Galilean 1+1 group + lie algebra

Here we study the group most closely related to our intuition of spacetime, the galilean group.

We’ll be considering a two-dimensional space made of 1 dimension of space and 1 of time. We can then write (x, t) for the coordinates of an event. On this space, there are a few natural transformations:

• Space translations: $(x, t) \mapsto (x+a, t)$, for any $a \in \mathbb{R}$
• Time translations: $(x, t) \mapsto (x, t+s)$, for any $s \in \mathbb{R}$
• Boosts: $(x, t) \mapsto (x+vt, t)$, for any velocity $v \in \mathbb{R}$

We can rewrite these operations as the action of a group, that we describe first here:

The Galilean Group

The galilean group is generated by matrices of the form: \(D_x(a) = \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, D_t(s) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & s \\ 0 & 0 & 1\end{pmatrix}, B(v) = \begin{pmatrix} 1 & -v & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\)

From which we can deduce the following commutation relationships: \(D_xD_t = D_tD_x \\ BD_x=D_xB \\ B(v)D_t(s)B(v)^{-1} = D_x(-vs)D_t(s)\)

From this we can conclude that the subgroup generated by $D_x, D_t$ is free abelian, and that the galilean group $\mathrm{SGal}$ is a semi-direct product of the subgroup generated by uniform boosts. So that we can write $\mathrm{SGal}$ as follow: \(\mathrm{SGal} = \left \{ \begin{pmatrix} 1 & -v & a \\ 0 & 1 & s \\ 0 & 0 & 1\end{pmatrix}, v, a,s \in \mathbb{R} \right \}\)

Action on 2D spacetime coordinates

Galilean lie algebra

The Lie algebra is generated by:

From which we get the structure coefficients of the Lie algebra:

We note that the center of $\mathfrak{gal}$ is $\lt D_X \gt$

Killing form

The killing form is null

Adjoint action in (Dx, Dt, B)

\(\begin{pmatrix} 1 & -v & a \\ 0 & 1 & s \\ 0 & 0 & 1 \end{pmatrix} \mapsto \begin{pmatrix} 1 & -v & s \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) TODO: verify this formula

Orbits of the adjoint action

Points of the form (x, 0, 0)

All this points are fixed points of the action. They correspond to spatial translations of a given speed x, which is a complete invariant for this orbit. Note that this speed x is not a movement in time.

The line $B = c\ \&\ D_t = t$ where $ct \neq 0$

This line is stable by the adjoint action. The matrix of the action on this line (x, t, c) is given by: \(\begin{pmatrix} 1 & -v & s \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ t \\ c \end{pmatrix} = \begin{pmatrix} x -vt + cs \\ t \\ c \end{pmatrix}\)

This line corresponds to transformations that move in time at a fixed speed s and accelerate at a fixed speed v.

Data for moving on 1+1 Galilean Space

• x,t coordinates of the point
• $D_t = \begin{pmatrix} 0 & 0 & v \ 0 & 0 & 1 \ 0 & 0 & 0 \end{pmatrix}$ (given by a single number)

From this data we can deduce the other components of the Lie basis:

• $D_x = \begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}$ which is constant
• $B = \begin{pmatrix} 0 & -1 & t \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}$

This construction insures that the Lie algebra basis structure coefficients are invariants throughout the action. It also insures that each of the Lie basis vector stays in the same conjugacy class. Finally, it guarantees that (x,t) is a fixed point of B.

The Lie action on points is just given by $X \cdot (x,t) \mapsto (I+X) \begin{pmatrix} x \ t \ 1 \end{pmatrix}$

Picture of the Lie Algebra

The Lie Algebra is globally invariant by translation in the Dx axis. In all the graphs below we’ll be identifying $\mathfrak{gal}$ with $\mathbb{R}^3$ using the $(D_x, D_t, B)$ basis.

The white cross shows the line of space translations <Dx>, which are all fixed points.

Computation of the Killing form

$k(X, Y) = \mathrm{tr}(\mathrm{ad}_X \circ \mathrm{ad}_Y)$

• k(B, B) = 0

  $\mathrm{ad}_B \circ \mathrm{ad}_B(X) = [B, [B, X]]$

  Which gives: \(\mathrm{ad}_B \circ \mathrm{ad}_B(Dx) = 0 \\ \mathrm{ad}_B \circ \mathrm{ad}_B(Dt) = 0 \\ \mathrm{ad}_B \circ \mathrm{ad}_B(B) = 0\)

• k(Dx, Dx) = 0

  $\mathrm{ad}_Dx \circ \mathrm{ad}_Dx(X) = [Dx, [Dx, X]] = 0$

  Because Dx anihilates the whole Lie algebra.

• k(Dt, Dt) = 0

  $\mathrm{ad}_Dt \circ \mathrm{ad}_Dt(X) = [Dt, [Dt, X]]$

  Which gives: \(\mathrm{ad}_Dt \circ \mathrm{ad}_Dt(Dx) = 0 \\ \mathrm{ad}_Dt \circ \mathrm{ad}_Dt(Dt) = 0 \\ \mathrm{ad}_Dt \circ \mathrm{ad}_Dt(B) = 0 \\\)

• k(Dx, Dt) = 0

  $\mathrm{ad}_Dx \circ \mathrm{ad}_Dt(X) = [Dx, [Dt, X]] = 0$

• k(B, Dx) = 0

  $\mathrm{ad}_B \circ \mathrm{ad}_Dx(X) = [B, [Dx, X]] = 0$

• k(B, Dt) = 0

  $\mathrm{ad}_B \circ \mathrm{ad}_Dt(X) = [B, [Dt, X]]$

  Which gives: \(\mathrm{ad}_B \circ \mathrm{ad}_Dt(Dx) = 0 \\ \mathrm{ad}_B \circ \mathrm{ad}_Dt(Dt) = 0 \\ \mathrm{ad}_B \circ \mathrm{ad}_Dt(B) = 0 \\\)