1
$\begingroup$

Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.

Thank you!

$\endgroup$

2 Answers 2

2
$\begingroup$

I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.

Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).

We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.

$\endgroup$
2
$\begingroup$

Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.

$\endgroup$

You must log in to answer this question.

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