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added information about the Finsler metric-measure case
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Robert Bryant
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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.

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.

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.

Source Link
Robert Bryant
  • 108.4k
  • 8
  • 340
  • 453

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.