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In "A Short Course on Spectral Theory", page 10, William Arveson asserts that the "$ax+b$ group", ie. the group generated by all dilations and translations of the real line, is isomorpic to the group of all (real) $2\times 2$ matrices of the form $ \begin{bmatrix} a & b\\ 0 & \frac{1}{a}\end{bmatrix}, \,\,\,\,\,\, a>0 \mbox{ and }, b \mbox{ real}$

It is very easy to check that the $ax+b$ group is isomorphic to the group of all matrices of the form

$ \begin{bmatrix} a & b\\ 0 & 1\end{bmatrix}, \,\,\,\,\,\, a>0 \mbox{ and }, b \mbox{ real}$

So these two matrix groups should be isomorphic. Is this correct? Can someone give me the isomorphism? I've tried for a while and can't seem to get it.

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    @Ross: See my answer, which illustrates the isomorphism at the function level.2010-11-18

2 Answers 2

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The isomorphism is given by mapping $\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$ to $\begin{pmatrix} a^2 & a b \\ 0 & 1 \end{pmatrix}$. The inverse is given by $\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \mapsto \begin{pmatrix} a^{1/2} & a^{-1/2} b \\ 0 & a^{-1/2}\end{pmatrix}.$

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The isomorphism arises simply from factoring $\rm\ \ aa\ X + b\ \ =\ \ a\ (ax+b/a)\:,\ $ namely

$\rm\ \ \ (aa\ X + b)\ \: \circ\: \ (AA\ X+B)\quad =\quad a\ a\ (A\ A\ \ X\ \ +\ \ B)\ \ +\ \ b\quad\ \ \ =\quad\ \ \ aa\ \ AA\ X\ +\ aaB + b$

$\rm\displaystyle\ a\bigg(aX+\frac{b}a\bigg)\circ \:A\:\bigg(AX+\frac{B}A\bigg)\ =\ aA\:\bigg(aX+\frac{b}{aA}\bigg)\circ\bigg(AX+\frac{B}A\bigg)\ =\ aA\bigg(aA\ X\ + \frac{aaB + b}{aA}\bigg)\ $