Continuous homomorphisms of Lie groups are smooth

Solution 1:

Here is a better (more elegant) proof for both questions using Cartan-Von Neumann's theorem on closed subgroups.

Let $H$ be the graph of $\gamma$, then $H$ is closed (and thus a Lie subgroup of $G\times \mathbb{R} \cong \mathbb{R}\times G$, so $H\hookrightarrow \mathbb{R}\times G$ is smooth) The projection from $H$ to $\mathbb{R}$ is a bijective lie group morphism, and therefore its inverse is smooth. Now just have a look at the following diagram

As for the second one, if $\varphi : G\to H$ is continuous, let $\Gamma_\varphi : G\to G\times H$ be the "graph map" which sends $g$ to $\big(g, \varphi(g) \big)$. Note that $\Gamma_\varphi $ is a group homeomorphism from $G$ onto its image $\mathrm{Graph}(\varphi)$ whose inverse is the restriction of the projection $G\times H \to G$. Now, $\mathrm{Graph}(\varphi)$ is clearly closed (therefore a Lie subgroup), and the inverse of $\Gamma_\varphi$ is smooth with constant rank, and thus $\Gamma_\varphi $ is a diffeomorphism. If $q: G\times H \to H$ is the second projection (smooth), you just have to note that $\varphi=q\circ \Gamma_\varphi $ and we are done :)

Solution 2:

Those dyadic rationals are dense in $[-1,1]$. So, near $0$, we have, by the continuity of $\gamma$, $\gamma(t\times t_0)=\exp(tX_0)$. In other words, $\gamma(t)=\exp\left(\frac t{t_0}X_0\right)$ and therefore $\gamma$ is differentiable.

For the other question, let $(X_1,\ldots,X_n)$ be a basis of $\mathfrak g$ and consider the map\begin{array}{rccc}\alpha\colon&\mathbb{R}^n&\longrightarrow&G\\&(t_1,\ldots,t_n)&\mapsto&\exp(t_1X_1)\ldots\exp(t_nX_n).\end{array}Then $\alpha$ is smooth. If $(x_1,\ldots,x_n)\in\mathbb{R}^n$, then\begin{align}\left.\frac{\mathrm d}{\mathrm dt}\alpha\bigl(t(x_1,\ldots,x_n)\bigr)\right|_{t=0}&=\left.\frac{\mathrm d}{\mathrm dt}\exp(tx_1X_1)\ldots\exp(tx_nX_n)\right|_{t=0}\\&=x_1X_1+\cdots x_nX_n.\end{align}and therefore $D\alpha_0$ is an isomorphism. So, $\alpha$ induces a diffeomorphism from a neighborhood $U$ of $(0,0,\ldots,0)$ onto a neighborhood $V$ of $e_G$. If $(t_1,\ldots,t_n)\in U$, then\begin{align}F\bigl(\alpha(t_1,\ldots,t_n)\bigr)&=F\bigl(\exp(t_1X_1)\ldots\exp(t_nX_n)\bigr)\\&=F\bigl(\exp(t_1X_1)\bigr)\ldots F\bigl(\exp(t_nX_n)\bigr),\end{align}which is smooth. Since $F|_V$ is smooth, $F$ is smooth.