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Math Help - Curvature With Parallel Surfaces

  1. #1
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    Curvature With Parallel Surfaces

    I have a two part question, one of which I have no clue of how to approach, the other has been giving me a difficult time for the last three days. The questions are as follow:

    Le M be an arbitrairy orientable surface, with U the unit normal vector. Define the following map: F: M \mapsto \mathhbb{R}^3 by F(p)=p+\varepsilon U(p)

    i) Show that the canonical isomorphisms of \mathhbb{R}^3 make U a unit normal on \overline{M}=F(M) for which \overline{S}(\overline{v})=S(v), where v are vectors, and S(v) indicates the shape operator.

    I have no idea what's being asked here.


    ii)Derive the following formulas of the Gaussian and Mean Curvature

    \overline{K}(F) = \frac{K}{J} where J = (1-\varepsilon k_1)(1-\varepsilon k_2) and k_i denote the principal vectors

    Before I begin I managed to deduce from a previous problem that (\overline{v})\times(\overline{w}) = J(p)v\times w.

    Now I would imagine that since K=det(k_1,k_2) that I would simply have F(k_1)\times F(k_2) wich would simply give  J(p)k_1 \times k_2 which isn't the desired result.

    So looking at both RHS and LHS I think I would get:

    F(k_1)\times F(k_2) = (k_1+\varepsilon k_1) \times (k_2+\varepsilon k_2)= k_1k_2(1-\varepsilon)^2

    then from the RHS I get :

    \frac{k_1k_2}{(1-\varepsilon k_1)(1-\varepsilon k_2)} which has lead nowhere.

    I'm not looking for the answer, but any hints would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I have a two part question, one of which I have no clue of how to approach, the other has been giving me a difficult time for the last three days. The questions are as follow:

    Le M be an arbitrairy orientable surface, with U the unit normal vector. Define the following map: F: M \mapsto \mathhbb{R}^3 by F(p)=p+\varepsilon U(p)

    i) Show that the canonical isomorphisms of \mathhbb{R}^3 make U a unit normal on \overline{M}=F(M) for which \overline{S}(\overline{v})=S(v), where v are vectors, and S(v) indicates the shape operator.

    I have no idea what's being asked here.
    So F "moves" M a short distance parallel to itself. What are "the canonical isomorphisms of \mathhbb{R}^3" and what is "the shape operator"?


    ii)Derive the following formulas of the Gaussian and Mean Curvature

    \overline{K}(F) = \frac{K}{J} where J = (1-\varepsilon k_1)(1-\varepsilon k_2) and k_i denote the principal vectors

    Before I begin I managed to deduce from a previous problem that (\overline{v})\times(\overline{w}) = J(p)v\times w.

    Now I would imagine that since K=det(k_1,k_2) that I would simply have F(k_1)\times F(k_2) wich would simply give  J(p)k_1 \times k_2 which isn't the desired result.

    So looking at both RHS and LHS I think I would get:

    F(k_1)\times F(k_2) = (k_1+\varepsilon k_1) \times (k_2+\varepsilon k_2)= k_1k_2(1-\varepsilon)^2

    then from the RHS I get :

    \frac{k_1k_2}{(1-\varepsilon k_1)(1-\varepsilon k_2)} which has lead nowhere.

    I'm not looking for the answer, but any hints would be greatly appreciated.
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  3. #3
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    The first question is to prove that M and \overline{M} have the same normal vectors, that is, \overline{U} \circ F=U. Note that the differential of F,
    d\mathbf{F}(v) = v + \varepsilon \cdot d\mathbf{U}(v), where v is any vector tangent to M at p.
    Also note that U is nothing but the Gauss map
    \mathbf{U} : M \mapsto S^2
    Thus -d\mathbf{U} is the Weingarten map \mathbf{S}.

    Identifying the tangent plane of M at \mathbf{p} and the tangent plane of S^2 at \mathbf{U}(\mathbf{p})( since they're parallel), we can see that
    d\mathbf{U}(v)=-\mathbf{S}(v) is tangent to M.
    Thus d\mathbf{F}(v) is orthogonal to U(p), we proved that the tangent plane of \overline{M} is orthogonal to U(p).
    We're done for i).
    I'm going to prove ii) in another post.
    Last edited by xxp9; August 12th 2010 at 06:04 PM.
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  4. #4
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    To see ii), note that for the two principle directions v_1 and v_2,
    \mathbf{S}(v_1) = k_1 \cdot v_1 and \mathbf{S}(v_2) = k_2 \cdot v_2, where k_1 and k_2 are the principle curvatures.
    So d\mathbf{F}(v_1) = v_1 - \varepsilon \mathbf{S}(v_1) = (1-\varepsilon k_1)\mathbf{v_1}
    d\mathbf{F}(v_2) = (1-\varepsilon k_2)\mathbf{v_2}
    Since v_1 and v_2 are orthogonal we can easily deduce that the pull back of the volume element
    \mathbf{F}^{*}(dV_{\overline{M}}) = (1-\varepsilon k_1)(1-\varepsilon k_2)dV_{M}, where dV_{\overline{M}} and dV_M are the volume elements of \overline{M} and M, respectively.
    Note that the Gaussian curvature K=\frac{\mathbf{U}^*(d\sigma)}{dV_M}, where d\sigma is the volume element of S^2 and the similar formula for \overline{M}, and \overline{U} \circ F = U we're done.
    Last edited by xxp9; August 12th 2010 at 06:06 PM.
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