# Is smoothness of multiplication redundant in the definition of Lie Group?

It is well known that in the definition of Lie groups, we actually only need that the multiplication be smooth, since this implies that inversion is smooth.

I'm now trying to solve the following exercise:

Show that in the definition of Lie group, it is enough to assume that the inverse map $$G \to G, g \mapsto g^{-1}$$ is smooth.

I don't know how to attack this problem, and also couldn't find any other references to it in the literature.

Could I get some hints or references on how to solve it?

• @coffeemath it's an exercise from some notes in my course, as far as I'm aware it's not in any other book Commented Jan 11 at 15:54
• The title of your question is misleading. Commented Jan 12 at 9:50
• @ImHackingXD Of course! Commented Jan 12 at 10:21
• I’m not sure it’s true: in the classical $\mathbb{R}$ endowed with the non-smooth multiplication $(x,y) \longmapsto \sqrt[3]{x^3+y^3}$, the inverse function is $x \longmapsto -x$, which is smooth? Commented Jan 12 at 13:10
• @Aphelli You should post this as an answer. Commented Jan 12 at 16:14

The statement is wrong. Consider the classical manifold $$\mathbb{R}$$, with the non-smooth “addition” law given by $$(x,y) \longmapsto \sqrt[3]{x^3+y^3}$$. This is a topological group law, and the inverse map is $$x \longmapsto -x$$, which is smooth.