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Let $X$ be a path-connected CW-complex and $x$, $y$ points in $X$. Any choice of a path between $x$ and $y$ provides maps (in both directions) between the space $L(x, y)$ of paths from $x$ to $y$ and the space $\Omega(X, x)$ of pointed loops based on $x$ (both spaces are endowed with compact-open topology). This pair of maps is easily seen to be homotopy equivalence. Obviously, neither of maps is a bijection. This does not rule out the possibility that $L(x, y)$ and $\Omega(X, x)$ are homeomorphic via some unnatural map. Are they?

Admittedly, checking whether two homotopy equivalent spaces are homeomorphic is hard and not particularly interesting. It would be surprising if the answer is known. As such, it looks like the smooth variant of the same question is more feasible (and more interesting).

Assume $X$ to be a smooth manifold, $L(x, y)$ a space of smooth paths (with the same compact-open topology), and $\Omega(X, x)$ a space of smooth loops. They are both Frechet manifolds (it's a standard statement for the loop space and the space of smooth paths is similar). Are they diffeomorphic?

Note that now $\Omega(X, x)$ is not $L(x, x)$, since smooth map of $S^1$ into $X$ is not the same thing as a smooth map from $[0, 1]$ to $X$ with coinciding images of $0$ and $1$. So there are more differences than in the topological case. What's more, we don't have any obvious maps between these two spaces at all: a concatenation of two smooth paths is not necessarily smooth, so homotopy equivalence is not clear (or maybe not even true).

(Clearly, the answer does not depend on a chosen pair of different points $x, y$: the group of diffeomorphisms acts 2-transitively, so $L(x,y)$ is diffeomorphic to $L(x', y')$ by post-composition).

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@GrigoryM Thank you! That answers both my questions. If you put this argument in an answer, I would accept it. – Dmitry Jan 12 '14 at 19:02
You're welcome. (I'm afraid this doesn't explain whether $L(x,x)$ and $\Omega(x)$ are diffeomorphic, though.) – Grigory M Jan 12 '14 at 19:08
It doesn't, but that question actually interested me only as a variant of the topological statement. – Dmitry Jan 12 '14 at 19:10
up vote 2 down vote accepted

Well, at least for manifolds spaces $L(x,x)$ and $L(x,y)$ are isomorphic (homeomorphic for topological version and diffeomorphic for smooth version): choose a family $F_t$ of automorphisms of the manifold s.t. $F_0(x)=x$ and $F_1(x)=y$; now $\gamma\mapsto(t\mapsto F_t(\gamma(t)))$ gives an isomorphism $L(x,x)\to L(x,y)$.

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