Proof. Parts (i), (ii) and (iii) of the proposition are respectively 1.1.2, 1.5.4 and 1.5.16 of [Reference EngelkingEng78]. To see (iv), note first that M is metrisable and hence paracompact, so we may choose a locally finite refinement $\mathcal {V}$ of $\mathcal {U}$ . Since open sets are $F_\sigma $ in a metrisable space, Theorem 4.1.10 of [Reference EngelkingEng78] gives that $\dim M = \sup _{V \in \mathcal {V}} \dim V$ . Since each $V \in \mathcal {V}$ is a subset of some $U \in \mathcal {U}$ , i gives $\dim V \leq \dim U$ , so $\dim M = \sup _{U \in \mathcal {U}} \dim U$ .

Proof. Let $\varepsilon> 1$ be such that $\varepsilon (a + 1)^{\text {d}(G)} < (a + 1)^{\text {d}(G)} + 1$ , fix a Haar measure m on G with the associated constant $c(\Omega )$ and set $\delta = c(\Omega )\frac {\varepsilon -1}{\varepsilon +1}> 0$ . By formula (2.4) there exists $N \in \mathbb {N}$ such that

(2.5) $$ \begin{align} (c(\Omega) - \delta) n^{\text{d}(G)} \leq m(\Omega^n) \leq (c(\Omega) + \delta) n^{\text{d}(G)} , \qquad \text{ for all } n \geq N . \end{align} $$

Let $n \geq N$ and let $\mathcal {S}$ be the set of subsets $S \subseteq \Omega ^{an}$ such that $g^{-1}g' \notin \Omega ^{2n}$ for all $g,g' \in S$ with $g \neq g'$ . Note that for every $S \in \mathcal {S}$ the sets $\{ g \Omega ^{n} \mid g \in S \}$ are pairwise disjoint. Indeed, if $h \in g \Omega ^{n} \cap g' \Omega ^{n}$ for distinct $g,g' \in S$ then $g^{-1}g' = (h^{-1}g)^{-1}(h^{-1}g') \in \Omega ^{n}\Omega ^{n} = \Omega ^{2n}$ , which is a contradiction.

We order $\mathcal {S}$ with respect to inclusion. Then every totally ordered subset of $\mathcal {S}$ has an upper bound (namely the union), hence $\mathcal {S}$ admits a maximal element S by Zorn’s lemma. We claim that $\Omega ^{an} \subseteq \bigcup _{g \in S} g \Omega ^{2n}$ holds. Otherwise we can find $h \in \Omega ^{an}$ with $h \notin g\Omega ^{2n}$ for any $g \in S$ , which implies that $S \cup \{ h \} \in \mathcal {S}$ . This contradicts the maximality of S, so we conclude that $\Omega ^{an}$ can be covered by translates of $\Omega ^{2n}$ by elements from S.

It remains to bound the size of S. Using the pairwise disjointness of $\{ g\Omega ^n \mid g \in S \}$ and the inclusion $\bigcup _{g \in S} g \Omega ^{n} \subseteq \Omega ^{an}\Omega ^n = \Omega ^{(a+1)n}$ we have that

$$ \begin{align*} |S| m(\Omega^{n}) = m \Big( \bigcup_{g \in S} g \Omega^{n} \Big) \leq m(\Omega^{(a+1)n}) . \end{align*} $$

Applying both bounds in (2.5) gives

$$ \begin{align*} |S| \leq \frac{m(\Omega^{(a+1)n})}{m(\Omega^{n})} \leq \frac{(c(\Omega) + \delta)((a+1)n)^{\text{d}(G)}}{(c(\Omega) - \delta)n^{\text{d}(G)}} = \varepsilon \cdot (a+1)^{\text{d}(G)} < (a+1)^{\text{d}(G)} + 1 , \end{align*} $$

by the choice of $\varepsilon $ . This implies that $|S| \leq (a+1)^{\text {d}(G)}$ .

Proof. Passing to the intersection $K \cap K'$ , we may assume that $K \subseteq K'$ and prove that $S \cap (K'S)^{\circ } \subseteq S \cap (KS)^{\circ }$ . Let $x \in S \cap (K'S)^{\circ }$ . Since $(K'S)^{\circ } \subset K'S$ is open and products of open subsets form a basis of the topology of $K' \times S$ , there are open subsets $U \subseteq K'$ and $V \subseteq S$ such that $x \in UV \subseteq (K'S)^\circ $ . Since $x \in S$ , we find that $e \in U$ and may hence assume without loss of generality that $U \subseteq K$ . Then $x \in UV \subseteq KS$ on the one hand and on the other hand we obtain inclusions where each term is open in the next one $UV \subseteq (K'S)^\circ \subseteq X$ . So $UV \subseteq X$ is open, which shows that $x \in (KS)^\circ $ .

Proof. Since G is second-countable we can multiply K by elements from a descending sequence of compact neighbourhoods of the identity to find a descending sequence $(L_n)_{n \in \mathbb {N}}$ of compact sets such that $\mathring {L_n} \supseteq K$ for all $n \in \mathbb {N}$ and $\bigcap _{n \in \mathbb {N}} L_n = K$ . Suppose for a contradiction that S is not an $L_n$ -slice for any $n \in \mathbb {N}$ . By Remark 3.3 we then find $g_n \in L_n^{-1}L_n \setminus K^{-1}K$ and $x_n,x_n' \in S$ such that $g_nx_n = x_n'$ for all $n \in \mathbb {N}$ . Passing to convergent subsequences, we may assume that $x_n \to x$ , $x_n' \to x'$ with $x,x' \in S$ and

$$ \begin{align*} g_n \to g \in \Big( \bigcap_{n \in \mathbb{N}} L_n^{-1}L_n \Big) \setminus (K^{-1}K)^{\circ} = \partial(K^{-1}K). \end{align*} $$

In particular, $g \neq e$ . Continuity of the action then implies $gx' = x$ , so $g S \cap S \neq \emptyset $ . By Remark 3.3 this is a contradiction to the fact that S is a K-slice.

Proof. We start by proving (3.1) and first show that $\mathring {K}\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S) \subseteq (KS)^{\circ }$ . Let ${x \in \mathring {K}\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)}$ , say $x=gy$ where $g \in \mathring {K}$ and $y \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ . Set $L = K \cap g^{-1}K$ and consider the open set $U = g(LS)^{\circ }$ in X. Then $U \subseteq gLS \subseteq gg^{-1}KS = KS$ , so that $U \subseteq (KS)^{\circ }$ follows. Furthermore, since $L \subseteq K$ , we know that S is also an L-slice and Lemma 3.7 shows that $y \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S) = S \cap (LS)^{\circ }$ . Thus $x = gy \in g(LS)^{\circ } = U \subseteq (KS)^{\circ }$ .

Now we show that $(KS)^{\circ } \subseteq \mathring {K}\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ . Let $x \in (KS)^{\circ }$ , say $x = gy$ with $g \in K$ and $y \in S$ . Then $g^{-1}K$ is a compact identity neighbourhood and S is a $g^{-1}K$ -slice, so by Lemma 3.7 we obtain

$$ \begin{align*} y = g^{-1}x \in S \cap g^{-1}(KS)^{\circ} = S \cap (g^{-1}KS)^{\circ} = \operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S) . \end{align*} $$

It remains to argue that $g \in \mathring {K}$ . For this, we apply Lemma 3.6 to find a compact identity neighbourhood L in G with $K \subseteq \mathring {L}$ such that S is an L-slice. By continuity of the group action, the set

$$ \begin{align*} W = \mathring{L} \cap \{ h \in G \mid hy \in (KS)^{\circ} \} \end{align*} $$

is open in G and it is clear that $g \in W$ . We show that $W \subseteq K$ . Indeed, if $h \in W$ then $hy \in (KS)^{\circ }$ so we can write $hy = h'y'$ with $h' \in K$ and $y' \in S$ . Since $h \in L$ and S is an L-slice, this forces $h = h'$ and $y = y'$ . In particular, we infer that $h \in K$ .

We next prove (3.2). First observe that

$$ \begin{align*} \mathring{K}\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S) = \mathring{K}S \cap K\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S) . \end{align*} $$

Indeed, the inclusion from left to right is obvious. Conversely, if $x = gy \in \mathring {K}S \cap K\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ for uniquely determined elements $g \in K$ and $y \in S$ , it follows from $x \in \mathring {K}S$ that $g \in \mathring {K}$ and from $x \in K\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ that $y \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ . Hence $x \in \mathring {K}\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ .

Applying the equality above and using (3.1), we get

$$ \begin{align*} (\partial_G K)S \cup K(\partial_{\operatorname{\mathrm{tube}}}(S)) & = (K\setminus \mathring{K})S \cup K(S\setminus \operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S)) \\ & = (KS\setminus \mathring{K}S) \cup (KS\setminus K\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S))) \\ & = KS\setminus (\mathring{K}S \cap K\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S))) \\ & = KS \setminus \mathring{K}\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(S) \\ & = KS \setminus (KS)^{\circ} \\ & = \partial_X (KS) . \end{align*} $$

Let us next show (3.3). First we show that $B \subseteq \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(\overline {B})$ . Using that $B \subseteq \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ and (3.1) we obtain $\mathring {K}B \subseteq \mathring {K}\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S) = (KS)^{\circ }$ . Since $\mathring {K}B$ is open in $KS$ and hence in $(KS)^{\circ }$ , it is also open in X. But then $B \subseteq \overline {B} \cap (K\overline {B})^{\circ } = \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(\overline {B})$ .

Now we show that $\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(\overline {B})\subseteq B$ . Let $x \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(\overline {B})$ . Since $(K\overline {B})^{\circ }$ is open in $KS$ and $x \in (K\overline {B})^{\circ }$ , we can by definition of tubes find open sets $W \subseteq K$ and $V \subseteq S$ such that $x \in WV \subseteq (K\overline {B})^{\circ }$ . The inclusion $WV \subseteq K\overline {B}$ then forces $V \subseteq \overline {B}$ , and since $x \in \overline {B} \subseteq S$ we get $x \in V$ . Since $x \in V \subseteq \overline {B}$ and V is open in S, this means that x lies in the interior of $\overline {B}$ inside S. By assumption the latter equals B, so $x \in B$ .

To show (3.4), we use (3.3) and obtain

$$ \begin{align*} \partial_{\operatorname{\mathrm{tube}}}(\overline{B}) = \overline{B}\setminus\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(\overline{B}) = \overline{B}\setminus B = \partial_S{B} . \end{align*} $$

Let us now show (3.5). Since $\overline {B}$ is a compact subset of S, it is a K-slice. Hence (3.2) together with (3.4) gives that

$$ \begin{align*} \partial(K\overline{B}) = (\partial_G K)\overline{B} \cup K(\partial_{\operatorname{\mathrm{tube}}}(B)) = (\partial_G K)\overline{B} \cup K(\partial_S B) . \end{align*} $$

Moreover, the definition of boundary together with (3.1) and (3.4) gives that

$$ \begin{align*} \partial(K\overline{B}) = K\overline{B} \setminus (K\overline{B})^{\circ} = K\overline{B} \setminus K\operatorname{\mathrm{int}}_{\operatorname{\mathrm{tube}}}(\overline{B}) = K\overline{B} \setminus KB . \end{align*} $$

This finishes the proof.

Proof. Since $KS \cap S'$ is compact and S is Hausdorff, it suffices to show that the map is injective. Let $g_1,g_2 \in K$ and $x_1,x_2 \in S$ such that $g_1x_1 , g_2x_2 \in S'$ . Suppose that the images of $g_1x_1$ and $g_2x_2$ under the projection $KS \to S$ are equal, that is, $x_1 = x_2$ . We then have that $g_1^{-1}(g_1x_1) = g_2^{-1}(g_2x_2) \in KS'$ and since $S'$ is a K-slice, it follows that $g_1^{-1} = g_2^{-1}$ . This implies injectivity.

Proof. Replacing K by $K \cup K^{-1}$ , we may assume that K is symmetric. Take any symmetric, relatively compact identity neighbourhood $V \subseteq G$ and define $U = VK \cup KV$ . Then U is symmetric and since K is symmetric it follows that

$$ \begin{align*} \bigcap_{k \in K} k U \supseteq \bigcap_{k \in K} k \left ( \bigcup_{k' \in K} k' V \right) \supseteq V . \end{align*} $$

Since V is relatively compact, also U is relatively compact.

Proof. Consider the linear map $T \colon {\text {C}}_{\text {c}}(A, \mathbb {R}) \to V$ given by $Tf = \int _U f(g^{-1}x) g v \, \text {d} g$ . Since V is finite dimensional and $\mathop {\mathrm {im}} T$ is a vector subspace of V, it suffices to show that for all convex neighbourhoods C of v we have $C \cap \mathop {\mathrm {im}} T \neq \emptyset $ .

