Almost normal surfaces with boundary

Embeddability in the 3-sphere is decidable

“…The next picture illustrates a meridional P in the case where X is embedded in S 3 as the complement of a solid torus neighborhood of the figure '8' knot: P X Next, by choosing P as above with suitable minimality properties, one can make sure that P is normal or almost normal 6 for the given triangulation. For the case of P essential, this is an old result going back to Haken and Schubert (and for our notion of complexity of P , a proof is given in Section 7), while for P strongly irreducible and boundary strongly irreducible this follows from [BDTS12]; also see [Sto00] for the case of a strongly irreducible surface in a closed manifold. It remains to show that, in this setting, at least one of the meridians in ∂P must be short.…”

“…The point has positive sign if the exchange connects the southwest quadrant to the northeast quadrant, and it has negative sign if it connects the northwest to the southeast; see Figure 4. This is equivalent to the definition given in [BDTS12]. The definition depends on the ordering of the pair of curves and on an orientation on the surface: reversing the order or the orientation reverses every sign.…”

“…To prove F is normal we must show that it meets each tetrahedron ∆ in a collection of disks whose boundaries are normal curves of length 3 or 4. We adopt the view taken in [BDTS12], showing F meets each tetrahedron in pieces that are incompressible and edge incompressible.…”

“…An edge compressing disk for a surface in ∆ is an embedded disk E whose boundary ∂E = e ∪ f , consists of two arcs, e ⊂ T 1 and f = E ∩ F = ∂E ∩ F ; see [BDTS12].…”

“…If P is strongly irreducible and boundary strongly irreducible, then the main result of [BDTS12] states that P is isotopic to an almost normal surface. Moreover, in the proof of Proposition 3.1 of [BDTS12] it is assumed that ∂P is least length (see Lemma 3.9), which is satisfied when ∂P is tight.…”