Results 1 to 6 of 6

Math Help - Regular Surface + Isometry

  1. #1
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347

    Regular Surface + Isometry

    Let  S_1 and S_2 be regular surfaces in \mathbb{R}^3. Let \phi: S_1 \rightarrow S_2 be an isometry, that is to say

    I_p^{S_1} (x,y) = I_{\phi(p)}^{S_2} (d_p \phi x, d_p \phi y) for all x,y \in T_p S_1, the tangent plane at  p\in S_1, where I is the first fundamental form on the appropriate tangent plane, and d_p is the differential map.


    Let (U_1,F_1,V_1) and  (U_2,F_2,V_2) be local parametrisations of S_1 at  p and S_2 at \phi(p) respectively.
    Let u = F^{1}(p).

    I am trying to show that the gaussian curvature is the same at p and \phi(p). I guess that this means I am trying to show that the first fundamental forms at p and \phi(p) are the same when i put the appropriate basis vectors in, i.e.

    I_p^{S_1} (X_i, X_j)= I_{\phi(p)}^{S_2} (X_i ', X_j '), where X_i = DF_1(u) e_i and X_i = DF_2 (F_2^{-1} \circ \phi \circ F_1 (u)) e_i. I figured showing this would be sufficient because then the gaussian curvature is completely determined by the first fundamental form (the notation I've used means you seperate the two columns of the jacobian and use them as basis vectors).

    So if I try to show this, then I end up with

     <br />
I_p^{S_1} (X_i, X_j) = I_{\phi(p)}^{S_2} (D(\phi \circ F_1)(u)e_i, D(\phi \circ F_1)(u)e_j ) <br />
and then I am totally stuck. Can anyone offer any advice? I understand this question is a bit long winded.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    since isometries keep the first fundamental form I and I determines the gaussian curvature, this is already done. what are you trying to prove? Re-writing an abstract expresion in specific coordinates does not make mush sense.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347
    why are the first fundamental forms the same? this is what I am trying to show, i.e.

    X_i \cdot X_j = X_i ' \cdot X_j '

    or am I fundamentally misunderstanding something here? The only definition of an isometry I have is I_p^{S_1} (x,y) = I_{\phi(p)}^{S_2} (d_p \phi x, d_p \phi y) . Why does this mean the first fundamental forms have the same matrix?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    do you understand the expression I_p(x,y) = I_f(p)( df(x), df(y) ) ? ( I'm using f instead of \phi)
    This expression tells: the first fundamental form I of S1 at p, corresponds to the first fundamental form I of S2 at f(p).
    "Corresponds to" here means: Given any two vectors x and y of S1 at p, their inner product I_p(x,y), or you can write as x . y, equals to the inner product of their images under the differential map df. This is exactly you Xi . Xj = Xi' . Xj'.

    Using any choice of coordinates won't change that. You need to understand that coordinates are only used for simplifying computation. The important thing is the underlying geometry.

    For your question, consider two vector spaces V and W, and a linear isomorphism f: V -> W. Also their are two extra structures, their inner products, are defined for each. Suppose that f is further an isometry, that is, f keeps inner products, <f(x), f(y)> = <x,y>. Then their two inner products have the same matrix, in any choice of basis v1,..., vn in V and f(v1),...,f(vn) in W.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347
    so if \phi:S_1 \rightarrow S_2 is an isometry is d_p \phi necessarily an isomorphism?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
    Posts
    347
    YES it is!

    Every tangent vector \alpha'(0) \in T_{\phi(p)} S_2 looks like d_p \phi (\phi^{-1} \circ \alpha) ' (0), so we have surjectivity \iff we have injectivity \iff bijectivity

    since the linear map is on a finite dimensional vector space. I guess there is a more general way of doing this
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 27th 2011, 03:42 PM
  2. Gauss Map on a Regular Surface
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: January 2nd 2011, 07:55 PM
  3. surface is regular if curvature is non zero
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 12th 2009, 02:26 AM
  4. regular surface
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 26th 2008, 03:36 PM
  5. regular surface
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 16th 2008, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum