What is Haar Measure?

Solution 1:

The question leaves your background a bit unclear, so I chat descriptively about the three Haar measures you are likely to be aware of without even knowing that they are Haar measures. Nothing in what follows is rigorous, but rather seeks to give you a taste of what Haar measure is about.

We can measure the size $m(S)$ of a subset $S$ of $\mathbb{R}$ simply by the integral $$ m(S)=\int_S 1\,dx. $$ If $S=[a,b]$ is an interval, this gives the length $m(S)=b-a$. Here we can think of that $dx$ as a measure (technically it's not, but I ignore that here).

What makes this into a Haar measure is the fact that it is translation invariant. If $c\in\mathbb{R}$ is a constant, and $c+S=\{c+s\mid s\in S\}$, then $$ m(c+S)=\int_{c+S}1\,dx=\int_S1\,dx=m(S), $$ because the substitution $t=x+c$ transforms one integral to the other. Effectively we use $$ d(x+c)=dx $$ and say that $dx$ is a translation invariant measure, i.e. a Haar measure of the additive group of reals.

What about the multiplicative group of positive reals? A Haar measure should be invariant under the group operation. So if $S\subseteq\mathbb{R}$ (is a measurable subset), we want $m(S)=m(cS)$ to hold for all $c>0$. Obviously the above measure $\int_S\,dx$ won't do. For example, the length of an interval is not invariant under scaling. This time we should use the definition $$ m(S)=\int_S\frac{dx}x $$ instead. Here the 1-form $x^{-1}dx$ works. Basically because $$ \frac{d(cx)}{cx}=\frac{c\,dx}{cx}=\frac{dx}x. $$ You can check that whenever $0<a<b$, we have $$ m([a,b])=\int_a^b\frac{dx}x=\ln b-\ln a=\int_{ca}^{cb}\frac{dx}x=m([ca,cb]). $$ Thus $x^{-1}\,dx$ is a Haar measure of the multiplicative group of reals.

A third instance of Haar measure that you have surely seen is that of the unit circle of the complex plane $C=\{e^{i\phi}\mid 0\le\phi <2\pi\}$. Here the group operation is multiplication. Multiplication by the number $e^{i\phi_0}$ amounts to rotating the circle counterclockwise by the angle $\phi_0$. Here the "measure" $d\phi$ will be invariat under such rotations as $$ d(\phi+\phi_0)=d\phi, $$ or, if $S\subseteq C$, then $$ m(S)=\int_Sd\phi=\int_{e^{i\phi_0}S}d(\phi+\phi_0)=\int_{e^{i\phi_0}S}d\phi=m(e^{i\phi_0}S). $$

In the last example (as the group $C$ is compact as a topological space), it is customary (but not necessary for all purposes) to divide the measure by $2\pi$ so that $m(C)=1$.

The fun facts are that for Lie groups and locally compact topological groups a Haar measure always exists. It is unique up to a constant multiplier in many important cases such as compact. For non-commutative groups (e.g. matrix groups) there is a distinction between invariance under group operation from the left or from the right. The two notions coincide in the compact case.

Sketching an example, where there is a difference between left and right invariant. Consider the group $G$ of real upper triagular matrices of the form $$ G=\left\{\left(\begin{array}{cc}\sqrt{y}&\frac{x}{\sqrt y}\\0&\frac1{\sqrt{y}} \end{array}\right)\mid x,y\in\mathbb{R},y>0\right\}. $$ Let us denote the above element of $G$ by $g(x,y)$. Consider the differential $$ dg(x,y)=\frac{\partial}{\partial y}g(x,y)\,dy+\frac{\partial}{\partial x}g(x,y)\,dx. $$ For any fixed element $g(r,s)\in G$ we see that $$ (g(r,s)g(x,y))^{-1}d(g(r,s)g(x,y))=g(x,y)^{-1}g(r,s)^{-1}g(r,s)dg(x,y) =g(x,y)^{-1}dg(x,y), $$ so the entries of the matrix of 1-forms $$ g(x,y)^{-1}dg(x,y)=\left( \begin{array}{cc} \frac{dy}{2y}&\frac{dx}y\\0&-\frac{dy}{2y}\end{array}\right) $$ are left invariant. Therefore the exterior product of the top row entries $$ \frac{dy}{2y}\wedge\frac{dx}y=\frac12\frac{dy\wedge dx}{y^2} $$ is a left-invariant 2-form. Number theorists will recognize this (up to a scalar factor) as the hyperbolic metric of the upper half plane. We can define an action of $G$ on the upper half plane by the recipe of fractional linear transformations. Let $\tau$ be an arbitrary element of the upper half-plane and define $g\cdot\tau=z_1/z_2$, where $$ \left(\begin{array}{c}z_1\\ z_2\end{array}\right) =g\left( \begin{array}{c}\tau\\1\end{array}\right). $$ Here $g(y,x)\cdot i=x+iy$, so the entire upper half plane is the orbit of $i$ under $G$.

We get a right invariant 2-form similarly by using $$ dg(x,y)g(x,y)^{-1}=\left( \begin{array}{cc} \frac{dy}{2y}&dx-\frac{x\,dy}y\\0&-\frac{dy}{2y}\end{array}\right). $$

Again the wedge product of the entries of the upper row gives a right invariant 2-form $$ \frac{dy}{2y}\wedge \left(dx-\frac{x\,dy}y\right)=\frac{dy\wedge dx}{2y}. $$ We see that this time the left and right invariant measures are not scalar multiples of each other.

Solution 2:

Haar measure on a locally compact topological group is a Borel measure invariant under (say) left translations, finite on compact sets. It exists and is unique up to multiple. On $\mathbb R,+$ it is the Lebesgue measure (up to multiple).

edit a simple example (for the simplest non-Abelian Lie group):

Let $G$ be the group of affine transformations of $\mathbb R$, $x\mapsto ax+b$, $a>0$. $G$ is the half-plane $(a,b);a>0$. A left-invariant Haar measure is $\mu(A)=\int_A \frac{1}{a^2}da\,db$, whereas a right-invariant Haar measure is $\mu'(A)=\int_A\frac{1}{a}da\,db$. Notice that they are different; $G$ is not unimodular.

In general, try to understand what happens for Lie (or matrix) groups.

Solution 3:

One of the most useful properties of the Lebesgue measure is "translation invariance". ie. For any Borel set $A$, the measure of $$ x+A = \{x+a : a\in A\} $$ is the same as the measure of $A$. This allows one to prove some very interesting properties (for instance, the fact that convolution of two $L^1$ functions is $L^1$)

One would like to do the same sort of thing for any topological group $G$. ie. one would like a measure $\mu$ defined on all Borel sets of $G$ such that $$ \mu(gA) = \mu(A) $$ for any Borel set $A$, where $gA = \{ga : a\in A\}$

As in the case of $\mathbb{R}$, the existence of such a measure allows you to do "harmonic analysis" on such groups. The fact that such a measure exists on a locally compact group was first proved by Haar, and hence the name.