# Why is the Galilean group not commutative?

As I understand it, the Galilean transformation is a matrix

$$\left[ {\begin{array}{ccccc} R_{11} & R_{12} & R_{13} & v_x & a_x\\ R_{21} & R_{22} & R_{23} & v_y & a_y\\ R_{31} & R_{32} & R_{33} & v_z & a_z\\ 0 & 0 & 0 & 1 & s\\ 0 & 0 & 0 & 0 & 1\\ \end{array} } \right]$$ that can be used to transform a vector $(x,y,z,t,1)$. I saw this group represented as a matrix here but I'm not really sure why it includes five components.

Anyways, I'm more familiar with this form of the transformation.

$$\mathbf{x}'=R\mathbf{x} + \mathbf{v}t+\mathbf{a}$$ $$t'=t+s$$

and we denote this $$G(R,\mathbf{v}\mathbf{,a},s)$$

Why is this just a group and not an abelian group? In other words, what makes this non-commutative?

• Rotations in three dimensions are famously not commutative. Commented Jun 27, 2017 at 4:24
• Related: Non commutative Rotations
– user154420
Commented Jun 27, 2017 at 5:22
• 1. Note that the Galilean transformations are affine, not linear, transformations of 4-dimensional vectors, so you can't represent them by 4-by-4 matrices. 2. It's unclear what you want as an answer to the question of "why" this group is non-commutative - you can just check that it is by computing the commutator of two Galilean transformation, so I'm not sure what further reason you're looking for here. Commented Jun 27, 2017 at 8:40

As mentioned in the comments, the Galilean group includes as a subgroup the rotations in 3D, and those are not commutative, so this isn't at all mysterious. Moreover, even if you restrict yourself to two dimensions (so the rotation group is $\rm SO(2)$ and thus commutative), it's pretty clear that "translate by two units along $x$" and "rotate $x$ into $y$" do not commute, so there's essentially no wriggle room here.