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In the studies of active contours they describe the set of all simple smooth closed curves on $\mathbb{R}^2$ to be a Riemannian Manifold $M$. The tangent space at a curve $c$, $T_cM$ is a set of vector fields defined on $c(t)$. The Riemannian metric is $\langle x,y \rangle_c = \int_c x(t)\cdot y(t)dt$. Is this in fact a manifold? If so how would you embed it in $\mathbb{R}^n$ i.e. what are the coordinate charts of this manifold?

edited Dec 7 at 5:36

asked Dec 7 at 2:36

mperez32
43●4

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It's not a traditional manifold, in that its not finite-dimensional. But it's a type of manifold, in that its locally homeomorphic to a topological vector space. – Ryan Budney Dec 7 at 2:59

And these are not all dimple closed curves. Some smoothness condition is needed to define vector fields. – Alexandre Eremenko Dec 7 at 3:16

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I think this is a perfectly good question, would those voting to close explain why? – David Roberts Dec 7 at 5:02

However, I do have one criticism, namely do you, mperez32, want closed simple curves or all curves? The former is a subset of the space of maps $S^1 \to \mathbb{R}^2$, probably open. In that case, the space $(\mathbb{R}^2)^{S^1}$ is a rather nice Fréchet manifold. – David Roberts Dec 7 at 5:08

I think it is in fact possible to embed $M$ into some (necessarily infinite-dimensional) Hilbert space, but I only base this on some comment I read somewhere online, rather than a paper or book I've read. – David Roberts Dec 7 at 6:07

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David C · Accepted Answer · 2012-12-07 08:41:14Z UTC

As Ryan Budney explained it is not a finite dimensional manifold. First let us notice that the space of all smooths loops $L\mathbb{R}^2=C^{\infty}(S^1,\mathbb{R}^2)$ is a topological vector space, it is Fréchet and it is an inverse Hilbert limit vector space. If you consider the subset $Emb(S^1,\mathbb{R}^2)\subset L\mathbb{R}^2$ of embeddings this is an open subset hence an infinite dimensional manifold of Fréchet type. Let us go further, so far we have considered parametrized curves, if we want to get rid of the parametrization we have to divide by the action of the infinite dimensional Lie group $Diff(S^1)$ and we get a smooth principal bunde $$Emb(S^1,\mathbb{R}^2)\rightarrow Emb(S^1,\mathbb{R}^2)/Diff(S^1)$$ I suppose that this is the base of this bundle you are interested in. Maybe a good reference for all this stuff is Kriegl-Michor "The Convenient Setting of Global Analysis." Mathematical Surveys and Monographs, Volume: 53 (Chapter IX, section 44).

If you are interested by a discussion of Riemannian metric on these spaces have a look at: Michor, P. ; Mumford, D. Riemannian geometries on spaces of plane curves. J. Eur. Math. Soc. 8 (2006), no. 1, 1–48.

Peter Michor · Answer 2 · 2012-12-07 07:43:20Z UTC

Look at:

Peter W. Michor; David Mumford: Riemannian geometries on spaces of plane curves. J. Eur. Math. Soc. (JEMS) 8 (2006), 1-48 (pdf)

Peter W. Michor, David Mumford: An overview of the Riemannian metrics on spaces of curves using the Hamiltonian approach. Applied and Computational Harmonic Analysis 23 (2007), 74-113. (pdf), (Erratum)

Thomas Rot · Answer 3 · 2012-12-07 10:09:15Z UTC

Another nice reference for this kind of stuff is Klingenberg: Riemannian geometry. If one imposes less smoothness on the loops (closed curves), and assume the loops are $H^1$ (in Sobolev sense), the space of loops on a Riemannian manifold is a Hilbert manifold. To get the charts, one uses the exponential map on the base, to map ($H^1$) vector fields along a loop (which is morally a tangent vector to this curve) to a loop nearby. There are some issues with differentiability of the loops which define the charts, which can be overcome by approximations of smooth loops. The details are in the above mentioned reference.

In this approach loops are allowed to self-intersect. It might be possible to study also a loop space of non-self intersecting loops, but I do not know what this space actually looks like, and if it is nicely embedded in the above mentioned space.

Is the set of all smoothed closed simple curves on $\mathbb{R}^2$ a manifold?

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