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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_P
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    You'll need the Lie group to be connected, at least. – Hew Wolff Aug 19 '19 at 00:07
  • Thanks. In the Gilmore book, he includes connectedness in the definition of a Lie group. Is connectedness sufficient? – A_P Aug 19 '19 at 00:30
  • My guess (see the comment on my answer) is yes, connectedness is sufficient, but I have only a sketch of a proof to support this guess. – Hew Wolff Aug 19 '19 at 00:33
  • There is ambiguity to your question. When you say "... single coordinate chart" you mean a single chart in a particular predetermined compatible atlas, or a chart in the maximal structure generated by such an atlas? If you just mean the second, then take disjoint neighborhoods of the two points in question, each homeomorphic to an open ball via a chart of your atlas. The disjoint union of these balls, and the disjoint union of their respective charts (plus possibly a translation to disconnect the Euclidean balls) is a chart in the smooth structure. The domain of a chart need not be connected. – Laz Aug 19 '19 at 03:27
  • @Laz Thanks! I don't understand compatibility and maximality very well (for example, what it means for a single predetermined atlas to be compatible), but I'm pretty sure I mean the second thing! So if I understand correctly, it is always possible to find a chart in the maximal atlas that maps disjoint neighborhoods of two points to disjoint Euclidean balls? The translation you mention is in case our charts mapped the neighborhoods onto overlapping regions, yes? I guess when you put it that way, it's obvious! – A_P Aug 19 '19 at 04:27
  • I didn't realize the ambiguity in my question, by the way. I thought that by asking if such a chart exists without referencing a particular atlas, it was implied that we are free to choose the atlas. It seems like when talking about a manifold, it's sometimes assumed that an atlas has already been chosen? – A_P Aug 19 '19 at 04:33
  • Well, to construct a manifold you first start with a compatible atlas (its domains cover the underlying space $M$, and the transition maps between every two charts is smooth, this is the "compatible" part), and then take the family of all charts in $M$ compatible with all those charts you started with. This is the maximal atlas or smooth structure I mean. Then, by a smooth manifold one usually means a topological manifold with a smooth structure. – Laz Aug 19 '19 at 04:42
  • Now, if you only meant charts in a compatible atlas, not necessarily maximal, your claim is false, as @HewWolff pointed out. If you allow charts to live in the smooth structure, then your claim is true, as I proved above in my first comment. – Laz Aug 19 '19 at 04:44
  • Ah okay, so "allow charts to live in the smooth structure" means "allow choosing any chart that is compatible with some compatible atlas." I think this is what I mean. I say "I think" because I didn't even realize smoothness was important: even for a non-smooth manifold, I imagine one can construct many atlases that have some weaker compatibility-like constraint between them? I was imagining choosing a chart amongst all such atlases. – A_P Aug 19 '19 at 04:52
  • Well, I only assumed smoothness because you were talking about Lie groups, which are required to be smooth by definition, and then generalized to smooth manifolds. But, if you don't mean smooth, what do you mean? Topological manifolds, topological groups? – Laz Aug 19 '19 at 05:08
  • Oh right, Lie groups must be smooth. Is there an equivalent of "compatibility" for differentiable manifolds or other constraints weaker than smooth? BTW, if you want to write this up as an answer I'll accept it. Thanks! – A_P Aug 25 '19 at 19:01

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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 $x,y$.

Jesus RS
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  • (And, if $\dim M = 1$, then $M$ is diffeo to either $S^1$ or $(0,1)$. For $S^1$, given any two distinct points $x,y$, pick a third distinct point $z$. Then $S^1\setminus{z}$ is a chart containing $x$ and $y$. For $(0,1)$, $(0,1)$ is a chart containing all points.) – Jason DeVito - on hiatus Nov 20 '19 at 14:15
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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 $b$.

Hew Wolff
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    But that's silly. If you tell me $a$ and $b$, I can always choose a chart on the circle that contains them both. My reading of the question is that one specifies the two points ahead of time. (If a fixed chart is supposed to contain any two arbitrary points, then of course it must contain the whole manifold.) – Ted Shifrin Aug 19 '19 at 00:18
  • I know almost no topology, so apologies if this is stupid, but: while your given charts of course do not cover both $a$ and $b$, isn't it true that we can pick some other chart corresponding to $S^1 - {c}$ (for any $c$ other than $a$ or $b$) that does contain both $a$ and $b$? (Edit: what @TedShifrin said.) – A_P Aug 19 '19 at 00:19
  • @A_P: Yes, of course you are correct. – Ted Shifrin Aug 19 '19 at 00:19
  • However, it's possible that you meant the following: can we find a new chart, compatible with all the existing charts, which contains both points? In that case, my guess is yes: we can take a path $\gamma$ between $x_0$ and $x_1$, take a small neighborhood $N$ of $\gamma$, and choose a smooth homeomorphism $N \to \Bbb{R}^n$ which is given by all the old charts that $\gamma$ passes through. – Hew Wolff Aug 19 '19 at 00:23
  • @TedShifrin, Hew Wolff's answer is not that silly. The OP never specified if he was referring to charts in a fixed compatible atlas, or charts in the maximal structure generated such a compatible atlas. Plus, nobody said anything about a fixed chart containing every two points ... – Laz Aug 19 '19 at 03:39