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Is there a classification for surfaces on which we can specify a grid that forms right-angles at every intersection without singularities? For example, we can do this on the plane, cylinder, torus, and mobius-strip, but not the cone nor sphere which (I think) will always have singularities when we try. It seems related to the idea of a developable surface but not quite.

examples

Do such surfaces have any special importance for this fact? I understand that coordinates are, by nature, quite arbitrary, but this is a question about the existence of coordinates with a certain property, so perhaps it can be more fundamentally related to the topology of the surface. Is there a formalization of this concept in terms of a manifold's atlas? Or perhaps something about an everywhere diagonal metric-tensor? (Just spitballing).

Edit: The reason the cone fails is obvious: it isn't a regular surface / manifold, i.e. it has a "sharp point." Meanwhile, the reason the sphere fails might be a result of the Hairy-Ball Theorem? I'm beginning to think that finding an orthogonal grid doesn't make a manifold special. What would make a manifold special is proving that no such grid can exist, which may require a Hairy-Ball-like theorem. Still open to answers!

jnez71
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  • Why does the non-existence of a grid on the sphere is a consequence of the hairy-ball theorem? Also, I think there is something here to do with Euler characteristic numbers. – Didier Jan 13 '21 at 23:04
  • @DIdier_ : I could be wrong, but it seems to imply one can't lay down any singularity-free coordinates because the corresponding tangent vector field must vanish somewhere. Definitely open to corrections! – jnez71 Jan 13 '21 at 23:09
  • @DIdier_ At least the Euler characteristic argument seems to work for closed surfaces. Which means the torus (and depending on your definitions the Klein bottle) are the only two that work. As for surfaces with boundary, or non-compact surfaces, or non-smooth surfaces, I am less certain. – Arthur Jan 13 '21 at 23:11
  • No, if you had local coordinates $(u,v)$ everywhere where the coordinate vectors $\partial/\partial u, \partial/\partial v$ were everywhere an orthonormal basis, then the surface would have to be flat (but, as you've seen with the torus, not necessarily ruled). As you've seen with the Möbius strip, you have a ruled, but not developable (flat) surface. If the grid were by geodesics everywhere, you could also prove the surface would have to be flat. – Ted Shifrin Jan 13 '21 at 23:31
  • But, yes, if you have a grid, you can take the unit tangent vector to each grid line and get both a nowhere-vanishing vector field and, in fact, an (orthogonal in your case) trivialization of the tangent bundle. (For surfaces, one nowhere-vanishing vector field does that anyhow.) – Ted Shifrin Jan 13 '21 at 23:31
  • I am not sure what you mean, but it sounds like you are interested in what are called flat surfaces. That link gives the classification. (However, that doesn't embrace surfaces like your cone, that aren't smooth. Also, this is not a topological property: the surface of the cone is homeomorphic to the sphere or a disc - depending on whether you are including the base of the cone.) – Rob Arthan Jan 13 '21 at 23:33
  • @RobArthan : "flat Riemannian manifold" might be what I was looking for. I am willing to exclude the cone and just address Riemannian manifolds. Feel free to leave an answer with some insight on how the formal definition of "flat" there relates to the ability to lay everywhere-orthogonal grid-lines. (And to clarify, I mean parameterizing the surface such that a step in the $u$-direction and a step in the $v$-direction will always appear perpendicular to the "ant" taking the steps). – jnez71 Jan 13 '21 at 23:49
  • Just found this highly related Q/A. – jnez71 Jan 14 '21 at 01:54
  • This paper seems to have the results I was looking for. Definition 2.6 seems like the right way to formalize the "orthogonal grid" idea, and Figure 1 confirms the intuition. The whole of the document is spent discussing when such coordinates are possible, and there are many other related works now that I know the key term isothermal coordinates (it's a shame they're not called orthogonal coordinates!). I'll leave the question open for further discussion / posterity though. – jnez71 Jan 14 '21 at 02:17

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Here's an answer for finite type oriented surfaces by which I mean any surface $S$ that is diffeomorphic to an oriented surface $F$ which is closed --- meaning compact, connected, and empty boundary --- minus a finite subset $A \subset F$. (I suspect something similar works in the nonorientable case, but I think I"ll restrict my answer to the orientable case)

First, if $A = \emptyset$ and hence $S=F$ is closed then a necessary and sufficient condition is that the genus equals $1$.

I claim that if $A \ne \emptyset$ then $S$ has such a grid as you ask for.

First let me describe all genus $0$ examples, i.e. all examples where $F$ is a sphere minus $1$ or more points. If $A$ has just $1$ point then $S=F-A$ is diffeomorphic to $\mathbb R^2$, in which the horizontal/vertical lines give such a grid, as depicted in your question. If $A$ has $k \ge 2$ points then $S=F-A$ is homeomorphic to $\mathbb R^2$ minus $k-1$ points, and again the horizontal/vertical lines (punctured at $k-1$ points) gives such a grid.

Also one can do all genus $1$ examples, i.e. where $F$ is a torus, because $F$ itself has such a grid, and then you can remove any finite subset of points.

You can see from these examples that if the genus is fixed and if you can describe a grid on the once punctured surface of genus $g$, then you obtain a grid on a genus $g$ surface punctured at any nonempty finite set.

So, what does one do if $F$ is the closed surface of genus $2$, and $S$ is $F$ minus $1$ point? In this case, consider a regular octagon in the plane, and glue every side with the opposite parallel side, using a gluing map identifies each opposite pair of sides by a Euclidean translation. The quotient under this gluing is a closed surface of genus $2$. All 8 vertices of the octagon are identified to a single point of $F$ which we take to be $A$. Now puncture $F$ at that one point to give the surface $S=F-A$, so we can think of $S$ as the quotient of the octagon with its 8 vertices removed. What one observes is that if one restricts the horizontal/vertical lines of $\mathbb R^2$ to the octagon with vertices removed, and then maps those lines to the quotient surface $S=F-A$, the result is a grid as you desire.

And then this generalizes to any genus: take a regular planar $4g$-gon, remove its vertices, and glue opposite side pairs; the horizontal\vertical grid on the plane descends to a grid such as you ask for, on the genus $g$ surface puncture at $1$ point.

Lee Mosher
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