Invariant Measures on Compact Homogeneous Spaces

From Haar Measure to a Measure on a Quotient

Let \(G\) be a compact Hausdorff topological group and \(H\) a closed subgroup. The quotient \(G/H\) consists of the left cosets \(gH\), equipped with the quotient topology. It is a compact Hausdorff space, and \(G\) acts on it continuously by left multiplication: \(a\cdot(gH)=(ag)H\). The goal is to put a probability measure on \(G/H\) that does not favor any point or direction of this action.

Let \(\mu_G\) denote normalized Haar measure on \(G\): it is a regular Borel measure with \(\mu_G(G)=1\) and \(\mu_G(aE)=\mu_G(E)\) for every \(a\in G\) and Borel \(E\subseteq G\). Since \(G\) is compact, normalized Haar measure is also right invariant. Write \(q:G\to G/H\) for the quotient map, \(q(g)=gH\). We define the quotient measure by pushing \(\mu_G\) forward through \(q\).

The formula says how to calculate with the quotient measure without first finding coordinates on the quotient: lift the function to \(G\), integrate there using Haar measure, and use the resulting number as the quotient integral. The measure itself is defined on the Borel sigma-algebra of the compact Hausdorff space \(G/H\); no completion is being assumed.

Existence and Invariance

Proof. The quotient map \(q\) is continuous, so \(q^{-1}(A)\) is Borel in \(G\) whenever \(A\) is Borel in \(G/H\). The pushforward definition gives a Borel measure, and \(\mu_{G/H}(G/H)=\mu_G(G)=1\). To verify regularity, let \(A\subseteq G/H\) be Borel. By inner regularity of \(\mu_G\), for every \(\varepsilon>0\) there is a compact \(K\subseteq q^{-1}(A)\) with \(\mu_G(K)>\mu_G(q^{-1}(A))-\varepsilon\). The set \(q(K)\) is compact, hence closed and Borel in the Hausdorff space \(G/H\), and \(q(K)\subseteq A\). Also \(K\subseteq q^{-1}(q(K))\), so \(\mu_{G/H}(q(K))\ge\mu_G(K)\). Since \(q(K)\subseteq A\), this proves inner regularity of \(\mu_{G/H}\). It is finite, so inner regularity applied to \(A^c\) also gives outer regularity: if compact \(K\subseteq A^c\) has \(\mu_{G/H}(K)> \mu_{G/H}(A^c)-\varepsilon\), then the open set \(U=(G/H)\setminus K\) contains \(A\) and \(\mu_{G/H}(U)<\mu_{G/H}(A)+\varepsilon\).

For invariance, let \(a\in G\) and let \(A\subseteq G/H\) be Borel. The quotient map satisfies \(q^{-1}(aA)=a\,q^{-1}(A)\): a point \(g\) maps into \(aA\) precisely when \(gH=a x\) for some \(x\in A\), or equivalently \(a^{-1}gH\in A\). Therefore left invariance of \(\mu_G\) gives \(\mu_{G/H}(aA)=\mu_G(a\,q^{-1}(A))=\mu_G(q^{-1}(A))=\mu_{G/H}(A)\). Finally, the pushforward integral formula follows first for indicators from the definition, then for simple functions by finite linearity, and then for bounded Borel functions by the standard construction of the integral from simple functions (or bounded convergence). The lift \(g\mapsto f(gH)\) is continuous, hence Borel, and bounded because \(G/H\) is compact. \(\square\)

Uniqueness Among Regular Invariant Probabilities

The regularity qualification in the uniqueness statement is important: the uniqueness clause of the Riesz representation theorem identifies regular Borel measures from their integrals against continuous functions. We use that standard theorem in its regular-measure form. We do not assert here that an arbitrary Borel probability is regular.

Proof. Fix a continuous real-valued function \(f\) on \(G/H\). The function \((g,x)\mapsto f(gx)\) is continuous on the compact product \(G\times(G/H)\), so it is bounded and measurable. The measures \(\mu_G\) and \(\nu\) are finite; thus Fubini's theorem applies to this integrand and their product measure. First integrate in \(x\). Invariance of \(\nu\) gives \(\int_{G/H}f(gx)\,d\nu(x)=\int_{G/H}f(x)\,d\nu(x)\) for each \(g\in G\). Integrating this equality over \(G\) yields \[ \int_G\int_{G/H} f(gx)\,d\nu(x)\,d\mu_G(g)=\int_{G/H}f\,d\nu. \]

Now fix \(x=aH\in G/H\). The map \(g\mapsto f(gx)\) is \(g\mapsto f(gaH)\). By right invariance of normalized Haar measure, changing \(g\) to \(ga\) shows that \(\int_G f(gx)\,d\mu_G(g)=\int_G f(gH)\,d\mu_G(g)\), independent of \(x\). Integrate this equality with respect to the probability \(\nu\), and use Fubini again, with the same boundedness and finiteness hypotheses. It follows that \[ \int_{G/H}f\,d\nu=\int_G\int_{G/H}f(gx)\,d\nu(x)\,d\mu_G(g)=\int_G f(gH)\,d\mu_G(g)=\int_{G/H}f\,d\mu_{G/H}. \] Thus \(\nu\) and \(\mu_{G/H}\) have the same integrals against every continuous real-valued function. The uniqueness clause of the Riesz representation theorem for regular Borel measures on a compact Hausdorff space implies \(\nu=\mu_{G/H}\). \(\square\)

Worked Quotient Integrals

Normalization and the Meaning of Uniqueness

A common error is to push forward an unnormalized Haar measure and then call the result a probability. For example, counting measure on \(\mathbb{Z}/12\mathbb{Z}\) assigns mass \(1\) to each element. Its pushforward to the four cosets of \(H=\{0,4,8\}\) assigns mass \(3\) to each coset, for total mass \(12\), not \(1\). The correction is to normalize first: Haar probability assigns mass \(1/12\) to each group element, and its pushforward assigns \(1/4\) to each coset. More generally, the pushforward of a finite Haar measure of total mass \(c>0\) has total mass \(c\); dividing by \(c\) is necessary to obtain a probability.

A second point is to keep the scope of the uniqueness claim precise. The proof above establishes uniqueness among regular invariant Borel probabilities. It uses equality of continuous-function integrals and the Riesz uniqueness theorem, whose measure-uniqueness conclusion is for regular Borel measures. Without a proof of regularity or another argument, that reasoning does not establish uniqueness among all Borel probabilities. The constructed quotient measure is regular, as verified above, and the stated class of competitors matches the theorem used.