Double branched covers of $S^3$ branched along prime knots

Question

Let $M(k)$ denote the double branched cover of $S^3$ along the knot $k$. It is known that $M(k_1\#k_2)=M(k_1)\#M(k_2)$. It is also known that the unknot is the only knot such that the corresponding double branched cover is $S^3$. These two facts together imply that the double branched cover of $S^3$ along a non-prime knot is a non-prime 3-manifold. I wonder if the converse is true. More precisely, is it true that if $k$ is a prime knot, then $M(k)$ is prime?

Answer 1

If you think of $(S^3, k)$ as an orbifold with singular set $k$ (and cone angle $\pi$ along $k$), then you should think about orbifold covers - eg the double branched cover - of this as analagous to covers of 3-manifolds themselves. And you know that covers of prime 3-manifolds are still prime (though the converse is not necessarily true). So you should try to mimic that proof. The way this usually works is by passing through irreducible 3-manifolds.

Suppose the branched double cover is not irreducible, so there is a sphere that does not bound a ball. Then (Meeks-Simon-Yau; Dunwoody) there is a sphere $S$ that does not bound a ball, intersects $\tilde k$ in as few points as possible, and the involution $\iota$ either has $\iota S = S$ or $\iota S \cap S = \varnothing$. Let's look at the two cases separately.

1) $\iota S \cap S = \varnothing$. Since $\tilde k$ is the fixed point set of the involution, clearly this implies $S$ is disjoint from $\tilde k$. Push the sphere $S$ downstairs; this implies its image is a sphere disjoint from $k$. Because spheres in $S^3$ bounds a ball on both sides, pick the ball disjoint from $k$. Then $f^{-1}(B) \to B$ is again a double cover, and thus $f^{-1}(B)$ is two disjoint balls, so $S$ bounded a ball after all. Whoops.

2) $\iota S = S$. Every involution of $S$ is either the identity or conjugate to the antipodal map, rotation by $\pi$ around an axis, or reflection across a plane. It cannot be either the identity or the reflection, since then it would not intersect the fixed point set $\tilde k$ transversely. If it were the antipodal map, the image would be an embedded $\Bbb{RP}^2$ in $S^3$, which is impossible. So all that's left is that $S$ intersects $\tilde k$ at exactly two points, and so $S$ passes downwards to a sphere that intersects $k$ at exactly two points. If $k$ is prime, then one of the balls $S/\iota$ bounds must intersect $k$ in an arc, and thus the double branched cover of that ball is a ball, contradicting the choice of $S$.

So the double branched cover of a prime knot is irreducible and thus prime.