Continuous homomorphisms of Lie groups are smooth

I'm working on problem 20-11 from Lee's "Introduction to Smooth Manifolds", which asks us to prove:

1. Every continuous homomorphism $\gamma : \mathbb R \to G$ is smooth ($G$ a Lie group).
2. Every continuous homomorphism $F : G \to H$ of Lie groups is smooth.

The first part comes with a hint: let $V \subseteq \mathrm{Lie}(G) = \mathfrak g$ be a neighborhood of $0$ such that $\exp: 2V \to \exp(2V)$ is a diffeomorphism (with $2V = \{2X : X \in V\}$). Choose $t_0$ small enough that $\gamma(t) \in \exp(V)$ whenever $|t| \leq t_0$, and let $X_0$ be the element of $V$ such that $\gamma(t_0) = \exp X_0$. Then one can show $\gamma(qt_0) = \exp(qX_0)$ whenever $q = m/2^n$ for some $m,n$.

I've been able to show all of this in the hint, but I'm not sure why that implies $\gamma$ is smooth. Is it because it now depends smoothly on $X_0$, which is in one-to-one correspondence with $t_0$? But why should that be true? And why do we care about the dyadic rational $q$?

Part 2 also comes with a hint: show that there's a map $\phi : \mathfrak g \to \mathfrak h$ so that the following diagram commutes: $\require{AMScd}$ \begin{CD} \mathfrak g @>\phi>> \mathfrak h\\ @V \exp V V @VV \exp V\\ G @>>F> H \end{CD} and then show $\phi$ is linear. But without knowing whether we can talk about $dF_e$, how could we construct such a $\phi$?

Any help with either of these problems would be greatly appreciated (or even a good resource on why continuous homomorphisms of Lie groups are automatically smooth).

• You asked for a reference: Here's one: Corollary 3.50 in Halls 2015 book Lie groups, Lie Algebras, and Representations. He only gives the proof for matrix Lie groups, but the argument holds for any (finite-dimensional?) Lie group. Dec 23, 2020 at 19:40

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 :)
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.
• Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $\phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $\mathfrak g$... Jul 21, 2018 at 18:30
• My guess is that $\phi$ is defined as follows: first, you find, for each $k\in\{1,\ldots,n\}$, a $Y_k\in\mathfrak h$ such that $t\mapsto\exp(tY_k)$ is equal to $t\mapsto F\bigl(\exp(tX_k)\bigr)$. Then you define $\phi$ as$$\exp(tX_1)\ldots\exp(tX_n)\mapsto\exp(tY_1)\ldots\exp(tY_n).$$But this only occurred to me only after answering your question. Jul 21, 2018 at 18:40