A technical detail in the definition of a Lie group

Question

I begin with the following common Definition:

A Lie group $ G $ is a differentiable (smooth, analytic) manifold equipped with a group structure, i.e. with an associative binary operation

$$ G\times G \longrightarrow G :\qquad\left\{x\,, y\right\} \longmapsto x\centerdot y $$

and an inversion operation

$$ G \longrightarrow G :\qquad g \longmapsto g^{-1} $$

both of which are differentiable (smooth, analytic).

For points of an open set $ U_{a} $ projected by a homomorphism $\alpha_a$ into an open subset of $R^N$, the inverse of the homeomorphism $\alpha_a$ can be written as

$$ x = g(\alpha_x), $$

where

$$ \alpha_x = (\alpha_x^1 , . . . , \alpha_x^N) $$

are the coordinates of the manifold point (group element) $ x\in U_a$.

When two group elements $ x $ and $ y $ and their product $ x y $ reside in the same open set $ U_a $, the coordinates of the product are functions of the coordinates of the two multipliers:

$$ x=g( \alpha_x ) ,\quad y=g( \alpha_y ) ,\quad xy = g( \alpha_{xy} ) \quad \Longrightarrow \quad g( \alpha_{xy} ) = g( \alpha_x ) \centerdot g( \alpha_y ) \quad \Longrightarrow \quad \alpha_{xy} = \alpha_{x y}( \alpha_x , \alpha_y). $$

This remains true also when the elements $ x $, $ y $, $ xy $ lie in different coordinate charts, in which case the dependence $ \alpha_{xy}(\alpha_x,\alpha_y) $ will have to incorporate coordinate transformations between charts.

Dependent on whether we postulate in the above Definition that our manifold is differentiable (smooth, analytic), these transformations must be assumed differentiable (smooth, analytic)

Now, my Question:

Does this assumption automatically warrant that the resulting functions $ \alpha_{xy}(\alpha_x,\alpha_y)$, too, come out differentiable (or smooth, or analytic)? Or should this be checked on each occasion?

Answer 1

Your notation is a bit odd.

So, assume $\alpha$ is a diffeomorphism from $U\subseteq G$ to $V\subseteq\Bbb R^n$, and let's denote its inverse by $\beta:V\to U$ and the multiplication map $G\times G\to G$ by $m$.

Then $\alpha_x$ is just $\alpha(x)$ and $\beta(v)$ is what you called $g(v)$, and the map you're looking at is a restriction of the below composition $$V\times V\underset{\beta\times\beta}\to U\times U\ \underset{m}\to \ U\ \underset{\alpha}\to \ V\,.$$ Namely, define $D:=\{(x,y)\in U\times U:xy\in U\}$, show that it's open, and restrict the above to $(\alpha\times\alpha)(D)$ to obtain a well defined map, which is automatically smooth (differentiable/analitical).

Note that, on the other hand, in general nothing guarantees that $D$ is nonempty.