Chapter 12: The Lie Correspondence: Lie’s Analogies with Galois Theory

In the early days of Lie theory after 1870, Sophus Lie imagined a Galois theory for continuous groups and strove for a theory along these lines. Indeed there is a beautiful bijection known as the Lie Correspondence that we are about to study that is in many ways a realisation of Lie’s original forethought, although it could be said that Lie theory is somewhat more complicated than Galois theory. The Lie algebra, as a linear space, simplifies the study of some properties of the nonlinear Lie group it belongs to and

Theorem 12.1 (Lie Correspondence)

For a connected Lie group $\G$, there is the one-to-one, onto correspondence between (i) the connected Lie subgroups $\H_\omega\subseteq\G$ of $\G$ which are also immersions in $\G$ (i.e. included by a differentiable map $\H_\omega \hookrightarrow \G$) and (ii) the Lie subalgebras $\h_\omega = \operatorname{Lie}(\H_\omega)\subseteq\g$ of its Lie algebra $\g = \operatorname{Lie}(\G)$. (to be proven)

just as the normal subgroups of a field extension’s Galois group correspond to one to one with all the extension fields contained within the particular extension, and we can therefore study extension fields by studying the Galois group of automorphisms on them. In the other direction, the Lie group and all of its connected Lie subgroups act on their own Lie algebras through their own adjoint representation, just as the Galois group of automorphisms act on the extension fields they are used to study. For a thorough introduction to Galois theory and the Galois correspondence, see [Artin], but read the most excellent summary [Stillwell] gives of the subject first.

The Forwards Lie Correspondence

Hereafter I shall call the functor $\Gamma$ defined by (Campbell Baker Hausdorff Series were universally convergent, the proof of this assertion would be simple and elegant indeed: we could then pull all group multiplications in the finite products between members of $\K$ back into the Lie algebra $\h$ and the logarithm of these products could never give anything outside $\h$, thanks to the Campbell Baker Hausdorff Theorem. Unfortunately, however, the CBH series diverges unless the logarithms $X$ and $Y$ of the Lie group members in question belong to a small enough neighbourhood of $\Or$ in $\g$, e.g. one defined by Lemma 8.4. The basic idea of the proof of the Lie correspondence is still the CBH formula, but we must now address the technicalities that arise when products are outside this neighbourhood of convergence and prove that we still don’t get anything new (i.e. not within $\exp(\h)$) if we follow a $C^1$ path in $\Gamma(\h)$ outside the neighbourhood of convergence for the CBH series, wander arbitrarily through $\Gamma(\h)$ and then come back inside $\g$.

The Irrational Slope Subgroup of the Torus

We shall soon prove the “Forwards Lie Correspondence” and in readiness for this proof, let us first look at how this correspondence might plausibly falter; that is, “plausibly” before knowledge of the proof! We begin with $\mathfrak{G} = \mathbb{T}^2$ the 2-torus and the one-parameter subgroup $\mathfrak{H} = \mathbb{P}(\kappa)$ when $\kappa$ is an irrational real:

\begin{equation}\label{2Torus}\begin{array}{lcl}\mathbb{T}^2 &=& \{ (e^{i\,a},\,e^{i\,b}) |\; a, b \in \R\}\\\mathbb{P}(\kappa) &=& \{(e^{i\,a},\,e^{i\,\kappa\,a})| \; a \in \R\}\end{array}\end{equation}

FailureModes

Figure 12.1: Plausible Failure Modes for the Lie Correspondence (a.) Looping back to the identity with a tangent outside $\h$; (b.) The image under $\exp$ of a neighbourhood of $\Or$ in the whole vector space spanned by $\h$ and the “rogue” tangent, indeed the smallest Lie algebra holding these two, must then be filled in

 

Torus

Figure 12.2: The Irrational Slope Subgroup of the Torus $\mathbb{T}^2$: $\mathbb{P}(\kappa)$ winds itself around $\mathbb{T}^2$ such that it is dense in $\mathbb{T}^2$, thus stacking a countably infinite number of copies of $\exp(\mathcal{H})$ and “threatening” to build a tangent orthogonal to $\mathbb{P}(\kappa)$! Inset: Zooming in on $\id$, where we see $\exp(\mathcal{H})$ and some of its copies, as well as the relationship between the neighbourhoods of $\id$ in the whole group topology, $\mathbb{P}(\kappa)$ s group topology and the relative topology in $\mathbb{P}(\kappa)$ inherited from the whole group. $\exp(\tau Y)$ is also shown, piercing $\mathbb{P}(\kappa)$ at a countable number of points, thus the tangent $Y$ cannot belong to $\mathbb{P}(\kappa)$’ s Lie algebra and the question is answered in the negative.

The Lie algebra of $\mathbb{T}^2$ is of course $\g = \{(i\,a,\;i\,b)|\;a, b \in \R\} \cong (\R^2, +)$. Now we consider the Lie subalgebra $\h = \{(i\,a,\;i\,\kappa\,a)|\;a \in \R) \cong (\R, +)$ and assume for the momement that we do not know that this the Lie algebra of $\mathbb{P}(\kappa)$. Let us choose $X = (i,\;i\,\kappa)$ as the one basis vector needed to span the Lie subalgebra $\h$. We then take a neighbourhood of $\Or$ in $\h$, say some small open interval $\mathcal{H} = \{\epsilon \, X|\; \epsilon \in \R, |\epsilon| > 0\}$ and form the smallest group containing $\exp(\mathcal{H})$, to wit $\mathfrak{F} = \bigcup\limits_{k=0}^\infty \exp(\mathcal{H})^k$; certainly any group whose Lie algebra contains $\mathfrak{h}$ must contain this subgroup and thus $\exp(\h)$. We now ask what the Lie algebra $\mathfrak{f} = \mathrm{Lie}(\mathfrak{F})$ of $\mathfrak{F}$ is, i.e. what is the vector space of all tangents to $C^1$ paths in $\G$ that are also contained in $\mathfrak{F}$? There are two ways wherein this vector space might be bigger than the original Lie subalgebra $\h$. Firstly, as we translate the set $\exp(\mathcal{H})$ in $\mathfrak{F}$ as $e^{i\,\tau} \exp(\mathcal{H})$ for $\tau \in \R$ and then bring it back to be centred on the group identity $\id$, it might get rotated. Otherwise put, a $C^1$ path in $\mathfrak{F}$ might come back to $\id$ without passing through the set $\exp(\mathbf{h})$, see Figure 12.1 a. Of course, if it does such a thing, we now have a new tangent $Y$ to the identity in $\mathfrak{F}$ that is linearly independent of $X$ and the group operation then “fills the vector space spanned by $X, Y$ in” (Figure 12.1 b.), i.e. all entities of the form $\exp(x\, X + y\, Y)$ for $x, \,y \in \R$ will also belong to $\mathfrak{F}$ and the latter’s Lie algebra would then the same as that of the whole torus $\mathbb{T}^2$. For this particular example, such a situation cannot arise, for the exponential map $\tau \mapsto (e^{i\,\tau},\,e^{i\,\kappa\,\tau})$ is globally one-to-one; if it were not, then $(e^{i\,\tau_1},\,e^{i\,\kappa\,\tau_1}) = (e^{i\,\tau_2},\,e^{i\,\kappa\,\tau_2})$ for some $\tau_1, \tau_2 \in \R$, thus $1 = e^{i\,\tau_0},\, 1 = e^{i\,\kappa\,\tau_0}$ with $\tau_0 = \tau_1-\tau_2$ and we must then have $\tau_0 = 2\,m\,\pi,\;\kappa\,\tau_0 = 2\,n\,\pi$ for $m, n \in \mathbb{Z}$, thus gainsaying $\kappa$’s irrationality. Therefore, the $C^1$ path $\mathbb{P}(\kappa)$ never comes back to $\id$. If $\kappa$ is rational, $\mathbb{P}(\kappa)$ does return to $\id$, but does so through the original $\exp(\mathcal{H})$.

Moreover:

Lemma 12.2

$\H$ is dense in $\mathbb{T}^2$

Proof: Show Proof

Suppose, to the contrary, that there is some open ball (open in the group topology of $\G$) $\mathcal{B}(\zeta,\,r)$ such that $\mathbb{P}(\kappa)\cap\mathcal{B}(\zeta,\,r)=\emptyset$. But, since $\gamma\,\mathbb{P}(\kappa)=\mathbb{P}(\kappa)$ ($\mathbb{P}(\kappa)$ is a group) we then see that:

\begin{equation}\label{ClearRegion}\mathbb{P}(\kappa)\cap\gamma\,\mathcal{B}(\zeta,\,r)=\mathbb{P}(\kappa)\cap\mathcal{B}(\gamma\,\zeta,\,r)=\emptyset;\,\forall\gamma\in\mathbb{P}(\kappa)\end{equation}

This means that, for example, the intersections of $\mathbb{P}(\kappa)$ with the loop $\{(1,\,e^{i\,\phi})|\,\phi\in\R\}$ must be (i) distinct (otherwise $\mathbb{P}(\kappa)$ closes on itself, gainsaying $\kappa$’s irrationality), i.e. there are countably infinitely many of them and (ii) spaced by at least $2\,r>0$, which is impossible, because the loop is compact, thus has the Bolzano-Weierstrass property, and the needfull limit point of the set of intersections means that the spacings between intersections has a greatest lower bound of nought. $\quad\square$

So this lemma brings us to the second way wherein $\mathfrak{f}$ might be bigger than the original Lie subalgebra $\h$. The whole group $\mathbb{P}(\kappa)$ contains an infinite number of copies of $\exp(\h)$ laid infinitely near together side by side about $\id$ in the whole torus $\mathbb{T}^2$; we ask the question, “can we choose a point from each of the copies and build a ‘shortcut’ path through $\mathbb{P}(\kappa)$ that is orthogonal to the copies?”. A tangent $Y$ to the “shortcut” would clearly be linearly independent from $X$; again the Lie algebra $\mathfrak{f} = \mathrm{Lie}(\mathfrak{F})$ of $\mathfrak{F}$ would then be the whole of $\g$. In this example, however, we are clearly safe from this second mode of failure (Figure 12.2), and what keeps us safe is the countability of the set of copies of $\exp(\h)$. We can’t build a path (i.e. something cardinally equivalent to a real interval) if our budget is only $\aleph _0$ points! The countability shields the group $\mathfrak{F}$ from acquiring “rogue” tangents orthogonal to the original Lie algebra $\h$ and thus $\mathrm{Lie}(\mathfrak{F}) = \mathfrak{f} = \h$ and, furthermore $\mathfrak{F} = \mathfrak{H}$. This example, although Abelian, shows us in simplified form pretty much all of the steps in a proof of the Forwards Lie Correspondence, and we shall now address this proof.


Example 12.3: Mathematica Simulation of the subgroup $\mathfrak{P}(\kappa)$ in the Abelian Torus Group $\mathbb{T}^2$

This is a Mathematica simulation that plots the subgroup $\mathbb{P}(\kappa)$ on the torus $\mathbb{T}^2$ to help with intuition and visualisation of the concept of the one dimensional Lie subgroup of the two dimensional torus.

The slope $\kappa$ is set by the “Thread Slope” field, the subgroup is plotted from $\theta=0$ up to a maximum value set by the field “Max Length”; “StepSsize” sets the speed of the plot and “Plot Res” sets the plotting resolution.


The Forwards Lie Correspondence Proof

For a connected Lie group $G$ with Lie algebra $\g$ containing proper Lie subalgebra $\h\subset\g$, the subgroup $\H = \bigcup \limits_{k=1}^\infty \gamma_k \exp(\mathcal{H}) \subset\G$, where $\mathcal{H}$ is a neigbourhood of $\Or$ in the Lie subalgebra $\h$, is the subject matter of Theorem 10.5 and is a connected Lie group in its own right as shown by Lemma 10.6 and the Lie subalgebra $\h$ is clearly the Lie algebra of $\H$ given the structure defined by $\Nid=\exp(\mathcal{H})$, $\lambda=\log$ and $\mathcal{V} = \mathcal{H}$. However, the point of the following theorem is to find out whether, with $\H$ as part of a larger whole $\G$, we can define a new topology with a new, higher dimensional $\Nid^\prime=\exp(\mathcal{H}^\prime)\supset\Nid$ with $\mathcal{H}^\prime\supset\mathcal{H}$ a neighbourhood in a higher dimensional Lie algebra $\mathfrak{f}$ such that $\h\subset\mathfrak{f}\subset\g$. The answer is in the negative, as follows.

Theorem 12.4: (Forwards Lie Correspondence)

Let $\G$ be a connected, finite-dimensional, connected Lie group with Lie algebra $\g$. Then for every Lie subalgebra $\h \subset \g$ of the Lie algebra $\g$ there is precisely one connected Lie subgroup $\H \subset \G$ whose Lie algebra is $\h$. Otherwise put, $\H = \bigcup\limits_{k=0}^\infty \exp(\mathcal{H})^k=\bigcup\limits_{k=0}^\infty \exp(\h)^k$ is the the smallest group containing $\exp(\h)$ where $\mathcal{H}$ is any neighbourhood of $\Or$ in the Lie subalgebra $\h$, and the Lie algebra of this group is $\h$.

Proof: Show Proof

Although there may seem to be many details, the proof is simple and can be summarised thus: if $\H$ in Theorem 10.5 contains a path of the form $\tau \in \R \mapsto e^{\tau\,Y}\in\G$ for $Y \notin \h$, then some wrangling of transition maps and implications of the CBH formula show that this path can have only a countable number of intersections with each of the patches $\gamma_k \exp(\mathcal{H})$. Thus the whole path, contained by Theorem 10.5 within a countable number of these patches, can itself comprise only a countable number of points, gainsaying its nature as a smooth path (unless $Y = 0$).

On to the details: suppose (hoping to provoke a contradiction) that the smallest group containing $\exp(\mathcal{H})$, to wit, the group $\H$ in Theorem 10.5, has a Lie algebra $\mathfrak{f} = \operatorname{Lie}(\H)$ that is strictly bigger than $\h$, say of dimension $\dim \mathfrak{f} = N + M$ where $N = \dim \h$. Then let the basis $\{\hat{X}_1,\,\hat{X}_2,\,\cdots,\,\hat{X}_N,\,\hat{Y}_1,\,\hat{Y}_2,\,\cdots,\,\hat{Y}_M\}$ span $\mathfrak{f}$, where $\{\hat{X}_1,\,\hat{X}_2,\,\cdots,\,\hat{X}_N\}$ are basis vectors spanning $\h$; here the $\hat{Y}_j$ are the “rogue tangents” spoken about in the informal description of the irrational slope subgroup of the torus.

The group topology for $\H$ is then generated by ball neighbourhoods of each group member $\gamma \in \H$ with the form:

\begin{equation}
\label{ForwardsLieCorrespondenceTheorem_1}
\begin{array}{l}
\mathcal{B}(\gamma,\,\varepsilon) =\{\gamma\,e^X|\; X \in \mathcal{b}(\varepsilon)\}\text{ with }\\\\
\mathcal{b}(\varepsilon)=\{X =\sum\limits_{j=1}^N x_j\,\hat{X}_j + \sum\limits_{j=1}^M y_j \,\hat{Y}_j \in \mathfrak{f}|\;\left\|X\right\| < \varepsilon\}
\end{array}
\end{equation}

for some suitable norm $\left\|\cdot\right\|$ defined on $\mathfrak{f}$.

Firstly, we choose a ball size $\varepsilon_0$ small enough to give the ball $\mathcal{b}(\varepsilon_0)$ the two crucial properties: (i) the map $X \mapsto \gamma\,e^X$ is one-to-one and onto between $\mathcal{b}(\varepsilon_0)$ in the Lie algebra $\mathfrak{f}$ and the ball $\mathcal{B}(\gamma,\,\varepsilon_0)$ in the Lie group; and (ii) the CBH series $\varphi_{CBH}(X, \,Y)$ converges for all $X \in \mathcal{H} = \mathcal{b}(\epsilon)$ and all $W \in \mathcal{b}(\varepsilon_0)$, where $\epsilon$ (take heed well: distinct from $\varepsilon_0$) was the ball size used in the proof of Theorem 10.5.

Property (ii) just stated has the following significance: for any $\zeta \in \exp(\mathcal{H})$, we can use the CBH formula to give us co-ordinates for the ball $\mathcal{B}(\zeta,\,\varepsilon_0)$ within the exponential (geodesic) chart centred centred on $\zeta$ in terms of the co-ordinates in the chart centred on on $\id$: the transition (chart-conversion) map is the bijective $X \mapsto \log(\zeta^{-1}\, e^X)$. Thus, if $\xi = \zeta\, \exp\left(\sum\limits_{j=1}^N x_j \hat{X}_j\right) \in \mathcal{B}(\zeta,\,\varepsilon_0)$ so that the geodesic co-ordinates of $\xi$ in the chart centred on $\zeta$ are $(x_1,\,x_2,\,\cdots,\,x_n)$, then the transition map shows that:

\begin{equation}
\label{ForwardsLieCorrespondenceTheorem_2}
\xi = \exp\left(\varphi_{CBH}\left(-\log \zeta,\,\sum\limits_{j=1}^N x_j \hat{X}_j\right)\right) =\exp\left(\sum\limits_{j=1}^N x_{\zeta,\,j}(x_1,\,x_2,\,\cdots,\,x_n) \hat{X}_j\right)
\end{equation}

i.e. there are no terms of the form $\sum\limits_{j=1}^M y_j \hat{Y}_j$ inside $\exp$ on the right hand side, because $\mathfrak{h}$ is a Lie algebra, thus closed under the operation $(W,\, V) \mapsto \varphi_{CBH}(W,\, V)$, and $\xi$ can still be labelled by a vector in $\h$ with components $x_{\zeta,\,j}(x_1,\,x_2,\,\cdots,\,x_n)$ which are then $\xi$’s co-ordinates in the geodesic chart centred on $\id$.

