Crowell’s state space is connected

Crowell’s state space is connected

Daniel Selahi Durusoy Michigan State University durusoyd@math.msu.edu
Abstract.

We study the set of Crowell states for alternating knot projections and show that for prime alternating knots the space of states for a reduced projection is connected, a result similar to that for Kauffman states. As an application we give a new proof of a result of Ozsváth and Szabó characterizing torus knots among alternating knots.

Key words and phrases:
State sum, Alexander polynomial, spanning trees

1. Introduction

One of the many definitions of the Alexander polynomial of a knot is through state sums. Kauffman has described and studied a state sum model for the Alexander polynomial in great detail [Kau83]. In an earlier paper Crowell has described another state sum model for the Alexander polynomial for the subclass of alternating knots ([Cro59], Theorem 2.12).

In the next section we will recall the definition of Crowell states and examine some of their properties. In Section 3 we will prove that

Theorem 1.

If is an alternating prime knot and is a reduced knot diagram for , then any two states differ by a finite sequence of terminal edge exchanges.

This theorem is similar in nature to the Clock Theorem of Kauffman [Kau83] which states that any two Kauffman states differ by a finite sequence of clockwise and counterclockwise moves, which was also proven in the language of graphs in section 4 of [GL86]. This work is independent of those mentioned because of the simple reason that Kauffman states and Crowell states do not correspond to each other in any natural way as observed from the fact that the space of Crowell states do not form a lattice in general (see Proposition 5).

In section 4, as an application we will give an alternative proof that torus knots are characterized by their Alexander polynomials among alternating knots, which was originally proven by Ozsváth and Szabó (Proposition 4.1 in [OS05]).

2. A state model

In this section we will review the definition of the state sum model for alternating links given by Crowell [Cro59] and investigate some properties of the states.

Given a knot and an oriented alternating diagram of with crossings we obtain a weighted labeled directed planar graph as follows: replace a small neighborhood of each crossing by a degree vertex according to the following figure ( is the vertex label):

Figure 1. From a knot diagram to a directed graph.
Proposition 1.

This definition orients all edges and these orientations are compatible with orientations coming from a checkerboard coloring of the regions in the complement of .

Proof.

Each edge gets an orientation since while traveling around a region the strands traveled alternate between under strand and over strand. Hence at each crossing if an edge is coming in, the boundary of the region will continue along the over strand, which becomes an under strand at its other end, hence we get an orientation on that edge as well, consistent with the previous edge. These orientations are compatible with a checkerboard coloring since crossing a region to another across an edge we get opposite orientations in the plane. ∎

Choose a vertex of . Spanning trees rooted at (edges are directed away from ) will be called states. Let be the space of states and be the product of weights of all edges in a state . According to [Cro59, Theorem 2.12] we get the renormalized Alexander polynomial as a sum of monomials corresponding to each state by

(1)

where is chosen so that the term with the least power of is a positive constant.

Proposition 2.

For any vertex , there is a rooted Hamiltonian path from to in .

Proof.

Since is a knot, as an unoriented graph is connected. Pick an unoriented path starting at , ending at . If is not oriented away from the root, pick edges that go around one of the two regions adjacent to . Due to Proposition 1, we get compatible orientations. At the end we get a rooted path which might visit some vertices more than once. For each vertex that is visited more than once, remove all edges between the first and last visits. ∎

Corollary 1.

Any rooted tree can be extended to a rooted spanning tree .

Proof.

For any vertex not in , find a rooted Hamiltonian path from to using Proposition 2. Add enough of the final segment of to the current tree so that the union will be connected. ∎

Proposition 3.

Given a reduced alternating diagram for a prime knot, any edge in except those ending at appear as a teminal edge in at least one state.

Proof.

Call the given edge starting at vertex , ending at vertex . We will find a directed Hamiltonian path from to , add the edge , and extend this path into a rooted spanning tree.

Use Proposition 2 to construct a rooted Hamiltonian path from to . There are two possible cases:

(a) doesn’t go through : Do nothing extra.

(b) goes through before reaching : Adding to produces a loop. To avoid this problem we will go around as follows. Assume contains an edge going from to and an edge going from to . We need to connect to by an oriented path avoiding and . To achieve this, let be the region bounded on two sides by and and be the union of regions adjacent to along edges other than and . Then starting at , following edges on the boundary of that are not on the boundary of , we reach . Assume that is prime and is reduced, then this new path does not include and since otherwise we could draw a separating circle passing through the or and another common edge of and . Replacing and by in , we get a rooted path from to avoiding . It could include loops, which can be eliminated as in the proof of Proposition 2.

Next we need to extend the rooted Hamiltonian path to a rooted spanning tree, keeping a terminal edge. Consider the two edges coming out from , call them and , with terminal vertices and . Using a similar argument as in the proof of case (b) above, consider being the region bounded by and , and the union of regions adjacent to except along and , and removing loops we get a directed path from to . Starting at , add enough edges from the final segment of so that is connected to a vertex in . Similarly do so for .

Now, use Corollary 1 to extend this tree to a rooted spanning tree. During this process stays a terminal edge since adding or would create a loop. ∎

The state sum in Equation 1 resembles the state sum defined by Kauffman [Kau83]. Kauffman has studied an operation called clock move that transforms a state to another that differ only at two crossings and showed that all states differ from one another by a sequence of clock moves. With that in mind we define the following operation for reduced alternating diagrams:

Definition 1.

A state is obtained from a state by a terminal edge exchange move (edge exchange for short) if replacing a terminal edge in by the other incoming edge at the terminal vertex gives .

Figure 2. Terminal edge exchange.
Proposition 4.

At any terminal edge, edge exchange gives a new state, except at edges ending at a kink.

Proof.

If there is a kink at , there is a unique edge that connects to in any spanning tree, since the opposite edge is a loop. At any other vertex, it is easy to check that one still gets a rooted spanning tree. ∎

Edge exchange gives a partial order on the set of states by defining the covering relation of the partial order as comes immediately before if is obtained from by one positive edge exchange.

Comparing these states with the black trees in Kauffman states, even though states are rooted spanning trees in both models, in Kauffman states the orientations on the edges are chosen after a spanning tree of the black graph is chosen, so the same edge can inherit different orientations in different states. Furthermore, consider the graph whose vertices are Crowell states and any two vertices are connected by an edge if there is a terminal edge exchange that takes one state to the other. The following proposition shows that edge exchanges do not correspond to clock moves under any bijection between the Crowell and Kauffman states since Kauffman states form a distributive lattice [Kau83].

Proposition 5.

The space of Crowell states is not a lattice in general with any choice of a partial order compatible with terminal edge exchanges.

Proof.

Figure 3 illustrates the graph of Crowell states for the knot in Rolfsen’s table. Let us assume that there is a partial order compatible with this graph, i.e., an edge between two states exist if one is an immediate successor of the other. Then any degree one vertex is either a local maximum or a local minimum. Since this graph has three degree 1 vertices and in a finite lattice there is only one local maximum and only one local minimum, this particular graph can not be a lattice. ∎

Figure 3. The knot , the chosen root, and its space of states.

3. Proof of Theorem 1

In this section we will assume that is a prime knot and is a reduced alternating projection for . Choose a root vertex in . We will provide an algorithm to go from one rooted spanning tree to another rooted spanning tree through a sequence of edge exchanges. We will label vertices of with distinct integers.

An initial segment of a rooted spanning tree is the sequence of vertices on the unique rooted path from the root to in . For , let denote the vertex that points to through an edge not in . Let be a small neighborhood of the set of vertices below in , i.e., those that can be reached from via directed paths in , the edges between them (not necessarily in ) and the elementary regions surrounded by those edges. Let be the connected component of containing the successor of in with the smaller label. When the tree is obvious from the context, we will suppress from these notations. The rooted meet of two rooted trees and is the connected component of the root in and will be denoted by .

Figure 4. The knot diagram and a state in .

In order to prove Theorem 1, we will show that for any given two states and , we can persistently enlarge .

Lemma 1.

Given a state and a vertex other than the root , if , then there is a sequence of edge exchanges that converts the incoming edge for into a terminal edge, removing edges only below .

Lemma 2.

Under the conditions of Lemma 1, there is a vertex with .

Figure 5. Thick edges belong to the spanning tree .
Proof of Lemma 2.

Assume is nonempty. Given , consider . Since is prime and is reduced, is not a separating circle, hence there are at least intersections. Since the orientations of adjacent regions alternate, we get at least two edges entering into .

