Interchanging the sides of a Heegaard splitting Suppose that $M$ is a closed orientable connected $3$-manifold and that $M = U \cup V$ is a Heegaard splitting of $M$ (i.e. $U$ and $V$ are both handlebodies with a common boundary).  What are some necessary/sufficient conditions for there to exist a homeomomrphism $h : M \to M$ where $h(U) = V$ and $h(V) = U$?

For what it's worth, your question can be translated into a group theoretic criterion, expressed in the mapping class group $\text{MCG}(S)$ of the surface $S = U \cap V$; this is the group of homeomorphisms of $S$ modulo the normal subgroup of homeomorphisms isotopic to the identity. 
A homeomorphism $h : M \to M$ which swaps the two sides $U$ and $V$ restricts to a homeomorphism $\Phi : S \to S$. Furthermore, if one considers all homeomorphisms $\Phi : S \to S$, the ones that extend to a side-swapping homeomorphism of $M$ are invariant up to composition by homeomorphisms of $S$ isotopic to the identity. Thus your question can be equivalently reformulated as follows: 


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*Question: What is a necessary and sufficient condition for the existence of $\phi \in \text{MCG}(S)$ that is represented by the restriction of a side-swapping homeomorphism of $M$?


Let $\text{MCG}_U(S) < \text{MCG}(S)$ denote the subgroup of all mapping classes that are represented by the restriction to $S$ of a homeomorphism of the handlebody $U$ (this is a well-studied subgroup of $\text{MCG}(S)$, used by Masur, Canary, and others in understanding hyperbolic structures on handlebodies with applications to the proof of Thurston's ending lamination conjecture). Similarly denote $\text{MCG}_V(S)$. These two subgroups are conjugate in $\text{MCG}(S)$: the set of conjugators is represented by the restrictions to $S$ of homeomorphisms $U \mapsto V$. Furthermore this set of conjugators is a left coset of the normalizer $N_U < \text{MCG}(S)$ of $\text{MCG}_U(S)$ which I'll denote $N^V_U$. Similarly let $N^U_V$ denote the left coset of the normalizer $N_V < \text{MCG}(S)$ of $\text{MCG}_V(S)$ represented by the restrictions to $S$ of homeomorphisms from $V \mapsto U$. Notice that $(N^U_V)^{-1} = N^V_U$.


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*Answer: $N^V_U \cap (N^V_U)^{-1} \ne \emptyset$

