Necessary and sufficient conditions for any two points on a manifold to share a coordinate chart? I'm sorry if this question isn't phrased properly, or has a trivial answer. While studying Lie groups, the following question arose for me: given any two elements, is there always a (single) coordinate chart that maps both of them? (Edit: according to Gilmore's definition, a Lie group is by definition connected.)
It's related to my question here, where Gilmore writes the group operation $\phi$ in terms of coordinates. Basically, I'm wondering whether, for any $\alpha, \beta$, they share a chart, so that $\phi^\mu$ can be defined in terms of this chart. I suppose this still leaves the question of whether the result $\gamma$ can also be mapped by the chart.
 A: The answer is yes, and that's just a manifold matter. In general, given any two points $a,b$ in a connected manifold $M$ with $\dim\ge2$, there is a diffeomorphism $f:M\to M$ such that $f(a)=b$ which is the identity off any given connected neighborhood $W$ of $a,b$. Then fix $x,y\in M$ and pick a neighborhood $U$ of $x$ and a diffeomorphism $h:U\to\mathbb R^m$ (a coordinate system at $x$). Pick any $b\in U$, $b\ne x$, and  a connected neighborhood $W$ of $a=y,b$ with $x\notin W$ (which exists because $\dim(M)\ge2$ implies $M\setminus \{x\}$ connected). We have our diffeo with $f(a)=b$ and $f(x)=x$. So $f^{-1}(U)$ is an open set that contains $x,y$ and $h\circ f:f^{-1}(U)\to\mathbb R^m$ is a diffeo, hence $f^{-1}(U)$ is a coordinate system containing both $x,y$. 
A: Your Lie group $G$ comes with an underlying smooth manifold $M$, which has a collection of charts. I think your question is, given points $x_0, x_1 \in M$, is there always one of the given charts that contains both points? The answer is no; each chart is generally much smaller than the whole manifold. For example, the standard charts for the circle $S^1$ correspond to the open subsets $S^1 - \{a\}$ and $S^1 - \{b\}$, where $a$ and $b$ are the north and south poles. Then clearly none of the given charts contains both $a$ and $b$.
