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Let $M$ be a smooth manifold (a topological manifold with a smooth structure) endowed with a transitive smooth action by a Lie group $G$. Can $M$ be a manifold with boundary, or necessarily without boundary?

Work/Confusions:

In a topological manifold $M$ (possibly with boundary), each point $p \in M$ must have an open neighborhood $p \in U$ which is homeomorphic either to an open set in $\mathbb{R}^n$ or to an open set in $\mathbb{H}^n := \{ (x^1, \dots, x^n) \in \mathbb{R}^n \mid x^n \geq 0 \}$. The set $\partial \mathbb{H}^n := \{ (x^1, \ldots, x^{n-1}, 0) \mid x^1, \dots, x^{n-1} \in \mathbb{R} \}$ is called the boundary of $\mathbb{H}$.

If $p\in M$ is in the domain of a chart that is a hormeomorphism to an open subset of $\mathbb{R}^n$, then $p$ is said to be an interior point, and if it is a domain of a boundary chart (a map $\phi\colon M\supset U\to \mathbb{R}^n$ such that $\phi(U)$ is an open subset of $\mathbb{R}^n$ and $\phi(U)\cap\partial\mathbb{H}^n\neq\emptyset$, where $U\subset M$ is an open subset).

Each point of $M$ is either a boundary point or an interior point, but not both.

$G$ as a manifold cannot have a boundary: every point is interior. Indeed, $G$ is a topological group, and thus the inversion and multiplications by $G$ on $G$ themselves are homeomorphisms from $G$ to itself. (reading Can a Lie group be a manifold with boundary?)

But if an action $\phi \colon G\times M \to M$ is continuous, then each map $\phi_g \colon M\to M$ is a homeomorphism of $M$ ($\phi_g$ is just $\phi_g(x)=gx$).

Now, $G$ actis on $M$ transitively. So if one point is in the interior, then every point is? But at least one point must be an interior point because otherwise all the points are on the boundary and we have $n-1$-dimensional (as opposed to $n$) manifold? So every point must be the interior?

Also, $M$ can be identified with $G/G_p$, where $G_p=\{g\in G\mid gp=p \}$ with an arbitrary $p\in M$. [J. M. Lee, smooth manifolds, Theorem 21.18 (Homogeneous Space Characterization Theorem).]

But I have heard when $G$ and $M$ are just topological spaces and the action $G\times M\to M$ is continuous, $G/G_p$ and $M$ are not necessarily homeomorphic under the quotient topology, so we need more structure in $G$ or $M$?

shall.i.am
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1 Answers1

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Being a boundary point of a manifold can be characterised purely topologically (in terms of local homology groups say). So for each group element $g$ as $x\mapsto gx$ is a homeomorphism of $M$ to itself, then $gx$ is a boundary point iff $x$ is. As $G$ is transitive, then either all points of $M$ are boundary points or none are. They can't all be boundary points.

Angina Seng
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  • I don't really understand the local homology group argument to be honest (Lee's topological manifolds one seems to have some content but haven't read it seriously). – shall.i.am Dec 10 '17 at 21:58
  • But basically as in the question itself, "if an action $\phi \colon G\times M \to M$ is continuous, then each map $\phi_g \colon M\to M$ is a homeomorphism of $M$. Now, $G$ acts on $M$ transitively. So if one point is in the interior, then every point is"? I wasn't too sure why all can't be boundary points, though. Is it because since it means $n-1$ dim and it contradicts with the manifold being $n$-dim? (I think this makes sense?) Or would I need to study the homology group to see this? – shall.i.am Dec 10 '17 at 21:59
  • @shall.i.am A neighbourhood of a boundary point in a manifold-with-boundary contains a nonempty open subset of interior points.... – Angina Seng Dec 11 '17 at 04:19
  • So sorry but could I have an explanation or a reference for this... – shall.i.am Dec 11 '17 at 05:46
  • Reading Lee's smooth manifold pp. 24--26, I can't really see why $\text{Int}(M)=\emptyset$ must be excluded from the definition, except for saying $M$ must have at least one point whose neighborhood is homeomorphic to $\mathbb{R}^n$ not $\mathbb{R}^{n-1}$ because it is $n$-dimensional manifold-with-boundary. Oh wait are we saying the same thing? – shall.i.am Dec 11 '17 at 09:44