ANNALES SCIENTIFIQUES DE L'É.N.S.

BERTRAM KOSTANT

**On convexity, the Weyl group and the Iwasawa decomposition**

*Annales scientifiques de l'É.N.S. 4ᵉ série*, tome 6, n°4 (1973), p. 413-455

<http://www.numdam.org/item?id=ASENS_1973_4_6_4_413_0>

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# ON CONVEXITY, THE WEYL GROUP AND THE IWASAWA DECOMPOSITION

BY BERTRAM KOSTANT (*)

## 1. Introduction

1.1. If $p$ is a positive definite matrix how does the spectrum of $p$ change when $p$ is multiplied by a unitary matrix ? C. Thompson in [14] (Theorem 1) and much earlier A. Horn in [16], proved the following theorem : Let $p$ and $q$ be any two positive definite $n \times n$ matrices and let $x_1 \ge x_2 \ge \dots \ge x_n$ and $y_1 \ge y_2 \ge \dots \ge y_n$ denote the respective sets of eigenvalues. Then there exists a unitary matrix $v$ such that $pv$ and $q$ have the same spectrum if and only if $\det p = \det q$ and

$$ (1.1.1) \qquad x_1 x_2 \dots x_i \ge y_1 y_2 \dots y_i \quad \text{for all } 1 \le i \le n. $$

This rather nice theorem may be cast in a form which makes sense for any semi-simple Lie group $G$. The point is that the theorem is then true for $G$. Let

$$ (1.1.2) \qquad G = \text{KAN} $$

be an Iwasawa decomposition for $G$. See e. g. p. 234 in [2]. If $g \in G$ and $g = \text{kan}$ where $k \in K$, $a \in A$ and $n \in N$ then $a = a(g)$ is called the $a$-component of $g$.

Now let $a, g$ ($a \subseteq g$) be the Lie algebras of $A$ and $G$ and let $W$ be the Weyl group of $(a, g)$ regarded as operating in $a$ (and $A$). For each $x \in a$ let $a(x)$ be the convex hull of the Weyl group orbit $W.x$. Correspondingly for each $b \in A$ let

$$ A(b) = \exp_a (\log b) $$

[so that, multiplicatively, $A(b)$ is the convex compact set having the Weyl group orbit $W.b$ as its extreme points].

<footnote>(*) This paper was partially supported by a grant from the National Science Foundation Grant N° GP 28969.</footnote>
Now let $b \in A$ be arbitrary. In doing analysis on semi-simple Lie groups or in representation theory, one is very often concerned with the $a$-component, $a(bv)$, of the product $bv$ where $v \in K$ is arbitrary. (Recall, for example, Harish-Chandra's well known formula [see (4.1.2)] for the elementary spherical functions on $G$.) The generalization which we will prove here of the Horn-Thompson theorem determines the set of all $a(bv)$ as $v$ runs through $K$. Not only is $a(bv) \in A(b)$ but in fact one has

**THEOREM 4.1.** — Let $b \in A$ be arbitrary. Then

$$A(b) = \{ a(bv) \mid v \in K \}.$$

The connection between the Weyl group convexity formulation in Theorem 4.1 and the Horn-Thompson theorem may be clarified by introducing a natural partial ordering in $G$. This partial ordering is different from (but was inspired by) a partial ordering defined by Thompson for $GL(n, \mathbb{C})$. Thompson's definition is based on the polar decomposition and hence depends on a particular choice of a Cartan decomposition of $\mathfrak{g}$. One obtains an invariant partial ordering by just noting that any element $g \in G$ may be uniquely written

$$(1.1.3) \qquad g = \text{ehu}$$

where $e$ is elliptic, $u$ is unipotent and $h$ is hyperbolic and all three elements, $e, h$ and $u$ commute. Write $h = h(g)$. An element is hyperbolic if and only if it is conjugate to an element in $A$. Now for any element $g \in G$ let $A(g) \subseteq A$ be defined by putting $A(g) = A(b)$ where $b \in A$ is conjugate to the hyperbolic component, $h(g)$, occurring in (1.1.3). It is easy to see that $A(g)$ is independent of $b$. Given $f, g \in G$ define

$$(1.1.4) \qquad g \ge f \quad \text{in case } A(f) \subseteq A(g).$$

The ordering is independent of $A$ since one has

**THEOREM 3.1.** — For any finite dimensional representation $\pi$ of $G$ and any $g \in G$ let $|\pi(g)|$ denote the spectral radius of $\pi(g)$. Then if $g, f \in G$ one has $g \ge f$ if and only if $|\pi(g)| \ge |\pi(f)|$ for all finite dimensional representations $\pi$ of $G$.

**REMARK 1.1.** — In case say $G \subseteq SL(n, \mathbb{C})$ for some $n$ the ordering in $G$ induced by that in $SL(n, \mathbb{C})$ is not necessarily the given ordering in $G$. Thus it is not sufficient just to define the ordering in $SL(n, \mathbb{C})$ (see Remark 3.1.1).

Theorem 4.1 now says that $a(bv)$ runs through all elements $a \in A$ where $a \le b$ as $v$ runs over $K$. That is, by Theorem 3.1, all $a \in A$ such that $|\pi(a)| \le |\pi(b)|$ for any finite dimensional representation $\pi$.
In case $G = SL(n, \mathbf{C})$ and $b = p, a = q$ are diagonal matrices, this just means (1.1.1) since it suffices to consider only the (fundamental) representations $\pi_1$, of $G$, $1 \le i \le n-1$, on the space $\Lambda^i \mathbf{C}^n$ of alternating $i$-vectors. But one has $|\pi_i(b)| = x_1 \dots x_i$ and $|\pi_i(a)| = y_1 \dots y_i$. This in essence establishes the connection of Theorem 4.1 with the Horn-Thompson theorem. Not quite, since the Horn-Thompson theorem deals with eigenvalues and Theorem 4.1 deals with the Iwasawa decomposition.

Theorem 5.4 is the direct generalization of the Horn-Thompson theorem to the arbitrary semi-simple case. Indeed for SL($n, \mathbf{C}$) it reduces to Horn's theorem which is stronger than Thompson's result in that Horn also proves the additional fact that if $v$ is any unitary matrix and the eigenvalues $z_1, \dots, z_n$, of $pv$, (no longer necessarily positive) are ordered so that $|z_1| \ge |z_2| \ge \dots \ge |z_n|$ then one has

$$x_1 \dots x_i \ge |z_1| \dots |z_i| \text{ for all } i.$$

1.2. One may apply Theorem 4.1 to obtain some results on K-double cosets. For any $a \in A$ let

$$G_a = \{ g \in G \mid p(g) \ge a \}.$$

Here $p(g)$ is the “positive definite” component of $g$ relative to the polar decomposition [see (4.2.6)].

Now one knows that

$$(1.2.1) \qquad G = \text{KAK.}$$

This fact is quite useful. It implies among other things that a spherical function on $G$ is determined by its restriction to $A$. If we say that, $f, g \in G$ are congruent when $KfK = KgK$ then (1.2.1) of course says that any element in $G$ is congruent to an element of $A$. Now if we replace $A$ by a coset of $N$ this is no longer true. Nevertheless one can now say exactly which elements in $G$ are congruent to this coset.

**THEOREM 5.3.** — Let $a \in A$. Then

$$G_a = K a N K.$$

Since $p(g) \ge 1$ for all $g \in G$ the special case where $a = 1$ yields

**THEOREM 5.1.** — One has

$$G = \text{KNK.}$$

**REMARK 1.2.** — Theorem 5.1 says that a spherical function is determined by its restriction to $N$ (as well as its restriction to $A$). It is suggestive therefore from Theorem 5.1 that it might be quite interesting to see
what the spherical functions look like on N. Geometrically, Theorem 5.1 says that if X is the symmetric space G/K then any two points on X lie on some horocycle. In case X is the unit disc this is of course clear since the horocycles are just the circles in the disc that are tangent to the boundary.

An element $a \in A$ is called regular if $\sigma a = a$, $\sigma \in W$, implies $\sigma$ is the identity.

Another corollary of Theorem 5.3 is

**THEOREM 5.5.** — Let $a \in A$ be regular and let $D_a$ be the conjugacy class of $a$. That is $D_a = \{ gag^{-1} \mid g \in G \}$. Then

$$G_a = K D_a K$$

(See Remark 5.5).

1.3. One of the main points of the paper [14] was to recast some generalizations of the Golden-Thompson inequality made by Lenard in [9]. The Golden-Thompson equality states that if $x$ and $y$ are two $n \times n$ Hermitian matrices then

$$(1.3.1) \qquad \operatorname{tr} e^x e^y \ge \operatorname{tr} e^{x+y}.$$

The generalization made in [9] replaces the trace by the character of any representation of $GL(n, \mathbb{C})$. But now this generalizes to an arbitrary semi-simple Lie group in two steps, Theorem 6.1 and Theorem 6.3.

If $g \in G$ is hyperbolic and $\pi$ is a finite dimensional representation of G then the eigenvalues of $\pi(g)$ are all positive. Thus $|\pi(g)|$ then is just the maximal value of $\pi(g)$. Hence if $f, g \in G$ are hyperbolic and $g \ge f$ then by Theorem 3.1 the maximal eigenvalue of $\pi(g)$ is greater or equal to the maximal eigenvalue of $\pi(f)$. But the minimal eigenvalue of $\pi(g)$ is less or equal to the minimal eigenvalue of $\pi(f)$. Thus one cannot immediately compare $\chi_\pi(f)$ and $\chi_\pi(g)$ if $\chi_\pi$ is the character of $\pi$. However one has

**THEOREM 6.1.** — Let $f, g \in G$ be hyperbolic and assume $g \ge f$. Then if $\pi$ is any finite dimensional representation one has

$$\chi_{\pi}(g) \ge \chi_{\pi}(f).$$

Now let $k$ be the Lie algebra of $K$ so that $g = k + p$ is a Cartan decomposition of $g$ where $p$ is the (Killing form) orthocomplement of $k$ in $g$. If $P = \{ \exp x \mid x \in p \}$ it is easy to see that $P^2$ is the set of all hyperbolic elements in $G$ (see Proposition 6.2). Thus if $x, y \in p$ then $e^x e^y$ and $e^{x+}$
are hyperbolic. As a dividend of the partial ordering based on (1.1.3) rather than on the polar decomposition one has

**THEOREM 6.3.** — For any $x, y \in \mathfrak{P}$,

$$ (1.3.2) \qquad e^x e^y \ge e^{x+y} $$

so that, by Theorem 6.1, $\gamma_{\pi}(e^x e^y) \ge \gamma_{\pi}(e^{x+y})$ for any finite dimensional representation $\pi$ of G.

1.4. The relation (1.3.2) has a nice geometric interpretation. In fact it is really a statement about geodesic triangles on Riemannian symmetric spaces of negative curvature.

An element $x \in \mathfrak{g}$ is called real semi-simple if ad $x$ is diagonalizable and has real eigenvalues. Let $l \subseteq \mathfrak{g}$ denote the set of such elements. One notes that the exponential map sets up a bijection between $l$ and all hyperbolic elements in G. Furthermore the partial order in G defines a partial order in $l$. That is if $x, y \in l$ then $y \ge x$ if $\exp y \ge \exp x$. This is equivalent to the condition that $a(x) \subseteq a(y)$ where for any $z \in l$, $a(z) = a(z')$ and $z' \in a$ is an element conjugate to $z$. The statement $y \ge x$ is also equivalent to the condition that the maximal eigenvalue of $\pi(y)$ is greater or equal to the maximal eigenvalue of $\pi(x)$ for all finite dimensional representations $\pi$ of G.

If $x \in l$ and $(x, x)$ denotes the inner product defined by the Killing form then $(x, x) \ge 0$ and one puts $|x| = (x, x)^{1/2}$. For $x, y \in l$ one easily has

$$ (1.4.1) \qquad y \ge x \text{ implies } |y| \ge |x|. $$

The converse is false in general. See Remark 7.1.

Now let $X = G/K$ so that X has the structure (normalized by the Killing form on $\mathfrak{g}$) of a Riemannian symmetric of negative curvature. For any two points $r, s \in X$ let $d(r, s)$ be the distance of $r$ to $s$, i. e. $d(r, s)$ is the length of the unique geodesic arc segment $(r, s)$ joining $r$ to $s$.

Now let $o, r$, and $s$ be any three points in $X$ and consider the corresponding geodesic triangle. Let $a = d(o, r)$, $b = d(o, s)$ and $c = d(r, s)$. Also let $\psi$ be the angle at $o$ made by $(o, r)$ and $(o, s)$. Then the Law of Cosines in flat space is replaced by the inequality

$$ (1.4.2) \qquad c^2 \ge a^2 + b^2 - 2ab \cos \psi $$

on the space $X$. See Lemma 4 in [11]. But the point is that a geodesic arc segment $(r, s)$ carries more information than just its length $d(r, s)$. In fact one can naturally associate to $(r, s)$ a real semi-simple element
$x(r, s) \in \mathfrak{l}$. In the language of Elie Cartan the motion of X defined by $\exp x(r, s)$ is that transvection along the geodesic containing $(r, s)$ which carries $r$ to $s$.

But now $d(r, s) = |x(r, s)|$ and (1.4.2) is just the statement that $x(r, o) + x(o, s) \in \mathfrak{l}$ and

$$ (1.4.3) \qquad |x(r, s)| \ge |x(r, o) + x(o, s)|. $$

What (1.3.2) amounts to (see the proof of Theorem 7.2) is that we can remove the absolute value signs in (1.4.3).

**THEOREM 7.2.** — Let $o, r, s \in X = G/K$ be any three points. Then

$$ x(r, s) \ge x(r, o) + x(o, s). $$

1.5. Now let $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n$ be any $n$ real numbers and let $\lambda = (\lambda_1, \lambda_2, \dots, \lambda_n) \in \mathbb{R}^n = \mathfrak{a}_0$. For any permutation $\sigma$ on $1, 2, \dots, n$ let $\sigma\lambda = (\lambda_{\sigma^{-1}(1)}, \lambda_{\sigma^{-1}(2)}, \dots, \lambda_{\sigma^{-1}(n)}) \in \mathfrak{a}_0$ and let $\mathfrak{a}_0(\lambda) \subseteq \mathfrak{a}_0 = \mathbb{R}^n$ be the convex hull of all the vectors $\{\sigma\lambda\}$ for all permutations $\sigma$.

Now let $x = (x_{ij})$ be any $n \times n$ Hermitian matrix with eigenvalues $\lambda_1, \dots, \lambda_n$. Then A. Horn in [3] proved that the “diagonal” $(x_{11}, x_{22}, \dots, x_{nn})$ of $x$, regarded as a vector in $\mathbb{R}^n$, lies in $\mathfrak{a}_0(\lambda)$ and that one obtains all vectors in $\mathfrak{a}_0(\lambda)$ this way, by considering all Hermitian matrices $x$ with the eigenvalues $\lambda_1, \dots, \lambda_n$. This result may also be generalized to all semi-simple groups.

This time, however, the generalization, Theorem 8.2 is a statement about the Lie algebra $\mathfrak{g}$ and adjoint representations of $K$ on $\mathfrak{p}$ rather than, as in the case of Theorem 4.1, a statement about $G$ and double $K$-cosets. However the techniques, at least in one direction, of proving Theorem 4.1 and Theorem 8.2 are similar. Theorem 8.3 is an application of Theorem 8.2.

# CONTENTS

1. Introduction.

2. Preliminaries on elliptic, hyperbolic and unipotent elements.

3. The partial ordering in $G$ and the spectral radius $\pi_\lambda(g)$, $\lambda \in \hat{G}$.

4. The convexity theorem.

5. Applications : Some K-double coset theorems.

6. Character values and the partial ordering; a generalized Golden-Thompson inequality.

7. Applications to the geometry of symmetric spaces of negative curvature.

8. A conjugation convexity theorem; a generalization of a theorem of A. Horn.

## 2. Preliminaries on elliptic, hyperbolic and unipotent elements

2.1. Let $\mathfrak{g}$ be any semi-simple Lie algebra over the real numbers $\mathbb{R}$. An element $x \in \mathfrak{g}$ is called real semi-simple (resp. nilpotent) if $\mathrm{ad} x$ is
diagonalizable over $\mathbb{R}$ (resp. ad $x$ is nilpotent). In particular ad $x$ has
real eigenvalues if $x$ is real semi-simple.

Now let $G$ be any Lie group having $\mathfrak{g}$ as its Lie algebra and let $\exp : \mathfrak{g} \to G$ denote the exponential map. An element $a \in G$ is called hyperbolic (resp. unipotent) if $a$ is of the form $a = \exp x$ where $x$ is real semi-simple (resp. nilpotent). In either case the element $x$ is easily seen to be unique and we write $x = \log a$. Since only the identity $1 \in G$ is both hyperbolic and unipotent there is no ambiguity in this definition.

An element $e \in G$ is called elliptic if Ad $e$ is diagonalizable over $\mathbb{C}$ with eigenvalues of norm 1. One has the following :

**PROPOSITION 2.1.** — Let $g \in G$ be arbitrary. Then $g$ may be uniquely written

$g = ehu$

where $e$ is elliptic, $h$ is hyperbolic and $u$ is unipotent and where the three elements $e, h$ and $u$ commute.

*Proof*. — Let $\mathfrak{g}_c$ be the complexification of $\mathfrak{g}$ and let $G_c \subseteq \text{Aut } \mathfrak{g}_c$ be the adjoint group of $\mathfrak{g}_c$. Thus the adjoint representation (Ad) of $G$ maps $G$ into $G_c$.

Now since $G_c$ is algebraic one has the multiplicative Jordan decomposition $\text{Ad} g = s_1 u_1$ where $s_1, u_1 \in G_c$, $s_1$ is semi-simple (i. e. diagonalizable), $u_1$ is unipotent and $s_1$ and $u_1$ commute (see e. g. [12], p. 4-11). On the other hand by embedding $s_1$ in a complex torus of $G_c$ it is clear that we uniquely write $s_1 = e_1 h_1$ where $e_1, h_1 \in G_c$ are two commuting semi-simple elements where the eigenvalues of $e_1$ have norm 1 and the eigenvalues of $h_1$ are positive. Thus one has the decomposition

$$
(2.1.1) \qquad \text{Ad} g = e_1 h_1 u_1.
$$

The uniqueness of $u_1$ and $s_1$ imply (after conjugating by $u_1$) that all
three elements, $e_1$, $h_1$ and $u_1$ commute.

