# Chapter 5: The Exponential Map

The aim of this post is to explore one parameter groups within a connected Lie group, namely those defined as follows:

Definition 5.1 (One Parameter Subgroup):

A one parameter subgroup of a connected Lie group is a homomorphic image $\mathfrak{P}\subset \G$ of the group $(\R,\,+)$ of reals with addition within $\G$. In other words, there is a homomorphism $\sigma:\R\to\G$ and its homomorphic image $\mathfrak{P}=\sigma(\R)\subseteq\G$ is a one parameter subgroup. By definition of a homomorphism:

\label{OneParameterGroupEquation}
\mathfrak{P}=\{\rho(\tau):\,\tau\in\R;\,\rho:\R\to\G;\,\rho(\tau)\,\rho(s) = \rho(s)\,\rho(\tau) = \rho(\tau+s)\}

Take Heed: We make this definition at first regardless of whether or not $\sigma(\tau)$ traces a $C^0$ path through $\Nid$.

This is a very long and complicated chapter, but its aims are simply stated, so I’ll first give an overview.

1. One parameter subgroups of a connected Lie group $\G$ are the images of an exponential function.
2. We first prove that $C^1$ one parameter subgroups exist by defining restrictions of exponential functions to $\Nid$ as solutions of certain differential equations in $\Nid$. These solutions are readily extended out of $\Nid$ by the flow property. Only the $C^1$ assumptions of axioms 1 through 5 are needed to do this;
3. Next we show how the exponential function as a bijective, $C^1$ mapping between at least one neighbourhood of $\Or\subseteq\g$ in the Lie algebra $\g$ and a nucleus $\K\subset \G$. Thus $\exp$ has a well defined, $C^1$ inverse $\log: \K\to\g$ which can be used to give any member of $\K$ unique geodesic coordinates.
4. We then look at an algebraic definition of the exponential function grounded on the functional equation $\exp(s+\tau)=\exp(s)\,\exp(\tau)$. Within some nucleus $\K\subset\G$, square roots are uniquely defined and so can be used to define powers $\gamma^q$ of any $\gamma\in\K$ where the power $q$ is of the form $m\,2^{-n}$ for $m,\,n\in\mathbb{Z}$. The set of powers $\gamma^q$ are readily shown to be the same as the powers computed as $\gamma^q=\exp(q\,\log\gamma)$ where $\exp$ is the mapping defined through differential equations above, thus $\exp(q\,X)$, where $X=\log \gamma$ is the same as $\gamma^q$ when the power $q$ belongs to the dense subset $\mathcal{P}=\{m\,2^{-n}|\,m,\,n\in\mathbb{Z}\} \cap (-1,\,1)$ of the interval $(-1,\,1)$. Because $\mathcal{P}$ is dense in the interval, there is only one possible continuous extension of the definition of the power $\gamma^q$ to any real power $q\in(-1,\,1)$. Therefore the exponential function $\exp(q\,X)$ defined for all real $q$ through the path differential equation definition above is this unique continuous extension.
5. Since $\exp$ “happens” to be $C^1$ as well as $C^0$, it follows that any one parameter subgroup $\mathfrak{P}\subset\G$ that is continuous in $\G$ must also be $C^1$ in $\G$, and, from results to follow in Chapter 7 and 9, it must also be $C^\omega$ (analytic) as well.

Unfortunately, although the last statement sounds very relevant to a proof of Hilbert’s Fifth Problem (for those you whom this idea is wonted to), it is not because all our proofs rest on our five axioms, which of course assume that the group product of two $C^1$ paths is a $C^1$ path.

The kernel of the homomorphism in Definition 5.1 of course defines $\mathfrak{P}$. We shall be most interested in continuous one parameter subgroups, i.e. such that the path traced by $\rho(\tau)$ is a $C^0$ path as it passes through $\Nid$.

Theorem 5.2 (Classification of the Kernel of One Parameter Subgroups):

For a one parameter subgroup in a connected Lie group $\G$ defined by the homomorphism $\rho:\R\to\G$ we have the following possibilities. These are exhaustive, i.e. there are no other possibilities:

1. The kernel $\ker(\rho) = \{\tau\in\R:\,\rho(\tau) = \id\}$ is the whole of $\R$ and the homomorphism is the trivial homomorphism $\R\to\G$;
2. The kernel $\ker(\rho) = \{\tau\in\R:\,\rho(\tau) = \id\}$ is dense in $\R$;
3. The kernel is the equispaced grid of numbers $\tau_0\,\mathbb{Z} = \{k\,\tau_0:\,k\in\mathbb{Z}\}$ for some constant spacing $\tau_0$;
4. The kernel is trivial and $\rho$ is an isomorphism.

Proof: Show Proof

Clearly the first possibility exists and needs no futher discussion.

Clearly $\rho(0)=\id$ (since $\rho(0+0) = \rho(0)\,\rho(0) = \rho(0)\Rightarrow\rho(0)=\id$). Now, suppose there is no smallest number $\tau_0>0$ such that $(0,\,\tau_0)\cap\ker(\rho)=\emptyset$, i.e. there is no nonzero length interval of non-kernel members. Then we can find an arbitrarily small number $\delta\in\ker(\rho)<\epsilon$, therefore, by the homomorphism, $\{k\,\delta:\,k\in\mathbb{Z}\}\subset\ker(\rho)$ for any $\epsilon>0$. Therefore, $\ker(\rho)$ is dense in $\R$: this is the second possibility.

So, if neither of the first two possibilities hold, we must have $(-\tau_0,\,\tau_0)\cap\ker(\rho) = \{0\}$ for some $\tau_0>0$.

Now we assume that there is a biggest positive number $\tau_0$ such that$(-\tau_0,\,\tau_0)\cap\ker(\rho) = \{0\}$, i.e. $\tau_0$ is the greatest lower bound on the set of nonzero, positive kernel members. Now suppose $s\in\ker(\rho)$ for some arbitrary real $s$, then $\sigma(s) = \sigma(s)^{-1}=\sigma(-s)=\id$ and $\sigma(\delta+s) =\sigma(-s)\,\sigma(\delta+s) = \sigma(\delta) \neq \id$ if $\delta\neq0$ and $|\delta|<\tau_0$ because $\ker(\rho)\cap\{\tau:|\tau|<\tau_0\} = \{0\}$. In other words, there are open intervals $(s,\,s+\tau_0)$ and $(s-\tau_0,\,\tau_0)$ which are free of kernel members on either side of any kernel member. So suppose that $\rho(\tau_0+\delta) = \id$ for some $\delta$ where $0<\delta<\tau_0$, then this would mean that the whole interval $\rho(\delta,\,\tau_0+\delta)$ would be free of kernel members, which in turn means that $\tau_0$ itself could not be a kernel member and, since $(0,\,\tau_0)$ would also be free of kernel members, the whole interval $(0,\,\delta)$ would be free of kernel members. This would gainsay our assumption that $\tau_0$ was the greatest lower bound on the set of nonzero, positive kernel members. So $(\tau_0,\,2\,\tau_0)$ must also be free of kernel members. Likewise for the interval $(-2\,\tau_0,\,-\tau_0)$. We can go forward likewise by induction to prove that the open interval $(n\,\tau_0,\,(n+1)\,\tau_0)$ is free of kernel members for every $n\in\mathbb{Z}$. Therefore, the only possible kernel members are numbers of the form $k\,\tau_0$ for $k\in\mathbb{Z}$. But if $\tau_0$ is not a kernel member, then a repetition of the above argument shows $,-\tau_0,\,\pm3,\tau_0,\,\pm5\tau_0,\,\cdots$ cannot be kernel members and so the only possible kernel members would be numbers of the form $2\,k\,\tau_0$ for $k\in\mathbb{Z}$, and our original spacing should have been $2\,\tau_0$. We can make a wholly analogous argument by induction: if the first $\pm\,n\,\tau_0$ are kernel members, then $\pm\,(n+1)\,\tau_0$ must be kernel members, otherwise we gainsay our assumption of $\tau_0$’s being the greatest lower bound on the set of nonzero, positive kernel members. Therefore all the $\tau_0\,\mathbb{Z}$ are kernel members, and all other numbers are not. This proves that the second possibility arises from the assumption of a $\tau_0$ that is the greatest lower bound on the set of nonzero, positive kernel members. It is also well to heed that in this case $\rho$ is a periodic function, with period $2\,\tau_0$ and that $\rho((-\tau_0,\,\tau_0) )$ is the whole of the one parameter subgroup.

