Let $E$ be a 3-dimensional metric space that isometrically isomorphic to $\mathbb R^3$ by the bijective isometry $x:\ E\longrightarrow\mathbb R^3$. Then, we can define a vectorspace structure for $E$ by \begin{align} \oplus:\ E\times E&\longrightarrow E \\ (p,q)&\longmapsto p\oplus q=x^{-1}\big(x(p)+x(q)\big) \\ \odot:\ \mathbb R\times E&\longrightarrow E \\ (k,p)&\longmapsto k\odot p=x^{-1}\big(kx(p)\big). \end{align} $\big(E,\oplus,\odot\big)$ then is a vector space on $\mathbb R$, or $\big(E,+,\cdot\big)$ without any confusion.
Further, we can construct a vectorspace structure for $\,\mathcal N=\mathbb R\times E\,$ by \begin{align} (s,p)+(r,q)&=\Big(s+r,x^{-1}\big(x(p)+x(q)\big)\Big) \\ k\cdot(s,p)&=\Big(ks,x^{-1}\big(kx(p)\big)\Big). \end{align} Here, we shall call $E$ a Newtonian space and $\mathcal N$ a Newtonian spacetime.
Now, with $0\in I\subseteq\mathbb R$, we define a straight line in $E$ is a map \begin{align} \alpha:\ I&\longrightarrow E \\ s&\longrightarrow \alpha(s)=sp+ q,\ \text{ for some }p,q\in E. \end{align} and a straight line in $\mathcal N$ is a map \begin{align} \gamma:\ I&\longrightarrow\mathcal N \\ s&\longrightarrow \gamma(s)=\big(as+b,sp+q\big),\ \text{ for some }p,q\in E. \end{align}
Assume a diffeomorphism that we call an inertial frame
\begin{align}
f=(t,x):\ \mathcal N&\longrightarrow\mathbb R^4
\\ (s,p)&\longmapsto \Big(t(s,p),x(s,p)\Big)=\big(t_s(p),x_s(p)\big)
\end{align}
that turns every straight line of the form $(s,ps+q)$ in Newtonian spacetime $\mathcal N$ to straight line in $\mathbb R^4$.
I knew that there's a result that if the Newtonian spacetime $\mathcal N=\mathbb R^4$, then $f$ is an affine transformation $f(x)=Ax+b$. The proof of this claim is followed by these steps
- By composing with an appropriate translation we can assume without loss of generality that f maps the origin to the origin. Therefore we just have to prove that f is linear.
- Give a plane $P\subset\mathbb R^4$ and $d_1,d_2,d_3\subset P$ are 3 straight lines which intersect pairwise. Then $f(d_1),f(d_2),f(d_3)$ are straight lines which intersect pairwise, thus they form a plane $Q$.
- We claim that $f(P)\subseteq Q$ and by a same argument, $Q\subseteq f(P)$. Hence, $f$ maps the planes to the planes.
- We show that $f$ takes parallelograms to parallelograms, this implies $f(u+v)=f(u)+f(v),\ \forall u,v\in\mathbb R^4$.
- Since $f$ is continuous, we conclude $f$ must be linear.
But I wonder if we can tell that $f$ is an affine map in the general context where $E$ needs not to be $\mathbb R^3$ ?
Could anyone enlight me with some idea or some counter example for this problem ? Thanks.
Every straight line can be represented parametrically $ P_{t} = A +vt $ where $A \in \mathbb{R}^n$ and $ v$ is from asociated vector space If $f$ is affine of course $f(P_{t})= f( A +vt) = f(A)+ g(vt)= f(A) + g(v)t $ so f maps lines into lines Now we consider arbitrary transiguration wich for every $A + vt $ gives $ B + ut $ To prove affinity we have just prove for distinct $s$ and $t$" $f(A + vt ) - f( A + vs ) = u(t-s) $
So we have just to prove for every vector $v$ $g(v) = vt $ is,linear trasformation becaulse $f(x) = f(x_{0}) + ut $ is affinic space for every $u$