# The Coulomb gauge in non-associative gauge theory

Sergey Grigorian

###### Abstract

The aim of this paper is to extend existence results for the Coulomb gauge
from standard gauge theory to a non-associative setting. Non-associative gauge
theory is based on smooth loops, which are the non-associative analogs of Lie
groups. The main components of the theory include a finite-dimensional smooth
loop $\mathbb{L}$, its tangent algebra $\mathfrak{l},$ a finite-dimensional
Lie group $\Psi$, that is the pseudoautomorphism group of $\mathbb{L}$, a
smooth manifold $M$ with a principal $\Psi$-bundle $\mathcal{P}$, and
associated bundles $\mathcal{Q}$ and $\mathcal{A}$ with fibers $\mathbb{L}$
and $\mathfrak{l}$, respectively. A configuration in this theory is defined as
a pair $\left(s,\omega\right)$, where $s$ is a section of $\mathbb{Q}$ and
$\omega$ is a connection on $\mathcal{P}$. The torsion
$T^{\left(s,\omega\right)}$ is the key object in the theory, with a role
similar to that of a connection in standard gauge theory. The original
motivation for this study comes from $G_{2}$-geometry, and the questions of
existence of $G_{2}$-structures with particular torsion types. In particular,
given a fixed connection, we prove existence of configurations with
divergence-free torsion, given a sufficiently small torsion in a Sobolev norm.

###### Contents

  1. 1 Introduction
  2. 2 Smooth Loops
  3. 3 Loop-valued maps
  4. 4 Gauge theory
  5. 5 $G_{2}$-manifolds
  6. A Appendix

##  1 Introduction

The main goal of this work is to extend results on existence of Coulomb gauge
transformations from standard gauge theory to a _non-associative_ setting. One
of highly successful areas at the intersection of differential geometry,
analysis, and mathematical physics is gauge theory, which, as it is well-
known, is the study of connections on bundles with particular Lie groups as
the structure groups. In [26], the author initiated a theory of smooth loops,
which are non-associative analogs of Lie groups, and began the development of
gauge theory based on loops, i.e. a non-associative gauge theory. The key
example of a non-associative smooth loop is the loop of unit octonions. A non-
associative gauge theory has the following components:

  1. 1.

A finite-dimensional smooth (right) loop $\mathbb{L}$, i.e. a smooth manifold
with a right multiplication diffeomorphism $R_{p}$ defined for every
$p\in\mathbb{L}$ and a distinguished identity element $1\in\mathbb{L},$ with
tangent algebra $\mathfrak{l}$ at identity.

  2. 2.

A finite-dimensional Lie group $\Psi$ that is the _pseudoautomorphism_ group
of $\mathbb{L}$, a non-associative generalization of an automorphism group.

  3. 3.

A smooth manifold $M$ with a principal $\Psi$-bundle $\mathcal{P},$ and
associated bundles $\mathcal{Q}$ and $\mathcal{A}$ with fibers $\mathbb{L}$
and $\mathfrak{l},$ respectively.

  4. 4.

A configuration is defined by a pair $\left(s,\omega\right),$ where $s$ is a
section of $\mathcal{Q}$ and $\omega$ is a connection on $\mathcal{P}.$
Together they define the _torsion_ $T^{\left(s,\omega\right)},$ which is an
$\mathcal{A}$-valued $1$-form on $M.$ The torsion is then the key object in
the theory, in the same way that a connection is the key object in standard
gauge theory.

  5. 5.

In addition to standard gauge transformations of $\omega$ by $\Psi,$ we now
also have transformations of $s$ induced by loop multiplication. Both of these
kinds of transformations induce transformations of the torsion.

The original motivation for studying non-associative gauge theories comes from
$G_{2}$-geometry [24]. A $G_{2}$-structure on a $7$-dimensional Riemannian
manifold is a reduction of the structure group of the orthornormal frame
bundle from $SO\left(7\right)$ to $G_{2},$ which is the automorphism group of
the octonions. A further relationship between $G_{2}$-structures and the
octonions is that unit norm sections of an octonion bundle parametrize
$G_{2}$-structures that are associated with the same metric, also known as
_isometric_ $G_{2}$-structures. A defining characteristic of a
$G_{2}$-structure is its torsion, and one of the unanswered questions in
$G_{2}$-geometry is which torsion types are admissible within a fixed metric
class. One of the main goals in the theory of $G_{2}$-structures is to obtain
existence results for torsion-free $G_{2}$-structures, similar to the Yau’s
Theorem [49], that settled the existence question for Calabi-Yau manifolds.
While this goal is formulated in terms of $G_{2}$-structures, the real
question is the existence of a Riemannian metric with holonomy group equal to
$G_{2}.$ The fact that for each metric there exists an entire family of
compatible $G_{2}$-structures presents a degeneracy in this problem. Some of
the existing approaches involve Laplacian flows of $G_{2}$-structures [4, 5,
7, 8, 9, 23, 21, 33, 38, 39, 40], with the hope of a flow eventually
converging to a torsion-free $G_{2}$-structure. As shown in [21], the
Laplacian flow of a generic $G_{2}$-structure has a component that moves
within a metric class, and that component is precisely given by $\mathop{\rm
div}T.$ Laplacian flows have been more successful for _closed_
$G_{2}$-structures, in which case $\mathop{\rm div}T$ automatically vanishes,
and thus the degeneracy is resolved. More generally, however, this degeneracy
is the source of non-parabolicity of Laplacian flows, such as in the case of
_co-closed_ $G_{2}$-structures [21]. Therefore, the condition $\mathop{\rm
div}T=0$ can be regarded as a _gauge-fixing condition_. Moreover, in [24], it
was found that on a compact manifold, $G_{2}$-structures with $\mathop{\rm
div}T=0$ are precisely the critical points of the $L^{2}$-norm of the torsion
when restricted to a fixed metric class. As shown in [24, 26], within the loop
bundle framework, this is the precise analog of the Coulomb gauge condition.

Existence of $G_{2}$-structures with divergence-free torsion has been studied
from different perspectives by several authors: Bagaglini in [3]; Dwivedi,
Gianniotis, and Karigiannis in [12]; the author in [25]; Loubeau and Sá Earp
in [41]. All these approaches relied on a flow of isometric $G_{2}$-structures
(or more generally, geometric structures in [41] and [13]), and have shown
existence of a $G_{2}$-structure with divergence-free torsion as a long-term
limit of the flow, given sufficiently small pointwise initial torsion or
another quantity, the entropy.

The interpretation of isometric $G_{2}$-structures as an octonionic non-
associative gauge theory allows to adapt some gauge theory techniques in this
setting. Moreover, without much additional effort, more general loops can be
considered, with potential wider-reaching applications.

In gauge theory there are a number of versions of local and global existence
results for connections in the Coulomb gauge, depending on the desired
regularity [11, 14, 15, 16, 17, 47, 48]. In this paper we use the Quantitative
Implicit Function Theorem for Banach Spaces, as given in [17], to prove the
following main result.

###### Theorem A

Suppose $\mathbb{L}$ is a smooth compact loop with tangent algebra
$\mathfrak{l}$ and pseudoautomorphism group $\Psi.$ Let $\left(M,g\right)$ be
a closed, smooth Riemannian manifold of dimension $n\geq 2,$ and let
$\mathcal{P}$ be a $\Psi$-principal bundle over $M$ and let $\mathcal{A}$ be
the associated vector bundle to $\mathcal{P}$ with fibers isomorphic to
$\mathfrak{l}$. Let $\omega$ be a smooth connection on $\mathcal{P}.$ Also,
suppose $k$ is a non-negative integer and $r\geq 0$ such that $kr>n.$ Then,
there exist constants $\delta\in(0,1]$ and $K\in\left(0,\infty\right),$ such
that if $s\in\Gamma\left(\mathcal{Q}\right)$ is a smooth defining section for
which

$\left\|T^{\left(s,\omega\right)}\right\|_{W^{k-1,r}}<\delta,$

then there exists a section $A\in W^{k,r}\left(\mathcal{Q}^{\prime}\right),$
such that

$\left(d^{\omega}\right)^{\ast}T^{\left(As,\omega\right)}=0$

and

$\left\|T^{\left(As,\omega\right)}\right\|_{W^{k-1,r}}<K\left\|T^{\left(s,\omega\right)}\right\|_{W^{k-1,r}}\left(1+\left\|T^{\left(s,\omega\right)}\right\|_{W^{k-1,r}}^{k-1}\right).$
(1.1)

If moreover, $\left(k-1\right)r\geq n,$ then $A$ is smooth.

For $G_{2}$-structures, this gives the following result for existence of
smooth $G_{2}$-structures with divergence-free torsion.

###### Theorem B

Suppose $M$ is a closed $7$-dimensional manifold with a smooth
$G_{2}$-structure $\varphi$ with torsion $T$ with respect to the Levi-Civita
connection $\nabla.$ Suppose $U\mathbb{O}M$ is the corresponding unit octonion
bundle. Also, suppose $k$ is a positive integer and $r$ is a positive real
number such that $kr>7.$ Then, there exist constants $\delta\in(0,1]$ and
$K\in\left(0,\infty\right),$ such that if $T$ satisfies

$\left\|T\right\|_{W^{k,r}}<\delta,$

then there exists a smooth section $V\in\Gamma\left(U\mathbb{O}M\right),$ such
that

$\mathop{\rm div}T^{\left(V\right)}=0$

and

$\left\|T^{\left(V\right)}\right\|_{W^{k,r}}<K\left\|T\right\|_{W^{k,r}}\left(1+\left\|T\right\|_{W^{k,r}}^{k}\right).$
(1.2)

The results presented in this paper are of interest and importance in their
own right, but perhaps even more crucially, they show that some well-known
results and techniques from classical gauge theory can be reinterpreted and
adapted in a non-associative setting. In particular, this may open the door to
some analogues of Uhlenbeck compactness and a better understanding of the
torsion of non-associative gauge theories. Furthermore, a non-associative
version of Yang-Mills equations can be considered. Moreover, any such advances
will give immediate results in $G_{2}$-geometry.

The structure of this paper is the following. In Section 2, we give an
overview of smooth loops, extending [26]. We give the basic properties of a
smooth loop $\mathbb{L}$, define the pseudoautomorphism group $\Psi$ and the
tangent algebra $\mathfrak{l}$ at identity. The algebra $\mathfrak{l}$ is a
generalization of a Lie algebra, but due to the non-associativity of
$\mathbb{L}$ does not satisfy the Jacobi identity. Similarly as for Lie
algebras, there is a notion of an exponential map. There is however a family
of brackets $\left[\cdot,\cdot\right]^{\left(s\right)}$ on $\mathfrak{l}$,
defined for each point $s\in\mathbb{L}.$ For later use, we also give estimates
for the exponential and adjoint maps. In particular, we analyze solutions of
the following initial value problem for $\mathfrak{l}$-valued maps
$\eta\left(t\right):$

$\left\\{\begin{array}[]{c}\frac{d\eta\left(t\right)}{dt}=\left[\xi,\eta\left(t\right)\right]^{\left(\exp_{s}\left(t\xi\right)s\right)}\\\
\eta\left(0\right)=\eta_{0}\end{array}\right.,$ (1.3)

where $\xi\in\mathfrak{l}$ and $s\in\mathbb{L}.$

In Section 3, we switch attention to loop-valued maps. In particular, given a
smooth manifold $M,$ consider a map $s:M\longrightarrow\mathbb{L}.$ Using this
map, we may define products of $\mathbb{L}$-valued maps and brackets of
$\mathfrak{l}$-valued maps. Then, using the right quotient, translating the
differential $ds$ to the tangent space at $1\in\mathbb{L},$ we obtain an
$\mathfrak{l}$-valued $1$-form $\theta_{s}$ on $M,$ which is the analogue of
the Darboux derivative of Lie group-valued maps [45]. The differentials of
various operations defined by $s$ are then expressed in terms of $\theta_{s}.$
Suppose $A\left(t\right)=\exp_{s}\left(t\xi\right)$ for some
$\mathfrak{l}$-valued map $\xi.$ We show that $\theta_{A\left(t\right)s}$
satisfies a non-homogeneous version of (1.3).

Further, we define Sobolev spaces of maps from $M$, and show in Lemma 3.9
that, similarly as for Lie groups, $s\in W^{k,r}\left(M,\mathbb{L}\right)$ if
and only if $\theta_{s}\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{l}\right).$ Using the evolution
equation satisfied by $\theta_{A\left(t\right)s}$ then allows us to obtain
Sobolev space estimates of $\theta_{A\left(t\right)s}$ and other quantities
that satisfy equations based on (1.3).

###### Theorem C

Let $M$ be a compact Riemannian manifold and $\mathbb{L}$ is a smooth compact
loop. Suppose $kr>n=\dim M.$ Let $s\in W^{k,r}\left(M,\mathbb{L}\right)$ and
$\xi\in W^{k,r}\left(M,\mathfrak{l}\right),$ and suppose
$A=\exp_{s}\left(\xi\right).$ Then,

$\left\|\theta_{As}\right\|_{W^{k-1,r}}\lesssim
e^{Ck\left\|\xi\right\|_{C^{0}}}\left(\Theta^{k}+\Theta\right),$ (1.4)

where
$\Theta=\left\|\theta_{s}\right\|_{W^{k-1,r}}+\left\|\xi\right\|_{W^{k,r}}.$

Similarly, if $X=X\left(1\right),$ where $X\left(t\right)$ is $1$-parameter
family of $\mathfrak{l}$-valued maps that satisfies

$\left\\{\begin{array}[]{c}\frac{dX\left(t\right)}{dt}=\left[\xi,X\left(t\right)\right]^{A\left(t\right)s}+Y\\\
X\left(0\right)=X_{0},\end{array}\right.$

for $\mathfrak{l}$-valued maps $X_{0}$ and $Y$, then,

$\left\|X\right\|_{W^{k,r}}\lesssim
e^{C\left(k+1\right)\left\|\xi\right\|_{C^{0}}}\left(\left\|X_{0}\right\|_{W^{k,r}}+\left\|Y\right\|_{W^{k,r}}\right)\left(\Theta^{k+1}+\Theta\right).$
(1.5)

In Section 4, we introduce a principal $\Psi$-bundle $\mathcal{P}$ over a
compact manifold $M$, and then apply the above results to $\Psi$-equivariant
maps from $\mathcal{P}$ to a loop $\mathbb{L}$ and other related spaces. This
immediately then allows to consider sections of bundles over $M$ that are
associated to $\mathcal{P}.$ In particular, suppose we have a connection
$\omega$ on $\mathcal{P}$ and suppose $s$ is a section of the associated loop
bundle $\mathcal{Q},$ with fibers diffeomorphic to $\mathbb{L}.$ It uniquely
corresponds to a $\Psi$-equivariant map
$\tilde{s}:\mathcal{P}\longrightarrow\mathbb{L},$ and thus we obtain the
equivariant $\mathfrak{l}$-valued $1$-form $\theta_{\tilde{s}}$ on
$\mathcal{P}.$ On the other hand, the connection $\omega$ defines a
decomposition of $T\mathcal{P}$ into vertical and horizontal subspaces.
Therefore, we may compose $\theta_{\tilde{s}}$ with the horizontal projection
to obtain a _basic_ , i.e. horizontal and equivariant $\mathfrak{l}$-valued
$1$-form on $\mathcal{P}.$ This then corresponds to a section of a bundle over
$M,$ and gives us the _torsion_ $T^{\left(s,\omega\right)}$ of the
configuration $\left(s,\omega\right).$ Defining fiberwise loop multiplication,
we see that all the possible configurations with a fixed $\omega$ may be
obtained by multiplying $s$ by some section $A.$ Therefore, the _loop gauge
transformations_ are precisely the transformations $s\mapsto As.$ Moreover, as
it was already known previously, [24, 25, 27, 26, 41], given appropriate
algebraic conditions on the loop, the critical points of the functional
$s\mapsto\int_{M}$ $\left|T^{\left(s,\omega\right)}\right|^{2}\mathop{\rm
vol}\nolimits$ are precisely the sections $s$ for which
$\left(d^{\omega}\right)^{\ast}T^{\left(s,\omega\right)}=0,$ which relates to
the previous discussion on divergence-free torsion and the Coulomb gauge.

Considering the transformations of $s$ of the form
$s\mapsto\exp_{s}\left(\xi\right)s$, for $\mathfrak{l}$-valued sections $\xi$,
and using the loop exponential map, as developed in Section 2, the quantity
$T^{\left(\exp_{s}\left(t\xi\right)s\,,\omega\right)}$ is then shown to
satisfy an ODE with the homogeneous part of the form (1.3). This allows to
rewrite $T^{\left(\exp_{s}\left(t\xi\right)s\,,\omega\right)}$ in terms of the
evolution operator for (1.3). The equation
$\left(d^{\omega}\right)^{\ast}T^{\left(\exp_{s}\left(\xi\right)s,\omega\right)}=0$
is then written as a second-order PDE for $\xi.$ This operator then defines a
smooth functions between appropriate Banach spaces of sections of vector
bundles, which allows to apply the Implicit Function Theorem from [17], to
show existence of solutions for sufficiently small initial
$T^{\left(s,\omega\right)}$ in an appropriate Sobolev norm, and thus prove
Theorem A.

In Section 5, we carefully apply the general theory of smooth loops to the
particular case of $G_{2}$-structures, and then Theorem B follows as an
immediate corollary of Theorem A.

### Acknowledgements

The author is supported by the National Science Foundation grant DMS-1811754.

##  2 Smooth Loops

For a detailed introduction to smooth loops, the reader is referred to [26].
The reader can also refer to [29, 34, 43, 44, 46] for a discussion of these
concepts.

###### Definition 2.1

A _loop_ $\mathbb{L}$ is a set with a binary operation $p\cdot q$ with
identity $1,$ and compatible left and right quotients $p\backslash q$ and
$p/q$, respectively.

In particular, existence of quotients is equivalent to saying that for any
$q\in\mathbb{L},$ the left and right product maps $L_{q}$ and $R_{q}$ are
invertible maps. Restricting to the smooth category, we obtain the definition
of a smooth loop.

###### Definition 2.2

A _smooth loop_ is a smooth manifold $\mathbb{L}$ with a loop structure such
that the left and right product maps are diffeomorphisms of $\mathbb{L}.$

###### Remark 2.3

In this paper we will not use the left quotient, so in fact everything that
follows also holds true for smooth _right_ loops, i.e. where only the right
quotient is defined, but the left product is not necessarily invertible.
However, for brevity, we will keep referring to _loops_ , rather than _right
loops_. As Example 2.4 below shows, smooth right loops are plentiful and easy
to construct.

###### Example 2.4

Suppose $G$ is a Lie group with a Lie subgroup $H$ and consider the left
quotient
$K=\scalebox{-1.0}[1.0]{$\nicefrac{{\scalebox{-1.0}[1.0]{$G$}}}{{\scalebox{-1.0}[1.0]{$H$}}}$}$.
Suppose $\sigma:K\longrightarrow G$ is a section of $G$, regarded as a bundle
over $K.$ In particular, $\sigma$ maps each right coset to a particular
representative of that coset. Suppose $\sigma\left(H\right)=1.$ A product
structure on $K$ is then given by

$\left(Ha\right)\circ\left(Hb\right)=Ha\left(\sigma\left(Hb\right)\right).$
(2.1)

Consider the equation $\left(Hx\right)\circ\left(Ha\right)=Hb,$ Since $\sigma$
is a section, we can see right away that we have a unique solution
$Hx=Hb\left(\sigma\left(Ha\right)\right)^{-1}.$ Thus,
$\left(\mathbb{L},\circ\right)$ has _right_ division, and is thus a _right
loop_[34, 43]. To define left division, and hence to obtain a full loop
structure, more structure is needed.

###### Definition 2.5

A _pseudoautomorphism_ of a smooth loop $\mathbb{L}$ is a diffeomorphism
$h:\mathbb{L}\longrightarrow\mathbb{L}$ for which there exists another
diffeomorphism $h^{\prime}:\mathbb{L}\longrightarrow\mathbb{L}$, known as the
partial pseudoautomorphism corresponding to $h,$ such that for any
$p,q\in\mathbb{L}$,

$h\left(pq\right)=h^{\prime}\left(p\right)h\left(q\right).$ (2.2)

In particular, $h^{\prime}=R_{h\left(1\right)}^{-1}\circ h$ and
$h^{\prime}\left(1\right)=1.$ The element $h\left(1\right)\in\mathbb{L}$ is
the _companion_ of $h^{\prime}.$ From (2.2), we also see the following
property of $h^{\prime}$ with respect to quotients:

$h^{\prime}\left(p/q\right)=h\left(p\right)/h\left(q\right).$ (2.3)

It is easy to see that the sets of pseudoautomorphisms and partial
pseudoautomorphisms are both groups. Denote the former by $\Psi$ and the
latter by $\Psi^{\prime}$. We also see that the _automorphism_ group of the
loop $\mathbb{L}$ is the subgroup $H\subset\Psi$ which is the stabilizer of
$1\in\mathbb{L}.$

###### Remark 2.6

To avoid introducing additional notation, but at the risk of some ambiguity,
we will use the same notation $\mathbb{L}$ to denote the underlying manifold,
the loop, and the $G$-set with the full action of $\Psi$. However, since
$\mathbb{L}$ also admits the action of $\Psi^{\prime},$ if a distinction
between the $G$-sets is needed, we will use $\mathbb{L}^{\prime}$ to denote
the set $\mathbb{L}$ with the action of $\Psi^{\prime},$

Let $r\in\mathbb{L}$, then we may define a modified product $\circ_{r}$ on
$\mathbb{L}$ via $p\circ_{r}q={\left(p\cdot qr\right)}/{r}$, so that
$\mathbb{L}$ equipped with product $\circ_{r}$ will be denoted by
$\left(\mathbb{L},\circ_{r}\right),$ the corresponding quotient will be
denoted by $/_{r}.$ We have the following properties [26].

