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$M$ be any Riemannian manifold, and $S^1$ is a circle. We can give Manifold structure to $C^\infty(S^1, M)$ modeled on nuclear frechet space.

Take $Imm(S^1, M):\{f\in C^\infty(S^1,M): f \text{ is an immersion} \}$. This is open subset of $C^\infty(S^1,M)$ hence it is also a nuclear frechet manifold.

Consider set of all isomoetric immersion of $S^1\to M$, can we give some differentiable structure here. Please provide the reference where people have already studied the isometric immersed loops over a manifold.

Edit: Can we see Set of isometric immersion as a manifold modeled over some Locally convex space.

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What do you mean by "differentiable structure"? If you work in a suitable category of generalised smooth spaces then you've already given it a smooth structure by embedding it in the loop space, so it inherits the ambient structure. –  Andrew Stacey Mar 15 '12 at 11:49
    
Edited.. Differentiable structure: I mean manifold structure. –  zapkm Mar 15 '12 at 12:03
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I suggest that you take a look at Peter W. Michor's "Manifolds of Differerentiable Mappings" and related work of his and his co-authors. They have many articles on the arXiv that address questions of smooth structures of various kinds on spaces of smooth mappings. –  Robert Bryant Mar 15 '12 at 14:05
    
You also might be interested in "A Kähler structure on the moduli space of isometric maps of a circle into Euclidean space" by John J. Millson and Brett Zombro, Inventiones, vol. 123 (1996) Number 1, 35-59. They consider both smooth and Lipschitz settings. –  Misha Mar 15 '12 at 14:50
    
thanks for the references. –  zapkm Mar 16 '12 at 4:34
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2 Answers

up vote 2 down vote accepted

The case where $\dim M = 3$ is considered in Brylinski's book Loop spaces, characteristic classes and geometric quantization (I don't know where it originates, presumably there are references in the book if it doesn't originate there - I don't have a copy to hand).

The method given there ought to generalise. Look at an isometric immersion, say $\alpha \colon S^1 \to M$. A chart at $\alpha$ for the full loop space, $L M = C^\infty(S^1, M)$, has domain $\Gamma_{S^1}(\alpha^* T M)$. So long as the chart map is chosen carefully (and I'm pretty sure it can be done so), the following should work. As $\alpha$ is an immersion, we have a non-zero section $\alpha' \in \Gamma_{S^1}(\alpha^* T M)$. The fibrewise orthogonal complement (in the induced metric from $T M$) of this defines a subbundle, say $E_\alpha$, of $\alpha^* T M$. We take sections of this, $\Gamma_{S^1}(E_\alpha)$, and restrict the chart map to this. The idea being that if $\beta$ is the image of a section $X$ then $\|\beta'\|^2 = \|\alpha'\|^2 + 2\langle \alpha', X \rangle + \| X\|^2$ and to first order, as $\langle \alpha', X \rangle = 0$ then this is $\|\alpha'\|^2 = 1$ (as $\alpha$ is an immersion). To make this precise one would need to choose the original chart map very carefully, but as I said I don't think that would be difficult and the details should generalise from the 3-manifold case as given in Brylinski's book.

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@Andew Stacey... So if i understand correctly, You are trying to give manifold structure to set of all immersion.... –  zapkm Mar 15 '12 at 12:02
    
Isometric immersions. By taking the orthogonal complement of the tangent vector in the chart domain, we don't disturb the isometric condition. –  Andrew Stacey Mar 15 '12 at 12:04
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Let $(M, g)$ be a smooth Riemannian manifold, and $f \colon S^1 \to M$ a smooth immersion. The metric $f^* g$ on $S^1$ is flat, since $S^1$ is one-dimensional. Since the flat metric is unique up to diffeomorphism then, given any fixed metric $h$ on $S^1$, there exists a diffeomorphism $\varphi \colon S^1 \to S^1$ such that the pull-back $f^* g$ under $\varphi$ is $h$. In particular, given any immersion $f$, there exists a $\varphi$ such that $f \circ \varphi$ is an isometry. As such, the space of isometric immersions may be identified with the quotient of the space of smooth immersions by isometric diffeomorphisms (i.e. rotations) of $S^1$.

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But how is this relevant to the question? –  Deane Yang Mar 15 '12 at 10:20
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All I am saying is that I believe any immersion is isometric, therefore the space of isometric immersions has the same differentiable structure, etc as the space of immersions. –  Arch Stanton Mar 15 '12 at 10:22
    
If i am not making mistake then, take $f:S^1\subset \mathbb R^2\to \mathbb R^2$ $f(e^{i\theta))= \frac{1}{2}e^{i\theta}$... This is immersion but not isometry. All metric are induced from $\mathbb R^2$. –  zapkm Mar 15 '12 at 10:26
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To clarify, as Prof. Yang points out, I was implicitly allowing myself to reparametrise the $S^1$. –  Arch Stanton Mar 15 '12 at 10:47
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Ah, ok, I get your point now. Yes, you're right. Sorry for confusing matters. –  Arch Stanton Mar 15 '12 at 20:04
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