Invariant Probabilities on Compact Group Orbits

From Haar Probability to an Orbit Probability

A compact group can move a point through a compact orbit, and normalized Haar measure records how much group mass maps into each part of that orbit. Pushing Haar measure forward gives a probability that is invariant under the action. More strongly, when the action is transitive on the orbit, this is the only invariant probability there. The measure is attached to the orbit: a space with several orbits can have several different invariant probabilities.

Let \(G\) be a compact Hausdorff topological group, let \(X\) be a compact Hausdorff space, and suppose \(G\) acts continuously on \(X\). Thus \(g\cdot x\in X\), \(e\cdot x=x\), and \((gh)\cdot x=g\cdot(h\cdot x)\). Fix \(x_0\in X\), and let \(O=G\cdot x_0=\{g\cdot x_0:g\in G\}\) be its orbit. The orbit map \(\pi:G\to O\), \(\pi(g)=g\cdot x_0\), is continuous, and \(O\) is compact. We use the normalized Haar probability \(m_G\) on \(G\), and consider regular Borel probabilities on \(O\). For compact metric spaces, Borel probabilities are regular automatically.

Definition: The pushforward of \(m_G\) under \(\pi\) is the Borel probability \(\nu=\pi_*m_G\) on \(O\), defined by \(\nu(B)=m_G(\pi^{-1}(B))\) for every Borel set \(B\subseteq O\). A Borel probability \(\lambda\) on \(O\) is \(G\)-invariant if \((T_h)_*\lambda=\lambda\) for every \(h\in G\), where \(T_h(y)=h\cdot y\). Equivalently, \(\lambda(T_h^{-1}(B))=\lambda(B)\) for every Borel \(B\subseteq O\).

The pushforward is a probability because \(\pi^{-1}(O)=G\), so \(\nu(O)=m_G(G)=1\). Its integration rule is \(\int_O\phi\,d\nu=\int_G\phi(\pi(g))\,dm_G(g)\) for bounded Borel \(\phi\). This follows first for indicator functions from the definition, then for simple functions by finite additivity, and then for bounded Borel functions by approximation with simple functions.

Existence and Uniqueness on a Transitive Orbit

Theorem: With \(G,X,O,\pi\), and \(m_G\) as above, \(\nu=\pi_*m_G\) is a \(G\)-invariant regular Borel probability on \(O\). It is the unique \(G\)-invariant regular Borel probability on \(O\).

Proof. First we prove invariance. For \(h\in G\) and Borel \(B\subseteq O\),

\nu(T_h^{-1}(B)) =m_G\{g\in G:h\cdot(g\cdot x_0)\in B\} =m_G\{g\in G:hg\in\pi^{-1}(B)\} =m_G(\pi^{-1}(B)) =\nu(B).

The third equality follows from left invariance of Haar measure: the set in the preceding term is \(h^{-1}\pi^{-1}(B)\). Thus \(\nu\) is invariant. It is regular: the pushforward of a Radon probability under a continuous map between compact Hausdorff spaces is a Radon measure.

Now let \(\lambda\) be any invariant regular Borel probability on \(O\), and let \(\phi:O\to\mathbb{R}\) be continuous. The function \(F(g,y)=\phi(g\cdot y)\) is bounded and continuous on the compact product \(G\times O\). The product probability \(m_G\times\lambda\) is finite, so Fubini's theorem applies to \(F\). For each \(g\), invariance and the pushforward integration rule give \(\int_O\phi(g\cdot y)\,d\lambda(y)=\int_O\phi(y)\,d\lambda(y)\). Therefore, integrating first in \(y\),

\int_O\phi\,d\lambda =\int_G\int_O\phi(g\cdot y)\,d\lambda(y)\,dm_G(g) =\int_O\int_G\phi(g\cdot y)\,dm_G(g)\,d\lambda(y).

For every \(y\in O\), transitivity gives \(y=k\cdot x_0\) for some \(k\in G\). Compact groups have bi-invariant normalized Haar measure, so right translation \(g\mapsto gk\) preserves \(m_G\). Consequently,

\int_G\phi(g\cdot y)\,dm_G(g) =\int_G\phi((gk)\cdot x_0)\,dm_G(g) =\int_G\phi(g\cdot x_0)\,dm_G(g) =\int_O\phi\,d\nu.

This value does not depend on \(y\). Integrating the constant over the probability \(\lambda\) shows \(\int_O\phi\,d\lambda=\int_O\phi\,d\nu\) for every continuous real-valued \(\phi\). To conclude that \(\lambda=\nu\), use regularity and the Urysohn lemma. If \(C\subseteq O\) is closed and \(U\subseteq O\) is open with \(C\subseteq U\), there is a continuous \(\psi:O\to[0,1]\) equal to \(1\) on \(C\) and \(0\) on \(O\setminus U\). Thus \(\lambda(C)\le\int\psi\,d\lambda=\int\psi\,d\nu\le\nu(U)\). Outer regularity, taking open \(U\supseteq C\) with \(\nu(U)\) arbitrarily close to \(\nu(C)\), gives \(\lambda(C)\le\nu(C)\). Reversing the measures gives equality on closed sets. Inner regularity then gives equality on every Borel set. Hence \(\lambda=\nu\). \(\square\)

Computing Orbit Probabilities

Worked Example: A Finite Transitive Action

Let \(G=S_3\), the group of six permutations of \(\{1,2,3\}\), act on \(O=\{1,2,3\}\) by permutation. Take \(x_0=1\). Haar probability on this finite group is normalized counting measure, so each permutation has mass \(1/6\). For each \(j\in O\), exactly two permutations send \(1\) to \(j\): the images of the other two points can be assigned in \(2!=2\) ways. Therefore the orbit map gives

\nu(\{j\})=\frac{|\{g\in S_3:g(1)=j\}|}{|S_3|} =\frac{2}{6}=\frac13 \qquad (j=1,2,3).

For example, \(\nu(\{1,3\})=1/3+1/3=2/3\). The three singleton masses sum to \(1\), as required. To check uniqueness directly, an invariant probability must give the same mass to all three singletons, since a permutation sends any chosen singleton to any other. Their masses sum to \(1\), so each must equal \(1/3\), agreeing with the pushforward.

Worked Example: Rotation of the Circle

Let \(G=O=\mathbb{T}=\{z\in\mathbb{C}:|z|=1\}\), with the group acting on itself by multiplication, and set \(x_0=1\). Write \(z=e^{it}\), where normalized Haar measure is \(dt/(2\pi)\) for \(0\le t<2\pi\). The orbit map is \(\pi(g)=g\), so the pushforward is normalized arc length. For the arc \(A=\{e^{it}:0\le t\le\pi/2\}\),

\nu(A)=\frac{1}{2\pi}\int_0^{\pi/2}dt =\frac{\pi/2}{2\pi} =\frac14.

Including or excluding either endpoint makes no difference: each point has Haar measure zero. Indeed, invariance gives every singleton the same mass \(c\); for any positive integer \(N\), \(N\) distinct singletons have total mass \(Nc\le1\), so \(c=0\). As an integral check, for the continuous complex-valued function \(\phi(z)=z\),

\int_{\mathbb{T}}z\,d\nu(z) =\frac1{2\pi}\int_0^{2\pi}e^{it}\,dt =\frac1{2\pi}\left[\frac{e^{it}}{i}\right]_0^{2\pi} =0.

This also agrees with rotational symmetry: multiplying the integral by any fixed \(h\in\mathbb{T}\) leaves it unchanged, so choosing \(h\ne1\) forces the integral to be zero.

Worked Example: Different Orbits, Different Probabilities

Let \(G=\mathbb{T}\) act on the compact annulus \(X=\{z\in\mathbb{C}:1\le|z|\le2\}\) by multiplication. For \(r\in[1,2]\), the orbit of \(x_r=r\) is the circle \(O_r=\{re^{it}:0\le t<2\pi\}\). Its orbit probability is the image of \(dt/(2\pi)\) under \(t\mapsto re^{it}\). The sector \(A_r=\{re^{it}:0\le t\le\pi/3\}\) has probability

\nu_r(A_r)=\frac{1}{2\pi}\int_0^{\pi/3}dt =\frac{\pi/3}{2\pi} =\frac16.

The value is the same for every radius, but the probabilities live on different orbits. Specifically, for \(B=\{z\in X:|z|<3/2\}\), every point of \(O_1\) lies in \(B\), whereas no point of \(O_2\) lies in \(B\). Hence \(\nu_1(B)=1\) and \(\nu_2(B)=0\). Both probabilities are invariant, so there is no unique invariant probability on all of \(X\). The uniqueness theorem applies to each transitive orbit separately.

What the Construction Does—and Does Not—Say

The orbit probability can be used to average a continuous observable \(\phi\) on the orbit: its mean is \(\int_G\phi(g\cdot x_0)\,dm_G(g)\). The uniqueness proof shows that this average is forced for every invariant probability on that orbit. This is useful when an orbit has no convenient coordinates: invariance and transitivity determine the probability even if an explicit density is unavailable.

A common mistake is to conclude that a compact group action always has a unique invariant probability on the whole space. The annulus example gives a concrete correction. The measures \(\nu_1\) and \(\nu_2\) are both invariant, yet \(\nu_1(B)=1\) and \(\nu_2(B)=0\) for the same Borel set \(B=\{|z|<3/2\}\). The missing hypothesis is transitivity on the space under consideration. Restricting to one orbit restores transitivity and makes the uniqueness argument apply.

Check Your Understanding

For each question, identify the orbit and the invariance or transitivity condition being used.

For a finite group \(G\) acting transitively on a finite set \(O\), express the pushforward mass of \(y\in O\) in terms of the number of elements mapping \(x_0\) to \(y\).
Why does left invariance of Haar measure prove that \(\pi_*m_G\) is invariant under the action?
In the uniqueness proof, where is transitivity used to show that the group average of \(\phi(g\cdot y)\) is independent of \(y\)?
For the circle rotation action, what is the pushforward probability of an arc subtending angle \(\alpha\), where \(0\le\alpha\le2\pi\)?
Why do the two orbit probabilities on the annulus not contradict uniqueness on a transitive orbit?

Invariant Probabilities on Compact Group Orbits

What You'll Learn