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 3manifolds. I am looking for an explicit one on the 3torus $\mathbb{T}^3$. In paritcular:
It is true that every niceenough 3manifold admits a selfindexing 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 selfindexing 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 $\mathbb{T}^3N$ for a small neighborhood $N$ about the generating 1complex $S^1\cup S^1\cup S^1\subset\mathbb{T}^3$, and splitting surface $\partial N$), what is the corresponding Morse function? I can't write one down.

2$\begingroup$ Technically there's a gap between having a Heegaard splitting and a selfindexing 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 "midpoint". As long as you don't consider that group to be "big" then you're okay. But it is a big group. $\endgroup$– Ryan BudneyApr 9, 2012 at 8:07
2 Answers
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 1skeleton 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
$$f(z_1,z_2,z_3) = z_11^2 + z_21^2 + z_31^2$$
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 selfindexing then you'll need the function $\frac{f(z_1,z_2,z_3)}{4}$

$\begingroup$ Ah so simple, and truly based off the 1skeleton! I am wondering what the relation this one is now to the one I just posted, do you know? $\endgroup$ Apr 9, 2012 at 22:17
After seeing Ryan Budney's function, I came up with this:
On an ntorus $\mathbb{T}^n$ 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 selfindexing 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 index1 critical points (two angles are $\pi$ and one angle is $0$) with value $1$, and 3 index2 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 index1 and 3 index2 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 threetorus is topologically equivalent to this one (the one in the original question).
[[Edit]]: This is precisely Ryan's function... so nevermind.

$\begingroup$ Aren't they essentially the same? Write $z \in S^1$ as $z = e^{i\theta}$ then $z1^2 = 22Re(z) = 22\cos(\theta)$. $\endgroup$ Apr 9, 2012 at 22:23

$\begingroup$ Haha ooops. Well, nice redundancy then. $\endgroup$ Apr 9, 2012 at 22:27