Now since $h_1$ and $u_1$ define automorphisms of $\mathfrak{g}_c$ and since all deri-
vations are inner we may uniquely write $h_1 = \exp \text{ad } x$, $u_1 = \exp \text{ad } z$
where $x, z \in \mathfrak{g}_c$, ad $z$ is nilpotent and ad $x$ is diagonalizable with real
eigenvalues. But then the commutativity of $h_1$ and $u_1$ implies the com-
mutativity of $h_1$ and ad $z$ and then the commutativity of ad $x$ and ad $z$
so that

$$
(2.1.2) \qquad [x, z] = 0.
$$

Now let $\sigma$ be the automorphism of $\mathfrak{g}_c$ defined by putting $\sigma = 1$ on $\mathfrak{g}$ and $-1$ on $i \mathfrak{g}$. Then $d \mapsto d^\sigma = \sigma d\sigma$ defines an automorphism of $G_c$. Clearly $\text{Ad} g$ is fixed under this automorphism. But then by the unique-
ness of the decomposition (2.1.1) it follows that $e_1$, $h_1$ and $u_1$ are fixed by the automorphism so that $\mathfrak{g}$ is stable under $e_1$, $h_1$ and $u_1$. But then $\mathfrak{g}$ is stable under ad $x$ and ad $z$ so that $x, z \in \mathfrak{g}$. But then $x$ is real semi-simple and $z$ is nilpotent. Thus $h, u \in G$ are respectively hyperbolic and unipotent where

$$ (2.1.3) \qquad h = \exp x, \quad u = \exp z. $$

Moreover they commute by (2.1.2). Furthermore they commute with $g$ since Ad $g$ commutes with $h_1$ and $u_1$ and hence, by the uniqueness of $x$ and $z$, fixes $x$ and $z$. Thus if

$$ (2.1.4) \qquad e = gh^{-1}u^{-1} $$

then $e$ commutes with $h$ and $u$. Also $g = ehu$. But clearly $\mathrm{Ad} \, e = e_1$ and hence $e$ is elliptic. Assume $g = e' h' u'$ is another decomposition satisfying the same conditions. Then by the uniqueness of (2.1.1) one has $\mathrm{Ad} \, e' = e_1$, $\mathrm{Ad} \, h' = h_1$ and $\mathrm{Ad} \, u' = u_1$. But by taking the differential of $\mathrm{Ad}$ one has

$$ \exp \mathrm{ad} \log h' = \exp \mathrm{ad} x = h_1 \quad \text{and} \quad \exp \mathrm{ad} \log u' = \exp \mathrm{ad} z = u_1. $$

Thus, by uniqueness, $x = \log h'$, $z = \log u'$ and hence $h' = h$, $u' = u$ and consequently $e' = e$.

Q. V. D.

Given $g \in G$ the components $e$, $h$ and $u$ of Proposition 2.1 will be written $e(g)$, $h(g)$ and $u(g)$ respectively and the decomposition

$$ (2.1.5) \qquad g = e(g)h(g)u(g) $$

will be called the complete multiplicative Jordan decomposition of $g$.

**REMARK 2.1.** — Note that by uniqueness if $f \in G$ then $h(fgf^{-1}) = fh(g)f^{-1}$ (similarly for the elliptic and unipotent components). In particular $f$ commutes with $g$ if and only if it commutes with all three components. Furthermore it is then easy to see that if $f, g \in G$ commute

$$ (2.1.6) \qquad h(fg) = h(f)h(g), \quad e(fg) = e(f)e(g) \quad \text{and} \quad u(fg) = u(f)u(g). $$

2.2. Now let $\mathfrak{g} = \mathfrak{k} + \mathfrak{p}$ be a fixed Cartan decomposition of $\mathfrak{g}$. Thus if $K$ is the connected subgroup of $G$ corresponding to $\mathfrak{k}$ then $\mathrm{Ad} K$ is a maximal compact subgroup of $\mathrm{Ad} G$. Also $\mathfrak{p}$ is the orthocomplement of $\mathfrak{k}$ in $\mathfrak{g}$ with respect to the Killing form. Let $\mathfrak{a} \subseteq \mathfrak{p}$ be a maximal commutative subalgebra contained in $\mathfrak{p}$ so that $\mathfrak{a}$ is a maximal commutative of real semi-simple elements. Let $x_0 \in \mathfrak{a}$ be any fixed element for which

$$ (2.2.1) \qquad \mathfrak{p}^{x_0} = \mathfrak{a} $$

where $\mathfrak{p}^{x_0}$ is the centralizer of $x_0$ in $\mathfrak{p}$ (see e. g. § 1.3, [8]), and let $\mathfrak{n} \subseteq \mathfrak{g}$ be the sum of all eigenspaces of ad $x$ belonging to positive eigenvalues.
Then $n$ is a Lie subalgebra, all of whose elements are nilpotent, and one has the Iwasawa decomposition

$$ (2.2.2) \qquad g = k + a + n $$

of $g$. If $A$ and $N$ are, respectively, the subgroups of $G$ corresponding to $a$ and $n$ then the map $K \times A \times N \to G$, $(k, a, n) \mapsto kan$ is a diffeomorphism (see Theorem 5.1, p. 234, [2]) and one has the Iwasawa decomposition

$$ (2.2.3) \qquad G = \text{KAN} $$

of $G$.

The elements in $K$ are clearly elliptic since $\text{Ad } K$ is compact. Also the elements in $A$ are hyperbolic and the elements in $N$ are unipotent (since the maps $a \to A$, $n \to N$ defined by $\text{exp}$ are bijective).

If $g \in G$ is arbitrary then its components in $K$, $A$ and $N$ will be denoted by $k(g)$, $a(g)$ and $n(g)$ and

$$ (2.2.4) \qquad g = k(g) a(g) n(g) $$

is called the Iwasawa decomposition of $g$.

**REMARK 2.2.** — Both the complete multiplicative Jordan decomposition and the Iwasawa decomposition write $g$ as a product of an elliptic, hyperbolic and unipotent element. One notes, however, that the Iwasawa decomposition is not invariantly defined. It obviously depends on the choice of $K$, $A$ and $N$.

2.3. If $Z$ is the center of $G$ then $Z$ is the kernel of $\text{Ad}$ and hence by definition the elements of $Z$ are elliptic. We recall the following fact.

**PROPOSITION 2.3.** — An element $e \in G$ is elliptic if and only if it is conjugate to an element in $K$. In particular $Z \subseteq K$. Furthermore any element $k \in K$ is of the form $k = \exp y$ for some $y \in k$.

*Proof.* — Let $G_1 = \text{Ad } G$ be the adjoint group of $G$ so that $G_1 \subseteq G_0$. If $K_1 = \text{Ad } K$, $A_1 = \text{Ad } A$ and $N_1 = \text{Ad } N$ then $G_1 = K_1 A_1 N_1$ is clearly an Iwasawa decomposition of $G_1$. In particular $G_1/K_1$ is simply connected since it is diffeomorphic to $A_1 \times N_1$ (see Theorem 5.1, p. 234, [2]). But then if $K_* = \text{Ad}^{-1} K_1$ one has $G/K_* \simeq G_1/K_1$ is simply connected. Thus $K_*$ is connected. But the identity component of $K_*$ is clearly $K$. Thus $K_* = K$.

We have already remarked that the elements of $K$ are elliptic (since $K_1$ is compact) and hence any element conjugate to an element in $K$ is elliptic. Now let $e \in G$ be any elliptic element. Then clearly the closure $L_1$ of the subgroup of $G_1$ generated by $\text{Ad } e$ is compact. But then $L_1$
is conjugate to a subgroup of $K_1$. Thus $e$ is conjugate to an element $k \in G$ such that $\text{Ad} k \in K_1$. But then $k \in K_* = K$. Hence an element is elliptic if and only if it is conjugate to an element in $K$. Now $k = k_0 \oplus c$ where $c$ is the center of $k$ and $k_0 = [k, k]$ is semi-simple. But if $K_0$ and $C$ are the connected subgroups of $G$ corresponding to $k_0$ and $c$ then $K = K_0 C$ since $K_0 C$ is clearly a Lie subgroup whose Lie algebra is $k$. Thus if $k \in K$ we may write $k = k_0 c$ where $k_0 \in K_0$ and $c \in C$. But $K_0$ is compact since $k_0$ is compact semi-simple. Thus $k_0 = \exp y_0$ for some $y_0 \in k_0$. Also $c = \exp z$ for some $z \in c_0$ since $c_0$ is abelian. Thus $k = \exp y_0 \exp z = \exp y$ where $y = y_0 + z \in k$, since $[y_0, z] = 0$.

Q. E. D.

**2.4.** Now the Weyl group $W$ associated with $(a, g)$ is the finite group defined as the quotient of the normalizer of $A$ in $K$ modulo the centralizer of $A$ in $K$. The Weyl group $W$ naturally operates in $a$ and $A$ and in such a manner that the isomorphism $a \to A$ defined by $\exp$ is a $W$-isomorphism.

Now for each real semi-simple element $x \in g$ (resp. hyperbolic element $h \in G$) let $w(x)$ [resp. $W(h)$] be the set of all elements in $a$ (resp. $A$) which are conjugate to $x$ (resp. $h$).

**PROPOSITION 2.4.** — An element $x \in g$ is real semi-simple if and only if it is conjugate to an element in $a$. Moreover in such a case $w(x)$ is a single W-orbit in $a$. Similarly an element $h \in G$ is hyperbolic if and only if it is conjugate to an element in $A$ and in such a case $W(h)$ is a single W-orbit in $A$.

Also if $x$ is real semi-simple then $W(\exp x) = \exp(w(x))$.

*Proof.* — Since $\exp$ sets up a bijection between the set of real semi-simple elements in $g$ and hyperbolic elements in $G$ and since $\exp$ commutes with conjugation it suffices only to prove that if $x$ is hyperbolic then $w(x)$ is a single W-orbit.

Let $x$ be a real semi-simple. Since $\text{ad } x$ is diagonalizable $x$ lies in a Cartan subalgebra of $g$. But since the eigenvalues of $\text{ad } x$ are real $x$ lies in the vector part of the Cartan subalgebra. But the vector part of any Cartan subalgebra is conjugate to a subalgebra of $a$ (see e. g. Theorem 2 (2), p. 383, [13]). Thus $x$ is conjugate to an element $y \in a$.

Now assume $z, y \in a$ are $(\text{Ad } G)$ conjugate. We have only to show that they are W-conjugate. But for some $g \in G$, $\text{Ad } g(z) = y$. Let $g = k$ be the Iwasawa decomposition of $g$. Then $\text{Ad } g(z) = z + u$ where $u \in u$ since $A$ is abelian and $A$ normalizes $N$. But then $\text{Ad } k(z + u) = y$. But $z, y \in p$ and $p$ is invariant under $\text{Ad } K$, and $p \cap u = 0$. Thus $u = 0$. Hence $\text{Ad } k(z) = y$ so that $z$ and $y$ are Ad $K$-conjugate. Thus any
polynomial function on $\mathfrak{p}$ which is invariant under Ad K takes the same values on $z$ and $y$. But then any polynomial function on $\mathfrak{a}$ invariant under W also takes the same values on $z$ and $y$ (see Theorem 6.10, p. 430, [2]). Thus $z$ and $y$ are W-conjugate, say, by the argument in Lemma 9.2 (p. 1029 in [6]).

Q. E. D.

Now extend the definition of $W(h)$ from hyperbolic elements to all elements as follows: For any $g \in G$ let $W(g) = W(h(g))$ where we recall, [see (2.1.5)], that $h(g)$ is the hyperbolic component of $g$. Thus we have associated to an arbitrary element $g \in G$ a single W-orbit in $A$.

**REMARK 2.4.** — Obviously $W(g) = W(f)$ if $f, g \in G$ are conjugate. See Remark 2.1.

2.5. Propositions 2.3 and 2.4 dealt respectively with elliptic and hyperbolic elements. An analogous statement for unipotent elements is that an element is unipotent if and only if it is conjugate to an element in $N$. More generally one has

**PROPOSITION 2.5.** — Let $g \in G$. Then $g$ is conjugate to an element in $AN$ if and only if $e(g) = 1$ (see (2.1.5) for the definition of the elliptic part $e(g)$ of $g$]. Moreover in such a case there exists $k \in K$ such that $kgk^{-1} \in AN$.

Furthermore if $g$ is conjugate to an $a \in AN$ where $a \in A$, $n \in N$ then $W(g) = W(a) = W.a$, the Weyl group orbit of $a$.

[**REMARK 2.5.** — Note $a$ is not necessarily equal to $h(an)$.]

*Proof*. — Let $S = AN$ so that $S$ is solvable and its Lie algebra is $\mathfrak{s} = \mathfrak{a} + \mathfrak{n}$.

Now let $f \in S$ so that $f = a$ where $a \in A$, $n \in N$. The decomposition $f = a$ is of course the Iwasawa decomposition of $f$. Let $f = \text{ehu}$ be the complete multiplicative Jordan decomposition of $f$. We wish to show that $e = 1$, $h \in S$ and $u \in N$. Towards this end define subspaces $\mathfrak{g}_i \subseteq \mathfrak{g}$, $i = 0, 1, \dots$ inductively as follows: Let $\mathfrak{g}_0 = \mathfrak{g}$ and if $\mathfrak{g}_i$ has been defined let $\mathfrak{g}_{i+1} = [\mathfrak{n}, \mathfrak{g}_i]$ so that $\mathfrak{g}_{i+1} = 0$ for some minimal $k$. Now if $g \in G$ is in the normalizer of $N$ then $\mathfrak{g}_i$ is clearly stable under Ad $g$ for all $i$ and Ad $g$ defines an operator on $\mathfrak{g}_i/\mathfrak{g}_{i+1}$ and hence an operator which we denote by $\widetilde{\text{Ad}} g$ on $\bigoplus_{i=0}^k \mathfrak{g}_i/\mathfrak{g}_{i+1} = \tilde{\mathfrak{g}}$. Furthermore the eigenvalues of $\widetilde{\text{Ad}} g$ on $\mathfrak{g}$ are the same as the eigenvalues of Ad $g$. But now $f, a, n$ are in the normalizer of $N$ and $\widetilde{\text{Ad}} n = 1$. Thus $\widetilde{\text{Ad}} f = \widetilde{\text{Ad}} a$. But the eigenvalues of Ad $a$ are positive. Hence the eigenvalues of $\widetilde{\text{Ad}} f$ and $\widetilde{\text{Ad}} a$ are the same and consequently the eigenvalues of Ad $f$ are positive. But now if Ad $f = e_1 h_1 u_1$ is the decomposition of Ad $f$ defined as
in (2.1.1) (where $g=f$) it follows that $e_1 = 1$. Thus $\mathrm{Ad} f = h_1 u_1$ is the multiplicative Jordan decomposition of $\mathrm{Ad} f$. Hence $h_1$ and $u_1$ are polynomials in $\mathrm{Ad} f$. But $\mathfrak{u}$ is stable under $\mathrm{Ad} f$ since $f = \mathrm{an}$. Thus $\mathfrak{u}$ is stable under $h_1$ and $u_1$. But, if, as in the proof of Proposition 2.1, $x$ and $z$ are respectively the unique real and nilpotent elements in $\mathfrak{g}$ such that $h_1 = \exp \mathrm{ad} x, u_1 = \exp \mathrm{ad} z$ then clearly $\mathrm{ad} x$ and $\mathrm{ad} z$ are respectively polynomials in $h_1$ and $u_1$ and hence it follows that $x$ and $z$ are in the normalizer of $\mathfrak{u}$. But now we recall that $\mathfrak{u}$ (see § 2.2) is the sum of the eigenspaces of $\mathrm{ad} x_0$ belonging to the positive eigenvalues of $\mathrm{ad} x_0$. If $\mathfrak{u}^-$ is the corresponding sum for the negative eigenvalues then $\mathfrak{g} = \mathfrak{u} \oplus \mathfrak{u}^- \oplus \mathfrak{g}^{x_0}$ is a linear direct sum where $\mathfrak{g}^{x_0}$ is the centralizer of $x_0$. But $\mathfrak{g}^{x_0} = \mathfrak{u} + \mathfrak{m}$ where $\mathfrak{m}$ is the centralizer of $\mathfrak{u}$ in $\mathfrak{k}$ (see e. g. Proposition 8, p. 771, in [8]). Clearly $\mathfrak{g}^{x_0}$ is in the normalizer of $\mathfrak{u}$. In fact one recalls that the normalizer $\mathfrak{b}$ of $\mathfrak{u}$ in $\mathfrak{g}$ is given by

$$ (2.5.1) \qquad \mathfrak{b} = \mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{u}. $$

For this it is enough to note that $\mathfrak{u}^- \cap \mathfrak{b} = 0$. But if we embed $\mathfrak{a}$ in a Cartan subalgebra $\mathfrak{h} \subseteq \mathfrak{a} + \mathfrak{m}$ then $\mathfrak{u}$ is spanned by root vectors $e_{\phi}$ for some set $\mathbb{R}$ of roots and $\mathfrak{u}^-$ is spanned by root vectors for the roots in $\mathbb{R}$. But since $0 \neq [e_{\phi}, e_{-\phi}] \in \mathfrak{h}$ and $[e_{\phi}, e_{-\phi}] \in \mathfrak{h}^{\perp}$, where $\mathfrak{h}^{\perp}$ is the orthocomplement of $\mathfrak{h}$ in $\mathfrak{g}$ for $\psi, \varphi \in \mathbb{R}$, $\varphi \neq \psi$, it follows that $\mathfrak{u}^- \cap \mathfrak{b} = 0$.

Thus by (2.5.1) $x, z \in \mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{u}$. Let

$$ x = x_1 + x_2 + x_3 \quad \text{and} \quad z = z_1 + z_2 + z_3 $$

be the respective components in $\mathfrak{m}$, $\mathfrak{a}$ and $\mathfrak{u}$. Also for any $y \in \mathfrak{b}$ let $\tilde{\mathrm{ad}} y$ be the operator on $\tilde{\mathfrak{g}} = \sum \mathfrak{g}_i/\mathfrak{g}_{i+1}$ defined by $\mathrm{ad} y$ in a manner similar to the definition of $\tilde{\mathrm{Ad}} g$ for any $g \in \mathbb{G}$ in the normalizer of $\mathbb{N}$. Clearly the eigenvalues of $\tilde{\mathrm{ad}} y$ are the same as the eigenvalues of $\mathrm{ad} y$. But now $\tilde{\mathrm{ad}} \mathfrak{u} = 0$ so that $\tilde{\mathrm{ad}} z = \tilde{\mathrm{ad}} z_1 + \tilde{\mathrm{ad}} z_2$. But $[\mathfrak{m}, \mathfrak{a}] = 0$ so that $\tilde{\mathrm{ad}} z_1$ and $\tilde{\mathrm{ad}} z_2$ commute. However $z_1 \in \mathfrak{m} \subseteq \mathfrak{k}$ so that the eigenvalues of $\tilde{\mathrm{ad}} z_1$ are pure imaginary and $z_2 \in \mathfrak{a}$ so that the eigenvalues of $\tilde{\mathrm{ad}} z_2$ are real. But the eigenvalues of $\mathrm{ad} z$ and hence of $\tilde{\mathrm{ad}} z$ are zero. Thus

$$ \tilde{\mathrm{ad}} z_1 = \tilde{\mathrm{ad}} z_2 = 0 $$

since $\mathrm{ad} z_1$ and $\mathrm{ad} z_2$ are diagonalizable. Thus $z_1 = z_2 = 0$ and hence $z \in \mathfrak{u}$. But again since $\tilde{\mathrm{ad}} \mathfrak{u} = 0$ one has $\tilde{\mathrm{ad}} x = \tilde{\mathrm{ad}} x_1 + \tilde{\mathrm{ad}} x_2$. Also $\tilde{\mathrm{ad}} x_1$ and $\tilde{\mathrm{ad}} x_2$ commute, since $[\mathfrak{m}, \mathfrak{a}] = 0$, and the eigenvalues of $\tilde{\mathrm{ad}} x_1$ and $\tilde{\mathrm{ad}} x_2$ are, respectively, pure imaginary and real. Since the eigenvalues of $\tilde{\mathrm{ad}} x$ are real this implies that $\tilde{\mathrm{ad}} x_1 = 0$ and also $x_1 = 0$
since $ad x_1$ is diagonalizable. Thus $x \in \mathfrak{a} + \mathfrak{n}$. But now the hyperbolic and unipotent components $h$ and $u$ of $f$ are given by $h = \exp x$ and $u = \exp z$ [see (2.1.3)]. Thus $h \in \text{AN}$ and $u \in \text{N}$ and hence $hu = a' n'$ where $a' \in \text{A}, n' \in \text{N}$. But $f = ehu = ea' n'$. However $\text{Ad } e = e_1 = 1$ as shown above. Thus $e \in \text{Z}$, the center of $\text{G}$. But then $e \in \text{K}$ by Proposition 2.3. Hence $f = ea' n'$ is the Iwasawa decomposition of $f = an$. By uniqueness one has $e = 1$ or

$$ (2.5.2) \qquad e(f) = 1. $$

Thus $f = hu = \exp x \exp z = \exp(x+z)$ by (2.1.2).

But now $x \in \mathfrak{a} + \mathfrak{n} = \mathfrak{s}$ is a real semi-simple element. Thus $x$ lies in a Cartan subalgebra of $\mathfrak{s}$. But $\mathfrak{a}$ is clearly a Cartan subalgebra of $\mathfrak{s}$ [since (see § 2.2) the existence of $x_0 \in \mathfrak{a}$ such that $ad x_0$ is non-singular on $\mathfrak{n}$ shows that $\mathfrak{a}$ is its own normalizer in $\mathfrak{s}$]. However any two Cartan subalgebras of a solvable Lie algebra are conjugate (see e. g. [5], Theorem 1, p. 58). Thus there exists $b \in \text{AN}$ such that $\text{Ad } b(x) = x_2 \in \mathfrak{a}$. Let $z_2 = \text{Ad } b(z)$. But then $z_2 \in \mathfrak{n}$ since $\mathfrak{n}$ is an ideal in $\mathfrak{a} + \mathfrak{n}$. But $[x_2, z_2] = 0$ by (2.1.2) and hence if $a_2 = \exp x_2 \in \text{A}$ and $n_2 = \exp z_2$ then $bfb^{-1} = a_2 n_2$ is both the Iwasawa and the complete multiplicative Jordan decomposition of $bfb^{-1}$. Thus (see Remark 2.4) one has $W(f) = W(a_2) = W.a_2$, the W-orbit of $a_2$. We assert that $a_2 = a$. Indeed $f = an$ so that $bfb^{-1} = bab^{-1} bnb^{-1}$. But $bab^{-1} = an_3$ where $n_3 \in \text{N}$ (since $\mathfrak{n} = [\mathfrak{s}, \mathfrak{s}$]). Thus $bfb^{-1} = an_3$ where $n_3 bnb^{-1} = n_3 \in \text{N}$. Hence $a_2 n_2 = an_3$ so that $a_2 = a$. Hence $W(f) = W.a$.

To finish the proof we have only to show that if $g \in \text{G}$ is any element such that $e(g) = 1$ then $g$ is K-conjugate to an element in $\text{AN}$. Let $g = hu$ be the complete multiplicative Jordan decomposition of $g$ and let $x = \log h$, $z = \log u$ so that $[x, z] = 0$ by (2.1.2). But then $z$ is in the centralizer, $\mathfrak{g}^r$, of $x$. But $\mathfrak{g}^r$ is reductive (see e. g. § 3, p. 352 in [7]). Furthermore $z$ is not only nilpotent in $\mathfrak{g}$ but also in $\mathfrak{g}^r$ which means that it is a nilpotent element in the semi-simple Lie algebra $[\mathfrak{g}^r, \mathfrak{g}^r]$ (see e. g. (3.1.3) p. 352, [1]). By the Jacobson-Morosov theorem there exists an S-triple $(y, z, w)$ in $\mathfrak{g}^r$ in the notation of [6]. (See § 4, [6] p. 988.) In particular there exists a real semi-simple element $y$ of $\mathfrak{g}$ contained in $\mathfrak{g}^r$ such that $[y, z] = z$. But $x$ and $y$ span a 2-dimensional abelian subalgebra $\mathfrak{d}$ whose elements are real semi-simple.

But then $\mathfrak{d}$ can be embedded in the vector part of a Cartan subalgebra of $\mathfrak{g}$ using the notation of [13]. However by Theorem 2 (2), p. 383 in [13] it follows that $\mathfrak{d}$ is conjugate, say by $\text{Ad } f$, $f \in \text{G}$, to a subalgebra of $\mathfrak{a}$. Furthermore by applying an element of the Weyl group $\text{W}$, if necessary, we may assume that $y$ is carried into the same Weyl chamber (see [1],
p. 5) as $x_0$ by Ad $f$. But since $[y, z] = z$, i. e. $z$ is an eigenvector for ad $y$ corresponding to the eigenvalue 1 it follows that from the definition of $\mathfrak{a}$ that $\mathrm{Ad} f(z) \in \mathfrak{a}$. Thus $\mathrm{Ad} f$ carries both $x$ and $z$ into $\mathfrak{a} + \mathfrak{u}$. Since $g = \exp (x + z)$ one has $fgf^{-1} \in \mathrm{AN}$. However if $f^{-1} = \mathrm{kan}$ is the Iwasawa decomposition of $f^{-1}$ then clearly $k^{-1} gk \in \mathrm{AN}$ since $\mathrm{AN}$ is obviously stable under $\mathrm{Ad} \mathfrak{a}$.
Q. E. D.

### 3. The partial ordering in $G$ and the spectral radius $|\pi_\lambda(g)|$, $g \in G$, $\lambda \in \hat{G}$.

3.1. For any real semi-simple element $x \in g$ let $\mathfrak{a}(x)$ be the convex hull of the Weyl group orbit $\omega(x) \subseteq \mathfrak{a}$. Thus $\mathfrak{a}(x)$ is the compact convex subset of $\mathfrak{a}$ having $\omega(x)$ as its subset of extremal points.

Now, group-wise, for any $g \in G$ let $A(g) \subseteq A$ be the compact subset defined by putting

$$ (3.1.1) \qquad A(g) = \exp_{\mathfrak{a}} (\log h(g)) $$

where we recall $h(g)$ is the hyperbolic component of $g$ [see (2.1.5)]. In particular note that $W(g) \subseteq A(g)$ [and in a multiplicative sense $A(g)$ is the “convex hull” of $W(g)$].

**REMARK 3.1.** — Note the sets $\mathfrak{a}(x) \subseteq \mathfrak{a}$ for $x$ real semi-simple or $A(g) \subseteq A$ for $g \in G$ arbitrary, are invariant under the Weyl group.

The following definition is different from, but was inspired by a definition made by Colin Thompson for $\mathrm{GL}(n, \mathbb{C})$ in [14]. His definition simplified a previous one made by A. Lenard in [9]. Thompson's definition has nothing to do with convexity and is based on the polar decomposition. For $\mathrm{SL}(n, \mathbb{C})$ both agree on P. See section 4.2. This will be clearer as a consequence of Theorem 3.1 below and the Horn-Thompson Theorem, p. 470 in [14]. The definition here although defined in terms of $A$ is easily seen to be independent of the maximal “$\mathbb{R}$-split torus” $A$. The invariance of the definition is nevertheless more cogently illustrated by Theorem 3.1.

Given $f, g \in G$ we say that $g \ge f$ if $A(f) \subseteq A(g)$ [or equivalently if $W(f) \subseteq A(g)$]. This obviously defines a partial ordering on $G$.

**REMARK 3.1.1.** — One should note that the partial order is not necessarily the same as the partial order on $G$ that would be induced by a possible embedding of $G$ in $\mathrm{SL}(n, \mathbb{C})$. Indeed it follows immediately from Remark 3.1.2 that if $G \subseteq G'$ where $G'$ is also semi-simple and $f, g \in G$ then if $g \ge f$ in $G$ one also has $g \ge f$ in $G'$. It is the converse which may be false. Indeed take for example the case where the $\mathbb{R}$-split rank (dim $A$) of $G$ and $G'$ are the same, so that $A$ is a maximal $\mathbb{R}$-split
torus in both G and G', but that the Weyl group W' of (A, G') is larger
than W. If $\sigma \in W' - W$ and, say, $a = \exp x_0$ (see § 2.2) then $a$,
$\sigma a \in A \subseteq G$ and $a \ge \sigma a$ in G' but $a \ne \sigma a$ in G. It follows therefore that
if $G \subseteq G' \subseteq SL(n, \mathbb{C})$ for some $n$ then the partial ordering on G is not the
one induced on G by that on SL(n, G).

Now let $\hat{G}$ be an index set for the set of all equivalence classes of irreducible finite dimensional representations of G.

Also for any $\lambda \in \hat{G}$ let

$$
(3.1.2) \qquad \pi_\lambda : G \to \operatorname{Aut} V_\lambda
$$

be a fixed representation in the class corresponding to $\lambda$.

Now for any $g \in G$ and let $\lambda \in \hat{G}$ let $|\pi_\lambda(g)|$ be the maximum of the
absolute values of the eigenvalues of $\pi_\lambda(g)$. That is (as Helgason
reminded me) $|\pi_\lambda(g)|$ is the spectral radius of $\pi_\lambda(g)$.

THEOREM 3.1. — Let $f, g \in G$. Then $g \ge f$ if and only if

$$
(3.1.3) \qquad |\pi_\lambda(g)| \ge |\pi_\lambda(f)|
$$

for all $\lambda \in \hat{G}$.

REMARK 3.1.2. — Using say, complete reducibility, note that one
may substitute for the set of irreducible representations $\pi_{\lambda}$, $\lambda \in \hat{G}$, in
Theorem 3.1 the collection of equivalence classes of all finite dimensional
representations, irreducible or not.

3.2. Before proving Theorem 3.1 (see § 3.5) we will need notation which will be used in the proof and elsewhere. Let $\mathfrak{a}'$ be the real dual to $\mathfrak{a}$ and let $\Gamma \subseteq \mathfrak{a}'$ be the set of restricted roots. That is $\Gamma$ is the set of non-zero weights for the adjoint action of $\mathfrak{a}$ on $\mathfrak{g}$. An account of the theory of restricted roots may be found in [1]. The set of positive restricted roots $\Gamma_+$ corresponding to $\mathfrak{a}$ may be given by

$$
(3.2.1) \qquad \Gamma_+ = \{ \gamma \in \Gamma \mid \langle \gamma, x_0 \rangle \ge 0 \}
$$

where $x_0$ is fixed as in section 2.2. Let $\Sigma = \{\beta_i, \dots, \beta_k\} \subseteq \Gamma_+$ be the set of simple, positive restricted roots so that every $\gamma \in \Gamma$ is an integral combination of the $\beta_i$ and $\Sigma$ is a basis of $\mathfrak{a}'$.

Now the Killing form on $\mathfrak{g}$ induces a positive definite bilinear form
on $\mathfrak{a}$ and by duality a positive definite bilinear form on $\mathfrak{a}'$. Then one
knows that for any $\gamma$, $\partial \in \Gamma$, $\frac{2(\partial, \gamma)}{(\gamma, \gamma)}$ is an integer. (See § 2.4, [1].)
That is, if $x_{\gamma} \in \mathfrak{a}$ is an element corresponding to $\frac{2\gamma}{(\gamma, \gamma)}$ under the isomor-
phism $\mathfrak{a} \to \mathfrak{a}'$ defined by the Killing form then $\langle \partial, x_{\gamma} \rangle \in \mathbb{Z}$ for all $\partial \in \Gamma$.
In particular $\langle \gamma, x_{\gamma} \rangle = 2$.
Moreover for any $\gamma \in \Gamma$ there is an element $\tau(\gamma) \in W$ (the reflection defined by $\gamma$) such that for any $x \in a$,

$$ (3.2.2) \qquad \tau(\gamma)x = x - \langle\gamma, x\rangle x_{\gamma} $$

and $W$ is generated by these reflections.

Now let

$$ (3.2.3) \qquad a_+ = \{ x \in a \mid \langle \beta_i, x \rangle \ge 0 \text{ for } i = 1, 2, \dots, r \} $$

so that $a_+$ is a Weyl chamber in $a$ (see p. 5 in [8]). One knows that any $x \in a$ is $W$-conjugate to a unique element in $a_+$ (see e. g. p. 52 in [4]).

**REMARK 3.2.** — Note that by Proposition 2.4 that any real semi-simple element $y \in \mathfrak{a}$ is conjugate to a unique element in $a_+$.

The subset $a_+$ is a cone in $a$. We define another cone $a_\rho$ in $a$ as follows. Let $x_i = x_{\beta_i}$, $i = i, 2, \dots, k$ and put

$$ (3.2.4) \qquad a_\rho = \left\{ x \in a \mid x = \sum_{i=1}^k r_i x_i, r_i \ge 0 \right\} $$

so that $a_\rho$ is the cone generated by the $x_i$. Since every positive root is a non-negative combination of positive simple roots one clearly has $x_\gamma \in a_\rho$ for any $\gamma \in \Gamma_+$.

Now by contragredience the Weyl group $W$ also operates on the dual $a'$. Thus if $a'_+$ corresponds to $a_+$ under the Killing form induced isomorphism $a \to a'$ then $a'_+$ is a fundamental domain for the action of $W$ in $a'$. The cone $a'_+$ may also be given by

$$ (3.2.5) \qquad a'_+ = \{ \lambda \in a' \mid \langle \lambda, x \rangle \ge 0 \text{ for all } x \in a_\rho \}. $$

The cone $a'_\rho \subseteq a'$ corresponding to $a_\rho$, on the other hand, is given by

$$ (3.2.6) \qquad a'_\rho = \left\{ \lambda \in a' \mid \lambda = \sum_{i=1}^k r_i \beta_i, r_i \ge 0 \right\} $$

and one has $\gamma \in a'_\rho$ for any $\gamma \in \Gamma_+$.

Finally a sequence of positive restricted roots $\gamma_1, \gamma_2, \dots, \gamma_n$ will be called a strongly positive sequence (of length $n$) if for any $x \in a_+$ one has

$$ \langle \gamma_i, \tau(\gamma_j) \tau(\gamma_{j-1}) \dots \tau(\gamma_1) x \rangle \ge 0 \quad \text{for all } 1 \le j \le i \le n. $$

Any element $1 \ne \sigma \in W$ may be written as a product $\sigma = \tau(\gamma_n) \dots \tau(\gamma_1)$ where $\gamma_1, \dots, \gamma_n$ is a strongly positive sequence. (One joins the interior of $a_+$ with the interior of $\sigma a_+$ by a suitable line segment and considers the root hyperplanes that are crossed in going from $a_+$ to $\sigma a_+$ along the segment.) Now one has
LEMMA 3.2. — Let $x \in a_+$. Then for any $\sigma \in W$ one has $x - \sigma x \in a_p$.
Furthermore for any $\lambda_\sigma \in a'_+$ one has

$$ \langle \lambda, x \rangle \geq \langle \lambda, \sigma x \rangle. $$

*Proof*. — We may assume $\sigma \neq 1$. Let $\gamma_1, \dots, \gamma_n$ be a strongly positive sequence such that $\sigma = \tau(\gamma_n) \dots \tau(\gamma_1)$. Put

$$ x_i = \tau(\gamma_i) \tau(\gamma_{i-1}) \dots \tau(\gamma_1) x \quad \text{and} \quad x_0 = x. $$

Then $x_i = x_{i-1} - \langle \gamma_i, x_{i-1} \rangle x_{i-1}$. But $\langle \gamma_i, x_{i-1} \rangle \geq 0$. Thus $x_{i-1} - x_i \in a_p$. Since $a_p$ is a cone this implies $x - \sigma x \in a_p$. But now if $\lambda_\sigma \in a'_+$ one has $\langle \lambda_\sigma, x \rangle \leq \langle \lambda_\sigma, \sigma x \rangle$ by (3.2.5).

Q. E. D.

3.3. Let $\nu_j, j = 1, 2, \dots, k$ be the basis of $a'$ such that $\langle \nu_j, x_i \rangle = \delta_{ij}$. Thus $\nu_j \in a'_+$ by (3.2.5) and in fact $a'_+$ is the cone generated by the $\nu_j$. Now recall that $a(x)$, for any $x \in a$, is the convex hull of the W-orbit, W.x, of x.

LEMMA 3.3. — Let $x \in a_+$. Then for any $y \in a$ :

(1) $y \in a(x)$ if and only if $x - \sigma y \in a_p$ for all $\sigma \in W$.

(2) If $y \in a_+$ then $y \in a(x)$ if and only if $x - y \in a_p$.

*Proof*. — Since the orbit W.y always meets $a_+$ and since $y - \sigma y \in a_p$ for any $y \in a_+$ and $\sigma \in W$, by Lemma 3.2, and since $a(x)$ is stable under W it suffices only to prove (2). Assume $y \in a(x)$. Then, by a property of extreme points of convex sets, given any $\nu \in a'$ there exists $\sigma \in W$ such that $\langle \nu, \sigma x \rangle \geq \langle \nu, y \rangle$. But if $\nu = \nu_j \in a'_+$ one has $\langle \nu_j, x \rangle \geq \langle \nu_j, \sigma x \rangle$ by Lemma 3.2. Thus $\langle \nu_j, x - y \rangle \geq 0$. But if $x - y = \sum_{i=1}^k r_i x_i$ then $\langle \nu_j, x - y \rangle = r_j$. Hence $r_j \geq 0$ for all $j$ so that $x - y \in a_p$.

Conversely assume $y \in a_+$ and $x - y \in a_p$. To prove that $y \in a(x)$ it suffices to show that $y$ cannot be separated from $a(x)$ by a hyperplane or equivalently it suffices to show that for any $\nu \in a'$ there exists $\sigma \in W$ such that $\langle \nu, \sigma x \rangle \geq \langle \nu, y \rangle$. But now given any $\nu \in a'$ there exists $\sigma \in W$ such that $\sigma^{-1} \nu \in a'_+$. But now if $x - y \in a_p$, then $x - \sigma^{-1} y \in a_p$ since $y - \sigma^{-1} y \in a_p$ by Lemma 3.2. Thus $\langle \sigma^{-1} \nu, x - \sigma^{-1} y \rangle \geq 0$ by (3.2.5). Applying $\sigma$ one has $\langle \nu, \sigma x - y \rangle \geq 0$ or $\langle \nu, \sigma x \rangle \geq \langle \nu, y \rangle$.

Q. E. D.

3.4. The proof of Theorem 3.1 reduces to the case of hyperbolic elements because of the following Proposition. Proposition 3.4 asserts that the spectral radius $|\pi_\lambda(g)|$ of $\pi_\lambda(g)$ depends only on the hyperbolic part of $g$.
**Proposition 3.4.** — Let $\lambda \in \hat{G}$ and let $g \in G$. Let $h = h(g)$ be the hyperbolic component of $g$ [see (2.1.5)]. Then all the eigenvalues of $\pi_\lambda(h)$ are positive and $|\pi_\lambda(g)|$ is the maximal eigenvalue of $\pi_\lambda(h)$.

*Proof.* — Let $\pi_\lambda$ also denote the (differential) representation of $\mathfrak{g}$ and $\mathfrak{g}_c$ defined by the group representation $\pi_\lambda$. If $x \in \mathfrak{g}$ is nilpotent then $\pi_\lambda(z)$ is a nilpotent operator. This follows, for example, from the representation theory of a TDS (see e. g. [5], § 2.1) and the Jacobson-Morosov theorem which asserts that any nilpotent element lies in a TDS of $\mathfrak{g}$. Thus if $u \in G$ is unipotent then all the eigenvalues of $\pi_\lambda(u)$ are equal to 1. If $x \in \mathfrak{g}$ is such that $ax$ is diagonalizable then $\pi_\lambda(x)$ is diagonalizable, using the theory of weights, since $x$ can be embedded in a Cartan subalgebra $\mathfrak{h}_c$ of $\mathfrak{g}_c$. Thus $\pi_\lambda(g)$ is diagonalizable if $g$ is either elliptic or hyperbolic by Proposition 2.3. But if $x$ is real semi-simple then all the roots of $(\mathfrak{h}_c, \mathfrak{g}_c)$ take real values on $x$. But since the weights of $\pi_\lambda$ are rational combinations of the roots, it follows that the eigenvalues of $\pi_\lambda(x)$ are real and hence the eigenvalues of $\pi_\lambda(h)$ are positive for any hyperbolic element $h \in G$. Similarly if $x \in K$ then all the roots take pure imaginary eigenvalues on $x$ and hence the same is true of the weights of $\pi_\lambda$. Thus $\pi_\lambda(k)$ has eigenvalues of norm 1 for any $k \in K$ by Proposition 2.3. But then, also by Proposition 2.3, $\pi_\lambda(e)$ has eigenvalues of norm 1 for all elliptic elements in $G$.

Now let $g = \text{ehu}$ be the complete multiplicative Jordan decomposition of $g$ [see (2.1.5)], so that $\pi_\lambda(g) = \pi_\lambda(e) \pi_\lambda(h) \pi_\lambda(u)$. Since the three operators on the right commute the eigenvalues of $\pi_\lambda(g)$ use just products of the eigenvalues of $\pi_\lambda(e)$, $\pi_\lambda(h)$ and $\pi_\lambda(u)$ and hence it is clear that $|\pi_\lambda(g)|$ is the maximal eigenvalue of $\pi_\lambda(h)$.

Q. E. D.

### 3.5. We now give the

*Proof of Theorem 3.1.* — Let $g, f \in G$ be arbitrary. Then there exists unique elements $x, y \in \mathfrak{a}_+$ such that $x$ is conjugate to $\log h(g)$ and $y$ is conjugate to $\log h(f)$. By Proposition 3.4 we have only to show that for any $\lambda \in \hat{G}$ the maximal eigenvalue of $\pi_\lambda(x)$ is not less than the maximal eigenvalue of $\pi_\lambda(y)$ if and only if $y \in \mathfrak{a}(x)$; or by Lemma 3.3 if and only if $x - y \in \mathfrak{a}_+$.

Now let $\mathfrak{m}$ be the centralizer of $\mathfrak{a}$ in $\mathfrak{k}$ and let $\mathfrak{h}_m$ be a maximal commutative subalgebra in $\mathfrak{m}$ so that $\mathfrak{h}_0 = \mathfrak{a} + \mathfrak{h}_m$ is a Cartan subalgebra of $\mathfrak{g}$ (see p. 221, [2]) and hence its complexification $\mathfrak{h}_c$ is a Cartan subalgebra of $\mathfrak{g}_c$. Let $\mathfrak{h} = \mathfrak{a} + i\mathfrak{h}_m$ so that $\mathfrak{h}$ is the set of all real semi-simple elements (of $\mathfrak{g}_c$) in $\mathfrak{h}_c$. Let $\Delta$ be the set of roots of $(\mathfrak{h}_c, \mathfrak{g}_c)$ so that $\Delta \subseteq \mathfrak{h}'$, the real dual of $\mathfrak{h}$. Furthermore if $\Delta_0$ is the set of roots which vanish
on $\mathfrak{a}$ then the set of restrictions to $\mathfrak{a}$ of the roots in $\Delta - \Delta_0$ is just $\Gamma$ (see § 2.4, p. 6, [1]). Furthermore we may choose a system of positive roots $\Delta_+ \subseteq \Delta$ so that $x_0$ [see (2.2.1)] is in the corresponding Weyl chamber, i. e. one has $\langle \varphi, x_0 \rangle \ge 0$ for all $\varphi \in \Delta_+$. Thus any root in $\Delta_+$ restricts either to zero or an element in $\Gamma_+$ as a linear functional on $\mathfrak{a}$.

Now let $\mathfrak{h}'_+ \subseteq \mathfrak{h}'$ be the co-Weyl chamber corresponding to $\Delta_+$. Thus the elements in $\mathfrak{h}'_+$ are the dominant integral linear forms on $\mathfrak{h}$. In particular we may regard $\hat{\mathbb{G}} \subseteq \mathfrak{h}'_+$ by using, as an index set for the classes of irreducible representations of $\hat{\mathbb{G}}$, the highest weights of the corresponding representation of $\mathfrak{g}_c$.

But now if $\lambda \in \hat{\mathbb{G}}$ and if $\Delta \subseteq \mathfrak{h}'$ denotes the set of weights of the representation $\pi_\lambda$, then the eigenvalues $\pi_\lambda(z)$ for any $z \in \mathfrak{a}$ are numbers of the form $\langle \mu, z \rangle$ where $\mu \in \Delta_+$. However for any $\mu \in \Delta_+$ one knows from representation theory that $\lambda - \mu$ is a sum of positive roots. Since positive roots restrict either to zero or elements of $\Gamma_+$ on $\mathfrak{a}$ it follows from (3.2.3) that

$$ (3.5.1) \qquad \langle \lambda, z \rangle \ge \langle \mu, z \rangle $$

for any $z \in \mathfrak{a}_+$. We assert first of all that (3.5.1) implies that

$$ (3.5.2) \qquad \lambda |_{\mathfrak{a} = \mathfrak{a}_+} $$

Indeed the set of restrictions of $\Delta_\lambda$ to $\mathfrak{a}$ is clearly stable under the Weyl group $W$. Thus if $\lambda_1 = \lambda |_{\mathfrak{a}}$ and $\sigma \in W$ is such that $\sigma\lambda_1 \in \mathfrak{a}'_+$ then there exists $\mu \in \Delta_\lambda$ such that $\mu |_{\mathfrak{a}} = \sigma\lambda_1$. But for any $z \in \mathfrak{a}_+$,

$$ \langle \mu, z \rangle = \langle \sigma\lambda_1, z \rangle \ge \langle \lambda_1, z \rangle = \langle \lambda, z \rangle $$

by Lemma 3.2. But then one has $\langle \sigma\lambda_1, z \rangle = \langle \lambda_1, z \rangle$ by (3.5.1). Since this holds for all $z \in \mathfrak{a}_+$ this implies $\lambda_1 = \sigma\lambda_1 \in \mathfrak{a}'$ establishing (3.5.2).

Now to prove the theorem we have to show that the maximal eigenvalue of $\pi_\lambda(x)$ is not less than the maximal eigenvalue of $\pi_\lambda(y)$ if and only if $x - y \in \mathfrak{a}_p$. But by (3.5.1) the maximal eigenvalues involved are $\langle \lambda, x \rangle$ and $\langle \lambda, y \rangle$ respectively. Thus we have to show $\langle \lambda, x - y \rangle \ge 0$ if and only if $x - y \in \mathfrak{a}_p$. But if $x - y \in \mathfrak{a}_p$ then $\langle \lambda, x - y \rangle \ge 0$ by (3.2.5) and (3.5.2). Conversely assume

$$ (3.5.3) \qquad \langle \lambda, x - y \rangle \ge 0 $$

for all $\lambda \in \hat{\mathbb{G}}$.

We must prove that $x - y \in \mathfrak{a}_p$.

Let $\Pi = \{\alpha_1, \dots, \alpha_l\} \subseteq \Delta_+$ be the set of simple positive roots. Then the co-Weyl chamber $\mathfrak{h}'_+$ is the cone generated by $\lambda_1, \dots, \lambda_l \in \mathfrak{h}'$ where
$\frac{2(\lambda_i, \alpha_j)}{(\alpha_j, \alpha_j)} = \partial_{ij}$, where the inner product on $\mathfrak{h}'$ is induced on $\mathfrak{h}'$ by the isomorphism $\mathfrak{h} \to \mathfrak{h}'$ defined by restricting the Killing form on $\mathfrak{g}_c$ to $h$. Furthermore as one knows from representation theory the $\lambda_i$ are the highest weights of the fundamental representations of a simply connected Lie group $G_c^s$ having $\mathfrak{g}_c$ as its Lie algebra (see e. g. [4], § 13.1, p. 67). But $G_c^s$ has a finite center, say of order $m$. Thus the center is in the kernel of the irreducible representation of $G_c^s$ having highest weight $m\lambda_i$. Hence $m\lambda_i$ is a highest weight of an irreducible representation of the adjoint group $G_c$. But Ad maps $G$ into $G_c$. By taking the composition it follows therefore that $m\lambda_i \in \hat{G}$. Hence $\mathfrak{h}_+'$ is a cone generated by $\hat{G}$. Thus

$$ (3.5.4) \qquad \langle \mu, x - y \rangle \ge 0 $$

for all $\mu \in \mathfrak{h}_+$ by (3.5.3).

On the other hand an element $\mu \in \mathfrak{h}'$ clearly lies in $\mathfrak{h}_+'$ if and only if $(\mu, \alpha_i) \ge 0$ for $i = 1, 2, \dots, l$. But now if $\nu_j \in \mathfrak{a}'$ is, as in section 3.3, the dual basis to the $x_i$ then, in $\mathfrak{a}'$, one has $(\nu_j, \beta_i) \ge 0$ for all $1 \le i, j \le k$ by definition of $x_i$. But now if we embed $\mathfrak{a}'$ in $\mathfrak{h}'$ by regarding the elements of $\mathfrak{a}'$ as having zero restrictions to $\mathfrak{h}_m$ then since $i \mathfrak{h}_m$ and $\mathfrak{a}$ are orthogonal with respect to the Killing form the inner product on $\mathfrak{a}'$ is just the restriction to $\mathfrak{a}'$ of the inner product on $\mathfrak{h}'$. Furthermore the map $\mathfrak{h}' \to \mathfrak{a}'$, defined by restriction to $\mathfrak{a}$, is just orthogonal projection. On the other hand one knows (see [1], § 2.8, p. 11) that any element in $\Pi$ restricts either to zero or to an element in $\Sigma$. Thus $(\nu_j, \alpha_i) \ge 0$ for $1 \le j \le k, 1 \le i \le l$ and hence

$$ (3.5.5) \qquad \nu_j \in \mathfrak{h}_+' \quad \text{or} \quad \mathfrak{a}_+ \subseteq \mathfrak{h}_+' $$

But if $x - y = \sum_{i=1}^k r_i x_i$ then $r_j = (\nu_j, x - y) \ge 0$ by (3.5.4). Thus $x - y \in \mathfrak{a}_p$ by (3.2.4). \qquad Q. E. D.

### 4. The convexity theorem

4.1. Let $b \in \mathcal{A}$. One of the questions that seems to arise quite often in doing analysis on semi-simple Lie groups or symmetric spaces (also in representation theory) is: what is the $\mathfrak{a}$-component (in the Iwasawa decomposition [see (2.2.4)]) when $b$ is multiplied on the right by an element in $\mathcal{K}$. That is, if $v \in \mathcal{K}$ what is $a = a(bv)$ when we write

$$ (4.1.1) \qquad bv = \text{kan} $$

for $k \in \mathcal{K}$, $\alpha \in \mathcal{A}$, $n \in \mathbb{N}$. In the notation of Harish-Chandra $\log a$ is written $H(bv)$. For example, the elementary spherical function $\phi_v$
corresponding to any $\nu \in \mathfrak{a}_G$ is determined by its restriction to A and its
value at any $b \in A$ is given by

$$
(4.1.2) \qquad \wp_\nu(b) = \int_K e^{\langle \nu, H(bv) \rangle} dv
$$

(see e. g. p. 428, [2]).

Our main theorem here, Theorem 4.1, says that H (bv) lies in the
convex hull of the Weyl group orbit of H (b) and that as v runs through K,
H (bv) runs through all the elements in this convex set. That is, in the
notation defined in section 3.1, one has

THEOREM 4.1. — Let $b \in A$ be arbitrary. Then

$$
A (b) = \{ a (bv) \mid v \in K \}.
$$

The proof of Theorem 4.1 will be given after Lemma 4.8.

REMARK 4.1. — With regard to multiplication of *b* by elements in K,
A, or N note that only for right multiplication by elements in K is there
a difficulty in finding the *a*-component. (Because A normalizes N.)

4.2. As a first corollary we generalize Theorem 4.1 to the case where
$b$ is replaced by any element $g \in G$. Recall that $\mathfrak{g} = k + p$ is a fixed
Cartan decomposition of $G$. But since all the elements in $p$ are real
semi-simple all the elements in

$$
(4.2.1) \qquad P = \{ p \in G \mid p = \exp x, x \in \mathfrak{p} \}
$$

are hyperbolic. The set P is closed, diffeomorphic to p by the expo-
nential map and the map P×K → G, (p, v) ↦ pv is a diffeomorphism.
In particular

$$
(4.2.2) \qquad G = \mathrm{PK}
$$

and every element $g$ can be uniquely written $g = pv$, the polar decom-
position of $g$, and we write $p = p(g)$ and $v = v(g)$ (see e. g. [10], p. 155).

More explicitly, the Cartan involution $\emptyset$ of $\mathfrak{g}$ ($\emptyset = 1$ on $k$, $\emptyset = -1$ on $\mathfrak{p}$) induces an automorphism of $G$, $g \mapsto \emptyset(g)$. This is clear since $\emptyset$ induces an automorphism of a simply connected group $G_\emptyset$ having $\mathfrak{g}$ as its Lie algebra. But $G = G_\emptyset/D$ where $D$ is a central subgroup. However $D$ is fixed by $\emptyset$ since by Proposition 2.3, $D$ is in the subgroup corresponding to $k$. Thus $\emptyset$ induces an automorphism of $G$. For any $g \in G$ let

$$
(4.2.3) \qquad g^* = \emptyset(g^{-1})
$$

so that

$$
(4.2.4) \qquad (fg)^* = g^* f^*.
$$
In particular if $g = pv$ is the polar decomposition of $g$ then

$$ (4.2.5) \qquad g^* = v^{-1} p. $$

Now since the hyperbolic elements in $G$ are in one-one correspondance with the real semi-simple elements in $g$ every hyperbolic element $h$ has a unique hyperbolic square root $h^{1/2}$. In fact $h^{1/2} = \exp 1/2 \log h$. If $h \in P$ so is $h^{1/2}$. From (4.2.5) one has that for any $g \in G$, $gg^* \in P$ and

$$ (4.2.6) \qquad p(g) = (gg^*)^{1/2} \in P. $$

Finally since the elements in $p$ are real semi-simple any element $y$ in $p$ is conjugate to a unique element in $a_+$ by Proposition 2.4. Furthermore the argument in the proof of Proposition 2.4. shows in fact that $y$ is $K$-conjugate to such an element. Thus if $A_+ = \exp a_+$ any element $p \in P$ can be

$$ (4.2.7) \qquad p = k \, dk^{-1} $$

for a unique $d \in A_+$. But then by (4.2.2) one has the familiar product relation

$$ (4.2.8) \qquad G = KA_+K $$

so that any $g \in G$ can be written

$$ (4.2.9) \qquad g = k \, dv $$

for $k \in K$, $d \in A_+$, $v \in K$.

**REMARK 4.2.** — The $k$ and the $v$ in (4.2.9) are not unique but the $d$ is. In fact $g = k \, dk^{-1} \, kv$ so that $p(g) = k \, dk^{-1}$. Thus $d$ is unique by (4.2.7). We write $d = d(g) \in A_+$.

Now as a corollary and a generalization of Theorem 4.1 one has the following. Note that the set in question $A(p)a(f)$ is still convex since it is just the translation of $A(p)$ by $a(f)$.

**COROLLARY 4.2.** — Let $g$, $f \in G$ be arbitrary. Then

$$ A(p) \cdot a(f) = \{ a(gvf) \mid v \in K \} $$

where $p = (gg^*)^{1/2}$, $A(p)$ is defined in section 3.1 and $a(g)$ is defined in section 2.2.

*Proof.* — Write $g = k_1 dk_2$ where $d = d(g) \in A_+$ and $k_1, k_2 \in K$. Then if $f = kan$ is the Iwasawa decomposition of $f$ so that $a = a(f)$, one has $gvf = k_1 dk_2 vkan$. But as $v$ runs over $K$ so does $k_2 vk$ and hence if $dk_2 vk = k' a' n'$ is the Iwasawa decomposition then as $v$ runs over $K$, $a'$ runs over $A(d)$ by Theorem 4.1. But $dk_2 vka = k' a' an''$, where $n'' \in N$,
since A normalizes N. Thus A (d) $a = \{ a (gvf) | v \in K \}$. But A (d) = A (p) by Remark 4.2. Also $a = a (f)$.

Q. E. D.

4.3. Now since $\mathfrak{g} = \mathfrak{k} + \mathfrak{p}$ is a Cartan decomposition of $\mathfrak{g}$ one knows that $\mathfrak{g}_u = \mathfrak{k} + i\mathfrak{p}$ is a compact real form of $\mathfrak{g}$. Thus for each $\lambda \in \hat{\mathbb{G}}$ there exists a (by irreducibility) unique, up to scalar, Hilbert space structure on $V_\lambda$ such that $\pi_\lambda(z)$ is skew-Hermitian for all $z \in \mathfrak{g}_u$. We will assume henceforth that V is given this structure. Thus for any $g \in G$ the operator norm $\| \pi_\lambda(g) \|$ is well defined. From standard properties of the operator norm one has that

$$ (4.3.1) \qquad \| \pi_\lambda(g) \| \cdot \| \pi_\lambda(f) \| \ge \| \pi_\lambda(gf) \| $$

for any $g, f \in G$. Also the operator norm is not less than the absolute value of any eigenvalue so that one always has the inequality

$$ (4.3.2) \qquad \| \pi_\lambda(g) \| \ge |\pi_\lambda(g)| $$

for any $g \in G$.

On the other hand since $\pi_\lambda(z)$ is skew-Hermitian for $z \in i\mathfrak{p}$ it follows that $\pi_\lambda(x)$ is Hermitian for $x \in \mathfrak{p}$ and hence $\pi_\lambda(p)$ is positive definite for any $p \in P$. For a positive definite operator, however, the spectral radius is the same as the operator norm so that one has

$$ (4.3.3) \qquad | \pi_\lambda(p) | = \| \pi_\lambda(p) \| $$

for all $p \in P$. Thus under certain circumstances we can replace the spectral radius by the operator norm in Theorem 3.1 obtaining

**PROPOSITION 4.3.** — Let $p \in P$ and $g \in G$ be arbitrary. If for every irreducible representation $\lambda \in \hat{\mathbb{G}}$ one has

$$ \| \pi_\lambda(p) \| \ge \| \pi_\lambda(g) \| $$

then $p \ge g$ (see § 3.1).

*Proof.* — Under the assumption in the Proposition one has

$$ | \pi_\lambda(p) | = \| \pi_\lambda(p) \| \ge \| \pi_\lambda(g) \| \ge | \pi_\lambda(g) | $$

for all $\lambda \in \hat{\mathbb{G}}$ by (4.3.3) and (4.3.2). The result then follows from Theorem 3.1.

Q. E. D.

But now $\pi_\lambda(z)$ is skew-Hermitian for any $z \in \mathfrak{k}$ and hence $\pi_\lambda(k)$ is unitary for any $k \in K$. Thus

$$ (4.3.4) \qquad \| \pi_\lambda(k) \| = 1 $$

for all $k \in K$.
Proposition 4.3 is applied to yield the following:

LEMMA 4.3. — Let $p \in P$ and let $k, v \in K$ be arbitrary. Then, with respect to the partial ordering of section 3.1,

$$p \geq kpv.$$

*Proof.* — For any $\lambda \in \hat{G}$ one has

$$\| \pi_{\lambda}(p) \| \geq \| \pi_{\lambda}(kpv) \|$$

by (4.3.1) since $\| \pi_{\lambda}(k) \| = \| \pi_{\lambda}(v) \| = 1$ by (4.3.4). The result then follows from Proposition 4.3.

We can now establish one half of Theorem 4.1, namely that

$$(4.3.5) \qquad a(bv) \in A \quad (b) \quad \text{for any } v \in K.$$

Indeed we write the Iwasawa decomposition $bv = kan$. Then $k^{-1}bv = an$. But by Lemma 4.3 one has $b \geq an$. That is $A(an) \subseteq A(b)$. But by Proposition 2.5 one has $A(an) = A(a)$. Thus $a(bv) = a \in A(b)$ proving (4.3.5).

4.4. We now use the Horn-Thompson result for $SL(2, \mathbf{R})$ (slightly modified with unipotent elements).

LEMMA 4.4 (Horn-Thompson). — Assume $\mathfrak{g}$ is isomorphic to the Lie algebra of $SL(2, \mathbf{R})$. Let $b \in A, n \in \mathbb{N}$. Then for any $a \in A$ ($b$) there exists $k, v \in K$ and $n' \in \mathbb{N}$ such that

$$kbnv = an'$$

*Proof.* — Let $G_1$, as in section 2.3, be $\mathrm{Ad} G$ (the real adjoint group). We first observe the lemma is true for $G$ if and only if it is true for $G_1$. Indeed $\mathrm{AN}$ maps bijectively onto its image by $\mathrm{Ad}$ (since $Z \subseteq K$). Thus, if the lemma is true in $G$, by applying $\mathrm{Ad}$, it is true for $G_1$. Conversely if it is true for $G_1$ then in $G$ it is immediate that $k, v, n'$ exist so that $kbnv = zan'$ where $z$ is in the center of $G$. But $z \in K$ by Proposition 2.3 so that one replaces $k$ by $z^{-1}k$.

But now since $G$ and $SL(2, \mathbf{R})$ having isomorphic adjoint groups it is enough to prove the result assuming that $G = SL(2, \mathbf{R})$. We may take $K = SO(2, \mathbf{R})$, $A$ is the set of all $2 \times 2$ unimodular diagonal matrices with positive entries and $N$ is the set of all lower triangular $2 \times 2$ matrices with 1's along the diagonal. Note that $W$ is the group of order 2 where if $1 \neq \sigma \in W$ then $\sigma d = d^{-1}$ for any $d \in A$. It follows then that if $d \in A$ is such that the eigenvalues (on $\mathbf{R}^2$) of $d$ are $x$ and $\frac{1}{x}$ where $x \geq 1 \geq \frac{1}{x} > 0$ then $A(d)$ is the set of all $c \in A$ with positive eigenvalues $y \geq \frac{1}{y}$ such that $x \geq y \geq 1 \geq \frac{1}{y} \geq \frac{1}{x} > 0$. [Note that for $c$ (resp. $d$) it is not specified
fied that $y$ (resp. $x$) should be in the upper left hand corner or in the lower right hand corner.]

Now write $bn = k_1 dk_2$ where $d \in A$ and $k_1, k_2 \in K$. Then $k_1^{-1} bn = dk_2$ so that $b = a (dk_2)$. By (4.3.5) one has $b \in A (d)$. But now if $a \in A (b)$ then certainly $a \in A (d)$. However by Lemma 3, p. 471 in [14] where $x_2 = x_1^{-1}$ there exists $k_3, k_4 \in K$ [Thompson says unitary but in fact the matrices he constructs are in $SO(2, \mathbf{R})$] such that $k_3 dk_4 = an''$ for some $n'' \in \mathbb{N}$. Now put $k = k_3 k_1^{-1} \in K$, $v = k_2^{-1} k_4 \in K$ then $kbnv = k_3 dk_4 = an''$.

Q. E. D.

4.5. Now return to the general case. The real form $\mathfrak{g}$ of $\mathfrak{g}_c$ is called a normal real form (or $\mathbf{R}$-split form, or Chevalley form) in case $a_c$, the complexification of $a$, is a Cartan subalgebra of $\mathfrak{g}$. In this case the restricted roots are the roots in the usual sense, and $W$ is the ordinary Weyl group, and the root spaces are 1-dimensional.

To finish the proof of Theorem 4.1 we assert it is enough to prove the theorem under the assumption that $\mathfrak{g}$ is a normal real form of $\mathfrak{g}_c$. Indeed let $\Gamma^* \subseteq \Gamma$ be the set of all restricted roots $\gamma \in \Gamma$ such that $\frac{\gamma}{2} \notin \Gamma$. For example $\Sigma \subseteq \Gamma^*$. Note that since $\gamma$ and $2\gamma$ define the same hyper-plane in $a$ the Weyl group $W$ is generated by $\tau(\gamma)$ for $\gamma \in \Gamma^*$. But now there exists a semi-simple subalgebra $\mathfrak{g}^* \subseteq \mathfrak{g}$ such that $a \subseteq \mathfrak{g}^*$ and such that $\mathfrak{g}^*$ is a normal real form of its complexification of $\mathfrak{g}_c^*$ and $a_c$ is a Cartan subalgebra of $\mathfrak{g}_c^*$. Furthermore $\Gamma^*$ is the set of roots of $(a_c, \mathfrak{g}_c^*)$ and $W$ is the usual Weyl group for $(a_c, \mathfrak{g}_c^*)$. Finally $\mathfrak{g}^*$ is stable under $\mathfrak{0}$ so that

$$ (4.5.1) \qquad \mathfrak{g}^* = k^* + \mathfrak{p}^* $$

is a Cartan decomposition where $k^* = k \cap \mathfrak{g}^*$ and $\mathfrak{p}^* = \mathfrak{p} \cap \mathfrak{g}^*$. For the proof of these statements see section 11.2, p. 786 in [8]. More specifically see Propositions 2.1 and 2.3 noting that now $\mathfrak{g}^* = \mathfrak{b}$ in the notation there since now center $\mathfrak{g} = 0$. See also Remark 14, p. 790.

Now let $K^* \subseteq G^*$ be the subgroup corresponding to $k^*$ and $\mathfrak{g}^*$ respectively. Since $\mathfrak{g}^*$ is semi-simple $K^*$ and $G^*$ are closed. Also $K^* \subseteq K$ and $A \subseteq G^*$. Also the definition of $A$ ($b \in A$) does not change if we replace $G$ by $G^*$ since $W$ is not changed. Also using $x_0$ to define $\mathfrak{n}^* \subseteq \mathfrak{g}^*$ in the manner of section 2.2 one has $\mathfrak{n}^* = \mathfrak{n} \cap \mathfrak{g}^*$ so that $N^* \subseteq N$ where $N^*$ is the subgroup corresponding to $\mathfrak{n}$. Thus for any $g \in G^*$ the $a$-component $a(g) \in A$ does not change if we replace $G$ by $G^*$.

Now returning to the notation of Theorem 4.1 put

$$ (4.5.2) \qquad C = \{ a(bv) \mid v \in K \}. $$
We have already shown [see (4.3.5)] that $C \subseteq A(b)$. To prove Theorem 4.1 we must prove that $A(b) \subseteq C$. But now if we put $C^* = \{a(bv) \mid v \in K\}$ then $C^* \subseteq C$. However if we prove Theorem 4.1 under the assumption that $\mathfrak{g}$ is a normal real form of $\mathfrak{g}_c$ we can conclude that $A(b) \subseteq C^*$. But this implies $A(b) \subseteq C$ proving our assertion.

Henceforth until Theorem 4.1 is proved we will assume that $\mathfrak{g}$ is a
real normal form of $\mathfrak{g}_c$.

4.6. Now for any root $\varphi \in \Gamma$ let $a_\varphi^\perp$ be the hyperplane in $\mathfrak{a}$ orthogonal to $\varphi$. Also for any $x \in \mathfrak{a}$ let $\{x, \tau(\varphi)x\} \subseteq \mathfrak{a}$ be the line segment joining $x$ to its reflected image $\tau(\varphi)x$ through the hyperplane $a_\varphi^\perp$. Thus $\{x, \tau(\varphi)x\}$ is perpendicular to the hyperplane $a_\varphi^\perp$ unless of course it reduces to a point (i. e. if $x \in a_\varphi^\perp$). Let

$$
(4.6.1) \qquad \varepsilon = \log C = \{ \log a(bv) \mid v \in K \} \subseteq \mathfrak{a}.
$$

LEMMA 4.6. — $\varepsilon$ is stable under the Weyl group. In fact for any $x \in c$
and any root $\gamma \in \Gamma$ we claim

$$
\{ x, \tau(\gamma) x \} \subseteq \varepsilon.
$$

Proof. — As in 3.2. let $\beta_i$, $i = 1, 2, \dots, l$ be the simple positive roots.
Fix $1 \le i \le l$ and write $a_i^\perp = a_i \beta_i$, so that

$$
(4.6.2) \qquad a = a_i \oplus a_i^{\perp}
$$

where $a_i$ is the one-dimensional subspace spanned by $x_i = x_{\beta_i}$ (see § 3.2).
Also let $e_i, f_i \in \mathfrak{g}$ be root vectors corresponding, respectively, to the roots $\beta_i$
and $-\beta_i$. The real 3-dimensional space, $\mathfrak{g}_i$, spanned by $x_i, e_i$ and $f_i$
is a Lie subalgebra isomorphic to the Lie algebra of SL(2, $\mathbb{R}$) (a real TDS).
Furthermore since $\mathfrak{g} = \mathfrak{g}_i$ and $\mathfrak{h} = \mathfrak{h}_i$ (because $\mathfrak{h} = -\mathfrak{g}$)
on $\mathfrak{a}$) it follows that $\mathfrak{g}_i$ is stable under $\mathfrak{h}$ and

$$
(4.6.3) \qquad \mathfrak{g}_i = k_i \oplus a_i \oplus \mathfrak{n}_i
$$

is an Iwasawa decomposition of $\mathfrak{g}_i$ where $k_i = k \cap \mathfrak{g}_i$, $\mathfrak{n}_i = \mathfrak{n} \cap \mathfrak{g}_i = \mathbb{R} e_i$
and $a_i$, defined above, is also given by $a_i = a \cap \mathfrak{g}_i$.

Thus if $K_i$, $A_i$ and $N_i$ are the subgroups of $G$ corresponding to $k_i$, $a_i$,
and $\mathfrak{n}_i$ respectively, then $G_i = K_i A_i N_i$ is an Iwasawa decomposition
of $G_i$ and if $g \in G_i$, then, by uniqueness, the Iwasawa decomposition
of $g$ is the same whether we regard $g$ as an element of $G_i$ or an element
of $G$. In particular the $a$-component, $a(g)$ is the same and $a(g) \in A_i$.
However, the Weyl group of $G_i$ operating on $A_i$ or $a_i$ is of order 2 where
the non-trivial element is defined by restricting $\tau_i = \tau(\beta_i)$ to $A_i$ or $a_i$.
Also for any $g \in G_i$ let $A_i(g) \subseteq A_i$ be defined as in section 3.1 but
with respect to $G_i$ not $G$.
Now let $n_i^{\perp} \subseteq g$ be the space spanned by all root vectors belonging to all positive roots $\gamma \in \Gamma_+$ where $\gamma \neq \beta_i$. Thus

$$ (4.6.4) \qquad n = n_i \oplus n_i^{\perp}. $$

Now $n_i$ is not only an ideal in $n$ but in fact one has

$$ (4.6.5) \qquad [g_i, n_i^{\perp}] \subseteq n_i^{\perp}. $$

This is clear since if $\gamma \in \Gamma_+$ where $\gamma \neq \beta_i$ and $\gamma - \beta_i$ is a root then $\gamma - \beta_i \in \Gamma_+$ (otherwise $\beta_i$ would not be simple). It follows therefore that if $N_i^{\perp}$ is the subgroup corresponding to $n_i$ then not only is

$$ (4.6.6) \qquad N = N_i N_i^{\perp} $$

a semi-direct product but

$$ (4.6.7) \qquad G_i \text{ normalizes } N_i^{\perp}. $$

Now let $y \in \mathfrak{c}$. Thus if $a = \exp y$ then there exists $v, k \in K$ and $n \in N$ such that

$$ (4.6.8) \qquad bv = kan. $$

Now we want to show that $\{y, \tau_i y\} \subseteq \mathfrak{c}$. But if we write $y = y_i + y_i^{\perp}$ where $y_i \in a_i$ and $y_i^{\perp} \in a_i^{\perp}$ then

$$ (4.6.9) \qquad \{y, \tau_i y\} = \{ty_i + y_i^{\perp} \mid -1 \le t \le 1\}. $$

But if $A_i^{\perp}$ is the subgroup of $A$ corresponding to $a_i^{\perp}$ then $a = a_i a_i^{\perp}$ where $\exp y_i = a_i \in A_i$ and $\exp y_i^{\perp} = a_i^{\perp} \in A_i^{\perp}$. But then since

$$ \begin{align*} A_i(a_i) &= \{ \exp ty_i \mid -1 \le t \le 1 \} \\ \exp \{ y, \tau_i y \} &= A_i(a_i) a_i^{\perp} \end{align*} $$

by (4.6.9). Thus to show $\{y, \tau_i y\} \subseteq \mathfrak{c}$ one must show $A_i(a_i) a_i^{\perp} \subseteq \mathfrak{c}$.

Now write $n = n_i n_i^{\perp}$ where $n_i \in N_i$, $n_i^{\perp} \in N_i^{\perp}$ according to the decomposition (4.6.6). But then by (4.6.8) one has

$$ (4.6.10) \qquad k^{-1}bv = a_i a_i^{\perp} n_i n_i^{\perp} = a_i n_i a_i^{\perp} n_i^{\perp}. $$

The commutativity of $n_i$ and $a_i^{\perp}$ follows from the relation

$$ (4.6.11) \qquad [a_i^{\perp}, g_i] = 0. $$

The relation (4.6.11) is immediate since $\beta_i$ vanishes on $a_i^{\perp}$.

But if $c_i \in A_i(a_i)$ then by Lemma 4.4 there exists $k_i, v_i \in K_i$ and $n'_i \in N_i$ such that

$$ (4.6.12) \qquad k_i a_i n_i v_i = c_i n'_i. $$
However by (4.6.10)

$$k_i k^{-1} b v v_i = k_i a_i n_i a_i^{\perp} n_i^{\perp} v_i.$$

But by (4.6.7) there exists $m_i^{\perp} \in N_i^{\perp}$ such that $n_i^{\perp} v_i = v_i m_i^{\perp}$. However $v_i$ commutes with $a_i^{\perp}$ by (4.6.11). Thus

$$ (4.6.13) \qquad \begin{aligned} k_i k^{-1} b v v_i &= k_i a_i n_i v_i a_i^{\perp} m_i^{\perp} \\ &= c_i n_i^{\perp} a_i^{\perp} m_i^{\perp} && \text{by (4.6.12)} \\ &= c_i a_i^{\perp} n_i^{\perp} m_i^{\perp} && \text{by (4.6.11)} \\ &= c_i a_i^{\perp} m \end{aligned} $$

where $m = n_i^{\perp} m_i^{\perp} \in N$. But if $k k_i^{-1} = k' \in K$ then by (4.6.13) :

$$b v v_i = k' c_i a_i^{\perp} m$$

and hence

$$(4.6.14) \qquad a (b v v_i) = c_i a_i^{\perp} \in A_i (a_i) a_i^{\perp}.$$

But $c_i$ is arbitrary in $A_i(a_i)$. Thus $A_i(a_i) a_i^{\perp} \subseteq C$ since $vv_i \in K$ so that

$$(4.6.15) \qquad \{ y, \tau_i y \} \subseteq c$$

for any $y \in c$ and $1 \le i \le l$. In particular $\tau_i y \in c$. Since W is generated by the reflections $\tau_i$, $1 \le i \le l$, this first of all implies that $c$ is stable under the action of W.

But now if $\gamma \in \Gamma$ is arbitrary (since $\gamma$ can be embedded in some simple set of roots) there exists $1 \le i \le l$ and $\sigma \in W$ such that $\sigma \tau_i \sigma^{-1} = \tau(\gamma)$. Now let $x \in c$ be arbitrary and put $y = \sigma^{-1} x$. Then by (4.6.15) one has $\{y, \tau_i y\} \subseteq c$. However since $c$ is stable under W one has $\sigma \{y, \tau_i y\} \subseteq c$. But by linearity $\sigma \{y, \tau_i y\}$ is the line segment joining $\sigma y = x$ to $\sigma \tau_i y = \sigma \tau_i \sigma^{-1} \sigma y = \tau(\gamma)x$. Thus $\{x, \tau(\gamma)x\} \subseteq c$.

Q. E. D.

4.7. Now for any root $\gamma \in \Gamma$ and any subset $d \subseteq a$ we define the subset

$$(4.7.1) \qquad \{ d, \tau(\gamma)d \} = \bigcup_{x \in d} \{ x, \tau(\gamma)x \}.$$

It is easy to give examples where, if $d$ is convex, then $\{d, \tau(\gamma)d\}$ is not necessarily convex. However if $d$ is on one side of the hyperplane $a_i^{\perp}$ then $\{d, \tau(\gamma)d\}$ is indeed convex.

**LEMMA 4.7.** — Assume $d \subseteq a$ is any convex subset such that $\langle \gamma, x \rangle \ge 0$ for all $x \in d$. Then $\{d, \tau(\gamma)d\}$ is convex.

*Proof.* — Let $y_i \in \{d, \tau(\gamma)d\}$, $i=1, 2$. If $r_i$, $i=1, 2$, are positive numbers such that $r_1 + r_2 = 1$ we must show that

$$y = r_1 y_1 + r_2 y_2 \in \{d, \tau(\gamma)d\}.$$
In fact if $y_i \in \{x_i, \tau(\gamma)x_i\}, i = 1, 2$, where $x_i \in \mathfrak{d}$ it suffices to show that $y_i \in \{x, \tau(\gamma)x\}$ where $x = r_1 x_1 + r_2 x_2$. This is clear since $x \in \mathfrak{d}$. But now $\tau(\gamma)x_i = x_i - \langle \gamma, x_i \rangle x_\gamma$, by (3.2.2) where by assumption $\langle \gamma, x_i \rangle \ge 0$. Thus

$$ \{ x_i, \tau(\gamma)x_i \} = \{ x_i - tx_\gamma \mid 0 \le t \le \langle \gamma, x_i \rangle \}. $$

Hence there exists

$$ (4.7.2) \qquad 0 \le t_i \le \langle \gamma, x_i \rangle $$

such that $y_i = x_i - t_i x_\gamma$. Thus $y = x - s x_\gamma$ where $s = r_1 t_1 + r_2 t_2$. But

$$ \langle \gamma, x \rangle = \langle \gamma, r_1 x_1 + r_2 x_2 \rangle = r_1 \langle \gamma, x_1 \rangle + r_2 \langle \gamma, x_2 \rangle. $$

Thus $0 \le s \le \langle \gamma, x \rangle$ by (4.7.2) and hence

$$ y \in \{ x, \tau(\gamma)x \} \subseteq \mathfrak{d}, \tau(\gamma)\mathfrak{d}. $$

Q. E. D.

4.8. The proof of Theorem 4.1 is now an immediate consequence of the following general fact about convexity and the Weyl group.

**LEMMA 4.8.** — Let $\mathfrak{d} \subseteq \mathfrak{a}$ be any subset such that for any $z \in \mathfrak{d}$ and root $\gamma \in \Gamma$ the line segment $\{z, \tau(\gamma)z\}$ lies in $\mathfrak{d}$. Then for any $x \in \mathfrak{d}$ the convex hull $\mathfrak{a}(x)$ of the Weyl group orbit $w(x) = W \cdot x$ of $x$ also lies in $\mathfrak{d}$.

*Proof.* — Since $\tau(\gamma)z \in \mathfrak{d}$ for any $\gamma \in \Gamma$ and $z \in \mathfrak{d}$ it is clear that $\mathfrak{d}$ is stable under the action of $W$. Thus it suffices to show that $\mathfrak{a}(x) \subseteq \mathfrak{d}$ for any $x \in \mathfrak{d} \cap \mathfrak{a}_+$. Let $x \in \mathfrak{d} \cap \mathfrak{a}_+$ and assume, inductively, for any $n$ elements, $\sigma_i \in W$, $i = 1, 2, \dots, n$, the convex hull, $c(\sigma_1, \dots, \sigma_n)$ of the vectors $x, \sigma_1 x, \dots, \sigma_n x$ lies in $\mathfrak{d}$. The assumption is obviously true if $n = 0$. We now wish to prove that $c(\sigma_1, \dots, \sigma_{n+1}) \subseteq \mathfrak{d}$ for any $n+1$ elements $\sigma_i \in W$. We will prove this statement by induction in the following way : In section 3.2 we defined what we meant by a strongly positive sequence $\gamma_1, \dots, \gamma_m$ (of positive roots). Assume inductively that for any such sequence of $m$ positive roots and any $n$ elements $\sigma_i \in W$ the convex hull $c(\sigma_1, \dots, \sigma_n, \gamma_1, \dots, \gamma_m)$ of $x, \sigma_1(x), \dots, \sigma_n(x)$ and the additional element $\tau(\gamma_m) \tau(\gamma_{m-1}) \dots \tau(\gamma_1) x = x_m$ lies in $\mathfrak{d}$. The statement is clearly true for $m = 0$ if $x_0 = x$. We will now prove that $c(\sigma_1, \dots, \sigma_n, \gamma_1, \dots, \gamma_{m+1}) \subseteq \mathfrak{d}$ for any $n$ elements $\sigma_i \in W$ and a strongly positive sequence $\gamma_1, \dots, \gamma_{m+1}$ of positive roots. Put $\gamma = \gamma_{m+1}$. Now by definition of a strongly positive sequence one has $\langle \gamma, x_m \rangle \ge 0$ since $x \in \mathfrak{a}_+$. One also has $\langle \gamma, x \rangle \ge 0$. On the other hand for any $1 \le j \le n$ one has $\langle \gamma, y \rangle \ge 0$ for either $y = \sigma_j x$ or for $y = \tau(\gamma) \sigma_j x$, the reflected image
of $\sigma_j x$. Put $\sigma'_j = \sigma_j$ or $\tau(\gamma)\sigma_j$, so that $\langle\gamma, \sigma'_j x\rangle \ge 0$. But now by induction $\mathfrak{d}_1 = c(\sigma'_1, \dots, \sigma'_n, \gamma_1, \dots, \gamma_m) \subseteq \mathfrak{d}$. On the other hand $\mathfrak{d}_1$ lies on one side of the hyperplane $a^\perp$. However if $z \in \mathfrak{d}$ then by assumption $\{z, \tau(\gamma)z\} \subseteq \mathfrak{d}$. Thus $\mathfrak{d}_1, \tau(\gamma)\mathfrak{d}_1 \subseteq \mathfrak{d}$. But by Lemma 4.7 $\mathfrak{d}_1, \tau(\gamma)\mathfrak{d}_1$ is convex. However $\sigma_j x \in \mathfrak{d}_1, \tau(\gamma)\mathfrak{d}_1$ for any $j$ since $\sigma_j$ is one of the elements $\sigma'_j$ or $\tau(\gamma)\sigma'_j$. But also

$$x_{m+1} = \tau(\gamma_{m+1}) x_m = \tau(\gamma) x_m \in \mathfrak{d}_1, \tau(\gamma) \mathfrak{d}_1.$$

Thus

$$c(\sigma_1, \dots, \sigma_n, \gamma_1, \dots, \gamma_{m+1}) \subseteq \mathfrak{d}_1, \tau(\gamma) \mathfrak{d}_1$$

since $\mathfrak{d}_1, \tau(\gamma)\mathfrak{d}_1$ is convex. Hence $c(\sigma_1, \dots, \sigma_n, \gamma_1, \dots, \gamma_{m+1}) \subseteq \mathfrak{d}$ proving the induction statement on $m$ for all $m$. But now given $\sigma_{n+1} \in W$ there exists a strongly positive sequence $\gamma_1, \dots, \gamma_m$ of positive roots such that $\sigma_{n+1} = \gamma_m \cdot \gamma_{m-1} \dots \gamma_1$. Thus $c(\sigma_1, \dots, \sigma_{n+1}) \subseteq \mathfrak{d}$ for any $n+1$ elements $\sigma_j \in W$ proving the induction assumption on $n$ for all $n$. Putting $n$ equal to the order of $W$ one obviously has $a(x) \subseteq \mathfrak{d}$.

Q. E. D.

**REMARK 4.8.** — Note that the argument at the end of the proof of Lemma 4.6 shows that one can weaken the assumption in Lemma 4.8 in that $\gamma$ can be restricted to be simple.

We now give the

*Proof of Theorem 4.1.* — Recalling the notation of Theorem 4.1 we must prove that if $C \subseteq A$ is defined by (4.5.2) then $C = A(b)$. As remarked after (4.5.2) we have only to show that $A(b) \subseteq C$ or if $c$ is defined by (4.6.1) and $x = \log b$ we have only to show that $a(x) \subseteq c$. But $x \in c$ [by choosing $v = 1$ in (4.5.2)]. On the other hand for any $z \in c$ and $\gamma \in \Gamma$ one has $\{z, \tau(\gamma)z\} \subseteq c$ by Lemma 4.6. But then by Lemma 4.8 one has $a(x) \subseteq c$.

Q. E. D.

## 5. Applications : Some K-double coset theorems

5.1. Now by (4.2.8) one has that $G = \text{KAK}$. This fact is important in analysis on Lie groups since, among other things, it says that a spherical function on $G$ is determined by its restriction to $A$. On the other hand a corollary to Corollary 4.2 is

**THEOREM 5.1.** — For any semi-simple Lie group $G$ one has

$$G = \text{KNK}.$$
*Proof.* — If $x \in a$ then the sum $y = \sum \sigma x$, over all $\sigma \in W$ equals zero. This is clear since $y$ is a $W$-invariant in $a$ and, since $\Gamma$ spans $a'$, only $0$ is a $W$-invariant. Thus $0 \in a(x)$ for any $x \in a$ and hence (5.1.1) $1 \in A(f)$

for any $f \in G$. But if $g \in G$ is arbitrary then $1 \in A(p)$ where $p = p(g)$ and hence by Corollary 4.2 there exists $v \in K$ such that $a(gv^{-1}) = 1$. That is $gv^{-1} = kn$ where $k \in K, n \in N$. Hence $g = knv \in KNK$.

Q. E. D.

Theorem 5.1 says that a spherical function on $G$ is now determined by its restriction to $N$.

**REMARK 5.1.1.** — It might be interesting to explore what the spherical functions look like on $N$. In the case of $A$ the elementary spherical functions appear in exponential form [see e. g. (4.1.2)]. Although we have not investigated this it seems likely that for $N$ these functions might have a natural power series expansion instead.

Of course just by dimension considerations it is clear that the spherical functions on $N$ do not separate points; thereby creating “level surfaces”. This is also true for $A$ but in that case the level surfaces are finite sets whose cardinality is at most the order of the Weyl group. One can approach this situation at least in the case where rank $K = \text{rank } G$ (existence of the discrete series case, by Harish-Chandra’s theorem). In fact if rank $K = \text{rank } G$ and $k = \text{dim } A$ then one can find a connected abelain subgroup $N_0 \subseteq N$ where $\text{dim } N_0 = k$ such that (5.1.2) $G = KN_0K$.

In fact one can find $k$ orthogonal restricted roots $\gamma_i$ where $\gamma_i | h_m = 0$. These define $k$ commuting TDS’s and (5.1.2) follows already using (4.2.8) and the Horn-Thompson theorem on $SL(2, \mathbf{R})$.

**REMARK 5.1.2.** — The relation (5.1.2) can be thought of as saying that $N_0$ carries the non-compact part of $G$. It should be easy to determine the measure decomposition of $G$ defined by (5.1.2). It might then be interesting to investigate what the representative functions of, say, the discrete series look like on $N_0$.

**5.2.** Theorem 5.1 can be given a geometric interpretation. If $X$ is the symmetric space (5.2.1) $X = G/K$

then a horocycle is an orbit of a conjugate $g N g^{-1} [= k N k^{-1}$ where $k = k(g) \in K]$ of $N$ on $X$. Theorem 5.1 says the same then as

**THEOREM 5.2.** — Any two points in $X$ can be embedded in a horocycle.
*Proof.* — If $o \in X$ is the coset of $K$ then the statement of Theorem 5.1 is the statement that every point of $X$ lies on a horocycle through $o$. Theorem 5.2 is just an application of the transitivity of $G$ on $X$.

Q. E. D.

REMARK 5.2. — One sees Theorem 5.2 immediately in the case of SL(2, $\mathbb{R}$). Here $X$ can be identified with the unit disc in $\mathbb{C}$ and the horocycles are just the circles in $X$ which are tangent to the boundary.

5.3. Theorem 5.1 says that every double coset of $K$ in $G$ contains a point of $N$. This, no doubt can be proved in a much simpler way than the one given here. In fact if rank $K = \text{rank } G$ we have already indicated such a proof [see the paragraph after (5.1.2)]. A stronger use of Corollary 4.2 is the following generalization of Theorem 5.1. In effect it says that if $N$ is replaced by a coset $fN, f \in G$, then it is no longer true. More explicitly

THEOREM 5.3. — Let $a \in A$ be arbitrary and let

$$ (5.3.1) \qquad G_a = \{ g \in G \mid p(g) \ge a \} $$

where $p(g)$ is defined by (4.2.6) and the order relation $p(g) \ge a$ is defined (in terms of convexity) in section 3.1. Then for any $f \in G$ one has

$$ G_a = K f NK $$

where $a = a(f)$.

*Proof.* — If we write $f = k_1 a n_1$ where $k_1 \in K, n_1 \in N$ then clearly $K f NK = K a NK$. Let $L$ denote this set. If $g \in L$ then there exists $k, v^{-1} \in K, n \in N$ such that $g = kanv^{-1}$ or $gv = kan$. Thus $a(gv) = a$. But then by Corollary 4.2 one has $a \in A(p(g))$ or $p(g) \ge a$. Thus $g \in G_a$. Conversely if $g \in G$ and $p(g) \ge a$ then by Corollary 4.2 there exists $v \in K$ such that $a(gv) = a$. That is $gv = kan$ for some $k \in K, n \in N$. Hence $g = kanv^{-1} \in L$.

Q. E. D.

5.4. The statement of Theorem 4.1 involves the Iwasawa decomposition and is expressed in terms of the $a$-component. The Horn-Thompson theorem on the other hand is an invariant formulation involving the complete multiplicative Jordan decomposition (since it is a statement about eigenvalues) and hence is expressed in terms of the hyperbolic component. The generalization of the Horn-Thompson theorem to the arbitrary semi-simple case follows, however, from Theorem 4.1.

THEOREM 5.4. — Let $p \in P = \exp p$. Then for any $k, v \in K$ one has $kpv \le p$. That is, $h(kpv) \le p$ using the notation of (2.1.5) and section 3.1. Conversely given any hyperbolic element $h \le p$ there exists $k$,
$v \in K$ such that $h(kpv)$ is conjugate to $h$. Moreover $k$ and $v$ can be chosen so that the elliptic component $e(kpv) = 1$.

*Proof*. — The statement that $kpv \le p$ is just Lemma 4.3. Assume $h \le p$ where $h$ is a hyperbolic element. Since the statement is only up to conjugacy and since any hyperbolic element is conjugate to an element in $A$ we may assume $h = a \in A$. But now by Corollary 4.2 there exists $v \in K$ such that $a(pv) = a$. That is $pv = k^{-1} an$ for some $k \in K$, $n \in N$. Thus $kpv = an$. But, by Proposition 2.5, $h(an)$ is conjugate to $a$. That is, $h(kpv)$ is conjugate to $a$. Also $e(kpv) = e(an) = 1$ by Proposition 2.5.

Q. E. D.

5.5. Theorem 5.4 amounts to a reformulation of Corollary 4.2, using Proposition 2.5. Similarly Theorem 5.3 can be reformulated using Proposition 2.5.

For any $a \in A$ let

$$ (5.5.1) \qquad D_a = \{ g \in G \mid h(g) \text{ is conjugate to } a \text{ and } e(g) = 1 \}. $$

*REMARK 5.5.* — In case $a \in A$ is regular hyperbolic in the sense that the centralizer of $x = \log a$ is just $a + m$ note that $D_a$ is exactly the conjugacy class of $a$. Indeed $a + m$ has no non-trivial nilpotent elements and hence no non-trivial unipotent element can commute with $a$. Thus in this case $u(g) = 1$ as well as $e(g) = 1$ for any $g \in D_a$.

*THEOREM 5.5.* — Let $a \in A$ be arbitrary and let $D_a$ be defined by (5.5.1). Let $G_a$, as in (5.3.1), be defined by putting $G_a = \{ g \in G \mid p(g) \ge a \}$. Then

$$ G_a = KD_a K. $$

*Proof.* — By Proposition 2.5 one clearly has

$$ D_a = \bigcup_{k \in K} ka N k^{-1}. $$

Thus $KD_a K = K a N k = G_a$ by Theorem 5.3.

Q. E. D.

## 6. Characters values and the partial ordering; A generalized Golden-Thompson inequality

6.1. Now by Theorem 3.1 if $g, f \in G$ then $g \ge f$ if and only if the spectral radius $|\pi_\lambda(g)|$ of $\pi_\lambda(g)$ is greater than or equal to the spectral radius $|\pi_\lambda(f)|$ of $\pi_\lambda(f)$ for every irreducible representation $\pi_\lambda, \lambda \in \hat{G}$, of $G$. Now if $g$ and $f$ are hyperbolic then by Proposition 3.4. the eigenvalues of $\pi_\lambda(g)$ and $\pi_\lambda(f)$ are positive so that $g \ge f$ if and only if the maximal eigenvalues of $\pi_\lambda(g)$ is greater than or equal to the maximal eigenvalues of $\pi_\lambda(f)$ for all $\lambda \in G$. But in such a case the minimal eigen-
value of $\pi_\lambda(g)$ is less than or equal to the minimal eigenvalue of $\pi_\lambda(f)$. Thus if $\chi_\lambda$ is the character of the representation $\pi_\lambda$ it is not immediately obvious whether or not $\chi_\lambda(g) \ge \chi_\lambda(f)$ in case $g \ge f$. However it is true. The question of comparing character values was inspired by the Golden-Thompson inequality and Thompson's proof of it (see § 3 [14]). But the point of view taken here is that a comparison of character values can be made whenever $g$ and $f$ are hyperbolic and $g \ge f$, not just in the special case where $g = e^r e^\gamma$ and $f = e^{r+\gamma}$ for $x, y \in \mathfrak{P}$.

**THEOREM 6.1.** — Let $f, g \in G$ be any two hyperbolic elements. Then if $\lambda \in \hat{G}$ is arbitrary and $\chi_\lambda$ is the character of the irreducible representation $\pi_\lambda$, one has

$$ \chi_\lambda(g) \ge \chi_\lambda(f) \\ \text{in case } g \ge f. $$

*Proof.* — Since character values and the order relation in $G$ are independent of conjugacy we may assume $g, f \in A$ and $g \ge f$. Let $y = \log g$ and $x = \log f$ so that $x \in A(y)$.

But now since $t \to e^t$ is a convex function on $\mathbb{R}$ it is immediate that if $\nu \in \nu'$ the multiplicative character $z \mapsto e^{\int_{\nu'} \nu}$, $z \in A$, is a convex function on $A$. However if $\tilde{\chi}_\lambda$ is the function on $A$ defined by $\tilde{\chi}_\lambda(z) = \chi_\lambda(\exp z)$ then $\tilde{\chi}_\lambda$ is a finite sum of such multiplicative characters and hence $\chi_\lambda$ is a convex function on $A$. But since $x \in A(y)$ there exists for each $\sigma \in W$ a positive scalar $c_\sigma$ such that $\sum_{\sigma \in W} c_\sigma = 1$ and $x = \sum_{\sigma \in W} c_\sigma \sigma y$. Thus

$$ (6.1.1) \qquad \tilde{\chi}_{\lambda}(x) \le \sum_{\sigma \in W} c_{\sigma} \tilde{\chi}_{\lambda}(\sigma y). $$

But $\tilde{\chi}_\lambda$ is clearly invariant under the Weyl group. Thus $\tilde{\chi}_\lambda(\sigma y) = \chi_\lambda(g)$ for any $\sigma \in W$. Also $\tilde{\chi}_\lambda(x) = \chi_\lambda(f)$. Thus (6.1.1) becomes

$$ \chi_\lambda(f) \le \left( \sum_{\sigma \in W} c_\sigma \right) \chi_\lambda(g) \text{ or } \chi_\lambda(f) \ge \chi_\lambda(g). $$

Q. E. D.

**REMARK 6.1.** — It is unknown to us whether or not the converse of Theorem 6.1 is true. That is whether or not $g \ge f$ if and only if $\chi_\lambda(g) \ge \chi_\lambda(f)$ for all $\lambda \in G$ where $g$ and $f$ are hyperbolic.

6.2. Now let $L \in G$ be the set of all hyperbolic elements in $G$. We know $P \subseteq L$ where $P$ is defined by (4.2.1). But also $P^2 \subseteq L$ and in fact

**PROPOSITION 6.2.** — One has

$$ P^2 = L. $$
*Proof.* — Let $g \in G$ and $p \in P$. Then first note that

$$ (6.2.1) \qquad gpg^* \in P $$

where $g^* \in G$ is defined by (4.2.3). Indeed if $f = gp^{1/2}$ then $gpg^* = ff^*$. But $ff^* \in P$ as noted in (4.2.6). Now if $q \in P$ then $q^{1/2}pq^{1/2} \in P$ by (6.2.1) since $q^{1/2} = (q^{1/2})^* \in P$. But $pq = q^{-1/2}(q^{1/2}pq^{1/2})q^{1/2}$ so that $pq$ is conjugate to $q^{1/2}pq^{1/2}$. Hence $pq$ is hyperbolic or $P^2 \subseteq L$. On the other hand if $h \in L$ then $h$ is conjugate to an element of $A \subseteq P$. That is there exists $g \in G$ and $p \in P$ such that $h = gpg^{-1} = (gpg^*)(g^*)^{-1}g^{-1}$. But $gpg^* = p_1 \in P$ by (6.2.1) and $q_1 = (g^*)^{-1}g^{-1} \in P$ by, say, (6.2.1), since clearly $(g^*)^{-1} = (g^{-1})^*$. Thus $h = p_1q_1 \in P^2$. Hence $L = P^2$.

Q. E. D.

6.3. Now if $x, y \in \mathfrak{p}$ then by Proposition 6.2 one has $e^r e^y$ is hyperbolic, writing $e^r$ for $\exp x$. But $e^{r+y}$ is also hyperbolic. The Golden-Thompson inequality states that in the case where $G = SL(n, \mathbb{C})$ [actually $GL(n, \mathbb{C})$ but the added scalars are trivially dealt with] and $\mathfrak{p}$ is the set of Hermitian matrices in $\mathfrak{g}$ then

$$ (6.3.1) \qquad \operatorname{tr} e^r e^y \ge \operatorname{tr} e^{r+y}. $$

This has been generalized by Lenard for the case where trace is replaced by the character of representations of $G$. But this follows from Theorem 6.1 as soon as one knows that $e^r e^y \ge e^{r+y}$ in the order relation. In fact Thompson's proof of (6.3.1) in effect proves $e^r e^y \ge e^{r+y}$. This very nice proof however, carries over to the case where $G$ is an arbitrary simple group and we repeat it word for word.

REMARK 6.3. — Even if we could embed $G$ in $SL(n, \mathbb{C})$ for some $n$ where $\mathfrak{p}$ corresponds to a subspace of Hermitian matrices the statement that $e^r e^y \ge e^{r+y}$ in $SL(n, \mathbb{C})$ does not imply this holds in $G$ (see Remark 3.1.1). It is just that Thompson's proof works for $G$.

Theorem 6.3. — Let $x, y \in \mathfrak{p}$ be arbitrary. Then $e^r e^y$ and $e^{r+y}$ are hyperbolic. Moreover

$$ (6.3.2) \qquad e^r e^y \ge e^{r+y} $$

so that, by Theorem 6.1, for any $\lambda \in \hat{G}$ one has

$$ (6.3.3) \qquad \chi_{\lambda}(e^r e^y) \ge \chi_{\lambda}(e^{r+y}). $$

*Proof.* — Let $g \in G$ be arbitrary. We first observe that

$$ (6.3.4) \qquad gg^* \ge g^2. $$

Indeed since $gg^* \in P$ it is enough to show by Proposition 4.3 that $\| \pi_\lambda(gg^*) \| \ge \| \pi_\lambda(g^2) \| = \| (\pi_\lambda(g))^2 \|$ for any $\lambda \in \hat{G}$. But since $\pi_\lambda(g^*)$
is the Hermitian adjoint of $\pi_\lambda(g)$ by (4.2.5) one has the familiar operator norm relation $\| \pi_\lambda(gg^*) \| = \| \pi_\lambda(g) \|^2$. However $\| \pi_\lambda(g) \|^2 \ge \|(\pi_\lambda(g))^2\|$ by (4.3.1) establishing (6.3.4).

More generally for any positive integer $n$ one has $(gg^*)^n \ge g^{2n}$. Indeed this follows from a similar argument as that above or from the general fact that

$$ (6.3.5) \qquad g \ge f \quad \text{implies} \quad g^n \ge f^n. $$

The relation (6.3.5) is an immediate consequence of the fact that if $x, y \in \mathfrak{a}$ and $x \in \mathfrak{a}(y)$ then clearly $nx \in \mathfrak{a}(ny) = n \mathfrak{a}(y)$. We are also implicitly using the obvious fact that

$$ [\text{see (2.1.6)}]. \qquad h(g^n) = h(g)^n $$

Now let $a, b \in P$ and put $g = ab$. Then $gg^* = ab^2a$ and $g^2 = abab$. It is then clear that $(gg^*)^n = ab^2(a^2b^2)^{n-1}a$. But then $a(gg^*)^n a^{-1} = (a^2b^2)^n$. But since $(gg^*)^n \ge g^{2n}$ one has $a(gg^*)^n a^{-1} \ge g^{2n}$ so that $(a^2b^2)^n \ge (ab)^{2n}$. Now put $n = 2^{k-1}$ for some positive integer $k$. Thus $(a^2b^2)^{2^{k-1}} \ge (ab)^{2k}$. Iterating this relation backwards on $k$ one gets the relation

$$ (6.3.6) \qquad a^{2k} b^{2k} \ge (ab)^{2k}. $$

But then if $a = e^{x/2k}$, $b = e^{y/2k}$ one has

$$ (6.3.7) \qquad e^x e^y \ge (e^{x/2k} e^{y/2k})^{2k}. $$

But one knows $\lim (e^{x/t} e^{y/t})^t = e^{x+y}$ as $t \to \infty$ (see e. g. [15], p. 110, 112). But then by taking the limit in (6.3.7) as $k \to \infty$ one has $e^x e^y \ge e^{x+y}$. The continuity of the order relation is an immediate consequence of Theorem 3.1 since the spectral radius is a continuous function on the space of operators.

Q. E. D.

## 7. Applications to the Geometry of Symmetric Spaces of Negative Curvature

7.1. The inequality (6.3.2) has a geometric interpretation which, I think, may be useful to elaborate upon. In effect (6.3.2) is a statement about geodesic triangles in a symmetric space of negative curvature.

In section 3.1 we defined a partial ordering in $G$. However it could have been in $\mathfrak{g}$ as well. In doing so now we limit the order relation to the set $\mathfrak{l}$ of all real semi-simple elements. If $x$ and $y$ are two such elements define $y \ge x$ if $\exp y \ge \exp x$. That is, if

$$ (7.1.1) \qquad a(x) \subseteq \mathfrak{a}(y). $$
Note that if $y \ge x$ then for any positive number $r$ one has

$$ (7.1.2) \qquad ry \ge rx, $$

[since clearly $a(rx) = r a(x)$] for any $x \in s$.

Now since $t \to e'$ is a monotonic increasing function on $\mathbb{R}$, Theorem 3.1 implies

**THEOREM 7.1.** — Let $x, y \in \mathfrak{l}$. Then $y \ge x$ if and only if for every $\lambda \in \hat{\mathbb{G}}$ the maximal eigenvalue of $\pi_\lambda(y)$ is greater or equal to the maximal eigenvalue of $\pi_\lambda(x)$.

Now for $x \in \mathfrak{l}$ let $(x, x)$ denote the inner product defined by the Killing form. Clearly $(x, x) \ge 0$. Put $|x| = (x, x)^{1/2}$. Since the Killing form is invariant under conjugation and since its restriction to $\mathfrak{a}$ is invariant under $W$ it is obvious that

$$ (7.1.3) \qquad |y| \ge |x| $$

in case $y \ge x$.

**REMARK 7.1.** — The converse is obviously false. That is, if, for example, $x, y \in \mathfrak{a}$ then saying that $y \ge x$ is giving more information than saying $|y| \ge |x|$. The latter refers only to the sphere generated by $y$ while the former refers to the smaller $W$-invariant convex set generated by $y$. An implication of this additional information is given in Theorem 7.1. But it is this additional information that (6.3.2) will give concerning geodesic triangles in symmetric spaces of negative curvature.

7.2. Let $X = G/K$ so that $X$ has a natural structure (normalized by using the Killing form) of a Riemannian symmetric space of negative curvature on which $G$ operates as a group of motions.

Now to every two points $r, s \in X$ we may associate a real semi-simple element $x(r, s) \in \mathfrak{l}$ as follows : Let $k_r$ be the Lie algebra of an isotropy subgroup at $r$. Thus if $\mathfrak{p}_r$ is the orthogonal complement of $k_r$ in $\mathfrak{g}$ then $\mathfrak{g} = k_r + \mathfrak{p}_r$ is a Cartan decomposition. But the exponential map $\sigma_r$ at $r$ defines a bijection of $\mathfrak{p}_r$ onto $X$, recalling that any two points of $X$ can be joined by a unique geodesic. Thus there exists a unique element $x(r, s) \in \mathfrak{p}_r \subset \mathfrak{l}$ such that $\sigma_r(x(r, s)) = s$. Since $G$ operates as motions on $X$ one clearly has

$$ (7.2.1) \qquad \mathrm{Ad} \, g(x(r, s)) = x(g.r, g.s) $$

for any $g \in G$ where $g.r$ and $g.s$ are, respectively, the image of $r$ and $s$ under the action of $g$.
In the language of Elie Cartan, the one parameter group of motions on $X$ generated by $x(r, s)$ is the group of transvections associated with the geodesic joining $r$ and $s$. But then if $d(r, s)$ is the distance of $r$ to $s$, one has

$$ (7.2.2) \qquad d(r, s) = |x(r, s)|. $$

Now let $o, r$, and $s$ be any three points in $X$ and consider the geodesic triangle generated by these three points, and consider the corresponding three real semi-simple elements $x(r, o)$, $x(o, s)$ and $x(r, s)$.

Now if instead of $X$ we were dealing with ordinary Euclidean space then we would have $x(r, o) + x(o, s) = x(r, s)$ so in particular

$$ |x(r, o) + x(o, s)| = |x(r, s)|. $$

However since we are dealing with a symmetric space of negative curvature then the geometry of such a space yields the fact that $d(r, s)$ is bigger than it would be in a flat space, or

$$ (7.2.3) \qquad |x(r, o) + x(o, s)| \le |x(r, s)| = d(r, s). $$

**REMARK 7.2.** — This relation is proved in [11]. In fact if $a = d(o, r)$, $b = d(o, s)$ and $\psi$ is the angle between $x(o, r)$ and $x(o, s)$ then

$$ |x(r, o) + x(o, s)|^2 = a^2 + b^2 - 2ab \cos \psi $$

since easily one has $x(r, o) = -x(o, r)$ [see e. g. (7.2.6)]. But then if $c = d(r, s)$, so that $a, b$ and $c$ are the lengths of the sides of the geodesic triangle made by $o, r$, and $s$, then (7.2.3) (by squaring) is just the statement that the law of cosines in flat spaces is replaced by the inequality

$$ (7.2.4) \qquad a^2 + b^2 - 2ab \cos \psi \le c^2 $$

is a symmetric space of negative curvature. But (7.2.4) is exactly Lemma 4, p. 35 in [11].

But the point now is that a geodesic arc in $X$ carries more information than just its length. Theorem 7.2 says that we may remove the absolute signs in (7.2.3) (see Remark 7.1).

**THEOREM 7.2.** — Let $o, r$ and $s$ be as above. Then $x(r, o) + x(o, s)$ is real semi simple and

$$ x(r, o) + x(o, s) \le x(r, s). $$

*Proof.* — Let $\pi: P \to X = G/K$ be the map defined by $\pi(q) = q^{1/2} K$. It is clear from (4.2.2) that $\pi$ is a bijection. But now if $f \in G$ let $T_f: X \to X$
be the motion induced by the action of $f$ on $X$. Thus $T_f(g K) = fg K$ for any $g \in G$. We now note that

$$ (7.2.5) \qquad T_f \pi(q) = \pi(fqf^*) $$

for any $q \in P$. Indeed $T_f \pi(q) = f q^{1/2} K$. But for any $g \in G$ one has $g K = p(g) K$ by (4.2.2). Thus $T_f \pi(q) = p(fq^{1/2}) K = (fqf^*)^{1/2} K = \pi(fqf^*)$ using (4.2.6).

Now for any $p, q \in P$ we next assert that

$$ (7.2.6) \qquad \exp x(\pi(p), \pi(q)) = (qp^{-1})^{1/2} $$

recalling that $qp^{-1}$ is hyperbolic by Proposition 6.2. Indeed by (7.2.5) one has $T_{p^{1/2}} \pi(1) = \pi(p)$. Now let $d \in P$ be defined by $T_{p^{-1/2}} \pi(q) = \pi(d)$ so that $T_{p^{1/2}} \pi(d) = \pi(q)$. Thus $T_{p^{1/2}}$ carries the unique geodesic joining $\pi(1)$ and $\pi(d)$ onto the unique geodesic joining $\pi(p)$ to $\pi(q)$ and hence by (7.2.1) Ad $p^{1/2} x(\pi(1), \pi(d)) = x(\pi(p), \pi(q))$ or

$$ (7.2.7) \qquad p^{1/2} \exp x(\pi(1), \pi(d)) p^{-1/2} = \exp x(\pi(p), \pi(q)). $$

But $\mathfrak{p}_{\pi(1)} = \mathfrak{p}$ and hence $x(\pi(1), \pi(d)) = 1/2 \log d$ so that (7.2.7) implies

$$ (7.2.8) \qquad \exp x(\pi(p), \pi(q)) = p^{1/2} d^{1/2} p^{-1/2}. $$

On the other hand since $T_{p^{1/2}} \pi(d) = \pi(q)$ one has $p^{1/2} dp^{1/2} = q$ by (7.2.5). Multiplying on the right by $p^{-1}$ one has $p^{1/2} dp^{-1/2} = qp^{-1}$ and hence $p^{1/2} d^{1/2} p^{-1/2} = (qp^{-1})^{1/2}$. Comparing with (7.2.8) then yields (7.2.6).

Now return to the points $o, r, s \in X$. Without loss [using (7.2.1) and the fact that the partial order is preserved by conjugation] we may assume $\pi(1) = o$. Let $p, q \in P$ be defined so that $\pi(p) = r$ and $\pi(q) = s$. Now by (7.2.6) $\exp x(r, o) = p^{-1/2}$ and $\exp x(o, s) = q^{1/2}$. But then $x(r, o)$ and $x(o, s)$ lie in $\mathfrak{p}$ and hence $x(r, o) + x(o, s) \in \mathfrak{p}$ is real semi-simple. But by (6.3.2) :

$$ \exp(2x(r, o) + 2x(o, s)) \le \exp 2x(o, s) \exp 2x(r, o) = qp^{-1}. $$

Hence

$$ 2(x(r, o) + x(o, s)) \le \log qp^{-1} = 2x(r, s), $$

using (7.2.6). But then $x(r, o) + x(o, s) \le x(r, s)$ [see (7.1.2)].

Q. E. D.

## 8. A Conjugation Convexity Theorem and a Generalization of a Theorem of A. Horn

8.1. Lemma 4.8 may be used also to prove sort of an additive version of Theorem 4.1. That is a theorem in $\mathfrak{g}$ rather than $G$ where the adjoint action of $K$ on $\mathfrak{p}$ replaces multiplication in $G$. The theorem we will
prove will bear a relation to a theorem of A. Horn (see [3], Theorems 4 and 5) that Theorem 5.4 bears to the Horn-Thompson theorem.

To state this result of Horn let $\mathfrak{p}_0$ denote the set of all $n \times n$ complex Hermitian matrices and let $\mathfrak{a}_0 \subseteq \mathfrak{p}_0$ denote the space of all diagonal Hermitian matrices in $\mathfrak{p}_0$. Consider the projection $\Gamma : \mathfrak{p}_0 \to \mathfrak{a}_0$ defined so that if $x \in \mathfrak{p}_0$ then $\Gamma x$ has the same diagonal entries as $x$. That is if $x_{ij}$ is the $i, j$-component of $x$ then $(\Gamma x)_{ij} = \delta_{ij} x_{ij}$.

Now let $x \in \mathfrak{p}_0$ and let $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n$ be the eigenvalues of $x$ and $y \in \mathfrak{a}_0$ be the diagonal matrix whose diagonal entries are the $\lambda_i$. That is $y_{ij} = \delta_{ij} \lambda_i$. Furthermore if $\sigma$ is any permutation of 1, 2, ..., $n$ let $\sigma y \in \mathfrak{a}_0$ be the diagonal matrix obtained by performing the permutation $\sigma$ on the diagonal entries of $y$. Thus $(\sigma y)_{ij} = \delta_{ij} \lambda_{\sigma^{-1}i}$. Next let $\mathfrak{a}_0(y) \subseteq \mathfrak{a}_0$ denote the convex hull of the finite set $\{ \sigma y^l \}$, over all permutations $\sigma$. Then not only does one have $x \in \mathfrak{a}_0(y)$ but in fact, the following is true : Let $O(y)$ denote the set of all Hermitian matrices with the eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$ [so that e. g. $x \in O(y)$], then Horn's theorem states that

$$ (8.1.1) \qquad \Gamma O(y) = \mathfrak{a}_0(y). $$

## 8.2. We will generalize (8.1.1) [more precisely (8.1.1) for the traceless Hermitian matrices. However the addition or subtraction of the scalar matrices is trivially dealt with] to the arbitrary semi-simple case $\mathfrak{g}$. Using the notation we have established so far let $\mathfrak{q}$ be the orthocomplement with respect to the Killing form of $\mathfrak{a}$ in $\mathfrak{p}$ so that

$$ (8.2.1) \qquad \mathfrak{p} = \mathfrak{a} + \mathfrak{q} $$

is an orthogonal direct sum. Let

$$ (8.2.2) \qquad \Gamma : \mathfrak{p} \to \mathfrak{a} $$

be the orthogonal projection of $\mathfrak{p}$ on $\mathfrak{a}$.

Now for any $y \in \mathfrak{p}$ let $O(y) = \text{Ad } K(y) \subseteq \mathfrak{p}$, the Ad K-orbit in $\mathfrak{p}$ defined by $y$.

**THEOREM 8.2.** — For any $y \in \mathfrak{p}$ one has

$$ \Gamma O(y) = \mathfrak{a}(y) $$

where $\mathfrak{a}(y)$ is defined in section 3.1.

*Proof.* — Since any element in $\mathfrak{p}$ is Ad K conjugate to an element in $\mathfrak{a}$ we can assume $y \in \mathfrak{a}$. Now if there exists $z \in O(y)$ such that $\Gamma z \notin \mathfrak{a}(y)$ then by the separation theorem there would exist a linear functional $\nu \in \mathfrak{a}'$ such that for all $\sigma \in W$ :

$$ (8.2.3) \qquad \langle \nu, \Gamma z \rangle = \langle \nu, \sigma y \rangle. $$
Furthermore we can choose $\nu$ so that

$$ (8.2.4) \qquad \langle \nu, x_\nu \rangle \neq 0 \quad \text{for all } \gamma \in \Gamma $$

(see § 3.2) since such linear functionals form a dense set in $\mathfrak{a}'$.

Now assume such a $z$ and, hence $\nu$, exist. But then since $O(y)$ is compact the function $x \mapsto \varphi(x) = \langle \nu, \Gamma x \rangle$ on $O(y)$ has a maximal value. Hence we can choose $z$ so that the maximal value of $\varphi$ is taken at $z$, and (8.2.3) is satisfied. But now if $\nu$ corresponds to the element $w \in \mathfrak{a}$ under the isomorphism $\mathfrak{a} \to \mathfrak{a}'$ defined by the Killing form then

$$ (8.2.5) \qquad (w, x) = \langle \nu, \Gamma x \rangle = \varphi(x) $$

for any $x \in O(y)$ where round brackets denote the Killing form inner product. On the other hand by (8.2.4) one has $\langle \gamma, w \rangle \neq 0$ for all $\gamma \in \Gamma$ so that [using, say (2.2.2)] the

$$ (8.2.6) \qquad \text{centralizer of } w \text{ in } \mathfrak{p} \text{ is } \mathfrak{a}. $$

But now for any $x \in \mathfrak{k}$ the function $\psi$ on $\mathbb{R}$ given by

$$ \psi(t) = (w, \mathrm{Ad}(\exp tx) z) $$

has zero derivative at $t=0$ since $\varphi$ takes its maximal value at $z$. But $\psi'(0) = (w, [x, z]) = ([z, w], x)$ by the invariance of the Killing form. Thus $([z, w], x) = 0$ for all $x \in \mathfrak{k}$. But $[z, w] \in [\mathfrak{p}, \mathfrak{p}] \subseteq \mathfrak{k}$ and the Killing form is non-singular on $\mathfrak{k}$. Thus $[z, w] = 0$ which implies $z \in \mathfrak{a}$ by (8.2.6). But since $z$ is conjugate to $y$ this implies $z = \sigma y$ for some $\sigma \in \mathbb{W}$, by Proposition 2.4. This however contradicts (8.2.3). Thus we have proved

$$ (8.2.7) \qquad \Gamma(O(y)) \subseteq \mathfrak{a}(y). $$

Now let $c = \Gamma(O(y))$. We must show that $\mathfrak{a}(y) \subseteq c$. To prove this, the argument of section 4.5 reducing the problem to the case where $\mathfrak{a}$ is a normal real form of $\mathfrak{a}_c$ also applies here. One needs only to remark that if $q^*$ is the orthogonal complement of $\mathfrak{a}$ in $\mathfrak{p}^*$, with respect to the Killing form on $\mathfrak{a}^*$ using the notation of section 4.5, then $q^* \subseteq q$ (so that the projection $\Gamma^* : \mathfrak{p}^* \to \mathfrak{a}$ is the restriction of $\Gamma$ to $\mathfrak{p}^*$). But this is clear since one easily shows that, using the Killing form, that $q^* = [\mathfrak{k}^*, \mathfrak{a}] \subseteq [\mathfrak{k}, \mathfrak{a}] = \mathfrak{q}$. Thus we may assume $\mathfrak{a}$ is a normal real form of $\mathfrak{a}_c$. Now clearly $y \in c$ since $\Gamma y = y$. Thus by Lemma 4.8 and Remark 4.8 $\mathfrak{a}(y) \subseteq c$ if it can be shown that one has $\{w, \tau(\beta_i) w\} \subseteq c$ for any $w \in c$ and any simple root $\beta_i$. (The reduction to the simple root case was a significant reduction in the group case. Here it is not necessary. We do so now only for notational convenience).
Let $w \in \mathfrak{c}$ and let $\mathfrak{g}_i$ be the real TDS defined as in the proof of Lemma 4.6. Using the notation of the proof of Lemma 4.6 one has

$$ (8.2.8) \qquad \{ w, \tau(\beta_i) w \} = \{ tw_i + w_i^{\perp} \mid -1 \le t \le 1 \}, $$

where $w_i$ and $w_i^{\perp}$ are the components of $w$ in $\mathfrak{a}_i$ and $\mathfrak{a}_i^{\perp}$ [see (4.6.9)]. But now $w = \Gamma(v.y)$ for some $v \in \mathbb{K}$, writing $v.y$ for $\mathrm{Ad} v(y)$. Now let $K_i \subseteq G_i$ be defined as in the proof of Lemma 4.6 and let the map

$$ \sigma: K_i \to \mathfrak{a} $$

be defined by putting $\sigma(k) = \Gamma(kv.y)$, so that

$$ (8.2.9) \qquad \sigma(K_i) \subseteq \mathfrak{c}. $$

Now if $w_i = 0$ then $\{w, \tau(\beta_i)w\}$ reduces to the point $w$ and one trivially has $\{w, \tau(\beta_i)w\} \subseteq \mathfrak{c}$. Hence we may assume $w_i \neq 0$. But then there exists unique continuous functions $\rho: K_i \to \mathbb{R}$ and $\psi: K_i \to \mathfrak{a}_i^{\perp}$ such that

$$ \sigma(k) = \rho(k)w_i + \psi(k). $$

We assert $\psi$ is the constant function given by $\psi(k) = w_i^{\perp}$ for all $k \in K_i$. Since $\psi(1) = w_i^{\perp}$ to prove this it suffices to show that the inner product $(z, \sigma(k))$ is constant as a function of $k$ for any fixed $z \in \mathfrak{a}_i^{\perp}$. But

$$ (z, \sigma(k)) = (z, \Gamma(kv.y)) = (z, kv.y) = (k^{-1}.z, v.y) $$

by the invariance of the Killing form. But $k^{-1}.z = z$ for any $k \in K_i$ since [see (4.6.11)] $[\mathfrak{a}_i^{\perp}, \mathfrak{g}_i] = 0$. Thus $(z, \sigma(k))$ is constant as a function on $K_i$ proving that $\sigma(k) = \rho(k)w_i + w_i^{\perp}$. But $\rho(1) = 1$. On the other hand since there is an element $\tau$ in the Weyl group $W_i$ of $(\mathfrak{a}_i, \mathfrak{g}_i)$ such that $\tau w_i = -w_i$ there exists $k_i \in K_i$ such that $k_i w_i = -w_i$. But $\mathrm{Ad} k_i = 1$ on $\mathfrak{a}_i^{\perp}$. Thus $\mathfrak{a}$ is stable under $\mathrm{Ad} k_i$ and hence $\mathfrak{q}$ is stable under $\mathrm{Ad} k_i$. Hence

$$ \sigma(k_i) = \Gamma(k_i v.y) = k_i \cdot \Gamma(v.y) = -w_i + w_i^{\perp}. $$

Thus $\rho(k_i) = -1$. But then since $K_i$ is connected $\rho(K_i)$ take all values in $[-1, 1]$. Thus $\{w, \tau(\beta_i)w\} \subseteq \sigma(K_i)$ by (8.2.8). But then $\{w, \tau(\beta_i)w\} \subseteq \mathfrak{c}$ by (8.2.9).

Q. E. D.

8.3. Now one can prove a K-conjugation theorem which is analogous to the K-double coset result, Theorem 5.3. For any $x \in \mathfrak{a}$ let

$$ (8.3.1) \qquad \mathfrak{p}_x = \{ y \in \mathfrak{p} \mid y \ge x \} $$

where the order relation is defined by (7.1.1).
THEOREM 8.3. — For any $x \in \mathfrak{a}$ one has

$$ (8.3.2) \qquad \mathfrak{p}_x = \bigcup_{k \in \mathbb{K}} \mathrm{Ad} \, k(x + \mathfrak{q}) $$

where $\mathfrak{q}$ is the orthocomplement of $\mathfrak{a}$ in $\mathfrak{p}$.

*Proof*. — Let $\mathfrak{d}$ denote the right side of (8.3.2) and let $y \in \mathfrak{d}$. Then there exists $k \in \mathbb{K}$ such that $y \in \mathrm{Ad} \, k(x + \mathfrak{q})$. Thus $k^{-1} \cdot y \in x + \mathfrak{q}$ or $\Gamma(k^{-1} \cdot y) = x$. But then $x \in \mathfrak{a}(y)$ by Theorem 8.2 and hence $y \ge x$ or $y \in \mathfrak{p}_x$. Conversely assume $y \in \mathfrak{p}_x$. Thus $x \in \mathfrak{a}(y)$ and hence there exists $k \in \mathbb{K}$ such that $x = \Gamma(k^{-1} \cdot y)$ by Theorem 8.2. That is, $k^{-1} \cdot y \in x + \mathfrak{q}$ or $y \in \mathrm{Ad} \, k(x + \mathfrak{q}) \subseteq \mathfrak{d}$.

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[6] B. KOSTANT, The principal three dimensional subgroups and the Betti numbers of a complex semi-simple Lie group (Amer. J. Math., vol. 81, 1959, p. 973-1032).

[7] B. KOSTANT, Lie group representations on polynomial rings (Amer. J. Math., vol. 86, 1963, p. 327-402).

[8] B. KOSTANT and S. RALLIS, Orbits and representations associated with symmetric spaces (Amer. J. Math., vol. 93, 1971, p. 753-809).

[9] A. LENARD, Generalization of the Golden-Thompson Inequality, Tr e$^{\lambda}$ e$^{\mu}$ tr e$^{\lambda+\mu}$ (Indiana Univ. Math. J., vol. 21, 1971, p. 457-467).

[10] O. Loos, Symmetric Spaces, I. General Theory, Benjamin, 1969.

[11] G. D. Mostow, Some new decomposition theorem for semi-simple Lie groups (Memoirs of the A. M. S., N° 14, 1955).

[12] Classification de Groupes de Lie algebriques, (Séminaire C. Chevalley 1956-1958, vol. I).

[13] M. SUGIURA, Conjugate classes of Cartan subalgebras in real semi-simple Lie algebras (J. Math Soc., vol. 11, 1959, p. 374-434).

[14] C. THOMPSON, Inequalities and partial orders on matrix spaces (Indiana Univ. M th. J., vol. 21, 1971, p. 469-480).

[15] F. WARNER, Foundations of Differential Manifolds and Lie Groups, Scott, Foresman and Company.

[16] A. HORN, On the eigenvalues of a matrix with prescribed singular values, Proceedings of the Amer. Math. Soc., vol. 5, 1954, p. 4-7.

(Manuscrit reçu le 23 avril 1973.)

Bertram KOSTANT,
M. I. T.

Departement of Mathematics 2155,
Cambridge, Mass. 02139,
U. S. A.

ANNALES SCIENTIFIQUES DE L'ÉCOLE NORMALE SUPÉRIEURE