The fourth possibility must hold if the first three do not, $0$ is then the only kernel member, and the homomorphism is then an isomorphism between $\R$ and a subgroup of $\G$. $\qquad\square$

The second and third possibilities above indeed beget a possibly new Lie group: once we have a homomorphism with $(\R,\,+)$ with either equispaced or trivial kernel, we clearly have an entity which fulfills our five Lie group defining axioms if we take $\V = (-\tau_0,\,\tau_0)$ or $\V=\R$ for the second or third possibility above, respectively. The homomorphism $\rho$ when restricted to $\V$ is thus one-to-one and onto, therefore has an inverse $\log$ which we take as our Labeller map $\lambda$, and then we take $\log(\V) = \mathrm{N}_\id$. The Lie group then has a Lie algebra of dimension 1. It is an important but subtle point that this new Lie group might be “unrelated” to the Lie group structure $\G$, i.e. the one dimensional Lie algebra we create through $\rho$ may not be a one dimensional linear subspace of $\g$. The one-parameter group, when embedded in the whole group $\G$ can sometimes have a completely different topology than the topology inherited from $\G$. I shall come back to this point below and explore it more fully in later posts, although this can happen only in fairly restricted ways. Once we give an abstract group a Lie group structure by finding $\V$, $\Nid$ and a labeller $\lambda$ such that our basic five axioms are fulfilled, there are only fairly restricted ways wherein we can give the absract group a Lie group structure.

Henceforth, we shall mainly be interested in continuous one parameter subgroups, i.e. such that the path traced by $\rho(\tau)$ is a $C^0$ path as it passes through $\Nid$. These will turn out to be automatically $C^1$. Such continuous one parameter subgroups are generated by paths $\sigma_{\exp}:[-\tau_0,\,\tau_0]\to\G$ for some $\tau_0>0$ within the group $\G$ that fulfill the defining property:

\label{ExpGermEquation}
\sigma_{\exp}(\tau)\,\sigma_{\exp}(\varsigma) = \sigma_{\exp}(\varsigma)\,\sigma_{\exp}(\tau) =\sigma_{\exp}(\tau+\varsigma),\,\forall \tau,\,\varsigma, \tau+\varsigma \in [-\tau_0,\,\tau_0]

Therefore, we define a one parameter group as the set of all finite products of entities of the form $\sigma_{\exp}(\tau),\, \tau\in[-\tau_0,\,\tau_0]$ where $\sigma_{\exp}$ is such a path. It is then readily shown, if we define inductively on the positive integer $n$, that

\label{ExtendingExpGermEquation}
\exp(\pm n\,\tau_0 + \tau) \stackrel{def}{=} \exp(\pm n\,\tau_0) \sigma_{\exp}(\tau);\;|\tau|\leq \tau_0

then we define a function:

\label{ExtendedExpEquation}
\exp:\R\to\G;\;\exp(0) = \id;\;\exp(\tau)\exp(\varsigma)=\exp(\varsigma)\exp(\tau) = \exp(\varsigma+\tau),\forall \tau,\varsigma\in\R

such that the set $\H_{\exp}=\{\exp(\tau):\tau\in\R\}$ is a subgroup of $\G$.

Through any one-parameter group we define a flow, that is, a group action of the additive group $(\R,\,+)$ of real numbers on the Lie group $\G$:

\label{FlowDefinitionEquation}
\varphi:\G\times\R\to\G;\; \varphi(\varphi(\gamma,\,\varsigma),\,\tau) = \varphi(\varphi(\gamma,\,\tau),\,\varsigma)=\varphi(\gamma,\,\tau+\varsigma);\;\varphi(\gamma,\,\tau) = \exp(\tau)\,\gamma

Clearly, we can make an in general different, but valid, flow through the right-handed definition $\varphi(\gamma,\,\tau) = \gamma\,\exp(\tau)$.

It should not be surprising that there are many different exponential maps $\exp$ and one-parameter groups, indeed, there is precisely one $C^1$ such group for each ray through the Lie algebra $\g$, i.e. where we define:

Definition 5.3 (Ray through Lie Algebra):

The ray defined by a Lie algebra member $X\in\g$ is the set $\{x\,X:\,x\in\R\}$

Indeed every nonzero $X\in\g$ belongs to precisely one ray. We could say there is precisely one $C^1$ exponential function for each point on the unit sphere $\{X\in\g:\,\left\|X\right\|=1\}$, where $\left\|\cdot\right\|$ is any norm defined on $\g$. Recall that a norm $\left\|\cdot\right\|:\g\to\R$ (i) fulfils the triangle inequality$\left\|X+Y\right\|\leq \left\|X\right\|+\left\|X\right\|,\,\forall X,\,Y\in\g$, (ii) is nondegenerate i.e. $\left\|X\right\|=0\Rightarrow X=0,\,\forall X\in\g$ and (iii) is scalable $\left\|x\,X\right\| =|x| \left\|X\right\|,\,\forall X\in\g,\,x\in\R$. The use of norms in the following does not depend on the particular norm used, as long as it has these three properties; for concreteness, you should simply assume the Euclidean norm if you like.

I shall now look at several approaches to defining exponential functions.

## Exponential Functions Defined by Differential Equations

### General $C^1$ Paths Defined by Differential Equations

A general $C^1$ path in $\Nid$ has a tangent defined by:

\label{PathDefintionDifferentialEquation}
\d_\tau \lambda(\sigma(\tau))= \mathbf{M}(\id,\,\sigma(\tau))\,X(\sigma(\tau),\,\tau) \stackrel{def}{=}\Sigma(\sigma(\tau),\,\tau)

for some Lie algebra member $X(\sigma,\,\tau)\in\g$ that varies with the parameter $\tau$, wandering about the Lie algebra to account for the twists and wends of the path. Here, $\mathbf{M}(\id,\,\sigma(\tau))$ is the tangent map induced by left translation from $\id$ to $\sigma(\tau)$.

However, with our axioms, we can only be sure that the tangent map $\mathbf{M}(\id,\,\sigma)$ is a $C^0$ function of the co-ordinates defining $\sigma$. In such a case, the Peano existence theorem guarantees the existence of a solution within $\Nid$ to a general equation of the form $\d_\tau \lambda(\sigma(\tau))= \mathbf{M}(\id,\,\sigma(\tau))\,X(\tau)$ with any initial conditions. The problem that the Peano Existence Theorem does not guarantee uniqueness of the solution and such uniqueness is extremely helpful to the simple and elegant description of Lie groups.

### The Wonted Solution: Assuming Lipschitz Continuity of the Group Product

The wonted solution to this problem in most treatments of Lie groups is to strengthen the group product continuity axioms, so that we would replace $C^1$ by $C^{1,1}$ (i.e. Lipschitz continuity) in the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4. Most Lie group texts assume the group is an analytic manifold, therefore $\mathbf{M}(\id,\,\sigma)$ is an analytic function of the co-ordinates defining $\sigma$. Therefore, in particular, $\mathbf{M}(\id,\,\sigma)$ is Lipschitz Continuous. I shall quickly retell this wonted tale before showing that, given the way we have defined the group axioms, that we do not need to make this assumption and the $C^1$ behaviour of the group product actually enforces uniqueness of solution of the differential equations even though we can only appeal to the Peano Existence Theorem to show existence of the solution.

Definition 5.4 (Lipschitz Continuity):

The function $\Sigma(\sigma)$ defined for $\sigma\in\Nid$ is called Lipschitz Continuous iff:

\label{LipschitzContinuityDefinition_1}
\exists K\in\R\,\ni\,\forall\,\gamma\in\Nid,\; \left\|\Sigma(\sigma)-\Sigma(\gamma)\right\|\leq K \left\|\lambda(\sigma)-\lambda(\gamma)\right\|

Lipschitz continuity of $\Sigma(\sigma,\,\tau)$ lets us use the Picard–Lindelöf theorem, an argument which runs as follows. With the initial conditions $\sigma(0) = \sigma_0$, we rewrite the basic differential equation as:

\label{PathDefintionDifferentialEquation_2}
\lambda(\sigma(\tau) )= \lambda(\sigma_0)+\mathcal{L}(\sigma) \stackrel{def}{=} \int_0^\tau \,\Sigma(\sigma(s),\,s)\,\d\,s

and seek a fixed point (in this case a $C^1$ path $\sigma:[0,\,\tau_{max}]$) to the operator $\sigma\to\mathcal{L} \sigma$ by beginning with a guess to the solution, say $\sigma_X(\tau)$ and then reckonning Picard iterates $\sigma_X,\,\mathcal{L} \sigma_X,\,\mathcal{L}^2\,\sigma_X,\,\mathcal{L}^3\,\sigma_X,\,\cdots$. We bound $\left\|X(s)\right\|\leq X_{max}$, calculate the bound $\sup\limits_{\gamma\in\Nid} \left\|\mathbf{M}(\id,\,\sigma)\right\| = M_{max}$, choose an open ball $\{\gamma:\,\left\|\lambda(\gamma)\right\|<R\}\subseteq \Nid$ and then set $\tau_{max}$ so that $\tau_{max} X_{max} M_{max} < R$. This guarantees that $\sigma([0,\,\tau_{max})\subseteq\Nid$. Lastly we define the norm of our solution $\sigma$ by:

\label{LInfinityNormDefinitionEquation}
\left\|\sigma\right\|\stackrel{def}{=}\max\limits_{\tau\in[0,\,\tau_{max}]}\left\|\sigma(\tau)\right\|

i.e. the $\mathbf{L}^\infty$ norm and so that Lipschitz continuity of $\mathbf{M}(\id,\sigma(s))\,X(\sigma(s),\,s)$ shows that:

\label{PicardIterationEquation}
\begin{array}{lcl}
\left\|\mathcal{L}^n(\sigma_1)-\mathcal{L}^n(\sigma_2)\right\|&=&\max\limits_{\tau\in[0,\,\tau_{max}]}\left\|\int_0^\tau\int_0^{\tau_1}\cdots\int_0^{\tau_{n-1}}\,\left(\Sigma(\sigma_1(\tau_n),\,\tau_n)-\Sigma(\sigma_2(\tau_n),\,\tau_n)\right)\,\d\,\tau_n\,\d\,\tau_{n-1}\cdots\,\d\,\tau_1\right\|\\
&\leq&\max\limits_{\tau\in[0,\,\tau_{max}]}\int_0^\tau\int_0^{\tau_1}\cdots\int_0^{\tau_{n-1}}\,\left\|\Sigma(\sigma_1(\tau_n),\,\tau_n)-\Sigma(\sigma_2(\tau_n),\,\tau_n)\right\|\,\d\,\tau_n\,\d\,\tau_{n-1}\cdots\,\d\,\tau_1\\
&\leq&\max\limits_{\tau\in[0,\,\tau_{max}]}\int_0^\tau\int_0^{\tau_1}\cdots\int_0^{\tau_{n-1}}\,K\,\left\|\sigma_1(\tau_n)-\sigma_2(\tau_n)\right\|\,\d\,\tau_n\,\d\,\tau_{n-1}\cdots\,\d\,\tau_1\\
&\leq&\max\limits_{\tau\in[0,\,\tau_{max}]}K\,\left\|\sigma_1(\tau)-\sigma_2(\tau)\right\|\int_0^\tau\int_0^{\tau_1}\cdots\int_0^{\tau_{n-1}}\,\d\,\tau_n\,\d\,\tau_{n-1}\cdots\,\d\,\tau_1\\
&=&\frac{K}{n!}\,\left\|\sigma_1-\sigma_2\right\| \tau_{max}^n
\end{array}

which is contractive for some big enough $n>0$. Hence, by the Banach Fixed Point Theorem (the “Contraction Mapping Principle”), the Picard iterates must converge to the unique solution to the differential equation.

So, then one simply defines the exponential map $\exp_X:[0\,\tau_{max}]\to\G),\,\exp_X(\tau)=\exp(X\,\tau)$ to be the solution, existing and unique for $\exp_X(\tau)\in\Nid$ by the Picard-Lindelöf theorem, to the differential equation $\d_\tau \lambda(\sigma)= \mathbf{M}(\id,\,\sigma(\tau))\,X$ with $\sigma(0)=\id$.

### Getting Rid of Picard-Lindelöf

However, to define the exponential function, we only need the special equation $\d_\tau \lambda(\sigma)= \mathbf{M}(\id,\,\sigma(\tau))\,X$, i.e. with constant $X$, and in this case, we can use the Lie group product continuity axioms to show that a solution to this equation, whose existence is guaranteed by the more general Peano Existence Theorem, is indeed unique. A wholly equivalent way to write this equation is in a “sliding reference frame”:

\label{SlidingReferenceFrameEquation}
\d_\tau \lambda(\sigma)= \mathbf{M}(\id,\,\sigma(\tau))\,X \;\Leftrightarrow\;\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X

The equation on the right of the $\Leftrightarrow$ above is a differential equation that gives an instantaneous “velocity” $X$ relative to a co-ordinate frame that is a copy of the one fixed to the identity but continuously left translated so that it moves with the group element $\sigma(\tau)$. Intuitively, we can imagine sitting in a little boat in the group and $\sigma \left(\tau \right)$ describing our position. We are, in effect, switching, at every instant, to a new chart on the group that is centred on the boat and setting a steady heading $X$ in that chart. The Lie algebra then describes all valid helm or steering wheel settings of the boat, and as a wayfarer, we can only see the World from our boat, so the concept is very intuitive.

The form of the equation given above motivates the following “shorthand” for differential equations defining paths in Lie groups. One often sees written $\d_\tau \sigma = \sigma\,X$, an equation that does not have literal meaning in our general Lie group setting. The product between Lie group and algebra members is not generally defined and of course the equation really means $\d_\tau \lambda(\sigma)= \mathbf{M}(\id,\,\sigma(\tau))\,X$, where the tangent map $\mathbf{M}$ induced by the translation effects a one-to-one, onto correspondence between the Lie algebra and the tangent space at $\sigma \neq \id$. It is the tangent one gets by multiplying $\sigma$ by a path $\sigma_X(\tau)$ with tangent $X$ at the identity where $\tau=0$ and then working out the tangent $\left.\d_\tau (\sigma\,\sigma_X(\tau))\right\|_{\tau=0}$. However, in a matrix Lie group the equation is literally meaningful: both Lie group and algebra members are $N\times N$ matrices and $\mathbf{M}(\id,\,\sigma(\tau)) = \sigma$. I shall henceforth write $C^1$ path-defining differential equations as $\d_\tau \sigma = \sigma\, X$ unless the literal meaning needs to be emphasised. Naturally, there are wholly analogous “right-translated” versions of all these notations, such as $\d_\tau \sigma = X\,\sigma$, which literally means $\d_\tau \lambda(\sigma)= \tilde{\mathbf{M}}(\id,\,\sigma(\tau))\,X$ where $\tilde{\mathbf{M}}$ is the tangent map induced by right translation, or, equivalently, $\left.\d_\tau\left(\sigma(s+\tau)\,\sigma(s)^{-1}\right)\right|_{\tau = 0} = X$.

The equivalence of the two forms of our basic differential equation above follows from the nonsingularity $\det\mathbf{M}(\id,\,\sigma)\neq0\,\forall\,\sigma\in\Nid$. We can either use Theorem 3.19 or we can argue from first principles that $\det\mathbf{M}(\id,\,\sigma)\neq0$ within some nucleus $\K\subseteq\Nid$ (since $\mathbf{M}(\id,\,\sigma)=\id_N$ and $\mathbf{M}(\id,\,\sigma)$ is at least a $C^0$ function of the co-ordinates of $\sigma$) and restrict ourselves to this nucleus. For $\d_s\lambda(\sigma(s)) = \left.\d_\tau \lambda(\sigma(s + \tau))\right|_{\tau=0}=\left.\d_\tau \lambda(\sigma(s)\,\sigma(s)^{-1}\,\sigma(s + \tau))\right|_{\tau=0}$, which, by definition of the tangent map, is $\mathbf{M}(\id,\,\sigma(s))\,\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0}$, because $\sigma(s)^{-1}\,\sigma(s+\tau)$, through the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4, defines a $C^1$ path through the identity, passing through there at $\tau=0$. Since $\mathbf{M}(\id,\,\sigma)$ is nonsingular, the two forms of the equation above imply one another. We are now ready to prove our key theorem, without leaning on Picard-Lindelöf:

Theorem 5.5 (Existence and Uniqueness of Exponential Function Defined by Differential Equation):

The $C^1$ path $\sigma:[-\tau_{max},\,\tau_{max}]\to\Nid$ defined by the Cauchy Initial Value Problem:

\label{ExponentialFunctionExistenceAndUniquenessTheorem_1}\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X;\text{ together with }\sigma(0)=\sigma_0

exists and is unique so long as $\tau_{max}>0$ is small enough to keep $\sigma(\tau)\in\Nid,\,\forall\,\tau\in[-\tau_{max},\,\tau_{max}]$..

Proof: Show Proof

The equation $\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X$ is equivalent to $\d_\tau \lambda(\sigma)= \mathbf{M}(\id,\,\sigma(\tau))\,X$ by our discussion above. Hence, given $\mathbf{M}(\id,\,\sigma(\tau))$ is at least a $C^0$ function of the co-ordinates $\lambda(\sigma)$, by the Peano Existence Theorem the Cauchy initial value problem above has some solution $\sigma:[-\tau_{max},\,\tau_{max}]\to\Nid$ for a small enough $\tau_{max}>0$. However, the Peano Existence Theorem does not show that this path is unique. So, we suppose there are two such paths $\sigma_1,\,\sigma_2$ and show they must be the same.

To this end, let $\sigma_1:[-\tau_{max},\,\tau_{max}]\to\Nid$, $\sigma_2:[-\tau_{max},\,\tau_{max}]\to\Nid$ be any two solutions of $\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X$. Take heed that the following calculation works whether or not the two solutions are actually different. We form the function $\sigma_{2,-1}:[0,\,\tau_{max}]\to\Nid;\,\sigma_{2,-1}(\tau) = \sigma_2(\tau)\,\sigma_1(\tau)^{-1}$, taking heed that it defines a $C^1$ path by the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4. Then $\sigma_{2,-1}(s)^{-1}\,\sigma_{2,-1}(s+\tau)$ is a $C^1$ path through the identity, passing through there at $\tau = 0$ and:

\label{ExponentialFunctionExistenceAndUniquenessTheorem_2}
\begin{array}{rl}
&\left.\d_\tau\left(\sigma_{2,-1}(s)^{-1}\,\sigma_{2,-1}(s+\tau)\right)\right|_{\tau = 0}\\
=& \left.\d_\tau\left(\sigma_1(s)\,\sigma_2(s)^{-1}\,\sigma_2(s+\tau)\,\sigma_1(s+\tau)^{-1}\right)\right|_{\tau = 0}\\
=&\left.\d_\tau\left(\sigma_1(s)\,\sigma_2(s)^{-1}\,\sigma_2(s)\,\sigma_1(s+\tau)^{-1}\right)\right|_{\tau = 0} +\left.\d_\tau\left(\sigma_1(s)\,\sigma_2(s)^{-1}\,\sigma_2(s+\tau)\,\sigma_1(s)^{-1}\right)\right|_{\tau = 0}\\
=&\left.\d_\tau\left(\sigma_1(s)\,\sigma_1(s+\tau)^{-1}\right)\right|_{\tau = 0} + \left.\d_\tau\left(\sigma_1(s)\,\sigma_2(s)^{-1}\,\sigma_2(s+\tau)\,\sigma_1(s)^{-1}\right)\right|_{\tau = 0}\\
\end{array}

the second term on the last line following because it is equal to the tangent of the $C^1$ path $\sigma_1(s)^{-1}\, \tilde{\sigma}_2(\tau)\,\sigma_1(s)$ at $\tau = 0$ where the path passes through the identity and $\tilde{\sigma}_2(\tau) = \sigma_2(s)^{-1}\,\sigma_2(s+\tau)$ with tangent $X$ at $\tau=0$ because $\sigma_2$ fulfills the stated Caucy initial value problem. So now, for $\sigma_{1,-1}(\tau) = \sigma_1(\tau)\,\sigma_1(\tau)^{-1} = \id$, we do the same calculation as above showing:

\label{ExponentialFunctionExistenceAndUniquenessTheorem_3}

an equation whose right hand side is the same as the right hand side of the last line of Equation \eqref{ExponentialFunctionExistenceAndUniquenessTheorem_2} because both $\sigma_1,\,\sigma_2$ fulfill $\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X$. Therefore:

\label{ExponentialFunctionExistenceAndUniquenessTheorem_4}
\begin{array}{rl}
&\left.\d_\tau\left(\sigma_{2,-1}(s)^{-1}\,\sigma_{2,-1}(s+\tau)\right)\right|_{\tau = 0} = 0\\
\Leftrightarrow&\d_\tau\,\sigma_{2,-1}(\tau) = \mathbf{M}(\id,\,\sigma(\tau))\,\Or = \Or
\end{array}

a differential equation with $C^\omega$ right hand side whose unique solution is:

\label{ExponentialFunctionExistenceAndUniquenessTheorem_5}
\sigma_2(\tau)^{-1}\,\sigma_1(\tau) = \sigma_2(0)^{-1}\,\sigma_1(0)\,\forall\,\tau\in[-\tau_{max},\,\tau_{max}]

for some $\tau_{max}>0$. But, since $\sigma_1,\,\sigma_2$ both fulfill $\sigma_1(0)=\sigma_2(0)=\sigma_0$ we get $\sigma_2(\tau)=\sigma_1(\tau),\,\forall\,\tau\in[-\tau_{max},\,\tau_{max}]$. $\quad\square$

Note that we can equally well use a “time varying” $X(\tau)$ in the above and the above proof will work. Therefore we have the following generalised version of the above that will be useful later on:

Theorem 5.6 (Existence and Uniqueness of $C^1$ Path Defined by Cauchy Initial Value Problem)

Let $X:[-\tau_{max},\,\tau_{max}]\to\R^N$ be a $C^0$ function of $\tau$. Then $C^1$ path $\sigma:[-\tau_{max},\,\tau_{max}]\to\Nid$ defined by the Cauchy Initial Value Problem:

\label{C1PathExistenceAndUniquenessTheorem_1}
\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X(\tau);\text{ together with }\sigma(0)=\sigma_0

exists and is unique so long as $\tau_{max}>0$ is small enough to keep $\sigma(\tau)\in\Nid,\,\forall\,\tau\in[-\tau_{max},\,\tau_{max}]$. $\quad\square$

So now we have proven the existence and uniqueness of solutions to our basic equation $\d_\tau \sigma = \sigma\,X;\,\sigma(0)=\id$ within some nucleus $\K$. There is one such solution for every Lie algebra member $X\in\g$. We meet the first basic property:

Lemma 5.7 (Flow Equation):

Any exponential function $\exp : [-\tau_{max},\,\tau_{max}]\to\K\subseteq \Nid$ defined by $\d_\tau \exp(\tau) = \sigma(\tau)\, X,\,\exp(0) = \id$ fulfills:

\label{FlowEquationLemma_1}
\exp(s)\,\exp(\tau) = \exp(\tau)\,\exp(s)=\exp(s+\tau),\,\forall\,s,\,\tau,s+\tau\in[-\tau_{max},\,\tau_{\max}]

Proof: Behold the path $\sigma_1(\tau) = \exp(s)\,\exp(\tau)$ for $\tau\in[0,\tau_{max}-s]$. It fulfills $\d_\tau \sigma_1(\tau)= \exp(s)\,\exp(\tau)\,X = \sigma_1(\tau)\,X$ and $\sigma_1(0) = \exp(s)$. Likewise, the path $\sigma_2(\tau) = \exp(s+\tau)$ fulfils the same Cauchy initial value problem. By Theorem 5.5, these two solutions must be equal, whence $\exp(s+\tau) = \exp(\tau+s) = \exp(s)\,\exp(\tau) = \exp(\tau)\,\exp(s)$.$\quad\square$

For any exponential function $\exp : [-\tau_{max},\,\tau_{max}]\to\K\subseteq \Nid$ defined by $\d_\tau \exp(\tau) = \sigma(\tau)\, X,\,\exp(0) = \id$ we have:

Proof: Show Proof

We use equation \eqref{ExponentialFunctionExistenceAndUniquenessTheorem_3} (which holds for any path fulfilling $\left.\d_\tau\left(\sigma(s)^{-1}\,\sigma(s+\tau)\right)\right|_{\tau = 0} = X$) in the proof of Theorem 5.5 and simplify with the flow equation so that

\begin{array}{lcl}
&=& \left.\d_\tau\left((\sigma(s)\,\sigma(s+\tau)^{-1})\,(\sigma(s+\tau) \sigma(-(s+\tau)))\right)\right|_{\tau = 0}+ \Ad(\sigma_1(s))\,X\\
&=& \left.\d_\tau\left(\sigma(s)\,(\sigma(s+\tau)^{-1}\,\sigma(s+\tau))\, \sigma(-(s+\tau))\right)\right|_{\tau = 0}+ \Ad(\sigma(s))\,X\\
\end{array}

Putting $s=0$ into the last line above we get $0=\left.\d_\tau\left(\sigma(-\tau)\right)\right|_{\tau = 0}+ X$ so, on putting this back into the last line of Equation \eqref{AdjointImageEigenvectorLemma_2} we get the claimed result $\eqref{AdjointImageEigenvectorLemma_1}$.  $\quad\square$

Now or any such exponential defined in some interval $[-\tau_{max},\,\tau_{max}]$ mapping into some nucleus $\K$ we can broaden the definition from $[0,\,\tau_{max}]$ to $\R$ straighforwardly.

Theorem 5.9 ($C^1$ One Parameter Subgroups):

Any exponential function $\exp : [-\tau_{max},\,\tau_{max}]\to\K\subseteq \Nid$ defined as the unique solution to the differential equation $\d_\tau\,\sigma = \sigma\,X = X\,\sigma$ can be broadened uniquely to an exponential function $\exp:\R\to\G$ that fulfils the flow equation $\exp(s)\,\exp(\tau) = \exp(\tau)\,\exp(s)=\exp(s+\tau),\,\forall\,s,\,\tau,\in\R$ by the following definitions:

\label{C1OneParameterSubgroupsTheorem_1}
\begin{array}{lcl}\exp(-\tau) &=& \exp(\tau)^{-1}\\
\exp(\pm n \tau_{max} + \tau) &=& \exp(\tau_{\max})^n\,\exp(\tau)
\end{array}

and the set $\{\exp(\tau):\,\tau\in\R\}$ is a one parameter subgroup of $\G$ isomorphic either to (i) the group of reals with addition $(\R,\,+)$ or (ii) the circle group $U(1)$ of unit magnitude complex numbers $\{e^{i\,\theta}:\,\theta\in[0,\,2\,\pi)\}$. The operation of multiplication of a member of $\G$ by this broadened defintion exponential function is then a flow. Moreover:

Proof: That the broadened definition fulfills the flow equation follows by simple induction on $n$ for positive $\R$ given the induction base that the flow equation is fulfilled for $|\tau|\leq\tau_{max}$. Likewise the adjoint image eigenvector equation follows from $\Ad(\exp(\tau_1))\, \Ad(\exp(\tau_2)) =\Ad(\exp(\tau_1)\,\exp(\tau_2)) = \Ad(\exp(\tau_1+\tau_2))$. All other results then follow straighforwardly from first principles and the global flow equation. $\quad\square$

So now we have defined exponential functions for any Lie algebra member $X\in\g$ and for all real values of the path parameter $\tau$. These exponential functions define $C^1$ paths (the defining differential equation defines their tangents at all points). The first thing to take good heed of is that, since we can define an $\exp$ in this way for all members of any basis for $\g$, we can label a nucleus $\K$ by exponential second kind canonical co-ordinates:

Theorem 5.10 (Exponential Canonical Co-ordinates of the Second Kind):

For any connected Lie group $\G$, there is a nucleus $\K\subseteq{N}_\id$ whose members $\gamma\in\K$ are uniquely labelled by co-ordinates $\tau_1,\,\tau_2,\,\cdots,\,\tau_N$ as follows:

\label{ExponentialSecondKindCanonicalCoordinatesTheorem_1}
\gamma(\tau_1,\,\tau_2,\,\cdots,\,\tau_N) = \prod\limits_{j=1}^N\,e^{X_j\,\tau_j}

where we write the unique $C^1$ path defined by $\d_\tau\,\sigma = \sigma\,X = X\,\sigma,\,\sigma(0)=\id$ as $\sigma(\tau)=e^{X\,\tau}$.

Proof: This follows straight away on heeding that $e^{X_j\,\tau}$ define $C^1$ paths through the identity with $e^{X_j\,0}=\id$ and with tangents $X_j$ at the identity. Then the $e^{X_j\,\tau_j}$ can work as the $\sigma_{X_j}$ in the proof of Theorem 3.9 and the theorem follows. $\quad\square$

The notation $e^{X\,\tau}$ will be justified shortly.

