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Is there an $f:\mathbb{R}^2 \to \mathbb{R}^2$ such that:

  • $(0,0)\mapsto (0,0)$; and
  • for any $a,b,c$ with $a^2 + b^2 >0$, the set $A=\{(x,y):ax+by=c\}$ is mapped onto f(A)=\{(x,y):a'x+b'y=c'\} for some a',b',c' with a'^2 + b'^2 >0; and
  • $f$ non-linear?
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    If continuity can be dropped, then see my answer below.2012-02-09

3 Answers 3

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Well, no. If lines go to lines, it is linear. If lines and circles go to lines or circles, it is a linear fractional or Möbius transformation, written in one complex coordinate as $ f(z) = \frac{\alpha z + \beta}{\gamma z + \delta}, $ or $\bar{f}$ if orientation reversing.

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    @WNY, fair enough. I will leave it in place for a few hours. The conclusion does not require analyticity, something like$C^2$is plenty. I left out orientation reversing, take $\bar{f}$2012-02-09
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A baby version of the fundamental theorem of projective geometry asserts that any bijection of $\mathbb{R}^n$ taking lines to lines must be an affine map, and so if you fix the origin, such a map must be linear. One impressive thing about this theorem is that no assumption of continuity of $f$ is required, which addresses WNY's comment above. Is it implicit in your formulation that this map has to be a bijection? Otherwise you could flatten $\mathbb{R}^2$ onto the $x$-axis in some unpleasant non-linear way, and take lines to lines that way without being linear; the condition that $f$ be a bijection is essential.

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    @RahulNarain: Yes it does work, but so does another one suggested by Noah Stein below which is surprisingly simple!2012-02-10
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Without surjectivity or some equivalent assumption, such an $f$ can exist. Take $f(x,y) = (x^3+y,0)$. If you assume that in addition to some line the image of $f$ contains a point off that line, then $f$ must be linear and invertible. I posed and proved this for myself at some point before finding a reference and proof on (according to my notes) p.107 of Lyndon's Groups and Geometry. If you cannot locate this book I can send you a scanned copy of my notes on the proof.

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    This is indeed very nice, and also much simpler than I expected! Will be reading on the proof on my next visit to the library.2012-02-10