The Unknotting Problem and Fixed-Parameter Tractability Adele Jackson Supervised by Dr Benjamin Burton Australian National University Vacation Research Scholarships are funded jointly by the Department of Education and Training and the Australian Mathematical Sciences Institute. 1 Introduction Informally, the unknot recognition problem is to, given a knot, recognise whether or not it can be untangled. A substantial open question in computational topology is whether or not there exists a polynomial time algorithm to answer this question. We investigated the question of whether unknot recognition is fixed-parameter tractable (FPT) in the cutwidth of a triangulation of the knot complement manifold, which would be trivially implied by unknot recognition being in P. In this report, we give an exposition of unknot recognition being in NP, and describe the approaches we explored in trying to show unknot recognition is FPT. 2 The unknot recognition problem 1 3 Definition 2.1. A knot K is a smooth embedding fK : S ! S . We can write this knot as an injective continuous map K : S1 ! S3. We say that two knots, K1 and K2, are equivalent if there exists some continuous 3 3 3 3 map f : S × [0; 1] ! S such that f(s; 0) = s, f(s; t0): S ! S for fixed t0 and 3 3 s 2 So is a homeomorphism on S , and f ◦ K1 = K2. Then f is an ambient isotopy taking K1 to K2. Then the unknot recognition problem is, given a knot, to check whether it is equivalent to the circle. To answer this with a computation, we give another criterion for a knot being unknottable. 3 A criterion for unknot recognition Theorem 3.1 ([3]). A knot K can be unknotted if and only if there exists a disk D with smooth boundary embedded in S3, such that its boundary, @D, is K traversed once. Alternately, we can view this disk as living in the complement of the knot, by setting up a criteria for this disk that does not directly reference traversing the knot's boundary. 3 Definition 3.2. The knot complement manifold of a knot K, MK , is S − fa solid torus neighbourhood of Kg. Definition 3.3. An essential disk in a manifold M is a disk D embedded in M such that the boundary of D lies in the boundary of M, and D cannot be homotoped into @M while holding @D fixed. 2 Figure 1: A solid torus with (a) a non-essential disk, and (b) an essential disk. For example, disk A in Figure 1 is not essential, as it can be homotoped onto the section of the boundary of the manifold cut out by @A, while B is essential. Theorem 3.4 ([3]). A knot K can be unknotted if and only if MK , the knot com- plement manifold, contains an essential disk. So now we can express the unknot recognition problem as, given a knot comple- ment manifold, checking whether or not that manifold contains an essential disk. We wish to describe how one would do this, and assess how effective that algorithm is. 4 Computational complexity When given an algorithm for a problem, we can describe how fast it is by how its running time grows as the size of the input n grows. This is effective as it does not vary with the hardware, language or details of implementation used to run the algorithm. Let the maximum number of steps taken for the algorithm to run for an input size n be M(n). We write that an algorithm is O(f(n)), for f a function of M(n) the input size, if lim supn!1 f(n) < 1. Complexity classes let us classify how difficult problems are. A decision problem is in class P if there is some algorithm to solve the problem that is O(P (n)) for P (n) a polynomial function. A decision problem is in NP if, if for some input the answer is `yes', we can give a certificate to prove this is correct that can be verified in polynomial time. For example, consider the decision problem of checking whether there exists a timetable with no clashes for some set of rooms, students, lecturers and classes. This 3 Figure 2: A one tetrahedron triangulation of the solid torus. is a difficult problem to solve. However, if such a timetable does exist, the certificate that consists of that timetable can be checked in polynomial time to indeed have no clashes. Unknot recognition is known to be in NP, as we will show in Section 5. 5 Normal surfaces To check whether or not a given knot complement manifold contains an essential disk, we need a way of concisely describing surfaces in a manifold. Normal surfaces are one method of doing this. Note that we can triangulate the knot complement manifold { that is, we can divide it into tetrahedra. For example, the knot complement manifold of S1 in S3 is again a solid torus. We can triangulate this using one tetrahedron by Figure 2, with face (023) joined to face (312). Definition 5.1. A normal surface in a triangulated 3-manifold M is a surface S embedded in M, with boundary on the boundary of M, such that the intersection of S with any tetrahedron in the triangulation is a finite union of triangles and quadrilaterals. For example, Figure 3 shows a normal surface within the triangulated torus, and how it looks in a drawing of the solid torus. The normal surface caps off the one vertex on the boundary of the torus. Given a normal surface S in a triangulation of M, we can describe this surface by listing how many of the four triangle types and three types of quadrilaterals it 4 Figure 3: A normal surface, (a) presented as triangles in the triangulated solid torus, and (b) drawn in the solid torus. 5 forms in each tetrahedron. For example, the surface in Figure 3 consists of one of each of the types of triangles in the one tetrahedron. For a surface S in a triangulation consisting of t tetrahedron, this defines a function v : normal surfaces ! Z7t giving a vector describing each surface. For example, the surface in Figure 3 can be described by the vector (1; 1; 1; 1; 0; 0; 0), as it has one of each triangle and no quadrilaterals. This description is defined for a fixed ordering of tetrahedron and ordering of types of triangle and quadrilateral. Theorem 5.2 ([3]). Any vector v 2 Z7t describes a valid (possibly disconnected) normal surface S so long as: 1) all entries in v are non-negative, 2) each tetrahedron contains only one type of quadrilateral (as different types of quadrilateral must intersect), and 3) if two faces are connected in the triangulation, the edges intersecting these faces must match up. The non-negativity and matching conditions define a cone of potential surfaces in Z7t, while the quadrilateral condition picks out a certain lattice of points within this cone. Then we can define an integer basis for this lattice. Definition 5.3. We call a normal surface S fundamental if it cannot be written as the sum of two other normal surfaces. That is, if there exist no other surfaces S0, S00 such that v(S) = v(S0) + v(S00). Now, it remains to show that normal surfaces are a useful way of detecting an essential disk in a knot complement manifold. Theorem 5.4 ([5], strengthening [3]). If MK is a knot complement manifold, with an associated triangulation, which contains an essential disk, then it contains an essential disk that is a fundamental normal surface. So if we can check all normal surfaces for an essential disk, we can solve the unknot recognition problem. 6 Unknot recognition is in NP Hass, Lagarias and Pippenger used this approach to show that unknot recognition is in NP, by a bound on the size of the fundamental normal surfaces. 6 Theorem 6.1 ([4]). Let M be a triangulated compact 3-manifold, that contains t tetrahedra in the triangulation. Then the entries of the vector describing any funda- mental normal surface in M are bounded by t27t+2 − 1. This then gives the following algorithm for checking whether a given knot K is unknottable, with input size n the number of crossings in K: 1) Produce a triangulation of the knot complement manifold MK . This can be done in polynomial time in n to have a number of tetrahedra t that is bounded by a polynomial function of n. 2) Generate all possible descriptions of fundamental normal surfaces in MK { that is, all vectors in Z7t with entries between 0 and t27t+2 − 1. 3) For each of these vectors, check if: a) the vector satisfies the quadrilateral and matching conditions, so is a valid normal surface, b) the vector is a disk, for which it is sufficient to check the surface has one puncture and Euler characteristic 1 as all fundamental normal surfaces are connected, and c) the vector describes an essential disk, which is a condition on the ho- mology class of the boundary of the surface. 4) If one of these vectors did describe an essential disk, the knot is the unknot; otherwise, it is not. All of these steps, aside from looking at every possible fundamental normal sur- face, take polynomial time in the crossing number of the original knot. Now, if the knot is equivalent to the unknot, we can give a polynomial time certificate for this: describe the triangulation and give the vector describing the essential disk.