# Finding a metric to make a certain curve a circle

Given a simple closed, regular $$C^\infty$$ curve $$\phi$$ in $$U\subset\mathbb{R}^n$$ naturally parametrized (by it's arc length), is there any way to obtain a Riemaniann manifold $$(S,g)$$ of dimension 2 without boundary (isometrically embedded in $$\mathbb{R}^m$$ equipped with the standard metric) such that there is a geodesic circle in this surface that is equal to the curve (meaning that it is mapped by the embedding to $$\phi$$)?

Two examples:

$$1)\gamma(t)=\begin{pmatrix} \left(\frac{\sin(20\pi t)}{10}+1\right)\sin(2\pi t)\\ \left(\frac{\sin(20\pi t)}{10}+1\right) \cos(2\pi t)\\ \sin(2\pi t)\end{pmatrix}\\ 2)\gamma(t)=\begin{pmatrix} \left(\frac{\sin(20\pi t)}{10}+1\right)\sin(2\pi t)\\ \left(\frac{\sin(20\pi t)}{10}+1\right) \cos(2\pi t)\end{pmatrix}$$

One possibility, considering $$\gamma \in \mathbb{R}^2$$, is to use the Riemann smooth mapping theorem in such a way to obtain a complex diffeomorphism $$\phi$$ between $$\gamma\bigcup \text{Int}(\gamma)$$ and the closed unitary disk $$D$$. In this way, we might define the metric tensor on $$S=\phi^{-1}(D)$$ as the pullback of the euclidean metric tensor restricted to the unitary disk, but that leaves us with a manifold with boundary. We may try to extend it, but such a subject is quite technical, and I would not know how to proceed. Even if this idea was succesful This method would work only in $$\mathbb{R}^2$$, leaving open the question for $$n>2$$.

The questions are thus: 1) Is my idea efficient to solve the problem in $$\mathbb{R}^2$$? If so, how to remove the boundary?

2)How to attack the problem if $$\gamma \subset \mathbb{R}^n$$ with $$n>2$$ (as an example, see the first example)?

• I guess you mean $\mathrm{dist}_g (x,p) = r$ instead of $g(x,p)=r$? And $\mathrm{image}(\phi)$ rather than $\mathrm{graph}(\phi)$? Since you're only talking about curves, it seems you should also restrict to $n=2.$ Apr 23, 2018 at 1:47
• If you assume additionally that $U$ is simply-connected then this should be true: the idea is to "fill in" the curve $\phi : S^1 \to U$ with a map $\psi : B^2 \to U,$ then transfer the Euclidean metric from $B^2$ across $\psi.$ Apr 23, 2018 at 2:08
• Your updated definition doesn't seem correct, either - $\langle x-p,x-p \rangle$ is the distance induced by an inner product, but this is not the same as the Riemannian distance. (If $x,p$ are points in an arbitrary Riemannian manifold, $x-p$ doesn't even make sense.) I'm also not sure that you really mean the graph - if you do, this is a subset of $U \times I$ where $I$ is the parameter domain of the curve $\phi$ - is that really what you want? Apr 23, 2018 at 6:26
• As for higher dimension, think about the 3D case: the set of points of distance $r$ from the origin is a sphere, which is not a curve. In $n$ dimensions you should expect an $(n-1)$-dimensional sphere (at least for small $r$). Apr 23, 2018 at 6:27
• By "isometric embedding", do you mean that the $g$ in $(S,g)$ has to be the pullback of the standard metric on $\mathbb{R}^n$? Nov 1, 2019 at 17:58

This is not a complete answer; just a few potentially useful thoughts.

Throughout, I'll use $$\mathbb{R}^n$$ for the manifold $$(\mathbb{R}^n,g)$$ to denote an arbitrary Riemannian metric on $$\mathbb{R}^n$$ and $$\mathbb{E}^n$$ to denote $$\mathbb{R}^n$$ with the standard metric.

I'll interpret the question (with some further restrictions) as follows.

Def. Given a closed curve $$\gamma:S_1\to\mathbb{E}^n$$, we say $$\gamma$$ is an extendable disc boundary there exists an open 2-ball $$B$$ containing a concentric closed 2-disc $$D$$ and an embedding $$\iota:B\to\mathbb{E}^n$$ such that $$\iota(\partial D)=\gamma(S_1)$$. Further, it is a extendable geodesic disc boundary if $$B$$ and $$D$$ can be made geodesic balls/discs with the induced metric.