###### Lemma 2.7

Let $h\in\Psi$. Then, for any $p,q,r\in\mathbb{L},$

$h^{\prime}\left(p\circ_{r}q\right)=h^{\prime}\left(p\right)\circ_{h\left(r\right)}h^{\prime}\left(q\right)\
\ \ \ \
h^{\prime}\left(p/_{r}q\right)=h^{\prime}\left(p\right)/_{h\left(r\right)}h^{\prime}\left(q\right).$
(2.4)

Also, for any $A\in\mathbb{L},$

$\left(p\circ_{r}\left(q\circ_{r}A\right)\right)/_{r}A=p\circ_{Ar}q.$ (2.5)

###### Lemma 2.8

Suppose $A\left(t\right)$ and $B\left(t\right)$ are smooth curves in
$\mathbb{L}$ with $A\left(t_{0}\right)=A_{0}$ and $B\left(t_{0}\right)=B_{0}$,
then

$\left.\frac{d}{dt}A\left(t\right)/B\left(t\right)\right|_{t=t_{0}}=\left.\frac{d}{dt}A\left(t\right)/B_{0}\right|_{t=t_{0}}-\left.\frac{d}{dt}\left(A_{0}/B_{0}\cdot
B\left(t\right)\right)/B_{0}\right|_{t=t_{0}}.$ (2.6)

Consider the tangent space $\mathfrak{l:=}T_{1}\mathbb{L}$ at
$1\in\mathbb{L}.$ For any $q\in\mathbb{L}$, the pushforward
$\left(R_{q}\right)_{\ast}$ of the right translation map $R_{q}$ defines a
linear isomorphism from $\mathfrak{l}$ to $T_{q}\mathbb{L}.$ In particular,
let us denote the linear map
$\left.\left(R_{q}\right)_{\ast}\right|_{1}:\mathfrak{l}\longrightarrow
T_{q}\mathbb{L}$ by $\rho_{q}$, and correspondingly,
$\left.\left(R_{q}^{-1}\right)_{\ast}\right|_{1}:T_{q}\mathbb{L}\longrightarrow\mathfrak{l}$
by $\rho_{q}^{-1}.$ Similarly, for left multiplication, define
$\lambda_{q}=\left.\left(L_{q}\right)_{\ast}\right|_{1}:\mathfrak{l}\longrightarrow
T_{q}\mathbb{L}.$ On a smooth right loop, $\lambda_{q}$ will not necessarily
be invertible. The corresponding maps with respect to the product $\circ_{r}$
will be denoted by $R_{q}^{\left(r\right)},$ $\rho_{q}^{\left(r\right)},$
$\lambda_{q}^{\left(r\right)}.$

###### Definition 2.9

For any $\xi\in\mathfrak{l,}$ define the _fundamental vector field_
$\rho\left(\xi\right)$ for any $q\in\mathbb{L},$
$\rho\left(\xi\right)_{q}=\rho_{q}\left(\xi\right).$

The above definition of the fundamental vector field is the analog of a right-
invariant vector field in Lie theory. However, in the loop case, although this
vector field is canonical in some sense, it is not _invariant_ under right
translations. We use fundamental vector fields to define the loop exponential
map.

###### Definition 2.10

Suppose $\mathbb{L}$ is a smooth loop and suppose $q\in\mathbb{L}.$ Then,
given $\xi\in\mathfrak{l},$ for sufficiently small $t,$ define

$p\left(t\right)=\exp_{q}\left(t\xi\right)q.$ (2.7)

to be the solution of the equation

$\left\\{\begin{array}[]{c}\frac{dp\left(t\right)}{dt}=\left.\rho\left(\xi\right)\right|_{p\left(t\right)}\\\
p\left(0\right)=q\end{array}\right..$ (2.8)

Equivalently, $\tilde{p}\left(t\right)=\exp_{q}\left(t\xi\right)$ satisfies

$\left\\{\begin{array}[]{c}\frac{d\tilde{p}\left(t\right)}{dt}=\left.\rho^{\left(q\right)}\left(\xi\right)\right|_{\tilde{p}\left(t\right)}\\\
\tilde{p}\left(0\right)=1\end{array}\right..$ (2.9)

###### Remark 2.11

In general, the solution $\exp_{q}$ will only be defined in a neighborhood of
$0\in\mathfrak{l},$ however as shown in [36, 42], if the loop $\mathbb{L}$ is
power-associative, so that powers of an element $p\in\mathbb{L}$ associate,
then $p\left(nh\right)=p\left(h\right)^{n}$ can be defined unambiguously. We
will show this from a different perspective further below. This can then be
used to define the solution $p\left(t\right)$ for all $t,$ and thus this
extends $\exp_{q}$ to all of $\mathfrak{l}.$

Let us consider $d\exp_{q}.$ From the definition of $\exp_{q}$, for any
$\xi\in\mathfrak{l,}$ have

$\left.d\exp_{q}\right|_{0}\left(\xi\right)=\left.\frac{d}{dt}\exp_{q}\left(t\xi\right)\right|_{t=0}=\xi.$
(2.10)

In particular, $\exp_{s}$ is smooth and since the identity map is a linear
isomorphism, by the Inverse Function Theorem, we have the following.

###### Lemma 2.12

For any $q\in\mathbb{L},$ the map
$\exp_{q}:\mathfrak{l}\longrightarrow\mathbb{L}$ is a local diffeomorphism
around $0\in\mathfrak{l}$.

###### Remark 2.13

To distinguish the exponential map on $\mathfrak{l}$ from the exponential map
on $\mathfrak{p,}$ we will use a subscript to denote with respect to which
element of $\mathbb{L}$ the exponential map is used. The exponential map on
$\mathfrak{p}$ will be without the subscript.

On smooth loops, we can define an analog of the Lie group Maurer-Cartan form.

######  Definition 2.14 ([26])

The Maurer-Cartan form $\theta$ is an $\mathfrak{l}$-valued $1$-form on
$\mathbb{L}$, such that for any vector field $X$, and any $p\in\mathbb{L},$
$\left.\theta\left(X\right)\right|_{p}=\rho_{p}^{-1}\left(X_{p}\right)\in\mathfrak{l}.$
Equivalently, for any $\xi\in\mathfrak{l},$
$\theta\left(\rho\left(\xi\right)\right)=\xi.$

The loop Maurer-Cartan form allows us to define brackets on $\mathfrak{l.}$
For each $p\in\mathbb{L}$ define the bracket
$\left[\cdot,\cdot\right]^{\left(p\right)}$ given for any
$\xi,\eta\in\mathfrak{l}$ by

$\left[\xi,\eta\right]^{\left(p\right)}=-\left.\theta\left(\left[\rho\left(\xi\right),\rho\left(\eta\right)\right]\right)\right|_{p}.$

As shown in [26, Theorem 3.7], we can equivalently define

$\displaystyle\left[\xi,\gamma\right]^{\left(p\right)}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dt}\left(\mathop{\rm
Ad}\nolimits_{\exp\left(t\xi\right)}^{\left(p\right)}\gamma\right)\right|_{t=0}$
$\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{dtd\tau}\exp\left(t\xi\right)\circ_{p}\exp\left(\tau\gamma\right)\right|_{t,\tau=0}$
$\displaystyle-\left.\frac{d^{2}}{dtd\tau}\exp\left(\tau\gamma\right)\circ_{p}\exp\left(t\xi\right)\right|_{t,\tau=0},$

where, for $p,q\in\mathbb{L}$, $\mathop{\rm
Ad}\nolimits_{q}^{\left(p\right)}:\mathfrak{l}\longrightarrow\mathfrak{l}$ is
the differential at $1\in\mathbb{L}$ of the conjugation map
$r\mapsto\left(q\circ_{p}r\right)/_{p}q\in\mathbb{L}.$

###### Remark 2.15

In [26], the conjugation map $r\mapsto\left(q\circ_{p}r\right)/_{p}q$ was
denoted by $\mathop{\rm Ad}\nolimits_{q}^{\left(p\right)},$ and its
differential as $\left(\mathop{\rm
Ad}\nolimits_{q}^{\left(p\right)}\right)_{\ast}.$ However here we adopt
notation that is more in line with standard usage in Lie theory.

###### Definition 2.16

The vector space $\mathfrak{l}$ equipped with the bracket
$\left[\cdot,\cdot\right]^{\left(p\right)}$ is known as the _loop tangent
algebra_ $\mathfrak{l}^{\left(p\right)}.$

Define the _bracket function_
$b:\mathbb{L}\longrightarrow\mathfrak{l}\otimes\Lambda^{2}\mathfrak{l}^{\ast}$
to be the map that takes
$p\mapsto\left[\cdot,\cdot\right]^{\left(p\right)}\in\mathfrak{l}\otimes\Lambda^{2}\mathfrak{l}^{\ast}$,
so that $b\left(\theta,\theta\right)$ is an $\mathfrak{l}$-valued $2$-form on
$\mathbb{L}$, i.e.
$b\left(\theta,\theta\right)\in\Omega^{2}\left(\mathfrak{l}\right)$.

###### Definition 2.17

For any $\eta,\gamma,\xi\in\mathfrak{l},$ and $p\in\mathbb{L},$ the
_associator_ $\left[\cdot,\cdot,\cdot\right]^{\left(p\right)}$ on
$\mathfrak{l}^{\left(p\right)}$ given by

$\displaystyle\left[\eta,\gamma,\xi\right]^{\left(p\right)}$ $\displaystyle=$
$\displaystyle\left.\frac{d^{3}}{dtd\tau
d\tau^{\prime}}\exp\left(\tau\eta\right)\circ_{p}\left(\exp\left(\tau^{\prime}\gamma\right)\circ_{p}\exp\left(t\xi\right)\right)\right|_{t,\tau,\tau^{\prime}=0}$
$\displaystyle-\left.\frac{d^{3}}{dtd\tau
d\tau^{\prime}}\left(\exp\left(\tau\eta\right)\circ_{p}\exp\left(\tau^{\prime}\gamma\right)\right)\circ_{p}\exp\left(t\xi\right)\right|_{t,\tau,\tau^{\prime}=0}.$

Moreover, define mixed associators between elements of $\mathbb{L}$ and
$\mathfrak{l}$. An
$\left(\mathbb{L},\mathbb{L},\mathfrak{l}\right)$-associator is defined for
any $p,q\in\mathbb{L}$ and $\xi\in\mathfrak{l}$ as

$\left[p,q,\xi\right]^{\left(s\right)}=\left(L_{p}^{\left(s\right)}\circ
L_{q}^{\left(s\right)}\right)_{\ast}\xi-\left(L_{p\circ_{s}q}^{\left(s\right)}\right)_{\ast}\xi\in
T_{p\circ_{s}q}\mathbb{L}$ (2.13)

and an $\left(\mathbb{L},\mathfrak{l},\mathfrak{l}\right)$-associator is
defined for an $p\in\mathbb{L}$ and $\eta,\xi\in\mathfrak{l}$ as

$\displaystyle\left[p,\eta,\xi\right]^{\left(s\right)}$ $\displaystyle=$
$\displaystyle\left.\frac{d}{dtd\tau}\left(p\circ_{s}\left(\exp\left(t\eta\right)\circ_{s}\exp\left(\tau\xi\right)\right)\right)\right|_{t\,,\tau=0}$
(2.14)
$\displaystyle-\left.\frac{d}{dtd\tau}\left(\left(p\circ_{s}\exp\left(t\eta\right)\right)\circ_{s}\exp\left(\tau\xi\right)\right)\right|_{t,\tau=0},$

where we see that $\left[p,\eta,\xi\right]^{\left(s\right)}\in
T_{p}\mathbb{L}.$ Similarly, for other combinations. Also define the _left-
alternating associator_
$a:\mathbb{L}\longrightarrow\mathfrak{l}\otimes\Lambda^{2}\mathfrak{l}^{\ast}\otimes\mathfrak{l}^{\ast},$
given by

$a_{p}\left(\eta,\gamma,\xi\right)=\left[\eta,\gamma,\xi\right]^{\left(p\right)}-\left[\gamma,\eta,\xi\right]^{\left(p\right)}.$
(2.15)

which we can call the _left-alternating associator_.

###### Remark 2.18

From the definitions of the associators, it is easy to see that if
$\left(\mathbb{L},\circ_{s}\right)$ is power-associative, given
$\xi\in\mathfrak{l},$ associators with any combinations of $\xi$ and
$\exp_{s}\left(t\xi\right),$ for any values of $t$, in the three entries, will
vanish. For example,

$\displaystyle\left[\xi,\xi,\xi\right]^{\left(s\right)}$ $\displaystyle=$
$\displaystyle 0$ (2.16a)
$\displaystyle[\xi,\xi,\exp_{s}\left(t\xi\right)]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle 0$ (2.16b)
$\displaystyle\left[\xi,\exp_{s}\left(t\xi\right),\exp_{s}\left(\tau\xi\right)\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle 0,$ (2.16c)

as well as any permutations.

Similarly, if $\left(\mathbb{L},\circ_{s}\right)$ is left-power-associative,
then associators with any combination of $\xi$ and $\exp_{s}\left(t\xi\right)$
in the first two entries will vanish, for example

$\displaystyle\left[\xi,\xi,\eta\right]^{\left(s\right)}$ $\displaystyle=$
$\displaystyle 0$ (2.17a)
$\displaystyle[\xi,\exp_{s}\left(t\xi\right),\eta]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle 0$ (2.17b)
$\displaystyle\left[\exp_{s}\left(t\xi\right),\exp_{s}\left(\tau\xi\right),\eta\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle 0,$ (2.17c)

for any $\eta\in\mathfrak{l}$ and similarly with the third entry replaced by
an element of $\mathbb{L}.$

From [26] we cite several useful properties of these brackets and associators.

######  Theorem 2.19 ([26, Theorem 3.20])

Suppose $p,s\in\mathbb{L}$, and $\xi,\eta\in\mathfrak{l}.$ Then the bracket
$\left[\cdot,\cdot\right]^{\left(ps\right)}$ is related to
$\left[\cdot,\cdot\right]^{\left(s\right)}$ via the expression

$\left[\xi,\eta\right]^{\left(ps\right)}=\left[\xi,\eta\right]^{\left(s\right)}+\left(\rho_{p}^{\left(s\right)}\right)^{-1}a_{s}\left(\xi,\eta,p\right).$
(2.18)

######  Theorem 2.20 ([26, Theorem 3.10])

The form $\theta$ satisfies

$d\theta=\frac{1}{2}b\left(\theta,\theta\right),$ (2.19)

where wedge product of $1$-forms is implied. Also, for any
$\xi,\eta\in\mathfrak{l},$ we have

$db\left(\xi,\eta\right)=a\left(\xi,\eta,\theta\right).$ (2.20)

It follows that $\xi,\eta,\gamma\in\mathfrak{l,}$ the generalized Jacobi
identity is satisfied:

$\mathop{\rm
Jac}\nolimits^{\left(s\right)}\left(\xi,\eta,\gamma\right)=a_{s}\left(\xi,\eta,\gamma\right)+a_{s}\left(\eta,\gamma,\xi\right)+a_{s}\left(\gamma,\xi,\eta\right),$
(2.21)

where

$\mathop{\rm
Jac}\nolimits^{\left(s\right)}\left(\xi,\eta,\gamma\right)=\left[\xi,\left[\eta,\gamma\right]^{\left(s\right)}\right]^{\left(s\right)}+\left[\eta,\left[\gamma,\xi\right]^{\left(s\right)}\right]^{\left(s\right)}+\left[\gamma,\left[\xi,\eta\right]^{\left(s\right)}\right]^{\left(s\right)}.$
(2.22)

###### Remark 2.21

Equation (2.19) is the loop Maurer-Cartan equation. The key difference from
the Maurer-Cartan equation on Lie groups is that on non-associative loops,
$b\left(s\right)$ is non-constant on $\mathbb{L},$ unlike on Lie groups, where
there is a unique bracket on the Lie algebra, and hence $b\left(s\right)$ is
constant. In particular, the non-constant $b$ leads to a non-trivial
associator (2.20) and the failure of the standard Jacobi identity to hold.

With respect to the action of $\Psi,$ the bracket and the associator satisfy
the following properties.

###### Lemma 2.22

If $h\in\Psi\mathbb{\ }$and $q\in\mathbb{L}$, then, for any
$\xi,\eta,\gamma\in\mathfrak{l}$,

$\displaystyle h_{\ast}^{\prime}\left[\xi,\eta\right]^{\left(q\right)}$
$\displaystyle=$
$\displaystyle\left[h_{\ast}^{\prime}\xi,h_{\ast}^{\prime}\eta\right]^{h\left(q\right)}$
$\displaystyle h_{\ast}^{\prime}\left[\xi,\eta,\gamma\right]^{\left(q\right)}$
$\displaystyle=$
$\displaystyle\left[h_{\ast}^{\prime}\xi,h_{\ast}^{\prime}\eta,h_{\ast}^{\prime}\gamma\right]^{h\left(q\right)}.$

If $A\left(t\right)$ is a path on $\mathbb{L},$ with
$A\left(t_{0}\right)=A_{0},$ and
$\left.\frac{d}{dt}A\left(t\right)/A_{0}\right|_{t=t_{0}}=\xi\in\mathfrak{l,}$
then for any $p,q\in\mathbb{L}$,

$\left.\frac{d}{dt}p\circ_{A\left(t\right)}q\right|_{t=t_{0}}=\left[p,q,\xi\right]^{\left(A_{0}\right)}\in
T_{p\circ_{A_{0}}q}\mathbb{L}.$ (2.23)

Also, for any $\eta,\gamma\in\mathfrak{l},$

$\left.\frac{d}{dt}\left[\eta,\gamma\right]^{A\left(t\right)}\right|_{t=t_{0}}=a_{A_{0}}\left(\eta,\gamma,\xi\right)$
(2.24)

Proof. The first part is given in [26, Lemma 3.17]. To show (2.23), consider

$\displaystyle\left.\frac{d}{dt}p\circ_{A\left(t\right)}q\right|_{t=t_{0}}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt}\left(p\left(qA\left(t\right)\right)\right)/A\left(t\right)\right|_{t=t_{0}}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt}\left(p\left(qA\left(t\right)\right)\right)/A_{0}\right|_{t=t_{0}}-\left.\frac{d}{dt}\left(\left(p\circ_{A_{0}}q\right)\cdot
A\left(t\right)\right)/A_{0}\right|_{t=t_{0}},$

where we’ve used (2.6). Now,

$\displaystyle\left(p\left(qA\left(t\right)\right)\right)/A_{0}$
$\displaystyle=$ $\displaystyle\left(p\left(q\left(A\left(t\right)/A_{0}\cdot
A_{0}\right)/A_{0}\cdot A_{0}\right)\right)/A_{0}$ $\displaystyle=$
$\displaystyle
p\circ_{A_{0}}\left(q\circ_{A_{0}}\left(A\left(t\right)/A_{0}\right)\right)$
$\displaystyle\left(\left(p\circ_{A_{0}}q\right)\cdot
A\left(t\right)\right)/A_{0}$ $\displaystyle=$
$\displaystyle\left(p\circ_{A_{0}}q\right)\circ_{A_{0}}\left(A\left(t\right)/A_{0}\right).$

Hence,

$\displaystyle\left.\frac{d}{dt}p\circ_{\left(t\right)}q\right|_{t=t_{0}}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{dt}p\circ_{A_{0}}\left(q\circ_{A_{0}}\left(A\left(t\right)/A_{0}\right)\right)\right|_{t=0}$
$\displaystyle-\left.\frac{d}{dt}\left(p\circ_{A_{0}}q\right)\circ_{A_{0}}\left(A\left(t\right)/A_{0}\right)\right|_{t=0}.$

To show (2.24), we could use (2.20), but more directly, we can obtain it from
the definition, using (2):

$\displaystyle\left[\xi,\gamma\right]^{\left(A\left(t\right)\right)}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\tau
d\tau^{\prime}}\exp\left(\tau\eta\right)\circ_{A\left(t\right)}\exp\left(\tau^{\prime}\gamma\right)\right|_{\tau,\tau^{\prime}=0}$
$\displaystyle-\left.\frac{d^{2}}{d\tau
d\tau^{\prime}}\exp\left(\tau^{\prime}\gamma\right)\circ_{A\left(t\right)}\exp\left(\tau\eta\right)\right|_{\tau,\tau^{\prime}=0}.$

Then, from (2.23),

$\displaystyle\frac{d}{dt}\left.\exp\left(\tau\eta\right)\circ_{A\left(t\right)}\exp\left(\tau^{\prime}\gamma\right)\right|_{t=t_{0}}$
$\displaystyle=$
$\displaystyle\left[\exp\left(\tau\eta\right),\exp\left(\tau^{\prime}\gamma\right),\xi\right]^{\left(A_{0}\right)}$
$\displaystyle\frac{d}{dt}\left.\exp\left(\tau^{\prime}\gamma\right)\circ_{A\left(t\right)}\exp\left(\tau\eta\right)\right|_{t=t_{0}}$
$\displaystyle=$
$\displaystyle\left[\exp\left(\tau^{\prime}\gamma\right),\exp\left(\tau\eta\right),\xi\right]^{\left(A_{0}\right)}$

and from the definition (2.17), we obtain (2.24).

