Chapter 9: The Lie Group Topology

The group operation clones the topological and framing structure defined on $\Nid$ as discussed so far, so that every element in the Lie group can have its own local co-ordinates relative to some nearby reference point $\gamma \in \G$. Small variations in any element $\gamma \in \G$ of the group, i.e. derivatives, can be studied with the left-translated set $\gamma\,\Nid$ where we now use the same set $\V\subset \R^N$ to give co-ordinates to elements of $\gamma \,\Nid$; for this new usage of $\V$ we think of $\V$ as a new chart for the region in the group near $\gamma$; the element $\gamma \,\zeta\in \gamma \,\Nid$ then gets the same co-ordinates $Z =\lambda(\zeta)\in \V$ as $\zeta$ gets in the chart $\V$ for the region $\Nid$ near $\id$ defined in the fundamental axioms 1 through 5. The new labeller is then $\lambda \,\circ \,\gamma ^{-1}:\gamma \,\Nid\to \V$whereby $\gamma^\prime\mapsto \lambda(\gamma^{-1}\gamma^\prime)$. Every $C^1$ path in the group running through $\gamma$ is expressible as $\gamma \,\sigma(\tau)$, where $\sigma(\tau)$ is the corresponding $C^1$path passing through $\id$ and contrariwise. As in Theorem 3.21, we can choose $\V$ so that its every element can be represented by unique second kind canonical co-ordinates. The set of tangents to the group at $\gamma $is then isomorphic to the Lie algebra when the tangent to the path $\gamma \,\sigma(\tau)$ at $\gamma $ corresponds to the tangent to $\sigma(\tau)$ at $\id$. Note that a right-translated set $\Nid\gamma $with right-translated paths $\sigma(\tau)\,\gamma $ will do just as well for defining these ideas, as we see in the lemma below. If $X\in\g$ is a tangent at $\id$ in the Lie algebra, then one wontedly writes the corresponding tangent at $\gamma$ as $\gamma \,X$ (or as $X\,\gamma$ if right-translation is used instead) as a shorthand for the image of $X$ under the isomorphism between the Lie algebra and the tangents at $\gamma $ whereby the tangent to the path $\gamma \,\sigma(\tau)$ at $\gamma$ corresponds to the tangent to $\sigma(\tau)$ at $\id$. Note that in general this is shorthand for there is in general no other product defined between a Lie group member $\gamma\in\G$ and a Lie algebra vector $X\in\g$. When we deal with matrix Lie groups, however, both $\gamma$ and $X$ are square matrices of the same dimension, and in this case the “isomorphism image shorthands” $\gamma\,X$ and $X\,\gamma$ do indeed have also the less abstract, alternative interpretation as the simple matrix product. Now for the promised lemma.

Lemma 9.1 (The “Adjoint Shuffle”: Equal Validity of Left and Right Translated Charts):

Every left-translated tangent $\gamma \,X$ can be rewritten as a right-translated tangent $\gamma \,X_L =X_R \,\gamma;\quad X_R =\gamma \,X_L \,\gamma ^{-1}$ is a Lie algebra member with $\gamma \,\sigma(\tau)\,\gamma ^{-1}$ as defining $C^1$path through $\id$. Naturally, this transformation can be inverted, so that every right-translated tangent can be written as a left-translated one.

Proof: Show Proof

Most of this statement is obvious; what is not clear is that $\gamma \,\sigma(\tau)\,\gamma ^{-1}$ is $C^1$, since, in general $\gamma\,\sigma(\tau)\not\in\Nid$ so that we cannot use the Group Product Continuity Axiom 3 directly. However, by the Homogeneity Axiom 5, we can write $\gamma =\gamma_1 \,\gamma_2 \,\cdots \,\gamma_M $ for some finite $M$ and where $\gamma_j \in \Nid,\,j=1\cdots M$. Thus $\gamma_M \,\sigma(\tau)\,\gamma_M ^{-1}=\sigma_1(\tau)$ for $|\tau|<\tau_1 $ for some $\tau_1 >0$ is a $C^1$ path through 1 (by the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4) as inductively are $\gamma_{M-j}\,\sigma_j (\tau)\,\gamma _{M-j}^{-1}=\sigma_{j+1} (\tau),\,j=1\cdots M-1$ for $|\tau|<\tau _j $ for some $\tau_j >0$. Note, in particular, $\gamma_1\,\sigma _{M-1}(\tau)\,\gamma_1 ^{-1}=\gamma \,\sigma(\tau)\,\gamma ^{-1}$. $\quad\square$

The reason for my nickname “The Adjoint Shuffle” for this lemma will become clearer later. It is a technique that underpins discussions of the group’s topology, which we are soon to talk about, as well as being useful for talking about and proving a fundamental theorem to do with “homomorphic liftings”, to wit, the one-to-one correspondence between the homomorphisms of a simply connected Lie group and homomorphisms of its Lie algebra. A sketch below (Figure 4.1) tries to help with the intuition.

Adjoint ShuffleFigure 4.1: The “Adjoint Shuffle”

My personal favourite way (Figure 9.2) to define a tangent vector at a general point $\gamma$ is as an equivalence class of paths through $\gamma$, where two paths are equivalent iff they have the same derivative $\d_\tau \,\left(\gamma \,\sigma(\tau)\right)$ at $\gamma$. With this definition, the notation $\gamma\,X$ makes sense: the paths are shifted to other paths by the group product. Otherwise, if one concentrates on the derivative $\d_\tau \,\sigma(\tau)$ in $\V$ to define the tangent, the notation $\gamma \,X$is more mysterious and, taken literally, seems meaningless given that there is in general no product defined between group members and vectors in $\V$, thus $\gamma \,X$ must then be thought of more abstractly as “just a notation” for the isomorphic image discussed above. This is one of the many ways whereby the restriction of discussion to linear groups (matrix Lie groups) can be helpful: tangents, like group members, are square matrices and can be meaningfully multiplied by group members; moreover the operation of left translation on a tangent $X$ is truly the matrix product $\gamma \,X$ (likewise for right translation).

Tangent Vector Figure 9.2: Thinking of Tangent Vectors as Equivalence Classes of ${\bf C^1}$ Curves

Around every $\gamma_1 ,\,\gamma_2 \in \G$ we can attach either left or right translated $\Nid$, so that we can study the group multiplication and inversion elements near to $\gamma_1,\,\gamma_2 $ by studying multiplication the “splice region” in Figure 9.3, which are labelled by the same co-ordinate systems as the untranslated$\Nid$. $\G$, as far as its group product and inverse are concerned, made of shifted copies of the same set $\Nid$.

Product Continuity Figure 9.3: Studying the Product Continuity in Translated Neighbourhoods

We can define a topology on $\G$, by defining the complete system of neighbourhoods ([Mendelson]) of any point within $\G$. We first recall some elementary ideas and lemmas from point set topology.

Definition 9.2: (Topological Space)

For any set $\mathbb{X}$ we call a collection $\mathscr{T}\subset2^\mathbb{X}$ of its subsets a topology iff $\mathscr{T}$ fulfils:

    1. $\mathbb{X}, \emptyset \in \mathscr{T}$;
    2. $\mathscr{T}$ is closed under the operation of arbitrary unions of its members: if $\mathcal{A}_\omega\in\mathscr{T}, \,\forall\,\omega\in\Omega$ then $\bigcup\limits_{\omega\in\Omega}\mathcal{A}_\omega\in\mathscr{T}$;
    3. $\mathscr{T}$ is closed under the operation of intersection: if $\mathcal{A},\,\mathcal{B}\in\mathscr{T}$ then $\mathcal{A}\cap \mathcal{B}\in\mathscr{T}$; from this it follows by induction that all finite intersections of members of $\mathscr{T}$ also belong to $\mathscr{T}$.

