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The fundamental group of any two-bridge knot K in $\mathbb{S}^3$ has a presentation with two generators and one relation.

On the other hand, it's possible to provide a CW-complex with only one 0-cell and no 3-cell on which the knot complement deformation retracts.

Is it reasonable to hope that we should provide such a CW-complex with only two 1-cell and one 2-cell, with boundary given by the relator of the $\pi_1$, on which $\mathbb{S}^3 \setminus V(K)$ retracts (notice it's possible for torus knots)? Or at least homotopically equivalent?

Thanks, any help would be appreciate.

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Yes, there is in fact a handle decomposition with one 0-handle, two 1-handles, and a single 2-handle. In general, for a k-bridge knot, you'll get k 1-handles and $k-1$ 2-handles. The way that I learned this was to think of the knot (this works for any knot) as a solid wire that you are holding in a bathtub as the water fills in. At first you see a 0-handle, then at each minimum (for the height function on the knot) you see a 1-handle in the complement, then each maximum gives a 2-handle. Finally there should be a 3-handle. But since you are interested in the exterior of the knot (a manifold with boundary), it is easy to see that the 3-handle cancels one of the 2-handles.

In practice, you can see how the attaching circles for the 2-handle(s) flow down to the boundary of the handlebody given by the union of the 0-handle and 1-handles. For 2-bridge knots, this is most efficiently described using the Schubert normal form; see for instance the picture at the top of page 209 in Burde-Zieschang, Knots. The general principle about complements is discussed in Gompf-Stipsicz, chapter 6.

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