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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) 2x2 matrices of the form

a b

0 1/a

a>0, b real.

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

a b

0 1

a>0, b 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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Surely, if there is an isomorphism in the first case, the group operation can not be multiplication. Do you know what is the suggested operation, or is it precisely part of the game to find out? – Raskolnikov Nov 18 '10 at 17:51
What is the group operation on the matrices? If I compose ax+b with cx+d I get acx + (ad+b). If I multiply the first matrices it goes to acx+ (ad+b/c). For the second set of matrices, multiplication seems to be the operation. – Ross Millikan Nov 18 '10 at 17:56
@Ross: See my answer, which illustrates the isomorphism at the function level. – Bill Dubuque Nov 18 '10 at 19:25

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)\ $

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