How do you prove that a the limit of a sequence of Lie group automorphisms is continuous in the whole connected component of the identity?

Question

Let $G$ be a connected Lie group with Lie algebra $\mathfrak g$. The automorphism group Aut($\mathfrak g$) is a Lie group and a closed subgroup of GL$(\mathfrak g)$. The tangent map (at $e$) of any automorphism of $G$ is an automorphism of $\mathfrak g$. It is proven that $\DeclareMathOperator{\Aut}{\operatorname{Aut}} \Psi : \Aut(G) \to \Aut(\mathfrak g)$ is an injective group homomorphism

Now suppose $(f_n)_{n=1}^{\infty}$ is a sequence in $Aut(G)$, such that $(T_e(f_n))_{n=1}^{\infty} \to \psi$ converges in $Aut(\mathfrak g)$

I want to show that $ f := lim_{n\to \infty} f_n$ exists as an continuous automorphism of the abstract group $G$

First of all the $f_n$ are smooth, because they are in Aut(G), so they are Lie group isomorphisms, and therefore smooth by definition

Then by a known property of the exp:

$f_n(\exp(X))=\exp(T_ef_n X), \forall X \in T_eG$ taking the limit yields:

$\lim_{n\to \infty} f_n(\exp(X))=\exp(\psi(X))\in G, \forall X \in T_eG$

$\lim_{n\to \infty} f_n(g)=\exp(\psi(X))=:f(g)\in G, \forall g$ in a nbhd of the identity

I know that the image of the exponential map generates the connected component of the identity $G_e$,and by connectedness this coincides with G so: $G=G_e=<f(T_eG)>=<\exp(\psi(T_eG))>=<\exp(T_eG)>$

This means that convergence is actually valid in the whole group, becuase if I have convergence in the generating set, I must have convergence in the generated set.

I hope this is correct,if not please tell me.

And for my question How do I argue from here that $f$ is continuous?. First I thought it was automatic, but then I recalled from analysis that pointwise convergence of a sequence of functions does not imply continuity. So I am clueless about how to proceed here

Answer 1

If I get your question correctly, then the limit f is always smooth (not only continuous). Indeed, given a connected Lie group $G$ with Lie algebra $\mathfrak{g}$,

1. you consider a sequence $\{f_n\}_{n\in\mathbb{N}}$ in $\operatorname{Aut}_{\textsf{LG}}(G)$, so that linearizing you get the sequence $\{T_ef_n\}_{n\in\mathbb{N}}$ in $\operatorname{Aut}_{\textsf{LA}}(\mathfrak{g})$, and further
2. you assume that $\{T_ef_n\}_{n\in\mathbb{N}}$ converges to $\psi$ in $\operatorname{Aut}_{\textsf{LA}}(\mathfrak{g})$.

These initial data have a few immediate consequences.

• Since $f_n(g\cdot h)=f_n(g)\cdot f_n(h)$, for all $n\in\mathbb{N}$ and $g,h\in G$, the points $g\in G$ such that $\{f_n(g)\}_{n\in\mathbb{N}}$ converges in $G$ is a subgroup of $G$, let us denote it $H$, i.e. $$H:=\{g\in G\mid \{f_n(g)\}_{n\in\mathbb{N}}\ \text{converges in}\ G\}\subseteq G\ \text{is a subgroup}.\tag{1}$$
• since $f_n(\exp(X))=\exp((T_ef_n)X)$, for all $n\in\mathbb{N}$ and $X\in\mathfrak{g}$, and $\{T_ef_n\}_{n\in\mathbb{N}}$ converges to $\psi$, one gets that $\exp(\mathfrak{g})\subseteq H$ with, precisely, $$\lim_{n\to\infty}f_n(\exp(X))=\exp(\psi(X)),\ \text{for all}\ X\in\mathfrak{g}.\tag{2}$$ By the same argument, one gets that $\exp(\mathfrak{g})\cdot g\subseteq H$ for any $g\in H$ with, precisely, $$\lim_{n\to\infty}f_n(\exp(X)\cdot g)=\exp(\psi(X))\cdot \lim_{n\to\infty}f_n(g),\ \text{for all}\ X\in\mathfrak{g}\ \text{and}\ g\in H\tag{3}.$$

As you said, since $G$ is connected and the subgroup $H\subseteq G$ contains $\exp(\mathfrak{g})$, one concludes that $H=G$. Hence there exists a (unique) map $\Psi\colon G\to G$ such that $$\Psi(g)=\lim_{n\to\infty}f_n(g),\ \text{for all}\ g\in G.$$ Now, because of Equation (3), this map $\Psi\colon G\to G$ also satisfies the following identity $$\Psi(\exp(X)\cdot g)=\exp(\psi(X))\cdot\Psi(g),\ \text{for all}\ g\in G.\tag{4}$$ Since the right-hand side of the latter depends smoothly on $X$ and the exponential map $\exp\colon\mathfrak{g}\to G$ induces a diffeomorphism from a neighborhood $\mathcal{V}$ of $0$ in $\mathfrak{g}$ onto a neighborhood $\mathcal{U}$ of $e$ in $G$, Equation (4) tells us that $\Psi$ is smooth on $\mathcal{U}\cdot g$ for any $g\in G$. This proves that $\Psi\colon G\to G$ is smooth on $G$.