# Chapter 3: The Lie Algebra of the Lie Group and Canonical Co-ordinates

We now turn to the Lie algebra, which we shall define in the same way as [Rossmann] does in the first chapter.

Definition 3.1 (Lie Algebra of a Connected Lie Group):

For a connected Lie group $(\G,\,\bullet)$ as defined by Axioms 1 through 5, the Lie algebra $\g$ is the set of all tangent vectors at the identity $\id$ to $C^1$ paths through the identity.

In this definiton, we measure derivatives with an $N$-element real vector in $\V$, so strictly speaking the calculated derivative is $\d_\tau \,\lambda \left(\sigma(\tau)\right)=\d_\tau \,\lambda \circ \sigma(\tau)$, but the shorthand $\d_\tau\,\sigma(\tau)$ should be clear.

It will be shown later that for many important Lie groups (notably compact semisimple ones), the Lie group structure is unique up to automorphism, so that any Lie algebra found by applying the above definition is isomorphic to any other such algebra. It therefore makes sense to speak of “the” Lie algebra of a Lie group when so, unqualified by the defining structure $\Nid$, $\V$, $\lambda$. We shall, however, always specify the $\Nid$, $\V$, $\lambda$ which define the group as there are some cases where one can tear down the Lie group’s topology and replace it by another, thus creating an altogether different Lie group. We note that we have yet to defined the group’s topology, but take heed that it is uniquely defined by $\Nid$, $\V$, $\lambda$.

We have the first important Lemma:

Lemma 3.2 (Linearity of the Lie Algebra):

The Lie algebra $\g$ of the Lie group $\G$ is a vector space over the reals $\R$.

Proof: Show Proof

If $X$ and $Y$ are Lie algebra members then there are $C^1$ paths $\sigma_X(s),\,\sigma_Y(t)$ such that $\sigma_X (0)=\sigma_Y( 0)=\id$ and $X=\left.\d_t\sigma_X(t)\right|_{t=0},\,Y=\left.\d_s\sigma_X(s)\right|_{s=0}$. By the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4, $\sigma_X(u)\,\sigma_Y(u)$ is also a $C^1$ path in $\Nid$ for $u\in [-u_0,\,u_0]$ for some small enough $u_0 > 0$. Then:

\begin{equation}
\label{LieAlgebraLinearityLemma_1}
\begin{array}{l}
\left.\partial_s \lambda\left(\sigma_X(s)\,\sigma_Y(t)\right)\right|_{s=0,\,t=0} = X\\\\
\left.\partial_t\lambda\left(\sigma_X(s)\,\sigma_Y(t)\right)\right|_{s=0,\,t=0} = Y
\end{array}
\end{equation}

and now, for $\alpha,\,\beta \in \R$, let $s = \alpha\,u$ and $t = \beta\,u$ so that the tangent to the $C^1$ path $\sigma_X(\alpha\,u)\,\sigma_Y(\beta\,u)$ at the identity $\id$ (where $u=0$) is:

\begin{equation}
\label{LieAlgebraLinearityLemma_2}
\begin{array}{lcl}
\left.\d_u \lambda\left(\sigma_X(u) \sigma_Y(u)\right)\right|_{u=0} &=& \left. \left(\partial_s \lambda\left(\sigma_X(s) \sigma_Y(t)\right) \frac{\d s}{\d u} + \partial_t \lambda\left(\sigma_X(s) \sigma_Y(t)\right) \frac{\d t}{\d u}\right)\right|_{u=0}\\
&=& \alpha X + \beta Y
\end{array}
\end{equation}

so if $X,\,Y\in \g$, then $\alpha X+\beta Y\in \g,\,\forall\alpha,\,\beta\in\R$, thus proving that $\g$ is a vector space over the field of reals.$\quad\square$

Although the Lie algebra $\g$ is a vector space over the field of reals, it isn’t simply any old real vector space. The Lie algebra has further structure begotten by the Lie bracket, which encodes information about the group’s non-Abelianhood and which will be addressed in a later post. So, in particular, the above lemma does not mean that any vector subspace of the Lie algebra is the Lie algebra of a Lie subgroup of $\G$, unless we are talking about an Abelian Lie group. When talking about such groups, there is a Lie subgroup for every vector subspace of $\g$ with this particular subspace as its Lie algebra.

Example 3.3: (One Dimensional and other Abelian Lie Group Examples)

All the one dimensional Lie groups cited in Example 1.1 have the one dimensional real vector space $\R$ as their Lie algebra.

All the Lie groups of Example 1.2 have the vector space $\R^N$ as their Lie algebra. For every linear subspace within an Abelian Lie group’s Lie algebra, there corresponds an Abelian Lie subgroup under the Lie correspondence of Chapter 9. As noted above, this is certainly not true for general Lie groups; not every vector subspace of a Lie group’s algebra corresponds to a Lie subgroup.

Example 3.4: (The Slide Rule)

The slide rule is a 20th century and earlier calculating device, invented by William Oughtred (1574-1660) whose working principle is grounded on the isomorphism $\log: \R^+\setminus\{0\}\to\R$ / $\exp: \R\to \R^+\setminus\{0\}$ between the Lie group $(\R^+\setminus\{0\},\,*)$ of reals with multiplication and its Lie algbera (also a Lie group in its own right) $(\R,\,+)$ of reals with addition, which isomorphism could be said to have been discovered by John Napier (1550-1617). Indeed, it could be said that the slide rule is a physical embodiment of this isomorphism. As we know, both $(\R^+\setminus\{0\},\,*)$ and $(\R,\,+)$ are Lie groups, and both have $(\R,\,+)$ as their Lie algebras. So the slide rule could be said to be a physical embodiment of the invertible mapping between the neighbourhood of the identity $\Nid$ in the Lie group and $\V$ in the Lie algebra: in this case $\Nid$ comprises the whole of the Lie group. One is to imaging an infinitely long slide rule, with each of the lengths of any of the A/B/C/D scales shown below in Figure 3.1 spanning a decade of $\R^+\setminus\{0\}$. The Lie algebra is represented on the linear L scale (“L” for Lie algebra!; of course I’m kidding; it of course stands for “logarithm” but to say that it is the beginning letter in Lie’s name makes a good story!) and as the linearly additive displacements between the sliding logarithmic scales. You can play with the fullsized version of the virtual Hemmi 153 slide rule here.

Figure 3.1: The Slide Rule: A Physical Embodiment of the Isomorphism between $(\R^+\setminus\{0\},\,*)$ and its Lie Algebra $(\R,\,+)$

Example 3.6: (Rotation Group $SO(3)$ and its Lie Algebra $\mathfrak{so}(3)$)

