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Show that doubled knots have genus at most 1.

Doubled knots are constructed by replacing the knot with a thickened band all the way around, forming essentially a knotted band.

My thoughts for the proof: Let $K$ be a doubled knot. By a theorem, we have that $genus(K)=\frac{1}{2}(2-D+B-C)$, where $D$ is the number of disks, $B$ is the number of bands, and $C$ is the number of boundary components. *(What follows is what I want to say, but I am not sure that it is necessarily true.) For a doubled knot, we know that both $B=1$ and $C=1$, since the entire knot is essentially one band and thus forms one boundary. Suppose $genus(K)>1$. Then,$$\frac{1}{2}(2-D+B-C)=\frac{1}{2}(2-D)>1 \implies$$ $$1-\frac{D}{2}>1 \implies$$ $$-\frac{D}{2}>0 \implies$$ $$D<0$$ which is clearly impossible, and thus a contradiction. Thus, $genus(K)\leq1$.

Any help would be greatly appreciated. If my previous statements are not true, what direction should I go in to prove this?

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I'm not sure what your definition of doubled knot is.

Whitehead double. The Whitehead link has a Seifert surface contained entirely in the solid torus the pattern knot is embedded in, and that surface has genus $1$. It follows the Whitehead double of a non-trivial knot has genus $1$ (where the trivial knot's Whitehead double is a trivial knot, so genus $0$).

Cabling. This is a satellite link formed using a torus link as the pattern. If the chosen torus link is a pair of unknots, then the corresponding satellite link is the boundary of the knot thickened into an oriented band. Oriented bands have genus $0$ because gluing a disk to each boundary circle results in a sphere. It follows that the genus of the satellite link is $0$. (If you care about the possibility of unoriented bands, it appears the genus can be arbitrarily high.)

A few things about what you wrote:

  • No need for a contradiction proof. Just start with $D\geq 0$ then show that the genus is no more than $1$ by some direct inequality manipulation.
  • Why do you assume $C=1$? If it is an oriented band, there will be two parallel copies of the knot as the two boundaries.
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  • $\begingroup$ I am working with a very imprecise definition of a doubled knot; what is included in my question is exactly what my problem states. You are correct that $C$ does not have to be $1$. However, $C\geq1$, which I can use for my inequality. Thanks! $\endgroup$ Oct 31, 2017 at 19:02

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