TITLE: Why are Lie groups automatically analytic manifolds?
QUESTION [8 upvotes]: In the book by Kolar, Michor, and Slovak, it is shown that multiplication $\mu:G\times G\to G$ is analytic in some neighborhood of $e$. Specifically, they show that in the chart given by $\exp^{-1}$, with domain some neighborhood of the origin, multiplication is real analytic. They claim that it follows that $\mu$ is analytic on all of $G$. By this, I think they mean there always exists a real analytic structure on $G$ in which $\mu$ is everywhere analytic. 
Could someone please supply the details? I'm assuming the thought is to push this neighborhood of analyticity around the group by using the multiplication map and the fact that a Lie group is generated by any neighborhood of $e$. However, this is only true for connected Lie groups, so this approach would fail for e.g. $\mathrm{O}(n)$, $n>2$.

REPLY [2 votes]: Asssume that your group is linear (a closed subgroup of $GL(n,R)$), and let $\cal G \subset M(n, R)$ its Lie algebra. Then in a neigbourhood of the identity, $G= \{g\in M(n,R)/ \log g \in \cal G \}$, where $r>0$ is a small number, and $\log : B(Id, r)\subset M(n,R)$ is an analytic function. This proves that in the neigbourhood of the identity, $G$ is an analytic submanifold of $Gl(n,R)$, as $\cal G$ is just a linear vector space. But left translation by a matrix $A$ is certainly an analitic (even polynomial) automorphism of $Gl(n,R)$, so $G$ is an analytic submanifold of the algebraic hence analytic. So $G$ is an analytic submanifold of $Gl(n,R)$ and a subgroup, therefore an analytic group. For the general case, you use the fact that $\cal G$ can be embedded as a Lie subalgebra of $M(n,R)$, and apply the same sort of argument.