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A classical result is that solutions of the isoperimetric problem on the plane, the sphere, and the hyperbolic plane are circles. Are there any other Riemannian metrics on these spaces that share this property or does this characterize metrics of constant curvature ?

I should add that by circles I mean really standard circles, not geodesic circles for the new metric.

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Could you say more by what you mean by a "standard circle" versus a "geodesic circle"? Are you assuming that the space has a "canonical" embedding into a Euclidean space of higher dimension? – Deane Yang Jun 26 '12 at 9:09
I mean this in the same sense that Hilbert's fourth problem asks for metrics whose geodesics are straight lines. The space (the plane, the sphere, the disc) is given and so are the usual circles, then one looks for metrics for which these circles solve the isoperimetric problem. – alvarezpaiva Jun 26 '12 at 9:15
So you are fixing one of three possible models of "standard circles" and asking what Riemannian metrics make them isoperimetric extremals? I believe that an extremal circle must have constant (extrinsic) curvature. So a related question seems to be: For which Riemannian metrics do the standard circles have constant curvature? Is anything known about this? Is this maybe equivalent to your question? – Deane Yang Jun 26 '12 at 9:24
Well, the classic geometries (two dimensional, complete, constant curvature) have all the same circles. Yes, in particular, the question asks for what Riemannian metrics on the plane do the standard circles have constant geodesic curvature (which is the constant mean curvature in dimension two). I guess that this should mean that geodesics are circles and thus that the only such metrics are of constant curvature ... – alvarezpaiva Jun 26 '12 at 11:10
up vote 20 down vote accepted

It is known (and follows from an easy calculation) that solutions of the isoperimetric problem on a surface have constant geodesic curvature. In his 1887 classic Leçons Sur La Théorie Générale Des Surfaces Et Les Applications Géométriques Du Calcul Infinitésimal, G. Darboux states, without indicating a proof, that a surface for which all of the curves of constant geodesic curvature are closed must have constant Gaussian curvature. According to Blaschke, a proof of this was published in 1921, by B. Baule in Math. Ann. 83 (286-310) and 84 (202-215), but I have never checked it out myself. I believe that this answers your question, at least for Riemannian metrics.

Concerning the Finsler case: (I had a little time over the holiday to think about this, and, after a brief calculation, came up with the following answer, which is interesting if you are willing to widen the question to considering 'Finsler metric-measure' spaces.)

Consider the following question: For what choices of Finsler metric $F$ on the plane and area form (i.e., 'measure) $\Omega$ can the solutions of the isoperimetric problem be the standard circles in the $xy$-plane? (Here, the isoperimetric problem is to be understood as finding domains of a given area with minimal perimeter where the 'perimeter' is measured by integrating $F$ over the boundary of the domain and the 'area' is measured by integrating $\Omega$ over the domain.)

The answer turns out to be this: Let $a$, $b$, $c$, $e$, $p$ and $q\not=0$ be constants, and let $D$ be the domain in the $xy$-plane where $a(x^2+y^2) + bx + cy + e >0$. Define the Finsler function $$ F = \phi + \frac{\sqrt{dx^2+dy^2}}{a(x^2+y^2) + bx + cy + e} $$ where $\phi$ is any $1$-form that satisfies $$ d\phi = \frac{p\ dx\wedge dy}{\bigl(a(x^2+y^2) + bx + cy + e\bigr)^2} $$ and has sufficiently small norm that $F$ is positive on nonzero tangent vectors in $D$. Meanwhile, take $$ \Omega = \frac{q\ dx\wedge dy}{\bigl(a(x^2+y^2) + bx + cy + e\bigr)^2} $$

Then the isoperimetric curves in $D$ for the metric-measure structure $(F,\Omega)$ are the standard circles in $D$. Moreover, any metric-measure structure on a domain in the plane that has this property is of the above form.

Note that the case $\phi=0$ is exactly the metrics of constant curvature that are conformal to the standard metric on the plane, as we already knew. The more general case is that of a so-called Randers metric of a particular kind, one whose 'associated' Riemannian metric is also a solution.

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Thanks Robert. Baule's paper looks interesting. It then seems likely that something non-trivial can be said about metrics that share the same solutions of the isoperimetric problem. Do you know if there are conditions on a third order differential equation on the plane (I'm thinking here of the work of Cartan and Chern on the invariants for these equations) so that its solutions are the constant curvature curves of some Riemannian metrics? – alvarezpaiva Jun 26 '12 at 15:31
These conditions can certainly be read off of Chern's work on third-order equations in the plane, but I think that that approach is overkill. Also, I think it would be an interesting problem to generalize this to the Finsler setting, or rather, the Finsler-area setting. I.e., which pairs of Finsler structures and area forms on the plane have the property that the solutions of the isoperimetric problem are the standard circles? This should be a straightforward calculation using Cartan's machinery, but it might be interesting, especially if it turned out that there are non-Riemannian solutions. – Robert Bryant Jun 26 '12 at 18:45
Thanks again. I had indeed the Finsler setting in mind, although intuitively it seems to me that if all sufficiently small circles are solutions of the isoperimetric problem, then the metric would be Riemannian and even conformal to the standard one. The idea being that the circles would approach the isoperimetrices of the unit tangent discs. – alvarezpaiva Jun 26 '12 at 21:18
@alvarerpaiva: See my comment above for the Finsler case. It turns out that there are nontrivial Finsler solutions. – Robert Bryant Jul 5 '12 at 15:59
@Robert: Thanks for the answer, it's very pretty. It was a good idea to "uncouple" the length and the area element! – alvarezpaiva Jul 5 '12 at 20:50

It is rather a comment and a question to the author than the answer (but it is too long for the comment window).

What is the (solutions of the) isoperimetric problem? If it is a closed curve of a fixed length such that it bounds the region of a maximal volume, or a shortest closed curve that bounds the region of certain fixed volume, I believe that the solutions of isoperimetric problem are very rare.

In order to confirm my suggestion, I would like to mention that for metrics of constant positive curvature the length of the circle bounding the ball of volume 1 is less than that of for metrics of constant negative curvature. This suggests that for the sphere of revolution such that the maximums of the curvature are the poles the isoperimentric circles (of sufficiently small length) are circles (= boundary of the balls) centered at the poles. Then, in a certain coordinate system they are usual circles.

(To see that the solutions of the isoperimetric problem are rare you could also combine two statements from the answer of Robert Bryant: curves of constant geodesic curvature are usually not closed and isoperimetric curves must have constant geodesic curvature.)

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Thanks Vladimir. I have yet to understand Baule's paper, but it is a bit unsettling. – alvarezpaiva Jun 29 '12 at 6:17
Juan Carlos, if all the curves of constant geodesic curvature of a metric are circles, then the geodesics are circles as well which implies (in dimension 2) that the metric has constant curvature by the classical result of Segre, see the link below… – Vladimir S Matveev Jun 29 '12 at 16:36
@Vladimir: You don't want to require that 'isoperimetric' curves be closed, since, as you know, there will be very few of these on a general surface, especially one for which the Gauss curvature has no critical points. Instead, you should define 'isoperimetric' locally, as in Dido's Problem: A curve $C$ is 'isoperimetric' if each point of $C$ lies in a small open arc $A\subset C$ such that all small area-neutral deformations compactly supported within $A$ do not decrease the length of $A$. In the case of a Riemannian surface, this is equivalent to having constant geodesic curvature. – Robert Bryant Jun 29 '12 at 22:13
Thanks indeed for the explanation, Robert. – Vladimir S Matveev Jul 3 '12 at 7:05

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