Four mutually equidistant points on a 2D manifold?

Question

In a certain math joke, the humor comes from the insight that having four people who are each 6ft away from one another is absurd because it violates the Pythagorean theorem.

This made me wonder whether such a setup is actually possible in a certain space. I wondered whether you could find four points, all separated by the same geodesic distance, in some smooth manifold---maybe an elliptical space or a hyperbolic space where the Pythagorean theorem no longer applies.

I tried on a sphere (as a representative of an elliptic space), since I'm familiar enough with polar coordinates to do the calculations—but I couldn't find a configuration.

Is it possible after all? I'm looking for a hopefully elementary manifold (such as a sphere or hyperboloid) with four mutually equidistant points. If it is possible, I hope to construct an example where I could draw a picture and compute the geodesic lengths. (I think this is equivalent to saying that the surface can be embedded in 3D space.) Exotic topological spaces, e.g. discrete spaces, aren't useful here. I think this is about geometry, which is an area I'm less familiar with. I also know you can always embed $n+1$ points equidistantly in an $n$ dimensional space via a simplex— but here I'm hoping for a smooth 2D surface in 3D space.

How about the vertices of a regular tetrahedron on a sphere? — Michal Adamaszek, Sep 13 '22 at 12:26
@MichalAdamaszek That sounds right. What are the spherical coordinates of a tetrahedron on a sphere? (I don't know the geometry well enough.) — user326210, Sep 13 '22 at 12:28

score 1 · Accepted Answer · answered Sep 13 '22 at 16:29

Besides the tetrahedron on the sphere, which certainly works for 4 points, you can construct a manifold with any specified set of non-zero distances between any number of points. You can do this by gluing ribbons of manifold together.

For $n$ points, cut out $n(n-1)/2$ long narrow strips of paper, all of the same length. Then glue the ends together into a complete graph on $n$ points, one strip for each edge of the graph, allowing the strip to bend in the middle to accomodate the extra length. Paths along each strip between the end points are all the same length, there is no shorter route, and we have $n$ points on a manifold, all the same distance apart.

If you wanted a compact manifold, you can wrap each strip into a tube and glue long thin cylinders together instead.

Four mutually equidistant points on a 2D manifold?

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