Existence of criteria for extension of action of embedded circles in a topological d-manifold.

Question

Let $M$ be a topological d-manifold viewed as an embedded subspace $M \hookrightarrow \mathbb{R}^{m}$ for some $m$. Let $End(\mathbb{S^1}) = Homeo(\mathbb{S^1})$ be the group of homeomorphisms of the circle. Let $i: \mathbb{S^1} \hookrightarrow M$ be an embedding. For $\phi \in End(\mathbb{S_1})$, define $\phi': i(\mathbb{S^1}) \to i(\mathbb{S^1})$ by setting $\phi'(x) = i(\phi(i^{-1}(x)))$. Do there exist criteria that guarantee the existence of a map $f \in End(M) = Homeo(M)$ such that $f|_{i(\mathbb{S^1})} = \phi'$? In general, given an action of $End(\mathbb{S^1})$ of $\mathbb{S^1}$, do there exist criteria that answer when this action can be lifted in the way described above to an action of $End(M)$ on $M$?

Answer 1

Suppose that $i(S^1)$ admits a product neighborhood $S^1\times D^{d-1}$ in $M$. In other words, I am assuming that $i(S^1)$ is a "tame topological circle" and preserves orientation in $M$. Assume also that $\phi$ is orientation-preserving.

I claim that such $\phi$ extends to a homeomorphism of $M$. Below are the steps of a proof.

1. It suffices to show that if $\phi: S^1\to S^1$ is an orientation-preserving homeomorphism, then for every $k$, $\phi$ extends to a homeomorphism of $D^k\times S^1$ which equals the identity on the boundary. Here I am identifying $S^1$ with $0\times S^1$ and use $D^k$ to denote the closed $k$-dimensional disk.

2. $D^k$ is homeomorphic to the quotient of $S^{k-1}\times [0,1]$ obtained by collapsing $S^{k-1}\times \{1\}$ to point.

3. The homeomorphism $\phi: S^1\to S^1$ gives rise to a homeomorphism $\Phi: S^1\times S^{k-1}\to S^1\times S^{k-1}$ which is the identity on the second component of the product and equals $\phi$ on the first component. After we collapse $S^{k-1}$ to point, $\Phi$ descends to $\phi: S^1\to S^1$.

4. Every orientation-preserving homeomorphism $\phi: S^1\to S^1$ is isotopic to the identity map. Let $\phi_t, t\in [0,1]$ denote the corresponding isotopy, where $\phi_1=\phi, \phi_0=id$.

5. Define a homeomorphism $$ \Psi: S^{k-1}\times [0,1]\times S^1\to S^{k-1}\times [0,1]\times S^1 $$ by the formula $$ \Psi(x, t, s)= (x, t, \phi_t(s)), x\in S^{k-1}, t\in [0,1], s\in S^1. $$ Then $\Psi(x, 0, s)=(x,s)$ for all $x, s$.

6. Collapsing $S^{k-1}\times \{1\}$ to point, we project $\Psi$ to a self-homeomorphism $$ \psi: D^k\times S^1 $$ which equals the identity on the boundary (which is the projection of $S^{k-1}\times \{0\}\times S^1$) and equals $\phi$ on $0\times S^1$.

Edit. Here is one example which shows what can go wrong for a general $i$ and $\phi$. Suppose that $i$ is a wild embedding of $S^1$ in $S^3$, i.e. an embedding defining a wild knot, as in the linked article, with exactly one "wild point" $p=i(q)\in K=i(S^1)$, and the rest of points "tame". Every self-homeomorphism of $S^3$ preserving $K$ has to send $p$ to itself. Hence, every extendible homeomorphism $\phi: S^1\to S^1$ has to fix $q$. In particular, no nontrivial rotation $\phi$ will extend homeomorphically to $S^3$.

Here is another obstruction, for tame embeddings. Consider a compact connected manifold $M$ of dimension $\ge 3$, such that $G=\pi_1(M)$ contains an element $g$ such that there are no automorphisms $\alpha: G\to G$ which send $g$ to $g^{-1}$. For instance, let $G$ be the integer Heisenberg group with the presentation $$ \langle a, b, t| [a,b]=t, [a,t]=1, [b,t]=1\rangle. $$ The central element $g=t$ has this property with respect to automorphisms of $G$. This group is isomorphic to the fundamental group of a certain compact 3-dimensional manifold $M$.

Now, represent $g$ by a simple oriented loop $c=i(S^1)$ in $M$. If there exists a homeomorphism $\Phi: M\to M$ sending $c$ to itself reversing the orientation on $c$, then $\Phi_*(g)$ is (up to conjugation) equals $g^{-1}$, which contradicts the choice of $g$. Therefore, an orientation-reversing homeomorphism $\phi: S^1\to S^1$ cannot extend to a self-homeomorphism of $M$.