TITLE: Can every manifold be turned into a Lie group?
QUESTION [30 upvotes]: I am studying Lie theory and just thought of this random question out of curiosity. Can any manifold be turned into a Lie group?
More precisely, given a manifold $G$, can we always construct (or prove the existence of) some smooth map $m:G\times G\to G$ that makes $G$ into a Lie group? If not, is there an easy counterexample?
I could imagine a construction going something like this: pick an arbitrary point $e\in M$ to be the identity, and define $m(e,g)=m(g,e)=g$ for all $g\in G$. Then we already have the elements of the Lie algebra given as the tangent space at the identity $T_eG$, and maybe we can use these to extend $m$ to all of $G$?

REPLY [15 votes]: The answers so far are great, but I wanted to add some more obstructions.  Suppose $M$ is a manifold which can be given the structure of a Lie group.  Then $M$ has the following properties...

$\pi_1(M)$ acts trivially on $\pi_n(M)$
Each $\pi_n(M)$ is finitely generated.
$\pi_2(M) = 0$.
$\pi_{2k}(M)$ is a finite abelian group for all $k\geq 1$.
$\pi_3(M)$ contains no torsion.
If $M$ is compact, then at least one of $\pi_1(M)$ and $\pi_3(M)$ contains $\mathbb{Z}$ as a subroup.
If $M$ is non-compact, then $M$ must be diffeomorphic to $\mathbb{R}^k\times N$ for some compact Lie group $N$.
If $M$ is simply connected, then it can only torsion of order $2$, $3$, or $5$ in its cohomology groups.

There are still many manifolds which pass all these obstructions (as well as all the obstructions in the other answers!) - for example, $M = S^3\times S^5$.  However, this $M$ isn't a Lie group (though the only way I know to show this is using the classification.  It's simply connected and dimension $8$, so the only Lie group $M$ could be diffeomorphic to is $SU(3)$.  However, $\pi_4(M) = \mathbb{Z}/2\mathbb{Z}$ while $\pi_4(SU(3)) = 0$.)