Download Possible Trajectories - Differential Geometry - Solved Exam and more Exams Computational Geometry in PDF only on Docsity!
Math 5378, Differential Geometry Solutions to practice questions for Test 2
- Find all possible trajectories of the vector field w(x, y) = (−y, x) on R^2. Solution: A trajectory would be a curve (x(t), y(t)) satisfying x′^ = −y, y′^ = x, and hence x′′^ = −x. Therefore, we would have x(t) = a cos(t) + b sin(t) for some constants a, b, and y = −x′^ = a sin(t) − b cos(t).
- If the first fundamental form in coordinates is given by E = eu, F = 0 , G = ev, find a vector field of unit length perpendicular to the vector field xu − xv. Solution: We apply the first fundamental form. For a vector field axu + bxv to be perpendicular, we require:
0 = 〈xu − xv , axu + bxv 〉 = aeu^ − bev^ ,
so b = aeu−v. To be unit length, we require:
1 = ||axu + aeu−vxv||^2 = a^2 eu^ + a^2 e^2 u−^2 vev,
so a = (eu^ + e^2 u−v^ )−^1 /^2. Plugging back in gives us the desired vector field.
- If f : S 1 → S 2 is an isometry between surfaces and α(s) : (a, b) → S 1 is a geodesic parametrized by arc length, show that f (α(s)) is also a geodesic parametrized by arc length. Solution: An isometry preserves the first fundamental form, and hence the lengths of vectors, so f (α(s)) has the same speed as α(s). Moreover, we have shown that the Christoffel symbols in coordinates depend only on the first fundamental form, and an isometry preserves the first fun- damental form, so the requirements that α(s) and f (α(s)) be geodesics are given by the same differential equations (which we state in a later problem).
- Suppose x is a coordinate chart on a surface, with coefficients E, F, and G of the first fundamental form. Prove the following identities.
〈xuu, xu〉 =
Eu
〈xuu, xv〉 = Fu −
Ev
Use these to show the matrix identity [ (^1) 2 Eu Fu − 12 Ev
]
[
E F
F G
] [
Γ^111
Γ^211
]
Solution: We recall that by definition, E = 〈xu, xu〉, F = 〈xu, xv〉, so their derivatives (by the Leibniz rule) are given by
Eu = 2 〈xuu, xu〉 , Ev = 2 〈xuv, xu〉 , Fu = 〈xuu, xv〉 + 〈xu, xuv〉.
Solving produces the first required formulas. To get the matrix equation, we take the equation
xuu = Γ^111 xu + Γ^211 xv + eN
and take dot products with xu and xv (both perpendicular to N) to get formulas as follows:
〈xuu, xu〉 = Γ^111 E + Γ^211 F 〈xuu, xv〉 = Γ^111 F + Γ^211 G
Plugging in for the inner products and rewriting this in matrix form gives the second desired formula.
- Prove that the sphere of radius R > 0 centered at the origin has con- stant Gaussian curvature 1/R^2 and mean curvature 1/R. Solution: A normal vector field on this sphere is given by N(x, y, z) = (x/R, y/R, z/R) (which is a unit normal vector), or N(v) = (^) R^1 v, which
and the tangent vectors at the point α(s) are
xu = (f ′(s) cos θ, f ′(s) sin θ, g′(s)), xv = (−f (s) sin θ, f (s) cos θ, 0).
The first coincides with α′(s), which is perpendicular to α′′(s) because α is parametrized by arc length. The second we can see is perpendicular to α′′(s) by direct calculation. Therefore, since α′′(s) is perpendicular to the tangent space, the co- variant derivative Dα′/ds is zero.
- Explain the sequence of steps (without calculating anything) taken to derive the Mainardi-Codazzi equations relating Christoffel symbols to e, f, and g from the formulas for xuu, xuv, and xvv. Solution: This is a vague question, but one basic idea is the following: - We start with the equations
xuu = Γ^111 xu + Γ^211 xv + eN, xuv = Γ^112 xu + Γ^212 xv + f N.
- We apply the identity (xuu)v = (xuv)u, and plug these equations into both sides.
- We take the dot product with the unit normal vector N (or equiva- lently ignore the xu and xv components of the result) remembering that Nu and Nv are perpendicular to N.
The resulting equation is one of the Mainardi-Codazzi equations; we get the other one by looking at (xuv)v = (xvv )u and comparing normal components.
- Find the absolute value of the geodesic curvature of the curve (cos t cos θ, sin t cos θ, sin θ) on S^2 for any fixed value of θ. Solution: We first note that this curve moves at speed cos θ, and so we reparametrize by arc length as
β(s) = (cos(s/ cos θ) cos θ, sin(s/ cos θ) cos θ, sin θ).
This curve has tangent vector
β′(s) = (− sin(s/ cos θ), cos(s/ cos θ), 0),
and second derivative
β′′(s) = (− cos(s/ cos θ)/ cos θ, − sin(s/ cos θ)/ cos θ, 0),
The length 1/ cos θ of this vector is the curvature k. The unit normal vector at β(s) is β(s), and so the length of the normal component is the normal curvature
kn = |β(s) · β′′(s)| = | − cos^2 (s/ cos θ) − sin^2 (s/ cos θ)| = 1.
Since the geodesic and normal curvatures satisfy k^2 = k^2 n + k^2 g , we find
|kg| =
cos θ
- On a sphere of radius R > 0, suppose that we have a triangle with three geodesic sides, with interior angles θ 1 , θ 2 , and θ 3. Find the area of the triangle. Solution: The sphere has constant Gaussian curvature 1/R^2. There- fore, applying the Gauss-Bonnet theorem to this triangle, we find
Area(T )/R^2 = θ 1 + θ 2 + θ 3 − π,
or Area(T ) = R^2 (θ 1 + θ 2 + θ 3 − π).
- Show that on a surface of nonpositive curvature, there are no simple closed geodesics that bound simple regions. Solution: If such a simple closed geodesic existed bounding a simple region R (which has Euler characteristic 1) then both the geodesic curvature and angle terms would vanish from the equation in the Gauss- Bonnet theorem. We would find
0 ≥
R
K = 2π,
a contradiction.
We parametrize the circle x^2 + y^2 = 1 as (cos t, sin t) for 0 ≤ t ≤ 2 π. Then in these coordinates, we find the following.
w(x, y) = (cos^2 t − sin^2 t, 2 cos t sin t) = (cos(2t), sin(2t)).
In the interval [0, 2 π], this vector field rotates by an angle of 4π, and therefore the index is 2.
