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?

  • 10
    $\begingroup$ Rotations in three dimensions are famously not commutative. $\endgroup$
    – Ryan Unger
    Jun 27, 2017 at 4:24
  • 1
    $\begingroup$ Related: Non commutative Rotations $\endgroup$
    – user154420
    Jun 27, 2017 at 5:22
  • 2
    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Jun 27, 2017 at 8:40

1 Answer 1


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.


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