Let $M$ and $W$ be smooth manifolds such that $\partial W=M$. Let $G$ be a group acting on $M$.

Can one generally extend the action of $G$ to $W$? If not, under which conditions on $W$ and/or $G$ does it work? How do you construct such an extension?

EDIT: suppose in addition that $G$ is a connected Lie group acting smoothly and that the extension should be smooth as well.

EDIT: suppose furthermore that the Lie group is acting by isometries if a Riemannian metric is given on $M$ and the extension should again be acting isometrically.

  • 3
    $\begingroup$ If you don't care about any properties of the action, then let $G$ act trivially on $W \setminus M$. Together with the existing action on $M$, this will give you an action on all of $W$. But if, say, $G$ is a Lie group acting smoothly on $M$ and you want to preserve this smoothness, then this construction of course fails horribly. Perhaps edit the question to clarify what properties you want the extension to satisfy? $\endgroup$
    – Sophie M
    Apr 14, 2020 at 15:13
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    $\begingroup$ Indeed certainly you want a continuous extension, but the conclusion might differ according to the degree of differentiability. In the case of the circle boundary of the disc, the topological extension is obvious but the differentiability issue seems non-trivial. The purely topologic question is interesting too. Also restricting to $G$ cyclic is very rich. Maybe give more context and focus a bit... $\endgroup$
    – YCor
    Apr 14, 2020 at 15:30
  • $\begingroup$ Thank you @SophieMacDonald for your comment. I edited my question to add the more specific question of Lie group actions, but I would also be interested in a more general answer. $\endgroup$
    – Kafka91
    Apr 15, 2020 at 9:27
  • $\begingroup$ @YCor I would actually also like to know if there is some reference discussing as many cases as possible. $\endgroup$
    – Kafka91
    Apr 15, 2020 at 9:29
  • $\begingroup$ @Kafka91 To elaborate on Sophie MacDonald and Ycor's comments, you can also ask the action on $W$ to be by isometries if $W$ is given a Riemannian metric structure. The case of an action on the circle extending as an action on the Poincaré disk is already very interesting. $\endgroup$
    – M. Dus
    Apr 15, 2020 at 13:14

2 Answers 2


Edit: I might have misinterpreted the question, which asks whether the action on $M$ extends to a specific bounding manifold $W$. The answer below concerns whether the action extends to some bounding manifold.

Your question can be rephrased in the language of algebraic topology as: If $M$ is a null-bordant manifold with a $G$-action, is $M$ equivariantly null-bordant? Equivariant (co)bordism was a very active subject in the 1960s and 70s, dating back at least as far as the monograph by Conner and Floyd:

Conner, P. E.; Floyd, E. E., Differentiable periodic maps, Ergebnisse der Mathematik und ihrer Grenzgebiete. Neue Folge. 33, Reihe: Moderne Topologie. Berlin-Göttingen-Heidelberg: Springer-Verlag. VII, 148 p. (1964). ZBL0125.40103.

Part I gives an introduction to bordism theory, and Part II studies $G$-equivariant bordism with $G=\mathbb{Z}/p$. Some of the results apply to general Lie groups $G$. For example: if $G$ acts freely on a closed manifold $M$ of dimension $m$, the bordism class of the pair $(M/G,f)$ in the bordism group $\Omega_m(BG)$, where $f:M/G\to BG$ classifies the principal $G$-bundle $M\to M/G$, provides an obstruction to extending the action to a free action on an $(m+1)$-dimensional manifold with boundary. This can be studied using characteristic numbers.

  • $\begingroup$ Thank you very much for your answer! However I was indeed asking under what conditions on $W$ the action can be extended on it. I will edit my question, but your answer is very instructive anyway. $\endgroup$
    – Kafka91
    Apr 23, 2020 at 6:44

Here is an example of a connected Lie group acting on $M$, such that the action does not extend to $W$ but extends to some $W'$ with $\partial W'=M$. All the existing actions are analytic isometric, and all obstructions are continuous, so the problem is not regularity.

Take $G=M=\mathbb T^1$ the 1-torus, acting on itself in the obvious fashion, by translations. Clearly, if $M$ is seen as the boundary of the disc $W'=\mathbb D^2$, then the action on $M$ extends to an action on $W'$, again by rotations. (In the following I prefer to think of $\mathbb T^1$ as $\mathbb R/\mathbb Z$.)

However, let $W$ be a 2-torus with a disc removed, and identify $M$ with the boundary of $W$. Assume by contradiction that the induced action on $\partial W$ is the restriction of an action on $W$.

Let $Z$ be the manifold constructed as the gluing of $[0,1]\times M$ with $W$ along $\lbrace1\rbrace\times M$ and $\partial W$. Define an action of $\mathbb R$ on $Z$ in such a way that the boundary of $Z$ (which corresponds to $\lbrace0\rbrace\times M$) is fixed, and the restriction to $W$ is the action induced by that of $\mathbb T^1$ on $W$. On $[0,1]\times M$, we have to continuously Dehn-twist the collar to match the rotation of $\lbrace1\rbrace\times M$.

Now the action $f$ at time $1$ is homotopic to the identity of $Z$ relative to its boundary (in the strong sense), but $f$ is just a Dehn twist of $[0,1]\times M$, since $f$ is the identity in $W$ (1 has image 0 in $\mathbb T^1$). It is probably classical that this Dehn twist is not homotopic to the identity relatively to the boundary, personally I convinced myself on a picture that the induced map $f_*$ on $\pi_1(M,x)$ for $x$ on the boundary is not the identity; in any case, this is a contradiction.

  • $\begingroup$ Let me try a (very) wild guess: the action $\rho$ of a connected Lie group $G$ on a manifold $M$ extends to every $W$ such that $\partial W\simeq M$ if and only if there is an action on the cylinder $M\times[0,1]$ such that the action restricted to $M\times\lbrace0\rbrace$ (resp.$M\times\lbrace1\rbrace$) is $\rho$ (resp. the trivial). $\endgroup$
    – Pierre PC
    Apr 26, 2020 at 1:01

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