We hearken back to Example 1.3, recalling that every group member can be written as $\exp\left(x\,\hat{S}_x + y\,\hat{S}_y + z\,\hat{S}_z\right)$, where the $\hat{S}_j$ are as defined in Example 1.3. So a $C^1$ path through $\Nid$ is of the form $\sigma:[-\epsilon,\,\epsilon]\to\Nid;\, \sigma(\tau) = \exp\left(x(\tau)\,\hat{S}_x + y(\tau)\,\hat{S}_y + z(\tau)\,\hat{S}_z\right)$ where $x(\tau), \,y(\tau),\, z(\tau)$ are $C^1$ functions of $\tau$ with $x(0)=y(0)=z(0)=0$. Therefore the tangent to the identity, i.e. the most general Lie algebra member, is the matrix $\dot{x}(0)\,\hat{S}_x + \dot{y}(0)\,\hat{S}_y + \dot{z}(0)\,\hat{S}_z$. Of course, as we have so far defined it, the “Lie algebra” is simply the vector space $\R^3$: we compute the rate of change of $\lambda:\Nid\to\mathcal{V}$, so the basis vectors for the space of tangents are simply the wonted unit vectors spanning $\R^3$. However, with $N\times N$ matrix Lie groups it is wonted to think of the Lie algebra not simply as the “bare” vector space, but as a vector subspace of the $2\,N^2$ dimensional real linear (vector) space $\mathcal{M}(N)$ of $N\times N$ complex matrices kitted with the metric induced by the e.g. Frobenius norm $\left\|X\right\| = \mathrm{tr}(X^\dagger\,X)$. In this way of thinking, the Lie algebra $\mathfrak{so}(3)$ vector space is still isomorphic to $\R^3$, but now its basis vectors are the matrices $\hat{S}_j$ and we can characterise the Lie algebra as the “linear space of $3\times3$, skew-Hermitian, real-element matrices”; this is a three dimensional space and one commonly cited basis for it is:

\begin{equation}\label{so3LieAlgebraExample_1}\hat{S}_x = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 0 & -1 \\0 & 1 & 0 \end{array}\right);\quad \hat{S}_y = \left(\begin{array}{ccc}0 & 0 & 1 \\0 & 0 & 0 \\-1 & 0 & 0\end{array}\right);\quad \hat{S}_z = \left(\begin{array}{ccc}0 & -1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)\end{equation}

One advantage for thinking of the Lie algebra in this way for matrix groups will become clear when we talk about the adjoint representations of a Lie group and its algebra: we define an operation on the space of tangents to the identity defined by the group element $\gamma$ by thinking of the operation $\sigma(\tau)\mapsto\gamma\,\sigma(\tau)\,\gamma^{-1}$ on a $C^1$ path $\sigma(\tau)$ through the identity. The image is also a path through the identity, and if $\sigma(0)=\id$, the tangent to the path $\gamma\,\sigma(\tau)\,\gamma^{-1}$ can be computed as the literal matrix product $\gamma\,\left.\d_\tau\sigma(\tau)\right|_{\tau=0}\,\gamma^{-1}$ in a matrix Lie group if we think of the form the Lie algebra members have as $N\times N$ matrices.

A different and effective way of finding the Lie algebra is to begin with the definition of $SO(3)$ as the set of $3\times3$ real, orthogonal matrices with unity determinant. That is, for every $\gamma\in SO(3)$ we have $\gamma^T\,\gamma = \gamma\,\gamma^T = \id$ and $\det\,\gamma = 0$. We now consider a $C^1$ path $\sigma:[-1,\,1]\to\Nid$ with $\sigma(0)=\id$. We know, as a matrix equation, $\sigma(\tau)^T\,\sigma(\tau)=\id$, so, when we differentiate at $\tau=0$ we get $\dot{\sigma}^T(0) + \dot{\sigma}(0) =0$, i.e. $\dot{\sigma}^T(0) = – \dot{\sigma}(0)$ every tangent to the identity is a real, skew-Hermitian matrix. The condition that $\det \sigma =1$ does not bear on the space of tangents, because $\sigma(\tau)^T\,\sigma(\tau)=\id\,\Rightarrow\, \det\sigma^2 = 1\,\Rightarrow\,\det\sigma(\tau)=\pm1$ (since $\det \sigma^T=\det\sigma$), and $\det\sigma(\tau)$ is a continuous function of $\tau$, therefore, since $\det\sigma$ can only take on the discrete values $\pm1$, $\det\sigma$ cannot change on a $C^1$ path, so that $\det\sigma(\tau) = 1,\,\forall\sigma(\tau)\in\Nid$. So the condition $\sigma(\tau)^T\,\sigma(\tau)=\id$ together with $\sigma$’s being a $C^1$ path connected to the identity automatically upholds $\det\sigma=1$.

Example 3.7: ($2\times2$ Special Unitary Group $SU(2)$ and its Lie Algebra $\mathfrak{su}(2)$)

This example for the Lie group $SU(2)$ is very like Example 3.6. A general Lie group element is of the form $\exp\left(x\,\hat{s}_x + y\,\hat{s}_y + z\,\hat{s}_z\right)$, where the $\hat{s}_j$ are as defined in Example 1.4. So the Lie algebra is the linear space of all matrices $\{\left.x\,\hat{s}_x + y\,\hat{s}_y + z\,\hat{s}_z\right|\;x,\,y,\,z\in\R \}$ isomorphic to $\R^3$.

Using the second technique in Example 3.6 to characterise the Lie algebra, we find from the relationship $\gamma^\dagger\,\gamma=\id$ implies $\dot{\gamma}^\dagger = -\dot{\gamma}$, so that all tangents to $C^1$ paths through the identity, at the identity, are $2\times2$ skew-Hermitian matrices. We must also look at the condition $\det\gamma=1$. Unlike Example 3.6, the relationship $\gamma^\dagger\,\gamma=\id$ implies that $\det\gamma = e^{i\,\phi}$ for $\phi\in\R$, the set of determinants allowed by $\gamma^\dagger\,\gamma=\id$ is no longer discrete but a full circle, so here $\dot{\gamma}^\dagger = -\dot{\gamma}$ does not fully characterise the Lie algebra. To be sure, the set of all complex $2\times2$ skew-Hermitian matrices is a vector space over the reals of dimension 4, so the Lie algebra $\mathfrak{su}(2)$ must be contained within this space. A unitary matrix, being a finite dimensional normal (i.e. commuting with its adjoint operator) operator always has a spectral diagonalisation by a unitary matrix. The determinant of $\gamma$ is thus $\det\gamma = \prod\limits_{k=1}^M \lambda_k$ where $\lambda_k$ are the eigenvalues, so that, if $\det\gamma = 1$ then:

\begin{equation}\label{su2Example_1}\left.\d_\tau \gamma(\tau)\right|_{\tau=0} = 0 = \prod\limits_{k=1}^M \lambda_k \,\sum\limits_{k=1}^M \frac{\left.\d_\tau \lambda_k(\tau)\right|_{\tau=0}}{\lambda_k(\tau)}=\sum\limits_{k=1}^M \left.\d_\tau \lambda_k(\tau)\right|_{\tau=0}=\mathrm{tr}(\dot{\gamma}(0))\end{equation}

given that $\lambda_k = 1$ when $\tau=0$ and the path $\gamma(\tau)$ passes through the identity. So we have a further condition for $\mathfrak{su}(2)$: its members must all be traceless. So $\mathfrak{su}(2)$ is the three-dimensional vector space over the reals of skew-Hermitian, traceless, $2\times2$ complex matrices and one commonly used basis is:

Example 3.8: (General Linear Group and its Lie Algebra $\mathfrak{gl}(M,\,\mathbb{C})$)

In Example 1.8 it was shown that the general member of $GL(M,\,C)$ near the identity (i.e. one inside $\Nid$) is of the form $e^X$, where $X\in\mathcal{M}(M)$ is any $M\times M$ complex matrix with norm small enough such that $\left\|e^X-\id\right\|<1$, i.e. such that the matrix logarithm Taylor series converges. Therefore, the space of tangents to $C^1$ paths at the identity is the set of all $\mathcal{M}(M)$ complex matrices.

Example 3.9: (Unimodular Group and its Lie Algebra Of Traceless Matrices)

In Example 1.5 with $\Nid=\{\gamma\in SL(2,\,\mathbb{C})|\;\left\|\gamma -\id\right\|<1\}$ and labeller function $\lambda = \log$ equal to the matrix logarithm, it was shown that$\gamma$ lies on the $C^1$ path $\sigma_\gamma(\tau) = \exp(\tau\,X)$ for some traceless $2\times 2$ complex matrix $X$. All $C^1$ paths through $\Nid$ are therefore of the form $\exp(X\,\tau + \tau\,\mathscr{D}(\tau))$ where $X$ and $\mathscr{D}(\tau)$ are traceless matrices and $\mathscr{D}(\tau)\to0$ as $\tau\to0$. Therefore, the set of all tangents to $C^1$ paths is precisely the set $\mathfrak{sl}(2,\,\mathbb{C})$ of traceless $2\times 2$ complex matrices. One often cited basis for this vector space over the reals is the six matrices:

\begin{equation}\label{SL2CLieAlgebraExample_1}\begin{array}{lclclclclcl}\hat{s}_x&=&\left(\begin{array}{cc}0&i\\i&0\end{array}\right)& &\hat{s}_y&=&\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)& &\hat{s}_z=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)\\\hat{t}_x&=&\left(\begin{array}{cc}0&1\\1&0\end{array}\right)& &\hat{t}_y&=&\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)& &\hat{t}_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\end{array}\end{equation}

For the group $SL(2,\,\R)$ of Example 1.6, the basis is the three dimensional vector space of real, traceless $2\times2$ matrices, and a basis for this linear space is the three matrices:

\begin{equation}\label{SL2CLieAlgebraExample_2}\hat{U}=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right);\;\hat{V}=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right);\;\hat{W}=\left(\begin{array}{cc}0&1\\1&0\end{array}\right);\end{equation}

More generally, the Lie algebras of the unimodular matrix groups $SL(M,\,\mathbb{C})$ and $SL(M,\,\mathbb{R})$ for $M\in\,\mathbb{N}$ are clearly the sets of traceless, $M\times M$ complex and real matrices, respectively.

Example 3.9: (Classical Groups, in Particular Proper Lorentz Group and Their Lie Algebras)

By now the pattern for matrix groups should be well wonted to the reader. The Lie algebra is the matrix logarithm of matrices within the group that are “near” the identity. In Example 1.10 we considered the classical group of $M\times M$ matrices (either real or complex) defined by the condition $\gamma^\dagger\,\eta\,\gamma = \eta$ where $\eta$ defines the quadratic function $Q:\mathbb{C}^M\times \mathbb{C}^M\to\R;\, Q(u,\,v) = u^\dagger\,\eta\,v$ (or, for a group of real matrices, $Q:\R^M\times \R^M\to\R;\, Q(u,\,v) = u^T\,\eta\,v$) of vectors that is conserved by all members of the group (i.e. $Q(\gamma\,u,\,\gamma\,v) = Q(u,\,v)$ ; in particular we considered the Lorentz group $O(1,\,3)$ for which $\eta = \operatorname{diag}[-1,\,1,\,1,\,1]$. We showed that tangents $X$ to $C^1$ paths through the identity at the identity were defined by the linear equation $X^\dagger \,\eta = – \eta\,X$ and so the Lie algebra is the vector space (over the reals) of all such matrices. For the particular case of the proper Lorentz group $SO^+(1,\,3)$, the Lie algebra is the vector space spanned by the six real matrices:

\begin{equation}\label{ClassicalLieAlgebraExample_1}\begin{array}{lcllcllcl}\hat{J}_x&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\end{array}\right)&\hat{J}_y&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&-1&0&0\end{array}\right)&\hat{J}_z&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{array}\right)\\\hat{K}_x&=&\left(\begin{array}{cccc}0&10&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)&\hat{K}_y&=&\left(\begin{array}{cccc}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{array}\right)&\hat{K}_z&=&\left(\begin{array}{cccc}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{array}\right)\end{array}\end{equation}

Take heed how in all these matrix group examples, the matrix logarithm maps a neighbourhood of $\id$ in the group $\G$ bijectively to a neighbourhood of $\Or$ in the Lie algebra $\g$, and the matrix exponential maps the Lie algebra back into the group. This behaviour will be shown to be general, given befitting generalisations of the definitions of $\exp$ and $\log$.

Now we add a “removable” Axiom to our five. It is redundant if we use Convention 1.2. However, we may want to relax that convention sometimes. However, even when we relax the dimension convention, the following axiom is “removable”, because I shall later show that, given a mathematical system fulfilling the other five axioms alone, one can define a mathematical subsystem of the given one that again fulfills the five axioms as well as the “removable” one below. That is, there are systems fulfilling the five axioms but not the sixth below, but one can always derive a system fulfilling all six given the other five. So, although the sixth below is not dependent on the other five, a given system fulfilling the other five can be reinterpreted in a “natural” way to fulfill all six. This point will be important in defining matrix groups.

Removable Axiom 6 (Full Dimension Tangent Space):

The Lie Algebra of the Lie group has full dimension, i.e. $\dim(\g) = \dim(\V) = N$.

It will later be shown that, if $\dim(\g) < \dim(\V) = N$, then one can redefine $\V$ and $\Nid$ so that $\dim(\g) = N$.

Now we have a new way to define co-ordinates in some open neighbourhood $\W \subset \V$ of $\mathbf{0}$. Here we mean, of course, open with respect to the wonted topology in $\R^N$. For now we can prove the following theorem.

Theorem 3.9 (Canonical Co-ordinates of the Second Kind):

Let $\sigma _j :[-\tau_0\,\tau_0]\to \Nid ;\;j=1\cdots N$ be $C^1$ paths in $\Nid$ through the identity ($\sigma _j \left( 0\right)=\id$ such that $\hat{X}_j =\left. {{\rm {d}}_\tau \sigma _j \left( \tau\right)} \right|_{\tau =0}$ span the Lie algebra $\g$. Then there is an open ball $\W\subseteq\V\subset\R^N$ neighbourhood of $\mathbf{0}\in\R^N$ wherein $\forall X\in\W \exists \gamma\in\Nid$ such that $\lambda(\gamma) = X$. In other words, the Labeller map $\lambda$ is onto (surjective) some open ball neighbourhood of $\mathbf{0}\in\R^N$.

Proof: Show Proof

By N-fold application of the Nontrivial Continuity Axiom 4, we can choose $\tau_0 >0$ small enough that $\sigma_1 (x_1 )\,\sigma_2 ( x_2)\,\cdots \,\sigma_N( x_N)\in \Nid ,\,\forall x_j \in[-\tau_0,\,\tau_0]$. Then we define the function:

\begin{equation}
\label{CanonicalCoordinatesSecondKindTheorem_1}
\mu :[-\tau_0,\,\tau_0]^N\to \V;\;\mu(x_1 ,\,x_2,\,\cdots ,\,x_N)=\lambda\left(\sigma_1( x_1)\,\sigma_2( x_2)\,\cdots \,\sigma_N( x_N)\right)
\end{equation}

Whenever any of the $x_j$ vary alone, $\mu$ follows a $C^1$path, by the Group Product Continuity Axiom 3, so that $\mu$ is continuously differentiable in all its $N$ real arguments. Moreover, its first derivative (an $N\times N$ Jacobi matrix) is non-singular at $x_1 =x_2 =\,\cdots =x_N =0$; indeed, if we use Cartesian co-ordinates for $\V$ with basis vectors $\hat{X}_j$ (the tangents at $\id$ to the $\sigma _j$) then this derivative is the identity matrix. Therefore, by the Inverse Function Theorem, $\mu$ has an inverse, also continuously differentiable in all its arguments, when restricted to some open ball $\U\subset [-\tau_0,\,\tau_0]^N$ containing $\mathbf{0}$ and $\mu$ is a homeomorphism between $\U$ and some open neighbourhood $\W\subseteq \V$ of $\mathbf{0}$. Thus, every inverse image $\gamma \in\G=\lambda^{-1}(X);\,X\in \W$ can be uniquely written as an $N$-fold product $\sigma_1 (x_1)\,\sigma_2( x_2)\,\cdots \,\sigma _N(x_N)$ and the $x_j$can be used as co-ordinates labelling all members of $\W \subset \V$.$\quad\square$

When the $\sigma _j$ are exponential paths as defined later on, these co-ordinates are called canonical co-ordinates of the second kind (even though in the present treatment these co-ordinates are the ones we define first) because this co-ordinate scheme was historically found after the first kind canonical co-ordinates when I talk about the exponential map and geodesic co-ordinates, hence the name. We have yet to show that the group operations are $C^1$ in these new co-ordinates, something I shall do when I talk about the exponential map, but for now the important fact is that the existence of the co-ordinates shows that the labeller map is needfully onto some open subset of $\V$. We conceive of the Lie group as a set of elements linked to the identity by paths and thus instead find out that “all the spaces between the paths are needfully filled in” and belong to the group by dint of the $C^1$ group operations. That is, although we foremostly think of the group as elements and threads, a continuous “cloth” is naturally woven from these threads. The most important conclusion of the foregoing theorem for the purposes of this post is a topological one: the fact of a “continuous cloth” needfully woven from the axiomatically postulated threads. Figure 3.2: Canonical Co-ordinates of the Second Kind: For some small enough ball, the group product “fills all the spaces in” between the limbs of our spider of $C^1$ paths

So there is some open ball neighbourhood $\W\subset\V$ of $\mathbf{0}$ in $\V$ all of whose members $X$ are images $\lambda(\gamma)$ of some $\gamma\in\Nid$. Let us call the inverse image $\K$, i.e. $\K=\left\{\gamma\in\Nid:\,\lambda(\gamma) \in \W\right\} = \lambda^{-1}(\W)$. But what about members $\gamma^\prime\in\Nid$ such that $\lambda(\gamma^\prime)\not\in\W$, i.e. members of $\Nid\setminus\K$? We can show three important things:

1. These outlier elements can nonetheless be written as finite products of members of $\K$ and therefore, by the Homgeneity Axiom 5, every member of $\G$ can be written as a finite product of members of $\K$;
2. That the inverse image $\K^\prime = \lambda^{-1}(\W^\prime)$ of any open ball neighbourhood $\W^\prime\subset\W\subset\V$ of $\mathbf{0}$ in $\V$ can play the role of the set $\K$ above, so that every member of $\G$ can be written as a finite product of members of $\K^\prime$;
3. That every member $\gamma\in\Nid$ of $\Nid$ is inside a set set $\mathcal{N}_\gamma \subseteq \Nid$ such that $\lambda(\mathcal{N}_\gamma)\subseteq\V$ is an open ball in $\V$ centred on $\lambda(\gamma)$.

To help us explore these ideas, we need a definition and a lemma:

Definition 3.10 (Nucleus of a Connected Lie Group):

A nucleus of a connected Lie group $\G$ is any subset $\K\subseteq\G$ of $\mathcal{G}$ containing a set of the form $\lambda^{-1}\left(\mathcal{O}_\mathbf{0}\right)$ where $\mathcal{O}_\mathbf{0}\subseteq\V\subset\R^N$ is an open neighbourhood of $\mathbf{0}$ in $\V$ (with respect to the wonted topology for $\R^N$).

So a nucleus is the inverse image under the Labeller function $\lambda$ of a neighbourhood of $\mathbf{0}$ in $\V$ (with respect to the wonted topology for $\R^N$). We take heed that, by dint of Theorem 3.9 above, there is indeed at least one nucleus $\K$, namely any open neighbourhood $\K\subseteq\lambda^{-1}(\W)$ of $\mathbf{0}$ where $\W$ is the open neighbourhood of $\mathbf{0}$ in $\V$ as defined in Theorem 3.9, because the theorem shows that the every member of $\W$ is an image under $\lambda$ of some $\gamma =\prod\limits_{j=1}^N \sigma_j(\tau_j) \in \Nid$, where all symbols are defined as in the theorem. Given any nucleus, we can always define a “fancier one”: a symmetric nucleus.

Definition 3.11 (Symmetric Nucleus):

A Symmetric Nucleus $\tilde{\K}\subset\G$ in a connected Lie group $\G$ is a nucleus wherein every element’s inverse is also in that nucleus, i.e. $\tilde{\K}=\tilde{\K}^{-1}$.

Lemma 3.12 (Symmetric Nucleus Lemma):

Every [nucleus] $\tilde{\K}\subset\G$ in a connected Lie group $\G$ contains a symmetric nucleus.

Proof:Show Proof

Given that $\tilde{\K}$ is a neighbourhood of $\id$ and given that fhe function $\gamma\mapsto \gamma^{-1}$ is $C^0$ at $\id$, there must be a neighbourhood $\U$ of $\id$ such that $\U^{-1}\subseteq\tilde{\K}$. Now choose the neighbourhood $\W=\U\cap\U^{-1}$ of the indentity $\id$; we have $\W\subseteq\U^{-1}\subseteq\tilde{\K}$ and also $\W^{-1} = \W$. So $\W\subseteq\tilde{\K}$ is a symmetric nucleus. $\qquad\square$

Now we apply the concept of a nucleus.

Lemma 3.13 (Behaviour of $C^1$ path in the small):

Let $\sigma:\mathcal{I}\to\Nid$ be a $C^1$ path and $\mathcal{I}\subset\R$ an open real interval. For every $\zeta = \sigma(\tau_\zeta) \in \sigma(\mathcal{I})$ (with $\tau_\zeta\in\mathcal{I}$) there is a nonzero length interval $[\tau_\zeta-\epsilon_\zeta,\,\tau_\zeta+\epsilon_\zeta]\subset\mathcal{I}$ such that $\zeta^{-1}\,\sigma\left([\tau_\zeta-\epsilon_\zeta,\,\tau_\zeta+\epsilon_\zeta]\right) \in \K$ where $\K$ is any nucleus of $\G$. Otherwise put, there is a nonzero length interval $[\tau_\zeta-\epsilon_\zeta,\,\tau_\zeta+\epsilon_\zeta]\subset\mathcal{I}$ such that $\sigma(\tau) = \zeta\, \kappa(\tau)$ where $\kappa(\tau)\in\K$, i.e. the path can be thought of as a path in the nucleus $\K$ translated by $\zeta$.

Proof:Show Proof

By the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4, $\zeta^{-1} \sigma(\tau)$ defines a nonzero length $C^1$ path $\sigma\left([\tau_\zeta-\eta_\zeta,\,\tau_\zeta+\eta_\zeta]\right)$ for $\eta_\zeta>0$ through the identity $\id$; here $\tau_\zeta$ is such that $\zeta = \sigma(\tau_\zeta)$. So $\lambda\left(\sigma\left([\tau_\zeta-\eta_\zeta,\,\tau_\zeta+\eta_\zeta]\right)\right)$ is a $C^1$ path through $\mathbf{0}$ in the namespace set $\V$ and as such some nonzero length path $\lambda\left(\sigma\left([\tau_\zeta-\epsilon_\zeta,\,\tau_\zeta+\epsilon_\zeta]\right)\right)$ for $\eta_\zeta \geq \epsilon_\zeta > 0$ must be contained in any open neighbourhood $\U$ of $\mathbf{0}$ in $\V$, so that $\sigma\left([\tau_\zeta-\epsilon_\zeta,\,\tau_\zeta+\epsilon_\zeta]\right)$ must be contained in the nucleus $\K = \lambda^{-1}(\U)$. $\quad\square$

One can take this last statement in the proof, as clear, but to clarify even further, the last statement is true simply by the continuity of the $C^1$ (hence $C^0$) map:

\begin{equation}
\label{LeftTranslatedCoordinates}
\mu_\zeta:[\tau_\zeta-\eta_\zeta,\,\tau_\zeta+\eta_\zeta]\to\V;\,\mu_\zeta(\tau) = \lambda\left(\zeta^{-1} \sigma(\tau)\right)
\end{equation}

so that $\K^\prime = \lambda(K) \cap \mu_\zeta\left([\tau_\zeta-\eta_\zeta,\,\tau_\zeta+\eta_\zeta]\right)$ is an open neigbourhood of $\mathbf{0}$ in the relative topology inherited from $\R^N$ and induced by $\mu_\zeta\left([\tau_\zeta-\eta_\zeta,\,\tau_\zeta+\eta_\zeta]\right)$. Therefore there must be an closed sub-interval $\mathcal{I}_\zeta = [\tau_\zeta-\epsilon_\zeta,\,\tau_\zeta+\epsilon_\zeta]$ included in an open sub-interval $\mathcal{I}_\zeta^\prime$ of $[\tau_\zeta-\eta_\zeta,\,\tau_\zeta+\eta_\zeta]$ such that $\mu(\mathcal{I}_\zeta^\prime)\subseteq \K^\prime$. Figure 2: The $C^1$ Path Path Behaviour In-The-Small Lemma

Now we can show that our connected Lie group $\G$ is generated by any nucleus.

Theorem 3.14 (Generation of Group by any Nucleus):

A connected Lie group is generated by any nucleus $\K\subset\Nid$.

Proof: Show Proof

We may as well assume the nucleus $\K\subseteq \lambda^{-1}\left(\W\right)$ in question is a subset of $\lambda^{-1}\left(\W\right)$: if we prove the theorem for a subset of $\lambda^{-1}\left(\W\right)$ then the theorem is certainly true for any other open set containing this set.

Let $\gamma\in\Nid$ be any member of $\Nid$. By the Connectedness Axiom 2, there is a $C^1$ path $\sigma:[0,\,1]\to\Nid$ linking $\id$ and $\gamma$, i.e. $\sigma(0) = \id$ and $\sigma(1) = \gamma$.

So now we apply Lemma 3.13 inductively as follows. Suppose we have shown that every element of the sub-path $\sigma([0,\,\tau_m])$ can be written as a product of elements of $\K$. Then we apply the lemma to the point on the far end of the sub-path $\zeta_m = \sigma(\tau_m)$ to see that there is some $\epsilon_m>0$ such that $\sigma(\tau_m)^{-1} \sigma(\tau_m + \delta)\in\K\,\forall \delta\in[0,\,\epsilon_m]$ . Therefore we see that the new sub-path $\sigma([0,\,\tau_m+\epsilon_m])$ can be written as a product of elements of $\K$. We begin the induction at $\tau_0 = 0$ so that, after $n$ induction steps, we have shown that sub-path $\sigma([0,\,\tau_n])$ where $\tau_n = \sum_{j = 1}^n \epsilon_j$ can be written as a product of elements of $\K$. Now one of two things can happen. Either (i) $\tau_n \geq 1$ after a finite number of induction steps and we have shown that the whole path and in particular $\gamma = \sigma(1)$ can be written as required or (ii) the series $\sum_{j = 1}^\infty\epsilon_j$ converges to a limit point $\tau_\infty < 1$. But if the second possibility arises, we can now apply the lemma to the limit point $\zeta_\infty = \sigma(\tau_\infty)$, which is not equal to any of the $\zeta_j$, itself and show that there is a finite interval $[\tau_\infty – \epsilon_\infty, \tau_\infty+\epsilon_\infty]$ where $\epsilon_\infty > 0$ such that $\sigma(\tau_\infty)^{-1} \sigma([\tau_\infty – \epsilon_\infty, \tau_\infty+\epsilon_\infty])\subseteq\K$. This finite interval, being a neighbourhood of the limit point $\tau_\infty$, contains some partial sums $\sum\epsilon_j$ converging to that limit point. That is, there is some partial sum $\tau_M$ such that $\sigma(\tau_\infty)^{-1}\sigma(\tau_M)\in\sigma(\tau_\infty)^{-1} \sigma([\tau_\infty – \epsilon_\infty, \tau_\infty+\epsilon_\infty])\subseteq\K$. That is, we have $\sigma(\tau_\infty)^{-1} \sigma(\tau_M)\in\K$ and also $\sigma(\tau_\infty)^{-1} \sigma(\tau_\infty + \epsilon_\infty)\in\K$, thus $\sigma(\tau_\infty + \epsilon_\infty) = \varsigma_\infty\sigma(\tau_M)$ for some $\varsigma_\infty\in\K$. $\sigma(\tau_M)$ is a finite product of members of $\K$, so then $\sigma(\tau_\infty + \epsilon_\infty)$ is also such a finite product of members of $\K$, thus gainsaying the assumption of there being a limit point. (In everyday words: if there is a limit point, we can apply the lemma to “jump beyond it”, contradicting the assumption of the limit point). So only possibility (i) stands and $\gamma = \sigma(1)$ can be written as the product of a finite number of members of $\K$.

So now we have shown that every element in $\Nid$ can be written as a finite product of members of $\K$. So now of course we can apply the Homgeneity Axiom 5, so show that any group member can be written as a finite product of members of $\K$. Take heed that the method of this proof works for any nucleus in $\G$.$\quad\square$.

We now get further insight into how $\Nid$ looks by proving a generalisation of Theorem 3.9 to open balls around any group member within $\Nid$:

Theorem (Open Ball Around Any Member of the Fundamental Neighbourhood $\Nid$):

For every $\gamma\in\Nid$ there is a set $\mathcal{N}_\gamma \subseteq \Nid$ such that $\lambda(\mathcal{N}_\gamma)\subseteq\V$ is an open ball centred on $\lambda(\gamma)$. Every member of $\mathcal{N}_\gamma$ can be uniquely represented in the form $\gamma \prod\limits_{j=1}^N\sigma_j(\tau_j)$ for $|\tau_j| < \epsilon_\gamma$ for some $\epsilon_\gamma > 0$.

But first we need a lemma, a definition and a “stepping stone” theorem, which shows that the Jacobian of the transformation of $C^1$ paths induced by left-translating them from the identity to $C^1$ paths through some other $\gamma\in\Nid$ is nonsingular. We can then use the inverse function theorem to prove the result just as we did in the proof of Theorem 3.9 to show the existence of the said open ball.

Firstly, for any $\gamma\in\Nid$, we study the space of tangents $\g_\gamma$ to all $C^1$ paths through $\gamma$. By the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4, there are indeed $C^1$ paths through any $\gamma\in\Nid$, defined by $\gamma\, \sigma_j(\tau)$, where the $\sigma_j:[-a,\,a]\to\Nid$ are $C^1$ paths with $a > 0$ through the identity with $\sigma_j(0) = \id$ (note that we can equivalently use “right-translated” paths $\sigma_j(\tau)\,\gamma$). Conversely, any $C^1$ path $\sigma_\gamma(\tau)$ through $\gamma$ is mapped, by the same axioms 3 and 4, into a nonzero length $C^1$ path through the identity $\gamma^{-1} \sigma_\gamma(\tau)$ when multiplied by $\gamma^{-1}$.

