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 $G$?
Best Answer
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 $h$.