5
$\begingroup$

A moment's thought (or some careful examination of maps) reveals that any circle drawn on a globe is in fact a circle in real life. The same claim holds for circles drawn on flat surfaces (obviously).

Are these the only examples?

Formally, let $M$ be any Riemannian surface, (isometrically) embedded in $\mathbb{R}^n$. Moreover, suppose that any metric circle $C$ (i.e., a locus of points with constant geodesic distance from some fixed point) is in fact a circle in $\mathbb{R}^n$. Then must $M$ have constant nonnegative Gaussian curvature?

$\endgroup$
1
  • 1
    $\begingroup$ This is not quite the setting of your question, but I’ll comment anyhow that geodesic circles in the hyperbolic plane are Euclidean circles (but with different centers). $\endgroup$
    – Ted Shifrin
    Sep 20, 2022 at 20:08

1 Answer 1

Reset to default
1
$\begingroup$

Yes!

First, note that such a surface must have nonnegative curvature. For: fix a point $p$. Working in geodesic polar coordinates about $p$, the group $\mathrm{SO}(\mathbb{R}^2)\cong S^1$ acts on any connected surface by rotating each metric circle. Since our metric circles are genuine circles, the resulting action is an isometry on some geodesic ball around $p$. But rotation by $90^\circ$ interchanges the principal curvatures; thus both curvatures are equal. Since the Gaussian curvature is the product of the two curvatures, it is a perfect square and so nonnegative.

Second, since genuine circles have constant geodesic curvature, this Q&A thread proves that $M$ must have constant curvature.

$\endgroup$
16
  • 1
    $\begingroup$ "Since our metric circles are genuine circles, M is invariant under this action". I guess you wanted to say: "since our metric circles are genuine circles, $SO(2)$ acts by isometries". Because $M$ is invariant by definition of action. Maybe I am not understanding the point $\endgroup$
    – A. J. Pan-Collantes
    Sep 20, 2022 at 5:24
  • $\begingroup$ Why must the curvature be nonnegative? $\endgroup$
    – Didier
    Sep 20, 2022 at 7:54
  • $\begingroup$ In general your $S^1$ action is not defined on the whole surface. $\endgroup$
    – Arctic Char
    Sep 20, 2022 at 8:14
  • $\begingroup$ @AntonioJPan That's exactly what I meant; thanks for the proper phrasing. $\endgroup$
    – Jacob Manaker
    Sep 21, 2022 at 18:40
  • 1
    $\begingroup$ In more detail: working in $\mathbb{R}^3$, take a point $p$ in the submanifold $M$. It seems you're trying to claim that the $S^1$ action on $\mathbb{R}^3$ which fixes $p$ and rotates the tangent plane at $p$ is an isometry of a small neighborhood $U\subseteq M$ of $p$. But it's not clear to me that the plane containing each metric circle about $p$ in $U$ is parallel to $T_pM \subseteq T_p \mathbb{R}^3$. If any one of those planes isn't parallel, the $S^1$ action need not map $M$ to itself. $\endgroup$
    – Jason DeVito - on hiatus
    Sep 22, 2022 at 15:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .