# What makes a Lie Group a Differentiable Manifold?

I've recently been trying to glance at the definition of a Lie group, but I'm not clear as to why a Lie group is defined the way it is, and why this becomes a smooth manifold. For example, if we have a Lie group $G$, why do we have the requirement that $G \times G \to G$, $(x,y) \mapsto xy^{-1}$ is a smooth continuous mapping between the product manifold and the manifold $G$.

What is it that makes $G$ a topological manifold in the first place? How does it inherit differentiable structure?

• It's part of the definition! – Zhen Lin Jul 16 '14 at 21:22
• You have a set. On that set, you have a group structure, and a differentiable structure. If the two structures play together nicely, it is a Lie group. – Daniel Fischer Jul 16 '14 at 21:23
• This isn't a theorem, it's one of the Lie group axioms. – Adam Hughes Jul 16 '14 at 21:24
• @DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why? – Anthony Peter Jul 16 '14 at 21:25
• @AnthonyPeter: I'd recommend doing some reading about topological groups in general. – Kyle Jul 16 '14 at 21:27

When you ask that $G \times G \to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G \times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a smooth structure on $G$ and this property you ask of multiplication is extra.