That pretzel knot again
Remember P(-2,3,7), that pretzel knot? (Yeah, big edit there. Yikes!)
Because it sits on the fiber of a genus one fibered knot (and thus belongs to a family of Berge’s doubly primitive knots), it has a lens space Dehn surgery along the framing the fiber induces. Let’s get a glimpse of this.
We’re gonna make a genus 2 Heegaard surface from two fibers with the knot sitting on one of them.
Billow out the surface, and carry the knot along.
Then billow it out in the other direction too and we’ll have a nice genus 2 Heegaard surface with the knot on it.
Rather than looking for the primitivizing disks, we’ll find compressing disks for the handlebodies on each side of the Heegaard surface that are disjoint from the knot. See, the Dehn surgery on the knot transforms the handlebodies into two solid tori and these two disks will become meridional disks.
We’ll start by finding an essential arc on the blue surface that is disjoint from the knot. (If we wanted primitivizing disks, we would start with an essential arc that crossed the knot once.)
Now we’ll sweep this arc through the fibers to the red surface, though the two handlebodies. The easy way is through the billowing we did.
This gives one meridional curve.
To get the other… well, going through the fibers outside the billowing, we apply the monodromy of the trefoil (a Dehn twist along two curves) to the arc on the red surface. Together with the arc on the blue surface we obtain this loop:
If you start by pulling the bottom red bit down and unhook it from the blue twisting, you might be able to convince yourself that this is indeed an unknotted loop. With a bit more work you could even convince yourself that it actually bounds a disk disjoint from the surface. Drawing that disk would’ve made quite a mess. Here’s the curve as it sits on the surface.
Now we can look at both these curves on the Heegaard surface together.
Count that they intersect 19 times. You see 18 in the middle here. The last comes from the red arc they have in common, but that can be perturbed to give a single transverse intersection.
One may count that the twisting is 7 — plus or minus, and perhaps the inverse mod 19 depending on how you’re counting. Starting off with the common red arc of the two meridians as 0, then count the intersections in order along the light blue meridian and again in order along the dark blue meridian. The torsion is the number that you multiply by the light blue count of an intersection to get the dark blue count of the same intersection, mod 19. A portion of this counting is shown here:
There’s an intersection that the light blue counts as 2 and the dark blue counts as 5. So if 2q=5 mod 19 then we may take q=12. Since 7=-12 mod 19 we’ve got the lens space claimed (up to homeomorphism).
Here are those two meridians again with the knot.