continuous epimorphism of compact groups preserves Haar measure

right preserves

Theorem - Let $G,H$ be compact Hausdorff topological groups. If $\varphi :G⟶H$ is a continuous surjective homomorphism, then $\varphi$ is a measure preserving transformation, in the sense that it preserves the normalized Haar measure.

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: Let $\mu$ be the Haar measure in $G$ (normalized, i.e. $\mu (G)=1$). Let $\nu$ be defined for measurable subsets $E$ of $H$ by

$$\nu (E)=\mu ({\varphi }^{-1}(E))$$

It is easy to see that $\nu$ defines a measure in $H$. Let us now see that $\nu$ is invariant under right translations. For every $s\in G$ and every measurable subset $E\subset H$ we have that

${\varphi }^{-1}(\varphi (s)E)=s{\varphi }^{-1}(E)$ (1)

The inclusion $\supseteq$ is obvious. To prove the other inclusion notice that if $z\in {\varphi }^{-1}(\varphi (s)E)$ then $\varphi (z)=\varphi (s)t$ for some $t\in E$. Hence, $\varphi (s^{-1}z)=t$, i.e $s^{-1}z\in {\varphi }^{-1}(E)$. It now follows that $z=s(s^{-1}z)\in s{\varphi }^{-1}(E)$.

Thus, equality (1) and the fact that $\mu$ is a Haar measure imply that

$$\nu (\varphi (s)E)=\mu \left({\varphi }^{-1}(\varphi (s)E)\right)=\mu (s{\varphi }^{-1}(E))=\mu ({\varphi }^{-1}(E))=\nu (E)$$

Since $\varphi$ is surjective it follows that $\nu$ is right invariant. It is not difficult to see that $\nu$ is regular, finite on compact sets and $\nu (H)=1$. Hence, $\nu$ is the normalized Haar measure in $H$ and, by definition, we have that

$$\nu (E)=\mu ({\varphi }^{-1}(E))$$

Thus, $\varphi$ preserves the Haar measure. $\mathrm{\square }$

Title	continuous epimorphism of compact groups preserves Haar measure
Canonical name	ContinuousEpimorphismOfCompactGroupsPreservesHaarMeasure
Date of creation	2013-03-22 17:59:06
Last modified on	2013-03-22 17:59:06
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	9
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 37A05
Classification	msc 28C10
Classification	msc 22C05