Extend (Lie) group action from the boundary to the entire manifold

Question

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.

Answer 1

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.

Answer 2

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.