MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is easy to visualize some Morse functions on surfaces (such as the torus) via the height function, but this seemingly doesn't work for 3-manifolds. I am looking for an explicit one on the 3-torus $\mathbb{T}^3$. In paritcular:
It is true that every nice-enough 3-manifold admits a self-indexing Morse function $f:M\to[0,3]$ (with unique maximum and minimum), and from this we get a Heegaard diagram, with splitting surface $\Sigma=f^{-1}(\frac{3}{2})$. Now apparently, a converse also holds, so that given a Heegaard decomposition we can read off a self-indexing Morse function (edit: perhaps multiple Morse functions, as per Ryan Budney's comment).
With the known Heegaard decomposition of $\mathbb{T}^3$ (handlebodies $N$ and $\mathbb{T}^3-N$ for a small neighborhood $N$ about the generating 1-complex $S^1\cup S^1\cup S^1\subset\mathbb{T}^3$, and splitting surface $\partial N$), what is the corresponding Morse function? I can't write one down.

share|cite|improve this question
Technically there's a gap between having a Heegaard splitting and a self-indexing Morse function. There's the diffeomorphism group $Diff(M,H)$ where $H$ is the heegaard splitting surface for the manifold $M$, this acts on all Morse functions with that Heegaard surface as its "mid-point". As long as you don't consider that group to be "big" then you're okay. But it is a big group. – Ryan Budney Apr 9 '12 at 8:07
up vote 12 down vote accepted

The Morse function on $(S^1)^3$ that gives the genus 3 Heegaard splitting (the standard one) is basically just a smoothed "distance from the 1-skeleton function".

So if you think of $S^1$ as the unit circle in $\mathbb C$, then

$$f : (S^1)^3 \to \mathbb R$$

is given by

$$f(z_1,z_2,z_3) = |z_1-1|^2 + |z_2-1|^2 + |z_3-1|^2$$

where we're taking the norm/modulus squared of vectors in $\mathbb C$ in the above formula. The critical points are 8 triples $(z_1,z_2,z_3)$ of the form $(\pm 1, \pm 1, \pm 1)$, so you have the minimum $(1,1,1)$, maximum $(-1,-1,-1)$, index 1 critical points $(-1,1,1), (1,-1,1), (1,1,-1)$ and index two (the negatives of the index one critical points).

If you really demand self-indexing then you'll need the function $\frac{f(z_1,z_2,z_3)}{4}$

share|cite|improve this answer
Ah so simple, and truly based off the 1-skeleton! I am wondering what the relation this one is now to the one I just posted, do you know? – Chris Gerig Apr 9 '12 at 22:17

After seeing Ryan Budney's function, I came up with this:

On an n-torus $\mathbb{T}^n$ we have a Morse function $f=\sum^n_{i=1}\cos\theta_i$, which has critical points $\text{crit}(f)=\lbrace(\theta_1,\ldots,\theta_n)\;|\;\theta_i = 0\text{ or }\pi\;\;\forall\;i\rbrace$.

In particular, we get a self-indexing Morse function $f(\theta_1,\theta_2,\theta_3)=\frac{1}{2}\sum^3_{i=1}\cos\theta_i+\frac{3}{2}$ on $\mathbb{T}^3$.

Indeed, it has a unique maximum $(0,0,0)$ with value 3, and a unique minimum $(\pi,\pi,\pi)$ with value 0. There are 3 index-1 critical points (two angles are $\pi$ and one angle is $0$) with value $1$, and 3 index-2 critical points (one angle is $\pi$ and two angles are $0$) with value $2$, and the indices are evident from the Hessian $H_{(\theta_1,\theta_2,\theta_3)}f=-\text{diag}(\cos\theta_1,\cos\theta_2,\cos\theta_3)$.

Now this $f$ then provides a Heegaard diagram for $\mathbb{T}^3$ of genus $3$ (since there are 3 index-1 and 3 index-2 critical points, corresponding to the $\alpha,\beta$-curves). As Wikipedia states: It was proved by Frohman and Hass that any other genus 3 Heegaard splitting of the three-torus is topologically equivalent to this one (the one in the original question).

[[Edit]]: This is precisely Ryan's function... so nevermind.

share|cite|improve this answer
Aren't they essentially the same? Write $z \in S^1$ as $z = e^{i\theta}$ then $|z-1|^2 = 2-2Re(z) = 2-2\cos(\theta)$. – Ryan Budney Apr 9 '12 at 22:23
Haha ooops. Well, nice redundancy then. – Chris Gerig Apr 9 '12 at 22:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.