(i) G is relatively compact in End(E); (ii) G is relatively compact in Aut(E); (iii) G leaves invariant a nondegenerate$^1$ positive hermitian form on E.

(iii) $\Rightarrow$ (i): Suppose that $G$ leaves invariant a nondegenerate positive hermitian form $\Psi$. Let $(e_1, \ldots, e_n)$ be an orthonormal basis for $\Psi$ (Alg., Ch. IX, § 6, No. 1, Cor. 1 of Th. 1). For every $u \in G$, let $(u_{ij})$ be its matrix with respect to $(e_i)$. For any $j$, we have $\sum_{i=1}^n |u_{ij}|^2 = 1$, thus $|u_{ij}| \leq 1$ for all $i$ and $j$, which proves (i).

(i) $\Rightarrow$ (ii): This follows from GT, X, § 3, No. 5, Cor. of Th. 4, taking into account the fact that the topology of $\mathrm{End}(E)$ is that of compact convergence.

(ii) $\Rightarrow$ (iii): Suppose that the closure $\overline{G}$ of $G$ in $\mathrm{Aut}(E)$ is compact. Let $\Phi$ be a nondegenerate positive hermitian form on $E$. If the field of scalars is $\mathbf{R}$ or $\mathbf{C}$, the giving of $\Phi$ makes $E$ a finite-dimensional Hilbert space, and condition (iii) will result from the following lemma:

$^1$ Non dégénérée; in EVT, the term is replaced by séparante, subsequently translated as "separating" (TVS, V, § 1, No. 1).

Let $\alpha$ be a Haar measure on K. For any $x, y$ in F, the mapping $s \mapsto (U(s)x|U(s)y)$ is continuous. Set

It is immediate that $\varphi(x, y)$ is a sesquilinear form on F. Since the set of endomorphisms $U(s)$ is compact in $\mathscr{L}_s(F; F)$, there exists a constant M such that $\|U(s)\| \leq M$ for all $s \in K$. For every $x \in F$, we therefore have

whence the inequalities

which shows that $\varphi$ is positive and nondegenerate, and that the norm $\varphi(x, x)^{1/2}$ is equivalent to the norm $\|x\|$. Finally, for all $t \in K$,

When the field of scalars is $\mathbf{H}$, one argues exactly as before on replacing everywhere the function $s \mapsto (U(s)x|U(s)y)$ by the function $s \mapsto \Phi(sx, sy)$ defined on G, with values in $\mathbf{H}$. This completes the proof of the proposition.

One reduces immediately to the case that $n = 1$. Since the additive group $\mathbf{R}$ is isomorphic to the multiplicative group $\mathbf{R}_+^*$, the proposition is then an immediate consequence of Prop. 7 of §2, No. 4.

a) There exists a continuous mapping f of X into V such that $f(xv) = f(x) + v$ for all $x \in X$ and $v \in V$.

b) If f is a mapping satisfying the conditions of a), then the mapping $x \mapsto (\pi(x), f(x))$ is a homeomorphism of X onto $B \times V$.

The assertion a) results from Prop. 2 in which g is taken to be the identity mapping of V. Let f be a mapping satisfying the conditions of a). The mapping $x \mapsto x \cdot (-f(x))$ of X into X is continuous, and is constant on each orbit, hence is of the form $\varphi \circ \pi$, where $\varphi$ is a continuous mapping of B into X; for every $b \in B$, $\pi(\varphi(b)) = b$. The mappings $x \mapsto (\pi(x), f(x))$ of X into $B \times V$ and $(b, v) \mapsto \varphi(b) \cdot v$ of $B \times V$ into X are inverse to each other, because $\varphi(\pi(x)) \cdot f(x) = x \cdot (-f(x)) \cdot (f(x)) = x$, $\pi(\varphi(b) \cdot v) = \pi(\varphi(b)) = b$, and, if $b = \pi(y)$, then

Since these mappings are continuous, they are homeomorphisms.

is invariant under $G$.

For, for every $t \in G$,

a) There exists a closed subgroup $L$ of $G$ such that $G$ is the topological semi-direct product of $L$ and $H$.

b) If $M$ is a compact subgroup of $G$, there exists an element $x \in H$ such that $x^{-1}Mx \subset L$.

c) Every compact subgroup of $G$ is contained in a maximal compact subgroup.

d) The maximal compact subgroups of $G$ are the subgroups that are the transforms of $L$ by the inner automorphisms of $G$.

Let $\pi$ be the canonical homomorphism of $G$ onto $K = G/H$. By passage to the quotient, the mapping $(s, h) \mapsto shs^{-1}$ of $G \times H$ into $H$ defines a continuous mapping $(\sigma, h) \mapsto \sigma \cdot h$ of $K \times H$ into $H$ such that $shs^{-1} = \pi(s) \cdot h$. We shall identify $H$ with $\mathbf{R}^n$ (and will therefore employ, as the case may be, either the multiplicative or the additive notation for the group law in $H$). By the Cor. of Prop. 2, there exists a continuous mapping $f$ of $G$ into $H$ such that $f(xh) = f(x) + h$ for $x \in G, h \in H$. For every $x \in G$, let $p(x) = x \cdot (-f(x))$, which depends only on the coset of $x$ with respect to $H$. Set

We see that if $F(x, y) = 0$ for all $x, y$ in $G$, then $p(G) = L$ is a subgroup of $G$ that intersects each coset of $H$ in one and only one point. Since $p$ is continuous, $G$ is then the topological semi-direct product of $L$ and $H$ (GT, III, §2, No. 10).

Now, for any $h, h' \in H$,

Therefore $F$ defines, by passage to quotients, a continuous mapping $\varphi$ of $K \times K$ into $H$.

On the other hand, for all $x, y, z$ in $G$, we have

therefore, for all $x', y', z'$ in $K$,

Let us integrate with respect to $z'$ by means of the normalized Haar measure $\alpha$ of $K$. Setting $\psi(x') = \int \varphi(z', x')\, d\alpha(z')$, $\psi$ is a continuous function on K, and (on observing that the operations of K in $\mathbf{R}^n$ respect the vector space structure of $\mathbf{R}^n$ by GT, VII, §2, No. 1, Prop. 1), one obtains

In other words, setting $k - \psi \circ \pi$, which is a continuous function on G,

(2)

Comparing (1) and (2), one sees that if $f$ is replaced by the continuous function $f + k$ (which leaves verified the property $f(xh) = f(x) + h$), F is replaced by 0 and, as we saw earlier, this completes the proof of a).

For every $g \in G$, let $l_g$ (resp. $h_g$) be the unique element of L (resp. H) such that $g = h_g l_g$. If $h_1 \in H$ and $g \in G$, then

thus $h_g h_1 = h_g + l_g h_1 l_g^{-1}$. For every $g \in G$, let $\psi_g$ be the mapping of H into itself defined by

One sees that the mapping $(g, h_1) \mapsto \psi_g(h_1)$ of $G \times H$ into H is continuous and makes H a homogeneous space for G, in which the stabilizer of the origin is L. We observe moreover that when H is identified with $\mathbf{R}^n$, $\psi_g$ is an affine mapping of H into itself. This said, let M be a compact subgroup of G; by Lemma 2, there exists an $x \in H$ such that $\psi_m(x) = x$ for all $m \in M$. For $y \in H$, $\psi_y$ is the translation with vector $y$; it follows that for every $m \in M$, $\psi_{x^{-1}} \circ \psi_m \circ \psi_x$ transforms the origin of H into itself, therefore $x^{-1}mx \in L$. This proves that $x^{-1}Mx \subset L$, whence b).

Let $L'$ be a closed subgroup of G containing L. Then $L'$ is the topological semi-direct product of L and $L' \cap H$. If $L'$ is compact, then $L' \cap H$ is compact hence reduces to a point (GT, IV, §2, No. 2, Cor. 1 of Th. 2), therefore $L' = L$. This proves that L is a maximal compact subgroup of G; the same is therefore true of the subgroups that are the transforms of L by the inner automorphisms of G. The assertions c) and d) of Prop. 3 are then immediate consequences of b).

Let $L = G \times \mathbf{R}$ and let M be the set of $(\xi, -u(\xi))$, where $\xi$ runs over H. It is clear that M is a closed normal subgroup of L. Let $L' = L/M$ and let $\pi$ be the canonical mapping of L onto $L'$. The subgroup of L generated by M and R is H × R, hence is closed; therefore π(R) is a closed subgroup N of L'. The restriction ρ of π to R is a bijective continuous representation of R onto N. Lemma 2 of Appendix 1 proves that ρ is bicontinuous. Moreover, L'/N is isomorphic to L/(H × R) = G/H, hence is compact. By Prop. 3, and taking into account the fact that N is in the center of L', L' is the product of N with another subgroup. Therefore there exists a continuous representation of L' onto N that reduces on N to the identity mapping. Therefore there exists a continuous representation v of L onto R that is trivial on M and reduces on R to the identity mapping. For ξ ∈ H, one has v((ξ, 0)) = v((ξ, −u(ξ))(e, u(ξ))) = u(ξ), which completes the proof.

Let C be a compact set such that G = CH. Enlarging C if necessary, we can suppose that C generates G and that G = ĈH. Then C^2 is compact and is covered by the Ĉs (s ∈ H), which are open. Therefore there exist s_1, ..., s_n in H such that C^2 ⊂ Ĉs_1 ∪ ... ∪ Ĉs_n. Let Γ be the subgroup of H generated by the s_i. Then C^2 ⊂ CΓ. By induction, it follows that C^n ⊂ CΓ for every n, therefore G = CΓ. Every element of H may be put in the form ab with a ∈ C, b ∈ Γ, whence a ∈ H, whence a ∈ C ∩ H. Therefore H is generated by C ∩ H and the s_i, that is, by a compact set.

For, let d ∈ D. The image of G under the continuous mapping x ↦ xdx^{-1} is a connected subset of D, hence reduces to {d}, which proves that xd = dx for all x ∈ G.

The group G is locally isomorphic to K (GT, III, §2, No. 6, Prop. 19), hence is locally compact; since it is connected, it is generated by a compact neighborhood of e. By Lemmas 3 and 4, D is a finitely generated abelian group, hence is isomorphic to a group $\mathbf{Z}^r \times D_1$ with D_1 finite (A, VII, §4, No. 7, Th. 3). Suppose that r > 0. Then there exists a representation f of D onto $\mathbf{Z}$. By Prop. 4, f may be extended to a continuous representation g of G in R. By passage to quotients, g defines a continuous representation $g'$ of $K$ in $\mathbf{R}/\mathbf{Z}$; since $\mathbf{R}/\mathbf{Z}$ is abelian, the kernel of $g'$ contains the commutator subgroup of $K$, therefore $g'$ is trivial; in other words, $g(G) \subset \mathbf{Z}$. Since $G$ is connected, it follows that $g(G) = \{0\}$, which is absurd since $f(D) = \mathbf{Z}$. Thus $r = 0$ and $D$ is finite. Consequently $G$ is compact (GT, III, §4, No. 1, Cor. 2 of Prop. 2).

Such a mapping is obviously a continuous representation of $\mathbf{GL}(n, K)$ in $\mathbf{C}^*$. Conversely, suppose that $\psi$ is a continuous representation of $\mathbf{GL}(n, K)$ in $\mathbf{C}^*$. For $x \in K^*$, set

and $\chi(x) = \psi(\widetilde{x})$. Then, for every matrix $X \in \mathbf{GL}(n, K)$, we have $(\det X^{-1}) \sim \cdot X \in \mathbf{SL}(n, K)$. Since $\mathbf{SL}(n, K)$ is the commutator subgroup of $\mathbf{GL}(n, K)$ (A, III, §8, No. 9, Cor. of Prop. 17), $\psi((\det X^{-1}) \sim \cdot X) = 1$, whence

This established, Cor. 1 of Prop. 10 of §1, No. 8 proves that the relatively invariant measures on $\mathbf{GL}(n, K)$ are, up to a constant factor, the measures of the form where $\chi$ is a continuous representation of $K^*$ in $\mathbf{C}^*$.

Example 2. — Affine group. For every $X \in \mathbf{GL}(n, K)$ and every $x \in K^n$, let $(X, x)$ be the affine linear mapping $\xi \mapsto X\xi + x$ in $K^n$. The set of $(X, x)$ is the affine group $G$ of $K^n$ (A, II, §9, No. 4). The set $T$ of translations is a closed normal subgroup of $G$, canonically isomorphic to $K^n$; on the other hand, $\mathbf{GL}(n, K)$ is a closed subgroup of $G$, and $G$ is the semi-direct product of $\mathbf{GL}(n, K)$ and $T = K^n$. One equips $G$ with the (locally compact) topology for which $G$ is the topological semi-direct product of $\mathbf{GL}(n, K)$ and $T$ (GT, III, §2, No. 10). One has

On the other hand, if $X \in \mathbf{GL}(n, K)$ and $x \in T$ then, for every $\xi \in K^n$, therefore the automorphism $(1, x) \mapsto (X, 0)(1, x)(X, 0)^{-1}$ of $T$ has modulus $\mod(\det X)$ (§1, No. 10, Prop. 15). In view of Example 1 and §2, No. 9, Remark, the measure is a left Haar measure on $G$. On the other hand, by Prop. 14 of §2, No. 9, or

Thus, a right Haar measure on $G$ is given by

Example 3. — Strict triangular group.

Let $[1, n]$ be the set of integers $m$ such that $1 \leq m \leq n$. Let $J$ be a subset of $[1, n] \times [1, n]$ satisfying the following conditions: 1) if $(i, j) \in J$ then $i < j$; 2) if $(i, j) \notin J$ then, for every integer $k$ such that $i < k < j$, at least one of the two pairs $(i, k)$ and $(k, j)$ does not belong to $J$.

Let $T_J$ be the set of matrices $Z = (z_{ij})_{1 \leq i \leq n, 1 \leq j \leq n}$ with elements in $K$, such that $z_{ii} = 1$, and $z_{ij} = 0$ if $i \neq j$ and $(i, j) \notin J$. This is a closed subset of $\mathbf{GL}(n, K)$. The mapping $Z \mapsto (z_{ij})_{(i, j) \in J}$ is a homeomorphism of $T_J$ onto $K^s$ (where $s$ denotes the number of elements of $J$). If $Z' = (z'_{ij}) \in T_J$, then $Z'Z = (z''_{ij})$ with

whence $Z'Z \in T_J$. If $T_J$ is identified with $K^s$, then the mapping $Z \mapsto Z'Z$ (for fixed $Z'$) is identified with an affine mapping, and its determinant is 1, as one sees by ordering the pairs $(i, j) \in J$ lexicographically and applying the following lemma:

**Lemma 6. — Let $L$ be a totally ordered finite set. For every $\lambda \in L$, let $V_\lambda$ be a free module of finite dimension over a commutative ring $k$; for $\lambda, \mu$ in $L$ such that $\lambda \leq \mu$, let $f_{\lambda \mu} \in \mathrm{Hom}_k(V_\mu, V_\lambda)$. Then the linear mapping

from $\prod_{\lambda \in L} V_\lambda$ into $\prod_{\lambda \in L} V_\lambda$, has determinant $\prod_{\lambda \in L} \det f_{\lambda \lambda}$.

One reduces immediately to the case that $L$ is an interval of integers, and the lemma then follows from A, III, §8, No. 6, formula (31).

If $Z \in T_J$, one then sees that there exists $Z' \in T_J$ such that $Z'Z = I_n$, whence $Z' = Z^{-1}$. Thus, $T_J$ is a closed subgroup of $\mathbf{GL}(n, K)$. On the other hand, Prop. 15 of §1, No. 10 shows that the measure

is a left Haar measure on $T_J$. By calculating $ZZ'$ one sees in the same way that this measure is a right Haar measure on $T_J$.

There is an analogous result if, in the definition of $T_J$, the roles of rows and columns are interchanged.

When $J$ is the set of pairs $(i, j)$ such that $i < j$, the group $T_J$ is called the upper strict triangular group of order $n$ over $K$, and is denoted $T_1(n, K)$. Its transpose is called the lower strict triangular group.