The pair $\left(\mathbb{X},\,\mathscr{T}\right)$ is then called a topological space. A set $\U\in\mathscr{T}$ belonging to the topology is called an open set or, if there is likelihood of confusion, open with respect to the topology $\mathscr{T}$.

Note the “arbitrary” in point 2 and “finite” in point 3. The sets within the topology are meant to capture the notion of neighbourhood and nearness: two points $x, y\in \mathbb{X}$ are deemed “near” to one-another in some way if they belong to the same open set. If $\mathcal{A},\,\mathcal{B}$ are open, $\mathcal{A}\subset \mathcal{B}$ and if $x, y\in \mathcal{A}\cap \mathcal{B}=\mathcal{A}$ and $z\in \mathcal{B}$ but $z\not\in \mathcal{A}$, then we can talk about $y$’s being nearer to $x$ than $z$.

Definition 9.3 (Neighbourhood):

A subset $\mathcal{N}_x\subset \mathbb{X}$ of a topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is called a neighbourhood of a point $x\in \mathbb{X}$ iff there is an open set $\O_x$ containing $x$ which is a subset of $\mathcal{N}_x$, i.e. $x\in\O_x\subset \mathcal{N}_x$ and $\O_x\in \mathscr{T}$.

Lemma 9.4 (Openness as Neighbourhood of All Members):

A set is open in a topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ iff it is a neighbourhood of all its member points.

Proof: Show Proof

If a set $\O$ is a neighbourhood of all its points, then $\forall x\in\O\,\exists\,\O_x\subset \O$ where $\O_x\in\mathscr{T}$. It is then readily shown that $\O=\bigcup\limits_{x\in\O} \O_x$ since $y\in\O\Rightarrow y\in \O_y\subset \bigcup\limits_{x\in\O} \O_x$, so $\O\subseteq \bigcup\limits_{x\in\O} \O_x$, on the other hand $\O_x\subseteq\O\,\forall\,x\in \O\Rightarrow\bigcup\limits_{x\in\O} \O_x \subseteq \O$. So $\O=\bigcup\limits_{x\in\O} \O_x$ is open, by topology axiom 2 above. Conversely, if a set is open, then $\forall x\in\O$, $\O$ is a neighbourhood of $x$ because $\O$ is open and also contains $x$.$\qquad\square$

Definition 9.5 (Base of a Topology):

For a topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ we call a collection of open sets $\mathscr{B}=\left\{\O_\omega\right\}_{\omega\in\Omega},\,\O_\omega\in\mathscr{T}\,\forall\omega\in\Omega$ indexed by some arbitrary index set $\Omega$ a base for $\mathscr{T}$ if every member of $\mathscr{T}$ can be derived from members of $\mathscr{B}$ by a finite sequence of abritrary union operations or finite intersection operations.

Otherwise, given any collection of subsets $\{\O_\omega\}_{\omega\in\Omega}\subset \mathbb{X}$ of a space $\mathbb{X}$, we can define a topology as the smallest topology containing the collection, i.e. the set of all members in the collection together with all arbitrary unions of its members and finite intersections of its members.

We are now ready to define the topology for our Lie groups.

Definition 9.6 (Topology of a Lie Group):

A base of the group topology for a connected Lie group $\G$ is the set

\begin{equation} \label{LieGroupTopologyBaseDefinition_1} \left\{\O_\gamma|\, \gamma\in\G;\,\O_\gamma \subseteq \gamma\,\Nid;\,\lambda\left(\gamma^{-1}\,\O_\gamma\right)\;\mathrm{open\,in\,}\V\subset\mathcal{R}^N\right\} \end{equation}

Alternatively, we can define the base of the group topology for a connected Lie group $\G$ to be the collection of all sets of the form $\gamma\,\exp(\O)$, where $\gamma\in\G$ and $\O\subseteq\g$ is open in the Lie algebra $\g$ when the latter is kitted with the wonted topology for $\R^N$.

An equivalent definition arises from defining the complete system of neighbourhoods of all members of $\G$:

Definition 9.7 (System of Neighbourhoods for a Lie Group Member):

A set $\mathcal{G}_\gamma$ is a neighbourhood of $\gamma \in \G$ iff it contains a set of the form $\gamma \,\K$ where $\K$ is a nucleus, i.e. $\id\in \K \subset \Nid$, $\Or\in \lambda(\mathcal{G})\subseteq \V$ and $\lambda(\mathcal{G})$ is a neighbourhood of $\Or \in \R^N$.

It should be clear that this definition is equivalent if we replace left-translated sets $\gamma \,\mathcal{G} $ with right-translated ones (although the individual neighbourhoods are transformed; we must replace $\mathcal{G} $ by $\gamma \,\mathcal{G} \,\gamma^{-1}$ in the above).

SkullWarning: It must be heeded that the topology defined above is the only one defined on the group $\G$ in this general Lie theory setting. Linear groups (matrix Lie groups) also have a topology defined on them through the metric $d(m_1,\,m_2)=\left\|m_1-m_2\right\|$ which may or may not be the same as the one just defined. On the one hand, this second topology defined in terms of the metric $d(m_1,\,m_2)$ on linear groups can make discussion of such groups easier, but it can also lead to mistakes when the topologies differ. We shall discuss this point more when we talk about the irrational angle helix one-parameter group as well as so called “virtual” Lie subgroups.

Definition 9.8 (Symmetric Neighbourhood of Identity):

A symmetric neighbourhood $\mathcal{S}\subset\Nid$ of the identity $\id$ in a connected Lie group is one for which every member has its inverse in the same neighbourhood: $\gamma\in\mathcal{S}\Rightarrow \gamma^{-1}\in\mathcal{S}$.

Lemma 9.9 (Existence of Symmetric Neighbourhoods)

Every neighbourhood $\K$ of $\id\in\G$ in a connected Lie group $\G$ contains a symmetric neighbourhood $\mathcal{S}$ of $\id$, i.e. $\id\in\mathcal{S}\subset \K$.

Proof : Show Proof

We can construct a symmetric neigbourhood in the Lie group $\G$ very readily as follows. With the notation of Theorem 3.21 we consider two sets:

\begin{equation} \label{ExistenceOfSymmetricNeighbourhoodLemma_1} \begin{array}{lcl} \O_1(\tau_0) &=& \left\{\prod\limits_{j=1}^N \sigma_j(x_j):\,-\tau_0\leq x_j\leq\tau_0\right\}\\ \O_2(\tau_0) &=& \left\{\prod\limits_{j=1}^N \sigma_{N-j}^{-1}(x_{N-j}):\,-\tau_0\leq x_j\leq\tau_0\right\} \end{array} \end{equation}

By the method of proof for Theorem 3.21, we see that both of these are open neighbourhoods of the identity for some small enough $\tau_0>0$; to apply the method to the second set, take heed that, by the Group Product Continuity Axiom 3 the Nontrivial Continuity Axiom 4, $\sigma_j(\tau)^{-1}\in\Nid$ and defines a $C^1$ path for $|\tau|<\tau_0$ for some $\tau_0>0$, then note that the tangent to the path $\sigma_j^{-1}$ at the identity is $-X_j$ when the tangent to $\sigma_j$ is $X_j$, so that the $x_j$ define valid second kind canonical co-ordinates in $\O_2(\tau_0)$.

The open set $\O_2(\tau_0)$ is simply the set of inverses of the members of $\O_1(\tau_0)$. Moreover, for any nucleus $\K\subseteq\Nid$ contained within $\Nid$, we can choose $\tau_0>0$ small enough that $\O_1(\tau_0)\subset\K$ and $\O_2(\tau_0)\subset\K$. So then the set $\mathcal{S} = \O_1(\tau_0)\cup \O_2(\tau_0)$ is the sought after symmetric neighbourhood of the identity, for some small enough $\tau_0 > 0$ and an open set of this form is contained within any nucleus.$\qquad\square$

Definition 9.10 (Hausdorff Space):

A topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is called Hausdorff iff for every pair of distinct points $x,\,y\in \mathbb{X}$ (i.e. $x\neq y$) there are two neighbourhoods $\mathcal{N}_x,\,\mathcal{N}_y$ of $x$ and $y$, respectively, that do not intersect ($\mathcal{N}_x\cap \mathcal{N}_y=\emptyset$). $x$ and $y$ are said to be separated or sundered by $\mathcal{N}_x$ and $\mathcal{N}_y$.

Many spaces of importance to mathematics are Hausdorff: $\R^N,\,\mathbb{C}^N$ are Hausdorff, as are all open subsets thereof. The idea is a generalisation of the metric space axiom for the distance function that $d(x,\,y) =0$ only if $x=y$.

Theorem 9.11 (All Connected Lie Groups are Hausdorff):

A Connected Lie Group $(\G,\,\bullet)$ is Hausdorff.

Proof: Show Proof

We know that there is some symmetric nucleus $\mathcal{S}$. With this nucleus we reason as follows. Suppose $\alpha,\,\beta\in\G$. If $(\alpha \mathcal{S}) \cap (\beta\mathcal{S}) = \emptyset$ then $\alpha \mathcal{S},\, \beta\mathcal{S}$ are two neighbourhoods of $\alpha,\,\beta$ respectively that sunder $\alpha$ and $\beta$ and we are done.

So instead now assume $(\alpha \mathcal{S}) \cap (\beta\mathcal{S}) \neq \emptyset$ and that $\gamma \in (\alpha \mathcal{S}) \cap (\beta\mathcal{S})$. So $\exists \zeta_\alpha\in \mathcal{S})$ such that $\alpha\,\zeta_\alpha = \gamma$ (since $\gamma\in \alpha \mathcal{S}$) and also $\exists \zeta_\beta\in \mathcal{S})$ such that $\beta\,\zeta_\beta = \gamma$ (since $\gamma\in \beta \mathcal{S}$). Therefore, since $\mathcal{S}$ is symmetric, there are $\zeta_\alpha^{-1},\,\zeta_\beta^{-1}\in\mathcal{S}$ such that $\gamma \,\zeta_\alpha^{-1} = \alpha$ and $\gamma \,\zeta_\beta^{-1} = \beta$. We look at the images $\lambda(\gamma^{-1}\,\alpha) = \lambda(\zeta_\alpha^{-1}), \lambda(\gamma^{-1}\,\beta) = \lambda(\zeta_\beta^{-1})\in\V$; $\V$ is Hausdorff, so there are sets $\O_\alpha,\,\O_\beta\subset\V$ where $\O_\alpha\cap\O_\beta=\emptyset$ and $\lambda(\gamma^{-1}\,\alpha) = \lambda(\zeta_\alpha^{-1})\in\O_\alpha,\,\lambda(\gamma^{-1}\,\beta)= \lambda(\zeta_\beta^{-1})\in\O_\beta$. The sets $\gamma \lambda^{-1}(\O_\alpha),\,\gamma \lambda^{-1}(\O_\beta)$ are thus nonintersecting open neighbourhoods of $\alpha$ and $\beta$ respectively: these open sets sunder $\alpha$ and $\beta$, thus proving that $\G$ is Hausdorff.$\quad\square$

Definition 9.12 (Closedness):

A subset of a topological space $\mathbb{X}$ is called closed iff its complement is open in that space.

There is a more practical, equivalent notion of closedness and that is that of “containing all its limit points”.

Definition 9.13 (Limit Point and Limit):

A point $x_\infty$ in a topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is said to be a limit point of a subset $\U\subset\mathbb{X}$ iff every neighbourhood of $x_\infty$ contains points of $\U$. A sequence $\{x_n\}_{n=1}^\infty\subset\mathbb{X}$ of points is said to converge to the limit $x_\infty\in\mathbb{X}$ iff, for every neighbourhood $\mathcal{N}$ of $x_\infty$ we can find $N(\mathcal{N})\in\mathbb{N}\,\ni\,x_n\in\mathcal{N}\,\forall\,n>N(\mathcal{N})$ and we write $\lim\limits_{n\to\infty} x_n=x_\infty$.

Lemma 9.14 (Closedness as Containing All Limit Points):

A subset $\U\subset \mathbb{X}$ in a topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is closed if and only if it contains all its limit points.

Proof: Show Proof

Suppose $\U$ is closed then $\V = \mathbb{X}\setminus\U$ is open and therefore is a neighbourhood of all its points. Therefore, any $x\in\V$ has a neighbourhood $\mathcal{N}_x\subseteq\V$, i.e. $\mathcal{N}_x$ contains no points of $\U = \mathbb{X}\setminus\V$ and so $x$ cannot be a limit point of $\U$, because otherwise $\mathcal{N}_x$ would contain points of $\U$. Therefore any limit point of $\U$ must be contained in $\U$. Conversely, suppose $\U$ contains all its limit points. Then for every $x\in\V = \mathbb{X}\setminus\U$ must have at least one neighbourhood $\mathcal{N}_x\subseteq\V$ without points in $\U$, otherwise $x$ would be a limit point of $\U$. Therefore, there is an open set $\O_x\subset\V$ for every $x\in\V$ and so $\bigcup\limits_{x\in\V}\O_x\subseteq\V$. But certainly $\V\subseteq\bigcup\limits_{x\in\V}\O_x$, thus $\V=\bigcup\limits_{x\in\V}\O_x$, so $\V$ is a union of open sets, hence open, hence its complement $\U$ is closed.$\qquad\square$

Definition 9.15 (Continuity):

A function $\varphi:\mathbb{X}\to \mathbb{Y}$ between two topological spaces $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ is called continuous at a point $x\in \mathbb{X}$ if, for every neighbourhood $N_y$ of $y = \varphi(x)\in \mathbb{Y}$ there is a neighbourhood $\mathcal{N}_x$ of $x$ such that $\varphi\left(\mathcal{N}_x\right)\subset \mathcal{N}_y$.

The definition is the same if we replace neighbourhoods $\mathcal{N}_x$ and $\mathcal{N}_y$ as open neighbourhoods $\O_x\in\mathscr{T}_X,\,\O_y\in\mathscr{T}_Y$ of $x$ and $y$, i.e. the two versions of the definition can be derived from each other. Witness that this definition becomes the $\epsilon,\,\delta$ definition of continuity in metric spaces.

Take heed that there are the notions of continuity at a lone point $x\in\mathbb{X}$ as in the definition above as well as continuity in a set $\mathcal{A}\subseteq\mathbb{X}$, open or otherwise, where $\varphi:\mathcal{A}\subseteq\mathbb{X}\to \mathcal{B}\subseteq{Y}$ is said to be continuous over $\mathcal{A}\subseteq\mathbb{X}$ or simply “continuous” without qualifiers iff it is continuous at each and every point $x\in\mathcal{A}$.

The order of taking a limit and the application of a continuous map can be freely swapped.

Lemma 9.16 (Switching Order of Limit and Continuous Map):

Let $\varphi:\mathbb{X}\to \mathbb{Y}$ be a continuous map between two topological spaces $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ and $\{x_n\}_{n=1}^\infty$ a convergent sequence in $\mathbb{X}$ with $\lim\limits_{n\to\infty}x_n = x_\infty$. Then $\{\varphi(x_n)\}_{n=1}^\infty$ is a convergent sequence and $\lim\limits_{n\to\infty}\varphi(x_n) = \varphi(x_\infty)$

Proof: Show Proof

Let $\O_Y\subset\mathbb{Y}$ be any neighbourhood of $\varphi(x_\infty)$; since $\varphi$ is continuous, there is a neighbourhood $\O_X\subset\mathbb{X}$ of $x_\infty$ and, since $x_n$ converges to $x_\infty$, there is an $N(\O_X)\in\mathbb{N}\,\ni\,x_n\in\O_X\,\forall\,n>N(\O_X)$, i.e. we get $\varphi(x_n)\in\O_Y\,\forall\,n>N(\O_X)$. Thus, by definition, this means that $\varphi(x_n)$ converges to $\varphi(x_\infty)$.$\qquad\square$

Definition 9.17 (Open Map):

A function$\varphi:\mathbb{X}\to \mathbb{Y}$ between two topological spaces $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ is called open iff it maps open sets to open sets i.e.

$$\mathcal{A}\in\mathscr{T}_X \Rightarrow \varphi(\mathcal{A})\in\mathscr{T}_Y$$

Lemma 9.18 (Openness of Continuous Preimages):

Let $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ be topological spaces and $\varphi:\U\subset\mathbb{X}\to \mathbb{Y}$ be continuous throughout its domain $\U$. Then the preimage or inverse image $\varphi^{-1}(\O_Y) = \{x\in\mathbb{X}|\,\varphi(x)\in\O_Y\}$ of any open set $\O_Y\in\mathscr{T}_Y$ is open in $\mathscr{T}_Y$.

Proof: Show Proof

Since $\O_Y$ is open, thus a neighbourhood of all its points, $\O_Y = \bigcup\limits_{y\in \O_Y} \O_y$, where $\O_y\subset\O_Y$ is an open neighbourhood of $y$. Therefore, by continuity of $\varphi$, if $\varphi(x)=y$, then there is an open neighbourhood $\O_x$ of $x$ such that $\varphi(\O_x)\subseteq \O_y$. Since $\O_y\subseteq \O_Y$, we have $\varphi(\O_x)\subseteq\O_Y, \,\forall\,x\in\varphi^{-1}(\O_Y)$, so that $\varphi(\O_x)\subseteq \O_Y\,\forall\,x\in\varphi^{-1}(\O_Y)$ so that $\varphi^{-1}(\O_Y)$ is a neighbourhood of all its points, hence open. $\qquad\square$

Lemma 9.19 (Inverse of a Continuous Map is Open):

Let $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ be topological spaces and $varphi:\U\subset\mathbb{X}\to \V\subset\mathbb{Y}$ be continuous throughout its domain $\U$ and be bijective (onto $\V$ with the inverse map $\varphi^{-1}$ defined. Then $\varphi^{-1}:\V\to\U$ is open.

Proof: Clear by application of Lemma 9.18, which shows that $\varphi^{-1}(\O)$ is open for every $\O\in\mathscr{T}_Y$ open in $\mathbb{Y}$.

Definition 9.20 (Homeomorphism):

A function $\varphi:\mathbb{X}\to \mathbb{Y}$ between two topological spaces $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ is called a homeomorphism if $\varphi:$ has an inverse, i.e. $\varphi:$ is bijective, and both $\varphi:$ and $\varphi:^{-1}$ is continuous. The two topological spaces are then said to be homeomorphic.

Theorem 9.21 (Homeomorphisms Preserve Topology)

A homeomorphism $\varphi:\mathbb{X}\to \mathbb{Y}$ between two topological spaces $\left(\mathbb{X},\,\mathscr{T}_X\right)$ and $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ is a one-to-one, onto correspondence between all the open sets in each topology $\mathscr{T}_X,\,\mathscr{T}_Y$ and thus, through $\varphi$, the two spaces and their respective topologies can abstractly be thought of as equivalent.

Proof: Show Proof

By applying Lemma 9.18 to both $\varphi$ and $\varphi^{-1}$, $\varphi$ maps open set $\O_X\in\mathscr{T}_X$ to an open set $\O_Y\in\mathscr{T}_Y$ (apply Lemma 9.18 to $\varphi^{-1}$ thus showing ${\varphi^{-1}}^{-1}(\O_X)$), and contrariwise, $\varphi^{-1}$ maps the open set $\O_Y\in\mathscr{T}_Y$ precisely to open set $\O_X\in\mathscr{T}_X$ (apply Lemma 9.18 to $\varphi$ thus showing $\varphi^{-1}(\O_Y)$ is open).$\qquad\square$

Lemma 9.22 (Group Operations are Homeomorphisms):

For a connected Lie group, the maps, all $\G\to\G$, $\zeta\mapsto\zeta\,\gamma$ and $\zeta\mapsto\gamma\,\zeta$ and $\zeta\mapsto\zeta^{-1}$ are all homeomorphisms.

Proof: Show Proof

We have seen (Lemma 3.17) that the group product and inverse are both $C^1$, thus $C^0$ (continuous). $\zeta\mapsto\gamma\,\zeta$ is, by the group laws, one to one and onto; it has the inverse map $\zeta\mapsto\gamma^{-1}\,\zeta$, both of which are continuous over the whole group $\G$, hence both are homeomorphisms. The inverse $\zeta\mapsto\zeta^{-1}$ being involutive (self inverse), it is thus automatically a homeomorphism. $\qquad\square$

We now look at a very important property of the topology of a connected Lie group.

Lemma 9.23 (Consistency of Openness and Neighbourhood):

Given a connected Lie group $\G$, let $\gamma_1\,\K$, $\gamma_2\,\K$ be two translated versions of an open nucleus i.e. $\lambda(\K)\subseteq \V$ is an open neighbourhood of the origin $\Or\in\V$ such that $\gamma_1 \,\K \cap \gamma_2\,\K\neq\emptyset$ then:

  1. Any set $\O$ that is open as a subset of $\gamma_1\,\K$ is also open as a subset of $\gamma_2\,\K$ and contrariwise, i.e. both $\lambda(\gamma_1^{-1}\,\O)$ and $\lambda(\gamma_2^{-1}\,\O)$ are open in $\V\subset\R^N$;
  2. The intersection $\gamma_1 \,\K \cap \gamma_2\,\K$ itseld is open as a subset of $\gamma_1\,\K$ and also as a subset of a subset of $\gamma_2\,\K$ i.e. $\lambda(\gamma_1^{-1}\,(\gamma_1 \,\K \cap \gamma_2\,\K))$ and $\lambda(\gamma_2^{-1}\,(\gamma_1 \,\K \cap \gamma_2\,\K))$ are both open in $\V\subset\R^N$.

Proof: Show Proof

Without loss of generalness we can assume $\gamma_1 = \id,\, \gamma_2=\gamma$. Then any open $\O\subseteq\K\cap\gamma\,\K$ that is open “as reckonned from $\K$” (in the sense that $\lambda(\O)$ is open in$\V$) is also open “as reckonned from $\gamma\,\K$” (in the sense that $\lambda(\gamma^{-1}\O)$ is open in$\V$), because $\zeta\mapsto\gamma^{-1}\zeta$ is a local homeomorphism. The converse also applies; this proves item 1. above.

Now suppose $\omega\in\K\cap\gamma\,\K$. Then $\omega\in\gamma\,\K\,\Leftrightarrow\,\gamma^{-1}\,\omega\in\K$, so, by Theorem 3.20, there is an $R_1$ such that:

\begin{equation} \label{ConsistencyOfOpennessAndNeighbourhoodLemma_1} \begin{array}{rrclcl} &\gamma^{-1}\,\omega\in\mathcal{B}(\gamma^{-1}\,\omega,\,R_1)&=&\{\zeta|\,\left\|\lambda(\zeta) – \lambda(\gamma^{-1}\,\omega)\right\|<R_1\}&\subset&\K\\ \Leftrightarrow&\omega&\in&\gamma\,\mathcal{B}(\gamma^{-1}\,\omega\,R_1)&\subset&\gamma\,\K \end{array} \end{equation}

Now, since $\omega\in\K$, by Theorem 3.20, there is an open ball:

\begin{equation} \label{ConsistencyOfOpennessAndNeighbourhoodLemma_2} \mathcal{B}(\omega,\,R_2)\subset\K \end{equation}

Because $\zeta\mapsto\gamma\,\zeta$ is a local homeomorphism (both $\zeta\mapsto\gamma\,\zeta$ and its inverse $\zeta\mapsto\gamma^{-1}\,\zeta$ are continuous when both the maps input and output lie inside $\K$), $\zeta\mapsto \gamma\,\zeta$ is continuous at $\zeta = \gamma^{-1}\,\omega$, so that we can, by dent of this continuity, always choose $R_1>0$ small enough that:

\begin{equation} \label{ConsistencyOfOpennessAndNeighbourhoodLemma_3} \gamma\,\mathcal{B}(\gamma^{-1}\,\omega,\,R_1)\subset\mathcal{B}(\omega,\,R_2)\subset\K \end{equation}

We now assume that we do so. Therefore, by $\eqref{ConsistencyOfOpennessAndNeighbourhoodLemma_1}$ and $\eqref{ConsistencyOfOpennessAndNeighbourhoodLemma_3}$:

\begin{equation} \label{ConsistencyOfOpennessAndNeighbourhoodLemma_4} \omega\in\gamma\,\mathcal{B}(\gamma^{-1}\,\omega,\,R_1)\subset\K\cap\gamma\,\K \end{equation}

Furthermore, the homeomorphism means that $\zeta\mapsto \gamma\,\zeta$ is an open map, so that $\gamma\,\mathcal{B}(\gamma^{-1}\,\omega,\,R_1)$, the image of an open set under $\zeta\mapsto \gamma\,\zeta$, is also open. So now we have shown that $\gamma\,\mathcal{B}(\gamma^{-1}\,\omega,\,R_1)$ is an open neighbourhood of $\omega$ contained wholly within $\K\cap\gamma\,\K$ ($\eqref{ConsistencyOfOpennessAndNeighbourhoodLemma_4}$) for an arbitrary $\omega \in \K\cap\gamma\,\K$, therefore $\K\cap\gamma\,\K$ is a neighbourhood of all its points and is therefore an open set “as reckonned wholly by the topology defined on $\K$”. The last statement simply means that $\lambda(\K\cap\gamma\,\K)\subset\V$ is an open subset of $\R^N$.

By doing the above calculation for the set $\gamma^{-1}\,\K\cap\K$ and then translating the results to $\gamma\,(\gamma^{-1}\,\K\cap\K) = \K\cap\gamma\,\K$, we analogously show that $\K\cap\gamma\,\K$ is an open set “as reckonned wholly by the topology defined on $\gamma\,\K$”, i.e. $\lambda(\gamma^{-1}\, (\K\cap\gamma\,\K))\subset\V$ is an open subset of $\R^N$. $\qquad\square$

Definition 9.24 (Compact Set)

A collection of open sets $\{\O_\omega\}_{\omega}$, where $\O_\omega\in\mathscr{T}\,\forall\omega\in\Omega$ (here $\Omega$ is an arbitrary index set) in a topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is called an open covering for a subset $\U\subseteq\mathbb{X}$ if $\U\subset\bigcup\limits_{\omega\in\Omega} \O_\omega$. A subset $\U\subseteq\mathbb{X}$ is called compact if every one of its open coverings has a finite open subcovering, i.e.

\begin{equation} \label{CompactSetDefinition_1} \U\subset\bigcup\limits_{\omega\in\Omega} \O_\omega;\,\O_\omega\in\mathscr{T}\,\forall\omega\in\Omega\,\Rightarrow\,\U\subset\bigcup\limits_{k=1}^N \O_{\omega_k} \end{equation}

for some finite subset $\{\omega_1,\,\omega_2,\,\cdots,\,\omega_N\}\subset \Omega$ of indices.

The notion of compactness has a long and interesting history. It was essentially begotten of two notions:

  1. The Bolzano-Weierstrass property of a set whereby any infinite sequence in a set with this property must have at least one limit point; for $\R^N$, a subset has this property if and only if it is closed and bounded (the Bolzano-Weierstrass theorem); this is the notion of sequential compactness;
  2. The Heine-Borel property of a set, which is essentially our modern idea of compactness. This was found by Eduard Heine to be essential to the proof that a continuous function defined on a closed and bounded interval is indeed uniformly continuous

The modern notion of compactness for a general topological space was formulated by the Russian point set topology school under the leadership of Pavel Alexandrov and Pavel Urysohn in the late 1920s, and the sequential compactness notion was shown to follow from the general one under appropriate and reasonable conditions. Many papers right through the 1930s still meant “having the Bolzano-Weierstrass property” by the word “compact”; van der Waerden and Freudenthal are two authors I have personally found to use the word in this way. For Lie theory, either notion works as well as the other in most cases. Aside from the notion of compactness, two properties of compactness that I shall need are:

Lemma 9.25 (Continuous Image of Compact Sets)

The continuous image of a compact set is compact, i.e. let $\left(\mathbb{X},\,\mathscr{T}_X\right)$, $\left(\mathbb{Y},\,\mathscr{T}_Y\right)$ be two topological spaces and $\varphi:\W\subset(\mathbb{X}\to\mathbb{Y}$ a function continuous over its whole domain. Then $\varphi(\U)$ is compact whenever $\U$ is compact.

Proof: Show Proof

Let $\{\O_\omega\}_{\omega\in\Omega}$ be an open cover for $\varphi(\U)$ so that $\varphi(\U)=\bigcup\limits_{\omega\in\Omega}\O_\omega$. Since $\varphi$ is continuous, by Lemma 9.18, the preimage $\varphi^{-1}(\O_\omega)\in\mathscr{T}_X$ is open as well as $\O_\omega\in\mathscr{T}_Y$. Therefore:

\begin{equation} \label{ContinuousImageOfCompactSetCompactLemma_1} \U =\bigcup\limits_{\omega\in\Omega} \varphi^{-1}(\O_\omega) \end{equation}

is an open cover for $\U$, which is compact, by hypothesis. Therefore we have $\U =\bigcup\limits_{k=1}^N\varphi^{-1}(\O_{\omega_k})$ so that $\varphi(\U) = \bigcup\limits_{k=1}^N \O_{\omega_k}$ and so we have found a finite open subcover for $\varphi(\U)$, so $\varphi(\U)$ is compact. $\qquad\square$

The next theorem is very standard, but one of the most beautiful all of mathematics. Let’s look at the proof first and then I’ll have quite a few comments on the intuition behind it all, and talk about why I think this is a truly stunning theorem.

Definition 9.26 (Heine-Borel Theorem)

A subset $\U\subset\R^N$ is compact if and only if it is closed and bounded.

Proof: Show Proof

[Stillwell, 2010] has a beautifully written proof in §8.4 using the standard bisection argument. He is my go at it: it is somewhat more abstract than Stillwell’s because I want to illustrate how tightly and necessarily linked other seemingly different notions that this theorem binds together.

Firstly for the $\Rightarrow$, so $\U\subset\R^N$ is closed and bounded. Suppose that $\U\subset\R^N$ has an infinite open cover $\bigcup\limits_{\omega\in\Omega} \O_\omega$ which has no finite open subcover. Since $\U\subset\R^N$ is bounded, we can inscribe it within a closed hypercube $\mathcal{H}_0$ of sidelength $L_0$. We now bi$^N$-sect this hypercube, i.e. split it into $2^N$ equal closed hypercubes of sidelength $L_0/2=L_1$, such that the only intersection between the sub-hypercubes is their common faces. By hypothesis, at least one of these sub-hypercubes can only be covered by a needfully infinite subcover of the $\O_\omega$. We then repeat this argument iteratively, yielding a nested sequence of closed hypercubes $\mathcal{H}_0\supset\mathcal{H}_1\supset\mathcal{H}_2\supset\,\cdots$ of sidelengths $L_0,\,L_0/2,\,L_0/4,\,\cdots$, all of which a covered by a needfully infinite subset of the $\O_\omega$. Taking points $X_0,\,X_1\,X_2\,\cdots$ on the common faces such that $X_j$ is a point on the face of the $j^{th}$ hypercube, $\{X_j\}_{j=0}^\infty$ is a convergent sequence (indeed $\left\|X_j – X_{j+1}\right\|\leq2^{-j}\,\sqrt{N}\,L_0$). Since $\U$ is by hypothesis closed, it contains all its limit points, therefore $x_j\to x_\infty\in\U$. But now $x_\infty$ must belong to at least one $\O_{\omega_0}$ of the $\O_\omega$. But $\O_{\omega_0}$ is open, therefore it contains a hypercube $\mathcal{H}_\infty$ of some nonzero length $L_\infty>0$ which contains at least one $\mathcal{H}_{j_0}$ of the $\mathcal{H}_j$ and therefore contains $\mathcal{H}_{j_0+1},\,\mathcal{H}_{j_0+2},\,\cdots$ because there is an arbitrarily small sidelength hypercube amongst the $\mathcal{H}_j$. So $O_{\omega_0}$ is an open cover of $\mathcal{H}_{j_0+1},\,\mathcal{H}_{j_0+2},\,\cdots$ and it is a finite open cover, gainsaying our initial assumption that none of the sub-hypercubes has a finite open subcover from the $\O_{\omega}$. Witness how directly this proof follows from the bounded and closed hypotheses; the former allows the inscribing of $\U$ inside some hypercube, the latter shows that there must be a point common to all the nested sub-hypercubes and therefore an open set that covers at least some “tail” in the sequence of hypercubes.

Now for the $\Leftarrow$, so we assume that $\U\subset\R^N$ is compact. $\R^N$ itself is covered by a countably infinite cover of open hypercubes of unit sidelength spaced on a regular lattice with a half unit periodicity in all directions, so that they overlap, therefore this cover also covers $\U$. Therefore a finite number of these hypercubes cover $\U$, which means that there is a hypercube whose centre is of maximal distance $L$ from the origin. Therefore, all hypercubes are inside the set $\{x|\,\left\|x\right\|\leq L+\sqrt{N}\}$, and so $\U$ is bounded.

Now assume that $x_0$ is a limit point of $\U$ but that $x_0\not\in\U$, but that $\U$ is compact. For each $x\in\U$ we choose an open set $\O_x$ such that $x\in\O_x$ but $x_0\not\in\O_x$ and, moreover, there is another open set $\tilde{\O}_x$ such that $x\not\in\tilde{\O}_x$ but $x_0\in\tilde{\O}_x$; this can always be done because $\R^N$ is Hausdorff and, since $x_0\not\in\U$, $x$ and $x_0$ are always different points. Therefore:

\begin{equation} \label{HeineBorelTheorem_1} \U\subseteq \bigcup\limits_{x\in\U} \O_x;\;x_0\not\in \O_x,\forall\,x\in\U \end{equation}

So now we choose any finite subcover $\{\O_{x_j}\}_{j=1}^N$ and take heed that the limit point $x_0$ belongs to all the tilde-sets $x_0\in\tilde{\O}_{x_j}$, which tilde-sets are all outside the sets $\O_{x_j}$ and thus outside the subcover , i.e. $x_0\in\bigcap\limits_{j=1}^N\tilde{\O}_{x_j}$ but $\left(\bigcap\limits_{j=1}^N\tilde{\O}_{x_j}\right)\cap\left(\bigcup\limits_{j=1}^N \O_{x_j}\right)=\emptyset$. Given that the subcover is finite, and finite intersections of open sets are open, $\bigcap\limits_{j=1}^N\tilde{\O}_{x_j}$ is a neighbourhood of $x_0$. But $x_0$ is a limit point, therefore $\bigcap\limits_{j=1}^N\tilde{\O}_{x_j}$ must contain at least one point $y$ of $\U$, but $y$ cannot be in any of the finite $\{\O_{x_j}\}_{j=1}^N$ and so the latter cannot cover $\U$.$\qquad\square$

Take Heed: It’s worth noting that the Heine-Borel theorem does not need to call on the Axiom of Choice. One might think a countable axiom of choice needs to be invoked to pick out the sequence $X_0,\,X_1\,X_2\,\cdots$ from the nested hypercubes, but we can explicitly construct a choice function as follows. Choose a ray $\{X|\, \left<X,\,\hat{Y}\right> \geq 0\}\subset\R^N$ through the origin $\Or$ by choosing a unit vector $\hat{Y}\in\R^N$. Then we can choose $X_N$ to be the unique point achieving the maximum inner product in the hypercube $\mathcal{H}_N$, i.e. $\left<X_N,\,\hat{Y}\right> = \max\limits_{X\in\mathcal{H}_N} \left<X,\,\hat{Y}\right>$. In the finite dimensional $\R^N$, the famous simplex algorithm of Dantzig does the trick for explicitly finding $X_N$.

For me the Heine-Borel Theorem is quite stunning; it takes what seems to be a definition of unworkably high complexity and makes it, in Euclidean space, into a much more concrete and readily tested for definition. At first glance, the notion of compact is unworkably thorny: how on earth do we check that every open cover has or has not a finite subcover? The class of all possible subcovers for a given set would seem to be impossibly big: how could we run a finite sum covered tests on every class member? Not only does the Heine-Borel Theorem tame this fiendish complexity, it does so in a way that shows directly how tightly linked the two equivalent notions are. Boundedness lets us build a hypercube, closedness make sure that our hypercubes nested by recursive bisection have a limit point within the set in question: these two simple statements are really all there is to the “closed and bounded implies compact” statement. The converse statement is also readily derived from elementary topology. Many proofs of the Heine-Borel theorem call on the Cantor Intersection Theorem to show that there must be an $x_\infty$ inside all the nested hypercubes, but here I prefer to leave it more to first principles, resting on the abstract property of closedness: it is the closedness, i.e. inclusion of all limit points, that is crucial to the existence of at least one $x_\infty$ inside all the closed hypercubes. This calling on the abstract property of closedness is equivalent to the Cantor Intersection Theorem, so that this latter can be taken as proving that a hypercube with all its edges included is closed in the abstract sense.

Definition 9.27 (Relative Topology; Induced Topology; Inherited Topology)

Give a topological space $(\mathbb{X},\,\mathscr{R})$ and a subset $\U\subset\mathbb{X}$ we say the “relative topology” for $\U$ “inherited from” or “induced by” $\mathscr{T}$ is the topology $\mathscr{T}_\U$ where:

\begin{equation} \label{RelativeTopologyDefinition_1} \mathscr{T}_\U=\{\mathcal{A}\cap\U|\,\mathcal{A}\in\mathscr{T}\} \end{equation}

i.e. subsets of $\U$ are open in $\mathscr{T}_\U$ iff they are the intersection of some set open in $\mathscr{T}$ with the set $\U$.