Let $\xi\in\mathfrak{l}$ and $s\in\mathbb{L}.$ Also let
$A\left(t\right)=\exp_{s}\left(t\xi\right)$ for $t$ in some interval
$I\subset\mathbb{R}\ $that contains $0$. Then consider a family
$\eta\left(t\right)\in\mathfrak{l}$ that satisfies the following initial value
problem:

$\left\\{\begin{array}[]{c}\frac{d\eta\left(t\right)}{dt}=\left[\xi,\eta\left(t\right)\right]^{\left(A\left(t\right)s\right)}\\\
\eta\left(0\right)=\eta_{0}\end{array}\right..$ (2.26)

In other words, this is linear first-order ODE $\dot{\eta}=\mathop{\rm
ad}\nolimits_{\xi}^{\left(A\left(t\right)s\right)}\eta$ , so for all $t\in I$
there exists an evolution operator $U_{\xi}^{\left(s\right)}\left(t\right)\in
GL\left(\mathfrak{l}\right)$, with
$U_{\xi}^{\left(s\right)}\left(0\right)=\mathop{\rm
id}\nolimits_{\mathfrak{l}},$ such that

$\eta\left(t\right)=U_{\xi}^{\left(s\right)}\left(t\right)\eta_{0}.$ (2.27)

From standard ODE theory, recall that if $\tau^{\prime},\tau^{\prime\prime}\in
I,$ then
$U_{\xi}^{\left(s\right)}\left(\tau^{\prime\prime}\right)U_{\xi}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}$
is the evolution operator from $\tau^{\prime}$ to $\tau^{\prime\prime}$ and is
given by:

$\displaystyle
U_{\xi}^{\left(s\right)}\left(\tau^{\prime\prime}\right)U_{\xi}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}$
$\displaystyle=$ $\displaystyle\mathop{\rm
id}\nolimits_{\mathfrak{l}}+\int_{\tau^{\prime}}^{\tau^{\prime\prime}}\mathop{\rm
ad}\nolimits_{\xi}^{\left(\left(\exp_{s}t_{1}\xi\right)s\right)}dt_{1}$
$\displaystyle+\sum_{n=2}^{\infty}\int_{\tau^{\prime}}^{\tau^{\prime\prime}}\int_{\tau^{\prime}}^{t_{n}}...\int_{\tau^{\prime}}^{t_{2}}\mathop{\rm
ad}\nolimits_{\xi}^{\left(\left(\exp_{s}t_{n}\xi\right)s\right)}...\mathop{\rm
ad}\nolimits_{\xi}^{\left(\left(\exp_{s}t_{1}\xi\right)s\right)}dt_{1}...dt_{n}.$

The following properties of $U_{\xi}^{\left(s\right)}\left(t\right)$ follow
immediately.

###### Lemma 2.23

The evolution operator $U_{\xi}^{\left(s\right)}\left(t\right)$ satisfies the
following properties:

  1. 1.

$U_{\tau\xi}^{\left(s\right)}\left(t\right)=U_{\xi}^{\left(s\right)}\left(\tau
t\right),$ for any $t$ and $\tau,$ as long as $\exp_{s}\left(t\xi\right)$ and
$\exp_{s}\left(\tau t\xi\right)$ are both defined.

  2. 2.

$U_{\xi}^{\left(s\right)}\left(t\right)\xi=\xi.$

  3. 3.

If $\mathbb{L}$ is compact, and $\mathfrak{l}$ is equipped with an inner
product, then in a compatible operator norm, there exists a constant
$C=\sup_{s\in\mathbb{L}}\left|b_{s}\right|,$ such

$\left|U_{\xi}^{\left(s\right)}\left(\tau^{\prime\prime}\right)U_{\xi}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}-\mathop{\rm
id}\nolimits_{\mathfrak{l}}\right|\leq
e^{C\left|\tau^{\prime\prime}-\tau^{\prime}\right|\left|\xi\right|}-1.$ (2.29)

Proof. Item 1 follows from a change of variables in (2.26). For item 2,
consider

$X\left(t\right)=U_{\xi}^{\left(s\right)}\left(t\right)\xi-\xi.$

Then,

$\displaystyle\frac{dX\left(t\right)}{dt}$ $\displaystyle=$
$\displaystyle\frac{d\left(U_{\xi}^{\left(s\right)}\left(t\right)\xi\right)}{dt}=\left[\xi,U_{\xi}^{\left(s\right)}\left(t\right)\xi\right]^{\left(A\left(t\right)s\right)}$
$\displaystyle=$
$\displaystyle\left[\xi,X\left(t\right)\right]^{\left(A\left(t\right)s\right)},$

since $\left[\xi,\xi\right]^{\left(A\left(t\right)\right)}=0.$ Hence,
$X\left(t\right)=U_{\xi}^{\left(s\right)}\left(t\right)X\left(0\right),$ but
$X\left(0\right)=0,$ so $X\left(t\right)=0$ for all $t$.

For the estimate, from (2), we obtain

$\displaystyle\left|U_{\xi}^{\left(s\right)}\left(\tau^{\prime\prime}\right)U_{\xi}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}-\mathop{\rm
id}\nolimits_{\mathfrak{l}}\right|$ $\displaystyle\leq$
$\displaystyle\exp\left(\int_{\tau^{\prime}}^{\tau^{\prime\prime}}\left|\mathop{\rm
ad}\nolimits_{\xi}^{\left(\left(\exp_{s}t\xi\right)s\right)}\right|dt\right)-1$
$\displaystyle\leq$
$\displaystyle\exp\left(\left|\tau^{\prime\prime}-\tau^{\prime}\right|\left|\xi\right|\sup_{s\in\mathbb{L}}\left|b_{s}\right|\right)-1$

Now, $s\mapsto\left|b_{s}\right|$ is a smooth real-valued map on a compact
manifold, and is hence bounded. Therefore, there exists a constant
$C=\sup_{s\in\mathbb{L}}\left|b_{s}\right|$ and hence
$\sup_{t\in\left[0,1\right]}\left|b_{\left(\exp_{s}t\xi\right)s}\right|\leq
C.$ Thus,

$\left|U_{\xi}^{\left(s\right)}\left(\tau^{\prime\prime}\right)U_{\xi}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}-\mathop{\rm
id}\nolimits_{\mathfrak{l}}\right|\leq
e^{C\left|\tau^{\prime\prime}-\tau^{\prime}\right|\left|\xi\right|}-1.$

###### Remark 2.24

Since
$U_{\xi}^{\left(s\right)}\left(t\right)=U_{t\xi}^{\left(s\right)}\left(1\right),$
for brevity let us denote the operator
$U_{\xi}^{\left(s\right)}\left(t\right)$ by $U_{t\xi}^{\left(s\right)}.$

If $\mathfrak{l}$ is a Lie algebra, then $\mathop{\rm
ad}\nolimits_{\xi}^{\left(A\left(t\right)s\right)}=\mathop{\rm
ad}\nolimits_{\xi}$ is independent of $t,$ and then
$U_{t\xi}^{\left(s\right)}=\exp\left(t\mathop{\rm
ad}\nolimits_{\xi}\right)=\mathop{\rm Ad}\nolimits_{\exp t\xi}.$ In the non-
associative case, this is no longer true in general, but needs additional
assumptions, as Theorem 2.25 below shows.

###### Theorem 2.25

Let $s\in\mathbb{L}$, $\xi,\eta\in\mathfrak{l,}$ and
$A\left(t\right)=\exp_{s}\left(t\xi\right).$ Suppose
$U_{t\xi}^{\left(s\right)}$ is the evolution operator for the equation (2.26)
as in (2.27). Then,

$\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta=U_{t\xi}^{\left(s\right)}\eta+U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\left(\rho_{A\left(\tau\right)}^{\left(s\right)}\right)^{-1}\left(\left[\xi,A\left(\tau\right),\eta\right]^{\left(s\right)}\right)d\tau.$
(2.30)

Moreover,

  1. 1.

If $\mathbb{L}$ is compact, and $\mathfrak{l}$ is equipped with an inner
product, then in a compatible operator norm, there exists a constant $C$ that
depends only on $\mathbb{L},$ such that,

$\left|\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}-U_{t\xi}^{\left(s\right)}\right|\leq
C\left(e^{C\left|\xi\right|t}-1\right).$ (2.31)

  2. 2.

If $\left(\mathbb{L},\circ_{s}\right)$ is left-power-alternative, then

$\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}=U_{t\xi}^{\left(s\right)}.$
(2.32)

  3. 3.

If $\left(\mathbb{L},\circ_{s}\right)$ is both left-power-alternative and
right-power-alternative, then

$\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}=\exp\left(t\mathop{\rm
ad}\nolimits_{\xi}^{\left(s\right)}\right).$ (2.33)

Proof. Let $x\left(t\right)=\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta$ and note that
$x\left(0\right)=\eta.$ Then, consider the derivative of $x\left(t\right).$
Let $B\left(t\right)=\exp_{s}\left(t\eta\right).$ From (2.9) we have

$\frac{dA\left(t\right)}{dt}=\rho_{A\left(t\right)}^{\left(s\right)}\left(\eta\right)=\left.\frac{d}{d\tau}B\left(\tau\right)\circ_{s}A\left(t\right)\right|_{\tau=0}$

Then, using (2.6) we have

$\displaystyle\frac{dx\left(t\right)}{dt}$ $\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{dtd\tau}\left(A\left(t\right)\circ_{s}\left(B\left(\tau\right)\right)\right)/_{s}A\left(t\right)\right|_{\tau=0}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\tau
d\tau^{\prime}}\left(A\left(\tau^{\prime}\right)\circ_{s}A\left(t\right)\right)\circ_{s}\left(B\left(\tau\right)\right)/_{s}A\left(t\right)\right|_{\tau^{\prime},\tau=0}$
$\displaystyle-\left.\frac{d^{2}}{d\tau
d\tau^{\prime}}(\left(\left(A\left(t\right)\circ_{s}\left(B\left(\tau\right)\right)\right)/_{s}A\left(t\right)\right)\circ_{s}\left(A\left(\tau^{\prime}\right)\circ_{s}A\left(t\right)\right))/_{s}A\left(t\right)\right|_{\tau^{\prime},\tau=0}$
$\displaystyle=$
$\displaystyle-\left(\rho_{A\left(t\right)}^{\left(s\right)}\right)^{-1}\left[\xi,A\left(t\right),\eta\right]^{\left(s\right)}+\left.\frac{d}{d\tau^{\prime}}\left(A\left(\tau^{\prime}\right)\circ_{s}\left(\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta\circ_{s}A\left(t\right)\right)\right)/_{s}A\left(t\right)\right|_{\tau^{\prime}=0}$
$\displaystyle-\left.\frac{d}{d\tau}(\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta\circ_{s}\left(A\left(\tau^{\prime}\right)\circ_{s}A\left(t\right)\right))/_{s}A\left(t\right)\right|_{\tau=0}.$

Using (2.5), the second term in (2) becomes

$\left.\frac{d}{d\tau^{\prime}}\left(A\left(\tau^{\prime}\right)\circ_{s}\left(\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta\circ_{s}A\left(t\right)\right)\right)/_{s}A\left(t\right)\right|_{\tau^{\prime}=0}=\left.\frac{d}{d\tau^{\prime}}A\left(\tau^{\prime}\right)\circ_{A\left(t\right)s}x\left(t\right)\right|_{\tau^{\prime}=0}$

and similarly, the third term in (2) becomes

$\left.\frac{d}{d\tau}(\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta\circ_{s}\left(A\left(\tau^{\prime}\right)\circ_{s}A\left(t\right)\right))/_{s}A\left(t\right)\right|_{\tau=0}=\left.\frac{d}{d\tau^{\prime}}x\left(t\right)\circ_{A\left(t\right)s}A\left(\tau^{\prime}\right)\right|_{\tau^{\prime}=0}.$

Using (2) we then conclude that

$\frac{dx\left(t\right)}{dt}=\left[\xi,x\left(t\right)\right]^{\left(A\left(t\right)s\right)}-\left(\rho_{A\left(t\right)}^{\left(s\right)}\right)^{-1}\left[\xi,A\left(t\right),\eta\right]^{\left(s\right)}.$
(2.35)

This is an inhomogeneous linear first order ODE. The homogeneous part is
precisely (2.26), and hence we obtain precisely (2.30).

For the estimate, suppose $\mathbb{L}$ is compact. Then, using (2.30) and
(2.29), we have

$\displaystyle\left|\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}-U_{t\xi}^{\left(s\right)}\right|$
$\displaystyle\leq$
$\displaystyle\left|\xi\right|\int_{0}^{t}\left|U_{t\xi}^{\left(s\right)}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\right|\left|\left(\rho_{A\left(\tau\right)}^{\left(s\right)}\right)^{-1}\left(\left[\cdot,A\left(\tau\right),\cdot\right]^{\left(s\right)}\right)\right|d\tau$
$\displaystyle\leq$
$\displaystyle\left|\xi\right|\int_{0}^{t}e^{C\left|\xi\right|\left(t-\tau\right)}\left|\left(\rho_{A\left(\tau\right)}^{\left(s\right)}\right)^{-1}\left(\left[\cdot,A\left(\tau\right),\cdot\right]^{\left(s\right)}\right)\right|d\tau.$

However,
$\left(A,s\right)\mapsto\left|\rho_{A}^{-1}\left(\left[\cdot,A,\cdot\right]^{\left(s\right)}\right)\right|$
is a real-valued function on a compact manifold, and hence there exists a
constant $C^{\prime}$ which is the supremum of this function over
$\mathbb{L}\times\mathbb{L}.$ Hence,

$\displaystyle\left|\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}-U_{t\xi}^{\left(s\right)}\right|$
$\displaystyle\leq$ $\displaystyle
C^{\prime}\left|\xi\right|\int_{0}^{t}e^{C\left|\xi\right|\left(t-\tau\right)}d\tau$
$\displaystyle=$
$\displaystyle\frac{C^{\prime}}{C}\left(e^{C\left|\xi\right|t}-1\right).$

Renaming the constant $C,$ we get (2.31).

Now if $\left(\mathbb{L},\circ_{s}\right)$ is left-alternative, the second
term on the right hand side of (2.35) vanishes, since

$\left[\xi,A\left(t\right),\eta\right]^{\left(s\right)}=\left[\xi,\exp_{s}\left(t\xi\right),\eta\right]^{\left(s\right)}=0.$

Then, $\mathop{\rm Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta$
satisfies the homogeneous equation, so the solution is just $\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta=U_{t\xi}^{\left(s\right)}\eta.$

From (2.18), we have

$\left[\xi,x\left(t\right)\right]^{\left(A\left(t\right)s\right)}=\left[\xi,x\left(t\right)\right]^{\left(s\right)}+\left(\rho_{A\left(t\right)}^{\left(s\right)}\right)^{-1}a_{s}\left(\xi,x\left(t\right),A\left(t\right)\right).$

If $\left(\mathbb{L},\circ_{s}\right)$ is left-power-alternative, then first
of all, the associator is skew-symmetric in the first two entries, so

$a_{s}\left(\xi,x\left(t\right),A\left(t\right)\right)=-2\left[x\left(t\right),\xi,A\left(t\right)\right]^{\left(s\right)},$

however due to right-power-alternativity,
$\left[x\left(t\right),\xi,\exp_{s}\left(t\xi\right)\right]^{\left(s\right)}=0.$
Hence, in this case,

$\left[\xi,x\left(t\right)\right]^{\left(A\left(t\right)s\right)}=\left[\xi,x\left(t\right)\right]^{\left(s\right)}.$

Then, $x\left(t\right)$ satisfies the first order homogeneous ODE with
constant coefficients:

$\frac{dx\left(t\right)}{dt}=\mathop{\rm
ad}\nolimits_{\xi}^{\left(s\right)}\left(x\left(t\right)\right),$

so $U_{\xi}\left(t\right)=\exp\left(t\mathop{\rm
ad}\nolimits_{\xi}^{\left(s\right)}\right)$ and hence the solution is now

$x\left(t\right)=\exp\left(t\mathop{\rm
ad}\nolimits_{\xi}^{\left(s\right)}\right)\eta.$

###### Corollary 2.26

If $\left(\mathbb{L},\circ_{s}\right)$ is power-associative, then

  1. 1.

$\mathop{\rm
Ad}\nolimits_{\exp_{s}\left(t\xi\right)}^{\left(s\right)}\xi=\xi$,

  2. 2.

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

Proof. From (2.30),

$\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\xi=U_{t\xi}^{\left(s\right)}\xi+U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{t\xi}^{\left(s\right)}\right)^{-1}\rho_{A\left(\tau\right)}^{-1}\left(\left[\xi,A\left(\tau\right),\xi\right]^{\left(s\right)}\right)d\tau.$
(2.36)

However, by power-associativity (2.16),

$\left[\xi,A\left(\tau\right),\xi\right]^{\left(s\right)}=\left[\xi,\exp_{s}\left(\tau\xi\right),\xi\right]^{\left(s\right)}=0.$

Hence, using Lemma 2.23, we obtain

$\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\xi=U_{t\xi}^{\left(s\right)}\xi=\xi.$

For the second part, define

$r\left(t\right)=\exp_{s}\left(t\xi\right)\circ\exp_{s}\left(t\xi\right).$

Then, informally, we write

$\displaystyle\frac{dr\left(t\right)}{dt}$ $\displaystyle=$
$\displaystyle\exp_{s}\left(t\xi\right)\circ_{s}\left(\xi\circ_{s}\exp_{s}\left(t\xi\right)\right)$
$\displaystyle+\left(\xi\circ_{s}\exp_{s}\left(t\xi\right)\right)\circ_{s}\exp_{s}\left(t\xi\right).$

Now using power-associativity, we see that $\xi$ associates with
$\exp_{s}\left(t\xi\right)$ and since $\mathop{\rm
Ad}\nolimits_{\exp_{s}\left(t\xi\right)}^{\left(s\right)}\xi=\xi,$ moreover
$\xi$ and $\exp_{s}\left(t\xi\right)$ commute. Hence, we can rewrite

$\frac{dr\left(t\right)}{dt}=2\xi\circ_{s}\left(\exp_{s}\left(t\xi\right)\circ_{s}\exp_{s}\left(t\xi\right)\right)=2\xi\circ_{s}r\left(t\right).$

The solution is thus

$r\left(t\right)=\exp_{s}\left(2t\xi\right),$

so by uniqueness of solutions we then have the needed equality.

###### Remark 2.27

Corollary 2.26 thus shows that indeed, power-associativity allows to extend
$\exp_{s}\left(t\xi\right)$ for all $t$. This is a slightly different proof of
this fact compared to [36, 42]. The result from Corollary 2.26 also allows to
conclude that if $\left(\mathbb{L},\circ_{s}\right)$ is power-associative,
then

$\exp_{s}\left(t\xi\right)\circ_{s}\exp_{s}\left(\tau\xi\right)=\exp_{s}\left(\left(t+\tau\right)\xi\right).$
(2.37)

###### Theorem 2.28

Suppose $\xi\left(t\right)$ is a path in $\mathfrak{l,}$ with derivative
$\dot{\xi}\left(t\right)$, then

$\frac{d}{dt}\left(\exp_{s}\left(\xi\left(t\right)\right)\right)=\rho_{\exp_{s}\left(\xi\left(t\right)\right)}^{\left(s\right)}U_{\xi\left(t\right)}^{\left(s\right)}\int_{0}^{1}\left(U_{\tau\xi\left(t\right)}^{\left(s\right)}\right)^{-1}\dot{\xi}\left(t\right)d\tau.$

Moreover,

  1. 1.

If $\left(\mathbb{L},\circ_{s}\right)$ is left-power-alternative, then

$\frac{d}{dt}\left(\exp_{s}\left(\xi\left(t\right)\right)\right)=\lambda_{\exp_{s}\left(\xi\left(t\right)\right)}^{\left(s\right)}\int_{0}^{1}\left(\mathop{\rm
Ad}\nolimits_{\exp_{s}\tau\xi\left(t\right)}^{\left(s\right)}\right)^{-1}\dot{\xi}\left(t\right)d\tau.$

  2. 2.

If $\left(\mathbb{L},\circ_{s}\right)$ is both left-power-alternative and
right-power-alternative, then

$\frac{d}{dt}\left(\exp_{s}\left(\xi\left(t\right)\right)\right)=\lambda_{\exp_{s}\left(\xi\left(t\right)\right)}^{\left(s\right)}\int_{0}^{1}\exp\left(-\tau\mathop{\rm
ad}\nolimits_{\xi\left(t\right)}^{\left(s\right)}\right)\dot{\xi}\left(t\right)d\tau.$

Proof. Using a similar approach as in the Lie group case, let

$\Gamma\left(\tau,t\right)=\left(\rho_{\exp_{s}\left(\tau\xi\left(t\right)\right)}^{\left(s\right)}\right)^{-1}\frac{\partial}{\partial
t}\exp_{s}\left(\tau\xi\left(t\right)\right).$

We can write this (somewhat informally) as

$\Gamma\left(\tau,t\right)=\left(\frac{\partial}{\partial
t}\exp_{s}\left(\tau\xi\left(t\right)\right)\right)/_{s}\exp_{s}\left(\tau\xi\left(t\right)\right)$
(2.38)

Note that $\Gamma\left(0,t\right)=0.$ Then, consider

$\displaystyle\frac{\partial\Gamma}{\partial\tau}$ $\displaystyle=$
$\displaystyle\left(\frac{\partial}{\partial
t}\left(\xi\left(t\right)\circ_{s}\exp_{s}\left(\tau\xi\left(t\right)\right)\right)\right)/_{s}\exp_{s}\left(\tau\xi\left(t\right)\right)$
$\displaystyle-\left(\Gamma\left(\tau,t\right)\circ_{s}\left(\xi\left(t\right)\circ_{s}\exp_{s}\left(\tau\xi\left(t\right)\right)\right)\right)/_{s}\exp_{s}\left(\tau\xi\left(t\right)\right)$
$\displaystyle=$ $\displaystyle\frac{\partial\xi\left(t\right)}{\partial
t}+\left(\xi\left(t\right)\circ_{s}\frac{\partial}{\partial
t}\exp_{s}\left(\tau\xi\left(t\right)\right)\right)/_{s}\exp_{s}\left(\tau\xi\left(t\right)\right)$
$\displaystyle-\Gamma\left(\tau,t\right)\circ_{\exp_{s}\left(\tau\xi\left(t\right)\right)s}\xi\left(t\right)$
$\displaystyle=$ $\displaystyle\frac{\partial\xi\left(t\right)}{\partial
t}+\left[\xi\left(t\right),\Gamma\left(\tau,t\right)\right]^{\exp_{s}\left(\tau\xi\left(t\right)\right)s}.$

For each $t,$ the homogeneous part of ODE is precisely (2.26), and since the
initial condition is $\Gamma\left(0,t\right)=0,$ we find that the solution of
the inhomogeneous equation is

$\Gamma\left(\tau,t\right)=U_{\tau\xi\left(t\right)}^{\left(s\right)}\int_{0}^{\tau}\left(U_{\tau^{\prime}\xi\left(t\right)}^{\left(s\right)}\right)^{-1}\frac{\partial\xi\left(t\right)}{\partial
t}d\tau^{\prime}.$

Setting $\tau=1$, we obtain

$\left(\rho_{\exp_{s}\left(\xi\left(t\right)\right)}^{\left(s\right)}\right)^{-1}\frac{\partial}{\partial
t}\exp_{s}\left(\xi\left(t\right)\right)=U_{\xi\left(t\right)}^{\left(s\right)}\left(1\right)\int_{0}^{1}U_{\xi\left(t\right)}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}\frac{\partial\xi\left(t\right)}{\partial
t}d\tau^{\prime}.$

The special cases now follow immediately from (2.32) and (2.33).

###### Corollary 2.29

Let $\xi,\eta\in\mathfrak{l},$ then

$\left(\rho_{\exp_{s}\xi}^{\left(s\right)}\right)^{-1}\left.d\exp_{s}\right|_{\xi}\left(\eta\right)=U_{\xi}^{\left(s\right)}\left(1\right)\int_{0}^{1}U_{\xi}^{\left(s\right)}\left(\tau^{\prime}\right)^{-1}\eta
d\tau^{\prime}.$ (2.39)

Moreover, if $\mathbb{L}$ is compact, given a norm on $\mathfrak{l}$ and a
corresponding operator norm,

$\left|\left(\rho_{\exp_{s}\xi}^{\left(s\right)}\right)^{-1}\left.d\exp_{s}\right|_{\xi}-\mathop{\rm
id}\nolimits_{\mathfrak{l}}\right|\leq e^{C\left|\xi\right|}-1,$ (2.40)

where $C>0$ is a constant that depends on $\mathbb{L}.$

Proof. The expression (2.39) follows directly from Theorem 2.28. We thus have

$\left(\rho_{\exp_{s}\xi}^{\left(s\right)}\right)^{-1}\left.d\exp_{s}\right|_{\xi}-\mathop{\rm
id}\nolimits_{\mathfrak{l}}=\int_{0}^{1}\left(U_{\xi}^{\left(s\right)}\left(U_{\tau^{\prime}\xi}^{\left(s\right)}\right)^{-1}-\mathop{\rm
id}\nolimits_{l}\right)d\tau^{\prime}.$ (2.41)

From (2.41) we then obtain (2.40).

Let us now explore the dependence of $\exp_{s}$ on $s.$ In particular, suppose
we have a smooth $1$-parameter family $s\left(\tau\right),$ with
$s\left(0\right)=s\in\mathbb{L}$ and
$\left.\frac{d}{d\tau}s\left(\tau\right)\right|_{\tau=0}=\rho_{s}\left(\eta\right)\in
T_{s}\mathbb{L}\mathfrak{,}$ with $\eta\in\mathfrak{l}.$ For each $\tau,$
$p\left(t,\tau\right)=\exp_{s\left(\tau\right)}\left(t\xi\right)$ satisfies

$\left\\{\begin{array}[]{c}\frac{dp\left(t,\tau\right)}{dt}=\left.\rho^{\left(s\left(\tau\right)\right)}\left(\xi\right)\right|_{p\left(t,\tau\right)}\\\
p\left(0,\tau\right)=1\end{array}\right..$ (2.42)

Now for each $t,$
$\left.\frac{d}{d\tau}p\left(t,\tau\right)\right|_{\tau=0}\in
T_{p\left(t,0\right)}\mathbb{L},$ so define

$\displaystyle\sigma\left(t\right)$ $\displaystyle=$
$\displaystyle\left(\rho_{p\left(t,0\right)}^{\left(s\right)}\right)^{-1}\left.\frac{d}{d\tau}p\left(t,\tau\right)\right|_{\tau=0}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{d\tau}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\right|_{\tau=0},$

so that in particular, $\sigma\left(t\right)\in\mathfrak{l}.$

###### Lemma 2.30

Let $A\left(t\right)=\exp_{s}\left(t\xi\right)$, then the quantity
$\sigma\left(t\right)$ is given by

$\sigma\left(t\right)=\left(U_{t\xi}^{\left(s\right)}-\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\right)\eta.$ (2.44)

In particular, if $\left(\mathbb{L},\circ_{s}\right)$ is left-power-
alternative, then $\sigma\left(t\right)=0$ for all $t.$

Proof. From (2.42), we have

$\displaystyle\frac{dp\left(t,\tau\right)}{dt}$ $\displaystyle=$
$\displaystyle\left.\rho^{\left(s\left(\tau\right)\right)}\left(\xi\right)\right|_{p\left(t,\tau\right)}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{d\varepsilon}\exp_{s\left(\tau\right)}\left(\varepsilon\xi\right)\circ_{s\left(\tau\right)}p\left(t,\tau\right)\right|_{\varepsilon=0}$
$\displaystyle=$
$\displaystyle\left.\frac{d}{d\varepsilon}p\left(\varepsilon,\tau\right)\circ_{s\left(\tau\right)}p\left(t,\tau\right)\right|_{\varepsilon=0}\text{,}$

and therefore,

$\displaystyle\frac{d\sigma\left(t\right)}{dt}$ $\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{dtd\tau}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\right|_{\tau=0}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,\tau\right)\circ_{s\left(\tau\right)}p\left(t,\tau\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle-\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\circ_{s}\left(p\left(\varepsilon,0\right)\circ_{s}p\left(t,0\right)\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0},$

where we used the derivative of the quotient formula (2.6) and also the fact
that $p\left(0,\tau\right)=1$ for all $\tau.$ Noting that for any $\tau$,
$\left.\frac{d}{d\varepsilon}p\left(\varepsilon,\tau\right)\right|_{\varepsilon=0}=\xi,$
consider the first term of (2):

$\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,\tau\right)\circ_{s\left(\tau\right)}p\left(t,\tau\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,0\right)\circ_{s\left(\tau\right)}p\left(t,\tau\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,0\right)\circ_{s\left(\tau\right)}p\left(t,0\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle+\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,0\right)\circ_{s}p\left(t,\tau\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}.$

Now,

$\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,0\right)\circ_{s\left(\tau\right)}p\left(t,0\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle=$
$\displaystyle\left(\rho_{\exp_{s}\left(t\xi\right)}^{\left(s\right)}\right)^{-1}\left[\xi,\exp_{s}\left(t\xi\right),\eta\right]^{\left(s\right)}$
$\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,0\right)\circ_{s}p\left(t,\tau\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(\varepsilon,0\right)\circ_{s}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\circ_{s}p\left(t,0\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle=$ $\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}p\left(\varepsilon,0\right)\circ_{p\left(t,0\right)s}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\right|_{\varepsilon,\tau=0}.$

The second term of (2) gives

$\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\circ_{s}\left(p\left(\varepsilon,0\right)\circ_{s}p\left(t,0\right)\right)\right)/_{s}p\left(t,0\right)\right|_{\varepsilon,\tau=0}=\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\circ_{p\left(t,0\right)s}p\left(\varepsilon,0\right)\right|_{\varepsilon,\tau=0}.$

Overall, (2) becomes

$\displaystyle\frac{d\sigma\left(t\right)}{dt}$ $\displaystyle=$
$\displaystyle\left.\frac{d^{2}}{d\varepsilon
d\tau}p\left(\varepsilon,0\right)\circ_{p\left(t,0\right)s}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle-\left.\frac{d^{2}}{d\varepsilon
d\tau}\left(p\left(t,\tau\right)/_{s}p\left(t,0\right)\right)\circ_{p\left(t,0\right)s}p\left(\varepsilon,0\right)\right|_{\varepsilon,\tau=0}$
$\displaystyle+\left(\rho_{\exp_{s}\left(t\xi\right)}^{\left(s\right)}\right)^{-1}\left[\xi,\exp_{s}\left(t\xi\right),\eta\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle\left[\xi,\sigma\left(t\right)\right]^{\left(\exp_{s}\left(t\xi\right)s\right)}+\left(\rho_{\exp_{s}\left(t\xi\right)}^{\left(s\right)}\right)^{-1}\left[\xi,\exp_{s}\left(t\xi\right),\eta\right]^{\left(s\right)}.$

This is precisely the equation (2.35) satisfied by $-\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\eta,$ however with initial
condition $\sigma\left(0\right)=0.$ Therefore, the solution is

$\sigma\left(t\right)=\left(U_{t\xi}^{\left(s\right)}-\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}\right)\eta.$

If $\left(\mathbb{L},\circ_{s}\right)$ is left-power-associative, then from
Theorem 2.25, $\mathop{\rm
Ad}\nolimits_{A\left(t\right)}^{\left(s\right)}=U_{t\xi}^{\left(s\right)},$
and thus $\sigma\left(t\right)=0$ for all $t$.

We will assume that group $\Psi$ of pseudoautomorphisms of $\mathbb{L}$ is a
finite-dimensional Lie group, and suppose the Lie algebras of $\Psi$ and
$H_{s}=\mathop{\rm Aut}\nolimits\left(\mathbb{L},\circ_{s}\right)$ are
$\mathfrak{p}$ and $\mathfrak{h}_{s}$, respectively. In particular,
$\mathfrak{h}_{s}$ is a Lie subalgebra of $\mathfrak{p}$. Also, we will assume
that $\Psi$ acts transitively on $\mathbb{L}$. The action of $\Psi$ on
$\mathbb{L}\ $induces an action of the Lie algebra $\mathfrak{p}$ on
$\mathfrak{l}$, which we will denote by $\cdot.$

###### Definition 2.31

Define the map $\varphi:\mathbb{L}\longrightarrow$
$\mathfrak{l}\otimes\mathfrak{p}^{\ast}$ such that for each $s\in\mathbb{L}$
and $\gamma\in\mathfrak{p}$,

$\varphi_{s}\left(\gamma\right)=\left.\frac{d}{dt}{\left(\exp\left(t\gamma\right)\left(s\right)\right)}/{s}\right|_{t=0}\in\mathfrak{l.}$
(2.46)

######  Lemma 2.32 ([26, Theorem 3.25])

The map $\varphi$ as in (2.46) is equivariant with respect to corresponding
actions of $\Psi,$ in particular for $h\in\Psi,$ $s\in\mathbb{L}$,
$\gamma\in\mathfrak{p},$ we have

$\varphi_{h\left(s\right)}\left(\left(\mathop{\rm
Ad}\nolimits_{h}\right)_{\ast}\gamma\right)=\left(h^{\prime}\right)_{\ast}\varphi_{s}\left(\gamma\right).$
(2.47)

Moreover, the image of $\varphi_{s}$ is $\mathfrak{l}^{\left(s\right)}$ and
the kernel is $\mathfrak{h}_{s}$, and hence, $\mathfrak{p\cong
h}_{s}\oplus\mathfrak{l}^{\left(s\right)}$.

######  Lemma 2.33 ([26, Lemma 3.33 and Lemma 3.35])

Suppose $\xi\in\mathfrak{p}$ and $\eta,\gamma\in\mathfrak{l},$ then

$\displaystyle\xi\cdot\left[\eta,\gamma\right]^{\left(s\right)}$
$\displaystyle=$
$\displaystyle\left[\xi\cdot\eta,\gamma\right]^{\left(s\right)}+\left[\eta,\xi\cdot\gamma\right]^{\left(s\right)}+a_{s}\left(\eta,\gamma,\varphi_{s}\left(\xi\right)\right)$
(2.48a) $\displaystyle\xi\cdot\varphi_{s}\left(\eta\right)$ $\displaystyle=$
$\displaystyle\eta\cdot\varphi_{s}\left(\xi\right)+\varphi_{s}\left(\left[\xi,\eta\right]_{\mathfrak{p}}\right)+\left[\varphi_{s}\left(\xi\right),\varphi_{s}\left(\eta\right)\right]^{\left(s\right)}.$
(2.48b)

Similarly as for Lie groups, we may define a Killing form $K^{\left(s\right)}$
on $\mathfrak{l}^{\left(s\right)}$. For $\xi,\eta\in\mathfrak{l}$, we have

$K^{\left(s\right)}\left(\xi,\eta\right)=\mathop{\rm Tr}\left(\mathop{\rm
ad}\nolimits_{\xi}^{\left(s\right)}\circ\mathop{\rm
ad}\nolimits_{\eta}^{\left(s\right)}\right),$ (2.49)

where $\circ$ is just composition of linear maps on $\mathfrak{l}$ and
$\mathop{\rm
ad}\nolimits_{\xi}^{\left(s\right)}\left(\cdot\right)=\left[\xi,\cdot\right]^{\left(s\right)}$.
Clearly $K^{\left(s\right)}$ is a symmetric bilinear form on $\mathfrak{l}.$
In [26] it is shown that for $h\in\Psi,$ and $\xi,\eta\in\mathfrak{l}$ it
satisfies
$K^{\left(h\left(s\right)\right)}\left(h_{\ast}^{\prime}\xi,h_{\ast}^{\prime}\eta\right)=K^{\left(s\right)}\left(\xi,\eta\right)$
.

General criteria for a loop algebra to admit a non-degenerate Killing form are
currently not known, but it is known [37] that for a _semisimple_ _Malcev_
algebra, the Killing form is non-degenerate. A _Malcev_ algebra is the tangent
algebra of a Moufang loop and is an _alternative_ algebra that also satisfies
the following identity [36, 42]:

$\left[\xi,\gamma,\left[\xi,\eta\right]^{\left(s\right)}\right]^{\left(s\right)}=\left[\left[\xi,\gamma,\eta\right]^{\left(s\right)},\xi\right]^{\left(s\right)}.$
(2.50)

Moreover, in this case, $K^{\left(s\right)}$ is $\mathfrak{p}$-invariant and
$\mathop{\rm ad}\nolimits^{\left(s\right)}$-invariant [26]. Suppose
$s\left(t\right)=\exp_{s}\left(t\gamma\right)s,$ then from (2.24), we see that
generally,

$\displaystyle\frac{dK^{\left(s\left(t\right)\right)}}{dt}\left(\xi,\xi\right)$
$\displaystyle=$ $\displaystyle\frac{d}{dt}\left(\mathop{\rm
Tr}\left(\left[\xi,\left[\xi,\cdot\right]^{\left(s\left(t\right)\right)}\right]^{\left(s\left(t\right)\right)}\right)\right)$
$\displaystyle=$ $\displaystyle 2\mathop{\rm
Tr}\left(\left[\xi,a_{s\left(t\right)}\left(\xi,\cdot,\gamma\right)\right]^{\left(s\left(t\right)\right)}\right).$

In the special case of $\mathbb{L}$ being a Moufang loop, and thus every
$\mathfrak{l}^{\left(s\right)}$ being a Malcev algebra, we have the following.

###### Lemma 2.34

Suppose $\mathbb{L}$ is a Moufang loop. Then, $K^{\left(s\right)}$ is
independent of $s$ and for each $\gamma\in\mathfrak{l},$ then map $\mathop{\rm
ad}\nolimits_{\gamma}^{\left(s\right)}$ is skew-adjoint with respect to
$K^{\left(s\right)}.$

Proof. If $\mathbb{L}$ is Moufang, then any
$\left(\mathbb{L},\circ_{s}\right)$ is also a Moufang loop, and hence for any
$s$, $\mathfrak{l}^{\left(s\right)}$ is a Malcev algebra. Since a Malcev
algebra is alternative,
$a^{\left(s\right)}\left(\cdot,\cdot,\cdot\right)=2\left[\cdot,\cdot,\cdot\right]^{\left(s\right)},$
and is moreover totally skew-symmetric. In particular, the Malcev identity
(2.50) can be written as

$a_{s}\left(\xi,\gamma,\left[\xi,\eta\right]^{\left(s\right)}\right)=\left[a_{s}\left(\xi,\gamma,\eta\right),\xi\right]^{\left(s\right)}.$
(2.52)

In particular, taking the trace, we get

$\displaystyle\mathop{\rm
Tr}a_{s}\left(\xi,\gamma,\left[\xi,\cdot\right]^{\left(s\right)}\right)$
$\displaystyle=$ $\displaystyle\mathop{\rm
Tr}\left[a_{s}\left(\xi,\gamma,\cdot\right),\xi\right]^{\left(s\right)}$
(2.53) $\displaystyle=$ $\displaystyle-\mathop{\rm
Tr}\left[a_{s}\left(\xi,\gamma,\left[\xi,\cdot\right]\right)\right]^{\left(s\right)}$
$\displaystyle=$ $\displaystyle 0.$

Then, (2) gives
$\frac{dK^{\left(s\left(t\right)\right)}}{dt}\left(\xi,\xi\right)=0.$ This
shows that $K^{\left(s\right)}$ is constant on $\mathbb{L}.$

For the second part, from the generalized Jacobi identity (2.21), we obtain

$\displaystyle
K^{\left(s\right)}\left(\left[\gamma,\eta\right]^{\left(s\right)},\xi\right)$
$\displaystyle=$
$\displaystyle-K^{\left(s\right)}\left(\eta,\left[\gamma,\xi\right]^{\left(s\right)}\right)$
$\displaystyle+\mathop{\rm
Tr}\left[\eta,a_{s}\left(\cdot,\xi,\gamma\right)+a_{s}\left(\xi,\gamma,\cdot\right)+a_{s}\left(\gamma,\cdot,\xi\right)\right]^{\left(s\right)}$
$\displaystyle+\mathop{\rm
Tr}\left[\xi,a_{s}\left(\cdot,\eta,\gamma\right)+a_{s}\left(\eta,\gamma,\cdot\right)+a_{s}\left(\gamma,\cdot,\eta\right)\right]^{\left(s\right)}.$

However, for an alternative algebra, this simplifies to

$\displaystyle
K^{\left(s\right)}\left(\left[\gamma,\eta\right]^{\left(s\right)},\xi\right)$
$\displaystyle=$
$\displaystyle-K^{\left(s\right)}\left(\eta,\left[\gamma,\xi\right]^{\left(s\right)}\right)$
$\displaystyle+3\mathop{\rm
Tr}\left[\eta,a_{s}\left(\cdot,\xi,\gamma\right)\right]^{\left(s\right)}+3\mathop{\rm
Tr}\left[\xi,a_{s}\left(\cdot,\eta,\gamma\right)\right]^{\left(s\right)}.$

The second line is symmetric in $\xi$ and $\eta$, so it is sufficient to
consider the case $\xi=\eta.$ Indeed, for $\xi=\eta,$using (2.53), this
vanishes, so we get

$K^{\left(s\right)}\left(\mathop{\rm
ad}\nolimits_{\gamma}^{\left(s\right)}\eta,\xi\right)=-K^{\left(s\right)}\left(\eta,\mathop{\rm
ad}\nolimits_{\gamma}^{\left(s\right)}\xi\right).$ (2.56)

###### Remark 2.35

Note that in Lemma 2.34, we only used the trace of the Malcev identity. The
non-degeneracy of the Killing form in a semi-simple Malcev algebra also hinges
on the property (2.56), same as for semi-simple Lie algebras. This suggests
that weaker conditions could be sufficient for these key properties.

