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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.

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

1 Answer 1

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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.

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