,\xi,\eta\in\mathfrak{l},$

$\displaystyle\left[\gamma,\left[\xi,\eta\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle\left[\gamma,\xi\right]^{\left(s\right)}\left\langle\eta,x\right\rangle-\left[\gamma,\eta\right]^{\left(s\right)}\left\langle\xi,x\right\rangle$
$\displaystyle=$
$\displaystyle\gamma\left\langle\xi,x\right\rangle\left\langle\eta,x\right\rangle-\xi\left\langle\gamma,x\right\rangle\left\langle\eta,x\right\rangle$
$\displaystyle-\gamma\left\langle\eta,x\right\rangle\left\langle\xi,x\right\rangle+\eta\left\langle\gamma,x\right\rangle\left\langle\xi,x\right\rangle$
$\displaystyle=$
$\displaystyle\eta\left\langle\gamma,x\right\rangle\left\langle\xi,x\right\rangle-\xi\left\langle\gamma,x\right\rangle\left\langle\eta,x\right\rangle$
$\displaystyle\left[\xi,\left[\eta,\gamma\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle\gamma\left\langle\xi,x\right\rangle\left\langle\eta,x\right\rangle-\eta\left\langle\xi,x\right\rangle\left\langle\gamma,x\right\rangle$
$\displaystyle\left[\eta,\left[\gamma,\xi\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle\xi\left\langle\eta,x\right\rangle\left\langle\gamma,x\right\rangle-\gamma\left\langle\eta,x\right\rangle\left\langle\xi,x\right\rangle$

Adding these terms together shows that indeed the Jacobi identity is
satisfied. Now consider the linear map
$F:\mathfrak{l}\longrightarrow\mathbb{R}^{n}\oplus\mathbb{R}$, given by

$F\left(\xi\right)=\left(\xi-\left\langle\xi,x\right\rangle
x,\left\langle\xi,x\right\rangle\right).$

This is clearly a linear isomorphism. Then,

$\displaystyle F\left(\left[\xi,\eta\right]^{\left(s\right)}\right)$
$\displaystyle=$ $\displaystyle
F\left(\xi\right)\left\langle\eta,x\right\rangle-F\left(\eta\right)\left\langle\xi,x\right\rangle$
$\displaystyle=$ $\displaystyle\left(\xi-\left\langle\xi,x\right\rangle
x,\left\langle\xi,x\right\rangle\right)\left\langle\eta,x\right\rangle$
$\displaystyle-\left(\eta-\left\langle\eta,x\right\rangle
x,\left\langle\eta,x\right\rangle\right)\left\langle\xi,x\right\rangle$
$\displaystyle=$
$\displaystyle\left(\xi\left\langle\eta,x\right\rangle-\eta\left\langle\xi,x\right\rangle,0\right)$
$\displaystyle=$ $\displaystyle\left[\left(\xi-\left\langle\xi,x\right\rangle
x,\left\langle\xi,x\right\rangle\right),\left(\eta-\left\langle\eta,x\right\rangle
x,\left\langle\eta,x\right\rangle\right)\right]$ $\displaystyle=$
$\displaystyle\left[F\left(\xi\right),F\left(\eta\right)\right].$

Hence, this is a Lie algebra homomorphism, and thus the Lie algebra
$\left(\mathfrak{l},\left[\cdot,\cdot\right]^{\left(s\right)}\right)$ is
isomorphic to $\mathbb{R}^{n}\oplus_{S}\mathbb{R}.$

Taking the derivative of the bracket in the direction of $f_{i}$ gives us

$\displaystyle d_{f_{i}}\left(\left[\xi,\eta\right]^{\left(s\right)}\right)$
$\displaystyle=$ $\displaystyle\xi\eta^{j}c_{ij}-\eta\xi^{j}c_{ij}$
$\displaystyle=$
$\displaystyle\xi\eta_{i}-\eta\xi_{i}-x_{i}\left[\xi,\eta\right]^{\left(s\right)}$

In particular, this shows that

$d_{f_{i}}b_{jk}^{\left(s\right)}=:a_{ijk}^{\left(s\right)}=e_{k}\delta_{ij}-e_{j}\delta_{ik}-x_{i}b_{jk}^{\left(s\right)}.$
(4.8)

Hence, for $\xi,\eta,\gamma\in\mathfrak{l}$,

$a^{\left(s\right)}\left(\xi,\eta,\gamma\right)=\gamma\left(\left\langle\xi,\eta\right\rangle+\left\langle
x,\xi\right\rangle\left\langle
x,\eta\right\rangle\right)-\eta\left(\left\langle\xi,\gamma\right\rangle+\left\langle
x,\xi\right\rangle\left\langle x,\gamma\right\rangle\right).$ (4.9)

Recall the notion of a _Lie triple system[8]: _this is a vector space equipped
with a skew-symmetric triple product $\left[\cdot,\cdot,\cdot\right]$, that is
skew-symmetric in two entries, has a vanishing sum of cyclic permutations(i.e.
satisfies the Jacobi identity), and also that the linear map defined by fixing
two of the entries of $\left[\cdot,\cdot,\cdot\right]$ is a derivation of the
triple product. Consider now $\mathfrak{l}$ equipped with the triple product
$a^{\left(s\right)}.$ We know that $a^{\left(s\right)}$ is skew-symmetric in
the last two entries. From (2), we also see that $a^{\left(s\right)}$
satisfies the Jacobi identity if and only if
$\left(\mathfrak{l},b^{\left(s\right)}\right)$ is a Lie algebra. Therefore, in
this case, it just suffices to check the derivation property to figure out if
$\left(\mathfrak{l},a^{\left(s\right)}\right)$ is a Lie triple system.

###### Theorem 4.3

For each $s\in\mathbb{L}$, the vector space $\mathfrak{l}$ equipped with the
triple product $a^{\left(s\right)}$ forms a Lie triple system.