##  3 Loop-valued maps

Let $M$ be a smooth, $n$-dimensional manifold and let
$s:M\longrightarrow\mathbb{L}$ be a smooth map. The map $s$ can be used to
define a product on $\mathbb{L}^{\prime}$-valued maps from $M$ and a
corresponding bracket on $\mathfrak{l}$-valued maps. Indeed, let
$A,B:M\longrightarrow\mathbb{L}^{\prime}$ and
$\xi,\eta:M\longrightarrow\mathfrak{l}$ be smooth maps, then at each $x\in M$,
define

$\displaystyle\left.As\right|_{x}$ $\displaystyle=$ $\displaystyle
A_{x}s_{x}\in\mathbb{L}$ (3.1a) $\displaystyle\left.A\circ_{s}B\right|_{x}$
$\displaystyle=$ $\displaystyle A_{x}\circ_{s_{x}}B_{x}\in\mathbb{L}$ (3.1b)
$\displaystyle\left.A/_{s}B\right|_{x}$ $\displaystyle=$ $\displaystyle
A_{x}/_{s_{x}}B_{x}\in\mathbb{L}$ (3.1c)
$\displaystyle\left.\left[\xi,\eta\right]^{\left(s\right)}\right|_{x}$
$\displaystyle=$
$\displaystyle\left[\xi_{x},\eta_{x}\right]^{\left(s_{x}\right)}\in\mathfrak{l.}$
(3.1d)

In particular, the bracket $\left[\cdot,\cdot\right]^{\left(s\right)}$ defines
the map
$b_{s}:M\longrightarrow\Lambda^{2}\mathfrak{l}^{\ast}\otimes\mathfrak{l.}$ We
also have the corresponding associator
$\left[\cdot,\cdot,\cdot\right]^{\left(s\right)}$ and the left-alternating
associator map
$a_{s}:M\longrightarrow\Lambda^{2}\mathfrak{l}^{\ast}\otimes\mathfrak{l}^{\ast}\otimes\mathfrak{l}.$
Similarly, define the map
$\varphi_{s}:M\longrightarrow\mathfrak{p}^{\ast}\otimes\mathfrak{l}.$

Then, similarly as for maps to Lie groups, we may define the (right) _Darboux
derivative_ $\theta_{s}$ of $s,$ which is an $\mathfrak{l}$-valued $1$-form on
$M$ given by the pull-back $s^{\ast}\theta$ of the Maurer-Cartan form on
$\mathbb{L}$ [45]. In particular, at every $x\in M$,

$\left.\left(\theta_{s}\right)\right|_{x}=\rho_{s\left(x\right)}^{-1}\left.ds\right|_{x},$
(3.2)

and for any vector $X\in T_{x}M$

$\left.\left(\theta_{s}\right)\right|_{x}\left(X\right)=\left.\theta\right|_{s\left(x\right)}\left(\left.ds\right|_{x}\left(X\right)\right)\in\mathfrak{l.}$

It is then clear that $\theta_{s}$, being a pullback of $\theta$, satisfies
the loop Maurer-Cartan structural equation (2.19). In particular, for any
vectors $X,Y\in T_{x}M$,

$d\theta_{s}\left(X,Y\right)-\left[\theta_{s}\left(X\right),\theta_{s}\left(Y\right)\right]^{\left(s\right)}=0.$
(3.3)

We can then calculate the derivatives of these maps (3.1).

######  Theorem 3.1 ([26, Theorem 3.51])

Let $M$ be a smooth manifold and suppose $s\in
C^{\infty}\left(M,\mathbb{L}\right)\ $and $A,B\in
C^{\infty}\left(M,\mathbb{L}^{\prime}\right),$ then

$\displaystyle d\left(As\right)$ $\displaystyle=$
$\displaystyle\rho_{s}\left(dA\right)+\lambda_{A}\left(ds\right)$ (3.4a)
$\displaystyle d\left(A\circ_{s}B\right)$ $\displaystyle=$
$\displaystyle\rho_{B}^{\left(s\right)}\left(dA\right)+\lambda_{A}^{\left(s\right)}\left(dB\right)+\left[A,B,\theta_{s}\right]^{\left(s\right)}$
(3.4b) $\displaystyle d\left(A/_{s}B\right)$ $\displaystyle=$
$\displaystyle\left(\rho_{B}^{\left(s\right)}\right)^{-1}dA-\left(\rho_{B}^{\left(s\right)}\right)^{-1}\left(\rho_{A/_{s}B}^{\left(s\right)}dB\right)$
(3.4d)
$\displaystyle-\left(\rho_{B}^{\left(s\right)}\right)^{-1}\left[A/_{s}B,B,\theta_{s}\right]^{\left(s\right)}$

Suppose now $\xi,\eta\in C^{\infty}\left(M,\mathfrak{l}\right)$, then

$d\left[\xi,\eta\right]^{\left(s\right)}=\left[d\xi,\eta\right]^{\left(s\right)}+\left[\xi,d\eta\right]^{\left(s\right)}+a_{s}\left(\xi,\eta,\theta_{s}\right).$
(3.5)

The $\mathfrak{l}\otimes\mathfrak{p}^{\ast}$-valued map
$\varphi_{s}:M\longrightarrow$ $\mathfrak{l}\otimes\mathfrak{p}^{\ast}$
satisfies

$d\varphi_{s}=\mathop{\rm
id}\nolimits_{\mathfrak{p}}\cdot\theta_{s}-\left[\varphi_{s},\theta_{s}\right]^{\left(s\right)},$
(3.6)

where $\mathop{\rm id}\nolimits_{\mathfrak{p}}\ $is the identity map of
$\mathfrak{p}$ and $\cdot$ denotes the action of the Lie algebra
$\mathfrak{p}$ on $\mathfrak{l}$ .

The Killing form $K^{\left(s\right)}:M\longrightarrow\mathop{\rm
Sym}\nolimits^{2}\mathfrak{l}^{\ast}$ satisfies

$dK^{\left(s\right)}\left(\xi,\xi\right)=2\mathop{\rm
Tr}\nolimits\left(\left[\xi,a_{s}\left(\xi,\cdot,\theta_{s}\right)\right]^{\left(s\right)}\right).$
(3.7)

Given $A:M\longrightarrow\mathbb{L}^{\prime}$ and
$s:M\longrightarrow\mathbb{L},$ as shown in [26], we have the following
expression for $\theta_{As}$

$\theta_{As}=\mathop{\rm
Ad}\nolimits_{A}^{\left(s\right)}\theta_{s}+\left(\rho_{A}^{\left(s\right)}\right)^{-1}dA.$
(3.8)

Moreover, let us consider the evolution equation satisfied by
$\theta_{A\left(t\right)s}$ for $A\left(t\right)=\exp_{s}\left(t\xi\right)$
for some $\xi:M\longrightarrow\mathfrak{l}.$ This gives us the following.

###### Lemma 3.2

Let $A\left(t\right)=\exp_{s}\left(t\xi\right),$ for $\xi\in
C^{\infty}\left(M,\mathfrak{l}\right),$ then

$\frac{d\theta_{A\left(t\right)s}}{dt}=\left[\xi,\theta_{A\left(t\right)s}\right]^{A\left(t\right)s}+d\xi$
(3.9)

and hence

$\theta_{A\left(t\right)s}=U_{t\xi}^{\left(s\right)}\theta_{s}+\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}d\tau\right)d\xi.$
(3.10)

Moreover, if $\mathbb{L}$ is compact and given a metric on $M$ and an inner
product on $\mathfrak{l},$

$\left|\theta_{A\left(t\right)s}\right|\leq
e^{Ct\left|\xi\right|}\left(\left|\theta_{s}\right|+t\left|d\xi\right|\right).$
(3.11)

Proof. We will write symbolically

$\frac{dA}{dt}=\xi\circ_{s}A,$

so

$\displaystyle\frac{d\theta_{A\left(t\right)s}}{dt}$ $\displaystyle=$
$\displaystyle\frac{d}{dt}\left(d\left(As\right)/As\right)$ $\displaystyle=$
$\displaystyle d\left(\frac{dA}{dt}s\right)/As$
$\displaystyle-\left(\left(d\left(As\right)/As\right)\cdot\left(\frac{dA}{dt}s\right)\right)/As$
$\displaystyle=$ $\displaystyle d\left(\left(\xi\circ_{s}A\right)s\right)/As$
$\displaystyle-\left(\theta_{As}\cdot\left(\left(\xi\circ_{s}A\right)s\right)\right)/As$
$\displaystyle=$ $\displaystyle d\left(\xi\left(As\right)\right)/As$
$\displaystyle+\left(\theta_{As}\cdot\left(\xi\left(As\right)\right)\right)/As$
$\displaystyle=$ $\displaystyle d\xi+\left(\xi\cdot
d\left(As\right)\right)/As$ $\displaystyle-\theta_{As}\circ_{As}\xi$
$\displaystyle=$ $\displaystyle
d\xi+\left[\xi,\theta_{As}\right]^{\left(As\right)}$

Solving this ODE, with $\theta_{A\left(0\right)s}=\theta_{s}$, we find (3.10).
To obtain the estimate, we first have

$\left|\theta_{A\left(t\right)s}\right|\leq\left|U_{t\xi}^{\left(s\right)}\right|\left|\theta_{s}\right|+\left|d\xi\right|\int_{0}^{t}\left|U_{t\xi}^{\left(s\right)}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\right|d\tau,$

but from (2.29), $\left|U_{t\xi}^{\left(s\right)}\right|\leq
e^{Ct\left|\xi\right|}\ $and
$\left|U_{t\xi}^{\left(s\right)}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\right|\leq
e^{C\left(t-\tau\right)\left|\xi\right|},$ so

$\int_{0}^{t}\left|U_{t\xi}^{\left(s\right)}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\right|d\tau\leq
te^{Ct\left|\xi\right|},$ (3.12)

since $t\geq 0,$ and hence,

$\left|\theta_{A\left(t\right)s}\right|\leq
e^{Ct\left|\xi\right|}\left|\theta_{s}\right|+te^{Ct\left|\xi\right|}\left|d\xi\right|,$

and thus indeed, we obtain (3.11).

In a very similar fashion we obtain the same results for $\varphi_{s}.$

###### Lemma 3.3

Let $A\left(t\right)=\exp_{s}\left(t\xi\right)$ and $\gamma\in\mathfrak{p},$
then

$\frac{d\varphi_{A\left(t\right)s}\left(\gamma\right)}{dt}=\left[\xi,\varphi_{A\left(t\right)s}\left(\gamma\right)\right]^{A\left(t\right)s}+\gamma\cdot\xi$
(3.13)

and hence

$\varphi_{A\left(t\right)s}\left(\gamma\right)=U_{t\xi}^{\left(s\right)}\varphi_{s}\left(\gamma\right)+\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}d\tau\right)\left(\gamma\cdot\xi\right).$
(3.14)

Moreover, if $\mathbb{L}$ is compact and given a metric on $M$ and an inner
product on $\mathfrak{l},$

$\left|\varphi_{A\left(t\right)s}\left(\gamma\right)\right|\leq
e^{Ct\left|\xi\right|}\left(\left|\varphi_{s}\left(\gamma\right)\right|+t\left|\gamma\cdot\xi\right|\right).$
(3.15)

Thus we see that three important quantities $\theta_{A\left(t\right)s}$ ,
$U_{t\xi}^{\left(s\right)},$ $\varphi_{A\left(t\right)s}$ satisfy similar
ODEs:

$\displaystyle\frac{d\theta_{A\left(t\right)s}}{dt}$ $\displaystyle=$
$\displaystyle\left[\xi,\theta_{A\left(t\right)s}\right]^{A\left(t\right)s}+d\xi$
$\displaystyle\frac{dU_{t\xi}^{\left(s\right)}}{dt}$ $\displaystyle=$
$\displaystyle\left[\xi,U_{t\xi}^{\left(s\right)}\right]^{A\left(t\right)s}$
$\displaystyle\frac{d\varphi_{A\left(t\right)s}}{dt}$ $\displaystyle=$
$\displaystyle\left[\xi,\varphi_{A\left(t\right)s}\right]^{A\left(t\right)s}+\mathop{\rm
id}\nolimits\cdot\xi.$

Suppose we have an affine connection $\nabla$ on $M$, then by differentiating
the above ODEs, we can obtain expressions for derivatives of $\theta_{As},\
U_{\xi}^{\left(s\right)}$, and $\varphi_{As}.$ However, first we have a
helpful technical lemma.

###### Lemma 3.4

Suppose $A:M\longrightarrow\mathbb{L}^{\prime}$ and
$s:M\longrightarrow\mathbb{L},$ and $\alpha_{A,s}$ is a $k$-linear map on
$\mathfrak{l}.$ Then, for $\xi_{1},...,\xi_{k}\in\mathfrak{l},$

$\displaystyle\left(d\alpha_{A,s}\right)\left(\xi_{1},...,\xi_{k}\right)$
$\displaystyle=$
$\displaystyle\alpha_{A,s}^{\left(1\right)}\left(\xi_{1},...,\xi_{k},\theta_{s}\right)$
$\displaystyle+\alpha_{A,s}^{\left(2\right)}\left(\xi_{1},...,\xi_{k},\theta_{As}\right),$

where $\alpha_{A,s}^{\left(1\right)}$ and $\alpha_{A,s}^{\left(2\right)}$ are
$\left(k+1\right)$-linear maps on $\mathfrak{l}.$

In particular, given a metric on $M$ and a norm on $\mathfrak{l,}$ we have the
following pointwise bound

$\left|\nabla^{k}db_{s}\right|\leq
f^{\left(k\right)}\left(s\right)\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}\left|\theta_{s}\right|^{i_{1}}\left|\nabla\theta_{s}\right|^{i_{2}}...\left|\nabla^{k}\theta_{s}\right|^{i_{k+1}}$
(3.17)

where $I=\left\\{\left(i_{1},...,i_{k+1}\right)\in\mathbb{N}_{0}^{k+1},\
\text{such that }\sum_{m=1}^{k+1}mi_{m}=k+1\right\\}$ and
$f^{\left(k\right)}:\mathbb{L}\longrightarrow\mathbb{R}_{+}\ $is a continuous
function for each $k$.

Proof. Since $\alpha$ depends on $s$ and $A,$ $d\alpha_{A,s}$, by chain rule
we have

$\displaystyle d\alpha_{A,s}$ $\displaystyle=$
$\displaystyle\frac{\partial\alpha}{\partial s}\circ
ds+\frac{\partial\alpha}{\partial A}\circ dA$ $\displaystyle=$
$\displaystyle\left(\frac{\partial\alpha}{\partial
s}\circ\rho_{s}\right)\left(\rho_{s}^{-1}ds\right)+\left(\frac{\partial\alpha}{\partial
A}\circ\rho_{A}^{\left(s\right)}\right)\left(\left(\rho_{A}^{\left(s\right)}\right)^{-1}dA\right)$
$\displaystyle=$ $\displaystyle\left(\frac{\partial\alpha}{\partial
s}\circ\rho_{s}+\frac{\partial\alpha}{\partial
A}\circ\rho_{A}^{\left(s\right)}\circ\mathop{\rm
Ad}\nolimits_{A}^{\left(s\right)}\right)\theta_{s}$
$\displaystyle+\left(\frac{\partial\alpha}{\partial
A}\circ\rho_{A}^{\left(s\right)}\right)\theta_{As},$

where we have used (3.8). So now we can set

$\displaystyle\alpha_{A,s}^{\left(1\right)}\left(\xi_{1},...,\xi_{k},\xi_{k+1}\right)$
$\displaystyle=$
$\displaystyle\left(\frac{\partial\alpha\left(\xi_{1},...,\xi_{k}\right)}{\partial
s}\circ\rho_{s}\right)\xi_{k+1}$
$\displaystyle+\left(\frac{\partial\alpha\left(\xi_{1},...,\xi_{k}\right)}{\partial
A}\circ\rho_{A}^{\left(s\right)}\circ\mathop{\rm
Ad}\nolimits_{A}^{\left(s\right)}\right)\xi_{k+1}$
$\displaystyle\alpha_{A,s}^{\left(2\right)}\left(\xi_{1},...,\xi_{k},\xi_{k+1}\right)$
$\displaystyle=$
$\displaystyle\left(\frac{\partial\alpha\left(\xi_{1},...,\xi_{k}\right)}{\partial
A}\circ\rho_{A}^{\left(s\right)}\right)\xi_{k+1}.$

Thus, we obtain (3.4).

From (3.5), we know that for $\xi,\eta\in\mathfrak{l},$

$\left(db_{s}\right)\left(\xi,\eta\right)=a_{s}\left(\xi,\eta,\theta_{s}\right),$

where the alternating associator $a_{s}$ is a trilinear form on
$\mathfrak{l.}$ Hence, we have the following estimate

$\left|db_{s}\right|\leq C\left|a_{s}\right|\left|\theta_{s}\right|,$

where $C$ is some universal constant. However $a_{s}$ is smooth in $s$, so
$\left|a_{s}\right|$ is in particular a continuous real-valued function on
$\mathbb{L}.$ Hence we can write

$\left|db_{s}\right|\leq
f^{\left(0\right)}\left(s\right)\left|\theta_{s}\right|$ (3.18)

for some positive real-valued function $f^{0}.$ Now, as we have just shown,

$da_{s}\left(\xi_{1},\xi_{2},\xi_{3},\xi_{4}\right)=a_{s}^{\left(1\right)}\left(\xi_{1},\xi_{2},\xi_{3},\theta_{s}\right),$

for some $4$-linear form $a_{s}^{\left(1\right)}.$ Therefore, for a vector
field $X_{1}$ on $M$,

$\displaystyle\left(\nabla_{X_{1}}db_{s}\right)\left(\xi,\eta\right)$
$\displaystyle=$
$\displaystyle\left(d_{X_{1}}a_{s}\right)\left(\xi,\eta,\theta_{s}\right)+a_{s}\left(\xi,\eta,\nabla_{X_{1}}\theta_{s}\right)$
$\displaystyle=$ $\displaystyle
a_{s}^{\left(1\right)}\left(\xi,\eta,\theta_{s},\theta_{s}\left(X_{1}\right)\right)+a_{s}\left(\xi,\eta,\nabla_{X_{1}}\theta_{s}\right),$

so, we have the following estimate

$\displaystyle\left|\nabla db\right|$ $\displaystyle\leq$ $\displaystyle
C\left(\left|a_{s}^{\left(1\right)}\right|\left|\theta_{s}\right|^{2}+\left|a_{s}\right|\left|\nabla\theta_{s}\right|\right)$
(3.19) $\displaystyle\leq$ $\displaystyle
f^{\left(1\right)}\left(s\right)\left(\left|\theta_{s}\right|^{2}+\left|\nabla\theta_{s}\right|\right),$

where
$f^{\left(1\right)}\left(s\right)=C\max\left(\left|a_{s}^{\left(1\right)}\right|,\left|a_{s}\right|\right)$
is continuous.

