Results 1 to 2 of 2

Math Help - Isometric?

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Isometric?

    Let X=\{A,\ B,\ C, \ D\} with d(A,D)=2, but all other distances equal to 1.

    d is a metric.

    Prove that the metric space X is not isometric to any subset of \mathbb{E}^n for any n.

    Hint: can you realise X as a subset of a sphere S^2 of appropriate radius, with the spherical "great circle" metric?
    I think I have to take two points in X and show that any mapping from X to \mathbb{E}^n would result in a different distance between them.

    Here's my attempt:

    Take a sphere of radius 1 and the points A and D.

    Since d(A,D)=2, A and D are on opposite sides of the sphere.

    Suppose there exists an isometry \phi:X \rightarrow \sigma where \sigma is is the plane used in stereographic projection (also a subset of \mathbb{E}^n).

    Then \phi(A)=\phi(D). Hence d(\phi(A),\phi(D))=0 which is not equal to d(A,D)=2.

    Hence \phi cannot be an isometry.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Showcase_22 View Post
    I think I have to take two points in X and show that any mapping from X to \mathbb{E}^n would result in a different distance between them.

    Here's my attempt:

    Take a sphere of radius 1 and the points A and D.

    Since d(A,D)=2, A and D are on opposite sides of the sphere.

    Suppose there exists an isometry  \phi:X \rightarrow \sigma where \sigma is is the plane used in stereographic projection (also a subset of \mathbb{E}^n).

    Then \phi(A)=\phi(D). Hence d(\phi(A),\phi(D))=0 which is not equal to d(A,D)=2.

    Hence \phi cannot be an isometry.
    I'm not sure that the hint is very helpful, and I don't think that stereographic projection is needed.

    If such an isometry \phi exists then the points \phi (A),\ \phi (B),\ \phi (C) form an equilateral triangle with side 1, in some 2-dimensional subspace of \mathbb{E}^n. So do the points \phi (B),\ \phi (C),\ \phi (D). Thus the images of the four points lie in some 3-dimensional subspace of \mathbb{E}^n, so we may as well assume that n=3.

    The points \phi (A) and \phi (D) must both lie at a distance \sqrt3/2 from the midpoint of the line segment joining \phi (B) and \phi (C). So the distance from \phi (A) to \phi (D) is at most \sqrt3<2, contradicting the condition that \phi is an isometry.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Isometric
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: August 26th 2014, 04:33 PM
  2. 3d isometric projection question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 12th 2011, 02:14 PM
  3. Isometric to symmetric group on...
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: October 6th 2010, 01:21 AM
  4. Isometric imbedding
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 18th 2009, 02:18 AM
  5. isometric in metric space
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 13th 2009, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum