A Morse exhaustion function on the punctured plane

Question

I don't think I have a good intuitive understanding of the purpose of a Morse function. The classic illustrative example seems to be a genus 2 torus with the Morse exhaustion function being given by the height function. From that, I thought any point at which the topology of the sublevel set changes is "critical point". In that case, what would be the set of critical points of a Morse exhaustion function on the punctured plane (plane with origin removed)? I feel like saying that the origin must be a critical point, but surely we cannot say that, as the origin has actually been removed.

If I could be given some help with correcting my intuition, I'd really appreciate it.

Answer 1

Every smooth manifold $M$ admits a Morse exhaustion function. One way to prove this is first to find a proper embedding of $M$ into $\mathbb R^N$ for some $N$ (which exists by the Whitney embedding theorem), so we can consider $M$ to be a closed submanifold of $\mathbb R^N$. Then use the fact that for almost every $v\in\mathbb R^N$, the function $f\colon \mathbb R^N\to\mathbb R$ given by $f(x) = |x-v|^2$ restricts to Morse exhaustion function on $M$. (This is proved, for example, in Corollary 1.25 of Nicolaescu's An Invitation to Morse Theory.)

In the example of $M=\mathbb R^2 \smallsetminus \lbrace(0,0)\rbrace$, we can first embed $M$ into $\mathbb R^3$ by the map $$ (x,y) \mapsto \left( \log r, \frac{x}{r}, \frac{y}{r}\right), $$ where $r = \sqrt{x^2 + y^2}$. (The image of this embedding is the unit cylinder around the $x$-axis.) Then we can take $f\colon M\to \mathbb R$ to be the square of the distance to the point $(0,0,-1)$. Specifically, in terms of the original $(x,y)$ coordinates on $M$, this is $$ f(x,y) = (\log r)^2 + \left(\frac{x}{r}\right)^2 + \left( \frac{y}{r} + 1\right)^2 = (\log r)^2 + \frac{2y}{r} + 2. $$ This has two critical points: $(0,-1)$ is an absolute minimum, and $(0,1)$ is a saddle point.

You can visualize the level sets by looking at the intersection of the cylinder with spheres centered at $(0,0,-1)$. When the radius of the sphere is small, the intersection is just a simple closed curve in the bottom half of the cylinder. When the radius becomes larger than $2$, the intersection becomes two simple closed curves. As the radius gets very large, these two curves get closer and closer to vertical circles far out along the cylinder.

One problem with your attempt to understand intuitively what's going on here ("what would be the set of critical points of a Morse exhaustion function on the punctured plane?") is that you seem to be thinking that the critical points depend only on the manifold. But in fact, the critical points depend very heavily on what Morse function you choose. In fact, it's not hard to show in the case of the example above that given any two distinct points of $M$, there's a Morse exhaustion function whose critical points are exactly those two points. And you can get more critical points by smoothly deforming the embedding in $\mathbb R^3$ so that the distance function has lots of local maxima and local minima.

Answer 2

The problem with your example is that the punctured torus(or Plane) is no compact. The statement describing the topology of level set depend on those facts. For example important theorem fails:

Suppose $f$ is a smooth real-valued function on $M$ and $p$ is a non-degenerate critical point of $f$ of index $\gamma$, and that $f(p) = q$. Suppose $f^{−1}[q − \varepsilon, q + \varepsilon]$ is compact and contains no critical points besides $p$. Then $M^{q+\varepsilon}$ is homotopy equivalent to $M^{q−\varepsilon}$ with a $\gamma$-cell attached.

and another one

Suppose $f$ is a smooth real-valued function on $M$, $a < b$, $f^{−1}[a, b]$ is compact, and there are no critical values between $a$ and $b$. Then $M^a$ is diffeomorphic to $M^b$, and $M^b$ deformation retracts onto $M^a$.

This means that when $M$ is compact morse theory have powerful tools to describe the topology of the level sets of $M$. In your case $f^{-1}[-\varepsilon,\varepsilon]$ is not compact if the plane is the $xz$ plane in $\mathbb{R}^3$.