SkullWarning: When we speak of Lie subgroups $\H\subset\G$ of a Lie group $\G$, the appropriate Lie group topology as defined above as being generated by all the sets that are open in each translated neighbourhood of the form $\gamma_\H\,\mathcal{N}_{\id,\H},\,\gamma_\H\in\H$ and $\mathcal{N}_{\id,\H} = \exp(\O_\h)$ for some open neighbourhood of $\Or\in\h$ in the Lie subalgebra $\h\subset\g$, then this is not in general the same concept as the relative topology on $\H$ inherited from the group topology of $\G$, where, of course, the group topology for $\G$ is generated by sets which are open in each translated neighbourhood of the form $\gamma_\G\,\mathcal{N}_{\id,\G},\,\gamma_\G\in\G$ and $\mathcal{N}_{\id,\G} = \exp(\O_\g)$ for some open neighbourhood of $\Or\in\g$ in the Lie algebra $\g$. The two concepts (Lie group topology of the Lie subgroup and the relative topology) are the same in the special case where the whole subgroup $\H\subset\G$ is a closed subset of $\G$ (closed, of course, in the group topology for the whole group $\G$). I shall illustrate this point at length when I talk about the Lie correspondence as well as when I talk about the irrational angle helix one-parameter group as well as so called “virtual” Lie subgroups.

Definition 9.28 (Connectedness):

A topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is called connected if it is not the union of two sundered nonempty open sets, i.e. $\mathbb{X}\neq\U\cup\V$ whenever both $\U,\,\V\in\mathscr{T}$ are open.

Lemma 9.29 (Connectedness Argument):

A topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is connected iff $\mathbb{X}$ and $\emptyset$ are the only members of $\mathscr{T}$ which are both open and closed at once.

Proof: Show Proof

If $\mathcal{A} \subset \mathbb{X}$ is both open and closed its complement $\mathbb{X} \setminus \mathcal{A}$ is also both open and closed, thus $\mathbb{X} = \mathcal{A} \cup \left(\mathbb{X} \setminus \mathcal{A}\right)$ is a union of disjoint nonempty open sets and is thus not connected unless either $\mathcal{A}$ or $\mathbb{X}\setminus \mathcal{A}$ is empty, i.e. $\mathcal{A} = \emptyset$ or $\mathbb{X}\setminus \mathcal{A} = \emptyset\Rightarrow \mathcal{A}=\mathbb{X}$. On the other hand, if $\mathbb{X}$ is connected, then $\mathbb{X}$ and $\emptyset$ are the only members of $\mathscr{T}$ which are both open and closed at once (by definition).$\quad\square$

The connectedness argument is surprisingly powerful; to illustrate this let us prove that a Topological Group Generated by Any Neighbourhood of the Identity.

Definition 9.30 (Topological Group)

A topological group is a group $(\G,\,\bullet)$ kitted with a topology $\mathscr{T} = \{\O_\omega\}_{\omega\in\Omega}$ of open sets $\O_\omega\subset\G$ such that the group operations $\zeta\mapsto\gamma\,\zeta$ and $\zeta\mapsto\zeta^{-1}$ are continuous with respect to the topology $\mathscr{T}$ for all $\zeta,\,\gamma\in\G$.

Theorem 9.31 (Topological Group Generated by Any Neighbourhood of the Identity):

Any connected topological group $\G$ is generated by any neighbourhood $\K$ of the identity (i.e. by any nucleus), i.e.:

\begin{equation} \label{TopologicalGroupNuclearGenerationTheorem_1} \G = \bigcup\limits_{k=1}^\infty \K^k \end{equation}

Intuitively: you can’t have a valid “neighbourhood” in the connected topological space without its containing “enough inverses” of its members to generate the whole group in this way. I originally read this a proof like this one in [Sagle and Walde], reworded it slightly and became adept at retelling it from memory because it is an canny little proof.

Proof: Show Proof

We consider the entity $\mathcal{Y} = \bigcup\limits_{k=1}^\infty \K^k$, the set of all finite products of members of $\K$. For any $\gamma \in \mathcal{Y}$ the map $\varphi_\gamma : \mathcal{Y} \to \mathcal{Y};\, \varphi_\gamma(x) = \gamma^{-1} x$ is continuous, thus the inverse image ${\varphi_\gamma}^{-1}\left(\K\right) = \varphi_{\gamma^{-1}}\left(\K\right) = \gamma \, \K$ of $\K$ contains an open neighbourhood $\O_\gamma \subset \gamma\,\K$ of $\gamma$, thus $\mathcal{Z} = \bigcup\limits_{\gamma \in \mathcal{Y}} \O_\gamma$ is open. Certainly $\mathcal{Y} \subseteq \mathcal{Z}$, but, since $\mathcal{Y}$ is the collection of all products of a finite number of members of $\K$, we have $\mathcal{Z} \subseteq \mathcal{Y}$ (since $\K$ contains only such finite products, then so too $\gamma\,\K$ only contain finite products), thus $\mathcal{Z} = \mathcal{Y}$ is open.

We now repeat the above reasoning for members of the set $\G \setminus \mathcal{Y}$, aside from that this time we choose a symmetric open set (by Lemma 9.9) $\mathcal{S}_\gamma \subset\K$ for every $\gamma\in\G \setminus \mathcal{Y}$ such that $\gamma\,\mathcal{S}_\gamma\subset \G \setminus \mathcal{Y}$; otherwise, if there were a $\zeta\in\mathcal{Y}\cap\gamma\,\mathcal{S}_\gamma$ then $\zeta \in \gamma\, \mathcal{S}_\gamma$ would be a finite product of members of $\K$, so $\gamma = \zeta\, \eta^{-1}$ where $\eta\in\mathcal{S}_\gamma$, so that $\eta^{-1}\in\mathcal{S}_\gamma\subset\K$ so that $\gamma$ would also be a finite product of members of $\K$, hence in $\mathcal{Y}$. So we have $\bigcup\limits_{\gamma\in\G \setminus \mathcal{Y}} \mathcal{S}_\gamma \subset \G \setminus \mathcal{Y}$ but certainly $\G \setminus \mathcal{Y}\subset\bigcup\limits_{\gamma\in\G \setminus \mathcal{Y}} \mathcal{S}_\gamma$, hence $\G \setminus \mathcal{Y}=\bigcup\limits_{\gamma\in\G \setminus \mathcal{Y}} \mathcal{S}_\gamma$, a union of open sets, hence open. Therefore, the complement of $\mathcal{Y}$ is open, so $\mathcal{Y}$ is closed. thus $\mathcal{Y}$, being both open and closed, must be the whole (connected) space $\G$.$\qquad\square$

So now we see why I have been calling the objects defined by the five axioms “connected” Lie groups. Let’s explore the idea further.

Definition 9.32 (Path Connectedness):

A topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is called path-connected if, for every $x,\,y\in\mathbb{X}$ there is a continuous function $\varphi_{x,y}:[0,\,1]\to\mathbb{X}$ of one real variable such that $\varphi_{x,y}(0) = x;\,\varphi_{x,y}(1) = x$. We say that the $C^0$ path links $x$ and $y$. We can also refer to $C^k$-path-connectness, if we replace “continuous” (i.e. $C^0$) function by a stronger, $C^k$ criterion where $k\in\mathbb{N}$ or we speak of $C^\infty$- or $C^\omega$-path-connectedness.

Note that a subset of $\R^N$ is connected if and only if it is path connected.

Definition 9.33 (Arcwise Connectedness):

A topological space $\left(\mathbb{X},\,\mathscr{T}\right)$ is called arcwise-connected if, for every $x,\,y\in\mathbb{X}$ there is a continuous function $\varphi_{x,y}:[0,\,1]\to\mathbb{X}$ of one real variable with continuous inverse, i.e. a homeomorphism between the closed unit interval and a subset of $\left(\mathbb{X}$ (the arc) such that $\varphi_{x,y}(0) = x;\,\varphi_{x,y}(1) = x$. We say that the $C^0$ arc links $x$ and $y$. We can also refer to $C^k$-arcwise-connectness, if we replace “continuous” (i.e. $C^0$) function by a stronger, $C^k$ criterion where $k\in\mathbb{N}$ or we speak of $C^\infty$- or $C^\omega$-path-connectedness.