Pick a vertex among all terminal vertices of edges in entering into not originating from . This choice implies that . Furthermore the union contains an unoriented circuit of edges and vertices that separate from (see Figure 5), hence going from to a vertex in would take at least two edge exchanges. We conclude that as well. ∎

Proof of Lemma 1.

If is empty, then is already a terminal vertex. Otherwise, we will use induction on the depth of the tree .

For , could contain up to two vertices. If there is only one vertex, it is a terminal vertex and an edge exchange empties . If there are two vertices, has two components, which as in the proof of Lemma 2, are not adjacent to one another. Lemma 2 tells that an edge exchange at either vertex decreases the size of , and we are led to the case of one vertex.

Assume the hypothesis is true for all trees of depth and less. If has depth , use Lemma 2 to find a with the property , in particular . Therefore by the induction hypothesis becomes a terminal vertex after a finite sequence of edge exchanges only removing edges in . Then performing an edge exchange at decreases the size of . Hence repeating this process becomes a terminal vertex while only edges below being removed throughout the process. ∎

Proof of Theorem 1.

Given two distinct rooted spanning trees and , pick a vertex adjacent to along an edge in . Note that since and once is a terminal edge, after an edge exchange will be connected to along the same rooted Hamiltonian path as in .

Let be the vertices that lead to in and respectively. By definition, and . Hence , but , hence , which means .

Applying Lemma 1, we get a sequence of edge exchanges that ends in a state where is a terminal vertex without removing any edges from the rooted meet. Now perform an edge exchange at , this enlarges the rooted meet. Since the rooted meet only enlarges during this process, in finitely many repetitions of this process we reach . ∎

4. An application to torus knots

In this section we will provide a different proof of the following result originally proved by Ozsváth and Szabó:

Theorem 2.

The torus knots are characterized among alternating knots by the Alexander polynomial.

Proof.

Let be a reduced alternating projection for a knot with . Since all coefficients of powers of are , each state has a different weight and is not a product of two alternating knot polynomials (c.f. [OS05, Prop. 4.1]), hence is prime.

Let be the state with the least power. Since is prime and the fact that each edge exchange changes the power of by , using Theorem 1 we get a linear ordering on the states starting at , reaching each next state by exchanging an edge of weight with an edge of weight .

According to Proposition 4 and due to this linear order, and the top state have only one terminal edge each, hence they have no branching, whereas intermediate states have terminal edges.

Since Proposition 3 tells that each edge (except the two that point to ) can be extended to a state having as a terminal edge, and since we can reach that state from by positive edge exchanges, we see that all edges in have weight . We conclude that has edges since each edge of weight is used only once in an edge exchange and no new edges emerge with weight as we go from to .

Edge orientations and weights do not depend on the choice of the root vertex, hence, after moving the root from to , we still get a space of states with the same properties, in particular, there will be a new state containing a linear directed chain of vertices starting at , ending at . Hence we get a cycle of length of edges of weight . Similarly, all remaining edges have weight , form a loop and are used in , except the one pointing at the root.

Figure 6. Thick edges have weight . On the right, orientations at a typical node.

This information tells us that if there is an incoming edge of weight at a vertex , the next edge of weight has to be on the same side of the loop of edges due to the cyclic alternating orientation of edges at a vertex. Since these edges with weight form a loop as well, they have to go between consecutive vertices. This gives us the diagram for the torus knot. ∎

Acknowledgements. I would like to thank Bedia Akyar for the invitation to give a talk at Dokuz Eylül University, during which time period I started exploring the properties of the Crowell state space. Most of the work was done during my time at Ferris State University. I would also like to thank Mahir Bilen Can and Mohan Bhupal for providing feedback.

References

  • [Cro59] R. H. Crowell, Genus of alternating link types, Ann. of Math. (2) 69 (1959), 258–275.
  • [GL86] P. M. Gilmer, R. A. Litherland, The duality conjecture in formal knot theory, Osaka J. Math. 23 (1986), 229–247.
  • [Jon09] I. D. Jong, Alexander polynomials of alternating knots of genus two, Osaka J. Math. 46 (2009), no. 2, 353–371.
  • [OS05] P. Ozsváth and Z. Szabó, On knot Floer homology and lens space surgeries, Topology 44 (2005), 1281–1300.
  • [Kau83] L. Kauffman, Formal knot theory, Mathematical Notes, Vol. 30, Princeton University Press, 1983.
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