How to show that the $ax+b$ group is locally compact?

Question

Let $ax+b$ be the group of affine transformations $x\mapsto ax+b$ with $a>0$ and $b\in \mathbb{R}$. How do you topologize this group? As a group it is isomorphic to the set of $2$x$2$ matrices of the form \begin{bmatrix}a & b \\0 & 1\end{bmatrix}. But unless I am greatly mistaken this set of matrices connot be an open subset of $\mathbb{R}^{4}$ since two entries are fixed. So how do you argue that $ax+b$ is a locally compact group?

Answer 1

Let $G$ be your group and consider the bijection$$\begin{array}{rccc}\psi\colon&(0,\infty)\times\Bbb R&\longrightarrow&G\\&(a,b)&\mapsto&\begin{bmatrix}a&b\\0&1\end{bmatrix}.\end{array}$$Then, consider the distance defined on $G$ by $d(M,N)=\bigl\|\psi^{-1}(M)-\psi^{-1}(N)\bigr\|.$

Answer 2

$a>0, b\in \mathbb{R}$ gives you a pretty obvious homeomorphism with the (open) upper halfspace. Is that locally compact?

Answer 3

The topology on this group is inherited from $\mathbb{R}^2$, and when identified in $\mathbb{R}^2$ it is an open half-plane. Definitely locally-compact.