Topological Groups

Locally convex spaces

A locally convex space $E$ is a topological vector space (over $\R$ or $\C$) such that

the mappings $(x,y)\mapsto x+y$ and $(\l,x)\mapsto\l x$ are continuous,
the origin has a neighborhood base ${\cal U}$ consisting of convex and balanced sets, i.e. for all $U\in{\cal U}$, all $x,y\in U$, all $t\in[0,1]$ and all $|\l|\leq1$: $$ (1-t)x+ty\in U\quad \mbox{and}\quad \l x\in U~. $$

A linear functional $x^*$ on $E$ is continuous if and only if there exists $U\in{\cal U}$ such that for all $x\in U$: $|x^*(x)|\leq1$.

Suppose $E$ is a locally convex space with dual $E^*$, i.e. $x^*\in E^*$ if and only if $x^*$ is a continuous linear functional on $E$. For $x^*\in E^*$ and $x\in E$ we will use the notation $\la x,x^*\ra$ to denote the bi-linear map $E\times E^*$: $(x,x^*)\mapsto x^*(x)$. The weak topology $\s(E,E^*)$ on $E$ is the coarsest topology on $E$ making all mappings $x\rar\la x,x^*\ra$, $x^*\in E^*$, continuous. Similarly the weak * topology $\s(E^*,E)$ on $E^*$ is the coarsest topology on $E^*$ making all mappings $x^*\rar\la x,x^*\ra$, $x\in E$, continuous.

Probably the most important reason for the definition of locally convex spaces is the celebrated Hahn-Banach Theorem; one of it`s versions states that given $U\in{\cal U}$ and $x\notin U$, there is some $x^*\in E^*$ such that $\la x,x^*\ra=1$ and for all $u\in U$: $|\la u,x^*\ra| < 1$ - we say $x^*$ separates $x$ and $U$. This in particular implies that a convex and balanced subset $C$ of $E$ is closed if and only if it is weakly closed. If $C$ is a closed, convex and balanced subset of $E$, then it`s polar $$ C^\perp\colon=\{x^*\in E^*:\forall x\in C:\ |\la x,x^*|\leq1\} $$ is a weakly * closed, convex and balanced subset of $E^*$ and the Bi-Polar Theorem states that $$ C=C^{\perp\perp}\colon=\{x\in E:\forall x^*\in C^\perp:\ |\la x,x^*|\leq1\}~. $$ Finally the polar of any neighborhood $U$ of $0$ in $E$ is weakly * compact - that`s called the Banach-Alaoglu Theorem. In case $E$ is separable $U^\perp$ is also metrizable.

The locally convex space $C_c(G)$

In the sequel we make use of the notation $K\comp G$ meaning that $K$ is a compact subset of $G$. So let $K\comp G$ be any compact subset of the locally compact group $G$ and denote by $C_0(K)$ the space of all continuous functions, whose support is contained in $K$. Then $\norm f_{K}\colon=\sup\{|f(x)|:x\in K\}$ is a norm on $C_0(K)$ and it`s complete. The finest locally convex topology on $C_c(G)$, such that all inclusions $C_0(K)\rar C_c(G)$ are continuous, is the topology we are considering. If $G$ is separable and metrizable, then there exists a strictly increasing sequence $K_n\sbe G$ such that every compact subset of $G$ is contained in some set $K_n$; it follows that the space $C_c(G)$ is the strict inductive limit of the sequence $C_0(K_n)$. A basis for the neighborhood filter of zero is obtained as follows: Take any sequence $U_n$ of convex and balanced neighborhoods of zero in $C_0(K_n)$ and put $$ U\colon=\convex{\bigcup j_n(U_n)} $$ where $j_n:C_0(K_n)\rar C_c(G)$ is the canonical inclusion. These sets form a base for the neighborhood filter of zero. As a consequence a subset $M$ of $C_c(G)$ is bounded (compact) iff there is some $n$ such that $M$ is actually a bounded (compact) subset of $C_0(K_n)$. Moreover, a linear mapping $A:C_c(G)\rar E$ is continuous iff its restriction $C_0(K_n)\rar E$ is continuous for all $n$; in particular a linear functional $x^*$ on $C_c(G)$ is continuous iff for all $n\in\N$ there is some constant $C_n$ such that for all $f\in C_0(K_n)$: $|x^*(f)|\leq C_n\norm{f}_K$. By the Riesz representation theorem we can identify the dual of $C_c(G)$ with the space of all complex Radon measures on $G$, i.e. all complex Borel measures $\mu$ such that $\mu(K)$ is finite for all compact subsets of $G$: every continuous functional $x^*$ on $C_c(G)$ is of the form: $$ \la f,x^*\ra=x^*(f)=\int f\,d\mu $$ for some complex Radon measure $\mu$, i.e. $C_c(X)^*=M(G)$ - the space of all complex Radon measures on $G$, and thus we just write $\la f,\mu\ra$ for $\int f\,d\mu$ and use the same notation for $f\in L_1(\mu)$, in particular for the indicator $I_A$ of a Borel set $A$ satisfying $\mu(A) < \infty$: $$ \la I_A,\mu\ra=\int I_A\,d\mu=\mu(A)~. $$ $M_b(G)$ denotes the set of bounded complex Radon measures, i.e. $|\mu|(G) < \infty$ and $M_1(G)$ the set of all probability measures, i.e. $\mu(G)=1$.

