I am trying to solve the following exercise:

Prove that on a surface of constant curvature the geodesic circles have constant curvature.

"Constant curvature" in case of the surface I take to refer to the Gaussian curvature. Now, the geodesic curvature of a curve parameterized by arc length in orthogonal coordinates is given by

$$k_g(s) = \frac{1}{2 \sqrt{EG}} \left(G_u v'- E_v u' \right)+ \phi',$$

where $\cdot'$ denotes the derivative with respect to $s$, and $\phi$ is the angle the tangent of the curve makes with $x_u$.

Using geodesic polar coordinates (setting $u = \rho$ and $v = \theta$), a surface with constant Gaussian curvature $K$ satisfies

$$(\sqrt{G}_{\rho\rho}) + K \sqrt{G} = 0$$

Also, we get $E=1$, $F=0$, and a geodesic circle has the equation $\rho = \mathrm{const.}$ Therefore, the first equation above yields

$$ k_g(s) = \frac{G_\rho \theta'}{2\sqrt{G}} $$

It seems to prove that $k_g$ is constant, you would have to show that its derivative is 0. I tried that, but the derivative gets rather ugly and I don't see how to proceed.

  • $\begingroup$ The meaning is unclear. The geodesics are supposed to have constant curvature with respect to what ambient space? The embedding of the surface has not been specified. $\endgroup$ Aug 10, 2014 at 10:01

3 Answers 3


Consider the three cases K=0, K>0 and K<0 in your second equation. In each case, you can use ODE theory (and the values of E, F and limits of G) to solve for G, which in all cases is independent of theta. Plug in the first fundamental form coefficients into your first equation for the geodesic curvature.


If you have textbook, Differential Geometry of Curves and Surfaces, Do CARMO,

Then See p.289, the part of Theorem by Minding. You will get $E=1,F=0,G=constant,G_ρ=constant$ in geodesic circles. (i.e, ρ=constant)

And see p.254, the part of Theorem by Liouville, curvature of curve ρ=constant, $k_{g2}=\large\frac {G_ρ}{2G\sqrt{E}}$. So k is constant.

  • $\begingroup$ Thank you very much. I was having the same question, this was a very simple way to solve it. $\endgroup$
    – Kernel
    Nov 24, 2015 at 0:34
  • $\begingroup$ Like $ k_{g2}=G_ρ/2G \sqrt{E} ? $ $\endgroup$
    – Narasimham
    Nov 5, 2016 at 7:10

If you're willing to allow some high-power Riemannian geometry theorems, the classification of constant curvature surfaces tells you that your surface is isometric to a sphere, the Euclidean plane, or the hyperbolic plane. It can then be checked by computation that all geodesic circles are constant-curvature.

  • 2
    $\begingroup$ Do you mean "... you surface is locally isometric to ..."? $\endgroup$
    – Neal
    Dec 11, 2012 at 0:14

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