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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?

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  • $\begingroup$ Is the set still regarded as a line if ${\bf v}={\bf0}$? $\endgroup$
    – David
    Nov 29, 2016 at 2:14
  • $\begingroup$ @David Yes. In more formal language, I'm taking a "line" to be an affine subspace of dimension 1 or 0. $\endgroup$
    – eepperly16
    Nov 29, 2016 at 2:21
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    $\begingroup$ en.wikipedia.org/wiki/Collineation $\endgroup$ Nov 29, 2016 at 4:28
  • $\begingroup$ I think I pondered more or less the same question some time ago. I think the answer would be "yes" if you strengthened the hypothesis to "preserves all affine subspaces". Also, I think it is more natural to drop the assumption about fixing the origin and changing the conclusion to being affine. This question is really more about affine than linear properties, considering the origin here is superfluous. $\endgroup$
    – tomasz
    Nov 29, 2016 at 23:50
  • $\begingroup$ @tomasz Do you have a proof? $\endgroup$
    – eepperly16
    Nov 30, 2016 at 18:24

3 Answers 3

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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 $k\ge 3$.

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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.

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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.

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  • $\begingroup$ Do you know if by now there is a publication of the result by Vilmos Totik and Wen-Xiu Ma that you state in your answer? I would love to see the proof $\endgroup$
    – Inzinity
    Mar 14, 2023 at 17:33
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    $\begingroup$ If you send me your email address I will try to dig up the proof. I doubt if it was ever published but I have it in my old email files, I think. Send me a message to [email protected] Meanwhile I will also check with Ma, Totik and others whether or not it was published anywhere. $\endgroup$ Mar 16, 2023 at 1:52

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