13 May 2014 No Comments

# Chapter 15: How Unique is the Lie Group Structure for a Given Lie Group?

We can think of our fundamental axioms 1 through 5 as a “specification” that we can lay down on an abstract group to beget a Lie group and ultimately, as we have shown, an analytic manifold structure. Can we do this in more than one way? How unique is the Lie group structure? Once we have found one structure fulfilling axioms 1 through 5, what does this structure way about any other Lie group structure that we might find fulfilling the same axioms by choosing a different set to be our fundamental neighbourhood $\Nid$ and a different mapping to be labeller $\lambda$? We imagine the violent act of tearing down the topology, leaving the bare abstract group structure and then, if we’re not to squeamish at the skeletal sight of a bare abstract group with its flesh, sinew and muscle torn away, we imagine gathering some ($\beth_1=2^{\aleph_0}$ of them!) group elements and gluing them together into a new $\Nid$, thus weaving a new Lie group structure. For this violent deed to be “valid”, the group multiplication must behave following Group Product Continuity Axiom 3 and Nontrivial Continuity Axiom 4, and the whole group must still be the smallest group containing $\Nid$*. *So

*Just how unique is the Lie Group Structure for a Given Lie Group?*

Another way of putting this, or at least a very like question, is:

*How much is the Lie group’s structure set by the algebraic, abstract group structure alone?*

As Hans Freudenthal would put it: when is the topology of a Lie group a wholly algebraic phenomenon? He adds that this formulation makes these ideas meaningful to someone who “forswears (“*überhaupt ablehnt*“) discontinuous pictures – from somewhat intuitionist considerations”[Freudenthal, 1941].

We already have a weak result relevant to this question: the Lie correspondence, which can be reinterpreted as follows. If we have a dimension $N$ connected Lie group $\H$ with Lie algebra $\h=\operatorname{Lie}(\H)$, there is no way of embedding it / immersing it in a higher dimension Lie group $\G$ with algebra $\g=\operatorname{Lie}(\G)$ so that **( i) **$\h$ is a Lie subalgebra of $\g$ and, when immersed in $\G$, we find that the Lie algebra of $\H$ is some bigger algebra $\mathfrak{f}\supset\h$. There is no way of folding $\H$ in the bigger topology afforded by $\G$ such that we can draw a “rogue tangent” $Y\in\g\setminus\h$ wholly contained in $\H$, as we showed in Theorem 12.4, for this would imply that the set of folded points in the rogue tangent would have cardinality $\beth_1=2^{\aleph_0}$, whereas, owing to the Rossmann Snakeskin Construction.

Going back to our uniqueness idea, we can give an elementary counterexample showing that the Lie group structure is not unique: we give any Lie group $\G$ of dimension $N\geq1$ its *discrete topology* (*i.e. *the topology is the Lie group’s powerset), so that it becomes a Lie group of dimension nought. It is now of course totally disconnected, with connected components given by the group’s singletons, and identity component given by the trivial group $\{\id\}$. So maybe this example is not too interesting. Therefore we’ll need to impose some restrictions on our question to make it interesting. Let’s henceforth, therefore, ask whether the Lie group structure for a given *connected* Lie group that *keeps it connected *is unique.

Still our answer is in the negative, as the following surprising counterexample shows.

**Example 15.1****: (Nonuniqueness of the Lie Group Structure in $(\R,\,+)$)**

We 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 [Hewitt, 1965]; $\phi$ 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 in $\R$. This topology is of course totally disconnected in the group topology for $\G$; however $\phi$ is continuous, indeed $C^\omega$ when thought of as a map from $\g$ to $G^\prime$, the latter with the new topology. Indeed, the Lie groups arising from the two topologies are isomorphic, their topologies are homoemorphic.

A somewhat more down-to-Earth (*i.e. *not needing the axiom of choice for its construction) example is the many discontinuous field automorphisms of the complex numbers $\mathbb{C}$. A field automorphism is one that respects the field, not only the Lie group, structure so that $\phi(x\,y+z)=\phi(x)\,\phi(y)+\phi(z)$ for such an automorphism $\phi$.

**Example 15.2****: (Discontinuous Complex Field Automorphism)**

Consider the field $\mathbb{Q}(\sqrt{2})$, the smallest field containing both the rationals and $\sqrt{2}$, a field extension of degree 2 of $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})=\{p+\sqrt{2}\,q|\,p,\,q\in\mathbb{Q}\}$. The automorphism $\phi:\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{2});\,\phi(p+\sqrt{2}\,q)= p-\sqrt{2}\,q$ extends to any algebraically closed field extension of $\mathbb{Q}(\sqrt{2})$, in particular to $\mathbb{C}$. In $\mathbb{C}$ it is everywhere discontinuous. As an automorphism of the group $\G=(\mathbb{C},\,+)$ Lie group automorphism, $\phi$ is now an exponential map from the Lie algebra $\g=\mathbb{C}$ onto a new Lie group $\G^\prime$, to wit the same group $(\mathbb{C},\,+)$ but now given a new topology generated by images $\phi(\U)$ of open balls $\U$ in $\mathbb{C}$.

**Example 15.3****: (Discontinuous Automorphism of $SL(2,\,\mathbb{C})$)**

Let $\phi$ be a discontinuous automorphism of the field $(\mathbb{C},\,+,\,\times)$, for example, the one given in Example 15.2. Then the map:

\begin{equation}\label{SL2CDiscontinuousAutomorphismExample_1}\varphi:SL(2,\,\mathbb{C})\to SL(2,\,\mathbb{C});\;\varphi\left(\,\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\,\right)=\left(\begin{array}{cc}\phi(a)&\phi(b)\\\phi(c)&\phi(d)\end{array}\right)\end{equation}

is a discontinuous automorphism of the Lie group $SL(2,\,\mathbb{C})$ (because, being a field automorphism, it leaves the matrix determinant invariant). However, when $SL(2,\,\mathbb{C})$ is given the topology generated by images $\varphi(U)$ of open balls $\U$ in $SL(2,\,\mathbb{C})$, it again becomes the Lie group $SL(2,\,\mathbb{C})$, but this time with the exponential map $\exp^\prime = \varphi\circ\exp$ where $\exp$ is the wonted, continuous matrix exponential function.

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, as in all the above examples. So these are all rather “tame” and it is not always so simple. Things get pretty wild in the following:

**Example 15.4****: (A “Wild” Discontinuous Automorphism Example)**

The same Hamel basis ideas in [Hewitt, 1965] as applied in Example 15.1 are used again, but this time using a Hamel basis for $\R^N$ over $\mathbb{Q}$ for any integer $N > 0$. This 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.

This all looks disastrously complicated, and shows that in general it is not meaningful to speak of a group as *a* Lie group; one must in general specify which topology is being given to the group and thus which of its many Lie algebras it can have by being made into different Lie groups.

But there is a wonderful thread of sanity in the case of semisimple Lie groups. We first need some definitions and a lemma.

**Definition 15.5****: (Simple Group)**

An abstract group $\mathbb{G}$ is called *simple* if it has no nontrivial, proper *normal subgroups*, *i.e.* groups $\mathbb{S}\subset\mathbb{G}$ such that $\gamma\,\mathbb{S} = \{\gamma\,\zeta|\;\zeta\in\mathbb{S}\} = \mathbb{S}\,\gamma = \{\xi\,\gamma|\;\xi\in\mathbb{S}\}$ so that every *left coset* $\gamma\,\mathbb{S}$ of $\mathbb{S}$ is also a *right coset* $\mathbb{S}\,\gamma$ and contrariwise. Otherwise put, a *normal **subgroup* $\mathbb{S}\subset\mathbb{G}$ is one for which $\forall\zeta\in \mathbb{G},\,\forall\gamma\in\mathbb{S}\,\exists\,\gamma^\prime\in\mathbb{S}\,\ni\,\zeta\,\gamma = \gamma^\prime\,\zeta$. Still another, equivalent characterisation of a normal subgroup is any subgroup $\mathbb{S}\subset\mathbb{G}$ which is fixed under the mapping $\phi_\zeta:\G\to\G;\,\phi_\zeta(\gamma)=\zeta\,\gamma\,\zeta^{-1}$ for any $\zeta\in\mathbb{G}$. So the only normal subgroup of a simple group is the trivial group $\{\id\}$.

A Lie group $\G$ is called *simple* if contains no nontrivial, proper normal *Lie* subgroups of dimension 1 or greater. So it is either outright simple in the abstract group sense, or, if it is not, it can only be non-abstract-group-simple if it contains a discrete normal subgroup (which is thus central, by Schreier's Theorem). By convention, a one-dimensional Lie group is not simple.

The grounds for the one dimensional group convention will become clearer in the following.

**Definition 15.6****: (Ideals and Simple Lie Algebra)**

An *ideal* of a Lie algebra $\g$ over a field $\mathbb{K}$ is a Lie subalgebra $\h\subset\g$ such that $\left[\h,\,\g\right]=\left[\g,\,\h\right]\subseteq\h$, *i.e.* for all $X\in\g$ and $Y\in\h$, $\left[X,\,Y\right]=\ad(X)\,Y\in\h$.

This is not quite the same concept as an ideal of a ring, for the Lie bracket is not associative. However, it is of course analogous aside from the lack of associativity. An ideal in a connected Lie group’s Lie algebra is the image under the functor $\operatorname{Lie}$ of a connected normal subgroup of the corresponding Lie group, and contrariwise, as we now show.

**Lemma 15.7****: (Normal Lie Subgroups have Ideal Lie Algebras)**

A connected Lie subgroup $\H\subset\G$ of a connected Lie group $\G$ is a normal subgroup of $\G$ if and only if its Lie algebra $\h=\operatorname{Lie}(\H)\subset\g$ is an ideal of the Lie algebra $\g=\operatorname{Lie}(\G)$ of $\G$. A Lie group is simple if and only if its Lie algebra is simple.

**Proof: **Show Proof

Suppose $\H$ is a normal subgroup of $\G$. Then let $Y\in\h$ and $X\in\g$, then, by assumption, $\,e^{s\,X}\,e^{\tau\,Y}\,e^{-s\,X} = \exp\left(\tau\,e^{s\,\ad(X)}\,Y\right)$ lies in $\H$ for all $\tau,\,s\in\R$. That is, $e^{s\,\ad(X)}\,Y\in\h\,\forall\,s\in\R$ and by the same limiting arguments made after Definition 7.2 to show that the Lie algebra is closed under the Lie bracket, we know that $\left[X,\,Y\right] = \left.\d_s\,e^{s\,\ad(X)}\,Y\right|_{s=0}\in\h$. So $\h$ is an ideal of $\g$. Conversely, assume $\h$ is an ideal of $\g$ and let $\gamma=\prod\limits_{k=1}^{M_1}\,e^{Y_k}\in\H$ where $Y_k\in\h$. Let $\zeta=\prod\limits_{k=1}^{M_2}\,e^{X_k}\in\G$ where $X_k\in\g$. We find that $\zeta\,\gamma\,\zeta^{-1} = \prod\limits_{k=1}^{M_1}\,\exp\left(\prod\limits_{j=1}^{M_2}\,e^{\ad(X_j)}\,Y_k\right) = \prod\limits_{k=1}^{M_1}\,e^{\tilde{Y}_k}$ where $\tilde{Y}_k=\prod\limits_{j=1}^{M_2}\,e^{\ad(X_j)}\,Y_k\in\h$ because $e^{\ad(X_j)} Z$ is a universally convergent bracket series in $X_j$ and $Z$, thus must be contained in $\h$ if $Z\in\h$ and $\h$ is an ideal. Therefore, $\zeta\,\gamma\,\zeta^{-1}\in\H$ and $\H$ is a normal subgroup of $\G$.$\quad\square$

Another definition that will be useful to us is.

**Definition 15.8****: (Semisimple Lie Algebra and Group)**

A *semisimple* *Lie algebra* $\g$ over a field $\mathbb{K}$ is a direct sum of simple Lie algebras over $\mathbb{K}$, *i.e.* $\g = \g_1\oplus\g_2\oplus\cdots\oplus\g_M$ for some $M\in\mathbb{N}$. A *semisimple Lie group* is a Lie group $\G$ whose Lie algebra is semisimple.

For a compact, simple Lie group, there is only one possible Lie algebra and any automorphism of the group is needfully continuous (and can therefore readily be shown $C^1$ by Theorem 7.12 and thus $C^\omega$). Let’s now look at the proof of this amazing fact. In van der Waerden’s hands[van Der Waerden, 1933], this proof is dazzling and simple: it is hard to better the original as a piece of technical writing. I’ll write down a lemma first, even though it is pretty obvious, because I believe it has an historical importance when guessing intelligently about how exactly Bartel van der Waerden was thinking about his problem.

**Lemma 15.9****: (Bolzano-Weierstrass Property In Compact Lie Groups) **

A compact Lie group has the Bolzano-Weierstrass property; that is, any infinite set within a compact Lie group must have at least one limit point.

**Proof: **As shown in Corollary 10.9, a compact Lie group $\G$ is a finite union $\G=\bigcup\limits_{k=1}^M\,\gamma_k\,\exp(\mathcal{B})$ of open sets of the form $\gamma_k\,\exp(\mathcal{B})$ where $\mathcal{B}\subset\g$ is an open ball centred on $\Or\in\g$. So certainly $\G=\bigcup\limits_{k=1}^M\,\gamma_k\,\exp(\bar{\mathcal{B}})$ where now $\exp(\bar{\mathcal{B}})$ are all closed and bounded, thus each have the Bolzano-Weierstrass property since they are homeomorphic to closed, bounded subsets of $\R$ (recall the Heine-Borel Theorem). Since there are finitely many of these sets $\gamma_k\,\exp(\mathcal{B})$, at least one of them must hold an infinite subset of any infinite subset of $\G$. So at least one must have a limit point, by the Bolzano-Weierstrass property of closed, bounded subsets of $\R^N$. $\quad\square$

The point is that our modern idea of “compact” was finalised somewhere between 1925 and 1935, depending on how you look at the history. A Lie group is second countable and Regular Hausdorff ($T_3$), so the Urysohn metrisation theorem[Munkres, 2000], proven 1925, holds for it. This means that sequential compactness and compactness are the same thing for a Lie group. [Alexandrof and Hopf, 1935] proposed in 1935 that the modern definition of compactness, rather than of having the Bolzano-Weierstrass property should be the “fundamental meaning”; I believe this is the first time a major work laid down such a behest, several comments on MathOverflow seem to back my belief up but I certainly wouldn’t claim to be any kind of historian. It’s almost certainly the first major textbook laying down this definition. Whichever way you look at it, it is curious that van der Waerden, given the thoroughness and clarity of his exposition to a wide audience, should simply use the word “compact” without definition unless it were a concept well rooted and well known widely throughout the mathematical community at the time. So it is unlikely that the distinction between “sequential compactness” and the Alexandrof-and-Hopf-proposed modern notion compactness would be in the wide mathematical vernacular at the time. I very much think van der Waerden’s conception of compactness was wholly and purely “having the Bolzano Weierstrass property”. Otherwise he would have given a brief summary of the “lemma” above – to have done otherwise would be out of his thorough and proudly “write-to-be-understandable-and-clear” professional character.

We need another little lemma before proving van der Waerden’s theorem:

**Lemma 15.10****: (Identity is Limit Point Of One Paramter Subgroup of a Compact Lie Group)**

Let $\G$ be a compact, connected Lie group and let

\begin{equation}\label{OneParameterSubgroupIdentityLimitPointLemma_1}\mathfrak{A} = \left\{\sigma(\tau) | \tau \in \R;\;\sigma: \R \rightarrow \G;\,\sigma(\tau) \,\sigma(\varsigma) = \sigma(\tau + \varsigma)\right\}\end{equation}

be *any* one parameter subgroup of $\G$, whether continuous in $\G$’s Lie group topology or not. Then in *any* topology that makes $\G$ a connected, compact Lie group, the group identity $\id$ is a limit point of $\mathfrak{A}$. That is, there is a sequence $\{\alpha_k\}_{k=1}^\infty \subset \mathfrak{A}$ converging to $\id$ in *any* topology that makes $\G$ a connected, compact Lie group (although of course the exact values $\alpha_k$ may be different for different topologies).

**Proof: **Show Proof

$\G$ by assumption is compact, so $\mathfrak{A}$ has a limit point $\tilde{\alpha}_\infty$ in the particular topology in question, by Lemma 15.9. So there is a sequence of $\mathfrak{A}$-elements converging to this limit point in the topology. Given this is a Lie group topology, this means that same thing as there is a sequence $\{\tilde{\alpha}_k\}_{k=1}^\infty\,\ni\,\tilde{\alpha}_k\to\id$ converging to the identity and such that $\tilde{\alpha}_k\,\tilde{\alpha}_\infty\in\mathfrak{A};\;\forall\,k\in\mathbb{N}$. So now we form the sequence $\{\alpha_k\}_{k=1}^\infty$ where $\alpha_k=\tilde{\alpha}_k\,\tilde{\alpha}_\infty\,(\tilde{\alpha}_{k+1}\,\tilde{\alpha}_\infty)^{-1}=\tilde{\alpha}_k\,\tilde{\alpha}_{k+1}^{-1}\in\mathfrak{A}$ whose members also lie inside $\mathfrak{A}$, since the latter is a group. Since $\tilde{\alpha}_k\to\id$, therefore $\alpha_k\to\id$ in the topology in question. To make this clearer, if not already so, write $\tilde{\alpha}_k =\exp(X_k)$ with $X_k\in\g,\,X_k\to0$ and consider the subsequence with $k>N_0$, $N_0$ big enough to make $\left\|X_k\right\|$ small enough for the Campbell-Baker-Hausdorff series converge; then since $X_k\to0$, therefore $\varphi_{CBH}(X_k,\,X_{k+1})\to0$, for $\varphi_{CBH}$ is $C^\omega$, in particular continuous, in both arguments.$\quad\square$

We can also use the Rossmann Snakeskin Construction to drop the “compact” restriction of the above:

**Lemma 15.11**

**: (Identity is Limit Point Of One Paramter Subgroup of a Lie Group)**

Let $\G$ be any connected Lie group and $\mathfrak{A}\subset\G$ be as in Lemma 15.10. Then the group identity $\id$ is a limit point of $\mathfrak{A}$.

**Proof:** By the Rossmann Snakeskin Construction $\G=\bigcup \limits_{k=1}^\infty \gamma_k \exp(\mathcal{G})$ where $\mathcal{G}\subset\g$ is an open ball neighbourhood of $\Or$ in the Lie algebra $\g=\operatorname{Lie}(\G)$ small enough to allow the construction to work and the $\gamma_k$ are finite products of members of $\exp(\mathcal{G})$, all as in Theorem 10.5. Then, certainly, $\G=\bigcup \limits_{k=1}^\infty \gamma_k \exp(\bar{\mathcal{G}})$. Since the union is countable, at least one of the compact, bounded $\gamma_k \exp(\bar{\mathcal{G}})$ must contain an infinite number of points (indeed $\beth_1=2^{\aleph_0}$ of them) of $\mathfrak{A}\subset\G$. Now the rest of the proof of Lemma 15.10 runs exactly as above. $\quad\square$.

So here at last is the proof of our little jewel:

**Theorem 15.12****: (van der Waerden, 1933; Rigidity of Lie Group Structure for Simple, Compact Lie Groups)**

Let $\G$ be a simple, compact, connected Lie group with Lie algebra $\g$, and let $\phi:\G\to\G$ be any abstract group isomorphism (*i.e.* not needfully an isomorphism preserving Lie group structure). Then $\phi$ is needfully continuous and therefore, by Theorem 7.12 $\phi$ is $C^\omega$. Otherwise put: all topologies that make a simple group into a connected, compact Lie group must be homeomorphic to one another, *i.e.* the topology that makes a simple group into a connected, compact Lie group is unique up to continuous homeomorphism. Otherwise put: any two simple compact connected Lie groups which are isomorphic as abstract groups are needfully isomorphic as Lie groups, with isomorphic Lie algebras with nonsingular linear Lie map between them.

**Proof: **Show Proof

**Step 1: **For any $\alpha\in\G$ we define a subset $\mathcal{M}(\alpha)$ as follows:

\begin{equation}\label{VanDerWaerdenTheorem_1}\mathcal{M}(\alpha) =\left\{\left.\prod\limits_{k=1}^N \left(\gamma_k\, \beta_k\, \alpha\, \beta_k^{-1}\, \alpha^{-1}\, \gamma_k^{-1}\right)\right|\;\; \beta_k,\,\gamma_k \in \G\right\}\end{equation}

Here $N$ is the Lie group’s dimension. Take heed that this collection is wholly in terms of group theoretical ideas; no topology or Lie algebra is specified. Then it will be shown in step 2 that the following claims are true:

- For any nucleus (neighbourhood of the identity) $\mathcal{K}_{\id}$ in
*any*topology $\mathscr{T}$ making $\G$ into a connected, compact Lie group, and for*any*one parameter subgroup $\mathfrak{A}\subset\G$, whether the $\mathfrak{A}$ be continuous or not in the particular subgroup in question, then there is an $\alpha_0\in\mathfrak{A}$ such that $\mathcal{M}(\alpha_0) \subset \mathcal{K}_{\id}$ is wholly contained within the nucleus; and - For
*any*noncentral $\alpha\in\G\setminus\{\id\}$ and for*any*topology $\mathscr{T}^\prime$ making $\G$ into a connected Lie group, the set $\mathcal{M}(\alpha)$ contains a nucleus $\mathcal{K}_{\id}^\prime$ in this topology $\mathscr{T}^\prime$. Note that for this claim, $\G$ doesn’t need to be compact, but this relaxation does not afford any advantage in this proof; - Likewise, by left translation, for
*any*neighbourhood $\mathcal{K}_\zeta$ of any $\zeta\in\G$ in any topology $\mathscr{T}$ making $\G$ into a connected, compact Lie group topology and for any one parameter subgroup $\mathfrak{A}\subset\G$ whether continuous or not with respect to the topology $\mathscr{T}$, then $\,\exists\alpha_\zeta\in\mathfrak{A}\,\ni\, \zeta\, \mathbb{M}(\alpha_\zeta)\subseteq\mathcal{K}_\zeta$; and - For any noncentral $\alpha\in\G\setminus\{\id\}$ and for
*any*topology $\mathscr{T}^\prime$ making $\G$ into a connected Lie group, $\zeta\, \mathbb{M}(\alpha,\,\beta)$ contains a neighbourhood of $\zeta$$.

Clearly claims 3 and 4 follow simply from claims 1 and 2 and indeed claims 1 and 2 imply and are implied by claims 3 and 4.

This means the following in particular. Let $\mathscr{T}_1$ and $\mathscr{T}_2$ be two different Lie group topologies making $\G$ connected and compact. Let $\mathcal{O}_1\in \mathscr{T}_1$ be open in $\mathscr{T}_1$; then, since $\mathcal{O}_1$ is a neighbourhood of all its points, $\mathcal{O}_1=\bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}_{\xi}$ where $\mathcal{O}_\xi\in\mathscr{T}_1$ is an open neighbourhood of $\xi$ and $\mathcal{O}_\xi\subseteq\mathcal{O}_1$. Now, for any one parameter group $\mathfrak{A}$, by claim 3 above we can find an $\alpha_\xi\in\mathfrak{A}$ such that $\mathbb{M}(\alpha_\xi)\subseteq \mathcal{O}_\xi$ for each $\xi\in\mathcal{O}_1$. Now we look at the topology $\mathscr{T}_2$: by claim 4 we can find an open neighbourhood $\mathcal{O}^\sim_\xi\in\mathscr{T}_2$ of $\xi$ *in this second topology* such that $\xi\in\mathcal{O}^\sim_\xi\subseteq\mathbb{M}(\alpha_\xi)\subseteq \mathcal{O}_\xi$ for every $\xi\in\mathcal{O}_1$ ($\alpha_\xi$ is not central: $\alpha_\xi\not\in\mathscr{Z}(\G)$). Therefore $\bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}^\sim_{\xi}$, a union of open sets in $\mathscr{T}_2$, is open in the second topology. But now $\mathcal{O}_1\subseteq\bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}^\sim_{\xi}\subseteq \bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}_{\xi} = \mathcal{O}_1$ and $\mathcal{O}_1=\bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}^\sim_{\xi}$. Therefore $\mathcal{O}_1$ is open in the second topology $\mathscr{T}_2$. So every set open in $\mathscr{T}_1$ is also open in $\mathscr{T}_2$; by swapping the roles of $\mathscr{T}_1$ and $\mathscr{T}_2$, we show that every set open in $\mathscr{T}_2$ is open in $\mathscr{T}_1$ and *so the two topologies must be precisely the same topology*, so that the topology making the group a connected, compact Lie group is indeed unique.

**Step 2:** So now we must prove claims 1. and 2. (claims 3. and 4. follow from the other two simply by left translation) and van Der Waerden’s theorem is proven.

**Proof of claim 1:** By Lemma 15.10, we can choose a sequence $\{\alpha_k\}_{k=1}^\infty\subset\mathfrak{A}$ with $\lim\limits_{k\to\infty}\,\alpha_k = \id$. Now suppose we have *any* nucleus $\mathcal{K}$ in *any* Lie group topology fulfilling axioms 1 through 5 which also makes $\G$ compact. Then there must be at least one $\alpha_j$, say $\alpha_{j_0}$, such that $\mathbb{M}(\alpha_{j_0})\subseteq \mathcal{K}$. For, were this not so, then there would be sets $\{\beta_{k,\,j}\}_{k=1}^N$ and $\{\gamma_{k,\,j}\}_{k=1}^N$ for each $j\in\mathbb{N}$ in the construction of $\eqref{VanDerWaerdenTheorem_1}$ such that:

\begin{equation}\label{VanDerWaerdenTheorem_2}\zeta_j=\prod\limits_{k=0}^N\left(\gamma_{k,\,j}\,\beta_{k,\,j}\,\alpha_j\,\beta_{k,\,j}^{-1}\,\alpha_j^{-1}\,\gamma_{k,\,j}^{-1}\right);\;\zeta_j\not\in\mathcal{K}\end{equation}

lies outside our nucleus $\mathcal{K}$ for every $j\in\mathbb{N}$. By the Bolzano-Weierstrass property, the sequences $\{\beta_{k,\,j}\}_{j=1}^\infty$ and $\{\gamma_{k,\,j}\}_{j=1}^\infty$ for $k=1,\,2,\,\cdots,\,N$ each hold at least one limit point, and so there is a subsequence $\{j_\mu\}_{\mu=1}^\infty$ of indices such that the subsequences $\{\beta_{k,\,{j_\mu}}\}_{\mu=1}^\infty$ and $\{\gamma_{k,\,{j_\mu}}\}_{\mu=1}^\infty$ all simultaneously converge (see comments after proof) to limit points $\tilde{\beta}_k,\,\tilde{\gamma}_k\,k=1,\,2,\,\cdots,\,N$. So now we choose the subsequence of the $\zeta_j$ in $\eqref{VanDerWaerdenTheorem_2}$ corresponding to this subsequence of indices: it is a convergent sequence and:

\begin{equation}\label{VanDerWaerdenTheorem_3}\lim\limits_{\mu\to\infty}\,\zeta_{j_\mu} = \lim\limits_{\mu\to\infty}\,\prod\limits_{k=0}^N\left(\gamma_{k,\,j_\mu}\,\beta_{k,\,j_\mu}\,\alpha_{j_\mu}\,\beta_{k,\,{j_\mu}}^{-1}\,\alpha_{j_\mu}^{-1}\,\gamma_{k,\,j_\mu}^{-1}\right) = \prod\limits_{k=0}^N\left(\tilde{\gamma}_k\,\tilde{\beta}_k\,\id\,\tilde{\beta}_k^{-1}\,\id\,\tilde{\gamma}_k^{-1}\right) = \id\end{equation}

Therefore, owing to the convegent subsequence, there must be infinitely many $\zeta_j$ within $\mathcal{K}$. This gainsays our assumption that all the $\zeta_j$ lie outside $\mathcal{K}$, and so we see that there must be an $\alpha_{j_0}$ such that $\mathbb{M}(\alpha_{j_0}) \subseteq\mathcal{K}$ for any nucleus $\mathcal{K}$ in any compact, connected Lie group topology, thus proving claim 1.

If you are mystified at why there are both $\beta$s and $\gamma$s in the above, take heed that one set is further to the needs of the above proof, which would work just as well if all the $\gamma_k$ were set to the identity $\id$. The two sets are only needed for the proof of claim 2. We need to let $\beta$ to wander over the whole group $\G$, because we had to apply claims 3. and 4. to infinitely many open subsets in the unions $\bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}^\sim_{\xi}$ and $\bigcup\limits_{\xi\in\mathcal{O}_1}\,\mathcal{O}_{\xi}$ above, so that, although one $\beta$ will work for each single application of claims 3. and 4., the same $\beta$ won’t in general work for all of the $\mathcal{O}_{\xi}$ and $\mathcal{O}^\sim_{\xi}$.

**Proof of claim 2: **For any noncentral $\alpha\in\G$ and any topology $\mathscr{T}^\prime$ making $\G$ into a connected Lie group, we can choose a one parameter subgroup $\mathfrak{B}=\{\exp(\tau\,X_\mathfrak{B})|\;\tau\in\R\}$ that is continuous (thus $C^1$, thus $C^\omega$ by Theorem 6.10) with respect to this topology (*i.e.* $X_\mathfrak{B}\in\g$ belongs to the Lie algebra defined by this topology) so that the path:

\begin{equation}\label{VanDerWaerdenTheorem_4}\sigma_\mathfrak{B}:\R\to\G;\;\sigma_\mathfrak{B}(\tau) = \exp(\tau\,X_\mathfrak{B})\,\alpha\,\exp(-\tau\,X_\mathfrak{B})\,\alpha^{-1}\end{equation}

is a $C^1$ path with respect to the topology $\mathscr{T}^\prime$ with $\sigma_\mathfrak{B}(0)=\id$ and with nonzero $\tilde{X}_\mathfrak{B} = X_\mathfrak{B} – \Ad(\alpha)\,X_\mathfrak{B}$ tangent there. Then all products of the form:

\begin{equation}\label{VanDerWaerdenTheorem_5}\prod\limits_{k=1}^N\,\gamma_k\,\sigma_\mathfrak{B}(\tau_k)\,\gamma_k^{-1}\end{equation}

wjere $\gamma_k\in\G$ and $\tau_k\in\R$ are clearly contained in $\mathcal{M}(\alpha)$, where $\gamma_k\in\G$ are any group members. Now consider the vector space $\mathcal{W}\subset\g$ spanned by all entities of the form $\exp(s \,Y)\,\tilde{X}_\mathfrak{B}\,\exp(-s \,Y) = \exp(s\,\ad(Y))\,\tilde{X}_\mathfrak{B}$ where $Y\in\g$ and $s\in\R$. Being a vector space over the reals, all Cauchy sequences in $\mathcal{W}$ converge in $\mathcal{W}$ so that $\left.\d_s \exp(s\,\ad(Y))\,\tilde{X}_\mathfrak{B}\right|_{s=0}=\left[Y,\,\tilde{X}_\mathfrak{B}\right]\in\mathcal{W}$. Moreover, if $Z\in\mathcal{W}$, then by the same construction $\left[Y,\,Z\right]\in\mathcal{W},\,\forall\,Y\in\g$ so that **( i) **$\mathcal{W}$ is a Lie subalgebra of $\g$ and

**(**an ideal of $\g$, thus, under the Lie correspondence, is the Lie algebra of a Lie subgroup that is also a normal subgroup by Lemma 15.7. But our assumption is that $\G$ is a simple Lie group, $\g$ cannot contain any nontrivial proper ideals, therefore $\mathcal{W}$ must be the whole of $\g$ and so there are $\tilde{\gamma}_k\in\G$ such that the $N$ tangents to the paths $\tilde{\gamma}_k\, \sigma_\mathfrak{B}(\tau)\,\tilde{\gamma}_k^{-1}$ at $\tau=0$ span $\g$. Therefore:

*ii*)\begin{equation}\label{VanDerWaerdenTheorem_6}\mu:\R^N\to\G;\;\mu(\tau_1,\,\tau_2,\,\cdots,\,\tau_N) = \prod\limits_{k=1}^N\,\tilde{\gamma}_k\,\sigma_\mathfrak{B}(\tau_k)\,\tilde{\gamma}_k^{-1}\end{equation}

defines a system of canonical co-ordinates of the second kind, and therefore there are $\tau_j$ in $\eqref{VanDerWaerdenTheorem_6}$ that represent every member of some nucleus $\mathcal{K}_\alpha^\prime$ in the topology $\mathscr{T}^\prime$ and so there is nuclues $\mathcal{K}_\alpha^\prime\subseteq\mathcal{M}(\alpha)$ for every noncentral $\alpha\in\G$ and therefore claim 2. is proven. $\quad\square$

Another proof to the above theorem by a different method was published in 1930 by Élie Cartan[Cartan, 1930] before van der Waerden’s work. Cartan’s proof is quite a bit fiddlier than van der Waerden’s.

It is interesting that claim 1. needs only compactness (without simplehood) and claim 2. only simplehood (without compactness) to make them true.

**Note On Group Dimension $N$ in proof of Theorem 15.12: **In the proof of van der Waerden’s theorem, we assumed a dimension $N$ of the Lie group concerned. It may seem that this dimension is a topological, rather than pure group theoretic notion, but this is not so here. The proof above works with different dimensions of the Lie groups defined by assumedly different topologies: once we show that the topologies must be the same, it follows that the dimensions of the two Lie groups must be the same. Witness that in Step 1 in the proof, we show how to use claims 3. and 4. to express a set $\mathcal{O}_1$ open in topology $\mathscr{T}_1$ as a union of sets open in topology $\mathscr{T}_2$. When the steps of the argument are checked, it is seen that $N$ needs to be set to the dimension $N_2$ of the Lie group defined by topology $\mathscr{T}_2$ to make the argument work for both claims 3. and 4. When we go back the other way to show that a set $\mathcal{O}_2$ open in topology $\mathscr{T}_2$ is needfully also open in topology $\mathscr{T}_1$, we need to set $N$ to the dimension $N_1$ of the Lie group defined by topology $\mathscr{T}_1$. Of course, once we have shown that $\mathscr{T}_1=\mathscr{T}_2$, it must follow that $N_1=N_2$: the dimensions must be the same.

To come back to the statement that we can choose subsequences used in the proof of claim 1. that the $\beta$ and $\gamma$ sequences contain subsequences $\{\beta_{k,\,{j_\mu}}\}_{\mu=1}^\infty$ and $\{\gamma_{k,\,{j_\mu}}\}_{\mu=1}^\infty$ that make them all converge simultaneously. There are two ways to prove this: we can think of a $2\,N$ element vector of all the $\beta$s and $\gamma$s and use the fact that the Cartesian product $\G^N$ is compact in the product topology, thus has the Bolzano-Weierstrass property. Otherwise, we can work through a more first principles, iterative proof as follows. The set $\{\gamma_{1,\,j}\}_{j=1}^\infty$ must contain a limit point $\tilde{\gamma}_1$, and so there is a convergent subsequence $\{\gamma_{1,\,{j_{1,\mu}}}\}_{\mu=1}^\infty\subseteq \{\gamma_{1,\,j}\}_{j=1}^\infty$ converging to the limit point. Now we look at the set $\{\gamma_{2,\,{j_{1,\mu}}}\}_{\mu=1}^\infty$; this is an infinite set in itself with a limit point $\tilde{\gamma}_2$ and a subsequence $\{\gamma_{2,\,{j_{2,\mu}}}\}_{\mu=1}^\infty\subseteq\{\gamma_{2,\,{j_{1,\mu}}}\}_{\mu=1}^\infty$ of this one that converges to this second limit point. But the subsequence $\{\gamma_{1,\,{j_{2,\mu}}}\}_{\mu=1}^\infty$ still converges to the first limit point $\tilde{\gamma}_1$: we now have a subsequence $\{j_{2,\mu}\}_{\mu=1}^\infty$ of indices such that $\{\gamma_{1,\,{j_{2,\mu}}}\}_{\mu=1}^\infty\to\tilde{\gamma}_1,\,\{\gamma_{2,\,{j_{2,\mu}}}\}_{\mu=1}^\infty\to\tilde{\gamma}_2$ converge simultaneously. We work through the finite set of analogous steps to find the subsequence $\{j_{2\,N,\mu}\}_{\mu=1}^\infty$ of indices that makes all the sequences converge at once to their respective limit points: $\{\gamma_{k,\,{j_{2\,N,\mu}}}\}_{\mu=1}^\infty\to\tilde{\gamma}_k$ and $\{\beta_{k,\,{j_{2\,N,\mu}}}\}_{\mu=1}^\infty\to\tilde{\beta}_k$.

Once one has proven Theorem 15.12, the following follows straight away.

**Theorem 15.13****: (van der Waerden, 1933; Rigidity of Lie Group Structure for Semisimple, Compact Lie Groups)**

Any two compact connected *semisimple *Lie groups which are isomorphic as abstract groups are needfully isomorphic as Lie groups, with isomorphic Lie algebras with nonsingular linear Lie map between them. The statement of Theorem 15.12 also holds if “simple” is replaced by “semisimple” throughout. $\square$

Some wonderful insights into the personality of Bartel van der Waerden can be found in [Dold, 1997]. Particularly amusing and poignant is his account of being on honeymoon with his wife Camilla, and being sternly summonsed back to Göttingen by Emily Noether. One gets the impression that there was a rather romantic and nonmathematical side to van der Waerden, something one could imagine the single minded genius Noether might not appreciate! Indeed Bartel’s father, himself a teacher of mathematics, stressed on his son the importance of outdoor play as well as study for a child and restricted Bartel’s access to his mathematics library in the same way that a parent in the 2000s might keep their child away from too much iPad and computer screen time. Bartel’s father relented somewhat when he found that his son had rediscovered many results in trigonometry and even had his own names and definitions for trigonometric concepts.

**Definition 15.14****: (Absolutely Simple Lie Algebras and Groups)**

An *absolutely* *simple Lie algebra *is a simple Lie algebra $\g$ that stays simple when complexified by the following procedure: (1) $\g$ (say of dimension $N$) is transformed into an $N$-dimensional Lie algebra $\g_\mathbb{C}$ over the field $\mathbb{C}$ with the same structure co-efficients and commutation relationships as $\g$, (2) the algebra $\g_\mathbb{C}$ is made into a dimension $2\,N$ Lie algebra $\g_W$ over the field $\R$ by replacing each member $\hat{X}_j$ of a $\g_\mathbb{C}$-basis by the two basis members $\hat{X}_j, \,i\,\hat{X}_j$ (Freudenthal’s concept of *waiving*, as in Example 1.5)

**Example 15.15****: (Absolutely Simple Lie Group)**

The group $SU(2)$ and then consider what group we get if we let the three real Lie algebra dimensions be complex instead or real: we get two Lie algebra basis members $\hat{X}_j,\,\hat{X}_j^\prime = i\,\hat{X}$ for each $\hat{X}_j$ in a $\mathfrak{su}(2)$-basis. Thus the group concerned is $SL(2,\,\mathbb{C})$, with six dimensional Lie algebra, as in as in Example 1.5. $SL(2,\,\mathbb{C})$ is simple, therefore $SU(2)$ is absolutely simple.

One thus cannot divorce the topology from the group in the compact simple Lie group case. Other variations on this theme are possible: for example [Freudenthal, 1941] considers absolutely simple Lie groups instead of compact ones, builds a purely group theoretical definition of some compact neighbourhood of the group identity $\id$ and then runs van der Waerden’s construction with the $\beta_j, \gamma_j$ wandering through this wholly-group-theoretically-defined compact set rather than through the whole group. A modern statement of the question’s status is given in [Kramer, 2011], where it is proven that every isomorphism between *any* locally compact and $\sigma$-compact group $\Gamma$ and an absolutely simple Lie group $\G$ (given a particular Lie structure through definition of $\Nid$ and $\lambda$) is needfully a homeomorphism. So, as in Example 15.3, there can be discontinuous automorphisms of a simple Lie group, but when we define a new topology as the one generated by the image in the latter topology of open balls in the former topology, we recover a new Lie group that is isomorphic as a Lie group to the original. Corresponding results hold for semisimple Lie groups.

**References:**

- Edwin Hewitt and Karl R. Stromberg, “
*Real and Abstract Analysis (Graduate Texts in Mathematics)*“, Springer-Verlag, Berlin, 1965. Chapter 1, §5 - Barten Leendert van der Waerden, “
*Stetigkeitssätze für halbeinfache Liesche Gruppen*“, Math. Zeitschrift,**36**, 1933, pp780–786 - James R. Munkres, “Topology (2nd Edition)”, Prentice-Hall International Limited, London UK, 2000. Chapter 4, §34 “The Urysohn Metrization Theorem”
- Paul S. Alexandroff (Pavel in his mother tongue, Paul for German publisher) and Heinz Hopf, “
*Topologie; Erster Band*“, Springer-Verlag Berlin, 1935. - Élie Cartan, “
*Sur les représentations lineaires des groupes clos*” Comment. Helvet.**2**(1930),*pp*269-283 - Yvonne Dold-Samplonius, “
*Interview with Bartel Leendert van der Waerden*“, Notices of the Amer. Math. Soc.**44**, #3,*pp*313-320, March 1997 - Hans Freudenthal, “
*Die Topologie der Lieschen Gruppen als algebraisches Phänomen. I*“, Annals of Mathematics, Second Series,**42**#5, 1941,*pp*1051-1074 - Linus Kramer, “
*The topology of a semisimple Lie group**is essentially unique*“, Advances in Mathematics**228**(2011) pp2623–2633

You must log in to post a comment.