# How to see that the figure 8 knot has genus 1?

I am reading the knot book from Colin Adams, and after he discussed how one may construct a (Seifert) surface whose boundary is the knot, he went on and stated that the figure 8 knot has genus 1 (that is, it has a seifert surface which can be "capped" off with a disk to obtain the torus, and he stated it without any explanation as if that's an obvious fact - which is not to me.

In general how might one expect to "capp" off the disk of a seifert surface? Especially when the boundary is any non trivial knot, it seems really hard for me to visualize how one would obtain one of the orientable compact surfaces by capping off the boundary, heck It's hard for me to even visualize how we might get a surface in 3d without any self-intersections.

• “Visualization of the Genus of Knots" by van Wijk and Cohen says: “A trefoil or a figure eight knot has genus 1, hence the corresponding Seifert surfaces are homotopic to a torus with a hole in the surface. Via a number of steps in which the Seifert surface is deformed, cut, and glued, this equivalence can be shown, but this is not really intuitive.” So perhaps Adams doesn't mean to imply that it is obvious, but only that you may know it already.
– MJD
May 2, 2020 at 19:05

"Capping off," like Moishe Kohan says, is where, for each boundary component of a surface, you glue to it the boundary of a topological disk. This operation does not happen inside 3-space itself, but rather it's done abstractly.

It's worth knowing that the genus $$g(\Sigma)$$ of a connected surface $$\Sigma$$ with nonempty boundary satisfies the equation $$d-s+b=2-2g(\Sigma)$$, where, given a disk and strip presentation of the surface, $$d$$ is the number of disks, $$s$$ is the number of strips, and $$b$$ is the number of boundary components. This is from a considering Euler characteristics.

For example, here is a Seifert surface for the figure-8 knot, which I obtained by applying Seifert's algorithm to the usual diagram and then moving things around a little:

There is one disk, two strips, and one boundary component. Since $$1-2+1=0$$, it follows $$g(\Sigma)=1$$ is the genus.

If you want to visualize this somehow in 3-space, you may consider your surface up to a couple of moves. The first is allowing bands to pass through each other, and the second is allowing full twists to be inserted in the strips. After such manipulations for the surface above, we can put it into this form:

I drew a torus in gray, and the surface has been placed on this torus. The complement of the surface in the torus is a disk, so the torus is the result of capping off the surface.

Seifert's algorithm provides one, most popular, way to construct a Seifert surface of the given knot. You can find a description in many places, for instance: here, here, as well as section 4.3 of the Colin Adams' book that you are reading. It is a theorem, again mentioned in the book, that for an alternating knot diagram, Seifert's Algorithm produces a surface of the minimal genus. In general, it is known that even the unknotting problem in NP-hard, so do not expect easy computations of the genus either. As for "cupping a surface with a disk", it just means "attach the 2-dimensional disk along the boundary of the given surface." By comparison, this requires very little work.

• Unknotting in an arbitrary 3-manifold is NP-complete (Agol-Hass-Thurston 2006), but in $S^3$ it's in $\text{NP}\cap\text{co-NP}$, a class that's unknown whether it intersects NP-hard. I think Lackenby has a quasi-polynomial-time algorithm for unknot detection in $S^3$. May 2, 2020 at 21:14