Fix a convex neighbourhood C of v as above and let $U_0 \subseteq G$ be an identity neighbourhood satisfying $U_0 v \subseteq C$ . Since $\overline {U}x \subseteq X$ is closed, there is a neighbourhood $B \subseteq A$ of x satisfying $B \cap \overline {U}x \subseteq U_0x$ . By freeness of $G \stackrel {}{\curvearrowright } X$ , this implies that for every $g \in U$ , the statement $g x \in B$ implies $g \in U_0$ . Let $f \in {\text {C}}_{\text {c}}(A, \mathbb {R})$ satisfy $0 \leq f \leq 1$ , $f(x) \neq 0$ , $\int _U f(g^{-1}x) \, \text {d} g = 1$ and $\mathop {\mathrm {supp}} f \subseteq B$ . Then by choice of B, the map $g \mapsto \mathbb{1}_U(g) f( g^{-1} x)$ is a probability density on G supported in $U_0$ . Since $U_0v \subseteq C$ , this implies that $\int _U f(g^{-1}x) gv \, \text {d} g \in C$ .

Proof. By Lemma 3.11 there is a symmetric, relatively compact identity neighbourhood U in G such that $U_0 = \bigcap _{k \in K} Uk$ is an identity neighbourhood. Since G acts freely, we have $x_0 \notin (K \overline {U} \setminus U_0)x_0$ . Hence,

$$ \begin{align*} \bigcap_{\substack{A \text{ compact}\\ \text{neighbourhood of }x_0}} A \cap (K \overline{U} \setminus U_0)A = \emptyset . \end{align*} $$

By compactness, there is a neighbourhood A of $x_0$ such that for any $g \in K \overline {U} \setminus U_0$ and every $x \in A$ we have $gx \notin A$ . Fix such an A.

By Lemma 3.12, there is $f \in {\text {C}}_{\text {c}}(A, \mathbb {R})$ supported in A such that $\int _U f(g^{-1}x_0) gv \, \text {d} g = v$ . Put

$$ \begin{align*} F(x) = \int_U f(g^{-1}x) gv \, \text{d} g . \end{align*} $$

For $k \in G$ , $x \in X$ , left-invariance of the Haar measure shows that

$$ \begin{align*} F(kx) & = \int_U f(g^{-1}kx) gv \, \text{d} g \\ & = \int_{kU} f(g^{-1}x) kgv \, \text{d} g \\ & = k F(x) + \int_{kU \setminus U} f(g^{-1}x) kgv \, \text{d} g - \int_{U \setminus kU} f(g^{-1}x) kgv \, \text{d} g . \end{align*} $$

Let us simplify the last expression, assuming that $x \in A$ holds. If $g^{-1}x \in \mathop {\mathrm {supp}} f \subseteq A$ and $g \in kU \setminus U$ , then $x \in (kU \setminus U)A \subseteq (K\overline {U} \setminus U_0)A = \emptyset $ , which is absurd. Similarly, $g^{-1}x \in \mathop {\mathrm {supp}} f$ together with $g \in U \setminus kU$ implies $x \in (U \setminus kU)A \subseteq (K\overline {U} \setminus U_0)A = \emptyset $ . We infer that $F(kx) = kF(x)$ for all $k \in K$ and all $x \in A$ .

Proof. Fix an embedding with closed image $G \subseteq \text {GL}(n, \mathbb {R})$ and consider the vector ${v = 1 \in {\text {M}}_n(\mathbb {R}) = V}$ . Let $K \subseteq G$ be a compact identity neighbourhood. By Lemma 3.13, there is a neighbourhood $A \subseteq X$ of x and a continuous map $F\colon X \to V$ such that $F(x) = v$ and $F(ky) = kF(y)$ for all $k \in K$ and $y \in A$ . After shrinking it, we may assume that A is compact. Since $G \subseteq \text {GL}(n, \mathbb {R})$ is closed, its action on V is proper, so that [Reference PalaisPal61, Theorem 2.3.2] shows the existence of a G-slice $S_0$ at v, that is a subset $S_0 \subseteq V$ with $v \in S_0$ and such that $G \times S_0 \longrightarrow V$ , $(g, w) \mapsto gw$ , is a homeomorphism onto its open image. Let C be a compact neighbourhood of v contained in the locally compact space $GS_0$ . Let D be the slice projection of C onto $S_0$ , which is a compact set. Put $S = F^{-1}(D) \cap A$ . Then S is compact, since D is closed in X. Further, $x \in S$ , since $F(x) = v$ . We show that S is a K-slice and then that $x \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ .

Let $(k_1,x_1), (k_2,x_2) \in K \times S$ . If $k_1x_1 = k_2x_2$ , then

$$ \begin{align*} k_1F(x_1) = F(k_1x_1) = F(k_2x_2) = k_2F(x_2) , \end{align*} $$

implying that $k_1 = k_2$ , since $S_0$ is a G-slice. So $x_1 = x_2$ follows, proving injectivity of the map $K \times S \to KS$ .

Now fix a neighbourhood $B \subseteq A$ of x and a symmetric identity neighbourhood L such that $L^2 \subseteq K$ and $L B \subseteq A$ . We claim that the neighbourhood $F^{-1}(L^\circ D) \cap B$ of x is contained in $KS$ , which will prove that $x \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ . Take $y \in F^{-1}(L^\circ D) \cap B$ . There is $k \in L$ satisfying $kF(y) \in C$ . Since $k \in K$ and $y \in A$ , we find that $F(ky) = k F(y) \in D$ , that is $ky \in F^{-1}(D)$ . As we also have $ky \in LB \subseteq A$ , we infer that $ky \in S$ . Using the fact that L is symmetric, we now conclude that $y \in L^{-1}S \subseteq KS$ , which finishes the proof.

Proof. By [Reference LosertLos01, Theorem 2] G contains a maximal compact normal subgroup H. The quotient group $G/H$ is then a Lie group of polynomial growth containing no nontrivial compact normal subgroups, hence by [Reference LosertLos20, Corollary 3.6] $G/H$ is a matrix Lie group. Denote by $\pi \colon G \to G/H$ and $p \colon X \to X/H$ the quotient maps.

Let K be an identity neighbourhood in G and let $x \in X$ . We may assume that ${K = HK = KH}$ . Since $G/H$ acts freely on $X/H$ , we can apply Theorem 3.14 to get a $\pi (K)$ -slice $S' \subseteq X/H$ such that $p(x) \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S')$ . Set $T = p^{-1}(S') \subseteq X$ . Then T is H-invariant, so we can consider the action $H \curvearrowright T$ . Since H is compact there exists by [Reference MostowMos57, Theorem 2.1] an H-slice $S \subseteq T$ such that $x \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ , where the interior is taken in the H-space T.

We claim that S is a K-slice. To see this, let $g,g' \in K$ and $x,x' \in S$ satisfy $gx=g'x'$ . Then $(gH)[x] = (g'H)[x']$ , so since $gH,g'H \in \pi (K)$ and $[x],[x'] \in p(S) \subseteq p(p^{-1}(S')) = S'$ , we get that $gH = g'H$ and $[x]=[x']$ . Let $h_1,h_2 \in H$ be such that $g' = gh_1$ and $x' = h_2x$ . Then $gx = gh_1h_2x$ , so by freeness of the action $h_1h_2 = e$ . But $x' = h_2x$ also implies that $h_2=e$ since $h_2 \in H$ and $x,x' \in S$ , so we arrive also at $h_1 = e$ . We conclude that $g=g'$ . It remains to argue that $x \in \operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ inside X, that is $x \in (KS)^\circ $ . Since $K = HK$ , we have $p^{-1}(p(KS)) = KS$ . So it suffices to show that $p(x) \in p(KS)^\circ $ holds. Since S is an H slice at x in the H-space T, there is a subset $x \in U \subseteq HS$ which is open in T. As p is a quotient map, it follows that $p(U)$ is open in $p(T) = S'$ . Set $\tilde {U} = p(U) \cap (\pi (K)S')^{\circ }$ . Then $\tilde {U}$ is open in $X/H$ and $p(x) \in \tilde {U}$ . Furthermore $\tilde {U} \subseteq \pi (K)^\circ p(U) \subseteq \pi (K) p(HS) = p(KS)$ which finishes the proof.

Proof. First observe that $\tau $ is well-defined, as for $g,h \in K$ and $x \in S$ the computation

$$ \begin{align*} hx & = s(gh^{-1},hx) , \\ gx & = gh^{-1}hx = r(gh^{-1},hx) , \end{align*} $$

shows that $(gh^{-1},hx) \in \mathcal {G}|_{KS}$ .

We start by showing i. In order to show that $\tau $ is a groupoid homomorphism, we compute for $(g,h,x),(h,k,x) \in \mathcal {H}$ that

$$ \begin{align*} \tau((g,h,x),(h,k,x)) = \tau(g,k,x) = (gk^{-1},kx) = (gh^{-1},hx)(hk^{-1},kx) = \tau(g,h,x)\tau(h,k,x) . \end{align*} $$

Its defining formula shows that $\tau $ is continuous. To show injectivity, suppose that for a pair of elements $(g,h,x),(g',h',x') \in \mathcal {H}$ we have $\tau (g,h,x) = \tau (g',h',x')$ . In different terms, we suppose $(gh^{-1}, hx) = g'(h')^{-1}, h'x')$ . Then $hx=h'x'$ implies $h=h'$ and $x=x'$ by the definition of a slice, so that subsequently $gh^{-1} = g'h^{\prime -1} = g'h^{-1}$ forces $g=g'$ . This shows that $\tau $ is injective.

We next prove (ii). If $(g,h,x) \in \operatorname {\mathrm {Pair}}(K) \times S$ , its image under $\tau $ belongs to the claimed set, since $\mathrm {proj}(hx) = x = \mathrm {proj}(gx) = \mathrm {proj}(gh^{-1}hx)$ . Conversely, given an element $(g',x') \in \mathcal {G}|_{KS}$ that satisfies $\mathrm {proj}(x') = \mathrm {proj}(g'x')$ , we write $x' = hx$ and $g'x' = gx$ for certain $h,g \in K$ and $x \in S$ . Then it follows that $g'hx = gx$ so that freeness of the action $G \stackrel {}{\curvearrowright } X$ implies that $g' = gh^{-1}$ . Hence we obtain that $(g',x') = (gh^{-1},hx) = \tau (g,h,x)$ lies in the image of $\tau $ . Using equation (3.1) of Lemma 3.7, saying that $(KS)^\circ = K^{\circ }\operatorname {\mathrm {int}}_{\operatorname {\mathrm {tube}}}(S)$ , we infer that also $\tau (\mathcal {H}^{\mathrm {open}}) = \{ (g,x) \in \mathcal {G}|_{(KS)^\circ } : \mathrm {proj}(x) = \mathrm {proj}(gx) \}$ holds.

Let us finish by proving iii. Since $\mathrm {proj}$ is continuous, the description of $\tau (\mathcal {H}^{\mathrm {open}})$ obtained in (ii) shows that $\tau (\mathcal {H}^{\mathrm {open}})$ is closed in $\mathcal {G}|_{(KS)^{\circ }}$ . Further, since $\tau $ is a continuous injection from a compact space to a Hausdorff space, it is a homeomorphism onto its image. In particular $\tau (\mathcal {H}^{\mathrm {open}})$ is open in $\mathcal {G}|_{KS}$ since $\mathcal {H}^{\mathrm {open}}$ is open in $\mathcal {H}$ . To conclude, we use the fact that $\mathcal {G}|_{(KS)^{\circ }}$ is an open subset of $\mathcal {G}|_{KS}$ , so that it follows that $\tau (\mathcal {H}^{\mathrm {open}})$ is open in $\mathcal {G}|_{(KS)^{\circ }}$ .

Proof. Write $B = KS$ , denote by $\pi \colon B \to S$ the projection on the slice S and consider the map $\psi \colon L \to S$ given by $l \mapsto \pi (lx)$ . We have to show that the image of $\psi $ is a singleton. Since $\psi $ is continuous and L is connected its image $\psi (L)$ is connected too. Assume that $s \in \psi (L)$ . Since $Lx \subseteq B^\circ $ , (3.1) implies that there is $g \in L$ and $k \in K^\circ $ such that ${gx = ks}$ . Let U be a symmetric identity neighbourhood in G such that $Uk \subseteq K$ and ${Ug \subseteq L}$ . Then $ugx = uks \in Ks$ for all $u \in U$ , so that $\psi ^{-1}(\{s\})$ has non-empty interior. So $L = \bigsqcup _{s \in S} \psi ^{-1}(\{s\})$ is a partition of L into sets with non-empty interior. Since G is second-countable, this implies that the image of $\psi $ is countable. So $\psi (L)$ is a countable, connected Hausdorff space, and hence a singleton.

Proof. Let $a \in {\text {C}}_{\text {c}}(G \times X)$ and $f \in {\text {C}}_{\text {b}}(X)$ . Then we have