Prime Decomposition of Three-Dimensional Manifolds into Boundary Connected Sum

In 2003 Matveev suggested a new version of the Diamond Lemma suitable for topological applications. We apply this result to different situations and get a new conceptual proof of theorem on decomposition of three-dimensional manifolds into boundary connected sum of prime components.


Introduction
Since 2003 Matveev [1][2][3] had suggested a new version of the Diamond Lemma [4] of great importance for various fields of mathematics, which is suitable for and efficient solving topological problems. In this paper we apply this result to get a new conceptual proof of theorem on decomposition of threedimensional manifolds into boundary connected sum of prime components.

Definition, Lemma, and -Irreducible Manifolds
Diamond Lemma (see [5]). If an oriented graph Γ has the properties (FP) and (MF), then each of its vertices has a unique root.
Definition 1. Let be a proper disk in a compact connected 3-manifold . A disk reduction of along consists in cutting along the disk . Its result is a new manifold .
We apply nontrivial disk reductions to a given manifold as many times as possible. If this process stops, then we obtain a set of new manifolds, which is called a root of .
(1) If the disk is splitting, the manifold is obtained by gluing the manifolds 1 and 2 together along disks on their boundaries. Then is called a boundary connected sum of the manifolds 1 and 2 and denoted by = 1 # 2 .
(2) If the disk is nonsplitting, then is also connected.

Definition 2.
is said to be irreducible if every properly embedded disk in is trivial.
Lemma 3 (see [6]). Let be a -irreducible manifold. Let be the manifold obtained from by attaching a 1-handle to make the boundary connected. Then is a prime which is not irreducible.

Proof of Theorem 4
Theorem 4. Any connected irreducible compact 3-manifold different from a ball and with nonempty boundary is homeomorphism to a boundary connected sum = 1 # ⋅ ⋅ ⋅ # of prime manifolds. All the summands are defined uniquely up to reordering and, if is non-Orientale, replacing solid tori by solid Klein bottles 1× 2 .

ISRN Applied Mathematics
We apply the universal scheme [4] in two stages. First, by considering reductions along all disks we establish uniqueness of the -irreducible manifolds . Then we restrict ourselves to reductions only along splitting disks and by lemma [3] complete the proof of the theorem.
We construct the graph Γ [5] whose vertices are compact connected irreducible manifolds, considered up to addition or deletion of three-dimensional balls. The edges of the graph correspond to reductions along both splitting and nonsplitting disks.
Since every set of vertexes of nonnegative integers has a minimal vertex, by Lemma 5, the process stops and we get a root. Then Γ has the properties (FP).

Lemma 6.
The graph Γ has a mediator function; that is, Γ has the properties (MF).
be a pair of edges with common beginning. Following the universal scheme, we define the value ( → 1 , This minimum is taken over all pairs of such disks. As usual, we assume that the disks are in general position, so that their intersection consists of a finite number of circles and arcs.
We first consider the property (MF1), that is, the case ( → 1 , When the reduction is carried out along disjoint disks 1 and 2 , then each of these disks survives under the reduction along the other disk. We can therefore assume that ⊂ for ( , ) = (1, 2) and ( , ) = (2, 1). By reducing each of the manifolds along the disk . We obtain the same vertex ∈ (Γ). This proves the property (MF1).
Next, we verify the property (MF2), that is, the case ( → 1 , → 2 ) > 0, when any two disks 1 and 2 defining the edges → 1 and → 2 intersect. Then these disks lie in the same connected manifold ⊂ . According to the universal scheme, it is enough to show that there exists a mediator disk, that is, a nontrivial disk ⊂ satisfying | ∩ | < | 1 ∩ 2 | for = 1, 2.
Case 1. Among the circles in 1 ∩ 2 we choose one, denoted by , which is innermost with respect to the disk 1 . This means that the circle bounds a disk in 1 such that 1 ∩ 2 = . We cut 2 along and glue up the boundaries of the cut by two parallel copies of the disk . By applying a small perturbation we obtain a new disk and a sphere whose intersection with 2 is empty and whose intersection with 1 consists of a smaller number of circles (since the circle has disappeared). The disks must be nontrivial in , since otherwise the disk 2 would be trivial. Therefore, can be taken as a mediator disk.
Case 2. Among the arcs in 1 ∩ 2 we choose one, denoted by , which is outermost with respect to the disk 1 . We cut 2 along and glue up the boundaries of the cut by two parallel copies of the disk . By applying a small perturbation we obtain a new disk and a sphere whose intersection with 2 is empty and whose intersection with 1 consists of a smaller number of arcs (since the circle has disappeared). At least one of these two disks (say ) must be nontrivial in , since otherwise the disk 2 would be trivial. Therefore, can be taken as a mediator disk.
By applying the Diamond Lemma we prove that a root of any vertex of the graph exists and is unique.
We therefore obtain the following theorem, which is a particular case of Theorem 4 for manifolds without nonsplitting proper disks.

Theorem 7. Any connected irreducible compact three-dimensional manifold different from a ball and with nonempty boundary is decomposed by disk reductions into a union of -irreducible parts. These parts are defined uniquely up to reordering.
In order to finish the proof of Theorem 4 we change the graph Γ constructed by keeping the vertex set intact while disallowing reductions along nonsplitting disks. The edges of the new graph Γ ⊂ Γ therefore correspond to reductions only along splitting disks, so that the root of any vertex consists exactly of the prime summands of the manifolds corresponding to the vertex. The properties (FP) and (MF1) of Γ are automatically inherited by the graph Γ . The only difficulty with the proof of the property (MF2) is that after applying to the disk 2 a surgery along the innermost circle ⊂ 1 both new spheres and and disk ⊂ may be nonsplitting and therefore can not be taken as mediator disks. Case 1. Assume that 1 ∩ 2 consists of > 3 circles. Then we connect the spheres and by a tube which does not intersect 1 .
Since each of these spheres is obtained by connecting and by tubes contained in one of the parts into which ∪ ∪ divides , the disk 3 is splitting and nontrivial (the latter follows from the simple observation that each simple arc connecting the sphere and disk on the boundary is trivial). It is important to note that, after a suitable small perturbation, 3 intersects 1 in − 1 circles and intersects 2 in two circles. Thus, 3 is a mediator sphere.
The cases = 1 and = 2 need to be considered separately. We show that in each of these cases there exists a mediator disk. If = 1, then the disks 1 and 2 split into 4 parts , 1 ≤ ≤ 4. These parts are different, since they are separated by 1 and 2 : any two parts lie on different sides with respect to at least one of the disks. The boundary of each part consists of a single splitting disk. At least one of these disks is nontrivial, since both 1 and 2 are nontrivial. Hence, it can be taken as a mediator disk. Now let = 2. Then the disks 1 and 2 split either into 5 parts , 1 ≤ ≤ 5 (if the parts 1 and 3 are different), or into 4 parts (if the parts 1 and 3 coincide). None of the other parts can coincide, since only the parts 1 and 3 are not separated by 1 and 2 . Since 1 ̸ = 3 , then each of the parts 2 and 3 is bounded by a sphere 1 , 4 is bounded by a disk, and at least one of these disks is nontrivial, since 2 is nontrivial. This disk can be taken as a mediator. We also note that the part 5 is bounded by a torus.
Assume that 1 and 3 coincide and constitute a single connected part . Then we have three candidates for a mediator sphere: the spheres 2 and 4 and the new disk = 1 # 3 ⊂ . The latter is obtained by connecting 1 and 3 by a tube inside . We assert that at least one of the three disks is nontrivial and therefore is a mediator.
Arguing by contradiction, assume that all three disks are trivial. This means that 2 is ball and is homeomorphism to 2 × . Then the reduction along the disk 1 produces two manifolds 1 = ( ∪ 4 ) ∪ 1 and 1 = ( 5 ∪ 2 ) ∪ 1 , and the reduction along the disk 2 also produces two manifolds 2 = ( ∪ 2 ) ∪ 2 and 2 = ( 5 ∪ 4 ) ∪ 2 . Here the balls 2 are viewed as handle of index 1 (if they are attached to ) or as handle of index 2 (if they are attached to 5 ). The balls 1 , 1 , 2 , 2 , and 2 glue up the disks on the boundaries of the corresponding manifolds. Handles 2 of index 1 connect different disks on the boundary of the manifold ≈ 2 × , which implies that 1 ≈ 2 ≈ 1 × 2 . On the other hand, the bases of handles 2 of index 2 in the torus 5 are isotopic; hence 1 ≈ 2 . Therefore, the reductions along the disks 1 and 2 give the same result. This contradicts the fact that in our situation when ( → 1 , → 2 ) > 0, the vertices 1 and 2 must be different.
Case 2. Assume that 1 ∩ 2 consists of > 3 arcs. The proof can be finished by an argument similar to that used in the proof of Case 1.
The arguments above are also applicable in the case of prime decompositions of non-Orientale manifolds, with the only difference that handles 2 of index 1 can now be attached to the boundary of the manifold ≈ 2 × in two different ways. In one case the result is the direct product 1 × 2 , and in the other case we get the twisted product 1× 2 . Nevertheless, since the manifold is non-Orientale, all summands 1 × 2 can be replaced by 1× 2 .