This post is dedicated to my number one follower for her fourteenth birthday,which I spoiled…
Let $latex G$ be a compact abelian group. By this we mean that $latex G$ is at once both an abelian group and a compact Hausdorff topological space, and that the group operations are continuous, meaning that $latex g mapsto g^{-1}$ is continuous on $latex G$ and $latex (g,h) mapsto g+h$ is continuous as a map from $latex G times G$ to $latex G$. It is known that there exists a regular Borel measure $latex mu$ on $latex G$, called the Haar measure, which is non-negative, satisfies $latex mu(G) = 1$, and is translation invariant:
$latex forall g in G . mu(g+ E) = mu(E) ,$
for every Borel set $latex E subseteq G$. In fact, the Haar measure is known to exist in greater generality ($latex G$ does not have to be commutative and if one allows $latex mu$ to be infinite then $latex G$ can also be merely locally compact). The Haar measure is an indispensable tool in representation theory and in ergodic theory. In this post we will use the weak* compactness of the unit ball of the dual to give a slick proof of the existence of the Haar measure in the abelian compact case.
1. The Kakutani–Markov fixed point theorem
The following theorem can be stated and proved in greater generality with more–or–less the same proof as we present.
Theorem 1 (Kakutani–Markov fixed point theorem): Let $latex X$ be a Banach space, and let $latex mathcal{F}$ be a commuting family of weak-$latex *$ continuous linear maps on $latex X^*$. Suppose that $latex C$ is a non-empty, weak*–compact and convex subset of $latex X^*$ such that $latex T(C) subseteq C$ for all $latex T in mathcal{F}$. Then $latex mathcal{F}$ has a common fixed point in $latex C$, i.e., there is an $latex x in C$ such that $latex Tx = x$ for all $latex T in mathcal{F}$.
Proof: Let us first prove the theorem in the case where $latex mathcal{F} = {T}$. Choose some $latex y_0 in C$. Construct the averages
$latex y_N = frac{1}{N+1} sum_{n=0}^{N} T^n y_0 .$
Since $latex C$ is convex and $latex T(C) subseteq C$, we have that $latex y_N in C$ for all $latex C$. Since $latex C$ is compact, the sequence $latex {y_N}_{N=1}^infty$ contains a subnet $latex {y_{N(alpha)}}$ that converges to some $latex y in C$. To show that $latex Ty = y$, we will prove that $latex langle x, Ty-y rangle = 0$ for all $latex x in X$.
Fix $latex x in X$. Then $latex x$ is continuous on $latex X^*$ in the weak* topology, hence bounded on the weak*–compact set $latex C$, say $latex |langle x, c rangle| leq M_x$ for all $latex c in C$. Now for any $latex N$,
$latex |langle x , Ty_N – y_N rangle | = frac{1}{N+1} |langle x, T^{N+1} y_0 – y_0 rangle| leq frac{2M_x}{N+1} .$
Thus $latex |langle x, T y – y rangle| = lim_alpha |langle x, T y_{N(alpha)} – y_{N(alpha)} rangle | = 0$ (recall that part of the definition of subnet is that $latex N(alpha)$ goes to infinity with $latex alpha$). That completes the proof for the case where $latex mathcal{F}$ is a singleton.
Now let $latex mathcal{F}$ be arbitrary, and for every finite $latex F in mathcal{F}$, denote $latex A_F = {c in C : forall T in F . Tc = c}$. $latex A_F$ is evidently closed. We will show that the family $latex {A_F : F subseteq mathcal{F} textrm{ finite}}$ has the finite intersection property. It will follow from compactness that there is some $latex x in cap_F A_F$, which will be the sought after fixed point. Now, $latex A_{F_1} cap A_{F_2} = A_{F_1 cup F_2}$, so it follows that we only have to show that every $latex A_F$ is non-empty. This is done by induction on the cardinality $latex |F|$ of $latex F$. If $latex |F| = 1$, then $latex A_F neq emptyset$ by the first half of the proof. Suppose that $latex A_F$ is not empty, and let $latex T in mathcal{F}$. Then for every $latex y in A_F$, we have for all $latex S in F$
$latex STy = TSy = Ty .$
Therefore $latex T(A_F) subseteq A_F$, so by the first half of the proof there is some $latex z in A_{F}$ fixed under $latex T$. In other words, $latex z in A_{F cup {T}}$, so this set is not empty.
2. The existence of Haar measure for abelian compact groups
We can now prove the existence of Haar measure for compact abelian groups.
Theorem 2: Let $latex G$ be a compact Hausdorff abelian group. Then there exists a Haar measure for $latex G$. That is, there is a regular Borel probability measure $latex mu$ on $latex G$ that is translation invariant.
Proof: For every $latex g in G$, let $latex L_g : C(G) rightarrow C(G)$ be the translation operator given by $latex (L_g f)(h) = f(g-h)$. We will find a regular Borel probability measure $latex mu$ on $latex G$ such that for all $latex g in G$,
(*) $latex int f d mu = int L_g f d mu .$
The regularity of the measure together with Urysohn’s Lemma then implies that $latex mu$ satisfies $latex mu(g+E) = mu(E)$ for all Borel $latex E$ and all $latex g in G$ (this might be trickier than it first seems).
Consider the family $latex mathcal{F} = {T_g : g in G}$ of operators on $latex M(X) = C(X)^*$ given by $latex T_g = L_g^*$. Then by Exercise G in Notes 11 $latex T_g$ is weak* continuous for all $latex g$. Moreover, for all $latex f in C(X), nu in M(X)$ and $latex g,h in G$,
$latex langle T_g T_h nu, f rangle = langle nu, L_h L_g f rangle = langle nu, L_g L_h f rangle = langle T_h T_g nu , f rangle .$
Therefore $latex mathcal{F}$ is a commuting family. Now let $latex C$ be the subset of $latex M(X)$ consisting of all probability measures. Then it is easy to see that $latex C$ is weak* closed and convex, and that $latex mathcal{F}$ leaves $latex C$ invariant. By the Kakutani–Markov Theorem, $latex mathcal{F}$ has a fixed point $latex mu in C$. By definition of $latex C$, $latex mu$ is a regular Borel probability measure on $latex G$. By the fact that $latex mu$ is a fixed point for $latex mathcal{F}$ we have $latex langle T_g mu , f rangle = langle mu , f rangle$, which is just another way of writing (*). That completes the proof.
Exercise A: It may seem as if the same argument would give the existence of a translation invariant regular Borel probability measure on a locally compact Hausdorff space.
- Prove why the theorem fails for non–compact spaces.
- What part of the argument breaks down?
- Make sure you know why that part doesn’t break down in the compact case.