Haar measure

Haar measure on compact metric groups

In case of a finite group $G$ the normalized counting measure $\mu$ has the following property: for all subsets $A$ of $G$ and all $x\in G$: $\mu(xA)=\mu(A)=\mu(Ax)$ - we say that $\mu$ is left- and right-invariant. In this section we are going to prove, that on any compact group $G$ there exists a unique left-invariant probability measure on the Borel sets of $G$ - and it will turn out that this measure is also right-invariant - cf. proposition. To start with, let $G$ be locally compact and $\mu$ a left-invariant Radon measure, i.e. for all Borel sets $A$ and all $x\in G$: $\mu(xA)=\mu(A)$. In other words the image measure of $\mu$ under the map $y\mapsto x^{-1}y$ coincides with $\mu$; by the transformation theorem this means that $$ \forall x\in G\, \forall f\in C_c(G):\quad \int f(x^{-1}y)\,\mu(dy) =\int L_xf\,d\mu =\int f\,d\mu =\la f,\mu\ra, $$ where $L_x:C_c(G)\rar C_c(G)$ denotes the linear operator $L_xf(y)\colon=f(x^{-1}y)$. The adjoint $L_x^*:M(G)\rar M(G)$ is thus given by $$ \forall A\comp G:\quad L_x^*\mu(A) =\la I_A,L_x^*\mu\ra =\la L_xI_A,\mu\ra =\int I_{xA}\,d\mu=\mu(xA), $$ i.e. $L_x^*\mu(A)=\mu(L_x(A))$. Thus a Radon measure $\mu$ is left-invariant, iff for all $x\in G$: $L_x^*\mu=\mu$. Similarly $\mu$ is right-invariant if for all $x\in G$: $R_x^*\mu=\mu$, where $R_xf(y)\colon=f(yx^{-1})$. Finally we define $I:C_c(G)\rar C_c(G)$ by $If(x)=f(x^{-1})$. It follows that $$ I^2=1,\quad IR_x=L_xI,\quad L_xR_y=R_yL_x,\quad R_xR_y=R_{yx},\quad L_xL_y=L_{yx}~. $$ A Haar measure on $G$ is a left-invariant Radon measure $\mu$ on $G$ such that for all Borel sets $A$: $\mu(A)\geq0$. For the time beeing we establish the existance of a Haar measures on compact metrizable groups:

Let $G$ be a compact metrizable group. Then there exists a Haar probability measure on $G$.

$\proof$ For $\e > 0$ let $N(\e)$ be a minimal $\e$-net of $G$ with respect to some left-invariant metric $d$. For $f\in C(G)$ define the measure $\mu_\e$ by $$ \la f,\mu_\e\ra =\int f\,d\mu_\e \colon=\frac1{|N(\e)|}\sum_{t\in N(\e)}f(t)~. $$ Since the unit ball of $M(G)=C(G)^*$ is weakly * compact and metrizable, we infer that there exists a null-sequence $\e_j$ and a measure $\mu$, such that for all $f\in C(G)$: $\la f,\mu_{\e_j}\ra\to\la f,\mu\ra$ - obviuously $\mu(G)=1$. Now suppose that $M(\e)$ is a further minimal $\e$-net; put for $t\in N(\e)$: $$ \Phi(t):=\{t^\prime\in M(\e):\,B(t,\e)\cap B(t^\prime,\e)\neq\emptyset\}~. $$ We claim that for all $K\sbe N(\e)$: $|\Phi(K)|\geq|K|$. If not, then $L:=\Phi(K)\cup(N(\e)\sm K)$ would be an $\e$-net: indeed, for any $x\in G$ we can find some $t\in N(\e)$ such that $d(x,t) < \e$; suppose $t\in K$, then for all $t^\prime\in M(\e)\sm\Phi(K)$: $$ d(x,t^\prime)\geq d(t,t^\prime)-d(t,x) > 2\e-\e=\e~. $$ Hence there must be some $t^\prime\in\Phi(K)$ such that $d(x,t^\prime) < \e$, i.e. $L$ is an $\e$-net, which contradicts the minimality of $N(\e)$. By the marriage theorem there is a bijection $\phi:N(\e)\to M(\e)$, such that for all $t\in N(\e)$: $$ \phi(t)\in\Phi(t) \quad\mbox{i.e.}\quad d(t,\phi(t)) < 2\e $$ and therefore, denoting by $\o_f$ the modulus of continuity of $f$: $$ |\la f,\mu_\e\ra-\la f,\mu_\e^\prime\ra| \leq\frac1{|N(\e)|}\sum|f(t)-f(\phi(t)|\leq\omega_f(2\e)~. $$ Thus $\mu$ doesn`t depend on the particular minimal $\e$-net $N(\e)$. Finally, let $y$ be any point in $G$, then $M(\e):=\{yt:t\in N(\e)\}$ is by left-invariance of the metric a minimal $\e$-net and thus \begin{eqnarray*} \int f(yx)\,d\mu(x) &=& \lim_n\frac1{|N(\e_n)|}\sum_{t\in N(\e_n)}f(yt)\\ &=&\lim_n\frac1{|M(\e_n)|}\sum_{t\in M(\e_n)} f(t)=\int f\,d\mu~. \end{eqnarray*} $\eofproof$

Let $G$ be a compact metrizable group. Then there exists a bi-invariant metric on $G$ defining the same topology.

$\proof$ By theorem there is a left-invariant metric $d$ and by theorem there is a left-invariant probability measure $\mu$ on $G$. Now put $$ \r(x,y)\colon=\int d(xg,yg)\,\mu(dg)~. $$ Then $\r$ is clearly left-invariant; but it`s also right-invariant, because for all $h\in G$ we get for $f(g)\colon=d(xg,yg)$: $$ \r(xh,yh) =\int d(xhg,yhg)\,\mu(dg) =\int f(hg)\,\mu(dg) =\int f(g)\,\mu(dg) =\int d(xg,yg)\,\mu(dg)~. $$ Clearly $\r\leq d$. On the other hand if $d(x,y) > r$ there is a neighborhood $U$ of $e$ such that for all $u\in U$; $d(xu,yu) > r/2$. Hence $\r(x,y) > r\mu(U)/2$. $\eofproof$

Haar measure on compact groups

Actually the proof of theorem gives something more: Let $G$ be a group acting isometrically on a compact metric space $M$, i.e. for all $x,y\in M$ and all $g\in G$: $d(gx,gy)=d(x,y)$. Then there exists a $G$-invariant Borel probability measure $\mu$ on $M$. We can even generalize theorem to arbitrary compact groups $G$: Let ${\cal U}$ be the neighborhood filter of $e$ in $G$. As $G$ is compact we can find for every $U\in{\cal U}$ a smallest $n\in\N$ as well as $x_1,\ldots,x_n\in G$, such that $\bigcup x_jU=G$. We will call $N\colon=\{x_1,\ldots,x_n\}$ a minimal $U$-net. Suppose $M$ is another minimal $U$-net, then its cardinality is $n$.

Use the marriage theorem to prove that if $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$ are two minimal $U$-nets, then by renumbering we can achieve that for all $m\leq n$: $x_mU\cap y_mU\neq\emptyset$.

Define $\Phi:N\rar{\cal P}(M)$ by $$ \Phi(x)\colon=\{y\in M: xU\cap yU\neq\emptyset\}~. $$ Suppose there is some subset $K\sbe N(U)$ such that $|\Phi(K)| < |K|$. We claim that in this case $L\colon=\Phi(K)\cup(N(U)\sm K)$ is a $U$-net: For any $z\in G$ there is some $x\in N$ such that $x^{-1}z\in U$. If $x\notin K$, then $x\in L$. So assume $x\in K$; we need to show that there is some $y\in\Phi(K)$ such that $y^{-1}x\in U$. Assume there is some $y\in M\sm\Phi(K)$ such that $y^{-1}z\in U$, then $yU\cap xU=\emptyset$ and $$ y^{-1}x=y^{-1}zz^{-1}x\in UU^{-1} $$ i.e. for some $u,v\in U$: $y^{-1}x=uv^{-1}$ or $xv=yu$, which shows that $yU\cap xU\neq\emptyset$. This finishes the proof that $L$ is indeed a $U$-net, but this in turn contradicts the minimality. Hence by the marriage theorem there is a bijection $\phi:N\to M$, such that for all $x\in N$: $$ \phi(x)\in\Phi(x) \quad\mbox{i.e.}\quad xU\cap\phi(x)U\neq\emptyset~. $$ Consequently for all $x\in N$: $x^{-1}\phi(x)\in UU^{-1}$ and therefore $|f(x)-f(\phi(x))|\leq\o_f(UU^{-1})$, where $\o_f(U)\colon=\sup\{|f(x)-f(y)|:x^{-1}y\in U\}$. So we just need to replace the $\e$-nets by the $U$-nets to get the proof of theorem work for arbitrary compact groups!
More generally we have the following theorem for locally compact groups.

Let $G$ be a locally compact group. Then there exists a non trivial Haar measure on $G$.

Uniqueness of the Haar measure on locally compact groups

So let us turn to uniqueness. Clearly, if $\mu\neq0$ is a Haar measure and $c>0$, then $c\mu$ is also a Haar measure. We will prove, that any Haar measure is of this form - we say $\mu$ is essentially unique. At first we prove, that $\mu$ is strictly positive, i.e. for all open neighborhoods $U$ of the identity element $e$ we have $\mu(U)>0$: Assume to the contrary that $\mu(U)=0$, the for all $x\in G$: $\mu(xU)=0$; it follows that for all compact subsets $K$: $\mu(K)=0$, i.e. $\mu=0$. Consequently a continuous function, which vanishes $\mu$ a.e. vanishes identically!

Suppose $\mu$ is a Haar measure on the locally compact group $G$. Then for all $f\in L_p(G,\mu)$, all $1\leq p < \infty$, the mapping $G\rar L_p(\mu)$, $x\mapsto L_xf$ is uniformly continuous. If moreover $f\in C_c(G)$, then $x\mapsto R_xf$ is continuous form $G$ into $L_p(G)$.

$\proof$ Since $C_c(G)$ is dense in $L_p(G,\mu)$, we can find for every $\e > 0$ some $g\in C_c(G)$, $\supp(g)=K$, such that $\norm{f-g}_p < \e/3$. By uniform continuity of $g$ we infer that there is some neighborhood $U$ of $e$ in $G$, such that $$ \sup_{x\in U}\norm{g-L_xg}_\infty < \tfrac13(\mu(K))^{-1/p}\e~. $$ Up to that point the same argument applies to $R_xg$. But now we use left-invariance and conclude that $$ \norm{f-L_xf}_p \leq\norm{f-g}_p+\norm{g-L_xg}_p+\norm{L_x(g-f)}_p < \e~. $$ Finally again by left-invariance: $\norm{L_xf-L_yf}_p=\tnorm{f-L_{x^{-1}y}f}_p$ and the conclusion follows for $L_x$. $\eofproof$

Suppose $\mu$ and $\mu^\prime$ are left- and right-invariant measures respectively. Then there exists some number $c > 0$ such that: $I^*\mu^\prime=c\mu$, where $If(x)\colon=f(x^{-1})$.

$\proof$ Suppose $f,g\in C_c(G)$ and $\la f,\mu\ra\neq0$. Then we conclude by Fubini: \begin{eqnarray*} \la f,\mu\ra\la g,\mu^\prime\ra &=&\int f(x)\int g(yx)\,d\mu^\prime(y)\,d\mu(x) =\iint f(x)g(yx)\,d\mu(x)\,d\mu^\prime(y)\\ &=&\iint f(y^{-1}x)g(x)\,d\mu(x)\,d\mu^\prime(y) =\int\la L_xIf,\mu^\prime\ra g(x)\,\mu(dx) =\la f,\mu\ra\la gG_f,\mu\ra \end{eqnarray*} where $G_f(x)\colon=\la f,\mu\ra^{-1}\la L_xIf,\mu^\prime\ra$. Therefore $\la g,\mu^\prime\ra=\la gG_f,\mu\ra$ i.e. the density of $\mu^\prime$ with respect to $\mu$ is $G_f$. Hence the function $G_f$ does not depend on $f$, i.e. for any pair $f,h\in C_c(G)$: $G_f=G_h$ $\mu$-a.e. By lemma $G_f$ is continuous and since $\mu$ is strictly positive: $G_f(x)=G_h(x)$ for all $x\in G$. In particular for $x=e$: $$ \frac{\la If,\mu^\prime\ra}{\la f,\mu\ra}= \frac{\la Ih,\mu^\prime\ra}{\la h,\mu\ra}=c, \quad\mbox{i.e.}\quad I^*\mu^\prime=c\mu~. $$ $\eofproof$

Obviously, a Radon measure $\mu$ is left-invariant iff $I^*\mu$ is right-invariant; the previous proposition thus proves uniqueness of the Haar measure up to a multiplicative constant.

The modular function

Suppose $\mu$ is a Haar measure, then for all $x,y\in G$: $$ L_y^*(R_x^*\mu) =(R_xL_y)^*\mu =(L_yR_x)^*\mu =R_x^*(L_y^*\mu) =R_x^*\mu, $$ i.e. $R_x^*\mu$ is left-invariant. By uniqueness there is some strictly positive number $\D_G(x)$, such that $\D_G(x)R_x^*\mu=\mu$. $\D_G$ is said to be the modular function of the group $G$.

The modular function $\D_G$ is a continuous homomorphism from $G$ into the multiplicative group $\R^+$; thus \begin{equation}\label{haaeq1}\tag{HAA1} \forall f\in C_c(G)\qquad\int f(yx)\,\mu(dy)=\D_G(x)\int f\,d\mu~. \end{equation} Moreover, we have: $I^*\mu=\D_G\mu$, i.e. \begin{equation}\label{haaeq2}\tag{HAA2} \forall f\in C_c(G)\qquad\int If\,d\mu=\int f\D_G\,d\mu~. \end{equation}

$\proof$ Choose $f\in C_c(G)$ such that $\la f,\mu\ra\neq0$, then by definition of $\D$: $$ \frac1{\D(x)} =\frac{\la f,R_x^*\mu\ra}{\la f,\mu\ra}= \frac{\la R_xf,\mu\ra}{\la f,\mu\ra} $$ proving the continuity of $\D$. Moreover $$ \frac{\la f,\mu\ra}{\D(xy)} =\la f,R_{xy}^*\mu\ra =\la f,R_y^*R_x^*\mu\ra =\frac{\la f,R_y^*\mu\ra}{\D(x)} =\frac{\la f,\mu\ra}{\D(x)\D(y)}, $$ which shows that $\D:G\rar(\R^+,.)$ is a homomorphism. Finally for any $f\in C_c(G)$: $$ \la f,R_x^*(\D\mu)\ra =\la R_xf,\D\mu\ra =\D(x)\la R_x(f\D),\mu\ra =\la f,\D\mu\ra, $$ i.e. the measure $\D\mu$ is right-invariant; by proposition there is some number $a > 0$, such that $I^*\mu=a\D\mu$. Since $I^2=1$ we get: $$ \mu =I^*(I^*\mu) =I^*(a\D\mu) =\frac a{\D}\,I^*\mu =a^2\mu \quad\mbox{i.e.}\quad a=1~. $$ $\eofproof$

A locally compact group $G$ is said to be unimodular if for all $x\in G$: $\D_G(x)=1$.

Locally compact abelian groups and compact groups are unimodular. In particular on every compact group the unique Haar probability measure $\mu$ is also right-invariant, i.e. for all $f\in L_1(G,\mu)$ and all $y\in G$ we have $$ \int f(yx)\,\mu(dx)=\int f(xy)\,\mu(dx)=\int f(x^{-1})\,\mu(dx)=\int f(x)\,\mu(dx)~. $$

$\proof$ If $G$ is compact, then $\mu(G) < \infty$ and we may decree $\mu(G)=1$: this however makes $\mu$ unique. Putting $f=1$ in proposition, it follows that for all $x\in G$: $\D_G(x)=1$. $\eofproof$

Let us consider the special case $G=\Gl(n,\R)$: denote by $\l$ the Lebesgue measure on $\Ma(n,\R)=\R^{n^2}$. We claim that $\mu(dX)=|\det X|^{-n}\,\l(dX)$ is a Haar measure on $\Gl(n,\R)$ and that $\Gl(n,\R)$ is unimodular: So let $L_X:\Gl(n,\R)\to\Gl(n,\R)$ be the translation $T\mapsto XT$ and let $A$ be an open bounded subset of $\Gl(n,\R)$; since $L_X$ is linear and its determinante is $(\det X)^n$, we get by the change of variable formula: $$ \l(L_X(A))=|\det X|^n\,\l(A)~. $$ Thus it follows that \begin{eqnarray*} \mu(XA) &=&\int_{XA}|\det Y|^{-n}\,\l(dY) =\int_{A}|\det DL_X(Y)||\det(XY)|^{-n}\,\l(dY)\\ &=&\int_{A}|\det Y|^{-n}\,\l(dY) =\mu(A)~. \end{eqnarray*} proving left-invariance of $\mu$. A similar argument also shows that $\mu$ is right-invariant, hence unimodular.

Haar measure on symmetric spaces

There are several definitions of a symmetric space, cf. e.g. wikipedia. We will use the following: Suppose $G$ is a locally compact group and $H$ a closed sub-group. Then the space $G/H$ is said to be a symmetric space. Of course, if $H$ is normal, then $G/H$ is again a locally compact group. A Radon measure $\s$ on $G/H$ is said to be a Haar measure, if for all Borel sets $A$ and all $x\in G$: $\s(A)\geq0$ and $\s(xA)=\s(A)$.

Let $E,F$ be locally convex and $u:E\rar F$ linear, continuous, open and onto. Then $\im u^*$ is $\s(E^*,E)$-closed and: $(\ker u)^\perp=\im u^*$. If both spaces are metrizable and complete, then $u$ is open if $u$ is onto.

Let $X$ be locally compact, $Y$ Hausdorff and $f:X\rar Y$ continuous, open and onto. Then for every compact subset $K$ of $Y$ there is a compact subset $C$ of $X$, such that $K=f(C)$.

$\proof$ Let $V_\a$, $\a\in I$, be an open, relatively compact cover of $X$. Since $f$ is open and onto, $f(V_\a)$, $\a\in I$, is an open cover of $K$ and thus there is a finite subcover $f(V_1),\ldots,f(V_n)$. The set $C\colon=f^{-1}(K)\cap\bigcup\cl{V_j}$ is closed and therefore a compact subset of $X$ satisfying $f(C)=K$. $\eofproof$

If $H$ is a closed subgroup of the Hausdorff group $G$, then by proposition $G/H$ is Hausdorff and the quotient map $G\rar G/H$, $x\mapsto[x]$, is open.

Suppose $H$ is a closed subgroup of a locally compact group $G$ and $\nu$ a Haar measure on $H$. Then the mapping $$ \pi:f\mapsto\left([x]\mapsto\int_Hf(xy)\,d\nu(y)\right) $$ is a continuous linear open mapping from $C_c(G)$ onto $C_c(G/H)$.

$\proof$ Since $\nu$ is left-invariant on $H$ we have for all $h\in H$: $\int_Hf(xhy)\,d\nu(y)=\int_Hf(xy)\,d\nu(y)$. Hence $\pi(f)$ is well defined and continuous on $G/H$. Moreover if $x\notin\supp(f)H$, i.e. $[x]\notin[\supp(f)]$, then $\pi(f)([x])=0$.
$\pi:C_c(G)\rar C_c(G/H)$ is onto: For any $f^\prime\in C_c(G/H)$ put $K^\prime=\supp(f^\prime)$; by lemma there is a compact subset $K$ of $G$ such that $[K]=K^\prime$. Now choose another compact subset $C$ in $H$ such that $e\in C$ and $\nu(C) > 0$, then $[KC]=K^\prime$. Finally choose $\vp\in C_c^+(G)$ in such a way that $\vp|KC=1$ and $\vp\leq1$. It follows that $$ \forall x\in K:\quad \pi\vp([x])=\int_H\vp(xy)\,d\nu(y) \geq\int_C\,d\nu=\nu(C) > 0 $$ and thus for all $[x]\in K^\prime$: $\pi\vp([x]) > \nu(C)$. Defining a function $f$ on $G$ by $$ f(x)=\left\{\begin{array}{cl} \vp(x)\frac{f^\prime([x])}{\pi\vp([x])}& \mbox{if $[x]\in K^\prime$}\\ 0&\mbox{otherwise} \end{array}\right. $$ we get a continuous function $f$ on $G$ with $\supp(f)=\supp(\vp)\cap KH$. Since for all $y\in H$: $f^\prime([xy])=f^\prime([x])$ and $\pi\vp([xy])=\pi\vp([x])$ we conclude that for all $x\in K$ and $y\in H$: $|f(xy)|\leq|f(x)|$ and for all $x\in G$: $$ \pi f([x]) =\int_H\vp(xy)\frac{f^\prime([xy])}{\pi\vp([xy])}\,\nu(dy) =f^\prime([x])~. $$ Finally for each compact subset $D$ of $G$ \begin{eqnarray*} \sup\{|f(x)|:x\in D\} &=&\sup\{|f(x)|:x\in KH\cap D\}=\sup\{|f(x)|:x\in K\cap DH\}\\ &\leq&\frac1{\nu(C)}\sup\{|f^\prime([x])|:[x]\in[K\cap DH]\} \leq\frac1{\nu(C)}\sup\{|f^\prime([x])|:[x]\in[D]\} \end{eqnarray*} which proves that $\Pi:C_c(G/H)\rar C_c(G)$, $f^\prime\mapsto f$ is continuous at $0$. Suppose $U$ is a neighborhood of $0$ in $C_c(G)$, then choose a zero neighborhood $V$ in $C_c(G/H)$ such that $\Pi(V)\sbe U$. Then $\pi(U)\spe\pi(\Pi(V))=V$, i.e. $\pi$ is open. $\eofproof$

