Is a map that preserves lines and fixes the origin necessarily linear?

Let $V$ and $W$ be vector spaces over a field $\mathbb{F}$ with $\text{dim }V \ge 2$. A line is a set of the form $\{ \mathbf{u} + t\mathbf{v} : t \in \mathbb{F} \}$. A map $f: V \to W$ preserves lines if the image of every line in $V$ is a line in $W$. A map fixes the origin if $f(0) = 0$.

Is a function $f: V\to W$ that preserves lines and fixes the origin necessarily linear?

Solution 1:

Consider $V=W=\Bbb F_2^k$. Then, since a subset is a line if and only if it contains (at most) two points, any bijective map that sends $0$ to $0$ does the trick. However, there are $(2^k-1)!$ such maps, while the bijective linear maps are less than that for $k\ge 3$.

Solution 2:

Let me give an example in the Euclidean plane. The function $f\colon\mathbb R^2\to\mathbb R$ given by $f(x,y)=x^3$ maps lines to lines. A vertical line $x=a$ is mapped to the line $a^3$ — points are lines by the OP's definition. Any other line is of the form $\{(t,a+bt);t\in\mathbb R\}$ for some $a,b\in\mathbb R$. The image of any such line is $\mathbb R$. Thus $f$ maps lines to lines, and it clearly fixes the origin. Non-linearity is evident.

This $f$ can be promoted to a function $g\colon\mathbb R^2\to\mathbb R^2$ by letting $g(x,y)=(f(x,y),0)$. This inherits the desired properties and is a function between two-dimensional spaces.

Below is a previous, erroneous answer. I left it here as a warning example. My actual answer is above. I can delete this if it would be more appropriate.

Let me give an example with infinite fields. The real line $\mathbb R$ is an infinite dimensional vector space over $\mathbb Q$. Lines — other than the origin — are translations of the rationals ($r+\mathbb Q$ for some $r$). Take the function $g\colon\mathbb R\to\mathbb R$, $$ g(x) = \begin{cases} x, & x\in\mathbb Q\\ 0, & x\notin\mathbb Q. \end{cases} $$ The image of the line $r+\mathbb Q$ is the line $\mathbb Q$ if $r\in\mathbb Q$ and $\{0\}$ if $r\notin\mathbb Q$. Therefore $g$ preserves lines and fixes the origin. But it is not linear: $5=g(5)\neq g(5-\pi)+g(\pi)=0$.

This is not a valid example because I had misidentified lines. For example, $2+\pi\mathbb Q$ is a line but its image $\{0,2\}$ is not. A weaker statement is true: the image of every line is either a line or a set containing the origin and a non-zero rational number.

Solution 3:

You may be interested in this paper Affinity of a Permutation of a Finite Vector Space It discusses the problem of how many k-flats (cosets of a k-dimentional subspace) of an n-dimensional vector space over a finite field must be preserved by a permutation to force the permutation to preserve all k-flats. See the references for the history of this problem for other fields. And, by the way, Vilmos Totik and Wen-Xiu Ma proved (personal communication) that if f is a transformation of Euclidean n-space, n > 1, such that for all but countably many lines L the image f(L) is a line, then the image of any line is a line, hence f is an affine transformation.