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 squared of vectors in $\mathbb C^2$ 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).

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 $\frac{f(z_1,z_2,z_3)}{4}$

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

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

where we're taking the norm squared of vectors in $\mathbb C^2$ 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).

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 squared of vectors in $\mathbb C^2$ 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).