I’m a student with a pure math background starting to work through Arnold’s “Mathematical Methods...” and I’m struggling right of the bat with Section 1.2 on Galilean Structure. (pg 4 - 6)
So we have this affine space $A^4$ accompanied by a space of displacements $\mathbb{R}^4$. Fine.
On page 5, Arnold defined Time as a linear mapping $t:\mathbb{R}^4 \to \mathbb{R}$, and says two events $a,b\in A^4$ are simultaneous if $t(b-a) = 0$. Fine.
Then Arnold says the set of events simultaneous with a given event is a three dimensional subspace $A^3$, to which I say "Not necessarily". The mapping $t(a)= 0$ for all $a\in A^4$ satisfies Arnold's definition of a time mapping, yet clearly has a four-dimensional kernel. Is a three-dimensional kernel a requirement for a Time mapping $t$? If so, Arnold is certainly not clear about that.
But let's say I accept that for now, meaning I believe we have some Time mapping $t$ with a three dimensional kernel. The text then says that we can define the distance between two simultaneous events $a,b\in A^3$ as $\rho(a,b)=\sqrt{\langle a-b, a-b \rangle}$ where $\langle, \rangle$ is the dot product in $\mathbb{R}^3$. But vector $a-b$ still has the same representation as it did in $\mathbb{R}^4$, (something like $(x_1, x_2, x_3, x_4)$, perhaps) so it does not make sense to directly apply the three dimensional dot product. I feel we would need to choose a basis for $\text{Ker}(t)$ and then we could use the coordinate representation of $b-a$.
I hope my gripes make sense. What I could really use a extremely rigorous definition of Galilean structure.