Similarly, we obtain the expression for the second derivative of $db_{s},$ for
a multilinear maps $a_{s}^{\left(2\right)}$ on $\mathfrak{l:}$

$\displaystyle\left(\nabla_{X_{2}}\nabla_{X_{1}}db_{s}\right)\left(\xi,\eta\right)$
$\displaystyle=$ $\displaystyle
a_{s}^{\left(2\right)}\left(\xi,\eta,\theta_{s},\theta_{s}\left(X_{1}\right),\theta_{s}\left(X_{2}\right)\right)+a_{s}^{\left(1\right)}\left(\xi,\eta,\nabla_{X_{2}}\theta_{s},\theta_{s}\left(X_{1}\right)\right)$
$\displaystyle+a_{s}^{\left(1\right)}\left(\xi,\eta,\nabla_{X_{1}}\theta_{s},\theta_{s}\left(X_{2}\right)\right)+a_{s}^{\left(1\right)}\left(\xi,\eta,\theta_{s},\nabla_{X_{2}}\theta_{s}\left(X_{1}\right)\right)$
$\displaystyle+a_{s}\left(\xi,\eta,\nabla_{X_{2}}\nabla_{X_{1}}\theta_{s}\right),$

where $X_{1}$ and $X_{2}$ are vector fields on $M$. Hence,

$\displaystyle\left|\nabla^{2}db_{s}\right|$ $\displaystyle\leq$
$\displaystyle
C\left(\left|a_{s}^{\left(2\right)}\right|\left|\theta_{s}\right|^{3}+3\left|a_{s}^{\left(1\right)}\right|\left|\theta_{s}\right|\left|\nabla\theta_{s}\right|+\left|a_{s}\right|\left|\nabla^{2}\theta_{s}\right|\right)$
$\displaystyle\leq$ $\displaystyle
f^{\left(2\right)}\left(s\right)\left(\left|\theta_{s}\right|^{3}+\left|\theta_{s}\right|\left|\nabla\theta_{s}\right|+\left|\nabla^{2}\theta_{s}\right|\right),$

for a continuous function $f^{\left(2\right)}.$

Note that in these cases for $k=0,1,2,$ we can symbolically write

$\nabla^{k}db_{s}=\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}a_{i_{1}...i_{k+1}}\left(\left(\theta_{s}\right)^{i_{1}},\left(\nabla\theta_{s}\right)^{i_{2}},...,\left(\nabla^{k}\theta_{s}\right)^{i_{k+1}}\right)$
(3.20)

where $a_{i_{1}...i_{k+1}}$ are multilinear maps that depend on $s$ and
$I=\left\\{\left(i_{1},...,i_{k+1}\right)\in\mathbb{N}_{0}^{k+1},\ \text{such
that }\sum_{m=1}^{k+1}mi_{m}=k+1\right\\}.$ Proceeding by induction, consider

$\displaystyle\nabla^{k+1}db_{s}$ $\displaystyle=$
$\displaystyle\nabla\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}a_{i_{1}...i_{k+1}}\left(\left(\theta_{s}\right)^{i_{1}},\left(\nabla\theta_{s}\right)^{i_{2}},...,\left(\nabla^{k}\theta_{s}\right)^{i_{k+1}}\right)$
$\displaystyle=$ $\displaystyle\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}a_{i_{1}...i_{k+1}}^{\left(0\right)}\left(\left(\theta_{s}\right)^{i_{1}+1},\left(\nabla\theta_{s}\right)^{i_{2}},...,\left(\nabla^{k}\theta_{s}\right)^{i_{k+1}}\right)$
$\displaystyle+\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}a_{i_{1}...i_{k+1}}^{\left(1\right)}\left(\left(\theta_{s}\right)^{i_{1}-1},\left(\nabla\theta_{s}\right)^{i_{2}+1},...,\left(\nabla^{k}\theta_{s}\right)^{i_{k+1}}\right)$
$\displaystyle+...+\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}a_{i_{1}...i_{k+1}}^{\left(k+1\right)}\left(\left(\theta_{s}\right)^{i_{1}},\left(\nabla\theta_{s}\right)^{i_{2}},...,\left(\nabla^{k-1}\theta_{s}\right)^{i_{k}-1},\left(\nabla^{k}\theta_{s}\right)^{i_{k+1}+1}\right)$
$\displaystyle+\sum_{\left(i_{1},...,i_{k+1}\right)\in
I}a_{i_{1}...i_{k+1}}^{\left(k+2\right)}\left(\left(\theta_{s}\right)^{i_{1}},\left(\nabla\theta_{s}\right)^{i_{2}},...,\left(\nabla^{k}\theta_{s}\right)^{i_{k+1}-1},\nabla^{k+1}\theta_{s}\right)$

where $a_{i_{1}...i_{k+1}}^{\left(j\right)}$ are new multilinear forms. The
form $a_{i_{1}...i_{k+1}}^{\left(0\right)}$ is obtained from $\nabla
a_{i_{1}...i_{k+1}},$ and adds another $\theta_{s}.$ Note that since
$\sum_{m=1}^{k+1}mi_{m}=k+1,$ replacing $i_{1}$ with $i_{1}+1,$ increases this
sum by $1$.

The remaining terms in (3) are obtained from differentiating the derivatives
of $\theta_{s}.$ Symbolically,

$\nabla\left(\left(\nabla^{j}\theta\right)^{i_{j+1}}\right)\sim\left(\nabla^{j}\theta\right)^{i_{j+1}-1}\left(\nabla^{j+1}\theta\right),$

so differentiation of each term decreases the power of $\nabla^{j}\theta$ by
one, and adds another $\nabla^{j+1}\theta.$ Again, in the sum
$\sum_{m=1}^{k+1}mi_{m}=k+1,$ replacing $ji_{j}+\left(j+1\right)i_{j+1}$ with
$j\left(i_{j}-1\right)+\left(j+1\right)\left(i_{j+1}+1\right)$ increases the
sum by $1.$

Overall we can then rewrite (3) in the form

$\nabla^{k+1}db_{s}=\sum_{\left(i_{1},...,i_{k+1},i_{k+2}\right)\in
I^{\prime}}a_{i_{1}...i_{k+2}}\left(\left(\theta_{s}\right)^{i_{1}},\left(\nabla\theta_{s}\right)^{i_{2}},...,\left(\nabla^{k+1}\theta_{s}\right)^{i_{k+2}}\right)$

where
$I^{\prime}=\left\\{\left(i_{1},...,i_{k+2}\right)\in\mathbb{N}_{0}^{k+2},\
\text{such that }\sum_{m=1}^{k+2}mi_{m}=k+2\right\\},$ hence proving the
inductive step. The estimate then follows immediately.

###### Lemma 3.5

We have the following pointwise estimates

$\left|\nabla^{j}\theta_{A\left(t\right)s}\right|\leq
Ce^{C\left(j+1\right)t\left|\xi\right|}p_{j}\left(t\right),$

where

$p_{j}\left(t\right)=t\left|\nabla^{j}d\xi\right|+\sum_{J_{j}}t^{k_{1}+...+k_{j}}\left|\theta_{s}\right|^{i_{1}}\left|\nabla\theta_{s}\right|^{i_{2}}...\left|\nabla^{j}\theta_{s}\right|^{i_{j}}\left|d\xi\right|^{k_{1}}\left|\nabla
d\xi\right|^{k_{2}}...\left|\nabla^{j-1}d\xi\right|^{k_{j}},$ (3.22)

with
$J_{j}=\left\\{\left(i_{1},...,i_{j},k_{1},...,k_{j}\right)\in\mathbb{N}_{0}^{2j}:\sum_{m=1}^{j}mi_{m}+\sum_{m^{\prime}=1}^{j}m^{\prime}k_{m^{\prime}}=j+1\right\\}.$

In particular, for $k=1$ and $k=2,$ we have

$\displaystyle\left|\nabla\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{2Ct\left|\xi\right|}\left(\left|\theta_{s}\right|^{2}+t\left|d\xi\right|\left|\theta_{s}\right|+t^{2}\left|d\xi\right|^{2}+\left|\nabla\theta_{s}\right|+t\left|\nabla
d\xi\right|\right)$ (3.23)
$\displaystyle\left|\nabla^{2}\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{3Ct\left|\xi\right|}\left(\left|\nabla^{2}\theta_{s}\right|+\left|\nabla\theta_{s}\right|\left|\theta_{s}\right|+\left|\theta_{s}\right|^{3}+t\left|\nabla\theta_{s}\right|\left|d\xi\right|+t\left|\theta_{s}\right|^{2}\left|d\xi\right|+t\left|\theta_{s}\right|\left|\nabla
d\xi\right|\right.$
$\displaystyle\left.+t^{2}\left|\theta_{s}\right|\left|d\xi\right|^{2}+t^{3}\left|d\xi\right|^{3}+t\left|\nabla^{2}d\xi\right|\right).$

Proof. By differentiating (3.9), we see that the $k$-th covariant derivative
of $\theta_{A\left(t\right)s}$ satisfies the following initial value problem

$\displaystyle\frac{d\nabla^{k}\theta_{A\left(t\right)s}}{dt}$
$\displaystyle=$
$\displaystyle\sum_{k_{1}+k_{2}+k_{3}=k}\left(\nabla^{k_{1}}b_{A\left(t\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}\theta_{A\left(t\right)s}\right)+\nabla^{k}d\xi$
$\displaystyle\nabla^{k}\theta_{A\left(0\right)s}$ $\displaystyle=$
$\displaystyle\nabla^{k}\theta_{s}$

for $k_{1},k_{2},k_{3}\geq 0.$ In particular,

$\frac{d\nabla^{k}\theta_{A\left(t\right)s}}{dt}=\left[\xi,\nabla^{k}\theta_{A\left(t\right)s}\right]^{A\left(t\right)s}+\sum_{\begin{subarray}{c}k_{1}+k_{2}+k_{3}=k\\\
k_{3}<k\end{subarray}}\left(\nabla^{k_{1}}b_{A\left(t\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}\theta_{A\left(t\right)s}\right)+\nabla^{k}d\xi,$

and thus the solution of the ODE is

$\displaystyle\nabla^{k}\theta_{A\left(t\right)s}$ $\displaystyle=$
$\displaystyle
U_{t\xi}^{\left(s\right)}\nabla^{k}\theta_{s}+\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}d\tau\right)\nabla^{k}d\xi$
$\displaystyle+U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\sum_{\begin{subarray}{c}k_{1}+k_{2}+k_{3}=k\\\
k_{3}<k\end{subarray}}\left(\nabla^{k_{1}}b_{A\left(\tau\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}\theta_{A\left(\tau\right)s}\right)d\tau.$

To estimate $\left|\nabla^{k}\theta_{A\left(t\right)s}\right|$, consider

$\displaystyle\left|U_{t\xi}^{\left(s\right)}\nabla^{k}\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
e^{Ct\left|\xi\right|}\left|\nabla^{k}\theta_{s}\right|$
$\displaystyle\left|\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}d\tau\right)\nabla^{k}d\xi\right|$
$\displaystyle\leq$ $\displaystyle
te^{Ct\left|\xi\right|}\left|\nabla^{k}d\xi\right|$

using (2.29) and (3.12), and moreover,

$\left|\left(\nabla^{k_{1}}b_{A\left(\tau\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}\theta_{A\left(\tau\right)s}\right)\right|\leq\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\left|\nabla^{k_{2}}\xi\right|\left|\nabla^{k_{3}}\theta_{A\left(\tau\right)s}\right|.$

From (3.17), and using the fact that $\mathbb{L}$ is compact, we know that

$\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\leq
C\sum_{\left(i_{1},...,i_{k_{1}}\right)\in
I_{k_{1}}}\left|\theta_{A\left(\tau\right)s}\right|^{i_{1}}\left|\nabla\theta_{A\left(\tau\right)s}\right|^{i_{2}}...\left|\nabla^{k_{1}-1}\theta_{A\left(\tau\right)s}\right|^{i_{k_{1}}}$
(3.26)

where
$I_{k_{1}}=\left\\{\left(i_{1},...,i_{k_{1}}\right)\in\mathbb{N}^{k_{1}},\
\text{such that }\sum_{m=1}^{k_{1}}mi_{m}=k_{1}\right\\}$ and $C$ is a
constant that depends only on $\mathbb{L}.$

To proceed with an induction argument, we now need to complete the base step.
First, from (3.11), we know that

$\left|\theta_{A\left(t\right)s}\right|\leq
e^{Ct\left|\xi\right|}\left(\left|\theta_{s}\right|+t\left|d\xi\right|\right).$

Thus, for $k=1$,

$\displaystyle\left|\nabla\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
e^{Ct\left|\xi\right|}\left(\left|\nabla\theta_{s}\right|+t\left|\nabla
d\xi\right|\right)$
$\displaystyle+Ce^{Ct\left|\xi\right|}\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left(\left|b_{A\left(\tau\right)s}\right|\left|d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|+\left|db_{A\left(\tau\right)s}\right|\left|\xi\right|\left|\theta_{A\left(\tau\right)s}\right|\right)d\tau$
$\displaystyle\leq$ $\displaystyle
e^{Ct\left|\xi\right|}\left(\left|\nabla\theta_{s}\right|+t\left|\nabla
d\xi\right|\right)$
$\displaystyle+Ce^{Ct\left|\xi\right|}\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left(\left|d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|+\left|\theta_{A\left(\tau\right)s}\right|^{2}\left|\xi\right|\right)d\tau$

where we used the estimate $\left|db_{A\left(\tau\right)s}\right|\leq
C\left|\theta_{A\left(\tau\right)s}\right|$ (3.18). Moreover,

$\displaystyle\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left|\nabla\xi\right|\left|\theta_{A\left(\tau\right)s}\right|d\tau$
$\displaystyle\leq$
$\displaystyle\int_{0}^{t}\left|\nabla\xi\right|\left(\left|\theta_{s}\right|+\tau\left|d\xi\right|\right)d\tau$
$\displaystyle\leq$ $\displaystyle
t\left|\nabla\xi\right|\left|\theta_{s}\right|+t^{2}\left|d\xi\right|^{2}$
$\displaystyle\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left|\theta_{A\left(\tau\right)s}\right|^{2}\left|\xi\right|d\tau$
$\displaystyle\leq$
$\displaystyle\int_{0}^{t}e^{C\tau\left|\xi\right|}\left(\left(\left|\theta_{s}\right|+\tau\left|d\xi\right|\right)^{2}\left|\xi\right|\right)d\tau$
$\displaystyle\leq$
$\displaystyle\frac{1}{C}e^{Ct\left|\xi\right|}\left(\left|\theta_{s}\right|+t\left|d\xi\right|\right)^{2}.$

Hence, for some new overall constant $C$, we have

$\left|\nabla\theta_{A\left(t\right)s}\right|\leq
Ce^{2Ct\left|\xi\right|}\left(\left|\theta_{s}\right|^{2}+t\left|d\xi\right|\left|\theta_{s}\right|+t^{2}\left|d\xi\right|^{2}+t\left|\nabla
d\xi\right|+\left|\nabla\theta_{s}\right|\right).$ (3.27)

Let us also complete the $k=2$ case.

$\displaystyle\left|\nabla^{2}\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
e^{Ct\left|\xi\right|}\left(\left|\nabla^{2}\theta_{s}\right|+t\left|\nabla^{2}d\xi\right|\right)$
$\displaystyle+Ce^{Ct\left|\xi\right|}\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left(\sum_{k_{1}+k_{2}+k_{3}=2}\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\left|\nabla^{k_{2}}\xi\right|\left|\nabla^{k_{3}}\theta_{A\left(\tau\right)s}\right|\right)d\tau.$

More explicitly,

$\displaystyle\sum_{\begin{subarray}{c}k_{1}+k_{2}+k_{3}=2\\\
k_{3}<2\end{subarray}}\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\left|\nabla^{k_{2}}\xi\right|\left|\nabla^{k_{3}}\theta_{A\left(\tau\right)s}\right|$
$\displaystyle=$ $\displaystyle\left|\nabla
db_{A\left(\tau\right)s}\right|\left|\xi\right|\left|\theta_{A\left(\tau\right)s}\right|$
$\displaystyle+\left|db_{A\left(\tau\right)s}\right|\left|d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|$
$\displaystyle+\left|db_{A\left(\tau\right)s}\right|\left|\xi\right|\left|\nabla\theta_{A\left(\tau\right)s}\right|$
$\displaystyle+\left|b_{A\left(\tau\right)s}\right|\left|\nabla
d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|.$

Let

$\displaystyle p_{0}\left(\tau\right)$ $\displaystyle=$
$\displaystyle\left|\theta_{s}\right|+\tau\left|d\xi\right|$ $\displaystyle
p_{1}\left(\tau\right)$ $\displaystyle=$
$\displaystyle\left|\theta_{s}\right|^{2}+\tau\left|d\xi\right|\left|\theta_{s}\right|+\tau^{2}\left|d\xi\right|^{2}+\left|\nabla\theta_{s}\right|+\tau\left|\nabla
d\xi\right|$

so that

$\displaystyle\left|\theta_{A\left(\tau\right)s}\right|$ $\displaystyle\leq$
$\displaystyle e^{C\tau\left|\xi\right|}p_{0}\left(\tau\right)$
$\displaystyle\left|\nabla\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{2C\tau\left|\xi\right|}p_{1}\left(\tau\right).$

Using also (3.18) and (3.19) we have

$\displaystyle\left|\nabla
db_{A\left(\tau\right)s}\right|\left|\xi\right|\left|\theta_{A\left(\tau\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{3C\left|\xi\right|\tau}\left|\xi\right|p_{0}\left(p_{0}^{2}+p_{1}\right)$
$\displaystyle\left|db_{A\left(\tau\right)s}\right|\left|d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{2C\left|\xi\right|\tau}p_{0}^{2}\left|d\xi\right|$
$\displaystyle\left|db_{A\left(\tau\right)s}\right|\left|\xi\right|\left|\nabla\theta_{A\left(\tau\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{2C\left|\xi\right|\tau}\left|\xi\right|p_{0}p_{1}$
$\displaystyle\left|b_{A\left(\tau\right)s}\right|\left|\nabla
d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|$ $\displaystyle\leq$
$\displaystyle Ce^{C\left|\xi\right|\tau}\left|\nabla d\xi\right|p_{0}.$

Since $p_{0}$ and $p_{1}$ are non-decreasing functions of $\tau$, we can
evaluate them at $\tau=t$. In particular,

$\displaystyle
e^{Ct\left|\xi\right|}\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left|\nabla
db_{A\left(\tau\right)s}\right|\left|\xi\right|\left|\theta_{A\left(\tau\right)s}\right|d\tau$
$\displaystyle\leq$ $\displaystyle
C\left|\xi\right|p_{0}\left(p_{0}^{2}+p_{1}\right)e^{Ct\left|\xi\right|}\int_{0}^{t}e^{2C\tau\left|\xi\right|}d\tau$
$\displaystyle\leq$ $\displaystyle
Ce^{3C\tau\left|\xi\right|}p_{0}\left(p_{0}^{2}+p_{1}\right)$ $\displaystyle
e^{Ct\left|\xi\right|}\int_{0}^{t}e^{-C\tau\left|\xi\right|}\left|db_{A\left(\tau\right)s}\right|\left|d\xi\right|\left|\theta_{A\left(\tau\right)s}\right|d\tau$
$\displaystyle\leq$ $\displaystyle
Cte^{2\left|\xi\right|t}p_{0}^{2}\left|d\xi\right|\leq
Cte^{3\left|\xi\right|t}p_{0}^{2}\left|d\xi\right|,$

and similarly for other terms. Overall, we find

$\displaystyle\left|\nabla^{2}\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{3Ct\left|\xi\right|}\left(p_{0}\left(p_{0}^{2}+p_{1}\right)+tp_{0}^{2}\left|d\xi\right|+p_{0}p_{1}+t\left|\nabla
d\xi\right|p_{0}+\left|\nabla^{2}\theta_{s}\right|+t\left|\nabla^{2}d\xi\right|\right)$
$\displaystyle\leq$ $\displaystyle
Ce^{3Ct\left|\xi\right|}\left(\left|\nabla^{2}\theta_{s}\right|+\left|\nabla\theta_{s}\right|\left|\theta_{s}\right|+\left|\theta_{s}\right|^{3}+t\left|\nabla\theta_{s}\right|\left|d\xi\right|+t\left|\theta_{s}\right|^{2}\left|d\xi\right|+t\left|\theta_{s}\right|\left|\nabla
d\xi\right|\right.$
$\displaystyle\left.+t\left|\nabla^{2}d\xi\right|+t^{2}\left|\theta_{s}\right|\left|d\xi\right|^{2}+t^{3}\left|d\xi\right|^{3}\right).$

Suppose for each $j<k,$

$\left|\nabla^{j}\theta_{A\left(t\right)s}\right|\leq
Ce^{\left(j+1\right)Ct\left|\xi\right|}p_{j}\left(t\right),$

where

$p_{j}\left(t\right)=t\left|\nabla^{j}d\xi\right|+\sum_{J_{j}}t^{k_{1}+...+k_{j}}\left|\theta_{s}\right|^{i_{1}}\left|\nabla\theta_{s}\right|^{i_{2}}...\left|\nabla^{j}\theta_{s}\right|^{i_{j+1}}\left|d\xi\right|^{k_{1}}\left|\nabla
d\xi\right|^{k_{2}}...\left|\nabla^{j-1}d\xi\right|^{k_{j}},$

where
$J_{j}=\left\\{\left(i_{1},...,i_{j+1},k_{1},...,k_{j+1}\right)\in\mathbb{N}^{2j+2}:\sum_{m=1}^{j+1}mi_{m}+\sum_{m=1}^{j}mk_{m}=j+1\right\\}.$

Therefore,

$\displaystyle\left|\sum_{\begin{subarray}{c}k_{1}+k_{2}+k_{3}=k\\\
k_{3}<k\end{subarray}}\left(\nabla^{k_{1}}b_{A\left(\tau\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}\theta_{A\left(\tau\right)s}\right)\right|$
$\displaystyle\leq$ $\displaystyle
C\sum_{\left(i_{0},i_{1},...,i_{k}\right)\in
I_{k}^{\prime}}\left|\nabla^{i_{0}}\xi\right|\left|\theta_{A\left(\tau\right)s}\right|^{i_{1}}\left|\nabla\theta_{A\left(\tau\right)s}\right|^{i_{2}}...\left|\nabla^{k-1}\theta_{A\left(\tau\right)s}\right|^{i_{k}}$
$\displaystyle\leq$ $\displaystyle
C\sum_{\left(i_{0},i_{1},...,i_{k}\right)\in
I_{k}^{\prime}}e^{C\left(k+1-i_{0}\right)\tau\left|\xi\right|}\left|\nabla^{i_{0}}\xi\right|p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k_{1}-1}^{i_{k_{1}}}$
$\displaystyle\leq$ $\displaystyle
C\sum_{\begin{subarray}{c}\left(i_{0},i_{1},...,i_{k}\right)\in
I_{k}^{\prime}\\\
i_{0}>0\end{subarray}}e^{C\left(k+1-i_{0}\right)\tau\left|\xi\right|}\left|\nabla^{i_{0}}\xi\right|p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k-1}^{i_{k}}$
$\displaystyle+C\sum_{\left(0,i_{1},...,i_{k}\right)\in
I_{k}^{\prime}}e^{C\left(k+1\right)\tau\left|\xi\right|}\left|\xi\right|p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k_{1}-1}^{i_{k_{1}}},$

where
$I_{k}^{\prime}=\left\\{\left(i_{0},i_{1},...,i_{k}\right)\in\mathbb{N}_{0}^{k+1},\
\text{such that }\sum_{m=1}^{k}mi_{m}+i_{0}=k+1\right\\}.$

Now, from (3), we find

$\displaystyle\left|\nabla^{k}\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{Ct\left|\xi\right|}\left|\nabla^{k}\theta_{s}\right|+te^{Ct\left|\xi\right|}\left|\nabla^{k}d\xi\right|$
$\displaystyle+Ce^{Ct\left|\xi\right|}\int_{0}^{t}\sum_{\begin{subarray}{c}\left(i_{1},...,i_{k_{1}}\right)\in
I_{k_{1}}\\\ k_{1}+k_{2}=k,\
k_{2}>0\end{subarray}}e^{Ck_{1}\tau\left|\xi\right|}p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k_{1}-1}^{i_{k_{1}}}\left|\nabla^{k_{2}}\xi\right|d\tau$
$\displaystyle+Ce^{Ct\left|\xi\right|}\int_{0}^{t}\sum_{\left(i_{1},...,i_{k}\right)\in
I_{k}}e^{Ck\tau\left|\xi\right|}p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k-1}^{i_{k}}\left|\xi\right|d\tau$
$\displaystyle\leq$ $\displaystyle
Ce^{Ct\left|\xi\right|}\left|\nabla^{k}\theta_{s}\right|+te^{Ct\left|\xi\right|}\left|\nabla^{k}d\xi\right|$
$\displaystyle+C\sum_{\begin{subarray}{c}\left(i_{1},...,i_{k_{1}}\right)\in
I_{k_{1}}\\\ k_{1}+k_{2}=k,\
k_{2}>0\end{subarray}}te^{C\left(k_{1}+1\right)t\left|\xi\right|}p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k_{1}-1}^{i_{k_{1}}}\left|\nabla^{k_{2}}\xi\right|$
$\displaystyle+C\sum_{\left(i_{1},...,i_{k}\right)\in
I_{k}}e^{C\left(k+1\right)t\left|\xi\right|}p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k-1}^{i_{k}},$

where we have bounded $p_{i}\left(\tau\right)\leq p_{i}\left(t\right),$ since
these functions are non-decreasing. Also, in the first integral, we bounded
$\int_{0}^{t}e^{Ck_{1}\tau\left|\xi\right|}d\tau\leq
te^{Ck_{1}t\left|\xi\right|}$ and in the second integral, we used
$\int_{0}^{t}e^{Ck\tau\left|\xi\right|}\left|\xi\right|d\tau\leq
C^{\prime}e^{Ck\left|\xi\right|}$ for some new constant $C^{\prime}.$ Further,
we can bound

$\displaystyle\left|\nabla^{k}\theta_{A\left(t\right)s}\right|$
$\displaystyle\leq$ $\displaystyle
Ce^{C\left(k+1\right)t\left|\xi\right|}\left(\left|\nabla^{k}\theta_{s}\right|+\left|\nabla^{k}d\xi\right|\right.$
$\displaystyle+\sum_{\begin{subarray}{c}\left(i_{1},...,i_{k_{1}}\right)\in
I_{k_{1}}\\\ k_{1}+k_{2}=k,\
k_{2}>0\end{subarray}}tp_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k_{1}-1}^{i_{k_{1}}}\left|\nabla^{k_{2}}\xi\right|$
$\displaystyle\left.+\sum_{\left(i_{1},...,i_{k}\right)\in
I_{k}}p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k-1}^{i_{k}}\right)$ $\displaystyle\leq$
$\displaystyle Ce^{C\left(k+1\right)t\left|\xi\right|}p_{k}\left(t\right).$

###### Corollary 3.6

For $k>0,\ b_{A\left(t\right)s}\ \ $satisfies

$\left|\nabla^{k}b_{A\left(t\right)s}\right|\lesssim
e^{Ckt\left|\xi\right|}p_{k-1}\left(t\right),$

where $p_{k}\left(t\right)$ is given by (3.22).

Proof. From (3.17),

$\left|\nabla^{k}b_{As}\right|\lesssim\sum_{\left(i_{1},...,i_{k}\right)\in
I_{k}}\left|\theta_{As}\right|^{i_{1}}\left|\nabla\theta_{As}\right|^{i_{2}}...\left|\nabla^{k-1}\theta_{As}\right|^{i_{k}}$
(3.28)

where $I_{k}=\left\\{\left(i_{1},...,i_{k}\right)\in\mathbb{N}_{0}^{k},\
\text{such that }\sum_{m=1}^{k}mi_{m}=k\right\\}.$ However, from Lemma 3.5,

$\left|\nabla^{j}\theta_{A\left(t\right)s}\right|\lesssim
e^{C\left(j+1\right)t\left|\xi\right|}p_{j}\left(t\right),$

where

$p_{j}\left(t\right)=t\left|\nabla^{j}d\xi\right|+\sum_{J_{j}}t^{k_{1}+...+k_{j}}\left|\theta_{s}\right|^{i_{1}}\left|\nabla\theta_{s}\right|^{i_{2}}...\left|\nabla^{j}\theta_{s}\right|^{i_{j+1}}\left|d\xi\right|^{k_{1}}\left|\nabla
d\xi\right|^{k_{2}}...\left|\nabla^{j-1}d\xi\right|^{k_{j}},$

with
$J_{j}=\left\\{\left(i_{1},...,i_{j+1},k_{1},...,k_{j}\right)\in\mathbb{N}_{0}^{2j+1}:\sum_{m=1}^{j+1}mi_{m}+\sum_{m^{\prime}=1}^{j}m^{\prime}k_{m^{\prime}}=j+1\right\\}.\
$So

$\displaystyle\left|\nabla^{k}b_{A\left(t\right)s}\right|$
$\displaystyle\lesssim$ $\displaystyle
e^{Ckt\left|\xi\right|}\sum_{\left(i_{1},...,i_{k}\right)\in
I_{k}}p_{0}^{i_{1}}p_{1}^{i_{2}}...p_{k-1}^{i_{k}}$ $\displaystyle\lesssim$
$\displaystyle e^{Ckt\left|\xi\right|}p_{k-1}$

More generally, suppose we have $1$-parameter family $X\left(t\right)$ of
$\mathfrak{l}$-valued maps that satisfies

$\left\\{\begin{array}[]{c}\frac{dX\left(t\right)}{dt}=\left[\xi,X\left(t\right)\right]^{A\left(t\right)s}+Y\\\
X\left(0\right)=X_{0},\end{array}\right.,$ (3.29)

where $Y$ is also an $\mathfrak{l}$-valued map. We know that

$X\left(t\right)=U_{t\xi}^{\left(s\right)}X_{0}+\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}d\tau\right)Y,$
(3.30)

and in particular,

$\left|X\left(t\right)\right|\lesssim
e^{tC\left|\xi\right|}\left(\left|X_{0}\right|+t\left|Y\right|\right).$ (3.31)

Differentiating (3.29), we obtain estimates for higher derivatives of $X.$

###### Lemma 3.7

Suppose $X\left(t\right)$ is a $1$-parameter family $X\left(t\right)$ of
$\mathfrak{l}$-valued maps that satisfies (3.29). Then,

$\left|\nabla^{k}X\left(t\right)\right|\lesssim
e^{\left(k+1\right)Ct\left|\xi\right|}\sum_{k^{\prime}+k^{\prime\prime}=k}p_{k^{\prime}-1}\left(t\right)\left(\left|\nabla^{k^{\prime\prime}}X_{0}\right|+t\left|\nabla^{k^{\prime\prime}}Y\right|\right),$

with $p_{-1}\left(t\right)=1$ and for $k\geq 0$, $p_{k}\left(t\right)$ is
given by (3.22).

Proof. Differentiating (3.29), for $k\geq 1$, we get

$\frac{d\nabla^{k}X\left(t\right)}{dt}=\left[\xi,\nabla^{k}X\left(t\right)\right]^{A\left(t\right)s}+\sum_{\begin{subarray}{c}k_{1}+k_{2}+k_{3}=k\\\
k_{3}<k\end{subarray}}\left(\nabla^{k_{1}}b_{A\left(t\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}X\left(t\right)\right)+\nabla^{k}Y,$

and hence,

$\displaystyle\nabla^{k}X\left(t\right)$ $\displaystyle=$ $\displaystyle
U_{t\xi}^{\left(s\right)}\left(\nabla^{k}X_{0}\right)+\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}d\tau\right)\nabla^{k}Y$
$\displaystyle+\sum_{\begin{subarray}{c}k_{1}+k_{2}+k_{3}=k\\\
k_{3}<k\end{subarray}}\left(U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\left(\nabla^{k_{1}}b_{A\left(\tau\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}X\left(\tau\right)\right)d\tau\right)$

Let $q_{0}\left(t\right)=\left|X_{0}\right|+t\left|Y\right|$, so that

$\left|X\left(t\right)\right|\lesssim
e^{tC\left|\xi\right|}q_{0}\left(t\right).$

Suppose for all $j<k$,

$\left|\nabla^{j}X\left(t\right)\right|\lesssim
e^{\left(j+1\right)Ct\left|\xi\right|}q_{j}\left(t\right),$

where $q_{j}\left(t\right)$ is non-decreasing. Then,

$\displaystyle\left|U_{t\xi}^{\left(s\right)}\int_{0}^{t}\left(U_{\tau\xi}^{\left(s\right)}\right)^{-1}\left(\nabla^{k_{1}}b_{A\left(\tau\right)s}\right)\left(\nabla^{k_{2}}\xi,\nabla^{k_{3}}X\left(\tau\right)\right)d\tau\right|$
$\displaystyle\lesssim$ $\displaystyle
e^{Ct\left|\xi\right|}\int_{0}^{t}e^{k_{3}C\tau\left|\xi\right|}\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\left|\nabla^{k_{2}}\xi\right|q_{k_{3}}\left(\tau\right)d\tau$
$\displaystyle\lesssim$ $\displaystyle
e^{Ct\left|\xi\right|}\int_{0}^{t}e^{\left(k_{1}+k_{3}\right)C\tau\left|\xi\right|}p_{k_{1}-1}\left|\nabla^{k_{2}}\xi\right|q_{k_{3}}\left(\tau\right)d\tau$

For $k_{2}=0,$

$e^{Ct\left|\xi\right|}\int_{0}^{t}e^{k_{3}C\tau\left|\xi\right|}\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\left|\xi\right|q_{k_{3}}\left(\tau\right)d\tau\lesssim
e^{\left(k+1\right)Ct\left|\xi\right|}p_{k_{1}-1}q_{k_{3}}\left(t\right)$

For $k_{2}>0$,

$\displaystyle
e^{Ct\left|\xi\right|}\int_{0}^{t}e^{k_{3}C\tau\left|\xi\right|}\left|\nabla^{k_{1}}b_{A\left(\tau\right)s}\right|\left|\nabla^{k_{2}}\xi\right|q_{k_{3}}\left(\tau\right)d\tau$
$\displaystyle\lesssim$ $\displaystyle
e^{\left(k_{1}+k_{3}+1\right)Ct\left|\xi\right|}p_{k_{1}-1}\left|\nabla^{k_{2}}\xi\right|q_{k_{3}}\left(t\right)$
$\displaystyle\lesssim$ $\displaystyle
e^{\left(k+1\right)Ct\left|\xi\right|}p_{k_{1}+k_{2}-1}q_{k_{3}}\left(t\right).$

Thus,

$\left|\nabla^{k}X\left(t\right)\right|\lesssim
e^{\left(k+1\right)Ct\left|\xi\right|}\left(\left|\nabla^{k}X_{0}\right|+t\left|\nabla^{k}Y\right|+\sum_{k^{\prime}+k^{\prime\prime}=k-1}p_{k^{\prime}}q_{k^{\prime\prime}}\right)$

Therefore, for $k\geq 1$,

$q_{k}=\left|\nabla^{k}X_{0}\right|+t\left|\nabla^{k}Y\right|+\sum_{k^{\prime}+k^{\prime\prime}=k-1}p_{k^{\prime}}q_{k^{\prime\prime}}.$

Setting $x_{k}=\left|\nabla^{k}X_{0}\right|+t\left|\nabla^{k}Y\right|,$ it is
then easy to see that

$q_{k}=\sum_{L_{k}}p_{0}^{j_{0}}p_{1}^{j_{1}}...p_{k-1}^{j_{k-1}}x_{l}$

where

$L_{k}=\left\\{\left(j_{0},...,j_{k-1},x_{l}\right)\in\mathbb{N}_{0}^{k+2}:l+\sum_{m=0}^{j-1}\left(m+1\right)j_{m}=k\right\\},$

and thus

$q_{k}\lesssim
x_{k}+\sum_{k^{\prime}+k^{\prime\prime}=k-1}p_{k^{\prime}}x_{k^{\prime\prime}}.$

Therefore,

$\left|\nabla^{k}X\left(t\right)\right|\lesssim
e^{\left(k+1\right)Ct\left|\xi\right|}\sum_{k^{\prime}+k^{\prime\prime}=k}p_{k^{\prime}-1}\left(\left|\nabla^{k^{\prime\prime}}X_{0}\right|+t\left|\nabla^{k^{\prime\prime}}Y\right|\right),$

with $p_{-1}=1.$

We will need to be able to define loop-valued maps with Sobolev regularity.
First, let us recall Sobolev spaces $W^{k,p}$ of functions between manifolds.

###### Lemma 3.8

Suppose $M$ is a compact $n$-dimensional manifold and suppose $N$ is an
$l$-dimensional manifold. Let $k$ be a non-negative integer and $r\geq 0$ such
that $kr>n.$ Let $\Phi:N\longrightarrow\mathbb{R}^{2l}$ be a smooth embedding
(by Whitney Embedding Theorem) and suppose
$\left\\{\left(U_{\alpha},\phi_{\alpha}\right)\right\\}$ is an atlas for $N.$
Suppose $f:M\longrightarrow N$ is a continuous map. Then, the following are
equivalent:

  1. 1.

$f\in W^{k,r}\left(M,N\right)$

  2. 2.

$\Phi\circ f\in W^{k,r}\left(M,\mathbb{R}^{2l}\right)$

  3. 3.

$\phi_{\alpha}\circ f\in
W^{k,r}\left(\phi_{\alpha}^{-1}\left(U_{\alpha}\right),\mathbb{R}^{l}\right)$
for any chart $\left(U_{\alpha},\phi_{\alpha}\right)$

  4. 4.

$f^{\ast}\theta_{G}$ $\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{g}\right)$ in case when if $N=G,$ a
compact Lie group, with Lie algebra $\mathfrak{g}\ $and $\theta_{G}$ is the
Maurer-Cartan form on $G.$

In particular, conditions (2) and (3) are independent of the choice of the
embedding $\Phi$ and the atlas $\left(U_{\alpha},\phi_{\alpha}\right),$
respectively.

Note that the condition $kr>n$ is needed in Lemma 3.8 due to the Sobolev
embedding $W^{k,r}\subset C^{0}$ for $kr>n.$ We will prove a characterization
of loop-valued $W^{k,p}$-maps in terms of the loop Maurer-Cartan form that is
similar to item (4) in Lemma 3.8.

###### Lemma 3.9

Suppose $M$ is a compact $n$-dimensional manifold and suppose $\mathbb{L}$ is
a smooth loop of dimension $l,$ with tangent algebra $\mathfrak{l}$ and
$\mathfrak{l}$-valued Maurer-Cartan form $\theta$. Let $k$ be a non-negative
integer and $r\geq 0$ such that $kr>n.$ Suppose $s:M\longrightarrow\mathbb{L}$
is a continuous map. Then, $s\in W^{k,r}\left(M,\mathbb{L}\right)$ if and only
if $\theta_{s}\in W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{l}\right).$

Proof. Suppose $s\in W^{k,r}\left(M,\mathbb{L}\right).$ By Lemma 3.8, if
$\left\\{\left(U_{\alpha},\phi_{\alpha}\right)\right\\}$ is an atlas for
$\mathbb{L},$ then for each chart $\left(U_{\alpha},\phi_{\alpha}\right),$
$\phi_{\alpha}\circ s\in
W^{k,r}\left(\phi_{\alpha}^{-1}\left(U_{\alpha}\right),\mathbb{R}^{l}\right).$
Now, $\left\\{s^{-1}\left(U_{\alpha}\right)\right\\}$ is an open cover of $M,$
but using compactness of $M,$ let $\left\\{s^{-1}\left(U_{i}\right)\right\\}$
be an finite subcover, and suppose $\left\\{u_{i}\right\\}$ is a smooth
partition of unity subordinate to this subcover. Then, we can write

$\displaystyle\theta_{s}$ $\displaystyle=$ $\displaystyle\theta
s_{\ast}=\sum_{i}\left(u_{i}\theta\right)s_{\ast}$ $\displaystyle=$
$\displaystyle\sum_{i}\left(\left(u_{i}\theta\right)\left(\phi_{i}^{-1}\right)_{\ast}\right)\left(\left(\phi_{i}\right)_{\ast}s_{\ast}\right).$

For each $i$, $\left(\phi_{i}\right)_{\ast}s_{\ast}=\left(\phi_{i}\circ
s\right)_{\ast}\in
W^{k-1,r}\left(\phi_{i}^{-1}\left(U_{i}\right),\mathbb{R}^{l}\right).$ On the
other hand $\ \left(u_{i}\theta\right)\left(\phi_{i}^{-1}\right)_{\ast}$ is a
smooth function, and hence composition with it is a continuous map
$W^{k-1,r}\left(\phi_{i}^{-1}\left(U_{i}\right),\mathbb{R}^{l}\right)\longrightarrow
W^{k-1,r}\left(\phi_{i}^{-1}\left(U_{i}\right),T^{\ast}U_{i}\otimes\mathfrak{l}\right)$
(using [48, Lemma B.8]). Overall, we see that each term of this finite sum is
bounded in the $W^{k-1,r}$ norm, and thus $\theta_{s}\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{g}\right).$

Conversely, suppose now $\theta_{s}\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{g}\right).$ We will use item (2) in
Lemma 3.8 to show that $s\in W^{k,r}\left(M,\mathbb{L}\right).$ This adapts
the proof of [48, Lemma B.5]. Let
$\Phi:\mathbb{L}\longrightarrow\mathbb{R}^{2l}$ be a smooth embedding, so that
$\Phi\circ s$ is continuous. In particular, $\Phi\circ s\in
L^{r}\left(M,\mathbb{R}^{2l}\right).$ Now, let $x\in M$ and consider

$\displaystyle\left.d\left(\Phi\circ s\right)\right|_{x}$ $\displaystyle=$
$\displaystyle\left.d\Phi\right|_{s\left(x\right)}\left.ds\right|_{x}$
$\displaystyle=$
$\displaystyle\left(\left.d\Phi\right|_{s\left(x\right)}\rho_{s\left(x\right)}\right)\left(\rho_{s\left(x\right)}^{-1}\left.ds\right|_{x}\right)$
$\displaystyle=$ $\displaystyle
E\left(s\left(x\right)\right)\left(\theta_{s}\right)_{x},$

where, for each $p\in\mathbb{L},$ we have the linear map
$E\left(p\right)=\left.d\Phi\right|_{p}\rho_{p}:\mathfrak{l}\longrightarrow\mathbb{R}^{2l}$,
and the map $p\mapsto$ $E\left(p\right)$ is a smooth map from $\mathbb{L}$ to
$\mathop{\rm Hom}\nolimits\left(\mathfrak{l},\mathbb{R}^{2l}\right).$ Thus, we
can write

$d\left(\Phi\circ s\right)=\left(E\circ s\right)\theta_{s},$

with $E\circ s$ being bounded in the operator norm, since $s$ is continuous.
Hence, there exists a constant $C>0,$ such that

$\left\|d\left(\Phi\circ s\right)\right\|_{L^{r}}\leq
C\left\|\theta_{s}\right\|_{L^{r}}\leq
C\left\|\theta_{s}\right\|_{W^{k-1,r}}.$

This shows that $\Phi\circ s\in W^{1,r}\left(M,\mathbb{R}^{2l}\right).$ To
show further that $\Phi\circ s\in W^{k,r}\left(M,\mathbb{R}^{2l}\right),$
$k\geq 2,$ similar estimates are obtained by considering higher derivatives.

