# Parameter Curves are Geodesics

So let's suppose we have a surface $M$ that is embedded in $\mathbb{R}^3$ with an orthogonal parametrization. Further, assume that the parameter curves (i.e., $X(u$0$, v)$ and $X(u, v$0$)$ ) are geodesics that are unparametrized (i.e., not necessarily unit speed).

What can we say about the Gauss curvature of $M$?

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I do believe (entirely from intuition) that Gauss curvature at a specific point $X(u_0, v_0)$ has the same sign as the scalar product of the second derivatives of the two curves at that point. Also, in an informal way, proportional to that scalar product, and inverse proportional to the size of the first derivative of the curves. – Arthur Dec 10 '11 at 15:54
By "unparametrized" do you mean "affinely parametrized?" If you actually mean unparametrized then the parameter curves in an orthogonal parametrization are automatically geodesics, and so this is basically no restriction on the curvature. If you do mean "affinely parametrized," I would try writing down the geodesic equations for the geodesics pointing in the $u$ and $v$ directions to get a system of equations that the metric coefficients must satisfy and then go from there. – treble Mar 5 '12 at 0:35

The Gaussian curvature must be $0$.

Let us write $E = \langle X_u, X_u\rangle$ and $G = \langle X_v, X_v\rangle$ for the coefficients of the first fundamental form. We'll make use of two formulas:

Fact 1: For an orthogonal parametrization, the geodesic curvature of the $u$- and $v$-parameter curves are given by $$(\kappa_g)_{u = u_0} = \frac{-E_v}{2E\sqrt{G}}$$ $$(\kappa_g)_{v = v_0} = \frac{G_u}{2E\sqrt{G}}$$

Since the parameter curves are geodesics, it follows that $E_v = G_u = 0$.

Fact 2: For an orthogonal parametrization, the Gaussian curvature is given by $$K = \frac{-1}{2\sqrt{EG}}\left[ \frac{\partial}{\partial u}\left( \frac{G_u}{\sqrt{EG}} \right) + \frac{\partial}{\partial v}\left( \frac{E_v}{\sqrt{EG}} \right) \right].$$

From this, it follows that $K = 0$.

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A very nice question! My intuition tells me it has to be ZERO everywhere but I am not so sure. How to visualize it: take a square made out of paper and imagine the parameter curves. They form a grid. Having these things as geodesics after a diffeomorphic transformation (i.e. bend it in a manifold like manner) makes the surface resemble a cylinder in every neighbourhood of a point. And this makes sure at least one of the principal curvatures is zero, hence their product. What would happen if it weren't the case? well, any non zero gaussian curvature point is either a spherical or a hyperbolical point. what happens in that vicinity of that point? You can find at least two "perpendicular" $X(u_0, v)$ and $X(u, v_0)$ curves that.. well, won't really be geodesics. Why? If they were, you could find an $\epsilon$ such that $X(u_0 + \epsilon, v)$ is in the same vicinity/neighbourhood/ball on the surface centered at that point.. and since that vicinity resembles a sphere or a hyperbolic paraboloid.. you know it can't happen, they won't parallel transport vectors anymore.. hence not geodesics.

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$\int k_g ds +\int K dA + \Sigma Jumps Angles @ corners = 2 \pi$
The terms are: $0 + \int K dA + 4 \pi/2 = 2 \pi$ leaving $K =0$ and $\int K dA = 0$ as solutions to the question.
$K=0$ case: Flat planes, cylinders, cones and tangential developables along two lines of stiction and their two normals.
$\int K dA = 0$ case: It includes zero Euler characteristic $\chi$ toroidal segments spanning both inner negative and outer positive Gauss curvature regions contributing integral curvatures equal $\int K dA$ in magnitude but opposite in sign. Riemannian metric is flat,it is a toroidal cut segment included between any two meridional cuts of a circular Clifford Torus.