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Cross ratio is real

  1. Sep 5, 2006 #1
    Can some one help me out here, I'm stuck on the following problem:

    show that four distinct points [tex] z_i , i=1,\ldots,4[/tex] in [tex] \hat{\mathbb C} [/tex] lie on a circle in [tex] \hat{\mathbb C} [/tex] if and only if the cross-ratio [tex] \left[ z_1,z_2,z_3,z_4 \right] [/tex] is real.

    So, I know we can write the cross-ratio as
    [tex] \left[z_1,z_2,z_3,z_4 \right] = \frac{(z_1 - z_3) (z_2-z_4)}{(z_1-z_2)(z_3-z_4)} [/tex]

    I also know that I can make the problem simpler if I can reduce the problem to the case where
    [tex] z_1=1,z_2=0,z_3=\infty [/tex]
    but I have no idea even why if I could reduce the problem with to the above case, how that will help me. Can someone shed some light on this for me. Thanks in advance.
     
  2. jcsd
  3. Sep 5, 2006 #2

    AKG

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    Do you know that cross-ratios are invariant under fractional-linear transformations, in that [Tx,Ty,Tz,Tw] = [x,y,z,w] if T is an FLT? Do you know that T is an FLT iff for any circle C (in the Riemann sphere), TC is a circle (in the Riemann sphere)? Do you know that if the zi lie on a circle, then there's an FLT T such that F(z1) = 1, F(z2) = 0, and F(z3) = [itex]\infty[/itex]?
     
  4. Sep 5, 2006 #3
    We are just starting to cover that very material and we are just really glancing over most of this material without proof. So, we were basically told about it but we weren't shown proofs or examples. Basically, I'm trying to figure this out without much background work at all. I am familiar with the cross ratios from a projective geometry stand point, but that does not mean I understand it.
     
  5. Sep 5, 2006 #4
    I am aware that if one knows how a map transforms the points 0 and infinity one knows quite a bit. Though I do not know why or how? I guess just one more thing I don't understand.

    If I think of putting zero on the south pole and infinity on the north pole, I can see how a circle going through these points could enclose 1, but I do not know how this would help me.
     
  6. Sep 6, 2006 #5

    AKG

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    What's important is not whether you already know these facts, but whether you can prove them now that I've given them to you, and whether you know how to put them together to prove your claim.
     
  7. Sep 6, 2006 #6
    No, I really struggling with this one. What I've tried is
    [tex] \left[z_1,z_2,z_3,z_4\right] = \left[1,0,\infty,z_4\right] [/tex]
    So
    [tex] \left[1,0,\infty,z_4\right] = \frac{z_2-z_4}{z_2-z_1} = z_4 [/tex]
    Should I assume that the four distinct points lie on a circle to begin with? I am not sure if what I've done is correct, but if it is I'm not sure how this would be useful.
     
  8. Sep 6, 2006 #7

    AKG

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    I'm not sure what you're doing, but note that the only circle which passes through 1, 0, and infinity is the real line, so if 1, 0, infinity, and z4 are supposed to be on a circle, then z4 is real. But again, I don't know what you're trying to do.
    You know that you're proving an "if and only if" statement, right?
     
  9. Sep 6, 2006 #8

    AKG

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    There are three facts which I asked about in post #2. I recommend doing the following, in no particular order:

    Assume the facts I've given are correct, and prove the final things you want to prove using those assumed facts.
    Prove each of those facts themselves.
     
  10. Sep 6, 2006 #9
    For me I think this is easier said than done. I'm truly lost on this problem having stared at it for hours. I just do not see the connection. I understand the way the proof should go, it's the pieces I'm plugging in I do not understand. In particular, I just don't see how to apply the three things you mentioned in post 2, perhaps this is because I really don't even understand what the three things mean. Well thanks for taking the time to try and help, I appreciate it, but you don't need to keep trying I think this goes into the many problems of unanswerable for me.
     
  11. Sep 7, 2006 #10

    AKG

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    Okay, to make it easier to write without having to use LaTeX, let's replace z1, ..., z4, with q, r, s, and t, in that order. Let's define an FLT T by:

    Tz = [q, r, s, z]

    [q, r, s, t] is real
    iff Tt is real (since [q, r, s, t] = Tt)
    iff 1, 0, [itex]\infty[/itex], and Tt all lie on the real circle, and Tt is not infinity (since 1, 0, and [itex]\infty[/itex] all lie on the real circle, and since being real is equivalent to lying on the real circle and not being infinity)
    iff 1, 0, [itex]\infty[/itex], and Tt all lie on the real circle (since it's impossible for Tt to be infinity anyways, because Ts is already infinity and FLTs are bijections and s and t are distinct by hypothesis)
    iff 1, 0, [itex]\infty[/itex], and Tt all lie on a circle (since the only circle all three of 1, 0, and [itex]\infty[/itex] could all lie on is the real circle)
    iff Tr, Ts, Tq, and Tt all lie on a circle (since Tr, Ts, and Tq are 1, 0, and [itex]\infty[/itex] respectively)
    iff r, s, q, and t all lie on a circle (since FLTs send circles to circles, and since FLTs are invertible with their inverses being FLTs as well)

    Do you see how the pieces fit together? I apologize, the first fact mentioned in post #2 is inconsequential to this proof. The second fact in post #2 is the second last thing in bold here, and you need to prove it. The other stuff in bold are facts which I didn't mention earlier, but are needed in the above proof, and you have to prove them too. The third fact of post #2 is embodied in our FLT T, used throughout this proof.

    The key to proving all bold facts in this post is to realize that all FLTs can be decomposed into a composition of translations (maps of the form [itex]z \mapsto z + c[/itex] for some complex constant c), homotheties (maps of the form [itex]z \mapsto cz[/itex] for some non-zero complex constant c), and inversion (the map [itex]z \mapsto 1/z[/itex]).
     
    Last edited: Sep 7, 2006
  12. Sep 7, 2006 #11
    AKG, thank you very much for helping so much, and being patient with me. I tend to be very slow to understand, especially proofs of almost any form. So, again, thanks very much.

    Now I wasn't aware you had posted this, so what I'm writing is what I took from what you had said earlier. So, I'll post this and then I'm going to read carefully through your proof. Please, let me know what you think, and right now I have it only one way. I'm working on the other way. So, what I have is:

    I want to first show that given four distinct points in the extended complex plane, which lie on a circle, implies the cross ratio is real. Now since we know that for three distinct points there exists a Mobius transformation such that [tex] f(z_1)=1,f(z_2)=0,f(z_3)=\infty[/tex] Moreover, since we have a Mobius transformation we know that [tex] \left[z_1,z_2,z_3,z_4\right] =\left[ f(z_1),f(z_2),f(z_3),f(z_4)\right][/tex] But by assumption all four points lie on a circle, and under a Mobius transformation, circles are mapped to circles. Since the map is unique, there is only one circle which maps [tex] z_1 \rightarrow 1, \, z_2 \rightarrow 0,\, z_3 \rightarrow \infty , \, z_4 \rightarrow f(z_4)[/tex] which is the real number line. So, all elements under this map must be elements of the real number line, which implies that the cross ratio is necessarily real.


    The backward direction (trying to show that since the cross-ratio is real implies the four points must lie on a circle) I'm still working on, but I think this shows it one way.
     
  13. Sep 7, 2006 #12

    AKG

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    Although you appear, for the most part, to have the right idea, the wording is off and some details are missing so I can't tell whether you really understand the reasoning. In the first few sentences, it seems like a few minor word replacements will do the trick. The second last sentence seems to have the right idea, but really badly needs to be reworded. You have some unjustified sentences in there, but I assume that's because I told you to assume them for now, and piece them together, and you can prove them later. But the last sentence is a new fact; can you elaborate on it?
     
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