Let $U\subseteq \mathbb R^m$ open and suppose that a group structure is defined on $U$ with multiplication $\mu: U\times U \rightarrow U$ of class $C^k,k\geq 1$. Suppose that $V\subseteq \mathbb R^n$ another open wit another group structure defined by its multiplication $\tilde \mu:V\times V \rightarrow V.$ Let $f: U \rightarrow V$ be a differentiable homomorphism (i.e, $f(\mu(g,h)) = \tilde \mu(f(g),f(h))) $. Prove that $f$ has constant rank (i.e, $f'(x)$ has constant rank).
Let $\tilde f: U\times U \rightarrow V\times V, (g,h) \mapsto (f(g),f(h)).$ Since $f$ is an homomorphism, then $f\circ \mu = \tilde \mu \circ \tilde f.$ We have
$\tilde f'(g,h)\cdot (X,Y) = (f'(g)\cdot X, f'(h)\cdot Y),$ so by the chain rule it follows that
$f'(\mu(g,h))\circ \mu'(g,h)\cdot (X,Y) = \tilde \mu'(f(g),f(h))\cdot (f'(g)\cdot X, f'(h) \cdot Y). $
Now, I really don't know how to use these informations in order to show that $f'(g)$ has the same rank for any $g\in U$. It seems that I should consider all these derivatives at $(g,e)$.
Any help, insight? thanks.
Let $g \in U, h=f(g) \in V$, define $\mu_g,\nu_h$ to be the left multiplications by $g$ in $U$ and by $h$ in $V$, so that $f \circ \mu_g=\nu_h \circ f$. Since $\mu_g$, $\nu_h$ are diffeomorphisms, their differentials are invertible at each point.
Differentiate at the neutral element $e$ of $U$: $f’(g) \circ \mu_g’(e)=\nu_h’(f(e)) \circ f’(e)$. It follows that $f’(g)$ and $f’(e)$ are equivalent as linear homomorphisms, ie have the same rank.