# What is Haar Measure?

• Is there any simple explanation for Haar Measure and its geometry?
• how do we understand analogy Between lebesgue measure and Haar Measure?
• How to show integration with respect to Haar Measure?
• what do we mean by integrating with respect to Measure?

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.