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Shouldn't any sphere on any manifold always be perpendicular to geodesics through its center.

This seems obvious to me, since given any function of a distance from the center of a sphere is the equipotential surface, and the flow lines of the gradient of any function of distance should always be geodesics. The derivative of any such function along any curve on the sphere should be zero and since vector gradient is determined by dot product it should be orthogonal

Bernard
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Kugutsu-o
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1 Answers1

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Let $\Gamma$ be the group of isometries of $\mathbb{R}^2$ generated by horizontal translations of two units and vertical translations of one unit. Let $T$ be the torus obtained by taking the quotient of $\mathbb{R}^2$ by $G$. Note that $T$ can be represented by the rectangle $[-1, 1]\times[-\frac{1}{2},\frac{1}{2}]$ with edges identified by translations.

Consider the sphere $S$ in $T$ of radius $\sqrt{5}/2$ centered at $0$ in this representation. In fact, $S$ comprises exactly one point, so it is not meaningful to talk about whether it's perpendicular to any curves.

(For any radius $r\in (1,\sqrt{5}/2)$, the sphere of radius $r$ has points where it does not admit a smooth parametrization hence "perpendicular" is not meaningful at those points.)

Neal
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  • I think you might be mixing up representations of geometric objects and objects themselves. I don't see how could a sphere on a 2 dimensional surface meaningfully amount to just 2 points. – Kugutsu-o Jul 14 '20 at 16:29
  • To take to the extreme, the sphere with radius $100$ has no point at all. – Arctic Char Jul 14 '20 at 16:35
  • @Ezio the sphere of radius $2$ about $p$ is the set ${ y\in T\ |\ d(p,y) = 2}$. You can check it yourself, for every $p\in T$ the sphere of radius $2$ about $p$ contains precisely two such points. – Neal Jul 14 '20 at 16:41
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    @ArcticChar the issue with the sphere of radius 100 is that it is (vacuously) perpendicular to every line that intersects it. – Neal Jul 14 '20 at 16:43
  • As I said representations, coordinate representations, matrix representations could be misleading. Besides blowing up the radius is trivial. If I have a sprere with a circumference O, the radius of circles can't be bigger than O/2 thats obvious. How meaningful is it to talk about a sphere as an empty set? – Kugutsu-o Jul 14 '20 at 16:58
  • @Ezio Not sure what you mean by "could be misleading," if you wish, you can verify my calculation yourself. And if a sphere has no elements, then it is pretty clearly the empty set. – Neal Jul 14 '20 at 17:30
  • You plugged out a bunch of random rationals for one. That's what you learn in elementary school, work algebraic when you can ally not numerically. Second probably the misunderstanding stems from a deeper level, the very axiomatization of manifolds. Thirdly we are stuck on semantics. A 2 sphere is a 2 manifold itself. This intrinsic geometry. 1 sphere doesn't but it's still a manifold. A 0 sphere is not. So automatically all these degenerate cases of empty sets otherwise drop of, most trivial of which when you consider a set of equidistant points on a real line. That's not a manifold – Kugutsu-o Jul 14 '20 at 19:00
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    Let's be precise about "sphere," then. In the context of a question about distance functions, most people would take "sphere in $M$ of radius $r$ about $p$ to mean "${q\in M\ |\ d(p,q) = r}$." Do you have a different definition in mind? – Neal Jul 14 '20 at 19:39
  • OK, so is it safe to say that any smooth sphere (smoothly parametrisable) is orthogonal to every geodesic through its center? – Kugutsu-o Jul 14 '20 at 19:48
  • Af it is is correct, is it too trivial(weak) of a statement compared to the lemma in question – Kugutsu-o Jul 14 '20 at 19:51
  • If you have just a real line a sphere centered at a point P with radius R is just a set of two points . So it makes no sense to talk about orthogonality . So naturally I would assume to have any orthogonality you just need some sphere that is smooth. – Kugutsu-o Jul 14 '20 at 20:02
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    Yes, if you have a smooth distance sphere then it is orthogonal to geodesics emanating from its center. – Neal Jul 15 '20 at 14:27