## Analytic Path Differential Equations

### Images of $C^1$ One Parameter Groups in Adjoint Representation

Recall that the (big-A) adjoint representation $\Ad(\G)\to GL(\g)$ is defined by calling $\Ad(\gamma)$ the linear, invertible (nonsingular) transformation wrought on the tangent space $\g$ as the differential at $\id$ of conjugation by $\gamma$. Otherwise put, the $C^1$ path $\sigma_X(\tau)$ through the identity with $\sigma_X(0)=\id$ with tangent $X\in\g$ there is mapped into another $C^1$ path $\gamma\,\sigma_X(\tau)\,\gamma^{-1}$ through the identity, also with $\gamma\,\sigma_X(0)\,\gamma^{-1}=\id$. The latter path has tangent $\Ad(\gamma)\,X$, some transformed version of $X$. The map $\Ad(\gamma)$ is readily shown to be linear by thinking of its effect on the path $\sigma_X(\alpha,\tau)\,\sigma_Y(\beta\,\tau)$ where $\sigma_Y$ is a $C^1$ path through the identity with $\sigma_Y(0)=\id$ and tangent $Y\in\g$ there, so that $\sigma_X(\alpha,\tau)\,\sigma_Y(\beta\,\tau)$ has tangent $\alpha\,X+\beta\,Y$ at the identity. We then consider the path $\gamma\,\sigma_X(\alpha,\tau)\,\sigma_Y(\beta\,\tau)\,\gamma^{-1} = \gamma\,\sigma_X(\alpha,\tau)\,\gamma^{-1}\,\gamma\,\sigma_Y(\beta\,\tau)\,\gamma^{-1}$ which therefore has tangent $\alpha\,\Ad(\gamma) \,X + \beta\,\Ad(\gamma) \,Y$ at the identity. Moreover $\Ad(\gamma_1\,\gamma_2)$ is clearly the composition $\Ad(\gamma_2)\,\Ad(\gamma_1)$ of the two maps $\Ad(\gamma_1)$ and $\Ad(\gamma_2)$ and, given the image of $\Ad$ is a matrix in $GL(\g)$, this composition is the simple matrix product between the two $N\times N$ matrices. Thus $\Ad(\gamma_1\,\gamma_2) = \Ad(\gamma_1)\,\Ad(\gamma_2)$; in particular $\id_N=\Ad(\id) = \Ad(\gamma^{-1}\,\gamma) = \Ad(\gamma^{-1})\, \Ad(\gamma)$ so that $\Ad(\gamma^{-1})=(\Ad(\gamma))^{-1}$, so that the image of $\Ad$ is always invertible, a fact that is readily checked from first principles from the definition of $\Ad(\gamma)$. So therefore the image $\Ad(\G)\subset GL(\g)$ is a group and $\Ad$ is a homomorphism of our Lie group onto $\Ad(\G)$. It is not yet clear, however, what the kernel $\ker(\Ad)$ is. However, since by the fundamental homomorphism theorem, $\ker(\Ad)$ is a normal subgroup of $\G$, if $\G$ is simple, thus having no normal subgroups, then $\ker(\Ad)$ must be the trivial group $\{\id\}$ and $\Ad(\G)$ is isomorphic to $\G$. For now, one can readily see that $\mathscr{Z}(\G)\subseteq \ker(\Ad)$, where $\mathscr{Z}(\G)$ is the centre of $\G$ comprising all elements that commute with all elements of $\G$. For then we have $\gamma \, \sigma(\tau)\,\gamma^{-1} = \sigma(\tau)$ and all paths are unchanged by conjugation under any $\gamma \in \mathscr{Z}(\G)$. It will turn out in the following that $\ker(\Ad)=\mathscr{Z}(\G)$.

Witness the crucial fact: $\Ad(\G)\subset GL(\g)$ is a matrix group, whether or not $\G$ may be. This simple and readily overlooked fact has huge implications.

So now we consider a one-parameter subgroup $\mathfrak{E}_X=\{\exp(\tau\,X):\,\tau\in\R\}$ and its image $\mathfrak{F}_X=\{\Ad(\exp(\tau\,X)):\,\tau\in\R\}$. Given $\exp(\tau\,X)$ defines a $C^1$ path through $\G$ passing through $\id$, $\Ad(\exp(\tau\,X))$ must define at least a $C^0$ path through the vector space $M(N)\cong \R^{N^2}$ of $N\times N$ matrices: we loose one notch in differentiability class since we must calculate a derivatives to find tangents. However, by the homomorphism property above:

so that, as a function of the parameter $\tau$ for the original path $\mathfrak{E}_X\subset \G$, we have:

\zeta_X(\tau)\,\zeta_X(s) = \zeta_X(s)\,\zeta_X(\tau) = \zeta_X(\tau+s);\;\zeta_X(u)\stackrel{def}{=} \Ad(\exp(u\,X))

So now we have a continuous path of square matrices that fulfills the flow equation. Unless $\mathfrak{F}_X\subset \Ad(\G)$ is trivial, there must be some $\tau_0>0$ such that $\zeta_X(\tau)\neq\id_N$ for $0<\tau\leq\tau_0$. For, if otherwise, we could always find some $\zeta_X(\tau_\epsilon) = \id$ with $\tau_\epsilon < \epsilon$ for any $\epsilon>0$ and $\zeta_X(n\,\tau_\epsilon) = \id_N,\,\forall\,n\in\mathbb{Z}$ by the flow equation. The set $\{\tilde{\tau}:\zeta(\tilde{\tau})=\id_N\}$ would then be dense in the whole one parameter group, and, since $\mathfrak{F}_X$ is a $C^0$ path (with respect to, say, the spectral or Frobenius norm) in the linear space of $N\times\,N$ matrices, we would then have $\zeta_X(\tau)=\id_N,\forall\,\tau\in\R$.

So now, by dent of the flow equation, square roots are defined and are unique in the set $\{\zeta_X(\tau):\,|\tau|<\tau_0 / 4\}$ because if $\zeta_X(\tau_1)$ and $\zeta_X(\tau_2)$ were two square roots of $\zeta_X(\tau_3)$ with $|\tau_1|,\,|\tau_2|,\,|\tau_3| < \tau_0 / 2$, then $\zeta_X(2 (\tau_1-\tau_2))=\id$ by the flow equation, which gainsays the definition o$\zeta_X(\tau)\neq\id_N$ for $0<\tau\leq\tau_0$. Square roots thus being unique, and given that $\mathfrak{F}_X$ is Abelian, we can show $\zeta_X(\tau) = \zeta(1)^\tau\,\forall\,\tau\in\mathbb{Q}_2\cap[0,\,1]$, where $\mathbb{Q}_2$ is the set of rationals with finite binary (i.e. radix-2 positional-numeral) expansions, a set which, like $\mathbb{Q}\cap[0,\,1]$, is a dense subset of the interval $[0,\,1]$. The matrix exponential:

\label{MatrixExponentialTaylorSeries}
\exp(\tau\,A_X) = \id_N + \tau\,A_X + \frac{\tau^2\,A_X^2}{2!}+ \frac{\tau^3\,A_X^3}{3!}+\cdots

defines, through the relationship $\exp(\tau\,A_X) = \exp(A_X)^\tau,\,\forall\,\tau\in\mathbb{Q}$, a continuous (indeed $C^\omega$) function of $\tau$ which is also equal to the power $\exp(A_X)^\tau$ for $\tau$ belonging to the dense subset $\mathbb{Q}_2$ of $\R$ comprising all rationals with finite binary (i.e. radix-2 positional-numeral) expansions. Therefore, the only $C^0$ matrix function of $\tau$ which fulfils the flow equation must be of the form $\exp(\tau\,A_X)$ where here $\exp$ is the matrix exponential defined by the universally convergent Taylor series. Therefore, we must have:

\zeta_X(\tau) = \exp(\tau\,A_X)

where $A_X$ is some matrix that depends on the tangent $X\in\g$ to the original one parameter group $\mathfrak{E}_X\subset \G$. $A_X$ is some function of $X$ whose nature we are yet to fully understand. For now, we shall write:

to show that $A_X$ is wholly defined by $X$ in an as yet to be fully understood way.

So $Ad(\exp(\tau\,X)$ traces a $C^\omega$ path $\exp(\tau\,\ad(X))$ through the vector space $M(N)\cong \R^{N^2}$ of $N\times N$ matrices. We summarise all these thoughts into the theorem:

Theorem 5.11 (Exponentiation of Small “a” Adjoint)

The image of $\exp(\tau\,X)$ for $\tau\in\R,\,X\in\g$ under $\Ad(\G)\to GL(\g)$ is a matrix exponential function of $\tau$:

where the exponent matrix $\ad(X)$ is wholly defined by the original tangent $X\in\g$.

$\square$

The “Theorem” part of the above is the fact that $\Ad(\exp(\tau\,X))$ is a matrix exponential function of $\tau$. As for $\ad(X)$, we know that the exponent in this matrix exponential function is wholly determined by $X$, so that Equation $\eqref{ExponentiationOfAdjointTheorem_1}$ in the above can be taken as the definition of the map $\ad(X)$, an approach I shall take up in the next chapter.