Very obviously, but importantly, given suitable $\epsilon$ in Theorem 10.5, one such value of $\varepsilon _0$ will work for any $\gamma \in \H$ i.e. the map is bijective between $\mathcal{b}(\varepsilon_0)$ and $\mathcal{B}(\gamma,\,\varepsilon_0)$ if and only if the map $X \mapsto e^X$ is bijective between $\mathcal{b}(\varepsilon_0)$ and $\mathcal{B}(\id,\,\varepsilon_0)$: this works because neighbourhoods of $\gamma$ are simply left (or right) translated neighbourhoods of the identity. Furethermore we can make analogous conversions between co-ordinate charts to centre a chart for the whole of $\mathbf{B}(\gamma,\,\varepsilon_0)$ on any member of $\eta \,\exp(\mathcal{H})$ if $\gamma \in \eta\, \exp(\mathcal{H})$ and the chart will always label all of its members by an expression of the form $\gamma \,e^W$ for $W \in \h$ (simply put $\zeta = \gamma \,\eta^{-1}$ in $\eqref{ForwardsLieCorrespondenceTheorem_2}$ to see this).

With this background, we witness that the whole Lie group path:

\begin{equation}
\label{ForwardsLieCorrespondenceTheorem_3}
\mathfrak{P}(\hat{Y}_1) = \{\exp(\tau\,\hat{Y}_1) |\, \tau\in \mathbb{R}\} \subseteq \H
\end{equation}

defined by the tangent $\hat{Y}_1 \notin \h$ outside the original Lie algebra is inside $\H$. Suppose that this path meets the patch $\gamma_k \exp(\mathcal{H})$ (in Theorem 10.5) at some point $\eta= \exp(\tau_\eta \,\hat{Y}_1)$ and consider this path’s behaviour within the neighbourhood $\mathcal{B}(\eta,\,\varepsilon_0)$: all path members in this neighbourhood have the form $\eta\,\exp(\tau\,\hat{Y}_1)$ with $\left\|\tau\,\hat{Y}_1\right\| < \varepsilon_0$ and those members which are also in $\gamma_k \,\exp(\mathcal{H})$ have the form $\eta\,\exp\left(\sum\limits_{j=1}^N x_j\, \hat{X}_j\right)$, thanks to the chart transition maps discussed above (put $\zeta = \eta \,\gamma_k^{-1}$ in$\eqref{ForwardsLieCorrespondenceTheorem_2}$: there are no terms of the form $\sum\limits_{j=1}^M y_j \,\hat{Y}_j$ in this latter expression thanks again to the CBH theorem.

Thus, since $X \mapsto \eta\,e^X$ is a bijection within $\mathcal{B}(\eta,\,\varepsilon_0)$ we can find all other intersections inside this ball between the path and $\gamma_k \exp(\mathcal{H})$ by solving $\tau\,\hat{Y}_1 = \sum\limits_{j=1}^N x_j \,\hat{X}_j$; owing to the assumed linear independence of $\hat{Y}_1$ from the $\hat{X}_j$, this equation has only one solution: $\tau=0, x_j = 0;\;j=1 \cdots N$. Thus the path $\mathfrak{P}(\hat{Y}_1)$ “pierces” $\gamma_k \exp(\mathcal{H})$ precisely once inside the neighbourhood $\mathcal{B}(\eta,\,\varepsilon_0)$ of any intersection $\eta$, to wit at only $\eta$.

Any new intersection between the path and the patch $\gamma_k \exp(\mathcal{H})$ must therefore happen outside the neighbourhood $\mathcal{B}(\eta,\,\varepsilon_0)$, so that there is only one intersection for $\tau\in\left(\tau_\eta – \frac{\varepsilon_0}{\left\|\hat{Y}_1\right\|},\,\tau_\eta + \frac{\varepsilon_0}{\left\|\hat{Y}_1\right\|}\right)$, to wit, that at $\tau=\tau_\epsilon$. Likewise, at every value of $\tau$ where $\exp(\tau\,\hat{Y}_1)$ intersects $\H$, there is an interval of width $\frac{\varepsilon_0}{\left\|\hat{Y}_1\right\|}$ on either side of the intersection value wherein there are no other intersections. Therefore, $\mathfrak{P}(\hat{Y}_1)$ has at most a countable number of intersections with $\gamma_k \exp(\mathcal{H})$, therefore at most a countable number of intersections with $\H=\bigcup\limits_{k=1}^\infty \gamma_k\,\exp(\mathcal{H})$.

This is the hoped-for contradiction: $\mathfrak{P}(\hat{Y}_1) \subset \mathfrak{H}$, so this path comprises only its (countable) intersections with $\H$; therefore, given that $\exp : \tau \mapsto \exp(\tau\,\hat{Y}_1)$ is a bijection for $|\tau| < \frac{\varepsilon _0}{\left\|\hat{Y}_1\right\|}$, the real interval $-\frac{\varepsilon _0}{\left\|\hat{Y}_1\right\|} < \tau < \frac{\varepsilon _0}{\left\|\hat{Y}_1\right\|}$ would then have to be cardinally equivalent to a countable set of points.$\quad\square$

Thus our intuitive comments on the torus example are shown by the foregoing proof to be general. We might imagine a Lie group where we stack the little patches $\gamma_k \exp(\mathcal{H})$ tightly on top of one another so that a path could pass through them at right angles, but the countability of the patches thwarts such an undertaking and “shields the Lie algebra $\h$ from acquiring any further tangents”.

Now, given our comments before the theorem on the validity of $\h$ as a lie algebra for $\H$, we can reword the above theorem as follows:

Theorem 12.5: (Forwards Lie Correspondence, Version 2)

Given a connected Lie group $G$ defined by the open set $\V\subset\R^N$, neighbourhood $\Nid$ fulfilling axioms 1 through 5 and with Lie algebra $\g$,there is no $\Nid^\prime\supset\Nid$ with a higher-dimensional $\mathcal{V}^\prime\supset\V$ and extended labeller function which also fulfil axioms 1…5 such that the higher dimensional Lie algebra $\g^\prime\supset\g$ includes $\g$ as a linear subspace and Lie subalgebra. Otherwise put, the dimension N of the original namespace open set is indeed unique.

Proof: Otherwise, we could let $\G^\prime = \bigcup\limits_{k=0}^\infty \exp(\g^\prime)$ and $\G$ play the roles of $\G$ and $\H$ respectively in Theorem 12.4 and show that $\{\exp(\tau\,\hat{Y})|\,\tau\in\R\}$ cannot be in $\G$ for any tangent $\hat{Y}\in\g^\prime\setminus\g$ so that $\G$ would be a proper subset of $G^\prime$. $\quad\square$

The Baire Category Theorem as Applied to the Forwards Lie Correspondence

[Rossmann] adapts the Baire Category Theorem, which he calls the “Baire Covering Lemma”, to Lie groups:

Lemma 12.6: (Baire Covering Lemma)

If a connected Lie group $\mathfrak{H}$ is covered by a countable union, say $\mathfrak{H}=\bigcup_{k=1}^\infty\mathbf{A}_k$, then the closure $\bar{\mathbf{A}}_k$ of at least one of the $\mathbf{A}_k$ contains an open subset of $\mathfrak{H}$ in the latter’s group topology. (to be proven in Lemma 12.8)

and uses it in [Rossmann], §2.5 as an alternative to the proof in Theorem 12.4 above in the following way to complete the proof of the Forwards Lie Correpondence. Again we add a little more detail, formulated with the hindsight afforded by seeing Rossmann’s work finished on the page.

Proof of Theorem 12.4 using Baire Covering Lemma: Show Proof


With $\G$ and $\H$ as defined in Theorem 12.4, we already know from that $\H$ is a Lie group with the topology defined by deeming the system of neighbourhoods of any point $\gamma\in \H$ to be generated by all open balls of the form $\gamma\, \exp (\mathcal{H})$ where $\mathcal{H}$ is an open ball in $\h$ centred on $\Or$. Let us call this topology $\mathscr{T}_\mathfrak{h}$. But might there not be another Lie algebra $\mathfrak{f}$ with $\mathfrak{h} \subset \mathfrak{f} \subseteq \mathfrak{g}$ that would generate $\H$ as $\H = \Gamma(\mathfrak{f})$? If so, then we bestow another different topology on the group $\H$ by deeming the system of neighbourhoods of any point $\gamma\in \H$ to be generated by all open balls of the form $\gamma\exp(\mathcal{F})$ where $\mathcal{F}$ is an open ball in $\mathfrak{f}$ centred on $\Or$. Let us call this second topology $\mathscr{T}_\mathfrak{f}$. Thus, when we we apply the Baire Covering Lemma in this second topology to the countable collection $\bigcup_{k=1}^\infty\gamma_k \exp(\mathcal{H})$ and infer that the closure of some $\gamma_k \,\exp(\mathcal{H})$ must contain an open neigbourhood in $\mathscr{T}_\mathfrak{f}$; we must of course use the closure in the latter topology $\mathscr{T}_\mathfrak{f}$. But now $\h$ is a closed vector subspace of $\mathfrak{f}$, so if we choose $\mathcal{F} \supset \mathcal{H}$ to be a neighbourhood of $\Or$ in $\mathfrak{f}$ and small enough to make $\exp : \mathcal{F} \to \mathfrak{H}$ into a local homeomorphism, the closure $\gamma_k \exp (\mathcal{\bar{H}})$ in the first topology $\mathscr{T}_\mathfrak{h}$ is also closed in $\mathscr{T}_\mathfrak{f}$, therefore this closure contains our wished for open set: suppose it is an open neighbourhood of $\eta$, then it must contain a ball, i.e. $\eta\exp(\mathcal{F}_0) \subseteq \gamma_k \exp(\mathcal{\bar{H}})$ where $\mathcal{H}_0$ is an open ball in $\mathfrak{f}$ centred on $\Or$, by definition of topology $\mathscr{T}_\mathfrak{f}$. Now, by using the analytic transition map $X \mapsto \log(\eta^{-1}\,\gamma_k\, e^X)$ to convert from a chart centred on $\gamma_k$ to one centred on $\eta$ and heeding the implications of the CBH theorem discussed around $\eqref{ForwardsLieCorrespondenceTheorem_2}$, we see that this open ball is of the form $\eta\,\exp(\mathcal{H}_0)$ where $\mathcal{H}_0 \subseteq \h$, thus $\mathcal{F}_0 \subseteq \h$, an inclusion that can only arise if $\mathfrak{f} \subseteq \h$. But since we know $\h \subseteq \mathfrak{f}$, we have the required result $\h = \mathfrak{f}$. $\quad\square$

The Backwards Lie Correspondence

The backwards Lie correspondence is the following.

Lemma 12.7: (Backwards Lie Correspondence)

Let $\G$ be a connected, finite-dimensional, Lie group with Lie algebra $\g$. Then the Lie algebra $\h$ of every connected Lie subgroup $\H \subseteq \G$ of $\G$ that is also an immersion (included by a differentiable map in $\G$) is a Lie subalgebra of $\g$. Equivalently: The Lie algebra $\h$ of every connected Lie subgroup $\H \subseteq \G$ of $\G$ is a Lie subalgebra of $\g$, so long as that the logarithm map $\log : \H \to \h$ is given by the restriction of the map $\log : \G \to \g$ to the subset $\H$.

Proof: Show Proof

Qualified as above, the above statement is straighforwardly proven: any tangent to a $C^1$ path at the group identity $\id$ in a subgroup $\H$ belongs to $\g$ and the set of all tangents can readily be shown to be closed under the Lie bracket by considering, for $X, Y \in \h$ the family of $C^1$ paths:

\begin{equation}\label{LieBracket}\sigma_\varsigma : \mathbb{R} \to \mathfrak{H};\; \sigma_\varsigma(\tau) =e^{\varsigma\,X} \,e^{\tau\,Y}\,e^{-\varsigma\,X}\end{equation}

and, on taking heed that the Lie algebra, as a vector space over $\R$ or $\mathbb{C}$ is closed and complete, one then shows that limit $\left.\d_\varsigma \left(\left.\d_\tau\,\sigma_\varsigma(\tau)\right|_{\tau=0}\right)\right|_{\varsigma=0}$ of the family of tangents $\left.\d_\tau\,\sigma_\varsigma(\tau)\right|_{\tau=0}$ indexed by $\varsigma$ is also a tangent and is defined as the bracket $[X,\,Y]$. $\quad\square$

The statement of this theorem in [Rossmann] and all other sources I have seen leaves out the crucial qualification, “so long as that the logarithm map $\log : \H \to \h$ is given by the restriction of the map $\log : \G \to \g$ to the subset $\H$”. In [Rossmann], the statement is made for the case of linear groups when the exponential and logarithm maps between Lie groups and algebras are defined by convergent matrix series, so the qualification on the $\log$ function is implicit and the statement is thus sound in the context of [Rossmann], §2.5. All other sources assume the immersion property implicitly. However, in the general case, the qualification must be given otherwise the statement of Lemma 12.7 fails, for in general Lie groups contain subgroups which are also Lie groups when given topologies quite unrelated to those defined by Lie subalgebras $\h\subseteq \g$ as shown by the following examples.

Take the additive group $\G = (\R,\,+)$, it is its own Lie algebra $\g$ and the exponential map is the identity map. Now take one of the everywhere discontinuous, bijective solutions $\phi$ of the Cauchy functional equation $f(x) + f(y) = f(x + y)$ as detailed in [HewittAndStromberg]; $\phi$ is now an exponential map from $\g$ onto a new Lie group $\G^\prime$, to wit the same group $(\R, \,+)$ but now given a new topology generated by images $\phi(\U)$ of open intervals $\U$ in $\R$. This topology is of course totally imcomparable in the group topology for $\G$; however $\phi$ is continuous, indeed $C^\omega$ when thought of as a map from $\mathfrak{g}$ to $\mathfrak{G}^\prime$, the latter with the new topology. Indeed, the Lie groups arising from the two topologies are isomorphic, their topologies are homoemorphic. So we can shred the manifold and put it back together again to give a definitely different topology, but one that is homeomorphic to the original and which doesn’t give a new Lie algebra. But it is not always so simple; the same Hamel basis ideas as in [HewittAndStromberg] but this time using a Hamel basis for $\R^N$ over $\mathbb{Q}$ for any integer $N > 0$ shows that the vector spaces $\R^m$ and $\R^n$ are isomorphic as additive abstract groups by being isomorphic to $(\R, \,+)$. But they of course, as Lie groups, are altogether different; their topologies are not even comparable and they have different Lie albegras. Similar arguments hold for the compact one parameter group $\G = U(1) = (\{e^{i\, z}|\; z \in \R\}, \bullet)$ where $\bullet$ is complex multiplication and its Lie algebra $\g = (\R,\,+)$: different exponential functions (i.e. any fulfilling the functional equation $f(x) \bullet f(y) = f(x + y)$) can be used to make it isomorphic, as an abstract group (but not of course as a Lie group) to a torus $\mathbb{T}^n$ of any positive dimension $n$.

Therefore, we see that any noncompact (respectively, compact) one-parameter subgroup in a Lie group can be given a topology making it into the Lie group $(\R, \,+)$ (respectively, $\mathbb{T}^N$) for any $N > 0$; in particular, $N$ can be greater than the original Lie group’s dimension, thus obviously violating the statement of Lemma 12.7 if the latter lacks its immersion / log function qualification. This qualification forces the topology given to the Lie subgroup $\H$ to be the group topology of Definition 9.6. The discussion of this topology brings us naturally to the discussion of nomenclature.

SkullImmersions, Embeddings and Virtual Subgroups

This little section is a warning. The group topology of a Lie subgroup $\H \subseteq \G$ may or may not be the same as the relative (subspace) topology inherited by $\H$ from its parent $\G$ and thus the Lie subgroup $\H$ with group topology is not in general a topological embedding in the the parent (i.e. object whose topology is the relative topology from the embedding parent). Some definitions of “submanifold” would thus forbid calling $\H$ a submanifold if it fails to be an embedding in the way that the irrational slope torus subgroup does. For this reason, some definitions of “Lie subgroup” would forbid calling $\H$ a “Lie subgroup”‘ if it fails to be a topological embedding.

However, it most definitely is a manifold, in the light of Lemma 10.6 and indeed in the light of the Frobenius Integrability Theorem (see [Spivak], Chapter 7): the closure under the Lie bracket of left- (or right-) invariant vector fields assured by the Lie algebrahood $\h$ guarantees that a manifold, whose tangent space at the group identity is $\h$, can be built. Furthermore, it is most definitely a Lie group, also by Lemma 10.6 and Theorem 12.4; its group topology, following Definition 9.6, is the one defined by deeming $\{\gamma\,\exp(\mathcal{H})| \,\gamma\in\G;\,\mathcal{H}\subset\h \text{ open in }\h;\,\Or\in\mathcal{H}\}$ to be the base of the topology. Lastly, with both $\G$ and $\H$ thought of as an abstract groups, $\H$ is most assuredly a subgroup of $\G$, so it would seem natural to call $\H$ a Lie subgroup of $\G$.

But the reader is warned that this nomenclature is not universal: if $\H$ is an immersion but not a topological embedding in $\G$, many authors do not regard $\H$ to be a “Lie Subgroup” nor a “Submanifold” of $\G$ even though I, for the reasons just discussed, would be comfortable with both of these terms for $\H$ generally. The clash of usages here shows that one must carefully read each author’s usage and carefully state one’s own: it would seem safest therefore to explicitly specify whether “embedding” is meant by one’s usage.

[Gorbatsevich] uses the word “virtual Lie subgroup” for the case where $\H$ fails to be embedded in $\G$ and holds only embedded Lie subgroups to be “genuine”. I appreciate that this term could have mnemonic value, reminding the reader that the group topology might be different from the relative one, however, I find the word “virtual” ugly: there is nothing virtual about the Lie-hood of such a group.

Appendix: The Baire Category Theorem Adapted to Lie Groups

We now look at Rossmann’s proof of the Baire Covering Lemma in detail ([Rossmann], §2.5). Rossmann explicitly refrains from naming this Lemma the Baire Category Theorem, and he is right to do so: it is a specialisation of the more general theorem: most importantly, a specialisation that gets rid of the need to call on the axiom of choice.

Rossmann’s Proof of the Baire Covering Lemma:

Lemma 12.8: (Baire Covering Lemma)

If a connected Lie group $\mathfrak{H}$ is covered by a countable union, say $\mathfrak{H}=\bigcup_{k=1}^\infty\mathbf{A}_k$, then the closure $\bar{\mathbf{A}}_k$ of at least one of the $\mathbf{A}_k$ contains an open subset of $\mathfrak{H}$ in the latter’s group topology.

Proof: Show Proof

We take heed that clearly $\mathcal{H} = \bigcup_{k=1}^\infty {\bar{\mathcal{A}}_k } $ and consider $\mathcal{T}_1 = \mathcal{H} \setminus \bar{\mathcal{A}}_1$, an open set (its complement $\bar{\mathcal{A}}_1 $ is closed). Then either $\bar {\mathcal{A}}_1 $ contains an open set, in which case we’ve found the open set needed and stop, or we choose a closed ball $\bar{\mathcal{B}}_1 \subseteq \mathcal{T}_1 $ (we can do this because $\mathcal{T}_1$ is open) in the part $\mathcal{T}_1$ of $\mathcal{H}$ not in $\bar{\mathcal{A}}_1$. Let $\mathcal{B}_1$ be the open ball interior to $\bar{\mathcal{B}}_1 $. Then we ask whether $\bar{\mathcal{A}}_2 $ contains the open ball $\mathcal{B}_1$; if not, we consider $\mathcal{T}_2 = \mathcal{H} \setminus \bar{\mathcal{A}}_1 \setminus \bar{\mathcal{A}}_2$ (again open), form $\mathcal{B}_1 \cap \mathcal{T}_2 $ (also open – the part of the ball $\mathcal{B}_1$ outside $\bar{\mathcal{A}}_2 )$, and choose the next closed ball $\bar{\mathcal{B}}_2 \subseteq \mathcal{B}_1 \cap \mathcal{T}_2$.

At step $n$ of this algorithm, we ask whether $\bar{\mathcal{A}}_n $ contains the open ball $\mathcal{B}_{n-1} $ and, if not, consider $\mathcal{T}_n =\mathcal{H} \setminus \bigcup\nolimits_{k=1}^n {\bar{\mathcal{A}}_k}$ (still open, because the finite union of closed sets is closed), find $\mathcal{B}_{n-1} \cap \mathcal{T}_n $ (also open – the part of $\mathcal{B}_{n-1} $ outside $\bar{\mathcal{A}}_n )$ and choose the next closed ball $\bar{\mathcal{B}}_n \subseteq \mathcal{B}_{n-1} \cap T_n $. If the algorithm terminates, then we have some $\bar{\mathcal{A}}_j $ containing an open set as needed. We can prove that nontermination implies a contradiction as:

  1. If the algorithm does not end, we must have an infinite sequence of nested closed balls $\{\bar{\mathcal{B}}_n\}_{n=1}^\infty$ with $\bar{\mathcal{B}}_n \subset \bar{\mathcal{B}}_m $ whenever $n > m$.
  2. Moreover, these balls have nonempty intersection.

For suppose the first ball is $\bar{\mathcal{B}}_1 = \gamma_c \exp(\{X|\; X \in \h;\;\left\|X\right\|\leqslant r_1\})$. Then, since $\bar{\mathcal{B}}_n \subset \bar{\mathcal{B}}_1$, we can choose all the other balls to be sets of the form $\bar{\mathcal{B}}_n = \gamma_c \exp(\{X|\; X \in \h;\;\left\|X – X_n\right\|\leqslant r_n\})$ where $r_n$ is the radius of ball $\bar{\mathcal{B}}_n $ and $X_n $ its “centre” in $\h$. For the question of whether the intersection is nonemepty, we can now just as well now consider the sets $\mathcal{B}^\prime_n = \{X|\; X \in \h;\;\left\|X – X_n \right\|\leqslant r_n\}$, i.e. we are now using closed balls in the geodesic co-ordinate manifold chart centred on $\gamma _c $ for our further deliberations.

The radiusses $r_n$ dwindle monotonically with rising $n$; suppose they asymptote to some nonzero radius $r_\infty$, then the intersection of all the balls is clearly the ball $\bar{\mathcal{B}}^\prime_\infty =\{X\; X \in \h;\;\left\|X – X_\infty\right\|\leqslant r_\infty\}$ and manifestly not empty. The only other alternative is that the radiusses dwindle monotonically to nought. If so, choose a sequence with $Z_n \in \mathcal{B}^\prime_n$ (we do not need to call on the countable axiom of choice to do this, see the notes after the theorem) then $\left\|Z_j – Z_k\right\| \leqslant 2\;r_M$ (by the triangle inequality) for $j,\;k > M$ and every sequence $s_M =\{Z_k\}_{k=M}^\infty$ is (i) Cauchy and (ii) contained in $\bar{\mathcal{B}}_M$, which is complete by dent of its being a closed subset of $\R^N$, so that $s_M$ converges to a point$\lim\limits_{M\to\infty}s_M \in \bar{\mathcal{B}}_m$; moreover, all the Cauchy sequences converge to the same limit point: $s_\infty =\lim\limits_{M\to\infty}s_M \in \bar{\mathcal{B}}_N $ which is thus shown to belong to all the closed balls and thus to $\{s_\infty\} \subseteq \bigcap_{n=1}^\infty \bar {\mathcal{B}}_n$. But then $\bigcap_{k=1}^\infty {\mathcal{T}_k }$ is nonempty as well (since $\bar{\mathcal{B}}_n \subseteq \mathcal{B}_{n-1} \cap \mathcal{T}_n \subseteq \mathcal{T}_n$), thus gainsaying the assumption that the $\bar{\mathcal{A}}_j$ covered $\mathcal{H}$. Therefore, the algorithm must terminate and one of the $\bar{\mathcal{A}}_j$ contains the required open set. $\quad\square$

Example 12.7

We must choose closed balls rather than open ones at each point for this proof to work: witness the nested sequence of open balls

\begin{equation}\label{OpenBallIntersection}\bigcap_{n=1}^\infty \{\left. x \right|\;x\in \R;\;|x – 1 / n| < 1 / n\} = \emptyset\end{equation}

whereas

\begin{equation}\label{ClosedBallIntersection}\bigcap_{n=1}^\infty \{\left. x \right|\;x\in \R;\;|x – 1 / n| \leqslant 1 / n\} = \{0\}\end{equation}

[Royden] gives a similar proof, but valid for only for metric spaces, so it would not seem directly applicable to Lie groups. However, instead of the above proof, we can call on the Urysohn Metrization Theorem (see [Willard], [Munkres]} to show that any connected Lie group, as second countable (the Theorem 10.5 shows this by showing that the connected Lie group is a countable collection of second countable patches) regular topological space, is metrisable and thus Royden’s proof can be directly applied to the space rigged with the metric whose existence Urysohn’s theorem guarantees. The above more first principles proof does not need the Urysohn theorem to metrise the space. All other proofs of the Baire Category Theorem for arbitrary complete metric spaces call on the axiom of choice. However, [Rossmann]‘s specialised proof does not need any form of the axiom of choice for we can explicitly construct a choice function that picks the $Z_n$ from the countable nested sequence of closed balls in exactly the same way as is done for the proof of the Heine-Borel Theorem. That is, 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 $Z_n$ to be the unique point achieving the maximum inner product in the ball $\mathcal{B}^\prime_n$, i.e. $\left<X_n,\,\hat{Y}\right> = \max\limits_{X\in\mathcal{B}^\prime_n} \left<X,\,\hat{Y}\right>$.

There are likenesses between the [Rossmann] Baire Covering Lemma proof and my proof of Theorem 12.4 – they both centre on the impossibility of building the space $\R^N$ from a countable collection of subspaces of the form $\R^{N-1}$ unless the latter are sliced up and re-arranged. It is also interesting to witness the striking likeness between the construction $\mathcal{T}_n =\mathcal{H} \setminus \bigcup\nolimits_{k=1}^n {\bar{\mathcal{A}}_k}$ and the Cantor diagonalisation process to construct a real not on a countable list.

References:

  1. Emil Artin, “Galois Theory: Lectures Delivered at the University of Notre Dame”, Notre Dame Mathematical Lectures, Number 2, sixth printing 1971 see also version stored at project Euclid
  2. John Stillwell, “Galois Theory for Beginners“, The American Mathematical Monthly, Vol. 101, No. 1 (Jan., 1994), pp. 22-27.
  3. Wulf Rossmann: “Lie Groups: An Introduction through Linear Groups (Oxford Graduate Texts in Mathematics)”
  4. E. Hewitt and K. R. Stromberg, “Real and Abstract Analysis (Graduate Texts in Mathematics)”, Springer-Verlag, Berlin, 1965. Chapter 1, §5
  5. V. V. Gorbatsevich, A. L. Onishchik and E. B. Vinberg, “Foundations of Lie theory and Lie transformation groups”, Springer-Verlag, Berlin, 1997, §2.3 “Virtual Lie Subgroups” as well as Theorem 5.4
  6. Michael Spivak, “A Comprehensive Introduction to Differential Geometry, Vol. 1”, Publish or Perish, Houston, Texas, USA, 1999. Chapter 7 discusion of Frobenius Integrability Theorem
  7. H. L. Royden, “Real Analysis”, Collier-Macmillan, New York, 1968
  8. S. Willard, “General Topology”, Addison-Wesley, Reading, Massachussets, 1970. p. 166
  9. J. R. Munkres, “Topology (2nd Edition)”, Prentice-Hall International Limited, London, UK, 2000. Chapter 4, §34 “The Urysohn Metrization Theorem”