Given a closed 3-manifold $M$, a self-indexing Morse function $f:M\to[0,3]$ produces a Heegaard splitting by : $$M=f^{-1}([0,3/2])\cup f^{-1}([3/2,3]).$$ (This question discussed a bit on the topic.)

Now, we define :

Definition : Let $M$ be a closed 3-manifold. The Morse number $\mathfrak{m}(M)$ of $M$ is the minimal number of critical points of Morse functions. The Heegard number $\mathfrak{h}(M)$ of $M$ is the smallest genus of any Heegard splitting of $M$.

We have a theorem that Milnor used to create exotic spheres :

Theorem : (Reeb) Let $M$ be a closed $n$-manifold such that $\mathfrak{m}(M)=2$. Then $M$ is a topological $n$-sphere.

Moreover, by a result from Alexander, we have that $\mathbb{S}^3$ is the only manifold whose Heegaard number is $0$, that is we have, for a closed 3-manifold : $$\mathfrak{h}(M)=0\iff\mathfrak{m}(M)=2\iff M\cong\mathbb{S}^3.$$

This suggests that the two may be related. My questions are the following :

  • Are $\mathfrak{m}(M)$ and $\mathfrak{h}(M)$ related ?

  • Given a self-indexing Morse function $f:M\to[0,3]$, is the genus of the associated Heegaard splitting related to the number of critical points of $f$ ?

About the first question, I am wondering whether there is a somewhat-nice formula, like a linear relation $\mathfrak{m}(M)=A\mathfrak{h}(M)+B$ ? About the second question, I found no literature about this construction for a Heegaard splitting, the one I knew about was using Moise's triangulation theorem.

As a side note, about the first question, if a linear relation existed, taking $M=\mathbb{S}^3$ or $\mathbb{S}^1\times\mathbb{S}^2$ would yield (unless I made mistakes in my computations) $\mathfrak{m}(M)=2[\mathfrak{h}(M)+1]$. However, I didn't manage to compute the two numbers for a 3-torus $\mathbb{T}^3$ to check whether there is a contradiction. In any case, a polynomial relation $\mathfrak{m}(M)=P(\mathfrak{h}(M))$ would be respectable !

Edit :

I've been reading some stuff here and there, and I have noticed :

  • Given a self-indexing Morse function $f:M\to[0,3]$, assuming that there are only one index zero and one index three critical points, then the genus of the associated Heegaard splitting is $g=\# f^{-1}(1)$ the number of index one critical points.

  • It is always possible to construct a Morse function satisfying the hypotheses of the previous bullet.

However, my concern is that the function constructed this way is not necessarily minimal (in the sense that its number of critical points is not necessarily the Morse number of the manifold). Moreover, this construction is certainly not natural. At last, I am not sure that given a self-indexing Morse function $f:M\to[0,3]$, the construction of the new Morse function $\tilde{f}:M\to[0,3]$ with $\# f^{-1}(0)=\# f^{-1}(3)=1$ preserves the total number of critical points -- I haven't yet looked at the very details of this construction (see Milnor, Lectures on the h-cobordism theorem).

In any case, this partially answers my second question, at least in that special setting where we assume only one index zero/three critical point.

It would be interesting to see what you think about this !


1 Answer 1


Here is a somewhat sketchy answer, based on the creed that Morse functions and (smooth) handles should be thought about interchangeably (and that Heegaard splittings are a form of handle decomposition).

Suppose $M$ is a closed, connected, orientable 3-manifold and $f:M\rightarrow [0,3]$ is a self-indexing Morse function. For $k=0,1,2,3$, let $c_k$ be the number of critical points of $f$ having index $k$. We can see that $c_1-c_0=c_2-c_3$, because $$0=\chi(M)=c_0-c_1+c_2-c_3.$$ Since $f^{-1}[0,1.5]$ is formed from $c_0$ zero-handles and $c_1$ one-handles, we can see that it is a handlebody of genus $c_1-c_0+1$ (this uses the assumption that $M$ is connected and orientable). Considering the upside-down Morse function $-f$, we can similarly observe that $f^{-1}[1.5,3]$ is a handlebody of genus $c_2-c_3+1$. So the genus of this particular Heegaard splitting is $c_1-c_0+1=c_2-c_3+1.$ If $c_0=c_3=1$, then the genus of this particular Heegaard splitting is $c_1=c_2.$

Suppose that our Morse function is minimal, i.e. it has $\mathfrak m(M)$ critical points. If we have $c_0>1$, then there are multiple zero-handles, which need to be glued together by 1-handles in order for $M$ to be connected. Hence, there is an index-one critical point $q$ whose two descending flow lines limit to two different critical points $p$ and $p'$ of index zero. Since there is exactly one flow line from $q$ to $p$, we can cancel these critical points, which contradicts the minimality of $f$. Thus, we must have had $c_0=1$ to begin with. We can see that $c_3=1$ by applying the same argument to $-f$. This proves that $$\mathfrak h(M)\leq c_1=\frac{2c_1+2}{2}-1=\frac{c_0+c_1+c_2+c_3}{2}-1=\frac{\mathfrak m (M)}{2}-1.$$ To see the reverse inequality, we need to prove that a genus $g$ Heegaard splitting gives rise to a Morse function $f:M\rightarrow [0,3]$ with $c_0=c_3=1$ and $c_1=c_2=g$. Letting $H_g$ denote a handlebody of genus $g$, we can define a self-indexing Morse function $h:H_g\rightarrow[0,1.5]$ having one index-zero critical point and $g$ index-one critical points, and satisfying $h^{-1}(1.5)=\partial H_g$ (because $H_g$ can be formed from a single zero-handle and $g$ one-handles). Then $3-h:H_g\rightarrow[1.5,3]$ is also a self-indexing Morse function having one index-three critical point and $g$ index-two critical points, and satisfying $(3-h)^{-1}(1.5)=\partial H_g$. Being careful to preserve smoothness, we can glue these two functions along the splitting surface of the Heegaard splitting to define the desired Morse function $f:M\rightarrow [0,3].$

  • 1
    $\begingroup$ Thank you for resurrecting this long-forgotten question! $\endgroup$
    – Anthony
    Jun 8, 2023 at 12:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .