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For example, given a Lie group, its fundamental group must be Abelian. So $\Sigma_g$ ($g>1$) can't have Lie group structure. We also know for $S^n$ only $n=0,1,3$ can have Lie group structures.

In general, what's the sufficient or necessary conditions for a manifold to have Lie group structure?

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    I can think of a couple more necessary conditions: the second homotopy group must be trivial, and the third torsion-free, according to https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups?rq=1 . Also, the (co)homology with field coefficients must carry a Hopf algebra structure. – Mark Grant Sep 11 '17 at 06:09
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    Of course a Lie group is parallelisable. – Benoît Kloeckner Sep 11 '17 at 07:38
  • A related post could be the following https://mathoverflow.net/questions/5262/lie-groups-and-manifolds/5492#5492 – Ali Taghavi Sep 11 '17 at 07:53
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    I'll understand the question in the smooth setting. 1) A manifold has a Lie group structure iff all its components are diffeomorphic and if some of its component admits a Lie group structure. This reduces to the connected case. 2) A connected manifold has a Lie group structure iff it's diffeomorphic to $K\times\mathbf{R}^n$ for some $n$ and some compact Lie group $K$. This reduces to the connected compact case, for which there is a full classification. – YCor Sep 11 '17 at 08:21
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    Another remark about $\pi_3$: if a connected manifold carries a Lie group structure and has trivial $\pi_3$ then it's a torus. – YCor Sep 11 '17 at 08:24

1 Answers1

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Compact Lie groups are finite loop spaces, but Andersen, Bauer, Grodal, Pedersen (Invent. Math., 2004) give an example of a finite loop space that is not (even rationally) homotopy equivalent to any compact Lie group.

John Rognes
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