Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f: \mathbb R^n \rightarrow \mathbb R^n$, where $n> 1$ be a bijective map such that the image of every line is a line.

Is $f$ continuous?

I think it is, but the proof isn't immediately obvious to me. Related to this question on math.SE.

Feel free to retag.

share|improve this question
    
What do you mean by "every line is mapped to a line"? Do you mean that the restriction of $f$ to any line is linear and injective, or do you just mean that the image of every line is a line? If you mean the latter, then it is easy to construct counterexamples even for $n=1$. –  Andy Putman Nov 21 '10 at 20:15
    
Take the second meaning, and restrict to $n > 1$, please. The original question at math.SE is about $\mathbb{R}^2$. –  Willie Wong Nov 21 '10 at 20:17
    
Ah, I see. That is an interesting question! –  Andy Putman Nov 21 '10 at 20:19
    
Thank you Willie Wong for editing accordingly. –  trutheality Nov 21 '10 at 20:23
    
I'm sure it's standard---but what is a "line"? Is [0,1] or (0,1) a line in $\mathbf{R}$, or does a line go off to infinity in both directions? –  Kevin Buzzard Nov 21 '10 at 20:23

1 Answer 1

up vote 22 down vote accepted

This is called the fundamental theorem of affine geometry. Let $f : E \to E'$ be a map between affine spaces over a field $K$. Suppose that

  1. $f$ is bijective;

  2. $\dim E=\dim E'\ge 2$;

  3. If $a, b, c\in E$ are aligned, then so are $f(a), f(b), f(c)$.

Then $f$ is semi-affine: fix some $a_0\in E$, then there exists a field automorphism $\sigma$ of $K$ such that the map $h: v\mapsto f(a_0+v)-f(a_0)$ (which goes from the vector space attached to ${E}$ to that attached to $E'$) is additive and $h(\lambda v)=\sigma(\lambda)h(v)$ for all $v$ and all $\lambda \in K$. I don't have an URL for this theorem, I find it in Jean Fresnel: Méthodes Modernes en Géométrie, Exercise 3.5.7. But I think it is in any standard textbook on affine geometry.

When $K=\mathbb R$, it is known that $K$ has no non-trivial field automorphism. So your $f$ is an affine function, hence continuous. If $K=\mathbb C$, as pointed out by Kevin in above comments, take any non-trivial automorphism of $\mathbb C$, then you get a semi-affine map $\mathbb C^n \to \mathbb C^n$ which will not be affine, even not continuous (if $\sigma$ is not the conjugation).

share|improve this answer
1  
Another textbook reference: M.Berger "Geometry", vol.1, section 2.6 "The fundamental theorem of affine geometry". –  Sergei Ivanov Nov 21 '10 at 21:05
    
Yes, <a href="books.google.com/… is a link.</a> –  Qing Liu Nov 21 '10 at 21:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.