Suppose $H$ is a closed subgroup of a locally compact group $G$ and $\nu$ a Haar measure on $H$. If $\s$ is a Haar measure on $G/H$, then $\pi^*\s$ is a Haar measure on $G$. Conversely, if $G$ and $H$ are unimodular and $\mu$ and $\nu$ are Haar measures on $G$ and $H$ respectively, then there is exactly one Haar measure $\s$ on $G/H$, such that: $\pi^*\s=\mu$, i.e. for all $f\in C_c(G)$: $$ \int f\,d\mu=\int_H\int_{G/H}f(zy)\,\s(dz)\,\nu(dy)~. $$

$\proof$ 1. For $f\in C_c(G)$ we have \begin{eqnarray*} \la f,L_x^*\pi^*\s\ra &=&\la\pi L_xf,\s\ra= \int_{G/H}\int_H f(x^{-1}zy)\,\nu(dy)\,\s(dz)\\ &=&\int_H\int_{G/H}f(x^{-1}zy)\,\s(dz)\,\nu(dy) =\int_H\int_{G/H}f(zy)\,\s(dz)\,\nu(dy)\\ &=&\la\pi f,\s\ra=\la f,\pi^*\s\ra~. \end{eqnarray*} Hence $\pi^*\s$ is a Haar measure on $G$.
2. For the reverse conclusion let $\mu$ and $\nu$ be Haar measures on $G$ and $H$, respectively. We will show that $\ker\pi\sbe\mu^\perp$, i.e. $\pi f=0$ implies: $\la f,\mu\ra=0$. So let $g\in C_c(G)$, then we get by $\pi f=0$, unimodularity and Fubini: \begin{eqnarray*} 0&=&\int_G g(x)\int_H f(xy)\,d\nu(y)\,d\mu(x) =\int_H\int_G g(x)f(xy)\,d\mu(x)\,d\nu(y)\\ &=&\int_H\int_G g(xy^{-1})f(x)\,d\mu(x)\,d\nu(y) =\int_G f(x)\int_H g(xy^{-1})\,d\nu(y)\,d\mu(x)\\ &=&\int_G f(x)\int_H g(xy)\,d\nu(y)\,d\mu(x) =\int_G f(x)\pi(g)(x)\,d\mu(x) =\la f\pi(g),\mu\ra~. \end{eqnarray*} Now choose $g$ such that for all $x\in\supp(f)$: $\pi(g)([x])=1$, then: $0=\la f,\mu\ra$.
By lemma, lemma and the Bi-Polar Theorem in locally convex spaces we infer from $\ker\pi\sbe\mu^\perp$ that $$ \im\pi^*=(\ker\pi)^\perp\spe\mu^{\perp\perp}=\C\mu~, $$ which readily implies that there exists some $\s\in M(G/H)$, such that $\pi^*\s=\mu$.
Finally, from $[x[y]]=[xy]$ we infer that: $L_x\pi=\pi L_x$, and thus: $$ \la L_x\pi f,\s\ra =\la\pi L_xf,\s\ra =\la L_xf,\pi^*\s\ra =\la f,\pi^*\s\ra =\la\pi f,\s\ra~. $$ $\eofproof$

Let us look at the special orthogonal group $G=\SO(n)$. For a fixed point $x_0\in S^{n-1}$ the group $H\colon=\{U\in G: Ux_0=x_0\}$ is a sub-group of $G$ and $F:G\rar S^{n-1}$, $F(U)=Ux_0$, is continuous, onto and $F(U)=F(V)$ iff $V\in UH$. Thus there is a unique map $\wh F:G/H\rar S^{n-1}$ such that $F(U)=\wh F([U])$; moreover $\wh F$ is a continuous bijection. As $G/H$ is compact, it`s a homeomorphism. For any $U\in G$ and $[V]\in G/H$ we have $$ \wh F(U[V])=\wh F([UV])=F(UV)=UVx_0=U\wh F([V]) $$ and thus for any subset $A\sbe G/H$: $\wh F(UA)=U\wh F(A)$. Identifying $G/H$ and $S^{n-1}$ via $\wh F$, we see that $G$ operates on $S^{n-1}$ by sending a point $x\in S^{n-1}$ to $Ux$. Choosing $x_0=e_n$, $H$ itself can be identified with the group $\SO(n-1)$. The formula in theorem then reads as $$ \forall f\in C(\SO(n)):\quad \int_{\SO(n)} f\,d\mu=\int_{\SO(n-1)}\int_{S^{n-1}}f(U(x))\,\s(dx)\,\nu(dU) $$ where $(U(x_1e_1+\cdots+x_ne_n)\colon=U(x_1e_1+\cdots+x_{n-1}e_{n-1})+x_ne_n$.
Starting with this operation of $G$ on $G/H$, we see that $H$ is the stabilizer of the point $x_0$. We generalize to arbitrary compact groups $G$ and compact spaces $X$, on which $G$ operates transitively, i.e. there is a continuous map $G\times X\rar X$, $(g,x)\mapsto gx$ such that for all $x\in X$ and all $g,h\in G$: $$ ex=x,\quad g(hx)=(gh)x,\quad Gx=X, $$ then $X$ is always homeomorphic to $G/H$ for the stabilizer $H\colon=\{g\in G: gx_0=x_0\}$ of any point $x_0\in X$. The above theorem states that there is a unique probability measure $\s$ on $X$ such that for all $g\in G$ and all Borel subsets $A$ of $X$: $\s(gA)=\s(A)$ - $\s$ is called the normalized Haar measure on $X$. In the case $G=\SO(n)$ and $X=S^{n-1}$ this probability measure $\s$ is given by $$ \s(A) \colon=\frac{\l((0,1]A)}{\l((0,1]S^{n-1})} =\lim_{r\uar1}\frac{\l((r,1]A)}{\l((r,1]S^{n-1})}, $$ where $\l$ denotes Lebesgue measure on $\R^n$ and $(r,1]A\colon=\{tx:r < t\leq1,x\in A\}$. In this case $\s$ is also called the normalized surface measure on $S^{n-1}$.

Sampling uniformly distributed points

The normalized surface measure on $S^{n-1}$ can be obtained by generating a random gaussian vector in $\R^n$ and normalize it.
For e.g. the special orthogonal group $\SO(n)$ we generate a random gaussian matrix in $\Ma(n,\R)$ with independent and identically distributed entries and perform more or less Gram-Schmidt. In e.g. sage use the numerics library NumPy:

dim=4
Gauss=numpy.random.normal(0,1,size=(dim,dim))
U,R=numpy.linalg.qr(Gauss)
if(numpy.linalg.det(U) < 0):
for k in range(dim):
U[1][k] *= -1

Sample uniformly distributed points on $\SU(n)$.

Let $G$ be a locally compact, separable group and $X$ a Baire-space. If $G$ operates continuously and transitively on $X$, then for all $x\in X$ the mapping $\Phi_x:G/S_x\rar X$ is a homeomorphism.

Let $Q_x:G\rar G/S_x$ be the quotient mapping. As $\Phi_x(Q_xg)=gx$ is continuous, $\Phi_x$ is continuous. Moreover, by transitivity $\Phi_x$ is a bijection. It remains to prove that for each neighborhood $V$ of $e$ in $G$ the set $Vx$ is a neighborhood of $x$ in $X$. So let $U$ be a compact, symmetric neighborhood of $e$ such that $U^2\sbe V$ and let $g_n$ be dense in $G$. Then we have $$ X=\Phi_x\Big(\bigcup g_nU\Big)=\bigcup_n g_nUx~. $$ Since $X$ is a Baire-space, we can find some $n\in\N$ and an open subset $W$ of $X$, such that $W\sbe g_nUx$; in particular there is some $u\in U$, such that $x\in u^{-1}g_n^{-1}W$, i.e.: $x\in u^{-1}Ux\sbe Vx$.