Lemma 3.15 (Linearity of Any Tangent Space to $\Nid$):

The space $\g_\gamma$ of tangents at $\gamma\in\Nid$ to all $C^1$ curves through $\gamma$ is a vector space over the reals. When a $C^1$ path $\tilde{\gamma}_1(\tau)$ through $\gamma_1 = \tilde{\gamma}_1(0)$ is mapped to the corresponding $C^1$ path $\gamma_2\,\gamma_1^{-1}\,\tilde{\gamma}_1(\tau)$ through $\gamma_2$ by left-translation (i.e. by multiplication by $\gamma_2\,\gamma_1^{-1}$), then the tangent vector $X\in \g_{\gamma_1}$ to the former path at $\gamma_1$ is mapped to the tangent vector $X^\prime\in \g_{\gamma_2}$ to the left translated path at $\gamma_2$ by a linear mapping, which can be defined by an $N\times\,N$ square matrix.

Proof:Show Proof

Reasoning exactly as in Lemma 3.2, we can show that if $X_{\gamma,j}, \,\ X_{\gamma,k}$ are tangents to the $C^1$ paths $\gamma\, \sigma_j(\tau)$ and $\gamma \,\sigma_k(\tau)$ at $\gamma$ (i.e. when $\tau=0$), then the tangent to $\gamma\,\sigma_j(\alpha\,\tau)\,\sigma_k(\beta\,\tau)$ at $\gamma$ is $\alpha\,X_{\gamma,j}+\beta\,X_{\gamma,k}$. That is, the space $\g_\gamma$ of tangents to all $C^1$ curves through $\gamma$ is a vector space over the reals exactly as the Lie algebra of tangents to the identity is.

Now choose any Cartesian basis for $\R^N\supset\V$ and let $X_j$ be the tangents $X_j = \left.\d_\tau\left(\lambda\left(\sigma_j(\tau)\right)\right)\right|_{\tau=0}\in\R^N$ to the paths $\sigma_j(\tau)$ at the identity $\id$ and let $X_{\gamma,j} = \left.\d_\tau\left(\lambda\left(\gamma\, \sigma_j(\tau)\right)\right)\right|_{\tau=0}$ be the tangents to the $C^1$ paths $\gamma \,\sigma_j(\tau)$ through $\gamma$ at $\gamma$. Naturally, we have chosen the $C^1$ paths $\sigma_j(\tau),\,j=1\cdots N$ through the identity so that their tangents $X_j$ span the whole Lie algebra $\g$ of $\G$. So a $C^1$ path through the identitywith a general tangent is of the form $\prod\limits_{j=1}^N \sigma_j(\alpha_j\tau)$, it has tangent $\sum\limits_{j=1}^N\alpha_j\,X_j$. Reasoning exactly as in Lemma 3.2, the tangent to $\gamma \prod\limits_{j=1}^N \sigma_j(\alpha_j\tau)$ is readily shown to be $\sum\limits_{j=1}^N\alpha_j\,X_{\gamma,\,j}$. Therefore, this is a linear mapping. If we align the Cartesian basis with the $N$ linearly independent tangents $X_j$ to the identity, the matrix of the linear mapping is the $N\times N$ matrix $\mathbf{M}(\gamma_1,\,\gamma_2)$ whose $j^{th}$ column is the tangent $X_{\gamma,j}$ written with respect to the chosen basis. $\quad\square$

Note, however, we do not yet know that the matrix of this linear mapping is nonsingular: some of the $X_{\gamma,j} = \left.\d_\tau\left(\lambda\left(\gamma\, \sigma_j(\tau)\right)\right)\right|_{\tau=0}$ could be linearly dependent. We shall show that this is not so very soon.

Take Heed: Important Note:

Also, at this point, take heed that we can use “the same” $\R^N$ to measure all tangents in each of the tangent spaces $\g_\gamma$ at any $\gamma\in\Nid$, because the tangents are limits of scaled position vector increments in the same copy $\V$ of an open patch of $\R^N$ even though they might belong to different tangent spaces $\g_{\gamma_1},\,\g_{\gamma_2}$ for different $\gamma_1,\,\gamma_2\in\Nid$. We shall not generally have this luxury: there is in general no way to compare a tangent space $\g_{\gamma_2}$ to some $\gamma_2$ outside $\Nid$ with a tangent space $\g_{\gamma_1}$ to some $\gamma_1$ inside $\Nid$.

Now for a definition:

Definition 3.16 (Tangent Map Induced by Left Translation):

We call the linear mapping $\g_{\gamma_1}\to\g_{\gamma_2}$:

\begin{equation}
\label{TangentMapInducedByLeftTranslationEquation}
X_{\gamma_1,j} = \left.\d_\tau\left(\lambda\left(\gamma_1\,\sigma_j(\tau)\right)\right)\right|_{\tau=0}\in\R^N\mapsto X_{\gamma_2,j} = \left.\d_\tau\left(\lambda\left(\gamma_2\,\sigma_j(\tau)\right)\right)\right|_{\tau=0}\in\R^N
\end{equation}

studied in the Lemma above the tangent map induced by left translation from $\gamma_1$ to $\gamma_2$. This linear map is defined by an $N\times N$ matrix $\mathbf{M}(\gamma_1,\,\gamma_2)$ (implicitly, we must choose a basis to define this matrix). Naturally we must have $\gamma_1,\gamma_2\in\Nid$ for these concepts to be meaningful, i.e. we must have images $\lambda(\gamma_1),\lambda(\gamma_2)\in\V$ so that we can measure tangents to $C^1$ curves through the points as limits of scaled position vector increments in the same set $\V\subset\ R^N$.

In differential geometry, the liner map $\mathbf{M}(\gamma_1,\,\gamma_2)$ is called the pushforward map from tangent space $\g_{\gamma_1}$ to $\g_{\gamma_2}$.

The idea of “pushforward” is that one imagine’s grasping a vector in a tangent space and “pushing” it “forward” (presumably by thrusting on its hinder end, as though you’re fighting to get a stubborn cow or sheep into its pen) to a new tangent space. I’ve never found this thought picture very evocative and don’t like the name; it actually evokes for me something more like the idea of a geodesic where one does imagine thrusting a vector by its tail so that it slides forward in exactly the direction its head is heading to. The stubborn cow has a definite direction it wants to head off in!

By our discussion above, this tangent map (“pushforward”) is a linear map from (at most) an $N$-dimensional vector space over the reals into another at most $N$-dimensional vector space over the reals, so it can be defined by the matrix $\mathbf{M}(\gamma_1,\,\gamma_2)$ Clearly there is the altogether obvious and analogous concept of the tangent map $\tilde{\mathbf{M}}(\gamma_1,\,\gamma_2)$ induced by right translation.

It’s also worth stating the idea of tangent maps induced by translation in slightly different words, namely, that group $\gamma\mapsto\gamma \sigma$ multiplication is a $C^1$ function of group co-ordinates whenever $\sigma,\,\gamma,\,\gamma\,\sigma\in\K$ for some small enough [nucleus] $\K$, which we can see through an extremely elementary lemma..

Lemma 3.17 ($C^1$ Behaviour of the Group Product):

Group multiplication $\gamma\mapsto\gamma\,\sigma$ is a $C^1$ function of all the co-ordinates of both multiplicands $\gamma\, \sigma$ for $\gamma,\,\sigma,\,\gamma\,\sigma\in\K\subseteq\Nid$ for some small enough nucleus $\K$. The group inverse $\sigma\mapsto\sigma^{-1}$ is also a $C^1$ function of all the co-ordinates of $\sigma$ when $\sigma,\,\sigma^{-1}\in\Nid$.

