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Suppose $K^1 \subset S^3$ is a (connected) knot. First, I am wondering is there a Morse function $f$ on $S^3$ such that $f$ restricted to a neighborhood $D^2 \times S^1$ of $K^1$ have standard form, i.e the standard Morse function on $S^1$ with two critical points plus $r^2$, where $r$ is the radial coordinate on $D^2$. In particular, $f$ restricted to $D^2 \times S^1$ has two critical points of index 0,1 and $\nabla f$ is outward pointing along $\partial (D^2 \times S^1)$.

I am also wondering what the genus of a knot has to do with the number of critical points of such a Morse function. If the knot genus is $k$, is it true that there are at least $k+1$ critical points of $f$ of index $1$? If $K^1$ is the unknot, then $f$ can be a Morse function with 4 critical points of index $0,1,2,3$. Then by topology, there must be at least $k+1$ critical points of index 2 to cancel these critical points of index 1.

Finally, I am trying to understand the index 3 critical points, which are the most confusing to me. I think, in general, there need to be many index 3 critical points but I am not sure how they appear from the perspective of the genus/Seifert surface. It would be helpful if someone could explain or provide a reference to a concrete example, say the $(p,q)$-torus knot (which has genus $(p-1)(q-1)$). Maybe this has to do with Heegaard splittings of $S^3$. Essentially I want a concrete description of the knot complements in terms of Morse theory or cell decompositions.

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    $\begingroup$ Your function is Morse in the neighborhood of the knot, so you can extend it arbitrarily and then perturb away from the knot to make it Morse. The number of necessary critical points has more to do with the topology of the knot complement (how many cells it needs) than anything about the knot itself (eg genus). $\endgroup$ – user98602 Nov 20 '16 at 20:25
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    $\begingroup$ @MikeMiller The knot complement and the knot genus are both determined by the knot. One question I was asking is how they're related. $\endgroup$ – user39598 Nov 20 '16 at 20:33
  • $\begingroup$ The index 3 critical points are uninteresting. You can always cancel all but exactly one of them off with index 2 points. $\endgroup$ – PVAL-inactive Nov 22 '16 at 19:54
  • $\begingroup$ @PVAL-inactive Why is that? I agree that cancel algebraically but I'm not sure they cancel geometrically. The question is whether a 3-dimensional manifold with one boundary $S^1 \times S^1$ admits a Morse function with at most one critical point of index 3. In high dimensions, this is true but I am not sure this is true in dimension 3. Would the answer be different if I considered $K^1 \subset B^3$, the 3-dimensional ball? $\endgroup$ – user39598 Nov 22 '16 at 20:00
  • $\begingroup$ @user39598 The attaching (ascending) sphere for a 3-handle is just a copy of $S^3$, and the the belt (descending) sphere for a 2-handle is an $S^0$. If they intersect algebraically once, they automatically intersect geometrically once. (This is also true for extraneous 0 and 1 handles (in this case not all of them are extraneous)). None of this answers your question it just means that all of the interesting part of your question relies on the interactions between the 1 and 2-handles. $\endgroup$ – PVAL-inactive Nov 22 '16 at 20:20

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