don't understand your question. All geodesics have geodesic curvature 0, which is obviously constant. Do you mean "curvature" instead of "geodesic curvature"?
Hi everyone!
I am having problem in a classical question, I have to prove that in a regular surface with constant gaussian curvature the geodesics circles have constant geodesic curvature.
I don't konwm how to start...
thanks!
No, in this question we have to use geodesic polar cordinates. Choose in the plane TpS, p S a system of polar coordinates where r is the polar radius an is the polar angle. Using the exponential map, We may parametrize the points of the V S by the coordinates. By geodesics circles I mean: fix r and make vary in [0, . E.g the images by exp_p:U V of circles in U centered in 0.
Is of that circles I have to prove the afirmation.
Thanks!
The following argument is wrong and I am going to edit tomorrow.
Choose a point o on S and use as the parameterization. Denote , and as the coordinate vectors. From the definition of the exponential map, we have , and , and . Denote N as the unit vector at the direction of J, so that , where is smooth.
The Jacobi equation says that , which yields , where K is the Gaussian curvature as a constant. Solve the equation ( e.g. when K > 0 ), we find that f doesn't depend on at all, that is, is constant on the geodesic circle .
Let be the rotation of any angle around 0 at the tangent plane, then it is obvious that the map is an isometry since it keeps the length of both T and J. And we're done since geodesic curvature is invariant under the isometry.
Choose a point p on S and let be the polar coordinates in the tangent plane . Since is a diffeomorphism in a neighborhood of p, we can use as the polar coordinates of S. Denote and the coordinate vectors of the tangent plane, denote T and J the coordinate vectors of S, such that and .
Denote N as the unit vector at the direction of J. Let is smooth, , and . From the definition and properties of the exponential map, we have , and , , and .
The Jacobi field equation says that . It is easy to verify that so from the equation we'll get , where K is the Gaussian curvature as a constant. We have already known that f(0)=0.
To compute , we compute . Use normal coordinates around p one can get that so we get that .
So the initial conditions of f doesn't depend on . So solve the equation ( e.g. when K > 0 ), we find that f doesn't depend on at all, that is, is constant on the geodesic circle .
Let be the rotation of any angle around 0 at the tangent plane, then it is obvious that the map is an isometry since it keeps the length of both T and J. And we're done since geodesic curvature is invariant under the isometry.
The above argument skipped many technical details however you may verify all the assertions on the unit sphere. A standard book on Riemann geometry may be referred to for the skipped details. Actually a so-called space form is locally isometric to a portion of the n-sphere, or the plane, or the hyperbolic space, corresponding to whether that K>0, K=0 or K < 0.