$\DeclareMathOperator{\inv}{inv}$
I am trying to understand the proof of the following from this document:
Let $M$ be a smooth manifold which admits a group structure such that the multiplication map $m:G\times G\to G$ defined as $m(g, h)=gh$ for all $g, h\in G$ is a smooth map.
Then the inversion map $\inv:G\to G$ defined as $\inv(g)=g^{-1}$ for all $g\in G$ is a smooth map.
The proof in the document proceeds as follows:
First note that the multiplication map is a constant rank surjective smooth map, and is therefore a submersion. Therefore, the level set $\Delta=m^{-1}(e)$ of $e\in G$, $e$ being the identity of the group structure, is an embedded submanifold of $G\times G$ of dimension $n$.
Now let $\pi_1:G\times G\to G$ be the projection on the first coordinate.
Consider the map $\pi_1\circ i:\Delta\to G$, where $i:\Delta\to G\times G$ is the inclusion map, which we know is a smooth embedding.
It is claimed in the document that $\pi_1\circ i$ is a diffeomorphism.
The reasoning given seems to be this: The map $\pi_1\circ i$ is smooth and is a homeomorphism and hence by the inverse function theorem, it is also a diffeomorphism.
This seems to suggest that any smooth homeomorphism is a diffeomorphism. But this is not true since the map $x\mapsto x^3:\mathbf R\to \mathbf R$ is a smooth topological embedding but not an immersion at $x=0$.
How do I show that that composition $\pi_1\circ i$ is a diffeomorphism?
Best Answer
You are right, the argument isn't conclusive, it's missing a piece, but it's fixable. You need to note that the map $p_1=\pi_1\circ\iota:\Delta'\to G$ has constant rank equal to $\dim G=\dim\Delta'$.
Indeed, let $m=(g,g^{-1})\in\Delta'$ be a point in $\Delta'$, then \begin{array}{cccc} \theta_g&:&G\times G&\longrightarrow & G\times G\\ &&(x,y) & \longmapsto &(gx,yg^{-1}) \end{array} is a diffeomorphism of $G\times G$ and sends $\Delta'$ diffeomorphically to itself (let's call $\widetilde{\theta_g}$ the induced diffeomorphism of $\Delta'$). Then $$p_1\circ\widetilde{\theta_g}=\pi_1\circ\iota\circ\widetilde{\theta_g}=\pi_1\circ\theta_g\circ\iota=L_g\circ\pi_1\circ\iota=L_g\circ p_1$$ (where $L_g$ is left multiplication by $g$). If we calculate the differentials at $m_0=(e,e)\in\Delta'$, we get $$d_{m}p_1\circ d_{m_0}\widetilde{\theta_g}=d_eL_g\circ d_{m_0}p_1 $$ Since $\widetilde{\theta_g}$ and $L_g$ are diffeomorphisms, $p_1$ has the same rank at $m\in\Delta'$ as at $m_0\in\Delta'$, so that it has constant rank ($\leq\dim G$). It follows from the constant rank theorem (see the first theorem in these notes) and the bijectivity of $p_1$, that $p_1$ has constant rank equal to $\dim G$, and now you can apply the global inversion theorem.