# Find the geodesics of a surface with orthonormal parametrization

Suppose a regular surface $$S$$ admits an orthonormal parametrization $$x(u,v):$$ $$(u,v)\in U$$, describe it's geodesics that pass through a point $$p\in x(U)$$.

I begin by supposing that $$\gamma$$ is a geodesic on $$x(U)$$. As $$\{x_u, x_v\}$$ is an orthonormal basis of $$T_pS$$ I can write the following

$$0=\nabla_D\gamma'=\langle\gamma'',x_u\rangle x_u+ \langle\gamma'',x_v\rangle x_v$$

for this is the projection to $$T_pS$$ of $$\gamma''$$ which is my definition for the covariant derivative of $$\gamma'$$.

From that I deduce that $$\gamma''$$ is orthogonal to both $$x_u$$ and $$x_v$$ so it must be parallel to the normal vector $$N(p)$$.

From this I can also obtain the following:

1. The geodesic curvature of such a $$\gamma$$ must be $$0$$
2. $$||\gamma ' ||$$ must be constant
3. $$\gamma ' \perp \gamma''$$ hence $$\gamma$$ is planar (contained within a plane)
4. The plane within which $$\gamma$$ is contained must contain $$\gamma''$$ (and thus $$N(p)$$) and $$\gamma '$$
5. The plane within which $$\gamma$$ is contained is $$Span(N(p),\gamma')+p$$

But from here on I don't know what else to do to to further describe $$\gamma$$ either in terms of $$x_u, x_v$$ or not.

Also I believe There might be something wrong here because another result is that if all geodesics of a connected surface are planar then the surface is a section of a sphere or a plane. And for sure there are surfaces with orthogonal parametrizations not contained in planes or spheres.

Any help is greatly appreciated.

• If there’s an orthonormal parametrization, then the surface is locally isometric to the Euclidean plane. Not so for orthogonal. Nov 17, 2021 at 4:40

Solution:

Using the local expresion for $$\nabla_D \gamma '$$ and the fact that many Christoffel symbols vanish, we managed to prove that such a $$\gamma$$ must verify the equation:

$$\gamma'(t)=ax_u(x^{-1}\circ\gamma(t))+bx_v(x^{-1}\circ\gamma(t))$$ for some fixed $$a,b\in \mathbb{R}$$ and for all $$t$$.

This could also be expresed by $$\gamma'=dx|_{x^{-1}\circ\gamma}{{a}\choose{b}}$$

Now writing $$\gamma=x\circ\alpha$$ for some curve $$\alpha:I\to U$$, $$\gamma'$$ has local expression $$dx|_{x^{-1}\circ\gamma}{{a}\choose{b}}=\gamma'=(x\circ\alpha)'=dx|_{\alpha}\alpha'=dx|_{x^{-1}\circ\gamma}\alpha'$$

Using the fact that $$dx$$ is locally inyective we get that $$\alpha'={{a}\choose{b}}$$ constantly, thus $$\alpha$$ is a straight line within $$U$$. So every geodesic is the image under $$x$$ of a straight line.

Another, easier way to show the same fact is that as the first fundamental form of $$x$$ must, by the orthonormality hypothesis, be the identity matrix, $$x$$ results an isometry between $$U$$ and $$x(U)$$ and a result we have is that geodesics are preserved under isometries. So the geodesics of $$x(U)$$ must be the image under $$x$$ of the geodesics of $$U$$ which are straight lines.