Let $M$ be a compact manifold with boundary and suppose a compact group $G$ acts on it. Can one always extend the action beyond the boundary? More precisely, does there always exist a $G$-manifold with boundary $N$, with $\dim M=\dim N$, and a $G$-equivariant embedding of $M$ into the interior of $N$?
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2 Answers
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If you are asking for a topological action, then it would seem you can just double $M$ along the boundary. I am not sure this works in the smooth category..
answered Mar 3, 2016 at 19:30
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$\begingroup$ You can always take a metric such that the boundary has a collar neighbourhood which looks like a product. You have to make sure first that the action of $G$ respects the product structure. If you average over $G$, the action will be by isometries. Then you can double. $\endgroup$ Mar 3, 2016 at 20:49
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$\begingroup$ @IgorRivin Topological extension is easy to construct. But I need a smooth extension. $\endgroup$ Mar 3, 2016 at 21:46
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$\begingroup$ @MaximBraverman Sebastian tells you how to tweak a topological extension to a smooth extension. $\endgroup$ Mar 3, 2016 at 21:48
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1$\begingroup$ @SebastianGoette I found a paper of Palais "Equivalence of nearby differentiable actions of a compact group" which shows that any two actions which are closed to each other are conjugated. Moreover, given a continuous family of actions, one can find a continuous family of conjugations. It is not clear to me from his proof whether one can find a smooth family of conjugations. However, I found a much more recent paper of Kankaanrinta, which proves all what I need, see my answer to this question. $\endgroup$ Mar 5, 2016 at 22:08
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Yes, the action can be extended. It immediately follows from the equivariant collaring theorem proven in the paper Kankaanrinta, Marja Equivariant collaring, tubular neighbourhood and gluing theorems for proper Lie group actions. Algebr. Geom. Topol. 7 (2007), 1–27.
Let me formulate the equivariant collaring theorem. An equivariant collar of $\partial M$ is a smooth $G$-equivariant embedding
$$f:\partial{}M\times [0,\infty)\to M,$$
such that $f(x,0)=x$ for all $x\in \partial M$. The equivariant collaring theorem states that given a smooth proper action of a (possibly non-compact) Lie group an manifold $M$ with boundary $\partial{}M$, the boundary has an equivariant collar.
To extend the action of $G$ "beyond the boundary" we now simply consider the manifold $M\cup_{\partial M}(\partial M\times(-1,0])$ and extend the action of $G$ to $\partial M\times(-1,0]$ by $$g\cdot (x,t)\ := \ (g\cdot x,t).$$