My question is about a possible characterisation of transformations of affine spaces using simple ratios. I first recall the necessary definitions.
[Affine space] An affine space is a set endowed with a vector-space action "+" such that, given a point P and two vectors v and w, it is true that P+(v+w)=(P+v)+w, and furthermore for every two points P and Q there exists a unique vector v such that P+v=Q (in this case, one writes v=Q-P).
[Affine transformation] An affine transformation between two affine spaces S and T over vector spaces V and W respectively (on the same field K) is defined as a function f from S to T such that there exists a vector-space homomorphism g from V to W such that, for every point P of S and vector v in V, f maps P+v to f(P)+g(v).
One can also show that this is equivalent to the condition such that, for every A, B, C and D points, if A-B = k(C-D), then fA-fB = k(fC-fD).
[Collinear triplet] Given three distinct and collinear (i.e. laying on a 1-dimensional affine subspace) points A, B and C of an affine space S, the simple ratio (A,B,C) is defined as the (existent and only) element k of the field K such that C-A = k (C-B).
[Simple ratio preserving function] A function f between affine spaces (with vector spaces over the same field) is said to preserve simple ratios if, for every three distinct and collinear points A, B and C of the domain, their images are distinct and collinear and (A,B,C)=(fA,fB,fC)
My question is the following: given an injective function between affine spaces, is it true that it is an injective affine transformation if and only if it preserves simple ratios? I succeeded in proving the direct implication, and I was wondering if the converse is true and how to prove it. Thank you in advance.
Best Answer
Let $\mathcal A$ and $\mathcal A'$ be two affine spaces over the vector spaces $E$ and $E'$ respectively (both over the same field $\mathbb K$), and
$$ f \;:\;\mathcal A \;\longrightarrow\; \mathcal A' $$
an injective function which preserves simple ratios. We want to prove that $f$ is an affine transformation.
Let us consider the function $\;f^* \;:\; E \longrightarrow E'\;$ defined by
$$ f^*(P-O) = f(P)-f(O), \tag{1}$$
where $O$ is any fixed point of $\mathcal A$. Observe that $\;P-O\;$ describes $\,E\,$ as $\,P\,$ describes $\mathcal A$.
We have
$$ f^*(P-O)=f^*(Q-O) \implies f(P)-f(O) = f(Q)-f(O) \implies f(P) = f(Q) \implies P=Q. $$
We have, for $\lambda\in\mathbb K$,
$$ f^*\big(\lambda(P-O)\big) = f^*\Big(\big(O+\lambda(P-O)\big)-O\Big) = f\big(O+\lambda(P-O)\big)-f(O). $$
Now, the simple ratio [ Warning: I use a different definition of simple ratio $(A, B, C)$ of three aligned points not all coincident. It is the unique $k\in\mathbb K\cup\{\infty\}$ such that $\;C-A=k(B-A)\;$ if $\;A\neq B\;$, and $(A, A, C)=\infty$ ]
$$\big(O, P, O+\lambda(P-O)\big) =\lambda, $$
implies
\begin{align} &\big(f(O), f(P), f(O+\lambda(P-O))\big) = \lambda\\[1.5ex] \implies\;\; & f(O+\lambda(P-O))-f(O) = \lambda(f(P)-f(O))\\[1.5ex] \implies\;\; & f^*(\lambda(P-O)) = \lambda f^*(P-O). \end{align}
Let $\,P-O\,$ and $\,Q-O\,$ be two vectors belonging to $E$. Consider the points of $\mathcal A$
$$ M := \frac12(P+Q)\qquad \text{and}\qquad R := O+2(M-O), $$
where, for $M$, we used barycentric coordinates (see Note below). Obviously, $M$ is the midpoint of the segment $[P, Q]$, so $(P, M, Q)=2$ and then
$$ \big(f(P), f(M), f(Q)\big) = 2, \text{$\quad$ i.e.$\quad$} f(Q)-f(P) = 2\big(f(M)-f(P)\big). \tag{2 }$$
Similarly, $M$ is the midpoint of $[O, R]\;$ [because we have $(R-O)-(M-O)=M-O,\;$ i.e. $\;M=\dfrac12(O+R)]$, so, as above, we obtain
$$ \big(f(O), f(M), f(R)\big)=2, \text{$\quad$ i.e. $\quad$} f(R)-f(O) = 2\big(f(M)-f(O)\big). \tag{3} $$
Moreover, we have (see Note below):
\begin{align} R-O &= 2\left(\frac12(P+Q)-O\right) =\\[1.5ex] &= 2\left(\frac12(P-O)+\frac12(Q-O)\right) = \tag{$4$}\\[1.5ex] &= (P-O)+(Q-O). \end{align}
From ($2$), ($3$) and ($4$), we deduce
\begin{align} f(R)-f(O) &= 2\Big((f(M)-f(P))+(f(P)-f(O)\Big) =\\[1.5ex] &= f(Q)-f(P) + 2\big(f(P)-f(O)\big) =\\[1.5ex] &= \Big(\big(f(Q)-f(P)\big) +\big(f(P)-f(O)\big)\Big) + \big(f(P)-f(O)\big)=\\[2ex] &= \big(f(P)-f(O)\big) + \big(f(Q-f(O)\big) \end{align}
and then, by $(1)$, $\;f^*(R-O) = f^*(P-O) + f^*(Q-O),\;$ i.e.
$$ f^*\big((P-O)+(Q-O)\big) = f^*(P-O) + f^*(Q-O). \qquad\text{q.e.d.} $$
We conclude that $f^*$ is an isomorphism $E \to f^*(E)\subseteq E'$, and $(1)$ prove that $f$ is an (injective) affine transformation, with $f^*$ as its linear part. In fact
$$ f(P)-f(Q) = \big(f(P)-f(O)\big) - \big(f(Q)-f(O)\big) = f^*(P-O)-f^*(Q-O) = f^*\big((P-O)-(Q-O)\big) = f^*(P-Q).$$
$\phantom{bbb}$
Note.$\quad$ Let $P_1, P_2,\ldots, P_m\;$ be $\;m\;$ points of an affine space, and $\;\lambda_1, \lambda_2,\ldots, \lambda_m\;$ scalars such that $$\lambda_1+\lambda_2+\ldots+\lambda_m=1. \tag{5} $$
By the Chasles property, it is easy to prove that, for all $\; 1\leq h, k\leq m$:
$$ P_h + \sum_{i=1}^m \lambda_i(P_i-P_h) = P_k + \sum_{i=1}^m \lambda_i(P_i-P_k). \tag{6} $$
The common value in ($6$) is a point of $\mathcal A$ denoted
$$ \sum_{i=1}^m \lambda_iP_i $$
and called barycentric combination of the points $P_1, P_2,\ldots,P_m$ with scalars $\lambda_1, \lambda_2,\ldots,\lambda_m$.
And for all $Q$ belonging to the space, we have easily
$$ \sum_{i=1}^m \lambda_i P_i - Q = \sum_{i=1}^m \lambda_i(P_i-Q). $$