Freely Periodic map of $(S^{3} , K) $ and a fixed loop in the induced isomorphism of $\pi_{1} ( S^{3} \backslash K )$

Question

Let $K$ be a link in $S^{3}$ and $f: S^{3} \rightarrow S^{3} $ a freely periodic map of order $n$ with $f(K) = K$. Let $\psi_{f} : \pi_{1} ( S^{3} \backslash K ) \rightarrow \pi_{1} ( S^{3} \backslash K )$ be the induced fundamental group isomorphism when restricting to the complement of $K$. Can it be the case that $\psi_{f} (\alpha ) = \alpha $ for some loop $\alpha$ in the fundamental group $\pi_{1} ( S^{3} \backslash K )$, or will the fixed loop under $\psi_{f} $ imply a non-empty fixed point set of $f$ disjoint of $K$?

Answer 1

I think your question has a reinterpretation which has an answer, at least in certain cases.

Let $M=S^3-K$. Then $f_{|M}$ induces a finite-order diffeomorphism. There is a well-defined map $Diff(M) \to Out(\pi_1(M))$, given by choosing a basepoint $x\in M$, and a path $c:[0,1]\to M$, $c(0)=x, c(1)=f(x)$, and take the composition $f_\ast: pi_1(M,x)\to \pi_1(M,f(x))$ and $c^{-1}:\pi_1(M,f(x))\to \pi_1(M,x)$ given by conjugation (in the fundamental groupoid if you like), which gives an automorphism which is well-defined up to conjugation in $Aut(\pi_1(M))$. Moreover, since $f$ acts freely (here, I'm assuming that all powers of $f$ act freely as well), the quotient $S^3/\langle f \rangle$ is a lens space by the geometrization theorem. This implies that $M'=M/\langle f\rangle$ is $n$-fold covered by $M$, and therefore $f$ represents an element of order $n$ in $Out(\pi_1(M))$, with extension $\pi_1(M)\to \pi_1(M')\to \mathbb{Z}/n$.

The action of $f$ on $\pi_1(M)$ is only well-defined up to inner automorphisms. Choose some lift $\hat{f}$ of $f$ to $Aut(\pi_1(M))$, and consider the map $\hat{f}^n$, which is trivial in $Out(\pi_1(M))$, since $f^n=Id$. Thus, $\hat{f}^n$ is an inner automorphism.

On the other hand, if $f_{| M}$ had a fixed point $x$, then there is a well-defined action $\hat{f}: \pi_1(M,x)\to \pi_1(M,x)$ such that $\hat{f}^n=Id$. So I think maybe your question could be reposed as:

If there is lift $\hat{f} \in Aut(\pi_1(M))$ such that $\hat{f}^n=Id\in Aut(M)$, then does $f$ have a fixed point?

The connection to your question is made in that by the Smale conjecture, there will be an unknotted loop $U$ in $S^3-K$ containing $x$, such that $f$ fixes $U$. Then the element of $\pi_1(M,x)$ represented by $U$ will be fixed by $\hat{f}$.

I think this has a positive answer, at least in the case that $M$ admits a finite volume hyperbolic metric. In this case, $f_{|M}$ is homotopic to a unique isometry $F$ by Mostow rigidity (in fact, it is isotopic to a finite-order isometry by Hatcher's proof of the Smale conjecture for Haken manifolds). Choose a lift $\hat{F}$ of $F$ to the universal cover $\tilde{M} \cong \mathbb{H}^3$. Then $\hat{F}^n$ is an isometry of $\mathbb{H}^3$ lying in $\pi_1(M)$. Moreover, there is a lift $\hat{F}$ such that $\hat{F}^n=Id$ if and only if there is a lift $\hat{f}\in Aut(M)$ which is finite-order. Then $\hat{F}$ must be elliptic, so it fixes an axis, and therefore $F$ does as well.