Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose X is a loop space; by this we mean there is some space $Y$ with $\Omega Y \simeq X$.

Under what assumptions is (the homotopy type of) $Y$ unique?

As has been pointed out below, the homotopy type of $Y$ being determined uniquely is far from true in general. But for connected $Y$, are there conditions we can impose that make it so?

share|improve this question
2  
The question is, probably: «if $\Omega X_1$ and $\Omega X_2$ are homotopy equivalent, are $X_1$ and $X_2$ homotopy equivalent?» –  Mariano Suárez-Alvarez May 16 '11 at 3:12
5  
Don't have time to leave a full answer, but if the homotopy groups of X are concentrated in a narrow "stable" range (roughly the top nonzero group below about twice the dimension of the bottom nonzero group) then there is a unique homotopy type. Similarly if Y has very small dimension relative to its connectivity. One can compare the obstruction theory for $Map(Y_1, Y_2)$ with that for $Map(\Omega Y_1, \Omega Y_2)$, together with a comparison of the cohomology of $Y_1$ and $\Omega Y_1$, to get a fuller statement. –  Tyler Lawson May 16 '11 at 3:57
3  
Yes, of course you would want $Y$ to be connected to avoid trivialities. –  Dr Shello May 16 '11 at 6:19
1  
@Mariano: yes, right. The question is what assumptions are needed to make that true. –  Dr Shello May 16 '11 at 6:25
2  
You might also ask a slightly different question: "Given a homotopy equivalence of based loop spaces, how do I tell if it is a loop map?" –  S. Carnahan May 16 '11 at 8:31
show 4 more comments

2 Answers

As others have pointed out, the generic case (whatever that should mean in this case) is that the loop structure on a loop space is not unique. However, things get quite interesting whenever we have a space that actually does have a unique loop structure. I highly recommend looking at:

Dwyer, Miller, Wilkerson: The homotopic uniqueness of $BS^3$, LNM 1298

and

Dwyer, Miller, Wilkerson: Homotopical uniqueness of classifying spaces. Topology 31 (1992), no. 1, 29–45.

share|improve this answer
add comment

As Ryan points out, if Y is allowed to be disconnected, then there is no hope, since the loop-space construction sees only the connected component of the basepoint. But even if Y is assumed to be connected, it is not unique. For instance, let G and H be two discrete groups whose underlying sets are bijective, but which are not isomorphic. Then as (discrete) topological spaces, we have $G\simeq H$, and so both $K(G,1)$ and $K(H,1)$ are spaces Y such that $\Omega Y \simeq G \simeq H$. But $K(G,1)$ and $K(H,1)$ are not homotopy equivalent unless $G\cong H$ as groups.

What is true, however, is that if we remember the "up-to-coherent-homotopy" multiplication (i.e. "$A_\infty$-structure") on a loop space $\Omega Y$, then the connected space Y is characterized up to homotopy equivalence by $\Omega Y$ and this additional data. For there is a delooping functor "B" from $A_\infty$-spaces to connected spaces, which preserves homotopy equivalence, and such that $B\Omega Y \simeq Y$.

share|improve this answer
    
So this means in general for $Y$ connected, its homotopy type is not unique. But are there conditions that can be put on $Y$ to make it so? –  Dr Shello May 17 '11 at 6:05
add comment

Your Answer

 
discard

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.