7
$\begingroup$

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?

$\endgroup$
6
  • 14
    $\begingroup$ It's part of the definition! $\endgroup$
    – Zhen Lin
    Jul 16, 2014 at 21:22
  • $\begingroup$ 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. $\endgroup$ Jul 16, 2014 at 21:23
  • $\begingroup$ This isn't a theorem, it's one of the Lie group axioms. $\endgroup$ Jul 16, 2014 at 21:24
  • $\begingroup$ @DanielFischer But is it Hausdorff and Locally Homeomorphic to Euclidean space? If so, why? $\endgroup$ Jul 16, 2014 at 21:25
  • 3
    $\begingroup$ @AnthonyPeter: I'd recommend doing some reading about topological groups in general. $\endgroup$
    – Kyle
    Jul 16, 2014 at 21:27

2 Answers 2

6
$\begingroup$

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.

Hope that helps,

$\endgroup$
1
$\begingroup$

As you know that lie group is a structure where G×G→G, it that you can get any element in the same structure with other elements and an operation, now it is a manifold so it's chart is in a same structure which is a diffeomorphism of the elements, so the has to be smooth...

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .