Can every manifold be turned into a Lie group? 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 $G$?
 A: If $G$ is a Lie group, then $G$ is homogeneous, meaning that for any $g,h\in G$, there is a diffeomorphism $F:G\to G$ such that $F(g)=h$, namely $F(x)=hg^{-1}x$.  This gives an easy way to find manifolds that cannot be Lie groups.  For instance, let $G$ be the disjoint union of $\mathbb{R}$ and $S^1$.  Then no diffeomorphism of $G$ can map a point of $S^1$ to a point of $\mathbb{R}$ (since a diffeomorphism must preserve the property of being in a compact connected component), so $G$ cannot be a Lie group.
(Note though that actually any connected manifold is homogeneous, whereas most connected manifolds do not admit a Lie group structure.  So admitting a Lie group structure is actually much stronger than just being homogeneous.)
A: As many people are giving intersting counterexamples I thought I should also add one .
Any surface( compact orientable hausdorff 2 manifold) with non zero Euler characteristics cannot be a Lie group because from standard theorem in differential topology , Euler's characteristic of compact orientable lie group is zero. For instance it's 2 for 2 sphere so it can't be a Lie group.
A: There is an easy counterexample: $S^2$ cannot be given a Lie group structure (this is a consequence of the hairy ball theorem). The problem with your construction is that it doesn't offer how to define $m(g,h)$ for any two nonidentity elements $g$ and $h$.
A: Lie groups as manifolds, are very special, owing to the group operations. Basically, "what happens at the identity" determines what happens everywhere. And this means that the tangent bundle $T G$ is always trivializable: here is a sketch of the proof, based on what I remember from Lee's book.
Take any basis $\{v_i\}^n_{i=1}$ for $T_eG$. Since left multiplication $L_g:G\to G:h\mapsto gh$ is a diffeomorphism, it induces an isomorphism $dL_g:T_eG\to T_gG.$ Now, define vector fields $\{V_i\}^n_{i=1}$ by $(V_i)_g:=dL_g(v_i)$and show that they are smooth. Then, since $dL_g$ is an isomorphism, $\{dL_g(v_i)\}^n_{i=1}$ is a basis for $T_gG$, so the vector fields $\{V_i\}^n_{i=1}$ are a global frame for $TG$.
A: To add to the previous answers, topological groups have abelian fundamental groups.
$G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian
Orientable surfaces of genus at least two are not parallelizable, but this is another way to show that they can't be Lie (even topological) groups. The Klein bottle is parallelizable (edit: no it's not), but its fundamental group is not abelian, so it can't be a group either.
A: 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 $\pi_4(SU(3)) = 0$.)
