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)?

  • $\begingroup$ 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 $n=2.$ $\endgroup$ Apr 23, 2018 at 1:47
  • $\begingroup$ 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 $\psi.$ $\endgroup$ Apr 23, 2018 at 2:08
  • $\begingroup$ 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? $\endgroup$ Apr 23, 2018 at 6:26
  • $\begingroup$ 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 $r$). $\endgroup$ Apr 23, 2018 at 6:27
  • $\begingroup$ By "isometric embedding", do you mean that the $g$ in $(S,g)$ has to be the pullback of the standard metric on $\mathbb{R}^n$? $\endgroup$
    – Kajelad
    Nov 1, 2019 at 17:58

1 Answer 1


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 $d(c,x)<d(c,y)$. 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.

  • $\begingroup$ 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. $\endgroup$
    – Kajelad
    Nov 6, 2019 at 22:57

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