# The radially symmetric isoperimetric problem

A solution to the $l$-isoperimetric problem on a Riemannian surface $(M,g)$ is a smooth closed curve $\gamma \subset M$ of length $l$ which minimizes the isoperimetric constant: $$h(\gamma) = \frac{l}{\text{vol}(M_\gamma)}.$$ Where $M\setminus \gamma$ denotes the submanifold of minimal volume of $M$ with boundary $\gamma$.

Claim 1 Solutions to the $l$-isoperimetric problem on the closed disk $D$ equipped with a radially symmetric metric are radially symmetric.

My disappointment for not finding a proof for such an obvious statement has been dwarfed only by my surprise in finding a counterexample: Claim 1 is false. (The idea of the counterexample is to have a small flat metric in a ball centered in the origin and a large flat metric in a disjoint annulus, then take $l$ small enough.)

My counterexample seems to fundamentally rely on two factors: the metric can be increasing, and the length $l$ can be chosen to be small. Taking care of both problems at the same time leads to the next claim.

A solution to the Cheeger-isoperimetric problem on a Riemannian surface $(M,g)$ is a smooth closed curve $\gamma \subset M$ which realizes the Cheeger constant: $$h(M) = \inf_{\gamma \subset M}\left\{\frac{\text{length}(\gamma)}{\text{vol}(M_\gamma)}\right\}.$$

Claim 2 Solutions to the Cheeger-isoperimetric problem on the closed disk equipped with a monotone decreasing radially symmetric metric are radially symmetric.

Statements similar to Claim 2 have been discussed on Math.OF or in the literature, but the proposed claim doesn't seem to be covered by these sources. Is Claim 2 correct? Is it obvious? If correct, does it remain so dropping the monotonicity hypothesis?

Foreword. Given my unfamiliarity with the subject, and the amount of literature on the isoperimetric problem, it is highly likely that the answer can be shortened or improved. Moreover I will take some days to verify the argument before accepting the answer. I would be happy for any comment or additional contribution.

First, a minimizer of the Cheeger problem exists. To prove it we observe that the Cheeger function $$h: \left\{ \gamma \subset M\; \text{smooth}, \text{closed} \right\} \rightarrow \mathbb{R}_+$$ factorizes through $$\varphi: \left\{ \gamma \subset M \; \text{smooth}, \text{closed} \right\} \rightarrow \mathbb{R}_+^2$$ given by $\varphi(\gamma) = \left( l(\gamma), \mathrm{vol}(M_\gamma)\right)$ and $(x,y) \mapsto \frac{x}{y}$. We now provide bounds on the candidates to be minimizers. Every curve $\tau$ with $l(\tau) > l(\partial D)$ has Cheeger constant larger than $\partial D$. On the other hand, also curves with too small length have Cheeger constant away from the lower bound on the surface. Indeed, as implied by Theorem 2.2 of "Some Sharp Isoperimetric Theorems for Riemannian Manifolds (Morgan, Johnson)", minimal perimeter solutions for small volume are (nearly round) spheres, whose Cheeger constant is known to blow for small perimeter. We can therefore restrict to curves $\tau$ with $l(\tau) \geq c_M >0$. We can argue in a similar way for the volume: for each curve $\tau$ holds $\mathrm{vol}(M_\tau) \leq \mathrm{vol}(M)$, and we can restrict to $\mathrm{vol}(M_\tau) \geq d_M>0$. In particular we are reduced to prove that the function $\frac{x}{y}$ achieves a minimum on $$\mathrm{Im}(\varphi) \cap \left\{(x,y): c_M \leq x \leq \partial D, \; d_M \leq y \leq \mathrm{vol}(D)\right\}.$$ Since the function is continuous away from $y=0$, and the domain is bounded, it remains to prove that it is also closed, thus compact. Clearly the domain is the region $\left[c_M,\partial D\right] \times \left[d_M ,\mathrm{vol}(D)\right]$ with the upper left corner cut by a convex function. To prove it is closed is therefore enough to show that the isovolumetric problem of volume $V$ has a minimal perimeter solution for each $V\in\left[d_M ,\mathrm{vol}(D)\right]$. For $V\in\left[d_M ,\mathrm{vol}(D)\right)$ this is proven in Theorem 3.4 of "Geodesics and soap bubbles in surfaces (Hass, Morgan)", while for $V=\mathrm{vol}(D)$ it is obvious.

Let $\gamma$ be the minimizer of the Cheeger problem, whose existence we just proved.

Second, the origin must lie inside $\gamma$: $O \in M_\gamma$. To prove it we proceed by contradiction. We can trivially exclude the cases where $\gamma$ has more than one connected component. Moreover, to avoid pathological cases where the upcoming construction would not work, we replace $\gamma$ by its circular symmetrization, whose Cheeger constant is not worse than the one of $\gamma$. Now, let $l_0$ be a tangent to $\gamma$ passing through $O$, and define $\theta_0$ to be the smallest angle such that $\gamma$ is contained in the cone limited by $l_0$ and its $\theta_0$-rotation. We denote this second line by $l_1$, it also tangent to $\gamma$. If $\theta_0 \leq \pi$ we replace $\gamma$ by itself plus its reflection through the line $l_1$, this is clearly not changing $h(\gamma)$. Now repeat the same construction: the new angle $\theta_1$ is $\theta_1 = 2\theta_0$. We repeat the construction until $\theta_n >\pi$.

This new $\gamma$ is still a solution of the Cheeger problem, indeed this construction does not change the Cheeger constant or the distance from the origin, but is now concave! Since the metric is radially symmetric, and $\theta > \pi$, the geodesic connecting the two extremal points of $\gamma$ is strictly shorter than the shortest path connecting them along $\gamma$. Moreover also the area of $M_\gamma$ increases, yielding a better Cheeger constant. We can thus assume that $O\in M_\gamma$.

Third, the solution must be a circle, i.e. radially symmetric. (This is a minimal variation of an argument appearing in "The isoperimetric problem on surfaces (Howard, Hutchings, Morgan)") Indeed let $l$ be a line passing through the origin $O$ and splitting $\gamma$ into two, not necessarily connected, parts. We replace $\gamma$ by the part with the better Cheeger constant plus its reflection through $l$. Let $m$ be the line through $O$ and orthogonal to $l$, again we replace $\gamma$ by the part with the better Cheeger constant plus its reflection through $m$. Thus $\gamma$ can be assumed to be invariant by reflections through $l$ and $m$, thus by $\pi$-rotations. Specifically, every line through the origin splits $\gamma$ in two parts with equal area and perimeter, thus with equal Cheeger constant. If $\gamma$ is not a circle, there exists a line $s$ through the origin that intersects it not orthogonally. Replacing $\gamma$ by the one of the half in the splitting by $s$ plus its reflection does not vary the Cheeger constant, but yields a concave curve. Since the metric is monotone decreasing, the convex hull of this curve has smaller perimeter and larger area, thus a better Cheeger constant. Contradiction.

As a final remark I would like to point out that, in the last step, we are actually using the monotonicity of the metric.