Whitehead double of a non-trivial knot is non-trivial

Question

How can one show that if a Whitehead double of a knot is trivial, then the original knot must have been trivial? Since the Alexander polynomial and the signature vanishes, and since it is not clear what is happening to the jones polynomial under whitehead doubling, I dont know how to show this. Is there a good invariant which can do this? I would need that it "stays non-trivial" under the operation of Whitehead doubling. Thank you for your time and help.

Answer 1

Bridge number works here, I believe. The index of a satellite knot is the minimum number of times a disk in the companion torus must intersect the pattern. So, for a whitehead double, this is 2. $J$ is going to be the original knot and $K$ is the resulting satellite knot.

Part of Theorem 1 of Jennifer Schulten's paper: Suppose $K$ is a satellite knot with companion $J$, companion torus $\tilde{V}$, pattern $(\hat{V} , L)$ and index $k$. Then $b(K)\geq k·b(J),$

Then if you have that $b(K)=0$, then $b(J)$ must have been zero and the only knot that does that is the unknot.