What conditions are sufficient to prove a transformation in $\mathbb{R}^2$ is affine?

I currently have shown that the transformation is bijective, points and lines are preserved and also the following:

If $M$, $N$, $X$, $Y$ are points such that $MN \parallel XY$ and $MN$ is parallel to one of the coordinate axes, then the images $M'$, $N'$, $X'$, $Y'$ also satisfy $M'N' \parallel X'Y'$ and $M'N'$ is parallel to one of the coordinate axes.

Do I still need to show parallel-ness in general is preserved or not?


closed as off-topic by Mark Sapir, abx, Francois Ziegler, Chris Godsil, coudy Mar 17 '18 at 10:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – abx, Chris Godsil, coudy
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I think the question is OK. When we have just proved, with Chubarev, the theorem referred to in my answer below, we did not have any idea about the existence of the fundamental theorem of geometry -- we found it in the literature only after we proved our result. $\endgroup$ – Iosif Pinelis Mar 15 '18 at 2:29
  • $\begingroup$ Possible duplicate of Line-preserving bijection of ${\mathbb{R}}^n$ onto itself $\endgroup$ – Francois Ziegler Mar 15 '18 at 2:42
  • $\begingroup$ @FrancoisZiegler : As explained on page 2737 in the paper cited in my answer below, the proof in the book by Berger you cited at mathoverflow.net/questions/139978/… contains gaps/errors. Also, as noted in my answer, one only needs surjectivity, rather than the bijectivity condition assumed in Berger's book. $\endgroup$ – Iosif Pinelis Mar 15 '18 at 2:50
  • $\begingroup$ In my first comment here, I wrote "When we have just proved". Of course, I meant "When we had just proved". $\endgroup$ – Iosif Pinelis Mar 15 '18 at 3:35
  • $\begingroup$ @IosifPinelis Yes; I still wonder if everything here, including your interesting answer and comment, might not find a better home at the other question. Cf. also the references at MR 3471078. $\endgroup$ – Francois Ziegler Mar 15 '18 at 5:17

By the fundamental theorem of geometry, if a transformation of $\mathbb R^d$ with $d\ge2$ is bijective and maps lines onto lines, then it is affine. The main result in AMS Proc. 1999 implies that, more generally, if a transformation of $\mathbb R^d$ with $d\ge2$ is surjective and maps lines into lines, then it is affine.

So, you don't need the condition "If $M$, $N$, $X$, $Y$ are points such that $MN \parallel XY$ ..." at all. Also, you can relax the conditions of bijectivity and preservation of lines to the conditions of surjectivity and preservation of collinearity -- and you don't need any other conditions to get the affinity.


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