On the surface of the sphere $S_2$, every point has exactly one antipode (a point at the maximum distance from it). I believe this is also true in a circle $S_1$, and a flat torus $S_1 \times S_1$. Is it possible to have a surface in which every point $P$ has precisely two antipodal points - two points, $P_1$ and $P_2$, both the same distance from $P$, that are distinct from each other, and neither of which can be perturbed without decreasing the distance to $P$?
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3$\begingroup$ @CedricBrendel I think the OP wants surface in which every point has exactly two antipodes but their use of the word "any" is ambiguous. $\endgroup$Jack Edward Tisdell– Jack Edward Tisdell2026-06-11 17:26:12 +00:00Commented Jun 11 at 17:26
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2$\begingroup$ Your torus example would suggest that you mean (intrinsic) geodesic distance. Is this right? The notion of antipode may change if you instead mean, say, the restriction of the Euclidean distance in some embedding. E.g., an equilateral triangle in the plane is intrinsically isometric to $S^1$ but under the Euclidean distance in the plane, each corner has two antipodes (the other corners). (In the embedded distance case, I wonder of some fractal curve might have the property you want.) $\endgroup$Jack Edward Tisdell– Jack Edward Tisdell2026-06-11 17:34:09 +00:00Commented Jun 11 at 17:34
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2$\begingroup$ On $r^2 (2+\cos(3 \theta)) + z^2 = 1$ maybe almost every point has exactly two antipodes. $\endgroup$Matt F.– Matt F.2026-06-11 17:40:48 +00:00Commented Jun 11 at 17:40
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1$\begingroup$ It's something like a sphere that's cinched in along three lines of longitude at 0º, 120º, and 240º. I'm not sure it fulfills your requirement though $\endgroup$Jack Edward Tisdell– Jack Edward Tisdell2026-06-11 17:53:40 +00:00Commented Jun 11 at 17:53
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2$\begingroup$ Given that you mean intrinsic distance, I think including the circle $S^1$ as an example is confusing. Every closed curve with its geodesic distance is isometric to $S^1$ and has unique antipodes half way around. $\endgroup$Jack Edward Tisdell– Jack Edward Tisdell2026-06-11 18:12:36 +00:00Commented Jun 11 at 18:12
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The torus obtained as the quotient of $\mathbb C$ by the lattice of Eisenstein integers (spanned by 1 and a sixth root of unity) has that property. As a fundamental domain one can choose a parallelogram consisting of two equilateral triangles. Then the point on the torus corresponding to the vertices of the parallelogram has exactly two points at maximal distance from it, which are the centers of the two equilateral triangles.
It would interesting to see if there are higher-dimensional examples of this sort.
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8$\begingroup$ The same is true of every torus obtained by identifying opposite sides of a non-square rhombus. Cutting along the short diagonal, the circumcenters of the resulting triangles are the two maximal distance points from the (identified) vertices. (In the square case, the circumcircles coincide.) I suspect the OP's intuition about looping around in two directions led him to the torus but the square torus fails this property for a different reason! $\endgroup$Jack Edward Tisdell– Jack Edward Tisdell2026-06-11 18:32:11 +00:00Commented Jun 11 at 18:32
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4$\begingroup$ As for higher-dimensional examples of this sort: Generalizing Jack Edward Tisdell's comment, it appears that almost any flat n-dimensional torus — with fundamental domain an n-dimensional parallelepiped — has exactly n! points at maximal distance from each point. Except for some highly symmetrical ones. $\endgroup$Daniel Asimov– Daniel Asimov2026-06-11 20:31:17 +00:00Commented Jun 11 at 20:31
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1$\begingroup$ @DanielAsimov, how do you see that for example for the 3-torus? $\endgroup$Mikhail Katz– Mikhail Katz2026-06-12 07:46:09 +00:00Commented Jun 12 at 7:46
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$\begingroup$ Let me think about that. $\endgroup$Daniel Asimov– Daniel Asimov2026-06-12 14:04:11 +00:00Commented Jun 12 at 14:04