Proof. Since $b^{\left(s\right)}$ satisfies the Jacobi identity, from (2) we
know that

$a^{\left(s\right)}\left(\xi,\eta,\gamma\right)+a^{\left(s\right)}\left(\eta,\gamma,\xi\right)+a^{\left(s\right)}\left(\gamma,\xi,\eta\right)=0.$

To show that $\left(\mathfrak{l},a^{\left(s\right)}\right)$ forms a Lie triple
system, we just need to show that for any $\xi,\eta\in\mathfrak{l}$, the map
$a_{\xi,\eta}^{\left(s\right)}:\mathfrak{l}\longrightarrow\mathfrak{l}$ given
by
$a_{\xi,\eta}^{\left(s\right)}\left(\gamma\right)=a^{\left(s\right)}\left(\gamma,\xi,\eta\right)$
is a derivation for $a^{\left(s\right)}$. That is, for any
$\xi,\eta,\gamma,u,v\in\mathfrak{l},$ we need to show that

$\displaystyle
a_{\xi,\eta}^{\left(s\right)}\left(a^{\left(s\right)}\left(u,v,\gamma\right)\right)$
$\displaystyle=$ $\displaystyle
a^{\left(s\right)}\left(a^{\left(s\right)}\left(u,\xi,\eta\right),v,\gamma\right)$
$\displaystyle+a^{\left(s\right)}\left(u,a^{\left(s\right)}\left(v,\xi,\eta\right),\gamma\right)$
$\displaystyle+a^{\left(s\right)}\left(u,v,a^{\left(s\right)}\left(\gamma,\xi,\eta\right)\right).$

This is a straightforward, but tedious calculation, which we show in Appendix
A.

To define products on $\mathfrak{l}$ and $\mathbb{L},$ we need to consider
integral curves of fundamental vector fields. In (2.5) suppose the curve
$p\left(t\right)\in\mathbb{L}$ has coordinates
$\left(x\left(t\right),\phi\left(t\right)\right).$ Suppose $\xi\neq 0$, and
let us rewrite $\rho\left(\xi\right)$ in terms of the coordinate basis:

$\displaystyle\rho\left(\xi\right)$ $\displaystyle=$
$\displaystyle\xi^{i}f_{i}$ $\displaystyle=$
$\displaystyle\xi^{i}\partial_{x_{i}}-\xi^{i}x_{i}\sum_{i=1}^{m+1}x_{j}\partial_{x_{j}}+\xi^{i}x_{i}\partial_{\phi}.$

Setting $\hat{\xi}=\xi^{i}x_{i}$, we can thus rewrite the flow equation (2.5)
as

$\left\\{\begin{array}[]{c}\frac{dx\left(t\right)}{dt}=\xi-\hat{\xi}x\\\
\frac{d\phi\left(t\right)}{dt}=\hat{\xi}\end{array}\right.,$ (4.11)

with initial conditions $\left(x_{0},\phi_{0}\right)$. Without loss of
generality, assume $\left|\xi\right|=1.$

###### Lemma 4.4

The solution of (4.11) exists for all $t\in\mathbb{R}$ and, if $\xi=\pm
x_{0}$, the solutions are given by

$\displaystyle x\left(t\right)$ $\displaystyle=$ $\displaystyle x_{0}$ (4.12a)
$\displaystyle\phi\left(t\right)$ $\displaystyle=$ $\displaystyle\phi_{0}\pm
t.$ (4.12b)

If $\xi\neq x_{0}$, then the solution is given by

$\displaystyle x\left(t\right)$ $\displaystyle=$
$\displaystyle\xi\tanh\left(\sigma+t\right)+\tilde{x}_{0}\mathop{\mathrm{s}ech}\left(\sigma+t\right)$
(4.13a) $\displaystyle\phi\left(t\right)$ $\displaystyle=$
$\displaystyle\phi_{0}+\ln\left(\frac{\cosh\left(t+\sigma\right)}{\cosh\sigma}\right),$
(4.13b)

where $\tilde{x}_{0}=x_{0}\cosh\sigma-\xi\sinh\sigma$ and $\sigma$ is such
that $\hat{\xi}\left(0\right)=\tanh\sigma.$

Proof. From (4.11), we see $x\left(t\right)$ is constant if and only if
$\xi=\pm x_{0}.$In that case, $\hat{\xi}\left(t\right)=\pm 1$, and thus
$\phi\left(t\right)=\phi_{0}\pm t.$

More generally, taking the inner product of the first equation in (4.11) with
$\xi$, we obtain an equation for $\hat{\xi}\left(t\right):$

$\frac{d\hat{\xi}}{dt}=1-\hat{\xi}^{2}.$ (4.14)

Since $\hat{\xi}^{2}=1$ if and only if $\xi=\pm x_{0},$ suppose
$\hat{\xi}\left(t\right)=\tanh\sigma\left(t\right),$ for some function
$\sigma\left(t\right).$ Then, (4.14) immediately gives
$\sigma\left(t\right)=t+\sigma$, where $\sigma=\sigma\left(0\right),$ that is,
$\hat{\xi}\left(0\right)=\tanh\sigma$.

Let us integrate $\hat{\xi}$, because this will be needed as an integrating
factor for the first equation in (4.11) and also to solve for
$\phi\left(t\right)$ in the second equation in (4.11):

$\int^{t}\hat{\xi}\left(\tau\right)d\tau=\ln\left(\cosh\left(\sigma+t\right)\right).$
(4.15)

Hence, we can write the integrating factor as

$M\left(t\right)=\cosh\left(\sigma+t\right)$

and the general solution is

$x\left(t\right)=\xi\tanh\left(\sigma+t\right)+\tilde{x}_{0}\mathop{\mathrm{s}ech}\left(\sigma+t\right),$
(4.16)

where

$\tilde{x}_{0}=x_{0}\cosh\sigma-\xi\sinh\sigma.$

We also obtain $\phi\left(t\right)$ (4.13b) by directly integrating the second
equation of (4.11):

$\phi\left(t\right)=\phi_{0}+\ln\left(\frac{\cosh\left(t+\sigma\right)}{\cosh\sigma}\right)$

As expected on a compact manifold, the integral curve exists for all values of
$t$.

Overall, we see the geometric picture of how the integral curve evolves. For
the non-trivial case, suppose $\xi\neq\pm x_{0}$, then from (4.16), we see
that $x\left(t\right)$ always lies in the plane spanned by $x_{0}$ and $\xi$,
so the curve is along the great circle of $S^{m}$ given by the plane spanned
by $x_{0}$ and $\xi$. If $\alpha\left(t\right)\ $is the angle between
$x\left(t\right)$ and $\xi,$ then
$\hat{\xi}\left(t\right)=\cos\alpha\left(t\right)$. In particular, then

$\displaystyle\cos\alpha\left(t\right)$ $\displaystyle=$
$\displaystyle\tanh\left(\sigma+t\right)$
$\displaystyle\sin\alpha\left(t\right)$ $\displaystyle=$
$\displaystyle\mathop{\mathrm{s}ech}\left(\sigma+t\right).$

In particular, we see that $\cos\alpha\left(t\right)$ is a monotonic function,
with $\cos\alpha\left(t\right)\longrightarrow+1$ as $t\longrightarrow+\infty$
and $\cos\alpha\left(t\right)\longrightarrow-1$ as $t\longrightarrow-\infty$.
Moreover, $\sin\alpha\left(t\right)$ is always positive. Therefore,
$\alpha\left(t\right)\in\left(0,\pi\right)$, with
$\alpha\left(t\right)\longrightarrow 0$ as $t\longrightarrow+\infty$ and
$\alpha\left(t\right)\longrightarrow\pi$ as $t\longrightarrow-\infty$. The
equation (4.13a) can be rewritten as

$x\left(t\right)=\xi\cos\alpha\left(t\right)+\tilde{x}_{0}\sin\alpha\left(t\right).$
(4.17)

This shows that in the adapted $\xi$-$\tilde{x}_{0}$ plane, $x\left(t\right)$
stays in the upper half-plane, moving clockwise as $t$ increases. In
particular, as $t\longrightarrow\pm\infty$,
$x\left(t\right)\longrightarrow\pm\xi$.

###### Example 4.5

Since $a^{\left(s\right)}$ in this case is non-trivial (4.9), from Theorem
2.26 we see that $\circ_{s}$ is necessarily non-associative. To illustrate
this product $\circ_{s}$ on $\mathfrak{l}$ (2.8) more explicitly, let
$s=\left(x,\phi\right)$ and consider elements unit length
$\xi,\eta\in\mathfrak{l}$, such that they are orthogonal to each other and to
$x$. The more general case is computationally a bit more involved. Consider
first
$\left(x_{1}\left(t\right),\phi_{1}\left(t\right)\right)=\left(t\xi\right)\cdot
s.$ From (4.13), noting that $\xi\cdot x=0$ implies $\sigma=0,$ we have

$\displaystyle x_{1}\left(t\right)$ $\displaystyle=$ $\displaystyle\xi\tanh
t+x\mathop{\mathrm{s}ech}t$ (4.18) $\displaystyle\phi_{1}\left(t\right)$
$\displaystyle=$ $\displaystyle\phi_{0}+\ln\left(\cosh t\right).$ (4.19)

Next, consider
$\left(x_{2}\left(\tau\right),\phi_{2}\left(\tau\right)\right)=\left(\tau\eta\right)\cdot\left(\left(t\xi\right)\cdot
s\right).$ By assumption, the initial value $x_{2}\left(0\right)=\xi\tanh
t+x\mathop{\mathrm{s}ech}t$ is orthogonal to $\eta$, so $\sigma$ is again $0$.
Thus,

$\displaystyle x_{2}\left(\tau\right)$ $\displaystyle=$
$\displaystyle\eta\tanh\tau+\xi\tanh
t\mathop{\mathrm{s}ech}\tau+x\mathop{\mathrm{s}ech}t\mathop{\mathrm{s}ech}\tau$
$\displaystyle\phi_{2}\left(\tau\right)$ $\displaystyle=$
$\displaystyle\phi_{0}+\ln\left(\cosh t\cosh\tau\right).$

To find
$\left(\tau\eta\right)\circ_{s}\left(t\xi\right)=\left(\tau\eta\right)\cdot\left(\left(t\xi\right)\cdot
s\right)/s$, we need to find a positive $t_{2}\in\mathbb{R}$ and
$\gamma\in\mathfrak{l}$ of unit length, such that

$\displaystyle x_{2}\left(\tau\right)$ $\displaystyle=$
$\displaystyle\gamma\tanh\left(\sigma_{2}+t_{2}\right)+\left(x\cosh\sigma_{2}-\gamma\sinh\sigma_{2}\right)\mathop{\mathrm{s}ech}\left(\sigma_{2}+t_{2}\right)$
$\displaystyle\phi_{2}\left(\tau\right)$ $\displaystyle=$
$\displaystyle\phi_{0}+\ln\left(\frac{\cosh\left(t_{2}+\sigma_{2}\right)}{\cosh\sigma_{2}}\right)+2\pi
k\text{,}$

where $\tanh\sigma_{2}=\gamma\cdot x$ and $k$ is an integer. The freedom to
add an integer multiple of $2\pi k$ stems from the fact that $\phi$ is a
coordinate on the circle. We know that $\gamma$ is a linear combination of
$\xi,\eta,x,$ so let

$\gamma=a\xi+b\eta+\left(\tanh\sigma_{2}\right)x.$ (4.20)

Then, taking inner products of $x_{2}$ with $\xi,\eta,x$, and also comparing
$\phi_{2}$, we find the following equations

$\displaystyle\tanh t\mathop{\mathrm{s}ech}\tau$ $\displaystyle=$
$\displaystyle
a\left(\tanh\left(t_{2}+\sigma_{2}\right)-\sinh\sigma_{2}\mathop{\mathrm{s}ech}\left(t_{2}+\sigma_{2}\right)\right)$
(4.21a) $\displaystyle\tanh\tau$ $\displaystyle=$ $\displaystyle
b\left(\tanh\left(t_{2}+\sigma_{2}\right)-\sinh\sigma_{2}\mathop{\mathrm{s}ech}\left(t_{2}+\sigma_{2}\right)\right)$
(4.21b) $\displaystyle\mathop{\mathrm{s}ech}t\mathop{\mathrm{s}ech}\tau$
$\displaystyle=$
$\displaystyle\tanh\sigma_{2}\tanh\left(t_{2}+\sigma_{2}\right)$
$\displaystyle+\mathop{\mathrm{s}ech}\sigma_{2}\mathop{\mathrm{s}ech}\left(t_{2}+\sigma_{2}\right)$
$\displaystyle e^{2\pi k}\cosh t\cosh\tau$ $\displaystyle=$
$\displaystyle\frac{\cosh\left(t_{2}+\sigma_{2}\right)}{\cosh\sigma_{2}}$
(4.21d) $\displaystyle 1$ $\displaystyle=$ $\displaystyle
a^{2}+b^{2}+\tanh^{2}\sigma_{2}.$ (4.21e)

The last equation comes from the fact that $\gamma$ is unit length. However if
we require that $t_{2}\longrightarrow 0$ as $t,\tau\longrightarrow 0$, which
makes sense in order for the product $\circ_{s}$to be defined in the
neighborhood of $0$, (4.21d) shows that $k=0$. From (4.21) and (4.21d) with
$k=0$, we find

$\tanh\sigma_{2}\tanh\left(t_{2}+\sigma_{2}\right)+\left(\mathop{\mathrm{s}ech}\sigma_{2}-\cosh\sigma_{2}\right)\mathop{\mathrm{s}ech}\left(t_{2}+\sigma_{2}\right)=0$

and thus

$\tanh\sigma_{2}\left(\sinh\left(t_{2}+\sigma_{2}\right)-\sinh\sigma_{2}\right)=0.$

Since $t_{2}>0$, we find that $\tanh\sigma_{2}=0$, and hence $\sigma_{2}=0.$
This immediately gives us

$\displaystyle\cosh t_{2}$ $\displaystyle=$ $\displaystyle\cosh t\cosh\tau$
$\displaystyle a$ $\displaystyle=$ $\displaystyle\frac{\tanh
t\mathop{\mathrm{s}ech}\tau}{\tanh t_{2}}$ $\displaystyle b$ $\displaystyle=$
$\displaystyle\frac{\tanh\tau}{\tanh t_{2}}.$

Thus, we conclude that for an orthogonal triple $\eta,\xi,x$, we get

$\left(\tau\eta\right)\circ_{s}\left(t\xi\right)=t_{2}\left(a\xi+b\eta\right),$

with $t_{2}$, $a,$ and $b$ as given above.

Let us now consider the isomorphisms of $\mathbb{L}$. Suppose
$h\in\Psi\left(\mathbb{L}\right)$ and
$h^{\prime}\in\Psi^{\prime}\left(\mathbb{L}\right)$ is the map of
$\mathfrak{l}.$ Recall from Lemma 3.4 that for $\xi\in\mathfrak{l}$ and
$s\in\mathbb{L}$, $h\left(\xi\cdot s\right)=h^{\prime}\left(\xi\right)\cdot
h\left(s\right).$

Given $s=\left(x,\phi\right)\in\mathbb{L}$, let the corresponding components
of $h\left(s\right)$ be $h\left(s\right)_{x}$ and $h\left(s\right)_{\phi}$.
From Theorem 3.10, we then know that for any $\xi,\eta\in\mathfrak{l}$ and
$s=\left(x,\phi\right)\in\mathbb{L}$

$h^{\prime}\left(\left[\xi,\eta\right]^{\left(s\right)}\right)=\left[h^{\prime}\left(\xi\right),h^{\prime}\left(\eta\right)\right]^{h\left(s\right)}.$

However, from (4.7),

$h^{\prime}\left(\left[\xi,\eta\right]^{\left(s\right)}\right)=h^{\prime}\left(\xi\right)\left\langle\eta,x\right\rangle-h^{\prime}\left(\eta\right)\left\langle\xi,x\right\rangle$
(4.22)

and

$\left[h^{\prime}\left(\xi\right),h^{\prime}\left(\eta\right)\right]^{h\left(s\right)}=h^{\prime}\left(\xi\right)\left\langle
h^{\prime}\left(\eta\right),h\left(s\right)_{x}\right\rangle-h^{\prime}\left(\eta\right)\left\langle
h^{\prime}\left(\xi\right),h\left(s\right)_{x}\right\rangle.$ (4.23)

Hence, in particular, for any $\eta\in\mathfrak{l}$ and $x\in S^{n},$

$\left\langle\eta,x\right\rangle=\left\langle
h^{\prime}\left(\eta\right),h\left(s\right)_{x}\right\rangle.$ (4.24)

Suppose now $\xi$ is a unit vector in $\mathfrak{l}$. Noting that
$h\left(t\xi\cdot s\right)=\left(th^{\prime}\left(\xi\right)\right)\cdot
h\left(s\right)$, from (4.11) we see
$h\left(s\left(t\right)\right)=h\left(t\xi\cdot s\right)$ satisfies the
following equations

$\left\\{\begin{array}[]{c}\frac{dh\left(s\left(t\right)\right)_{x}}{dt}=h^{\prime}\left(\xi\right)-\hat{\xi}h\left(s\left(t\right)\right)_{x}\\\
\frac{dh\left(s\left(t\right)\right)_{\phi}}{dt}=\hat{\xi}\end{array}\right..$
(4.25)

From (4.24), we see that $\left\langle
th^{\prime}\left(\xi\right),h\left(s\left(t\right)\right)_{x}\right\rangle=\left\langle
t\xi,x\left(t\right)\right\rangle$, and hence $\hat{\xi}$ is unchanged from
(4.11). Taking inner product of the first equation of (4.25) with
$h^{\prime}\left(\xi\right)$ yields

$\frac{d\hat{\xi}}{dt}=\left|h^{\prime}\left(\xi\right)\right|^{2}-\hat{\xi}^{2}.$

However, comparing with (4.14), we see that
$\left|h^{\prime}\left(\xi\right)\right|=1.$ Therefore, $h^{\prime}\in
SO\left(n+1\right)$, and hence from (4.24) we see that
$h\left(s\left(t\right)\right)_{x}=h^{\prime}\left(x\left(t\right)\right).$

We now need to understand how $h$ transforms the $\phi$ component of $s$. From
(4.12) and (4.13), for $\xi\in\mathfrak{l}$, let us denote the $x$ and $\phi$
components of $\xi\cdot s\ $by $\xi\cdot x\ $and $\phi+\phi_{\xi}$,
respectively. Define $\tilde{h}_{x}\left(\phi\right)=h\left(s\right)_{\phi}$
to be the transformation of the $\phi$ component, so that
$h\left(s\right)=h\left(x,\phi\right)=\left(h^{\prime}\left(x\right),\tilde{h}_{x}\left(\phi\right)\right).$
In particular,

$\displaystyle h\left(\xi\cdot s\right)$ $\displaystyle=$ $\displaystyle
h\left(\xi\cdot x,\phi+\phi_{\xi}\right)$ $\displaystyle=$
$\displaystyle\left(h^{\prime}\left(\xi\cdot x\right),\tilde{h}_{\xi\cdot
x}\left(\phi+\phi_{\xi}\right)\right).$

On the other hand, from the definition of $h$,

$\displaystyle h\left(\xi\cdot s\right)$ $\displaystyle=$ $\displaystyle
h^{\prime}\left(\xi\right)\cdot h\left(s\right)$ $\displaystyle=$
$\displaystyle
h^{\prime}\left(\xi\right)\cdot\left(h^{\prime}\left(x\right),\tilde{h}_{x}\left(\phi\right)\right)$
$\displaystyle=$ $\displaystyle\left(h^{\prime}\left(\xi\right)\cdot
h^{\prime}\left(x\right),\tilde{h}_{x}\left(\phi\right)+\phi_{h^{\prime}\left(\xi\right)}\right).$

