## Exponential maps of Lie groups: applications

We first recall a theorem proved in “Exponential maps on Lie groups”:

Ken’s Theorem: Let G be a Lie groups with a bi-invariant metric (e.g. if G is compact). Then the exponential map $exp: \mathfrak g \to G$ is the same as the Riemannian exponential map

$Exp_e : T_eG \to G$ at the identity $e \in G$.

In this note we assume that G is compact, and we wish to derive some consequences of the above theorem. The most important one is the Cartan’s theorem proved below. Some information concerning the fundamental group of G can also be obtained. One might consult the two previous blogs by KKK on Lie groups for the background information needed.

First we state a lemma.

Lemma 1: The exponential map $\mathfrak g \to G$ is surjective.

The lemma is a direct consequence of Ken’s theorem, because any metric defined on a compact manifold is complete, and completeness is equivalent to the fact that

$\forall p\in G, Exp_p :T_pG \to G$ is surjective.

Next we introduce the notion of a maximal torus. By definition, a maximal torus T in G is a maximal connected abelian subgroup of G. Here maximal means that if T’ contains T and T’ is connected and abelian, then $T' = T$. One can check that T must be closed (because the closure of any subgroup is still a subgroup), and  T must be a torus

$T \cong \mathbb S ^1 \times \cdots \times \mathbb S^1$

by compactness of G (the proof is omitted, it depends on some analysis on the exponential map).

Note that maximal tours are not unique: if T is a maximal torus, then $xTx^{-1}$ is also a maximal torus. Cartan’s theorem states that all maximal torus are of this form:

Cartan’s theorem: Let T, T’ be two maximal torus. Then there is a $g\in G$ such that $T' = gTg^{-1}$.

We first state a proposition.

Proposition 1: Let T be a maximal torus and $g\in G$. Then $\exists x\in G$ such that $g \in xTx^{-1}$.

Proof of proposition 1: Let $g\in G$ and $h_0 \in T$ be a generator of T. Here a generator means that $\overline{\{h_0^n : n\in \mathbb N\}} =T$ (why exists?). Using Lemma 1, there is $X \in \mathfrak g$ and $H_0 \in \mathfrak t$ such that $e^X = g$ and $e^{H_0} = h_0$. Let $x\in G$ be choosen such that $\langle X, Ad_{x} H_0 \rangle$ is maximized.

Let $H = Ad_{x} H_0$ (Then $h = e^H$ is a generator of $xTx^{-1}$). For any $Y \in \mathfrak g$, the term $\langle X, Ad_{e^{tY}} H \rangle$ has a maximum at $t=0$. Thus

$0 = (\langle X, Ad_{e^{tY}} H \rangle)'_{t=0} = \langle X, [Y,H]\rangle = \langle Y, [H,X]\rangle$.

The proof for the second and thrid equalities can be found in KKK’s “Lie groups with bi-invariant Riemannian metric”. As Y is arbitrary, $[H,X]=0$ and $e^{tX}h =he^{tX}$. Thus $e^{tX}h^n = h^n e^{tX}$ for any n and so $e^{tX}$ commutes with every elements in $xTx^{-1}$. Let W be the connected abelian group generated by $e^{tX}$ and $xTx^{-1}$. As $xTx^{-1}$ is maximal abelian, we have $W = xTx^{-1}$ and $g = e^X\in xTx^{-1}$.

Proof of Cartan’s theorem: Let t’ be a generator of T’. Then $t' \in xTx^{-1}$ for some x, by proposition 1. Then $T' \subset xTx^{-1}$.  As T’ is maximal abelian, $T' = xTx^{-1}$ and the theorem is proved.

Indeed, Cartan’s theorem is quite important in representation theory of compact Lie group. As T is abelian, representation on T is well known (just Fourier analysis). One hopes to know all representations of G by looking at that of T, and Cartan’s theorem tells us that all maximal abelian connected subgroups T are the same.

To end this note, we give an application of Cartan’s theorem. Let T be a maximal torus and

$i : T \to G$

be the inclusion map.

Theorem 2: The induced map $i_{*}:\pi_1 (T) \to \pi _1 (G)$ is surjective. Thus the fundamental group of G is abelian and finitely generated.

Proof of theorem 2: Let $[\gamma] \in \pi_1(G)$. Then by compactness of G, $\gamma$ can be choosen to be a geodesic (see Jost for a detailed discription, at least it is intuitively clear…).  By a left multiplication (which is an isometry as the metric is bi-invariant) if neccesary, we can assumed that this geodesic starts at the identity e in G. Then $\gamma(t) = e^{tX}$ for some $X \in \mathfrak g$. Note that this subset forms an abelian subgroup of G, thus is contained in some maximal torus T’. By Cartan’s theorem, $T' = xTx^{-1}$ for some x and

$\gamma (t) = x \beta(t) x^{-1}$,

where $\beta(t) \in T$. Let $x(u)$ be a path joining e to x. Then obviously

$A(t,u) = x(u) \beta(t) x(u)^{-1}$

forms a homotopy between $\gamma$ and $\beta$. Thus $i_* ([\beta]) = [\gamma]$ and the theorem is proved.

To avoid confusion, in the proof of Proposition 1, $\langle X, Ad_{e(tY)}H\rangle$ should be written as $\langle X, Ad_{e^{tY}}H\rangle$ instead (there are two instances).