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I had a geometry class which was proctored using the Moore method, where the questions were given but not the answers, and the students were the source of all answers in the class. One of the early questions which we never solved is listed in the title.

In this case, use any reasonable definition of "between-ness". I believe the definition we used was "$B$ is between $A$ and $C$ if and only if $|AB|+|BC|=|AC|$". A collineation is a mapping where every line is mapped to a line. A mapping is a function that operates over the set of points within the given space and returns points in the given space.

When we were studying this question, we managed to get to the point that a between-ness mapping must map lines to line segments. In particular, we had managed to show that for every $A-B-C$ (read: "$B$ is between $A$ and $C$") in the mapped space by applying between-ness preserving mapping $m$, we could guarantee that pre-images $P_A, P_C$ such that $m(P_A)=A,m(P_C)=C$ implies the existence of pre-image $P_B$ such that $P_A-P_B-P_C$ and $m(P_B)=B$. I have never seen any full proof of the title statement.

I would like to read any hints that are known for solving this question. Feel free to completely solve it, but please hide the full solution in such a way that I can start with your hint(s) and have an opportunity to finish the solution for myself.

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A collineation is a mapping where each pair of parallel lines is mapped to a pair of parallel lines.

In my (projective) vocabulary, a collineation is a mapping which maps collinear point triples to collinear point triples. Preservation of parallelity is not implied. The only problem with transformations which don't preserve parallelity is that they might map points in the plane to infinity and vice versa, so you'd need a projective framework to properly express these.

In the common Euclidean (non-projective) plane, if three points are collinear, then one of them is between the other two. If that between-ness is preserved, then the equation you stated holds for the image points as well. But the only way for this equation to hold is if the image points are again collinear. Thus your map must bee a collineation, in my sense as written above.

If you require preservation of parallelity, then I guess the key to that is in what you consider a mapping. Every collineation in the real projective plane is a projective transformation. (The same isn't true for projective planes over fields with non-trivial automorphisms, e.g. over the complex numbers.) A projective transformation which will not send finite points to infinity must fix the line at infinity as a whole. And any projective transformation which fixes the line at infinity is an affine transformation, and as such will map parallel lines to parallel lines. So if the domain of the mapping is the set of all finite points, and its codomain does not include points at infinity, then preservation of parallel lines is guaranteed.

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  • $\begingroup$ I have corrected my definition, thank you for suggesting the need for it. I have recalled the correct definition from the class, and now the point of my asking this question is clear: how can I finish the proof from line segments to full lines? $\endgroup$ – abiessu Sep 28 '13 at 23:23
  • $\begingroup$ In particular, preservation of parallel-ness is not necessary to the question, I had simply misremembered the definition from 10 years ago... $\endgroup$ – abiessu Sep 29 '13 at 1:10
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In an affine plane, there is a unique line through every two points. You can cover any of the lines with line segments. What a line segment-preserving map is then doing is sending all the points on a line to points on another line, and that line can be considered the image of line itself under the mapping.

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  • $\begingroup$ I don't believe that we had a concept of "affine plane" within our definitions at the point of this question arising; the problem we kept having was that all we could guarantee was that for every $A-B-C$ (Read: "$B$ is between $A$ and $C$") in the range of the mapping $m$, we could guarantee a "pre-image" $P_B$ of $B$ such that $m(P_B)=B$. We could never guarantee a pre-image of any point outside the line segment. $\endgroup$ – abiessu Sep 28 '13 at 23:57
  • $\begingroup$ @abiessu I assumed you were doing this in Euclidean geometry, sans the betweenness axioms. You have to assume that every two points of a line share some line segment. You'll have to have declared any three $A, B, C$ such that $A - B - C$ holds to be collinear. You also have to assume that for all $[ABC]$ (read: "$A$, $B$, and $C$ collinear") either $A - B - C$ or $A - C - B$ or $B - A - C$. Then $(m(X) - m(Y) - m(Z) \implies [m(X)m(Y)m(Z)]) \implies (A - B - C \implies [m(A)m(B)m(C)])$. $\endgroup$ – Loki Clock Sep 29 '13 at 10:05
  • $\begingroup$ I will have to pull out my notes, to see whether we had any axioms that ended up making the geometry Euclidean. I think I'm suffering from just a minor disconnect between what you're saying and how it connects "a between-ness preserving mapping maps lines to line segments" to "a between-ness preserving mapping maps lines to lines". Could you clarify a bit more, perhaps in your answer? $\endgroup$ – abiessu Sep 29 '13 at 22:38
  • $\begingroup$ @abiessu If you have no definition of a point and line being incident, so you can't check whether something's a collineation - a collineation being a map $m$ such that if point $A$ and line $B$ are incident, then $m(A)$ and $m(B)$ are, too. You seem to be using the real plane to begin with, since you can measure and add lengths of line segments without defining these - that is, use the Euclidean metric. The major clarification is whether you're using the real affine plane or the real projective plane. If you're not using Euclidean geometry, it's okay, but you need axioms or proofs like I gave. $\endgroup$ – Loki Clock Sep 29 '13 at 23:00
  • $\begingroup$ @abiessu I'm thinking of how I can clarify. Did you recognize the implication $A-B-C \implies [m(A)m(B)m(C)]$ as inducing collinearity of the image of three points on a condition implying their collinearity? $\endgroup$ – Loki Clock Sep 29 '13 at 23:05

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