Using this algorithm and the ideas from this paper (page 3) I gather that I can present $S^3-K$ where $K$ is the figure eight knot as the gluing of two genus 5 handle bodies, $H_5$, i.e. $$ S^3-K=H_5\cup_h H_5 $$ where $h$ is an element of the mapping class group of $S^3-K$.
Two questions:
(1) Suppose I'm given an element $h$ in the mapping class group of $S^3-K$. How do I take $h$ and produce a Heegaard splitting? I mean, every $h$ should be a prescription for mapping the meridians of one handle body (not necessarily of genus 5) to the parallels of the other right?
(2) Every mapping class group has the identity as an element. Since an element of the mapping class group is a homeomorphism of handle bodies, does that mean every manifold can be presented as the identity map on two handle bodies of the same genus?
EDIT: Warning, I've learned enough now to realize my first question isn't entirely clear. I'll see if I can't fix it up.