Clearly every $C^k$ arcwise connected space is also path connected. Any Hausdorff space that is $C^k$ path connected is also $C^k$ arcwise connected. Therefore these concepts are the same in Hausdorff spaces; in particular, they are the same in $\R^N$. So connected, path connected and arcwise connected are all the same concepts for subsets of $\R^N$.

Theorem 9.34 (Sameness of Connectedness Definitions):

In a mathematical system $(\G,\,\Nid,\,\V\subset\R,\,\lambda:\Nid\to\V)$ fulfilling axioms 1 through 4, but not needfully the Homgeneity Axiom 5 of the grounding axioms, and if the group topology is defined as above in Definition 9.6, the following are all equivalent:

  1. $\G$ is connected;
  2. $\G$ is generated by any nucleus;
  3. In geodesic co-ordinates, $\G$ is generated by $\exp(\g)$, where $\g$ is the Lie algebra of $\G$;
  4. $\G$ is $C^k$-path-connected, where $k=0,\,1,\,2,\,\cdots,\,\infty, \omega$;

Proof: Show Proof

Given that $\G$ fulfills Definition 9.30, $1.\Rightarrow2.$ is clear by Theorem 9.31. Now choose a $\g$-basis $\{\hat{X}_j\}_{j=1}^N$ and label $\K=\exp(\mathcal{N}_0)$ with geodesic oc-ordinates, where $\mathcal{N}_0$ is any neighbourhood of $\Or\in\g$. $2.$ then implies $\exp(\g)\subset\G$, and, since $\K$ generates $\G$, so too does $\exp(\g)$. So we have proven $2.\Rightarrow3.$. Now every group member can be written as $\prod\limits_{j=1}^M e^{X_j}$ by $3.$, where $X_j\in\g$. Then define $\sigma:[0,\,1]\to\g;\;\sigma(\tau) = \prod\limits_{j=1}^M e^{X_j}$; $\sigma$ is readily shown to be a $C^\omega$ function with $\G$ kitted with the group topology. Then a path of the form $\gamma\,\sigma$ links any $\gamma\in\G$ to any other member of $\G$ by a $C^\omega$ path. So $3.\Rightarrow4.$ is proven. Lastly, suppose $\sim1.$, i.e. that $\mathcal{G}=\U\cup \V$ where $\U\cap\V=\emptyset$ and both $\U,\,\V$ are open in the group topology. Now assume $4.$ and link $x\in\U,\,y\in\V$ by a $C^\omega$, thus $C^0$, path $\sigma:[0,\,1]\to\G$ with $\sigma(0)=x,\, \sigma(1)=y$. Moreover, consider the relative topology for the path $\sigma([0,\,1])$ inherited from the group topology of $\G$; clearly $\sigma([0,\,1]) = (\sigma([0,\,1])\cap \U) \cup (\sigma([0,\,1])\cap \V)$ and both $\sigma([0,\,1])\cap \U, \,\sigma([0,\,1])\cap \V$ are open in the relative topology, so $\sigma([0,\,1])$ is the union of two disjoint open sets in the relative topology. But $\sigma$ is continuous, so the preimages of these two disjoint open sets are also open, which means that $[0,\,1]$ would have to be the union of two disjoint open sets in the wonted topology for $\R$. This is clearly not so, so $\sim1.\,\Rightarrow\,\sim4.$, whence $4.\Rightarrow1.$.$\qquad\square$

Outside the field of Lie groups, path connectedness is in general a stronger condition than simply connected, i.e. every path connected topological space is connected, but the converse does not needfully hold. For Lie groups, however, the above theorem shows that the two concepts are the same and each the same as “being generated by any neighbourhood of the identity”.

We have seen two separate proofs (Lemma 9.9 and Lemma 3.12) that any nucleus contains a symmetric nucleus $\K_S$ (i.e. one whose every element’s inverse is also in the same nucleus) so that therefore the set $\bigcup\limits_{k=1}^\infty \K_S^k$ of all finite products of members of $\K_S$ is automatically a group, independently of the Homgeneity Axiom 5 and, moreover, $\bigcup\limits_{k=1}^\infty \mathcal{N}^k = \bigcup\limits_{k=1}^\infty \K_S^k$ for any set $\mathcal{N}\subset\Nid$ that contains a nucleus and thus a symmetric nucleus. Furthermore, everything that I have talked about and proven applies to mathematical systems fulfilling axioms 1 through 4, and we can re-interpret any use of the [axiom 5] as specialising our discussion to the identity-connected subgroup of a possibly more general group. Thus we are led to the following:

Definition 9.35 (General Lie Group)

A (general) Lie group is a group $(\G,\,\bullet)$ fulfilling:

  1. the Labeller Axiom 1;
  2. the Connectedness Axiom 2;
  3. the Group Product Continuity Axiom 3 and;
  4. the Nontrivial Continuity Axiom 4
  5. if $\gamma\in\G$ is not a finite product of members of $\Nid$, and if $\sigma:[-a,\,a]\to\G$ where $a>0$ is a $C^1$ path through $\Nid$ with $\sigma(0)=\id$, then there is a $b>0$ such that $\sigma_\gamma:[-b,\,b]\to\Nid;\,\sigma_\gamma(\tau) = \gamma\,\sigma(\tau)\,\gamma^{-1}$ is a $C^1$ path through $\Nid$.

The identity connected component $\G_\id$ of the group is given by $\G_\id = \bigcup\limits_{k=1}^\infty \K^k$, where $\K\subset\Nid$ is any nucleus. A connected Lie group further fulfills the Homgeneity Axiom 5. The group topology for the general Lie group $\G$ is the smallest topology containing the base:

\begin{equation} \label{GeneralLieGroupDefinition_1} \left\{\O_\gamma|\, \gamma\in\G;\,\O_\gamma \subseteq \gamma\,\Nid;\,\lambda\left(\gamma^{-1}\,\O_\gamma\right)\;\mathrm{open\,in\,}\V = \lambda(\Nid)\subset\mathcal{R}^N\right\} \end{equation}

which topology makes the group operations continuous, i.e. a Lie group is a topological group when kitted with this topology.

$\G_\id$ is naturally a connected Lie group and, from point 5. above, it follows that $\G_\id$ is a normal subgroup of $\G$, i.e. $\gamma\,\G_id\,\gamma^{-1}=\G_{id}$.

So now I have defined a general Lie group and its topology thoroughly and studied most of a Lie group’s elementary properties. It is now time to look at a Lie group’s wonted modern definition and show it is equivalent to what I have so far studied. With the theoretical background I have worked through, this is now an easy task and, moreover, it is easy to show that every mathematical system defined by the Lie group’s modern definition is also a mathematical system fulfilling Definition 9.35, so that Definition 9.35 and the modern Lie group definition are logically the same thing. Naturally, any system fulfilling the axioms of Definition 9.35 in the restricted sense so that the Full Tangent Space Dimension Axiom 6 is not fulfilled can be made into a mathematical system fulfilling the Full Tangent Space Dimension Axiom 6 by Theorem 6.13 and Theorem 6.14.


  1. Mendelson, B, “Introduction to Topology“, Third Edition, Dover. Definiton 3.2, Chapter 3, “Topological Spaces”
  2. Stillwell, John, “Näive Lie Theory“, Springer, 2010
  3. Sagle, A. A. and Walde, R. E., “Introduction to Lie Groups and Lie Algebras“, Academic Press, New York, 1973. §3.3