If we have a one-parameter group $\mathfrak{E}_X=\{\exp(\tau\,X):\,\tau\in\R\}$ we can build another through conjugation by any element $\gamma\in\G$. Witness that the $C^1$ path $\theta_{\gamma,\,X}(\tau) = \gamma\,\exp(\tau\,X)\,\gamma^{-1}$ passes through the identity at $\tau=0$ and has tangent $\Ad(\gamma)\,X$ there, by definition of $\Ad$. Furthermore:

\label{ConjugatedPathEquation}
\begin{array}{clcl}
&\theta_{\gamma,\,X}(\tau)\,\theta_{\gamma,\,X}(s) &=& \gamma\,\exp(\tau\,X)\,\gamma^{-1}\,\gamma\,\exp(s\,X)\,\gamma^{-1}\\
& &=& \gamma\,\exp(\tau\,X)\,\exp(s\,X)\,\gamma^{-1} \\
&&=& \gamma\,\exp((\tau+s)\,X)\,\gamma^{-1}\\\\
\Rightarrow& \theta_{\gamma,\,X}(\tau)\,\theta_{\gamma,\,X}(s)&=&\theta_{\gamma,\,X}(s)\,\theta_{\gamma,\,X}(\tau) =\theta_{\gamma,\,X}(\tau+s)
\end{array}

and so $\theta_{\gamma,\,X}$ fulfills the flow equation. Moreover $\theta_{\gamma,\,X}$ is a $C^1$ path:

\label{ConjugatedPathEquation_2}
\begin{array}{lcl}
\d_\tau\theta_{\gamma,\,X}(\tau) &=& \left.\d_u\theta_{\gamma,\,X}(\tau+u)\right|_{u=0}\\
&=&\left.\d_u(\theta_{\gamma,\,X}(\tau) \theta_{\gamma,\,X}(u))\right|_{u=0}\\
&=& \left.\mathbf{M}(1,\,\theta_{\gamma,\,X}(\tau))\,\d_u(\theta_{\gamma,\,X}(u))\right|_{u=0}\\
\end{array}

and so $\theta_{\gamma,\,X}$ is the unique solution to our basic Cauchy initial value problem $\d_\theta \sigma = \sigma\,Y;\,\sigma(0)=\id$ with $Y=\Ad(\gamma)\,X$. Hence we have the following lemma:

Lemma 5.12 (Conjugation of One Parameter Group)

Under conjugation by $\gamma\in\G$, the one parameter group $\mathfrak{E}_X=\{\exp(\tau\,X):\,\tau\in\R\}$ is transformed to $\gamma\,\mathfrak{E}_X\,\gamma^{-1}$, which is indeed another one-parameter group $\mathfrak{E}_{\Ad(\gamma) X}$:

\label{ConjugationOfOneParameterGroupLemma_1}

and $\gamma\,\exp(\tau\,X)\,\gamma^{-1} = \exp(\tau\,\Ad(\gamma)\,X)$.

$\square$

### Wei Norman Equations Imply $C^\omega$ Path Differential Equations

We shall now show that the tangent map induced by left (or right) translation has an explicit, and locally analytic (i.e. $C^\omega$) form when working in exponential canonical co-ordinates of the second kind, even though, with our axioms so far, the best general statement we can make about $\mathbf{M}(\id,\,\sigma)$ is that it must be at least a $C^0$ function of the co-ordinates of $\sigma$: it is essentially a Jacobi matrix, and the derivatives one calculates to find this matrix lower the differentiability class of the differentiated entity by one step.

We work a “trick” Wei and Norman brought to bear on exponential second kind canonical co-ordinates. Witness that:

\label{WeiNormanTrick}
\begin{array}{cl}
&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{(\tau_j+s)\,X_j}\,\cdots\,e^{\tau_N\,X_N}\\
=&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet\,\\
=&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet \\
=&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet \\
=&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet \\
=&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet\\
=&e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet\\
\end{array}

where naturally we recall Lemma 5.8 to reach the last line. The above is worth writing down explicity as a lemma:

Lemma 5.13 (Wei-Norman Shuffle)

\label{WeiNormanShuffleLemma_1}
\begin{array}{l}
e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{(\tau_j+s)\,X_j}\,\cdots\,e^{\tau_N\,X_N}=e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\,\bullet\\

$\square$

Therefore, we have:

\label{WeiNormanTrick_2}
\begin{array}{cl}
&\partial_{\tau_j} \lambda\left(e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\right) \\
=&\left.\partial_s \lambda\left(e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{(\tau_j+s)\,X_j}\,\cdots\,e^{\tau_N\,X_N}\right)\right|_{s=0} \\
\end{array}

Now, if the $X_j$ are a basis $\{\hat{X}_j\}_{j=1}^N$ for the Lie algebra $\g$ and we label the appropriate a nucleus $\K\subseteq{N}_\id$ with exponential canonical co-ordinates of the second kind, then $\partial_{\tau_j} \lambda\left(e^{\tau_1\,X_1}\,e^{\tau_2\,X_2}\,\cdots\,e^{\tau_N\,X_N}\right)$ is simply $\hat{X}_j$, which is a column vector with noughts in all placeholders aside from the $j^{th}$, where there is a $1$. $\Ad\left(e^{-\tau_N\,\hat{X}_N}\right)\,\Ad\left(e^{-\tau_{N-1}\,\hat{X}_{N-1}}\right)\,\cdots\,\Ad\left(e^{-\tau_{j+1}\,\hat{X}_{j+1}}\right)\,\hat{X}_j$ is simply the $j^{th}$ column of the square $N\times N$ matrix $\Ad\left(e^{-\tau_N\,\hat{X}_N}\right)\,\cdots\,\Ad\left(e^{-\tau_{j+1}\,\hat{X}_{j+1}}\right)$. So we write down an equation of the above form for each $j\in1,\,\cdots,\,N$ and load the resulting columns into a system of simultaneous equations to find:

\label{WeiNormanTrick_3}
\begin{array}{l}
\id_M = \mathbf{M}\left(\id,\,e^{\tau_1\,X_1}\,\cdots\,e^{\tau_N\,X_N}\right)\,\bullet\\

i.e. the $j^{th}$ column on the matrix on the far right is the $j^{th}$ column of the square matrix $e^{-\tau_N\,\ad(\hat{X}_N)}\,\cdots\,e^{-\tau_{j+1}\,\ad(\hat{X}_{j+1})}$. This matrix equals the $N\times N$ identity when all the $\tau_j$ are nought and is a $C^0$ function of all the $\tau_j$, so that there is an appropriate sized nucleus $\K\subseteq{N}_\id$ wherein this matrix is nonsingular. Moreover, containing columns of products of exponential matrices, this matrix is locally an analytic function of all the $\tau_j$ inside this neighbourhood. Therefore, we can invert the above equation and we thus see that our tangent map induced by left translation for exponential canonical co-ordinates of the second kind is alocally analytic ($C^\omega$) function.

What is striking about these co-ordinates and results is:

1. The analyticity of $\mathbf{M}\left(\id,\,e^{\tau_1\,X_1}\,\cdots\,e^{\tau_N\,X_N}\right)$, even though our basic axioms suggest that in general we can only assume $C^0$ behaviour for $\mathbf{M}$;
2. That the form of $\mathbf{M}$ is built wholly of analytic algebraic combinations of analytic matrix exponential function whether or not the original Lie group is a matrix group.

We draw all these results together in a theorem:

Theorem 5.14 (Analytic Path Differential Equations in Canonical Co-ordinates of the Second Kind)

There exists a nucleus $\K\subseteq{N}_\id$ wherein the tangent map induced by left translation is analytic over the whole of $\K$ and is equal to:

\label{AnalyticPathDifferentialEquationTheorem_1}
\begin{array}{l}
\mathbf{M}\left(\id,\,e^{\tau_1\,X_1}\,\cdots\,e^{\tau_N\,X_N}\right) = \\

the tangent map induced by right translation is analytic over the whole of $\K$ and is equal to:

\label{AnalyticPathDifferentialEquationTheorem_2}
\begin{array}{l}
\tilde{\mathbf{M}}\left(\id,\,e^{\tau_1\,X_1}\,\cdots\,e^{\tau_N\,X_N}\right) = \\
\end{array}

and every $C^1$ path through $\K$ is uniquely defined by Cauchy initial value problem:

\label{AnalyticPathDifferentialEquationTheorem_3}
\d_\tau \sigma = \mathbf{M}(\id,\,\sigma)\,X(\sigma,\,\tau)\;\Leftrightarrow\;,\;\sigma(0)=\sigma_0\in\K

when $X(\sigma,\,\tau)$ is a Lipschitz continuous ($C^{0,\,1}$) function of $\tau$ and $\sigma$.

Proof: Equations $\eqref{AnalyticPathDifferentialEquationTheorem_1}$ and $\eqref{AnalyticPathDifferentialEquationTheorem_2}$ were derived in the foregoing discussion. Now, given the analytic dependence of $\mathbf{M}$ on all the co-ordinates of $\sigma$, the right hand side is a Lipschitz continuous function of these co-ordinates, and so we can now freely bring to bear the Picard-Lindelöf theorem to show that the path defined by a differential equation of the form in Equation $\eqref{AnalyticPathDifferentialEquationTheorem_3}$ exists and is unique. We have inferred the analyticity of the $\mathbf{M}$ from our basic axioms without assuming Picard-Lindelöf, so that the latter’s deployment is justified here. $\quad\square$

We proved (Theorem 3.9) the existence of canonical co-ordinates of the second kind from the basic Lie group axioms. More specifically, we proved that there is a neighbourhood of $\Or$ in $\V$ that is uniquely labelled by such co-ordinates. However, we have never explicitly proven that our group operations stay $C^1$ if we then use the canonical co-ordinates instead of the ones postulated by our axioms. We do this now for exponential canonical second kind co-ordinates; not surprisingly: when we measure the differentiability class of our paths in exponential canonical co-ordinates of the second kind and we combine, through the group operations, paths that are $C^1$ when measured in these co-ordinates, the combined paths are also $C^1$ paths when measured in these co-ordinates.

Theorem 5.15 (Group Operations are $C^1$ in Exponential Canonical Co-ordinates):

Within a nucleus $\K\subseteq\Nid$ small enough that it can be uniquely labelled by exponential canonical co-ordinates of the second kind, when the paths $\sigma_1,\,\sigma_2$ which are $C^1$ when measured by these co-ordinates are combined through the group operations into the combinations $\sigma_1^{-1}\,\sigma_2$ and $\sigma_1\,\sigma_2^{-1}$, then these combinations are also $C^1$ when measured by these co-ordinates whenever the combination path lies inside $\K$.

Proof: Show Proof

As justified by Theorem 5.10, we choose a $\g$-basis $\{\hat{X}_j\}_{j=1}^N$ and kit the appropriately small nucleus $\K\subseteq{N}_\id$ with canonical co-ordinates. Now suppose we have two paths that are $C^1$ with respect to these co-ordinates, i.e. they can be written $\sigma_\alpha(\tau) = \prod\limits_{j=1}^N e^{\alpha_j(\tau)\,\hat{X}_j}$ and $\sigma_\beta(\tau) = \prod\limits_{j=1}^N e^{\beta_j(\tau)\,\hat{X}_j}$ where $\alpha_j,\,\beta_j$ are all $C^1$ functions of “time” $\tau$. We must prove that the second kind canonical co-ordinates of the product $\sigma_\alpha\, \sigma_\beta^{-1}$ are also $C^1$ functions of time i.e. that we can find unique $\omega_j(\tau)$ where:

\label{C1GroupOperationsInCanconicalCoordinatesTheorem_1}
\sigma(\tau)=\prod\limits_{j=1}^N e^{\omega_j(\tau)\,\hat{X}_j} = \prod\limits_{j=1}^N e^{\alpha_j(\tau)\,\hat{X}_j} \prod\limits_{j=1}^N e^{-\beta_{N-j+1}(\tau)\,\hat{X}_{N-j+1}}

and that the $\omega_j$ are also $C^1$ functions of time. Existence and uniqueness of the $\omega_j(\tau)$ for every $tau$ for which $\sigma(\tau)=\sigma_\alpha(\tau)\,\sigma_\beta(\tau)^{-1}\in\K$ is guaranteed by Theorem 5.10. We now calculate $\d_\tau\sigma(\tau)$ of the right hand side of Equation $\eqref{C1GroupOperationsInCanconicalCoordinatesTheorem_1}$, bringing an iterated Wei-Norman shuffl to bear on it to work Equation$\eqref{C1GroupOperationsInCanconicalCoordinatesTheorem_1}$ into the form:

\label{C1GroupOperationsInCanconicalCoordinatesTheorem_2}
\left(\begin{array}{c}\d_\tau\omega_1\\\d_\tau\omega_2\\\vdots\\\d_\tau\omega_N\end{array}\right)=\mathbf{M}(\id,\,\sigma(\omega_1,\,\omega_2,\,\cdots,\,\omega_N)) \mathbf{P}(\alpha_1,\,\alpha_2,\,\cdots,\,\alpha_N,\,\beta_1,\,\beta_2,\,\cdots,\,\beta_N)\,\left(\begin{array}{c}\d_\tau\alpha_1\\\d_\tau\alpha_2\\\vdots\\\d_\tau\alpha_N\\\d_\tau\beta_1\\\d_\tau\beta_2\\\vdots\\\d_\tau\beta_N\end{array}\right)

where $\mathbf{P}$ is an $N\times2N$ matrix and is found, by following the procedure of Lemma 5.13 to comprise columns made up of products of matrices of the form $\Ad(e^{-\alpha_j\,\hat{X}_j})^{\pm1}$ and $\Ad(e^{-\beta_j\,\hat{X}_j})^{\pm1}$ which we have seen are matrix exponential functions of $\alpha_j$ and $\beta_j$, so that $\mathbf{P}$ is analytic in the co-ordinates $\alpha_j$ and $\beta_j$. For a small enough kernel $\mathrm{K}$ we have likewise seen, from Theorem 5.14 that $\mathbf{M}(\id,\,\sigma(\omega_1,\,\omega_2,\,\cdots,\,\omega_N))$ is analytic in all the $\omega_j$. Therefore, given an initial value $\omega_j(0)$ of the $\omega_j$, the solution of exists and is unique by the Picard-Lindelöf theorem and the solution is manifestly $C^1$. This proves that the Group Product Continuity Axiom 3 holds for the exponential second kind canonical co-ordinates. The Nontrivial Continuity Axiom 4 is readily shown to hold, as long as we take our kernel $\K\subseteq{N}_\id$ to be an open set, i.e. it is the set of all points of the form $\prod\limits_{j=1}^N e^{\tau_j\,\hat{X}_j}$ where the $\tau_j$ roam over open intervals, i.e. our co-ordinate space $\V$ is the topological product of open intervals. This means that if $\sigma_\alpha(\tau)\,\sigma_\beta(\tau)^{-1}$ lies inside $\K$, we have just shown that the path is $C^1$ there, therefore continuous there and so there a nonzero length open interval $\mathcal{I}$ with $\tau\in\mathcal{I}$ such that $\sigma_\alpha(\mathcal{I})\,\sigma_\beta(\mathcal{I})^{-1}\subset\K$.$\qquad\square$

With the theory of the exponential map that we have developed, we now use the exponential function to define what is maybe the most important co-ordinate system of all for a connected Lie group: the geodesic co-ordinate system.

1. Giuseppe Peano, Demonstration de l’intégrabilité des équations différentielles ordinaires, Mathematische Annalen, 37 (1890) 182–228.
2. J. Wei and E. Norman, Representations of the Solutions of Linear Differential Equations as a Product of Exponentials, Proc. Amer. Math. Soc. Vol. 15, No. 2 (Apr., 1964), pp. 327-334. Theorem 1, “The Local Theorem”, §3, p329