Classical question about constant gaussian curvature and geodesics circles.

**charles01** · December 2nd 2012, 08:59 AM

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!

**xxp9** · December 2nd 2012, 09:59 AM

don't understand your question. All geodesics have geodesic curvature 0, which is obviously constant. Do you mean "curvature" instead of "geodesic curvature"?

**charles01** · December 2nd 2012, 10:27 AM

No, in this question we have to use geodesic polar cordinates. Choose in the plane TpS, p $\in$ S a system of polar coordinates $(r, \theta)$ where r is the polar radius an $\theta$ is the polar angle. Using the exponential map, We may parametrize the points of the V $\in$ S by the $(r, \theta)$ coordinates. By geodesics circles I mean: fix r and make $\theta$ vary in [0, $2\pi)$ . E.g the images by exp_p:U $\rightarrow$ V of circles in U centered in 0.
Is of that circles I have to prove the afirmation.
Thanks!

**xxp9** · December 3rd 2012, 05:31 PM

The following argument is wrong and I am going to edit tomorrow.
Choose a point o on S and use $(r, \theta)$ as the parameterization. Denote $T = \frac{\partial}{\partial r}$ , and $J = \frac{\partial}{\partial \theta}$ as the coordinate vectors. From the definition of the exponential map, we have $\langle T, T \rangle = 1$ , and $\langle T, J \rangle = 0$ , and $[T, J] = 0$ . Denote N as the unit vector at the direction of J, so that $J = f N$ , where $f = |J|$ is smooth. $f(0)=0$
The Jacobi equation says that $D_T D_T J + R(J, T)T = 0$ , which yields $\frac{{\partial}^2f}{{\partial}r^2} + f K=0$ , where K is the Gaussian curvature as a constant. Solve the equation ( e.g. $f(r)=a sin(\sqrt{K} r)$ when K > 0 ), we find that f doesn't depend on $\theta$ at all, that is, $|J|=|\frac{\partial}{\partial \theta}|$ is constant on the geodesic circle $r = const$ .
Let $R_{\alpha}$ be the rotation of any angle $\alpha$ around 0 at the tangent plane, then it is obvious that the map $g: S \rightarrow S, g: p \mapsto exp_o \circ R_{\alpha} \circ exp_o^{-1}(p)$ 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.

**charles01** · December 4th 2012, 02:19 AM

Thank you for the help!!

**xxp9** · December 6th 2012, 02:06 PM

Choose a point p on S and let $(r, \theta)$ be the polar coordinates in the tangent plane $S_p$ . Since $exp_p: S_p \rightarrow S$ is a diffeomorphism in a neighborhood of p, we can use $(r, \theta)$ as the polar coordinates of S. Denote $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ the coordinate vectors of the tangent plane, denote T and J the coordinate vectors of S, such that $T_p=dexp_p(\frac{\partial}{\partial r}(0))$ and $J_p=dexp_p(\frac{\partial}{\partial \theta}(0))$ .
Denote N as the unit vector at the direction of J. Let $f = |J|$ is smooth, $f(0)=0$ , and $J = f N$ . From the definition and properties of the exponential map, we have $\langle T, T \rangle = 1$ , and $\langle T, J \rangle = 0$ , $[T, J] = 0$ , and $D_T T=D_T N=0$ .

The Jacobi field equation says that $D_T D_T J + R(J, T)T = 0$ . It is easy to verify that $\langle D_T D_T J, T \rangle = 0$ so from the equation we'll get $\frac{{\partial}^2f}{{\partial}r^2} + f K=0$ , where K is the Gaussian curvature as a constant. We have already known that f(0)=0.

To compute $\frac{\partial f}{\partial r}(0)$ , we compute $D_T J=D_T(fN)=\frac{\partial f}{\partial r} N$ . Use normal coordinates around p one can get that $D_{T(0)}(J)=\frac{\partial}{\partial \theta}|_{(1, \theta)}$ so we get that $\frac{\partial f}{\partial r}(0)=1$ .

So the initial conditions of f doesn't depend on $\theta$ . So solve the equation ( e.g. $f(r)=\frac{1}{\sqrt{K}}sin(\sqrt{K} r)$ when K > 0 ), we find that f doesn't depend on $\theta$ at all, that is, $|J|$ is constant on the geodesic circle $r = const$ .

Let $R_{\alpha}$ be the rotation of any angle $\alpha$ around 0 at the tangent plane, then it is obvious that the map $\hat{R_{\alpha}}: S \rightarrow S, \hat{R_{\alpha}}: q \mapsto exp_p \circ R_{\alpha} \circ exp_p^{-1}(q)$ 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.

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