1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linearity vs. takes straight lines to straight lines

  1. Nov 18, 2012 #1

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Linearity vs. "takes straight lines to straight lines"

    1. The problem statement, all variables and given/known data

    Prove that if ##\Lambda:\mathbb R^n\to\mathbb R^n## is a bijection that takes straight lines to straight lines, and is such that ##\Lambda(0)=0##, then ##\Lambda## is linear.

    2. Relevant equations

    Fock's theorem implies that the statement I want to prove is true. But I don't want to use that. I want to find a simpler proof of a less general theorem. It's also possible that there's a published proof of Fock's theorem that is easy to simplify then the assumptions are stronger.

    Fock's theorem says (roughly) that if ##\Lambda## is a map that takes straight lines to straight lines, there's there's a 4×4 matrix A, two 4×1 matrices y,z, and a number c, such that
    $$\Lambda(x)=\frac{Ax+y}{z^Tx+c}.$$ Note that if ##\Lambda## is defined on a vector space (rather than a proper subset of one), there's always an x that makes the denominator 0. So, since my ##\Lambda## has a vector space as its domain, we must have z=0, and we can redefine A and y to absorb c. But I'm also assuming that ##\Lambda(0)=0##, so the theorem also says that ##y=0##, which means that my ##\Lambda## is linear.

    The proof of Fock's theorem is very hard. It looks like it's proved in Appendix B of this article, but the proof is so badly written that it's hard to tell. The proof is discussed in this thread. My thoughts on the first part of the proof appears in post #61.

    3. The attempt at a solution

    I'm nowhere near a solution. I guess the following is better than nothing, but only very slightly: Let x be arbitrary. Let v≠0 be arbitrary. Since ##\Lambda## takes straight lines to straight lines, there's a unit vector u and a function ##s:\mathbb R\to\mathbb R## such that for all t,
    $$\Lambda(x+vt)=\Lambda(x)+s(t)u.$$ (The link to "post #61" above explains this in more detail). I guess I want to prove that ##t\Lambda(v)=s(t)u##, but I don't see how to proceed from here.

    Also, my post #50 describes a few more of my thoughts about the problem.

    One of the reasons why I think it that the simpler theorem has a simpler proof is that it's an exercise in a book I own. It's exercise 1.3.1(2) on page 9 of this book (pdf link). Unfortunately, the exercise is messed up in at least two ways. It assumes that the map is surjective onto W, but it doesn't define W. And it asks the reader to prove that the map is linear, but that can't be the correct conclusion since the exercise doesn't include the assumption that the map takes 0 to 0. The exercise also doesn't assume that the map is bijective, but including that assumption can only make things easier, right?

    Edit: By the way, I'm not entirely clear on whether some assumptions about smoothness or at least existence of partial derivatives must be included.
     
    Last edited: Nov 18, 2012
  2. jcsd
  3. Nov 18, 2012 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Linearity vs. "takes straight lines to straight lines"

    I got a tip that the "simple" theorem I want to prove is more or less "the fundamental theorem of affine geometry". It's quite hard to prove, so it really shouldn't be an exercise on page 9 of a book on functional analysis. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linearity vs. takes straight lines to straight lines
Loading...