1
$\begingroup$

Schematically, I understand the path space fibration $PX$ over some path-connected, pointed topological space $X$ with base point $x_o$ as: $$\Omega X \hookrightarrow PX \twoheadrightarrow X,$$ where the first arrow is the inclusion and the second is the evaluation map.

If we understand $PX$ as the space of paths in $X$ (i.e., continuous maps $p(t)$ from the unit interval to $X$, with $p(0) = x_o$), then it seems that the natural definition of $\Omega X$ is the space of paths $p(t)$ with $p(0) = p(1) = x_o$. However, I have found conflicting statements from various sources:

  • This MSE question says that $\Omega X$ is the space I have just described.
  • The Wikipedia article for path space fibration says that $\Omega X$ is the loop space of $X$; the Wikipedia article for loop space defines that space as the space of continuous pointed maps from $S^1$ (with base point) to $X$.
  • This MSE question points out that the space of continuous pointed maps from $S^1$ to $X$ is not the same as the space of paths in $X$ beginning and ending at the same point, which I think is correct due to the fact that elements of the latter may be discontinuous at $x_o$ between $t=1$ and $t=0$.

Having become thoroughly confused by everything I found on the internet, I turned to Hatcher, where I discovered a lengthy and fairly dense discussion on loop spaces which requires the use of "James reduced products," a concept I have not encountered before in my coursework. This leads me to believe that perhaps the whole situation is more complicated than I initially thought.

My question is, what is $\Omega X$, and what subtlety am I missing here that is leading to the confusion described above?

$\endgroup$
4
  • $\begingroup$ I think your third point is wrong, you may be misinterpreting part of the linked question. Since $S^1 \cong [0,1] / \{0,1\}$ any pointed map $S^1 \to X$ gives you a path $[0,1] \to X$ that starts and ends at the basepoint, and vice versa; this a homeomorphism between $Map_\bullet(S^1, X)$ and $L(x,x)$. The question asserts that if $X$ has the structure of a smooth manifold then the space of smooth pointed maps $S^1 \to X$ is not the same as smooth loops $[0,1]\to X$, since the latter might not induce a smooth function from the circle. $\endgroup$ – William Jan 26 '20 at 16:45
  • $\begingroup$ The smooth category is tricky to work in for this context, because the composition of two smooth loops/paths may not even be smooth. You would have to add an extra assumption that they loops/paths behave a certain way in some $\epsilon$ neighbourhood of the basepoint in order for the composition to again be smooth. However in the topological category there is no issue. $\endgroup$ – William Jan 26 '20 at 16:50
  • $\begingroup$ Paul Frost answered explicitly what $\Omega X$ is, and @William helped to up my confusion - I had not been careful enough to distinguish "smooth" and "continuous." Thanks! $\endgroup$ – k-t Jan 27 '20 at 5:28
  • $\begingroup$ *helped to clear up $\endgroup$ – k-t Jan 27 '20 at 17:45
1
$\begingroup$

The definition in your question is that for pointed spaces $(X,x_0)$. More precisely we should write $P(X,x_0) = (X,x_0)^{(I,0)}$ = set of all basepoint-preserving maps $(I,0) \to (X,x_0)$ with compact-open topology for the pointed path space, $p : P(X,x_0) \to (X,x_0), p(u) = u(1)$, and $\Omega(X,x_0) = p^{-1}(x_0)$ fot the pointed loop space. Both have as basepoint the constant path at $x_0$. Then $\Omega(X,x_0)$ is the fiber over the basepoint $x_0 \in X$.

You can do an analogous construction for unbased spaces $X$:

$PX = X^I$ = set of all maps $I \to X$ with compact-open topology is the free path space. The evaluation map $p : PX \to X, p(u) = u(1)$, is a fibration. Its fibers are the sets $p^{-1}(x) = \{u \in X^I \mid p(u) = u(1) = x \} =(X,x)^{(I,1)}$. The latter is homeomorphic to $P(X,x)$.

A third construction is the free loop space of a space $X$:

$$\mathcal L X = X^{S^1} .$$

It can be viewed as the unpointed version of $\Omega (X,x_0)$. There is a canonical embedding $\iota : \Omega (X,x_0) \to \mathcal L X$: Each $u \in \Omega (X,x_0)$ is a closed path $u : I \to X$ such that $p(0) = p(1) = x_0$ which determines a unique continuous $\hat u : I/\{0, 1\} \to X$ and via the identification $I/\{0, 1\} = S^1$ this gives us $\iota(u) \in \mathcal L X$.

Note that this construction also allows to identify $\Omega (X,x_0)$ with $(X,x_0)^{(S^1,*)}$. In fact, $\iota(\Omega (X,x_0)) = (X,x_0)^{(S^1,*)} \subset X^{S^1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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