Map $n(g,h) = gh^{-1}$ is smooth implies $G$ is a Lie Group.

701 Views Asked by At

$G$ is a smooth manifold with group structure. The map $n(g,h) = gh^{-1}$ is smooth implies $G$ is a Lie Group (exercise 2.8, John Lee). We are using the definitions of smooth and etc from John Lee's "Smooth Manifolds".

We are suppose to show this using the definition of a Lie group as a smooth manifold with group structure, such that the multiplicative group operation and inverse are smooth maps.

Attempt:

The inverse map $n(e,h) = h^{-1} $ is smooth, since the restriction of $n$ is also smooth.

$n(g,e) = g$ is smooth, since the restriction of a smooth map is smooth. Let $m(g,h) = n(g,e)n(e,h) = gh$. This is smooth since it is the multiplication of smooth maps.

Question: Is my attempt correct? Especially the assumptions that restriction and multiplication of a smooth map is smooth.

2

There are 2 best solutions below

0
On BEST ANSWER

For completeness: my attempt is correct. (see the comments)

0
On

I believe that your attempt is wrong for two reasons - one is the calculation error pointed out in the comments, but that one can be fixed by $$m(g,h) = n(g,e) \cdot n(e, \iota(h)).$$ The more important reason is that you use smoothness of the multiplication in the last step. Your proof boils down to showing that $$G \times G \to G \times G, \, (g,h) \mapsto (g,h)$$ is smooth, after which you apply the multiplication and use some type of "composition of smooth functions is smooth"-argument, but this already requires the multiplication to be smooth.

For completeness, here is an argument that should be correct: Let $\mu: G \times G \to G$ denote the multiplication and $\iota: G \to G$ the inversion on $G$. The map $$f: G \to G \times G, \, h \mapsto (e, h)$$ is smooth (since it is smooth in both components), so $\iota = \eta \circ f$ is smooth. Up to this point your argument works and this is just another formulation of it. The above now implies that the map $$\operatorname{id}_G \times \iota: G \times G \to G \times G, \, (g,h) \mapsto (g,h^{-1})$$ is smooth as well and we thus obtain smoothness of the composition $$\mu = \eta \circ (\operatorname{id}_G \times \iota).$$