# Why a left-invariant Haar measure is $\mu(A)=\int_A \frac{1}{a^2}da\,db$ and a right-invariant Haar measure is $\mu'(A)=\int_A\frac{1}{a}da\,db$?

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 $\mu'(A)=\int_A\frac{1}{a}da\,db$.

Why a left-invariant Haar measure on $G$ is $\mu(A)=\int_A \frac{1}{a^2}da\,db$ and a right-invariant Haar measure on $G$ is $\mu'(A)=\int_A\frac{1}{a}da\,db$?

What are the explicit formulas for the left action and right action?

Any help will be greatly appreciated!

• I don't know exactly why, but it's not surprising. The action of the group on itself is highly asymmetric. – Matt Samuel Nov 29 '15 at 2:47

Identify $(a,b):=(x\mapsto ax+b)\in G$ with the column vector $\begin{bmatrix}a\\b\end{bmatrix}$. The left multiplication by $(a,b)\in G$ is given by the affine map $(a',b')\mapsto \begin{bmatrix}a&0\\0&a\end{bmatrix}\begin{bmatrix}a'\\b'\end{bmatrix}+\begin{bmatrix}0\\b\end{bmatrix}$, whereas the right multiplication by $(a,b)$ is given by the affine map $(a',b')\mapsto \begin{bmatrix}a&0\\b&1\end{bmatrix}\begin{bmatrix}a'\\b'\end{bmatrix}$. For the left multiplication by $(a,b)$, the determinant of $\begin{bmatrix}a&0\\0&a\end{bmatrix}$ is $a^2$, which is the reason for the factor $\frac{1}{a^2}$ in the left Haar measure $\text{d}\mu(a,b)$. For the right multiplication by $(a,b)$, the determinant of $\begin{bmatrix}a&0\\b&1\end{bmatrix}$ is $a$, which is why we have the factor $\frac{1}{a}$ in the right Haar measure $\text{d}\mu'(a,b)$.