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Matrices of the form $$\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$$ can helpfully be represented by dual numbers $a+\epsilon b$ with $a,b \in \mathbb{R}$ and $\epsilon^2=0.$

My question is, is there a similarly helpful way of looking at matrices of the form $$\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$$ for real $a,b$ with $a \neq 0.$ These are the upper triangular 2x2 matrices with determinant 1, and they form a group with respect to matrix multiplication. My question is, is there an easy way to understand this group ?

In particular, matrix multiplication yields $$\begin{pmatrix} a_1 & b_1 \\ 0 & a_1^{-1} \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & a_2^{-1} \end{pmatrix} = \begin{pmatrix} a_1a_2 & a_1b_2 + b_1 a_2^{-1} \\ 0 & (a_1a_2)^{-1} \end{pmatrix}$$

However, the form of the resulting matrix does not seem very easy for me to understand. I was hoping somebody could share some simple algebraic or geometric way to look at this situation, so I can better understand the top right element of the resulting matrix. For example, is there a well known group that is equivalent to my matrix group. Hopefully this question is not too vague.

J.-E. Pin
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Richard Southwell
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1 Answers1

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Short Answer: $\boldsymbol{G \cong \mathrm{Aff}_+(1) \times Z_2}$, where $\boldsymbol{\mathrm{Aff}_+(1)}$ is the orientation-preserving affine group of $\boldsymbol{\mathbb{R}}$ and $\boldsymbol{Z_2 = \{ \pm 1 \}}$.

As in my comment, the classification of $2$-dimensional Lie groups says there are only two simply connected $2$-dimensional Lie groups, with one $\mathbb{R}^2$ and one yours (dropping the component $a < 0$). But the linked question (with its answer) do not give an intuitive understanding.

As we know, $\mathrm{Aff}(1)$ is a non-Abelian Lie group of dimension $2$, and it has $2$ components, just like yours. This is a strong evidence towards $\mathrm{Aff}(1)$ being isomorphic to yours.

But actually this is not the case. Let $G$ denote your group. We identify $\mathrm{Aff}(1)$ as $$ \mathrm{Aff}(1) = \left\{ \begin{bmatrix} a & b \\ & 1 \end{bmatrix} : a, b \in \mathbb{R} \text{, and } a \ne 0 \right\} \text{.} $$

Then their identity components are indeed isomorphic: Let $$ \begin{aligned} G_+ &= \left\{ \begin{pmatrix} a & b \\ & a^{-1} \end{pmatrix} : a, b \in \mathbb{R} \text{, and } a > 0 \right\} \text{, and} \\ \mathrm{Aff}_+(1) &= \left\{ \begin{bmatrix} a & b \\ & 1 \end{bmatrix} : a, b \in \mathbb{R} \text{, and } a > 0 \right\} \text{.} \end{aligned} $$

Let’s give a isomorphism directly: The map $\varphi \colon G_+ \to \mathrm{Aff}_+(1)$ given by $$ \begin{pmatrix} a & b \\ & a^{-1} \end{pmatrix} \mapsto \begin{bmatrix} a^2 & a b \\ & 1 \end{bmatrix} $$ is a group isomorphism: Its inverse is $\psi \colon \mathrm{Aff}_+(1) \to G_+$ given by $$ \begin{bmatrix} a & b \\ & 1 \end{bmatrix} \mapsto \begin{pmatrix} a^{1/2} & a^{-1/2} b \\ & a^{-1/2} \end{pmatrix} \text{,} $$ and the multiplication is preserved: $$ \begin{aligned} \varphi \left( \begin{pmatrix} a & b \\ & a^{-1} \end{pmatrix} \begin{pmatrix} c & d \\ & c^{-1} \end{pmatrix} \right) &= \varphi \left( \begin{pmatrix} a c & a d + c^{-1} b \\ & a^{-1} c^{-1} \end{pmatrix} \right) \\ &= \begin{bmatrix} a^2 c^2 & a c (a d + c^{-1} b) \\ & 1 \end{bmatrix} \\ &= \begin{bmatrix} a^2 c^2 & a^2 c d + a b \\ & 1 \end{bmatrix} \\ &= \begin{bmatrix} a^2 & a b \\ & 1 \end{bmatrix} \begin{bmatrix} c^2 & c d \\ & 1 \end{bmatrix} \\ &= \varphi \left( \begin{pmatrix} a & b \\ & a^{-1} \end{pmatrix} \right) \varphi \left( \begin{pmatrix} c & d \\ & c^{-1} \end{pmatrix} \right) \text{.} \end{aligned} $$

Although $G_+ \cong \mathrm{Aff}_+(1)$, $G$ and $\mathrm{Aff}(1)$ are not isomorphic. To see this, let’s calculate their centers:

  • For $G$, $(a, b) \cdot (c, d) = (a c, a d + c^{-1} b)$, solve $$ \begin{aligned} a d + c^{-1} b &= c b + a^{-1} d \\ \iff a^2 c d + a b &= a c^2 b + c d \\ \iff (a^2 - 1) c d - a b c^2 + a b &= 0 \\ \iff a^2 - 1 = 0 \text{ and } a b &= 0 \\ \iff (a, b) &= (\pm 1, 0) \text{,} \end{aligned} $$ that is, $Z(G) = \{ I, -I \}$.
  • For $\mathrm{Aff}(1)$, $(a, b) \cdot (c, d) = (a c, a d + b)$, solve $$ \begin{aligned} a d + b &= c b + d \\ \iff (a - 1) d - b c + b &= 0 \\ \iff a - 1 = 0 \text{ and } b &= 0 \\ \iff (a, b) &= (1, 0) \text{,} \end{aligned} $$ that is, $Z(\mathrm{Aff}(1)) = \{ I \}$.

We conclude that $G \cong \mathrm{Aff}_+(1) \times Z_2$ while $\mathrm{Aff}(1) \cong \mathrm{Aff}_+(1) \rtimes Z_2$ with a different semidirect product.

PinkRabbit
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