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In proving that all closed, compact 3-manifolds $M$ admit Heegaard splittings one often takes a triangulation of $M$ and then takes as one side of the Heegaard splitting a regular neighborhood of the 1-skeleton. My question is: do all Heegaard splittings arise in this way?

Namely, given such a 3-manifold $M$ together with a Heegaard splitting $U \cup V = M$, does there exist some triangulation of $M$ such that $U$ is a regular neighborhood of the 1-skeleton of the triangulation?

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    $\begingroup$ No. You have a lower bound on the genus in terms of the number of 1-simplices. You can't get the genus 1 Heegaard splitting of the sphere this way. $\endgroup$
    – user98602
    Jan 4 '17 at 3:14
  • $\begingroup$ @MikeMiller I see now that given a triangulation with $t$ 3-simplices, that the genus must be less than or equal to $t+1$. But how do you get a bound just in terms of the number of 1-simplices? $\endgroup$
    – user101010
    Jan 5 '17 at 20:34
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Any triangulation of a 3-manifold contains at least one 3-simplex. As the 1-skeleton of 3-simplex is homotopy equivalent to a wedge of 4-circles, a regular neighborhood of the 1-skeleton of the whole 3-manifold is at least genus 4.

In general the genus is bounded by the number of 2-simplices as the face condition implies any two adjacent simplices have a 1-skeleton which is two one simplices glued along a single edge, and hence for $n$ 2-simplices the one-skeleton is homotopy equivalent to a wedge of $n$ circles and a regular neighborhood has genus at least $n$.

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