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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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    $\begingroup$ I can think of a couple more necessary conditions: the second homotopy group must be trivial, and the third torsion-free, according to mathoverflow.net/questions/8957/… . Also, the (co)homology with field coefficients must carry a Hopf algebra structure. $\endgroup$
    – Mark Grant
    Commented Sep 11, 2017 at 6:09
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    $\begingroup$ Of course a Lie group is parallelisable. $\endgroup$
    – Benoît Kloeckner
    Commented Sep 11, 2017 at 7:38
  • $\begingroup$ A related post could be the following mathoverflow.net/questions/5262/lie-groups-and-manifolds/… $\endgroup$
    – Ali Taghavi
    Commented Sep 11, 2017 at 7:53
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Sep 11, 2017 at 8:21
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    $\begingroup$ Another remark about $\pi_3$: if a connected manifold carries a Lie group structure and has trivial $\pi_3$ then it's a torus. $\endgroup$
    – YCor
    Commented Sep 11, 2017 at 8:24

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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.

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