We already know that $h^{\prime}\left(\xi\cdot
x\right)=h^{\prime}\left(\xi\right)\cdot h^{\prime}\left(x\right),$ and
moreover, since $\left\langle\xi,x\right\rangle=\left\langle
h^{\prime}\left(\xi\right),h^{\prime}\left(x\right)\right\rangle$,
$\phi_{h^{\prime}\left(\xi\right)}=\phi_{\xi}$. Thus we see that

$\tilde{h}_{\xi\cdot
x}\left(\phi+\phi_{\xi}\right)=\tilde{h}_{x}\left(\phi\right)+\phi_{\xi}.$
(4.26)

Now from (4.12), we see that taking $\xi=x$, $\left(tx\right)\cdot
s=\left(x,\phi+t\right)$, so $h\left(\left(tx\right)\cdot
s\right)=\left(h^{\prime}\left(x\right),\tilde{h}_{x}\left(\phi+t\right)\right).$
However, from the definition of $h$,

$\displaystyle h\left(\left(tx\right)\cdot s\right)$ $\displaystyle=$
$\displaystyle th^{\prime}\left(x\right)\cdot h\left(s\right)$
$\displaystyle=$
$\displaystyle\left(h^{\prime}\left(x\right),\tilde{h}_{x}\left(\phi\right)+t\right).$

Hence, we see that for any $x,\xi,\ $and $t$,
$\tilde{h}_{x}\left(\phi+t\right)=\tilde{h}_{x}\left(\phi\right)+t.$ This
shows that $\tilde{h}_{x}\left(\phi\right)$ $=\phi+c\left(x\right).$ From
(4.26), we then see that

$\phi+\phi_{\xi}+c\left(\xi\cdot x\right)=\phi+c\left(x\right)+\phi_{\xi},$

which shows that $c\left(x\right)$ is independent of $x.$ Hence, the
transformation of $\phi$ is just a transformation of $S^{1}$ by an action of
$U\left(1\right)$. Overall, we then see that $\Psi\left(\mathbb{L}\right)\cong
SO\left(n+1\right)\times U\left(1\right)$ and
$\Psi^{\prime}\left(\mathbb{L}\right)\cong SO\left(n+1\right).$

Consider now more generally, $\mathbb{L=}S^{m}\times N$ where $N$ is any
$n$-dimensional parallelizable manifold. Suppose
$\left\\{T_{1},...,T_{n}\right\\}$ is a frame of vector fields on $N$ that
satisfy the following Lie bracket relations:

$\left[T_{A},T_{B}\right]=D_{AB}^{\ \ \text{\ }C}T_{C},$ (4.27)

where the $D_{AB}^{\ \ \text{\ }C}$ for $A,B,C=1,...,n$ are functions on $N$.
The idea is to take a non-trivial vector field $T$ on $N$ and add the trivial
rank 1 bundle generated by it to $TS^{m}$. The resulting rank
$\left(m+1\right)$-bundle then be trivialized in the same way as in the
previous example of $S^{m}\times S^{1}$ [3]. Without loss of generality, let
$T=T_{1}$. As before, consider $S^{m}\subset\mathbb{R}^{m+1}$ and for
$i=1,...,n+1$, define

$f_{i}=M_{i}+x_{i}T,$ (4.28)

where $M_{i}$ is given by (4.2). More explicitly, the $f_{i}$ are given by

$f_{i}=\partial_{x_{i}}-x_{i}\sum_{j=1}^{m+1}x_{j}\partial_{x_{j}}+x_{i}T,$
(4.29)

Then, the set $\left\\{f_{1},...,f_{n+1},T_{2},...,T_{n}\right\\}$ is global
frame on $S^{m}\times N.$ From (4.28), we see that

$T=x^{i}f_{i.}$ (4.30)

The basis elements $\left\\{f_{1},...,f_{n+1}\right\\}$ satisfy the following
bracket relation

$\left[f_{i},f_{j}\right]=x_{i}f_{j}-x_{j}f_{i}.$ (4.31)

Also, for $A,B=2,...,n$,

$\left[f_{i},T_{A}\right]=\left[x_{i}T_{1},T_{A}\right]=x_{i}D_{1A}^{\ \
1}T_{1}+x_{i}D_{1A}^{\ \ \ B}T_{B}\ $

Now let $\mathfrak{l}$ be a $\left(m+n\right)$-dimensional real vector space,
with an inner product $\left\langle\cdot,\cdot\right\rangle$ and an
orthonormal basis $\left\\{e_{i},e_{A}^{\prime}\right\\}$, where $i=1,...,m+1\
$and $A=2,...,n$. Then, for each $s=\left(x,y\right)\in\mathbb{L},$ define

$\begin{array}[]{c}\rho_{s}:\mathfrak{l}\longrightarrow T_{s}\mathbb{L}\\\
e_{i}\mapsto\left.f_{i}\right|_{s}\\\
e_{A}^{\prime}\mapsto\left.T_{A}\right|_{s}\end{array},$ (4.32)

Since $T_{1}=x^{i}f_{i}$, we see that
$\rho_{s}^{-1}\left(T_{1}\right)=x^{i}e_{i}.$ Hence define
$e_{1}^{\prime}=x^{i}e_{i}$. As before, using (2.12) and (4.31), we define the
bracket on $\mathfrak{l}$ between basis elements $e_{i}$ and $e_{j}$ as:

$\displaystyle\left[e_{i},e_{j}\right]^{\left(s\right)}$ $\displaystyle=$
$\displaystyle-\rho_{s}^{-1}\left(\left.\left[f_{i},f_{j}\right]\right|_{s}\right)$
(4.33) $\displaystyle=$ $\displaystyle e_{i}x_{j}-e_{j}x_{i}.$

Similarly, for $A=2,...,n$, and using (4.27), we see that

$\displaystyle\left[e_{i},e_{A}^{\prime}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle-\rho_{s}^{-1}\left(\left.\left[b_{i},T_{A}\right]\right|_{s}\right)$
(4.34) $\displaystyle=$ $\displaystyle-
x_{i}\rho_{s}^{-1}\left(\left[T_{1},T_{A}\right]\right)$ $\displaystyle=$
$\displaystyle-x_{i}D_{1A}^{\ \ \ \tilde{C}}f_{\tilde{C}}$

for $\tilde{C}=1,...,n$. Moreover,

$\displaystyle\left[e_{i},e_{1}^{\prime}\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle
x^{j}\left[e_{i},e_{j}\right]^{\left(s\right)}$ (4.35) $\displaystyle=$
$\displaystyle e_{i}-x_{i}e_{1}^{\prime}$
$\displaystyle\left[e_{1}^{\prime},e_{A}^{\prime}\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle
x^{i}\left[e_{i},e_{A}^{\prime}\right]^{\left(s\right)}$ (4.36)
$\displaystyle=$ $\displaystyle-D_{1A}^{\ \ \
\tilde{C}}e_{\tilde{C}}^{\prime}$

Now, for $B=2,...,n$,

$\displaystyle\left[e_{A}^{\prime},e_{B}^{\prime}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle-\rho_{s}^{-1}\left(\left.\left[T_{A},T_{B}\right]\right|_{s}\right)$
(4.37) $\displaystyle=$ $\displaystyle-D_{AB}^{\ \ \
\tilde{C}}e_{\tilde{C}}^{\prime}.$

For $\tilde{A},\tilde{B},\tilde{C}=1,..,n$, we can then summarize

$\left[e_{\tilde{A}}^{\prime},e_{\tilde{B}}^{\prime}\right]^{\left(s\right)}=-D_{\tilde{A}\tilde{B}}^{\
\ \ \tilde{C}}e_{\tilde{C}}^{\prime}.$ (4.38)

Overall, we see that for $\left[e_{i},e_{j}\right]^{\left(s\right)}$ we obtain
the same structure constants as in the case of $S^{m}\times S^{1}$, for
$\left[e_{\tilde{A}}^{\prime},e_{\tilde{B}}^{\prime}\right]^{\left(s\right)},$
the structure constants are defined by the structure constants on the
parallelizable space $N$, and only the mixed brackets (4.34) and (4.35) are
new. We can combine these both in a single expression for
$\tilde{A},\tilde{C}=1,...,n$ as

$\left[e_{i},e_{\tilde{A}}^{\prime}\right]^{\left(s\right)}=-x_{i}D_{1\tilde{A}}^{\
\ \
\tilde{C}}e_{\tilde{C}}^{\prime}+\delta_{1\tilde{A}}\left(e_{i}-x_{i}e_{1}^{\prime}\right).$
(4.39)

If we consider now the Jacobi identity for
$\left[\cdot,\cdot\right]^{\left(s\right)}$, we find from Lemma 4.2, that the
Jacobi identity is satisfied for $e_{i},e_{j},e_{k}.$ Similarly, for
$e_{\tilde{A}}^{\prime},e_{\tilde{B}}^{\prime},e_{\tilde{C}}^{\prime},$ the
Jacobi identity is given by (2). Now consider a mix of $e_{i}$ and
$e_{A}^{\prime}$:

$\displaystyle\left[e_{i},\left[e_{j},e_{A}^{\prime}\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle-\left[e_{i},x_{j}D_{1A}^{\ \ \
\tilde{C}}e_{\tilde{C}}^{\prime}\right]^{\left(s\right)}$ $\displaystyle=$
$\displaystyle x_{i}x_{j}D_{1A}^{\ \ \ \tilde{C}}D_{1\tilde{C}}^{\ \ \
\tilde{B}}e_{\tilde{B}}^{\prime}+x_{i}x_{j}e_{1}^{\prime}D_{1A}^{\ \ \
1}e_{i}-x_{j}D_{1A}^{\ \ \ 1}e_{i}$
$\displaystyle\left[e_{j},\left[e_{A}^{\prime},e_{i}\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle-\left[e_{j},\left[e_{i},e_{A}^{\prime}\right]^{\left(s\right)}\right]^{\left(s\right)}=-x_{i}x_{j}D_{1A}^{\
\ \ \tilde{C}}D_{1\tilde{C}}^{\ \ \
\tilde{B}}e_{\tilde{B}}^{\prime}-x_{i}x_{j}e_{1}^{\prime}D_{1A}^{\ \ \
1}e_{i}+x_{i}D_{1A}^{\ \ \ 1}e_{j}$
$\displaystyle\left[e_{A}^{\prime},\left[e_{i},e_{j}\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle\left[e_{A}^{\prime},e_{i}x_{j}-e_{j}x_{i}\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle-x_{i}x_{j}D_{1A}^{\ \ \
\tilde{C}}e_{\tilde{C}}^{\prime}+x_{i}x_{j}D_{1A}^{\ \ \
\tilde{C}}e_{\tilde{C}}^{\prime}=0$

Hence,

$\displaystyle\left[e_{i},\left[e_{j},e_{A}^{\prime}\right]^{\left(s\right)}\right]^{\left(s\right)}+\left[e_{j},\left[e_{A}^{\prime},e_{i}\right]^{\left(s\right)}\right]^{\left(s\right)}+\left[e_{A}^{\prime},\left[e_{i},e_{j}\right]^{\left(s\right)}\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle D_{1A}^{\ \ \
1}\left(x_{i}e_{j}-x_{j}e_{i}\right)$ $\displaystyle=$ $\displaystyle-
D_{1A}^{\ \ \ 1}\left[e_{i},e_{j}\right]^{\left(s\right)}$

We see that whenever $D_{1A}^{\ \ \ 1}\neq 0$, the Jacobi identity is not
satisfied, and the bracket $\left[\cdot,\cdot\right]^{\left(s\right)}$ does
not define a Lie algebra. In particular, even if on $N$, the bracket algebra
at $y\in N$ is a Lie algebra, if for some $A$, $D_{1A}^{\ \ \
1}\left(y\right)\neq 0$, then the bracket algebra at $s\in\mathbb{L}$ is not a
Lie algebra, and from (2), $\left(\mathfrak{l},a^{\left(s\right)}\right)$ is
not a Lie triple system.

As a particular example, consider $S^{m}\times S^{n}\times S^{1}.$ We know
$S^{n}\times S^{1}$ is parallelizable, and using the same parallelization on
$S^{n}\times S^{1}$ as we did earlier on $S^{m}\times S^{1}$, but labelling
the basis elements on $T\left(S^{m}\times S^{1}\right)$ as $T_{A}$ for
$A=1,...,n+1$, from (4.6), we find that in this case, for
$s=\left(x,y,\phi\right)$, with $x$ being the coordinates in
$\mathbb{R}^{m+1}\supset S^{m}$, $y$ being the coordinates in
$\mathbb{R}^{n+1}\supset S^{n}$, and $\phi$ being the coordinate on $\phi$,

$\left.\left[T_{A},T_{B}\right]\right|_{y}=y_{A}T_{B}-y_{B}T_{A}.$

in this case, for $A\neq 1$,

$D_{1A}^{\ \ 1}\left(y\right)=-y_{A}.$

Thus, as long as $y\neq\left(1,0,...,0\right),$ this will be non-zero for some
$A$, and the algebra of $\left[\cdot,\cdot\right]^{\left(s\right)}$ for
$s=\left(x,y\right)$ will not satisfy the Jacobi identity.

##  Appendix A Appendix

Proof of Theorem 4.3. To complete the proof of Theorem 4.3, we need to show
that given

$a^{\left(s\right)}\left(\xi,\eta,\gamma\right)=\gamma\left(\left\langle\xi,\eta\right\rangle+\left\langle
x,\xi\right\rangle\left\langle
x,\eta\right\rangle\right)-\eta\left(\left\langle\xi,\gamma\right\rangle+\left\langle
x,\xi\right\rangle\left\langle x,\gamma\right\rangle\right),$ (A.1)

for any $\xi,\eta,\gamma,u,v\in\mathfrak{l},$

$\displaystyle
a_{\xi,\eta}^{\left(s\right)}\left(a^{\left(s\right)}\left(u,v,\gamma\right)\right)$
$\displaystyle=$ $\displaystyle
a^{\left(s\right)}\left(a^{\left(s\right)}\left(u,\xi,\eta\right),v,\gamma\right)$
$\displaystyle+a^{\left(s\right)}\left(u,a^{\left(s\right)}\left(v,\xi,\eta\right),\gamma\right)$
$\displaystyle+a^{\left(s\right)}\left(u,v,a^{\left(s\right)}\left(\gamma,\xi,\eta\right)\right).$

Now, let

$A=a^{\left(s\right)}\left(u,v,\gamma\right)=\gamma\left(\left\langle
u,v\right\rangle+\left\langle x,u\right\rangle\left\langle
x,v\right\rangle\right)-v\left(\left\langle u,\gamma\right\rangle+\left\langle
x,u\right\rangle\left\langle x,\gamma\right\rangle\right)$

,

$a^{\left(s\right)}\left(A,\xi,\eta\right)=\eta\left(\left\langle
A,\xi\right\rangle+\left\langle x,A\right\rangle\left\langle
x,\xi\right\rangle\right)-\xi\left(\left\langle
A,\eta\right\rangle+\left\langle x,A\right\rangle\left\langle
x,\eta\right\rangle\right)\text{,}$

but

$\displaystyle\left\langle A,\xi\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\gamma,\xi\right\rangle\left\langle
u,v\right\rangle+\left\langle\gamma,\xi\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
v,\xi\right\rangle\left\langle u,\gamma\right\rangle-\left\langle
v,\xi\right\rangle\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle$ $\displaystyle\left\langle A,\eta\right\rangle$
$\displaystyle=$
$\displaystyle\left\langle\gamma,\eta\right\rangle\left\langle
u,v\right\rangle+\left\langle\gamma,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
v,\eta\right\rangle\left\langle u,\gamma\right\rangle-\left\langle
v,\eta\right\rangle\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle$ $\displaystyle\left\langle A,x\right\rangle$
$\displaystyle=$ $\displaystyle\left\langle\gamma,x\right\rangle\left\langle
u,v\right\rangle+\left\langle\gamma,x\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
v,x\right\rangle\left\langle u,\gamma\right\rangle-\left\langle
v,x\right\rangle\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\gamma,x\right\rangle\left\langle
u,v\right\rangle-\left\langle v,x\right\rangle\left\langle
u,\gamma\right\rangle.$

Thus,

$\displaystyle
a^{\left(s\right)}\left(a^{\left(s\right)}\left(u,v,\gamma\right),\xi,\eta\right)$
$\displaystyle=$
$\displaystyle\eta\left(\left\langle\gamma,\xi\right\rangle\left\langle
u,v\right\rangle+\left\langle\gamma,\xi\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
v,\xi\right\rangle\left\langle u,\gamma\right\rangle-\left\langle
v,\xi\right\rangle\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle\right)$
$\displaystyle-\xi\left(\left\langle\gamma,\eta\right\rangle\left\langle
u,v\right\rangle+\left\langle\gamma,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
v,\eta\right\rangle\left\langle u,\gamma\right\rangle-\left\langle
v,\eta\right\rangle\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle\right)$ $\displaystyle+\left(\eta\left\langle
x,\xi\right\rangle-\xi\left\langle
x,\eta\right\rangle\right)\left(\left\langle\gamma,x\right\rangle\left\langle
u,v\right\rangle-\left\langle v,x\right\rangle\left\langle
u,\gamma\right\rangle\right)$

Similarly,

$\displaystyle
a^{\left(s\right)}\left(a^{\left(s\right)}\left(u,\xi,\eta\right),v,\gamma\right)$
$\displaystyle=$
$\displaystyle\gamma\left(\left\langle\eta,v\right\rangle\left\langle
u,\xi\right\rangle+\left\langle\eta,v\right\rangle\left\langle
x,u\right\rangle\left\langle x,\xi\right\rangle-\left\langle
v,\xi\right\rangle\left\langle u,\eta\right\rangle-\left\langle
v,\xi\right\rangle\left\langle x,u\right\rangle\left\langle
x,\eta\right\rangle\right)$
$\displaystyle-v\left(\left\langle\gamma,\eta\right\rangle\left\langle
u,\xi\right\rangle+\left\langle\gamma,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle
x,\xi\right\rangle-\left\langle\xi,\gamma\right\rangle\left\langle
u,\eta\right\rangle-\left\langle\xi,\gamma\right\rangle\left\langle
x,u\right\rangle\left\langle x,\eta\right\rangle\right)$
$\displaystyle+\left(\gamma\left\langle x,v\right\rangle-v\left\langle
x,\gamma\right\rangle\right)\left(\left\langle\eta,x\right\rangle\left\langle
u,\xi\right\rangle-\left\langle\xi,x\right\rangle\left\langle
u,\eta\right\rangle\right)$

Now let

$\displaystyle B$ $\displaystyle=$ $\displaystyle
a^{\left(s\right)}\left(v,\xi,\eta\right)=\eta\left(\left\langle
v,\xi\right\rangle+\left\langle x,v\right\rangle\left\langle
x,\xi\right\rangle\right)-\xi\left(\left\langle
v,\eta\right\rangle+\left\langle x,v\right\rangle\left\langle
x,\eta\right\rangle\right)$ $\displaystyle\left\langle B,u\right\rangle$
$\displaystyle=$ $\displaystyle\left\langle\eta,u\right\rangle\left\langle
v,\xi\right\rangle+\left\langle\eta,u\right\rangle\left\langle
x,v\right\rangle\left\langle
x,\xi\right\rangle-\left\langle\xi,u\right\rangle\left\langle
v,\eta\right\rangle-\left\langle\xi,u\right\rangle\left\langle
x,v\right\rangle\left\langle x,\eta\right\rangle$ $\displaystyle\left\langle
B,x\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\eta,x\right\rangle\left\langle
v,\xi\right\rangle-\left\langle\xi,x\right\rangle\left\langle
v,\eta\right\rangle$

and

$\displaystyle
a^{\left(s\right)}\left(u,a^{\left(s\right)}\left(v,\xi,\eta\right),\gamma\right)$
$\displaystyle=$ $\displaystyle\gamma\left(\left\langle
u,B\right\rangle+\left\langle x,u\right\rangle\left\langle
x,B\right\rangle\right)-B\left(\left\langle u,\gamma\right\rangle+\left\langle
x,u\right\rangle\left\langle x,\gamma\right\rangle\right)$ $\displaystyle=$
$\displaystyle\gamma\left(\left\langle\eta,u\right\rangle\left\langle
v,\xi\right\rangle+\left\langle\eta,u\right\rangle\left\langle
x,v\right\rangle\left\langle
x,\xi\right\rangle-\left\langle\xi,u\right\rangle\left\langle
v,\eta\right\rangle-\left\langle\xi,u\right\rangle\left\langle
x,v\right\rangle\left\langle x,\eta\right\rangle\right)$
$\displaystyle+\gamma\left\langle
x,u\right\rangle\left(\left\langle\eta,x\right\rangle\left\langle
v,\xi\right\rangle-\left\langle\xi,x\right\rangle\left\langle
v,\eta\right\rangle\right)$ $\displaystyle-\eta\left(\left\langle
v,\xi\right\rangle+\left\langle x,v\right\rangle\left\langle
x,\xi\right\rangle\right)\left(\left\langle u,\gamma\right\rangle+\left\langle
x,u\right\rangle\left\langle x,\gamma\right\rangle\right)$
$\displaystyle+\xi\left(\left\langle v,\eta\right\rangle+\left\langle
x,v\right\rangle\left\langle x,\eta\right\rangle\right)\left(\left\langle
u,\gamma\right\rangle+\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle\right)$

Similarly,

$\displaystyle
a^{\left(s\right)}\left(u,v,a^{\left(s\right)}\left(\gamma,\xi,\eta\right)\right)$
$\displaystyle=$
$\displaystyle-a^{\left(s\right)}\left(u,a^{\left(s\right)}\left(\gamma,\xi,\eta\right),v\right)$
$\displaystyle=$
$\displaystyle-v\left(\left\langle\eta,u\right\rangle\left\langle\gamma,\xi\right\rangle+\left\langle\eta,u\right\rangle\left\langle
x,\gamma\right\rangle\left\langle
x,\xi\right\rangle-\left\langle\xi,u\right\rangle\left\langle\gamma,\eta\right\rangle-\left\langle\xi,u\right\rangle\left\langle
x,\gamma\right\rangle\left\langle x,\eta\right\rangle\right)$
$\displaystyle-v\left\langle
x,u\right\rangle\left(\left\langle\eta,x\right\rangle\left\langle\gamma,\xi\right\rangle-\left\langle\xi,x\right\rangle\left\langle\gamma,\eta\right\rangle\right)$
$\displaystyle+\eta\left(\left\langle\gamma,\xi\right\rangle+\left\langle
x,\gamma\right\rangle\left\langle x,\xi\right\rangle\right)\left(\left\langle
u,v\right\rangle+\left\langle x,u\right\rangle\left\langle
x,v\right\rangle\right)$
$\displaystyle-\xi\left(\left\langle\gamma,\eta\right\rangle+\left\langle
x,\gamma\right\rangle\left\langle x,\eta\right\rangle\right)\left(\left\langle
u,v\right\rangle+\left\langle x,u\right\rangle\left\langle
x,v\right\rangle\right)$

Adding together the terms
$a^{\left(s\right)}\left(a^{\left(s\right)}\left(u,\xi,\eta\right),v,\gamma\right),$
$a^{\left(s\right)}\left(u,a^{\left(s\right)}\left(v,\xi,\eta\right),\gamma\right),$
$a^{\left(s\right)}\left(u,v,a^{\left(s\right)}\left(\gamma,\xi,\eta\right)\right),$
we see that the contributions that are proportional to $v$ and $\gamma$
vanish. Consider the contributions of $\xi:$

$\displaystyle\left(\left\langle v,\eta\right\rangle+\left\langle
x,v\right\rangle\left\langle x,\eta\right\rangle\right)\left(\left\langle
u,\gamma\right\rangle+\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle\right)-\left(\left\langle\gamma,\eta\right\rangle+\left\langle
x,\gamma\right\rangle\left\langle x,\eta\right\rangle\right)\left(\left\langle
u,v\right\rangle+\left\langle x,u\right\rangle\left\langle
x,v\right\rangle\right)$ $\displaystyle=$ $\displaystyle\left\langle
v,\eta\right\rangle\left\langle u,\gamma\right\rangle+\left\langle
v,\eta\right\rangle\left\langle x,u\right\rangle\left\langle
x,\gamma\right\rangle+\left\langle x,v\right\rangle\left\langle
x,\eta\right\rangle\left\langle u,\gamma\right\rangle+\left\langle
x,v\right\rangle\left\langle x,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle x,\gamma\right\rangle$
$\displaystyle-\left\langle\gamma,\eta\right\rangle\left\langle
u,v\right\rangle-\left\langle\gamma,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
x,\gamma\right\rangle\left\langle x,\eta\right\rangle\left\langle
u,v\right\rangle-\left\langle x,\gamma\right\rangle\left\langle
x,\eta\right\rangle\left\langle x,u\right\rangle\left\langle x,v\right\rangle$
$\displaystyle=$ $\displaystyle\left\langle v,\eta\right\rangle\left\langle
u,\gamma\right\rangle-\left\langle\gamma,\eta\right\rangle\left\langle
u,v\right\rangle+\left\langle v,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle x,\gamma\right\rangle+\left\langle
x,v\right\rangle\left\langle x,\eta\right\rangle\left\langle
u,\gamma\right\rangle$
$\displaystyle-\left\langle\gamma,\eta\right\rangle\left\langle
x,u\right\rangle\left\langle x,v\right\rangle-\left\langle
x,\gamma\right\rangle\left\langle x,\eta\right\rangle\left\langle
u,v\right\rangle$

This is precisely the coefficient of $\xi$ in
$a^{\left(s\right)}\left(a^{\left(s\right)}\left(u,v,\gamma\right),\xi,\eta\right).$
The expression is skew-symmetric in $\xi$ and $\eta,$ so the same conclusion
follows.

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