Lie algebra of motion group of $\mathbb{R}^{2}$

Let the motion group of $\mathbb{R}^{2}$, that is the group pf affine linear transformations of the form

$$(x,y) \mapsto (x\cos\theta - y\sin\theta + a\,,\, x\sin\theta + y\cos\theta + b)$$

The motion group $G$ can be identified with the subgroup $G$ of $GL(3, \mathbb{R})$, whose elements are matrices

$$\begin{pmatrix} \cos\theta & -\sin\theta & a \\ \sin\theta & \cos\theta & b \\ 0 & 0 & 1 \end{pmatrix}$$

Why is $G$ a closed subgroup of $GL(3, \mathbb{R})$ and why is the Lie algebra of $G$

$\frak{g}$ = $\left\lbrace \begin{pmatrix} 0 & -a & b \\ a & 0 & c \\ 0 & 0 & 0 \end{pmatrix} : a,b,c\in\mathbb{R} \right\rbrace$ ?

3 Answers

To answer your first question, if $\vec{t}=(t_1,t_2)$ and $$T(\theta, \vec{t}):=\begin{pmatrix} \cos\theta & -\sin\theta & t_1 \\ \sin\theta & \cos\theta & t_2 \\ 0 & 0 & 1 \end{pmatrix}, \quad R_\theta:= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$ Show that $T(\phi, \vec{s})T(\theta, \vec{t})=T(\theta+\phi, R_\phi\vec{t}+\vec{s})$. This takes care of $G$ being a closed subgroup of $GL(3,\mathbb{R})$.

To find the generators of the Lie algebra of a matrix Lie group $G$, one needs to find an "infintesimal'' transformation. Let $\epsilon_1, \epsilon_2, \epsilon_3>0$ be a "small" real number. Then $$T(\epsilon_1, \epsilon_2, \epsilon_3)= I_3+ \begin{pmatrix} 0 & -\epsilon_1 & \epsilon_2 \\ \epsilon_1 & 0 & \epsilon_3 \\ 0 & 0 & 0 \end{pmatrix}+O(\epsilon^2)$$ where $I_3$ is the identity matrix. This should be enough to convince you why the Lie algebra is $$\mathfrak{g}=\left\{\begin{pmatrix} 0 & -\theta & a \\ \theta & 0 & b \\ 0 & 0 & 0 \end{pmatrix}: \theta,a,b\in \mathbb{R}\right\}$$

$\mathrm{GL}_3$ is an open subset of $\mathrm{Mat}_{3 \times 3} \cong \Bbb{R}^9$ with coordinates $a_{ij}$ for $1 \leq i,j \leq 3$ (representing the matrix entries). Then the subgroup $G$ is closed since it is the solution set of the equations:

\begin {align*} a_{11} - a_{22} &= 0 \\ a_{21} + a_{12} &= 0 \\ a_{11}^2 + a_{21}^2 - 1 &= 0 \\ a_{31} &= 0 \\ a_{32} &= 0 \\ a_{33} - 1 &= 0 \end {align*}

To see why the Lie algebra is what it is, realize that this group is $3$-dimensional, since it just depends on the parameters $a,b,\theta$. The Lie algebra is the tangent space at the identity, so to see what a generic tangent vector looks like, look at a generic curve in $G$, and look at its tangent vector.

So suppose $\gamma(t)$ is a curve in $G$. Then $a,b,$ and $\theta$ depend on $t$, and we can write

$$\gamma(t) = \left(\begin{array}{ccc} \cos(\theta(t)) & -\sin(\theta(t)) & a(t) \\ \sin(\theta(t)) & \cos(\theta(t)) & b(t) \\ 0 & 0 & 1 \end{array}\right)$$

Now compute the velocity vector by differentiating:

$$\dot{\gamma}(t) = \left(\begin{array}{ccc} -\sin(\theta(t)) \, \dot{\theta}(t) & -\cos(\theta(t)) \, \dot{\theta}(t) & \dot{a}(t) \\ \cos(\theta(t)) \, \dot{\theta}(t) & -\sin(\theta(t)) \, \dot{\theta}(t) & \dot{b}(t) \\ 0 & 0 & 0 \end{array}\right)$$

At the identity element, $\theta(t) = 0$, and so $\sin(\theta(t)) = 0$ and $\cos(\theta(t)) = 1$. So this simplifies to

$$\dot{\gamma}(t) = \left(\begin{array}{ccc} 0 & -\dot{\theta}(t) & \dot{a}(t) \\ \dot{\theta}(t) & 0 & \dot{b}(t) \\ 0 & 0 & 0 \end{array}\right)$$

Hint: Part $1$. Consider the $\det: G\rightarrow \mathbb{R}$.

For part $2$. Observe \begin{align} \lim_{t\rightarrow 0}t^{-1}[A(t\theta', ta', tb')-A(0, 0, 0)]=&\ \lim_{t\rightarrow 0}t^{-1} \begin{pmatrix} \cos(t\theta')-1 & -\sin(t\theta') & ta'\\ \sin(t\theta') & \cos(t\theta')-1 & tb'\\ 0 & 0 & 0 \end{pmatrix}\\ =&\ \begin{pmatrix} 0 & -\theta' & a'\\ \theta'& 0 & b'\\ 0 & 0 & 0 \end{pmatrix} \end{align}