Dear Sir/friends,
How to give manifold structure to set of all $C^2$ path over any manifold.
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1$\begingroup$ Please see mathoverflow.net/howtoask -as regarding your questtion, what you ask is easy in the smooth case, but the space of C^2 functions is much bigger. Work by Kriegl + Michor would be appropriate to consult: books.google.com.au/…, chapter III section 12 may help. Note this is only for spaces of mappings into certain topological vector spaces. This is a necessary ingredient for the manifold case. $\endgroup$– David Roberts ♦Commented Jun 27, 2011 at 6:58
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2 Answers
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If by "path" you mean a map with domain $[0,1]$ then this is a standard construction and is independent of the class of maps (providing it is contained in $C^0$). You can find it in many places, search MathSciNet for "manifold" and "mapping space", or you can almost find it in my paper Constructing Smooth Manifolds of Loop Spaces. I deal with maps from $S^1$ there but there's no difference in the construction.
If by "path" you mean a map with domain $\mathbb{R}$ then it is much, much more complicated. With the standard topology then it isn't a manifold. You can put a topology on it to make it a manifold, but it has uncountably many components. For more on this, look in Kriegl and Michor's book A Convenient Setting for Global Analysis (that also has the construction for $[0,1]$ for the smooth setting, which readily adapts to $C^2$).
answered Jun 27, 2011 at 6:58
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$\begingroup$ snap! (but your answer is of course better) $\endgroup$– David Roberts ♦Commented Jun 27, 2011 at 7:00
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I give such a manifold structure here: The tangent bundle to an infinite-dimensional manifold
answered Jun 27, 2011 at 10:05