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Please verify my differential geometry results

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data



    Q1)

    One way to define a system of coordinants for a Sphere [tex]S^2[/tex] given by [tex]x^2 + y^2 + (z-1)^2 = 1[/tex] is socalled stereographical projection [tex]\pi \thilde \{N} \rightarrow R^2[/tex] which carries a point p=(x,y,z) of the sphere minus the Northpole (0,0,2) onto the intersection of the xy plane with the straight line which connects N to p.

    Let (u,v) = pi(x,y,z) where [tex]S^2 \thilde {N}[/tex] and [tex]u,v \in xy[/tex]

    1) show that [tex]pi^-1 = R^2 \rightarrow S^2[/tex] is given by

    [tex]x = \frac{4u}{u^2+v^2+4}, y = \frac{4v}{u^2+v^2+4}, z = \frac{2(u^2+v^2)}{u^2+v^2+4}[/tex]

    3. The attempt at a solution
    Solution (1)

    The line passing through N and (u,v,0) can be given by l(t) = t(0,0,2) + (1-t)(u,v,0) = ((1-t)u, (1-t)v, 2t)) When l intersecs with the unit sphere, we have (1-t)u)^2+(1-t)v)^2+ (2t-1)^2 = 1 and thus t =1 or \frac{u^2+v^2}{u^2+v^2+4} = (\frac{4u}{u^2+v^2+4}, \frac{4v}{u^2+v^2+4}, \frac{2(u^2+v^2)}{u^2+v^2+4})

    is then pi^-1(u,v)

    2) Show if possible using stereographical projection the sphere with two coordinant neighbourhoods.

    3. The attempt at a solution
    solution 2)

    Define F: S^2 -> S^2 by F(x,y,z) = (x,y,2-z). Then F is rigid motion of S^2 which the upper hemisphere to the lower hemisphere and vice versa. Now pi^-1: R^2 -> S^2 is a parameterization of S^2 which covers the whole sphere except N. Similary F o pi^-1: R^2 -> S^2is a parameterisation of S^2 missing the origin. So pi^-1, F o pi^-1 together covers the whole S^2, and these are both regular parameterization by definition.

    Q2)

    Prove that the absolute vale of the torsion tau at a point on an assymptotical curve, whose curvature is nowhere zero, is given by

    |tau| = sqrt(-K)


    3. The attempt at a solution

    Let α(s) be an asymptotic curve on S parametrized with arc length, with curvature k ≠ 0,
    and with α(0)=p.
    First, note that N, the unit normal vector to S at p, is normal to the osculating plane of
    α(s) at p: Clearly, N is normal to the tangent vector α’(0), and, since α(s) is asymptotic:
    kn = 0 = k <n, N> (by definition of kn, p. 141).
    Since k ≠ 0, this means that N is also normal to n (the normal vector of α(s)).
    Thus, for some orientation of S, N coincides with the binormal b = α’ ^ n, then,
    N’(s) = b’ = τ n, and:
    |N’(s)|2 = τ2. (1)
    On the other hand, N’(s) is normal to N, and therefore belongs to the tangent plane of S
    at α(s), and can be expressed in terms of the unit vectors along the principal directions
    of S at α(s), e1 and e2 (which form an orthonormal basis of the tangent plane):
    N’(s)= < N’(s), e1> e1 + < N’(s), e2> e2
    Now let’s compute these componets: if θ is the angle formed by α’(s) and e1, we have:
    < N’(s), e1> = <dN(α’(s)), e1> = <dN(e1 cos θ + e2 sin θ), e1> =
    = <cos θ dN(e1) + sin θ dN(e2), e1> = cosθ <dN(e1), e1> + sin θ <dN(e2), e1> =
    = cosθ < – k1e1, e1> + sin θ < – k2e2, e1> = – k1 cosθ.
    Similarly, we get: < N’(s), e2> = – k2 sinθ, so that
    N’(s)= – k1 cosθ e1 – k2 sinθ e2
    and
    |N’(s)|2 = <N’(s), N’(s)> = k1
    2 cos2 θ + k2
    2 sin2 θ
    On the other hand, according to the Euler formula, the normal curvature of the
    asymptotic curve α(s) is:
    kn= 0 = k1 cos2 θ + k2 sin2 θ = k1 cos2 θ + k2 (1- cos2 θ)
    yielding:
    cos2 θ = − k2 /( k1 – k2),
    sin2 θ = k1 /( k1 – k2).
    Summarizing,
    |N’(s)|2 = – k1
    2 k2 /( k1 – k2) + k2
    2 k1 /( k1 – k2) = – k1 k2 = –K
    and using (1):
    |N’(s)| = |τ| = sqrt(–K).

    q3)

    Let x: U \subset R^2 -> R^3 where

    U = (omega, phi) \in R^2, 0 < \omega < pi, 0 <\phi <2pi

    x(omega, phi) = sin(omega)cos(phi), sin(phi), cos(omega)

    be a parameterisation of the unit sphere S^2 let

    log tan 1/2 omega = u, phi = v

    and show that the new parameterization of the coordinant neighbourhood x(U) = V

    is given by

    y(u,v) = (sech u cos v, sech u sin v tanh u)

    Solution Q3)

    We can solve log tan(omega/2) = u, phi = v, for omega, phi,

    omega = 2arctan(e^u), pi, = v

    note that

    0<omega<pi corresponds to -infinity < u < inftiny
    0<omega<pi corresponds to 0 < v < 2pi

    If w set Y(u,v) = X(2arctan e^u, v) where X(omega, phi) = (sin(omega) cos(phi), sin(omega)sin(ph), cos(omega)))

    then Y:U' -> \subseteq S^2 is a chart on the sphere S^2 where U' = {(u,v) : 0 < v < 2pi}; V is the imega of X, name S^2 with a half great circle from the pole to South Pole removed.

    under the changed variable we find

    sin(omega)= sin(2 arctan(e^u) = 2sin(arctan(e^u)cos(arctan(e^u) = 2eu/1+e^2u

    cos(omega) = -tanh u

    So Y(u,v) = (sech u cos(v), sech u sin v, -tanh(u)) (Mistake in do Carmo?)

    From this we find E=G=sech^2(u), F=0, for Y, so Y is conformal and therefore Y^-1: V->U' is conformal as well. The meridians on the sphere corespondes to the constant value of omega, and we realise by constant values of U in U'(vetical lines) while the latittudes on the sphere corresponds to the constant value of phi, giving horisontal lines in U'.
     
    Last edited: May 26, 2009
  2. jcsd
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