$S^3$ as cyclic branched cover of itself

Question

In Chapter One of his notes (March 2002) Thurston says:

If $K$ is the trivial knot the cyclic branched covers are $S^3$. It seems intuitively obvious (but it is not known) that this is the only way $S^3$ can be obtained as a cyclic branched covering of itself over a knot.

This sentence sounds a little bit enigmatic to me. Is he saying that (we expect that) "if $K \subset S^3$ is such that all its cyclic branched covers are $S^3$ then $K$ is the unknot", or that "if for some $k\geq 2$ the $k$-fold cyclic branched cover over $K$ is $S^3$ then $K$ is the unknot"?

I would like to know which is the state of the art this conjecture. In particular if $S^3$ can appear as a branched cover over a non-trivial knot I would like to see some (families of) examples.

Answer 1

The statement that for arbitrary K in $S^3$, if for some $n \ge 2$, the n-fold cyclic branched cover is $S^3$ (or in some versions, a homotopy 3-sphere) then K is the unknot, was known as the Smith conjecture. It was proved around 1979, combining work of many authors: Thurston, Meeks-Yau, Gordon-Litherland, Bass, Shalen, and perhaps others. Its proof was viewed as an early triumph of Thurston's hyperbolization theorem and more generally the power of geometric methods in 3-manifold topology.

The best reference is the 1984 book edited by Morgan and Bass, "The Smith Conjecture". It has lots of background. There are strong generalizations, due to Thurston; google "Thurston's orbifold theorem'.