Let $H$ be a (not necessarily Lie) subgroup of the connected Lie group $G$. Suppose that every two points in $H$ can be connected by a smooth path contained in $H$. Also suppose that there are curves $c_i:[0,1]\rightarrow H$ such that $c_i'(0)=X_i\in\mathfrak{g}$, where the $X_i$ form a vector-space basis for $\mathfrak{g}$.
I'm trying to show $H=G$. The hint I'm given is to consider a function $f:[0,1]^r\rightarrow G$ defined as
$$ (t_1,\ldots,t_r)\mapsto \prod c_i(t_i) $$
My idea was to find a neighborhood of the identity $V\subset G$, that was a diffeomorphic image of an open set $U\subset\mathfrak{g}$, under the map $\mathrm{exp}:U\rightarrow V$. Because if the image of $f$ contains any open neighborhood of the identity, I'm done, because it will generate $G$. So I restrict the domain of $f$ to some set $W$ where $f(W)\subset V$.
Then we have the map $F: W\rightarrow\mathbb{R}^r$, where $F(t_1,\ldots,t_r)=(s_1,\ldots,s_r)$ is the unique tuple with
$$ \prod c_i(t_i) = \mathrm{exp}\left(\sum s_iX_i\right) $$
To finish this argument, I need to show $F(W)$ contains an open subset of the origin. However, I can't figure out how to do that. Is there some way to finish this proof? Or is there an alternative way I'm missing?
Best Answer
Hint: Use the inverse function theorem. – Moishe Kohan
Of course! You have done 99.99% of the proof! Do you believe that $F$ from an open neighborhood of $0 \in \mathbb{R}^r$ to $\mathbb{R}^r$ is smooth?! Then all that remains to show is that $Df(0)$ is non-singular, i.e. is a full rank matrix. But this is immediate, because every $ \vec{b}$ corresponds to some linear combination of $X_i$'s, say $$ a_1 X_1 + \cdots + a_r X_r \ . $$ Now the derivative $Df(0)$ maps the tangent to the curve $ t \to (a_1t,\cdots,a_rt)$ to $\vec{b}$. (Equivalently, the slope of the curve $t \to F(a_1t,\cdots,a_rt)$ at $t=0$ is $\vec{b}$.)
To see why, just compute partial derivatives at $t=0$ and the fact that derivative of each $c_j(t)$ equals $X_j$.