By Victor Guillemin

This article suits any direction with the notice "Manifold" within the identify. it's a graduate point ebook.

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J M is of type IV. 2. By the induction hypothesis S' , hence S is disc decomposable. • Case 3B: Case 3A does not hold. Proof: It follows that v is of type V. 7. Simplifying Assumption: If v is of type II, then If v is of type III, then (v) = A (V) = A

C) If |n. +-. 6, then apply b ) . 6 d) If n. +, < 0 for all i mod k k ^ 4 and i) n. -j = -2 n. 7, then apply c ) . | > 2 all i then decompose in an analogous way to Case le). 7 -2 > 2 m §4. TAUT FOLIATIONS AND FLAT MINIMAL SURFACES In this section we give the precise criteria to determine whether or not a candidate surface S is disc decomposable. There are two types of obstructions that must vanish for S to be a surface of minimal genus. First the projection, hence the candidate surface, associated to T may have simplifications analogous to the situation for pretzel links.

12 v potentially + If t(v) = 1 , then TnB 3 decomposes to C . 38 DAVID GABAI THB: T'nB decomposes to .. 12 B) t(v) f 1 , then C or one of A' or B ' . v potentially t(v) = -1 , then T n B3 decomposes to If t(v) f -1 , then T n B can decompose to a choice of either B' or one of A' or C . If Remark: 3 T n B can decompose to a choice of either If When |t(v)| f 1 we can decompose corresponding to the case if t(v) one of the other 2 possibilities. just chopping P(v), T n B B\ in two ways, one and the other corresponding to Decomposition A or A' corresponds to K M off of T .