MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Dear Sir/friends,

How to give manifold structure to set of all $C^2$ path over any manifold.

share|cite|improve this question
Please see -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:…, 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. – David Roberts Jun 27 '11 at 6:58
up vote 5 down vote accepted

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 $C^2$).

share|cite|improve this answer
snap! (but your answer is of course better) – David Roberts Jun 27 '11 at 7:00
Thanks a lot...I will look upon this.. – zapkm Jun 27 '11 at 18:47

I give such a manifold structure here: The tangent bundle to an infinite-dimensional manifold

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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