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Let $G$ be a real connected Lie group. I am interested in its special homotopy properties, which distinguish it from other smooth manifolds

For example

  1. $G$ is homotopy equivalent to a smooth compact orientable manifold. In particular, Poincaré duality holds for $G$.
  2. $\pi_1(G)$ is abelian, $\pi_2(G) = 0$

The impossibly perfect answer to my question is a list of properties that make up a complete homotopy characterization of Lie groups (that is, in every homotopy type (of smooth manifolds) with such properties, there exists a smooth manifold admitting the structure of Lie groups).

P.S. In this question, I am not interested in the homotopy properties of manifolds that distinguish them from other CW-complexes, for this see resp. question on MO

Arshak Aivazian
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  • You seem to assume $G$ connected. Indeed $G$ is homotopy equivalent to a connected compact Lie group (which are far from arbitrary among compact smooth manifolds). These are classified and, for instance, their rational homotopy groups are known. – YCor Mar 02 '22 at 08:18
  • @YCor Yes, I assume $G$ is connected, added to question text, thanks. – Arshak Aivazian Mar 02 '22 at 09:09
  • This doesn't answer your question but you might be interested in the notion of H-space. – YCor Mar 02 '22 at 09:11
  • I know about it from Hatcher, thanks – Arshak Aivazian Mar 02 '22 at 09:12
  • Just riffing off @YCor's first comment: Looking at https://en.wikipedia.org/wiki/Compact_group#Classification, it seems that your manifold would have to be finitely covered by something of the homotopy type of a product of tori, $Sp(n)$'s, $SU(n)$'s, $\operatorname{Spin}(n)$'s and exceptional Lie groups. In particular it's universal cover is homotopy equivalent to a product of $Sp(n)$'s, $SU(n)$'s, $\operatorname{Spin}(n)$'s and exceptional Lie groups. – Mark Grant Mar 02 '22 at 10:02
  • Hmm, indeed, in particular, the nth homotopy groups of any Lie group are simply some products of the nth homotopy groups of compact, simply connected, simple Lie groups. – Arshak Aivazian Mar 02 '22 at 10:08
  • @AivazianArshak since these are only products up to covering, I'm not sure exactly how one deduces the homotopy groups from those of factors. But I guess it's known. – YCor Mar 02 '22 at 10:13
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    There are manifolds whose homotopy groups are those of a compact Lie groups, but which are not homotopy equivalent to any Lie group. (Neither does homology, cohomology, nor even unstable $\mathcal{A}_p$-algebra structure distinguish these objects.) – Tyrone Mar 02 '22 at 11:56
  • @YCor For every $n > 1$, the covering $p \colon X \to B$ establishes an isomorphism $\pi_n(p) \colon \pi_n(X) \to \pi_n(B)$ – Arshak Aivazian Mar 02 '22 at 11:57

2 Answers2

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The problem you mention has a long history. The best homotopy characterization is probably using the notion of finite loop spaces:

A finite loop space is a space $BG$ such that $\Omega BG$ is homotopy equivalent to a finite CW-complex. There are many of those, but one can give a precise homotopy characterization of which ones come from compact Lie groups: They are the ones admitting a "maximal torus", defined to be a map $({\mathbb C}P^\infty)^r \to BG$, such that the homotopy fiber is homotopy equivalent to a finite complex.

This was the so-called maximal torus conjecture, solved as a consequence of the classification of $p$-compact groups, which states that there is a 1-1-correspondence between connected $p$-compact groups and ${\mathbb Z}_p$-root data, parallel to the classification of connected compact Lie groups, but with $\mathbb Z$ replaced by the $p$-adic integers ${\mathbb Z}_p$.

As a consequence one can also give a classification of all finite loop spaces: If you pick a connected $p$-compact group for every prime $p$ agreeing over $ \mathbb Q$, there is an explicit double coset space of finite loop space structures with this $p$-local data.

A reference is my ICM survey The Classification of p–Compact Groups and Homotopical Group Theory.

Jesper Grodal
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A very useful fact is that every connected Lie group is rationally homotopy equivalent to the product of several odd dimensional spheres where the number of spherical factors is equal to the rank of the group. For example, $SU(n)$ is rationally equivalent to $S^3\times S^5 \ldots \times S^{2n-1}$ and $Sp(n)$ is rationally equivalent to $S^3\times S^7\times \ldots \times S^{4n-1}$. For simple Lie groups the first factor is always $S^3$ and there is only one $S^3$ in each simple group. Also, every simple $G$ has $\pi_3\cong\mathbb Z$. For reference see for example the book "Topology of transitive transformation groups" by Onishchik.

Vitali Kapovitch
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    More precisely, G is rationally homotopy equivalent to $S^{2d_1-1} \times \cdots \times S^{2d_r-1}$, where the $d_i$'s are the degrees of the Weyl group (as defined in eg Humphrey's book on reflection groups). Now, one may ask if all finite loop spaces have the degrees of some Weyl group, i.e., if every finite loop space is rationally equivalent to some compact Lie group. This however turns out not to be the case, though the first counterexample happens in rank 66. This example is given in https://arxiv.org/abs/math/0306234 and also mentioned in my survey linked in my answer above. – Jesper Grodal Mar 11 '22 at 08:45
  • @JesperGrodal That's an interesting example. I was not aware of the conjecture it disproves but it was certainly a natural one. – Vitali Kapovitch Mar 11 '22 at 14:43