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Isometry and inverse

  1. Oct 4, 2005 #1
    Does every isometry have an inverse?
     
  2. jcsd
  3. Oct 4, 2005 #2

    Hurkyl

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    Have you tried proving it? How far did you get?

    Have you tried coming up with a counterexample? What things have you tried?
     
  4. Oct 4, 2005 #3

    AKG

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    Do you know that in order to have an inverse, a function must be a bijection? Do you know that in order to be a bijection, a function must be an injection and a surjection? You should be able to easily prove that every isometry is an injection. Suppose x is some point that is in the codomain of an isometry but not in its range. Now choose a line passing through x which contains at least two points in the range of the isometry (if no line passing through x contains two points in the range of the isometry, you should be able to show that there is something severly wrong with this "isometry"). Since the isometry is injective, those two points have unique preimages. Can you see where to go from here? As an additional hint, think about when the triangle inequality gives an equality.
     
  5. Jul 30, 2008 #4
    I just noticed this old thread earlier today, and I would appreciate some clarification on one or two of the points above.

    A function need only be injective to have an inverse; bijectivity isn't required.

    Some authors do not define an isometry as a bijection, and state only that it is a function F which is distance-preserving: d(F(a),F(b)) = d(a,b). It is then easy to show that F is one-to-one, but the issue of surjection doesn't seem to be explicit or implicit -- at least to me. Others (say, Roe, Silvester) define an isometry as a bijection to begin with.

    I have wondered for some time about one of the early problems in George Jennings' text, Modern Geometry with Applications. In chapter 1 he asks for a proof that an isometry (defined as above) maps a Euclidean circle in ExE (all those points at a fixed distance from the center point P) into a circle as its image.

    Here too it is clear that all points in the image are at the proper distance/radius from the image of P -- d(F(p),F(P)) = d(p,P). But it is NOT clear (to me ..) what prohibits the image from having other points at the correct distance and which do not have preimages under the isometry.

    Short of invoking the (topological) completeness property of bounded and closed subsets of the plane -- and thus inferring surjection for the isometry -- is there another, easier method to prove the simple assertion above?

    B.
     
  6. Jul 30, 2008 #5

    Dick

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    You can certainly have maps with the property that d(F(a),F(b))=d(a,b) which are not surjective. Beyond that I'm not sure what your question is. What is it?
     
  7. Jul 30, 2008 #6
    Clarification (from Jennings):

    "A function F from ExE to ExE" (he uses E^n for this definition) "is an isometry if for all points p, q in ExE, d(F(p),F(q)) = d(p,q).

    Let C (in ExE) be the circle with center P and radius r.

    If F is an isometry from ExE to ExE, prove that F(C) is the circle with center F(P) and radius r."

    As in my earlier post, it is trivial to show that all points in F(C) have distance r from F(P). So F(C) is contained in the circle with center F(P) at distance r. But short of assuming or otherwise demonstrating that F is surjective, my question is,

    What is the proof that there are no other points at distance r from F(P) except those in F's image?

    B.
     
  8. Jul 31, 2008 #7

    Dick

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    Take the circle in ExE to be centered on (0,0) with radius r. Then the distance from (r,0) and the distance from (0,r) determine a unique point on the circle. Since this point is determined completely by distance relations, it also is in the image ExE. I was hoping for a really more elegant proof of this, but it doesn't seem to be coming. It's late. Sorry.
     
  9. Jul 31, 2008 #8
    I appreciate your response. At the same time, Modern Geometry with Applications distinguishes RxR (with a distinguished origin and coordinates), from ExE which has neither. In this sense "distance" in ExE appears to be a primitive term, not otherwise defined for the exercise at hand.

    So barring consideration of the compactness of the Euclidean plane over RxR, I am having a small problem finding the logic that demonstrates

    F^-1(F(C)) = C (true by the definition of isometry of F, since F is injective)

    F^-1(C') = C (where C' is the circle centered on F(P))

    Without surjection of F, the rationale for stmt. 2 is .. vague.

    I am trying to not reproduce too much of the text here, but the "problem hint" suggests proving that F(C) is contained in the circle C' centered on F(P) (easy task); and then prove that F^-1(C') is contained in C (which is where the path stops for me w/o surjection).

    I concur as well that it's getting late. Perhaps this will become a bit more clear in the daylight.

    B.
     
  10. Jul 31, 2008 #9

    Dick

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    Hmm. I guess the general point (aside from subtlties of definitions distinguishing ExE from RxR) is that I can pick three points in the plane a, b and c so that any other point x is uniquely identified by d(a,x), d(b,x) and d(c,x). Is there a problem with that program?
     
  11. Jul 31, 2008 #10
    I agree that point uniqueness in the manner you describe is a successful approach to proving that F^-1(C') = C, but I'd consider it an "analytic" vs. "synthetic" solution. It is entirely possible that I am guilty of over-analyzing this particular, simple problem. Perhaps the author expects readers to use the compactness of the reals, or to just plain "know" that there can't be holes in a circle in ExE, for the pre-image and the image of the isometry.

    B.
     
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