11
$\begingroup$

Is there an exmaple of a closed smooth connected manifold $M$ having a structure of $A_{\infty}$-space (with unit) but $M$ is not homeomorphic to a compact connectd Lie group as space ?

Edit: First, I would like to thank Fernando, Vladimir and Jesper for the excellent answers! My original motivation was the following question: Suppose that $\mathcal{A}_{\infty}$ is a cofibrant replacement of the associative operad $\mathcal{A}$ in the category $\mathsf{Op}$ of topological operads (non-symmetric) and suppose that $X$ is a finite connected $CW$-complex. Is it true that $$Hom_{\mathsf{Op}}(\mathcal{A}_{\infty},End(X))\neq \emptyset \Rightarrow Hom_{\mathsf{Op}}(\mathcal{A},End(X))\neq \emptyset ?$$

Here is a counterexample: Suppose that $M$ is a smooth closed connected manifold with a structure of $\mathcal{A}_{\infty}$-space (i.e, Loop space) such that $M$ is not homotopy equivalent to a Lie group. By definition $Hom_{\mathsf{Op}}(\mathcal{A}_{\infty},End(M))\neq \emptyset$. On another hand if $Hom_{\mathsf{Op}}(\mathcal{A},End(M))\neq \emptyset$, this will implie that $M$ has at least one structure of topological monoid. By Wallace Theorem any closed manifold with a structure of topological monoid is a Lie group, hence $M$ is a Lie group. Contradiction.

$\endgroup$
  • 2
    $\begingroup$ Why the cryptic way of asking this question? How about just a manifold which has the structure of a loop space up to homotopy? That's the same thing as a connected manifold with an $A_\infty$-structure. $\endgroup$ – John Klein Feb 18 '16 at 21:15
  • 3
    $\begingroup$ @JohnKlein I had precisely the same reaction, but one man's meat is another man's poison. $\endgroup$ – Vidit Nanda Feb 18 '16 at 23:42
  • 1
    $\begingroup$ Ilias, how do you make End(X) a reduced operad? $\endgroup$ – Fernando Muro Feb 19 '16 at 10:28
  • $\begingroup$ Fernando, just made an edit. $\endgroup$ – Ilias A. Feb 19 '16 at 14:20
  • $\begingroup$ @IliasAmrani you should maybe stress that $\mathcal A$ is the unital associative operad (all spaces can be endowed with topological semigroup structures). $\endgroup$ – Fernando Muro Feb 19 '16 at 14:26
12
$\begingroup$

Yes:

As has been pointed out, admitting the structure of a connected $A_\infty$-space is the same thing as being homotopy equivalent to a connected loop space.

The Hilton-Roitberg criminal, mentioned by Vladimir Dotsenko, was the first example of a closed manifold homotopy equivalent to a loop space, but not to a Lie group. The space has the rational homotopy type of the compact Lie group Sp(2).

In fact there exists "rational criminals", i.e., finite loop spaces that are homotopy equivalent to closed smooth manifolds (as always, by the Broto-Kitchloo-Notbohm-Pedersen theorem), but not even rationally homotopy equivalent to a Lie group. However, they only start occurring in rank 66 and onwards.

See this paper that I wrote with Andersen-Bauer-Pedersen

$\endgroup$
9
$\begingroup$

If you require just "not homeomeorphic", then there are very silly examples of all sorts. What you want to ask is "not homotopic", I suppose.

For that, I know some useful references in "Which H-spaces are manifolds? I" by Cappell and Weinberger (http://www.sciencedirect.com/science/article/pii/0040938388900171), e.g. you can look at the manifold appearing in Formula (1.4) in "On Principal S3-Bundles Over Spheres" by Hilton and Roitberg (http://www.jstor.org/stable/1970683).

$\endgroup$
8
$\begingroup$

Yes, any connected A-infinity space with finitely generated homology is homotopy equivalent to a closed smooth connected manifold, by results of Bauer-Kitchloo-Notbohm-Pedersen. Many of those are known not to be homotopy equivalent to Lie groups (see references therein).

$\endgroup$
  • $\begingroup$ Thanks for the reference! could you tell me exactly where are the examples in the reference ? $\endgroup$ – Ilias A. Feb 18 '16 at 19:21
  • $\begingroup$ @IliasAmrani as it's said in the first paragraph, examples are to be found in the paper by Hilton and Roitberg, which Vladimir links to in his answer. $\endgroup$ – Fernando Muro Feb 18 '16 at 21:20

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.