6
$\begingroup$

Given an embedded closed 2-manifold with intrinsic metric $g_{ab}(\sigma_1,\sigma_2)$ embedded in $\mathbb{R}^3$. The surface is analytic (not just $C^\infty$). We are given only the metric and not the equation of the surface.

What is the formula to calculate the extrinsic distance between two points $(\sigma_1,\sigma_2)$ and $(\sigma_1',\sigma_2')$. That is the straight line in 3 dimensions, written only in terms of the intrinsic metric?

I think it must have something to do with the second fundamental form which gives the local embedding or the Ricci tensor. My guess would be it would be an integral of some functional of $g$ on a curve between the two points on the surface. $L=\int F[g].d\sigma$. If such a formula could exist.

(The formula won't exist when the 2-manifold has zero intrinsic curvature.)

As an example, take a sphere with the sphere metric and then the distance we're looking for is the distance between two points on a sphere with a straight line going inside the sphere.)

Edit: The answer must apply to analytic surfaces only. (Not just $C^\infty$ continuously differentiable.) An analytic surface is one in which the surface can be expressed as functions which can be expanded as taylor series - not piecewise surfaces such as the bump function.

Edit: I know the second fundamental $I_{ab}$ form is connected to the Riemann tensor by $R_{abcd}[g] = I_{ab}I_{cd}-I_{ac}I_{bd}$ which gives the metric in terms of the fundamental form. But what is needed is the reverse, a way to get the fundamental form in terms of the metric in cases where there is a unique embedding up to translations and rotations.

$\endgroup$
5
$\begingroup$

First, you mention that you want something only in terms of the metric, but then mention the second fundamental form... the second fundamental form is an extrinsic quantity. You cannot infer it from the metric alone.

If you just have the metric, you have no hope. The Nash-Kuiper theorem guarantees that you can isometrically $C^1$-embed your surface so that the extrinsic distance between two points is anything you want, so long as it's less than the intrinsic distance (for a nice visualization, consider Borelli's flat torus: https://blogs.scientificamerican.com/blogs/assets/roots-of-unity/File/flat%20torus%20bumps.jpg). Even if you insist on smooth embeddings, the isometric embedding of a manifold is not unique (consider a "bumpy box" where you start with a cube, round the corners, then extrude a bump function on each face, a little bit like https://cdn.thingiverse.com/renders/bf/2c/0a/56/36/941dda9d318d11af889ebb716ec7a3fa_preview_featured.jpg. Clearly you can "pop" the bumps in and out to get $2^6$ possible different isometric embeddings of the surface).

Now, if you have both the first and second fundamental form, then the embedding can be uniquely recovered (up to isometries of space) and extrinsic distance can be computed. In fact, your idea is a good one: pick an intrinsic geodesic $\gamma$ parameterized by arc length connecting the two points. Let $\phi$ be the (hypothetical, unknown) embedding function. Since $\gamma$ is a geodesic, $\psi(s) = \phi \circ\ \gamma$ is an arclength parameterized curve on the surface of the manifold whose curvature direction aligns with the manifold normal. This means that the curvature of the curve is $$\kappa = \langle dn\gamma', d\phi \gamma'\rangle = - \mathrm {I\!I}(\gamma',\gamma')$$ and the torsion is $$\tau = \langle dn\gamma', d\phi J\gamma'\rangle = - \mathrm {I\!I}(\gamma', J\gamma')$$ where $J$ rotates a vector to form its perpendicular vector, i.e. $d\phi Jv = n\times d\phi v.$ Intrinsically $$Jv = \mathrm {I}^{-1}\left[\begin{array}{cc}0 & -1\\1&0\end{array}\right]v \sqrt{\frac{v^T\mathrm {I}v}{v^T\mathrm {I}^{-1}v}}.$$

Now choose an arbitrary starting point, tangent vector $T = \psi'(0)$, and normal $N$ and integrate the Frenet-Serret equations $$\left[\begin{array}{c}T'\\N'\\B'\end{array}\right]=\left[\begin{array}{ccc}0 & - \mathrm {I\!I}(\gamma'[s],\gamma'[s]) & 0\\ \mathrm {I\!I}(\gamma'[s],\gamma'[s]) & 0 & - \mathrm {I\!I}(\gamma'[s], J\gamma'[s])\\ 0 & \mathrm {I\!I}(\gamma'[s],J\gamma'[s]) & 0\end{array}\right]\left[\begin{array}{ccc}T\\N\\B\end{array}\right]$$ to recover $\psi$ and consequently the distance.


Edit: You are right, I missed the requirement that the surface is analytic; ignore the first few paragraphs above. However, except in the case of manifolds of everywhere postitive Gaussian curvature (which are isometrically rigid, even in the case of $C^2$ embeddings) I know of no global uniqueness results for isometric embeddings even in the analytic category.

I'm still not clear whether you know the second fundamental form, or are wanting to compute it given the metric (and some guarantee that an isometric embedding is unique). I know of no special technique for doing the latter: the isometry equations give you a PDE to solve to recover the surface embedding. In neighborhoods of positive Gaussian curvature this PDE is elliptic and uniqueness is easy. Where the curvature vanishes or is negative, to my knowledge the question of rigidity is still open.

If you want to approximate the extrinsic distance, you could triangulate the manifold, compute intrinsic lengths on the triangle edges, then computationally isometrically embed the simplicial complex. I don't know of any bulletproof algorithm for doing so either, though I know this problem has been studied in applications such as multidimensional scaling and sensor networks.

$\endgroup$
8
  • $\begingroup$ It is not true you have no hope. Only in certain cases. For example, there is only one way to analytically embed a sphere in 3D without creases. So given the metric of a sphere only, you should be able to calculate the extrinsic distances. Also, the Nash-Kuiper theorem only applies to $C^\infty$ not analytic surfaces. The bumpy box example is not an analytic surface although it is a $C^\infty$surface. $\endgroup$ – zooby Jan 10 '18 at 14:30
  • $\begingroup$ It is a good answer but I can't mark it as the correct answer since it doesn't apply to analytic surfaces. $\endgroup$ – zooby Jan 10 '18 at 14:42
  • $\begingroup$ Or if there is another way to analytically embed a sphere in 3D that would be an interesting counter example. $\endgroup$ – zooby Jan 10 '18 at 14:45
  • $\begingroup$ If you made an analytic surface similar to the bumpy box and then reversed some of the bumps the surface would no longer be analytic. Similar to the function y=|x|^3 which looks smooth but is not analytic. $\endgroup$ – zooby Jan 10 '18 at 14:48
  • $\begingroup$ @zooby You are right, I missed the part of your question where you state the manifold is analytic. However except for surfaces of everywhere positive Gaussian curvature, which as you say are isometrically rigid, I know of no uniqueness results for isometric embeddings even on the analytic category. $\endgroup$ – user7530 Jan 10 '18 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.