# Why are Lie groups automatically analytic manifolds?

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. $\mathrm{O}(n)$, $n>2$.

• Analyticity doesn't (obviously) make sense on a smooth manifold, so you should start by clarifying what you mean by saying that $\mu$ is analytic. Do you mean that there exists a chart on which it's analytic? (Separately, you mean $O(n)$ in your final paragraph.) – user98602 Jul 11 '16 at 22:12
• @MikeMiller Does this clarification make sense? – Ryan Unger Jul 11 '16 at 22:23
• Yes. ${}{}{}{}$ – user98602 Jul 11 '16 at 22:24

I would like to point out that the question in the title and the question in your post are different. Every smooth manifold $M$ admits a maximal analytic subatlas, and any two choice of analytic structures on $M$ compatible with the smooth structure are analytically diffeomorphic. So every Lie group does admit a smooth structure. The question, of course, is whether $G$ admits an analytic structure such that the multiplication is analytic (that is, that it's an analytic Lie group). It's also true that this analytic structure is unique up to analytic group isomorphism (group homomorphisms are automatically analytic).

As for your question: Suppose $(U,\varphi)$ is your chart centered at $0$. Then $(gU, \varphi \circ L_{g^{-1}})$ is a chart centered at $g$. On overlaps $hU \cap gU$, the transition function is $\varphi L_{g^{-1}h} \varphi^{-1}$. So all you need to know is why $L_{g^{-1}h}$ is analytic on $U \cap h^{-1}g U$; but the very assumption you gave is that it's analytic in $U \times U$. The key point here is that if $h$ and $g$ are too far apart, so that the analyticity of $\mu$ on $U$ doesn't help, then $hU \cap gU$ is empty.

So this provides an analytic atlas on $G$. The question, then, is whether or not $\mu$ is analytic everywhere. But this is basically true by definition of the atlas. I'd rather not write it down, but the rest is further symbol-pushing, exactly of the sort above.

(The fact that a Lie group's multiplication is analytic can be proved by the Baker-Campbell-Hausdorff formula, which you can find in eg Serre's book on Lie groups.)

• Well, the "real" question is why the standard definition of Lie group as a group which is a smooth manifold and where $\mu$ and $i$ are smooth and Helgason's definition where smooth is replaced by analytic are equivalent. – Ryan Unger Jul 12 '16 at 2:05
• Because you can use the Baker-Campbell-Hausdorff formula to get an analytic chart near the identity, as I assume your source uses. – user98602 Jul 12 '16 at 2:06
• How does your construction work for disconnected $G$? I don't think you can reach all of $G$ by multiplication in that case. – Ryan Unger Jul 12 '16 at 2:13
• Absolutely nothing changes. If $g$ and $h$ are in different components, then $gU \cap hU$ is empty, so you don't need to worry about checking analyticity on overlaps. – user98602 Jul 12 '16 at 2:14
• Ah, wonderful. ${}{}{}{}{}{}{}{}$ – Ryan Unger Jul 12 '16 at 2:15

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.