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Let $n>1$ be an integer. Is every map $A : \mathbb{R}^n \to \mathbb{R}^n$ that maps lines onto lines (the image of a line is a line) affine?

Popov Florino
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2 Answers2

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No, not without more assumptions. For instance, there exist maps $A:\mathbb{R}^n\to\mathbb{R}$ which are surjective when restricted to every line. You can construct such a map by transfinite induction: choose a length $\mathfrak{c}$ enumeration $(L_\alpha,x_\alpha)_{\alpha<\mathfrak{c}}$ of all pairs $(L,x)$ where $L\subset\mathbb{R}^n$ is a line and $x\in\mathbb{R}$. At the $\alpha$th stage of the induction, choose a point $y\in L_\alpha$ where you have not yet defined $A$, and define $A(y)=x_\alpha$. At the end, define $A$ arbitrarily on any points where it has not been defined. By construction, for every $x\in\mathbb{R}$ and every line $L\subset\mathbb{R}^n$ there is then some $y\in L$ such that $A(y)=x$.

Such an $A$ as above can be considered as a map $\mathbb{R}^n\to\mathbb{R}^n$ by composing with a surjection from $\mathbb{R}$ to some line $L_0\subset\mathbb{R}^n$. Then for every line $L\subset\mathbb{R}^n$, $A(L)$ is a line, namely $L_0$. But such an $A$ clearly cannot be affine.

Eric Wofsey
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  • "choose a point $y \in L_\alpha$ where you have not yet defined $A$" : But why is there always such a point? And why is such an $A$ clearly not affine? – Popov Florino Jul 23 '16 at 09:40
  • Such a point exists because you have so far defined $A$ at fewer than $\mathfrak{c}$ points. Such an $A$ isn't affine because any affine map $\mathbb{R}^n\to\mathbb{R}$ must map some line to a single point. – Eric Wofsey Jul 23 '16 at 14:22
  • Is there anything we can say if we assume also that $A : \mathbb R^n\to \mathbb R^n$ is bijective? – Arctic Char Nov 23 '19 at 05:44
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    @ArcticChar: Then $A$ must be affine; see https://math.stackexchange.com/questions/2035287/is-a-function-which-preserves-zero-and-affine-lines-necessarily-linear. This is sometimes called the "fundamental theorem of affine geometry". – Eric Wofsey Nov 23 '19 at 06:01
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I will give a more simple answer.

It suffices to consider a projective transform.

For example, if we define transformation $\varphi:(x,y)\rightarrow (X,Y)$ by

$$\begin{cases}X&=&\dfrac{x}{x+y+1}\\Y&=&\dfrac{y}{x+y+1}\end{cases}$$

whose inverse is :

$$\begin{cases}x&=&\dfrac{X}{-X-Y+1}\\y&=&\dfrac{Y}{-X-Y+1}\end{cases}$$

Caution: a convention has to be taken for "points at infinity". For each line there is a point that is sent to infinity : more precisely, we omit the point(s) $(x,y)$ belonging to line (L) with equation $x+y+1=0$ and $(X,Y)$ belonging to line (L') with equation $-X-Y+1=0$

Then transformation $\varphi^{-1}$ maps any straight line with equation $uX+vY+w=0$ onto a straight line line with equation $ux+vy=w(x+y+1)$, i.e., $(u-w)x+(v-w)y-w=0$.

But transformation $\varphi^{-1}$ does not map parallel lines to parallel lines as an affine transform should do ; for example, parallel lines $X=0$ ($u=1, v=0, w=0$) and $X=1$ ($u=1, v=0, w=-1$) are mapped onto intersecting lines whith equations $X=0$ and $2X+Y+1=0$.

Jean Marie
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