###### Theorem 3.10

Let $M$ be a compact Riemannian manifold. Suppose $kr>n=\dim M.$ Let $s\in
W^{k,r}\left(M,\mathbb{L}\right)$ and $\xi\in
W^{k,r}\left(M,\mathfrak{l}\right),$ and suppose $A=\exp_{s}\left(\xi\right).$
Then,

$\left\|\theta_{As}\right\|_{W^{k-1,r}}\lesssim
e^{Ck\left\|\xi\right\|_{C^{0}}}\left(\Theta^{k}+\Theta\right),$ (3.32)

where
$\Theta=\left\|\theta_{s}\right\|_{W^{k-1,r}}+\left\|\xi\right\|_{W^{k,r}}.$

Similarly, if $X=X\left(1\right),$ where $X\left(t\right)$ is $1$-parameter
family of $\mathfrak{l}$-valued maps that satisfies (3.29). Then,

$\left\|X\right\|_{W^{k,r}}\lesssim
e^{C\left(k+1\right)\left\|\xi\right\|_{C^{0}}}\left(\left\|X_{0}\right\|_{W^{k,r}}+\left\|Y\right\|_{W^{k,r}}\right)\left(\Theta^{k+1}+\Theta\right)$

Proof. From Lemma 3.5, for each $j\leq k,$ we have the pointwise estimate

$\left|\nabla^{j-1}\theta_{As}\right|\lesssim e^{Cj\left|\xi\right|}p_{j-1},$

where

$p_{j-1}=\sum_{J_{j-1}}\left|\theta_{s}\right|^{i_{1}}\left|\nabla\theta_{s}\right|^{i_{2}}...\left|\nabla^{j-1}\theta_{s}\right|^{i_{j}}\left|d\xi\right|^{k_{1}}\left|\nabla
d\xi\right|^{k_{2}}...\left|\nabla^{j-1}d\xi\right|^{k_{j}},$ (3.33)

with
$J_{j-1}=\left\\{\left(i_{1},...,i_{j},k_{1},...,k_{j}\right)\in\mathbb{N}_{0}^{2j}:\sum_{m=1}^{j}mi_{m}+\sum_{m=1}^{j}mk_{m}=j\right\\}.$
Thus,

$\left\|\nabla^{j-1}\theta_{As}\right\|_{L^{r}}\lesssim
e^{Cj\left\|\xi\right\|_{C^{0}}}\left\|p_{j-1}\right\|_{L^{r}}.$

Now, from Lemma A.1, if
$\sum_{i^{\prime}=1}^{k^{\prime}}q_{i^{\prime}}m_{i^{\prime}}\leq k,$ then

$\left\|\prod_{i^{\prime}=1}^{k^{\prime}}A_{i^{\prime}}^{m_{i^{\prime}}}\right\|_{L^{r}}\lesssim\prod_{i^{\prime}=1}^{k^{\prime}}\left\|A_{i^{\prime}}\right\|_{W^{k-q_{i^{\prime}},r}}^{m_{i^{\prime}}}.$
(3.34)

We can apply this to (3.33), with the weight $q_{i}=i$ for each
$\left|\nabla^{i}\theta_{s}\right|$ or $\left|\nabla^{i-1}d\xi\right|$ factor.
Then,

$\displaystyle\left\|p_{j-1}\right\|_{L^{r}}$ $\displaystyle\lesssim$
$\displaystyle\sum_{J_{j-1}}\prod_{i^{\prime}=1}^{j-1}\left\|\nabla^{i^{\prime}-1}\theta_{s}\right\|_{W^{k-i^{\prime},r}}^{i_{i^{\prime}}}\prod_{i^{\prime\prime}=1}^{j}\left\|\nabla^{i^{\prime\prime}-1}d\xi\right\|_{W^{k-i^{\prime\prime},r}}^{k_{i^{\prime\prime}}}$
$\displaystyle=$
$\displaystyle\sum_{J_{j-1}}\left\|\theta_{s}\right\|_{W^{k-1,r}}^{i_{1}}...\left\|\nabla^{j-1}\theta_{s}\right\|_{W^{k-\left(j-1\right),r}}^{i_{j-1}}\left\|d\xi\right\|_{W^{k-1,r}}^{k_{1}}...\left\|\nabla^{j-1}d\xi\right\|_{W^{k-j,r}}^{k_{j}}.$

Since for each $i^{\prime}\leq k,$
$\left\|\nabla^{i^{\prime}-1}\theta_{s}\right\|_{W^{k-i^{\prime},r}}\lesssim\left\|\theta_{s}\right\|_{W^{k-1,r}}$
and$\left\|\nabla^{i^{\prime}-1}d\xi\right\|_{W^{k-i^{\prime},r}}\lesssim\left\|\xi\right\|_{W^{k,r}}$,
we obtain

$\left\|p_{j-1}\right\|_{L^{r}}\lesssim\sum_{J_{j-1}}\left\|\theta_{s}\right\|_{W^{k-1,r}}^{i_{1}+...+i_{j-1}}\left\|\xi\right\|_{W^{k,r}}^{k_{1}+...+k_{j}}.$
(3.35)

The right hand-side of (3.35) is thus a polynomial in
$\left\|\theta_{s}\right\|_{W^{k-1,r}}$ and $\left\|\xi\right\|_{W^{k,r}},$
and from the definition of $J_{j-1},$ the degree of this polynomial is $j$ and
the lowest order terms are $\left\|\theta_{s}\right\|_{W^{k-1,r}}$ and
$\left\|\xi\right\|_{W^{k,r}}.$ Hence, we can write

$\left\|\nabla^{j-1}\theta_{As}\right\|_{L^{r}}\lesssim
e^{Cj\left\|\xi\right\|_{C^{0}}}\left(\Theta^{j}+\Theta\right)$

where
$\Theta=\left\|\theta_{s}\right\|_{W^{k-1,r}}+\left\|\xi\right\|_{W^{k,r}}.$
In particular,

$\left\|\theta_{As}\right\|_{L^{r}}\lesssim
e^{C\left\|\xi\right\|_{C^{0}}}\Theta\lesssim
e^{Cj\left\|\xi\right\|_{C^{0}}}\left(\Theta^{j}+\Theta\right).$

Now, since

$\left\|\theta_{As}\right\|_{W^{k-1,r}}\lesssim\left\|\theta_{As}\right\|_{L^{r}}+\left\|\nabla^{k-1}\theta_{As}\right\|_{L^{r}}\lesssim
e^{Ck\left\|\xi\right\|_{C^{0}}}\left(\Theta^{k}+\Theta\right),$

which gives us (3.32).

Now from Lemma 3.7, for each $j$,

$\left|\nabla^{j}X\right|\lesssim
e^{\left(j+1\right)C\left|\xi\right|}\sum_{j^{\prime}+j^{\prime\prime}=j}p_{j^{\prime}-1}\left(\left|\nabla^{j^{\prime\prime}}X_{0}\right|+\left|\nabla^{j^{\prime\prime}}Y\right|\right),$

Hence,

$\displaystyle\left\|\nabla^{j}X\right\|_{L^{r}}$ $\displaystyle\lesssim$
$\displaystyle
e^{\left(j+1\right)C\left\|\xi\right\|_{C^{0}}}\sum_{j^{\prime}+j^{\prime\prime}=j}\left\|p_{j^{\prime}-1}\left(\left|\nabla^{j^{\prime\prime}}X_{0}\right|+\left|\nabla^{j^{\prime\prime}}Y\right|\right)\right\|_{L^{r}}$
$\displaystyle\lesssim$ $\displaystyle
e^{\left(j+1\right)C\left\|\xi\right\|_{C^{0}}}\sum_{J_{j-1}^{\prime}}\left\|\left|\theta_{s}\right|^{i_{1}}\left|\nabla\theta_{s}\right|^{i_{2}}...\left|\nabla^{j-1}\theta_{s}\right|^{i_{j-1}}\left|d\xi\right|^{k_{1}}\left|\nabla
d\xi\right|^{k_{2}}...\left|\nabla^{j-1}d\xi\right|^{k_{j}}\times\right.$
$\displaystyle\left.\times\left(\left|\nabla^{j^{\prime\prime}}X_{0}\right|+\left|\nabla^{j^{\prime\prime}}Y\right|\right)\right\|_{L^{r}}$

where

$J_{j-1}^{\prime}=\left\\{\left(i_{1},...,i_{j},k_{1},...,k_{j},j^{\prime\prime}\right)\in\mathbb{N}_{0}^{2j+1}:\sum_{m=1}^{j}mi_{m}+\sum_{m=1}^{j}mk_{m}+j^{\prime\prime}=j\right\\}.$

Hence, from Lemma A.1,

$\displaystyle\left\|\nabla^{j}X\right\|_{L^{r}}$ $\displaystyle\lesssim$
$\displaystyle
e^{\left(j+1\right)C\left\|\xi\right\|_{C^{0}}}\sum_{J_{j-1}^{\prime}}\left(\left\|\nabla^{j^{\prime\prime}}X_{0}\right\|_{W^{k-j^{\prime\prime},r}}+\left\|\nabla^{j^{\prime\prime}}Y\right\|_{W^{k-j^{\prime\prime},r}}\right)\prod_{i^{\prime}=1}^{j-1}\left\|\nabla^{i^{\prime}-1}\theta_{s}\right\|_{W^{k-i^{\prime},r}}^{i_{i^{\prime}}}\times$
$\displaystyle\times\prod_{i^{\prime\prime}=1}^{j}\left\|\nabla^{i^{\prime\prime}-1}d\xi\right\|_{W^{k-i^{\prime\prime},r}}^{k_{i^{\prime\prime}}}$
$\displaystyle\lesssim$ $\displaystyle
e^{\left(j+1\right)C\left\|\xi\right\|_{C^{0}}}\left(\left\|X_{0}\right\|_{W^{k,r}}+\left\|Y\right\|_{W^{k,r}}\right)\sum_{J_{j-1}^{\prime}}\left\|\theta_{s}\right\|_{W^{k-1,r}}^{i_{1}+...+i_{j}}\left\|\xi\right\|_{W^{k,r}}^{k_{1}+...+k_{j}}$
$\displaystyle\lesssim$ $\displaystyle
e^{C\left(j+1\right)\left\|\xi\right\|_{C^{0}}}\left(\left\|X_{0}\right\|_{W^{k,r}}+\left\|Y\right\|_{W^{k,r}}\right)\left(\Theta^{j}+\Theta\right),$

similarly as before. Hence, we conclude that

$\left\|X\right\|_{W^{k,r}}\lesssim
e^{C\left(k+1\right)\left\|\xi\right\|_{C^{0}}}\left(\left\|X_{0}\right\|_{W^{k,r}}+\left\|Y\right\|_{W^{k,r}}\right)\left(\Theta^{k+1}+\Theta\right).$

###### Corollary 3.11

Suppose $A\in C^{0}\left(M,\mathbb{L}^{\prime}\right)$ and $s\in
W^{k,r}\left(M,\mathbb{L}\right)$, where $kr>n=\dim M.$ Then, $A\in
W^{k,r}\left(M,\mathbb{L}^{\prime}\right)$ if and only if $\theta_{As}\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{l}\right)$.

Proof. The map
$\mu:\mathbb{L}^{\prime}\times\mathbb{L}\longrightarrow\mathbb{L}$ given by
$\left(A,s\right)\mapsto As$ is smooth, hence the composition with $\mu$ is a
continuous map from
$W^{k,r}\left(M,\mathbb{L}^{\prime}\times\mathbb{L}\right)$ to
$W^{k,r}\left(M,\mathbb{L}\right).$ If $A\in
W^{k,r}\left(M,\mathbb{L}^{\prime}\right),$ then since $s\in
W^{k,r}\left(M,\mathbb{L}\right)\subset C^{0}\left(M,\mathbb{L}\right),$
$As\in W^{k,r}\left(M,\mathbb{L}\right),$ and hence from Lemma 3.9,
$\theta_{As}\in W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{l}\right),$ and
thus $\theta_{A}^{\left(s\right)}\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{l}\right).$

Conversely, if $\theta_{As}\in
W^{k-1,r}\left(M,T^{\ast}M\otimes\mathfrak{l}\right),$ then $As\in
W^{k,r}\left(M,\mathbb{L}\right).$ Since right division is a smooth map, and
$s\in C^{0}\left(M,\mathbb{L}\right),$ we conclude that $A\in
W^{k,r}\left(M,\mathbb{L}^{\prime}\right).$

##  4 Gauge theory

Let $M$ be a smooth, finite-dimensional manifold with a $\Psi$-principal
bundle $\pi:\mathcal{P}\longrightarrow M.$

###### Definition 4.1

Let $s:$ $\mathcal{P}\longrightarrow\mathbb{L}$ be an equivariant map. In
particular, given $p\in\mathcal{P},$ the equivalence class $\left\lfloor
p,s_{p}\right\rfloor_{\Psi}$ defines a section of the associated bundle
$\mathcal{Q}=\mathcal{P\times}_{\Psi}\mathbb{L}$, where $\left\lfloor
p,s_{p}\right\rfloor_{\Psi}$ is the equivalence class with respect to the
action of $\Psi:$

$\left(p,s_{p}\right)\sim\left(ph,l_{h^{-1}}\left(s_{p}\right)\right)=\left(ph,s_{ph}\right)\
\ \text{for any }h\in\Psi.$ (4.1)

We will refer to $s$ as _the defining map_ or _defining section_.

We will define several associated bundles related to $\mathcal{P}.$ As it is
well-known, sections of associated bundles are equivalent to equivariant maps.
With this in mind, we also give properties of equivariant maps that correspond
to sections of these bundles. Let $h\in\Psi$ and, as before, denote by
$h^{\prime}$ the partial action of $h$.

Bundle | Equivariant map | Equivariance property
---|---|---
$\mathcal{P}$ | $k:\mathcal{P}\longrightarrow\Psi$ | $k_{ph}=h^{-1}k_{p}$
$\mathcal{Q}^{\prime}=\mathcal{P}\times_{\Psi^{\prime}}\mathbb{L}^{\prime}$ | $q:\mathcal{P}\longrightarrow\mathbb{L}^{\prime}$ | $q_{ph}=\left(h^{\prime}\right)^{-1}q_{p}$
$\mathcal{Q}=\mathcal{P\times}_{\Psi}\mathbb{L}$ | $r:\mathcal{P}\longrightarrow\mathbb{L}$ | $r_{ph}=h^{-1}\left(r_{p}\right)$
$\mathcal{A}=\mathcal{P\times}_{\Psi_{\ast}^{\prime}}\mathfrak{l}$ | $\eta:\mathcal{P}\longrightarrow\mathfrak{l}$ | $\eta_{ph}=\left(h^{\prime}\right)_{\ast}^{-1}\eta_{p}$
$\mathfrak{p}_{\mathcal{P}}=\mathcal{P\times}_{\left(\mathop{\rm Ad}\nolimits_{\xi}\right)_{\ast}}\mathfrak{p}$ | $\xi:\mathcal{P}\longrightarrow\mathfrak{p}$ | $\xi_{ph}=\left(\mathop{\rm Ad}\nolimits_{h}^{-1}\right)_{\ast}\xi_{p}$
$\mathop{\rm Ad}\nolimits\left(\mathcal{P}\right)=\mathcal{P}\times_{\mathop{\rm Ad}\nolimits_{\Psi}}\Psi$ | $u:\mathcal{P}\longrightarrow\Psi$ | $u_{ph}=h^{-1}u_{p}h$
(4.2)

Given equivariant maps $q,r:\mathcal{P}\longrightarrow\mathbb{L}^{\prime}$,
define an equivariant product using $s$, given for any $p\in\mathcal{P}$ by

$\left.q\circ_{s}r\right|_{p}=q_{p}\circ_{s_{p}}r_{p}.$ (4.3)

Due to Lemma 2.7, the corresponding map
$q\circ_{s}r:\mathcal{P}\longrightarrow\mathbb{L}^{\prime}$ is equivariant,
and hence $\circ_{s}$ induces a fiberwise product on sections of
$\mathcal{Q}$. Analogously, we define fiberwise quotients of sections of
$\mathcal{Q}.$ Similarly, we define an equivariant bracket
$\left[\cdot,\cdot\right]^{\left(s\right)}$ and the equivariant map
$\varphi_{s}$. Similarly, the Killing form $K^{\left(s\right)}$ is then also
equivariant.

Suppose the principal $\Psi$-bundle $\mathcal{P}$ has a principal Ehresmann
connection given by the decomposition
$T\mathcal{P}=\mathcal{HP}\oplus\mathcal{VP}$ and the corresponding vertical
$\mathfrak{p}$-valued connection $1$-form $\omega.$ Given an equivariant map
$f:\mathcal{P}\longrightarrow S$, define

$d^{\omega}f:=f_{\ast}\circ\mathop{\rm
proj}\nolimits_{\mathcal{H}}:T\mathcal{P}\longrightarrow\mathcal{HP}\longrightarrow
TS.$ (4.4)

This is then a horizontal map since it vanishes on any vertical vectors. The
map $d^{\mathcal{\omega}}f$ is moreover still equivariant, and hence induces a
covariant derivative on sections of the associated bundle
$\mathcal{P}\times_{\Psi}S$. If $S$ is a vector space, then this reduces to
the usual definition of the exterior covariant derivative of a vector bundle-
valued function and $d^{\mathcal{\omega}}f$ is a vector-bundle-valued
$1$-form.

Following [26], let us define the torsion of the defining map $s$ with respect
to the connection $\omega$.

###### Definition 4.2

The torsion $T^{\left(s,\omega\right)}$ of the defining map $s$ with respect
to $\omega$ is a horizontal $\mathfrak{l}$-valued $1$-form on $\mathcal{P}$
given by
$T^{\left(s,\omega\right)}=\left(s^{\ast}\theta\right)\circ\mathop{\rm
proj}\nolimits_{\mathcal{H}}$, where $\theta$ is Maurer-Cartan form of
$\mathbb{L}$. Equivalently, at $p\in\mathcal{P}$, we have

$\left.T^{\left(s,\omega\right)}\right|_{p}=\left(R_{s_{p}}^{-1}\right)_{\ast}\left.d^{\omega}s\right|_{p}.$
(4.5)

Thus, $T^{\left(s,\omega\right)}$ is the horizontal component of
$\theta_{s}=s^{\ast}\theta.$ We also easily see that it is $\Psi$-equivariant.
Thus, $T^{\left(s,\omega\right)}$ is a _basic_ (i.e. horizontal and
equivariant) $\mathfrak{l}$-valued $1$-form on $\mathcal{P}$, and thus defines
a $1$-form on $M$ with values in the associated vector bundle
$\mathcal{A=P\times}_{\Psi_{\ast}^{\prime}}\mathfrak{l}.$

Recall that the curvature
$F^{\left(\omega\right)}\in\Omega^{2}\left(\mathcal{P},\mathfrak{p}\right)$ of
the connection $\omega$ on $\mathcal{P}$ is given by

$F^{\left(\omega\right)}=d\omega\circ\mathop{\rm
proj}\nolimits_{\mathcal{H}}=d\omega+\frac{1}{2}\left[\omega,\omega\right]_{\mathfrak{p}},$
(4.6)

where wedge product is implied. Given the defining map $s$, define
$\hat{F}^{\left(s,\omega\right)}\in\Omega^{2}\left(\mathcal{P},\mathfrak{l}\right)$
to be the projection of the curvature $F^{\left(\omega\right)}$ to
$\mathfrak{l}$ with respect to $s$, such that for any $X_{p},Y_{p}\in
T_{p}\mathcal{P},$

$\hat{F}^{\left(s,\omega\right)}=\varphi_{s}\left(F^{\left(\omega\right)}\right).$
(4.7)

######  Theorem 4.3 ([26, Theorem 4.19])

$\hat{F}^{\left(s,\omega\right)}$ and $T^{\left(s,\omega\right)}$ satisfy the
following structure equation

$\hat{F}^{\left(s,\omega\right)}=d^{\mathcal{\omega}}T^{\left(s,\omega\right)}-\frac{1}{2}\left[T^{\left(s,\omega\right)},T^{\left(s,\omega\right)}\right]^{\left(s\right)},$
(4.8)

where a wedge product between the $1$-forms $T^{\left(s,\omega\right)}$ is
implied.

In the case of an octonion bundle over a $7$-dimensional manifold, this
relationship between the torsion and a curvature component has been shown in
[24].

As discussed earlier, equivariant horizontal forms on $\mathcal{P}$ give rise
to sections of corresponding associated bundles over the base manifold $M.$ So
let us now switch perspective, and work in terms of sections of bundles. In