# Low Dimensional Topology

## December 18, 2012

### Morse-Novikov number and tunnel number

Filed under: 3-manifolds,Heegaard splittings,Knot theory,Thin position — Jesse Johnson @ 9:33 am

Someone recently pointed out to me a paper by A. J. Pajitnov  proving a very interesting connection between circular Morse functions and (linear) Morse functions on knot complements. (A similar result is probably true in general three-manifolds as well.) Recall that a (linear) Morse function is a smooth function from a manifold to the line in which there are a finite number of critical points (where the gradient of the function is zero), and each critical point has one of a number of possible forms. For a two-dimensional manifold the possible forms are the familiar local minimum, saddle or local maximum. This post is about three-dimensional Morse functions, in which case the possible forms are slight generalizations of local minima, maxima and saddles.  A circular Morse function is a function with the same conditions on critical points, but whose range is the circle rather than the line. For a three-dimensional manifold, the minimal number of critical points in a linear Morse function is twice the Heegaard genus plus two, and for knot complements it’s twice the tunnel number plus two. (In particular, one can construct a Heegaard splitting or unknotting tunnel system directly from a Morse function, but that’s for another post.) The minimal number of critical points in a circular Morse function is called the Morse-Novikov number, and is equal to the minimal number of handles in a circular thin position for the manifold (usually a knot complement). Pajitnov has a very clever argument to show that the (circular) Morse-Novikov number of a knot complement is bounded above by twice its (linear) tunnel number. Below, I want to outline a slightly different formulation of this proof in terms of double sweep-outs, though I should stress that the underlying idea is the same.

Given a genus $g$ Heegaard splitting for a three-manifold $M$, one can construct a Morse function $f : M \rightarrow \mathbf{R}$ with $2g + 2$ critical points, namely one index-zero critical point (a local minimum), $g$ index-one critical points, $g$ index-two critical points and one index-three critical point (a local maximum). If the first cohomology group of $M$ is infinite (and thanks to Poincare duality, this is equivalent to having infinite first and/or second homology groups), there is a map $g : M \rightarrow S^1$ that realizes a generator of this group. For a knot complement, the first cohomology group is $\mathbf{Z}$, so there is a canonical generator which is dual to any Seifert surface. The question is, how do we turn the $2g + 2$ critical points in the linear Morse function $f$ into critical points of a circular Morse function $g$?

Pajitnov’s proof shows that you can adjust the gradient field defined by $f$ to the gradient field of a circular Morse function without introducing any new critical points.  But we can do essentially the same thing with double sweep-outs. First let $g$ be any circular Morse function for $M$ and consider the map $f \times g : M \rightarrow \mathbf{R} \times S^1$. In other words, we define $(f \times g)(x) = (f(x),g(x))$. By employing tricks from singularity theory (such as in ), we can isotope $f$ and $g$ so that this map is stable, a generalization of the Morse conditions to higher dimensional range spaces. If $f \times g$ is stable then the set of critical points (where the discriminant of the function is not onto) is a one-dimensional smooth submanifold of $M$ called the discriminant set (or sometimes the Jacobi set). The image in $\mathbf{R} \times S^1$ of the discriminant set is a graph called that graphic. (There are no vertices in the discriminant set, but its map to $\mathbf{R} \times S^1$ will generally be two-to-one at finitely many points, which form vertices in the graphic.)

It turns out that you can read a lot of information about $f$ and $g$ from the graphic. For example, as I pointed out in , the critical points of $f$ and $g$ correspond to vertical tangencies and horizontal tangencies, respectively, of edges of this graph. (This was for two linear Morse functions, but the same is true in this setting.) When I write horizontal and vertical, I’m thinking of the circle factor of $\mathbf{R} \times S^1$ as being vertical and the $\mathbf{R}$ factor as horizontal. Also, note that the edges of this graph will not be straight lines, but curves with isolated tangents.

In this picture, each of $f$ and $g$ is a composition of $f \times g$ with a projection of $\mathbf{R} \times S^1$ onto $\mathbf{R}$ or $S^1$. In particular, $f$ is a composition with a vertical projection and $g$ is a composition with a horizontal projection. The critical points of the composition correspond to the points where edges of the graphic are tangent to the direction of the projection.  If we choose a different projection, we’ll get a different function and we can use this fact to characterize the critical points of the new function.

In particular, define a projection from $\mathbf{R} \times S^1$ to $S^1$ by $p(x,y) = Kx + y$ where $x \in \mathbf{R}$, $y \in S^1$, $K$ is a very large number and $Kx + y$ is taken modulo 1 (or rather modulo whatever the circumference of $S^1$ is). I like to think of this as applying a vertical twist to the circles in $\mathbf{R} \times S^1$ then projecting horizontally. (This is roughly what happens to a slinky if you hold the bottom steady and twist the top, except our slinky is on its side.) As $K$ becomes larger, the slope in $\mathbf{R}^2 \times S^1$ that corresponds to being tangent to the projection becomes closer to vertical. Because the graphic is smooth, there will be some slope $\epsilon$-close to vertical such that every point with that slope is adjacent to a unique vertical tangency, i.e. a critical point of $f$. So if we pick the corresponding $K$ for our projection then the composition of $f \times g$ with $p$ will be a circular Morse function with the same number of critical points as $f$.

Note that we can’t go the other way, i.e. find a linear Morse function whose number of critical points is bounded by the Morse-Novikov number, because there is no way to “spin” around the $\mathbf{R}$ factor. For example there are fibered knots (with Morse-Novikov number zero) with arbitrarily large tunnel number.

 Kobayashi, Tsuyoshi; Saeki, Osamu The Rubinstein-Scharlemann graphic of a 3-manifold as the discriminant set of a stable map. Pacific J. Math. 195 (2000), no. 1, 101–156.