We can it seems forget about the open 2-ball entirely, and just look at the closed disc.

Def. Given a closed curve $$\gamma:S_1\to\mathbb{E}^n$$, we say $$\gamma$$ is a disc boundary if there exists a closed 2-disc $$D$$ and an embedding $$\iota:D\to\mathbb{E}^n$$ such that $$\iota(\partial D)=\gamma(S_1)$$. If is a geodesic disc boundary if $$B$$ can be made a geodesic disc in the induced metric.

It turns out these two conditions are equivalent.

Prop. All (geodesic) disc boundaries are extendable.

Proof (sketch). We need only show that a closed disc boundary has an embedded extension. If the disc is isometrically embedded, equipping the extension with the pullback metric satisfies the extendable geodesic disc boundary condition. Let $$D$$ be an embedded disc parameterized by coordinates $$x,y$$ with $$x^2+y^2\le r^2$$. Schematically, I think one can construct an extension as follows:

• Construct a closed, embedded tubular neighborhood $$T\supset D$$ and extend the normal coordinates functions $$\nu^2,...,\nu^n$$ to all of $$\mathbb{E}^n$$
• Show there is an open neighborhood $$\mathcal{O}_T\supset T$$ such that the level set $$\nu^i=0$$ is an embedded submanifold $$\hat{D}$$ through level set theorem and rank arguments.
• Extending the coordinate functions $$x,y$$ to $$\hat{D}$$. We know that the function $$(x,y):\hat{D}\to\mathbb{R}^2$$ is invertible when restricted to $$D$$, so there is an open set in $$\hat{D}$$ on which the function is invertible (see this answer for an idea of how to show this). This open set contains an open neighborhood containing $$D$$ diffeomorphic to an open disc.

Still, there are topological, differential, and geometric obstructions to this construction. The general requirement is that the curve be an embedding of $$S_1$$, i.e. has nonvanishing velocity, since the boundary of a disk is precisely such an embedding. For additional restrictions, we can proceed by dimension.

In $$n=0,1$$, the disc embedding is trivially impossible by rank considerations.

In $$n=2$$, given any closed, embedded curve $$\gamma(S_1)$$, is automatically the boundary of exactly one closed disc, so it suffices to check if this disc with the pullback metric is a geodesic disc. If $$\gamma(S_1)$$ is already geodesic circle, this is automatically true, since circles in $$\mathbb{E}^2$$ are geodesically convex. If a curve is not a circle, it is not an isometrically embedded disc: we can choose a point $$c$$ encircled by the curve, find a point $$x$$ which minimizes the distance function, and find some other point $$y$$ with $$d(c,x). Restricting to the disc encircled by $$\gamma(S_1)$$ preserves this inequality.

In $$n=3$$, the curve cannot be knotted. There is a result from topology that a knotted curve cannot be the boundary of a contractible embedded surface (in particular, a disc).

For unknotted curves $$n\ge 3$$ (which includes all curves in $$n\ge 4$$), we can always find an embedded disc $$D$$, but this disc need not be geodesic.

To find a geodesic disc, we will need to "deform" $$D$$ to have the right induced metric. A few possibilities come to mind for how to do this. The most straightforward try is probably to work in a tubular neighborhood with coordinates $$r,\theta,\nu^1,\dots,\nu^n$$ such that $$D=\{r\le 1,\nu^i=0\}$$ and deform it to a "graph" $$D'=\{r\le 1,\nu^i=\nu^i(r,\theta)\}$$. From there, one can write the metric as a function $$\nu^i$$, find a family of deformations $$\nu^i(r,\theta)$$, and try to change the metric on $$D'$$ to satisfy sufficient conditions for a geodesic circle.

• Yes; these were typos. The open 2-ball and closed 2-disc are manifolds which exist independently of any ambient space. There are several ways of constructing them, including as embedded submanifolds of larger manifolds like $\mathbb{R}^n$. There's no need to specify metrics on these spaces if we require the embedding to be isometric; the only allowed metric is the pullback metric from $\mathbb{E}^n$. For the curves, injective arc-length parameterized curves whose endpoints connect smoothly will suffice. Nov 6, 2019 at 22:57