Proof:Show Proof

Let us study the mapping $\sigma\mapsto\gamma\sigma$ by assuming $\sigma,\,\gamma\in\K$ belong to a nucleus $\K\subseteq\Nid$ small enough to be kitted with second kind canonical co-ordinates, i.e. $\sigma = \prod\limits_{j=1}^N \sigma_j(\tau_j)$ and $\gamma = \prod\limits_{j=1}^N \sigma_j(\xi_j)$ where $\tau_j,\,\xi_j$ are the unique second-kind-canonical co-ordinates for $\sigma,\,\gamma$, respectively. Then $\lambda(\gamma\sigma)$ is a $C^1$ function of the $\tau_j$ and $\xi_j$ and so the derivatives $\partial_{\tau_j} \lambda(\gamma\sigma),\,\partial_{\xi_j} \lambda(\gamma\sigma)$ are at least $C^0$ functions of the $\tau_j$ and $\xi_j$. As is our wont, we choose the $\sigma_j$ so that their derivatives at $\id$ are aligned with the Cartesian co-ordinate axes of $\V$. Therefore the derivatives:

\begin{equation}
\label{TranslatedDerivativesEquation}
\begin{array}{lcl}
\partial_{\tau_j} \lambda(\gamma\sigma) &=& \mathbf{M}(\id,\,\gamma)\,X_j\\
\partial_{\xi_j} \lambda(\gamma\sigma) &=& \tilde{\mathbf{M}}(\id,\,\sigma)\,X_j
\end{array}
\end{equation}

are at least a $C^0$ functions of the $\tau_j,\,\xi_j$. Here we write the tangent map induced by right translation as $\tilde{\mathbf{M}}(\id,\,\gamma)$ and $\mathbf{M}(\id,\,\id) = \tilde{\mathbf{M}}(\id,\,\id)=\id_N$.

For the inverse mapping, $\gamma\sigma\mapsto\sigma^{-1}\gamma^{-1}$ the proof is the almost the same, but with opposite signed derivatives. $\qquad\square$

Naturally, the group operations, being $C^1$ are also needfully $C^0$, so that when we define the fully topology of the connected Lie group in a later post, the connected group is an example of the more general concept of a topological group.

As stated above, we do not yet know that this matrix is nonsingular, so we cannot yet say anything about the dimensionality of the tangent spaces, although we do know the Lie algebra, the tangent space at the identity, is of full dimension $N$. However, we now show that $\mathbf{M}(\gamma_1,\,\gamma_2)$ is at least locally nonsingular. Figure 3: The Tangent Map $\mathbf{M}(\id,\,\gamma)$ Induced by Left Translation

Lemma 3.18 (Local Nonsingularity of Tangent Map Induced By Left Translation):

Suppose we have a point $\gamma_1\in\Nid$ and let $\gamma_2:[-a,\,a]\to\Nid$ be a $C^1$ path through $\gamma_1$; here $\gamma_2(0) = \gamma_1$. Then the matrix $\mathbf{M}(\gamma_1,\,\gamma_2(\tau))$ of the tangent map induced by left translation from $\gamma_1$ to $\gamma_2(\tau)$ is nonsingular for some nonzero length interval $-\epsilon\leq\tau\leq+\epsilon$, $\epsilon > 0$.

Proof:Show Proof

Clearly $\mathbf{M}(\gamma_1,\,\gamma_1)$ is the identity matrix (the corresponding left translation is no translation). The elements of the matrix $\mathbf{M}(\gamma_1,\,\gamma_2(\tau))$ are at least $C^0$ functions of $\tau$ (this is because there are first derivatives involved in reckoning the matrix elements: we are essentially computing a Jacobi matrix). Therefore, the determinant $\det\left(\mathbf{M}(\gamma_1,\,\gamma_2(\tau))\right)$ is at least a $C^0$ (continuous) function of $\tau$, so there is some nonzero length interval $-\epsilon\leq\tau\leq+\epsilon$ wherein $\det\left(\mathbf{M}(\gamma_1,\,\gamma_2(\tau))\right) \geq \delta > 0$ for some strictly positive $\delta$; otherwise put, the matrix is nonsingular within a closed, nonzero length interval around $\gamma_1$ along any $C^1$ path through $\gamma_1$.$\quad\square$

Now we can strengthen this lemma into the theorem:

Theorem 3.19 (Nonsingularity of Tangent Map Induced By Left Translation in $\Nid$):

The tangent map $\mathbf{M}(\gamma_1,\,\gamma_2)$ induced by left translation from $\gamma_1\in\Nid$ to $\gamma_2\in\Nid$ is nonsingular, for any pair $\gamma_1,\,\gamma_2\in\Nid$. The tangent space $\g_\gamma$ to $\gamma\in\Nid$ has the same dimension $N$ for all $\gamma\in\Nid$.

Proof:Show Proof

Clearly $\mathbf{M}(\gamma_2,\,\gamma_3)\, \mathbf{M}(\gamma_1,\,\gamma_2)=\mathbf{M}(\gamma_1,\,\gamma_3)$, for any $\gamma_1, \gamma_2\, \gamma_3\in\Nid$: we can think about left translation from $\gamma_1$ to $\gamma_3$ as being composed of two left translations, so that the matrix for the tangent mapping induced by left translation from $\gamma_1$ to $\gamma_3$ is the product of the matrices for the two “hops”, first from $\gamma_1$ to $\gamma_2$; thence from $\gamma_2$ to $\gamma_3$.

So now we consider any $\gamma\in\Nid$; By the Connectedness Axiom 2, we link $\gamma$ to the identity by the $C^1$ path $\tilde{\gamma}:[0,\,1]$ where $\tilde{\gamma}(0) = \id$ and $\tilde{\gamma}(1) = \gamma$. Beginning at the identity we show by our reasoning above that $\mathbf{M}(\id,\,\tilde{\gamma}(\epsilon_1))$ is nonsingular, for some $\epsilon_1 > 0$. Then we show that $\mathbf{M}(\id,\,\tilde{\gamma}(\epsilon_1 + \epsilon_2)) = \mathbf{M}(\tilde{\gamma}(\epsilon_1),\,\tilde{\gamma}(\epsilon_1 + \epsilon_2))\,\mathbf{M}(\id,\,\tilde{\gamma}(\epsilon_1))$ is nonsingular, for some $\epsilon_2$. After $n$ iterations of this process, we shall have reached the point $\tilde{\gamma}\left(\tau_n=\sum\limits_{j=1}^n\epsilon_j\right)$. Now, either we reach $\tilde{\gamma}(1)=\gamma$ after a finite number of steps, or the sum $\sum\limits_{j=1}^n\epsilon_j$ converges to a limit point $\tau_\infty = \sum\limits_{j=1}^\infty\epsilon_j < 1$. But we can reason that there is no such limit point by a method exactly like that used in the proof of Theorem 3.14. To wit, we study the matrix for the tangent mapping induced by left translation from the limit point itself and we know that the matrix $\mathbf{M}(\gamma,\,\tilde{\gamma}(\tau))$ is nonsingular for some nonzero length interval $[\tau_\infty-\epsilon_\infty,\,\tau_\infty+\epsilon_\infty]$ around $\tau=\tau_\infty < 1$. This interval, being centred on the limit point, must contain at least one of the partial sums $\tau_n=\sum\limits_{j=1}^n\epsilon_j$, so the matrix for the tangent mapping induced by left translation from $\tilde{\gamma}(\tau_n)$ to $\tilde{\gamma}(\tau_\infty)$ is nonsingular. Therefore, it has a nonsingular inverse, namely, the matrix for the tangent mapping induced by left translation from $\tilde{\gamma}(\tau_n)$ to $\gamma(\tau_\infty)$. Indeed, there is then a nonsingular matrix for the tangent mapping induced by left translation from $\tilde{\gamma}(\tau_n)$ to $\tilde{\gamma}(\tau_\infty + \epsilon_\infty)$ and so we have proven that the tangent mapping induced by left translation from the identity to $\tilde{\gamma}(\tau_\infty + \epsilon_\infty)$ is nonsigular. This gainsays the assumption of a limit point $\tau_\infty$: by assuming the limit point $\tau_\infty$, we can show how to “leap over” it to $\tau_\infty+\epsilon_\infty$, and do this in a finite number ($n+1$) of steps. Therefore, our iterative process must reach $\tau = 1$ after a finite number of steps, and therefore the tangent mapping induced by left translation from the identity to $\tilde{\gamma}(1) = \gamma$ is nonsingular. So we have proven $\mathbf{M}(\id,\,\gamma)$ is nonsingular $\forall\,\gamma\in\Nid$.

So now an inverse tangent mapping induced by left translation from $\gamma$ to the identity is well defined and has matrix $\mathbf{M}(\gamma,\,\id) = \mathbf{M}(\id,\,\gamma)^{-1}$. Moreover, this shows that the tangent space $\g_\gamma$ has full dimension, namely $N$, for all $\gamma\in\Nid$.

It is now a simple matter to show that $\mathbf{M}(\gamma_1,\,\gamma_2)$ is nonsingular $\forall\,\gamma_1,\,\gamma_2\in\Nid$ by concatenating paths linking $\gamma_1$ and $\gamma_2$ to the identity; we find that $\mathbf{M}(\gamma_1,\,\gamma_2) = \mathbf{M}(\id,\,\gamma_2)\,\mathbf{M}(\id,\,\gamma_1)^{-1}$.$\quad\square$

Since the tangent map induced by left translation is nonsingular for any translation between members of $\Nid$, we see that the tangent spaces at all points of $\Nid$ are all isomorphic.

It is important to heed that, whilst we clearly have the composition-of-tangent-maps rule $\mathbf{M}(\gamma_1,\,\gamma_3) = \mathbf{M}(\gamma_2,\,\mathbf{\gamma_3})\,\mathbf{M}(\gamma_1,\,\gamma_2);\,\forall\,\gamma_1,\,\gamma_2,\,\gamma_3\in\Nid$ there is no particular relationship between $\mathbf{M}(\id,\,\gamma_1\,\gamma_2)$ and $\mathbf{M}(\id,\,\gamma_1),\,\mathbf{M}(\id,\,\gamma_2)$ or between $\mathbf{M}(\id,\,\gamma)$ and $\mathbf{M}(\id,\,\gamma^{-1})$: the matrices of tangents maps do not form a group, even though they are invertible. If you think about this carefully, this is not surprising, for one can take any bijective function $\phi:\V\to\V$ and use $\phi\circ\lambda$ as a labeller map that still fulfils all the fundamental axioms 1 through 5. This arbitrary (aside from being bijective) map can “warp” the local $\V$ differently at the points $\gamma_1,\,\gamma_2,\,\gamma$ and $\gamma^{-1}$ and the differences between these local warpings can be rather aribtrary.

So now, at last, we can prove the theorem:

Theorem 3.20 (Open Ball Around Any Member of the Fundamental Neighbourhood $\Nid$):

For every $\gamma\in\Nid$ there is a set $\mathcal{N}_\gamma \subseteq \Nid$ such that $\lambda(\mathcal{N}_\gamma)\subseteq\V$ is an open ball centred on $\lambda(\gamma)$. Every member of $\mathcal{N}_\gamma$ can be uniquely represented in the form $\gamma \prod\limits_{j=1}^N\sigma_j(\tau_j)$ for $|\tau_j| < \epsilon_\gamma$ for some $\epsilon_\gamma > 0$.

Proof: Show Proof

We reason almost exactly as in the proof of Theorem 3.9. That is, we now consider the $\mu$-function:

\begin{equation}
\label{TranslatedMuFunction}
\mu :[-\tau_0,\,\tau_0]^N\to \V;\;\mu(x_1 ,\,x_2,\,\cdots ,\,x_N)=\lambda\left(\gamma\sigma_1(x_1)\,\sigma_2( x_2)\,\cdots \,\sigma_N( x_N)\right)
\end{equation}

instead of the $\mu$ function used in that theorem. We know that the tangent space at $\gamma$ is of full dimension and spanned by the tangents $X_{\gamma,\,j}$ to $\gamma\,\sigma_j(\tau)$ at $\tau=0$. The first derivative of our new $\mu$ is then nonsingular, and then we reason exactly analogously to how we did in Theorem 3.9.$\quad\square$

So now that we know that (i) we don’t need the whole of $\Nid$ to generate the Lie group but simply any nucleus and (ii) there is an open neighbourhood $\W$ which $\lambda$ is surjective onto, we can go back and take the open set $\V$ in all the axioms above to be the open set $\W$ from the outset and also replace $\Nid$ by any nucleus of the form $\K = \lambda^{-1}( \U)$, where $\U$ is any open neighbourhood of $\mathbb{0}$ contained within the open ball $\W$ as defined in the proof of Theorem 3.9. Thus, through this observation, we gather together the results so far into the next theorem:

Theorem 3.21 (Canonical Co-ordinates of the Second Kind 2):

If axioms 1 through 5 hold, then the sets $\Nid$ and $\V$ in the Labeller Axiom 1 can indeed be chosen so that $\lambda :\Nid \to \V$ is both one-to-one and onto; moreover, every element in $\Nid$ can be written as a unique $N$-fold product $\sigma_1( x_1)\,\sigma_2( x_2)\,\cdots \,\sigma _N( x_N)$ for $\left( x_j \right)_{j=1}^N \in \mathcal{T}\subset [-\tau_0,\,\tau_0]^N$ for some $\tau_0 >0$, where $\mathcal{T}$ is an open ball neighbourhood of $\mathbf{0}$. The $x_j$ are then co-ordinates for $\V$. $\quad\square$

We shall therefore henceforth take the sets $\V$ and $\Nid$ to have the properties of the above theorem. We can now begin to define a topology on the group $\G$ by defining $\mathcal{O}\subseteq \Nid$ to be open in $\Nid$ iff $\lambda\left(\mathcal{O}\right)$ is open in $\R^N$ or iff $\mu^{-1}\left(\lambda \left(\mathcal{O}\right)\right)$ is open in $[-\tau_0,\,\tau_0]^N$; $\mathcal{T}=\mu^{-1}\left(\V\right)$ and $\V$ are homeomorphic, so both definitions are the same.

Convention 3.22 (Dimension N in the Labeller Axiom 1):

We now have another way to state the Dimension Convention for the Labeller Axiom 1. $N$, and not fewer, is the dimension of the Lie algebra $\g$ of the connected Lie group $\G$. We have used this convention in stating the theorems above: it is assumed that the Lie algebra indeed has $N$ linearly independent basis vectors $\left.(X_j)\right|_{j=1}^N$, otherwise the Jacobi matrix is singular.

We began this section with axioms defining what we thought was a “spider” of $C^1$ paths in $\Nid$ linking all members of $\Nid$ to the identity, because we did not assume that the labeller was onto $\V$; however, let us sketch below what we have found out so far. There are now no “threads”, i.e. isolated $C^1$ paths, unless the dimension $N$ of the Lie algebra is unity. Rather, every point within our fundamental neighbourhood is not only linked to the identity by a $C^1$ path, but it is contained within an open ball itself wholly contained in $\Nid$. The group product operation “filled” all the spaces between the spider’s legs in, and we thus end this section by finding that what we thought was like a brittle starfish is indeed more like a sand dollar or sea urchin, homeomorphic to a fully fledged open ball in $\V$. Figure 4: Fundamental Neighbourhood: Every member is within an open neighbourhood kitted with its own canonical co-ordinates

Reference: