Proposition 3

Let $G$ be a locally compact group. Let $H$ be a closed normal subgroup of $G$, isomorphic to $\mathbf{R}^n$ and such that $G/H$ is compact.

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).