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 CW-complex. The monoid of homotopy self-equivalences $M = hAut(X)$ is the subspace of $Map(X,X)$ consisting of those maps with a homotopy inverse. It is a union of path components. It obviously acts on $X$, and the homotopy type only depends on the homotopy type of $X$.

It is known that we can find maps $G \leftarrow M' \to M$ of topological monoids, all homotopy equivalences, with $G$ a topological group.

It is also known that we can use this to find a $G$-space $Y$ and maps of $M'$-spaces $Y \leftarrow X' \to X$ which are all homotopy equivalence. In other words, this rigidifies the action of $hAut(X)$ to an honest action of a topological group.

However, even in this situation we have a composite map $G \to Aut(Y) \to hAut(Y)$ that we know is a homotopy equivalence, but it is unlikely to be the case that $Aut(Y)$ is homotopy equivalent to $hAut(Y)$.

Can we rigidify this and find an $Y$ whose automorphism group is equivalent to its homotopy automorphism group? Or does there exist a space for which the map $Aut(Y) \to hAut(Y)$ is never a homotopy equivalence for any space homotopy equivalent to $X$?

share|improve this question
    
Just a naive remark: on the "yes" side one has it in special cases: if $X$ is an oriented surface for example. Furthermore, rigidity theorems such as the Mostow rigidity give it on $\pi_0$ for $X$ in the class of examples for which the rigidity theorem holds (such as hyperbolic manifolds). –  John Klein Jan 23 '11 at 6:27
2  
This is related to an aside from one of JK's drafts: if $Q$ is the Hilbert cube and $X$ is finite, then the homeomorphism type of $X\times Q$ is the same information as the simple homotopy type of $X$. If $X$ is simply connected, then $Aut(X\times Q)\to hAut(X)$ is 1-connected, but far from an equivalence. If there is torsion in the fundamental group, there may be self-equivalences with non-trivial Whitehead torsion, which thus never lift to homeomorphisms of compact models. I suspect that if you also cross by $R^n$, there are related obstructions for higher $\pi_k$. Maybe $R^\infty$ helps. –  Ben Wieland Jan 25 '11 at 4:12
    
@Ben: Another interesting sequence associated with the finite complex $Y$ is gotten by choosing a homotopy equivalence $Y \simeq M$, where the latter is a codimension zero open submanifold of some Euclidean space. Then in a range we have a fibration sequence $F(M,\text{Top}_n) \to \text{TOP}(M) \to G(M)$. The range tends to infinity when we stabilize $M$ by taking iterated products with the reals. –  John Klein Jan 26 '11 at 1:25
add comment

1 Answer 1

up vote 4 down vote accepted

This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.

  1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction $$ EG(Y) \times_{G(Y)} Y \to BG(Y) $$ is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.

  2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop of the total singular complex of $BG(Y)$.

  3. Let $EG \to BG$ be a universal $G$-principal bundle, and set $$ Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U) $$ Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction $$ EG\times_G Z \to BG $$ is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.

  4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.

share|improve this answer
1  
OK. So it seems that the hard question is really whether one can make alterations to the "kernel" group of homeomorphisms equipped with homotopies to the identity, even for finite CW objects. –  Tyler Lawson Jan 25 '11 at 3:23
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.