Catalan States of Lattice Crossing: Application of Plucking Polynomial
For a Catalan state of a lattice crossing with no returns on one side, we find its coefficient in the Relative Kauffman Bracket Skein Module expansion of . We show, in particular, that can be found using the plucking polynomial of a rooted tree with a delay function associated to . Furthermore, for with returns on one side only, we prove that is a product of Gaussian polynomials, and its coefficients form a unimodal sequence.
Keywords: Catalan States, Gaussian
Polynomial, Knot, Kauffman Bracket, Link, Lattice Crossing, Plucking
Polynomial, Rooted Tree, Skein Module, Unimodal Polynomial
Mathematics Subject Classification 2000: Primary 57M99; Secondary 55N, 20D
Our work finds its roots in the theory of skein modules of -manifolds. Let us recall some standard notation and terminology. denotes an oriented surface of genus with boundary components, , and . The KBSM111For an oriented -manifold Kauffman Bracket Skein Module (KBSM) was defined in by J. H. Przytycki (). of has a structure of an algebra (Skein Algebra of ), with the multiplication defined by placing a link above a link . The algebra appears naturally in a study of quantizations of -character varieties of fundamental groups of surfaces [1, 15]. Therefore, in several important cases , these algebras have been well studied and understood (see, for instance Theorem , Corollary , and Theorem in ). Furthermore, a very elegant formula for the product in was found by Charles Frohman and Razvan Gelca in . Inspired by these results, we started our quest for a formula of a similar type in 222For , the presentation of the algebra was found by D. Bullock and J.H. Przytycki (see Theorem in ). (see ). In this paper, we unravel a surprising connection between our skein algebra, Gaussian polynomials and rooted trees.
The Relative Kauffman Bracket Skein Module was defined in . As shown in , of with fixed points on is a free module with a basis consisting of all crossingless connections (called Catalan states) between boundary points. Let be an -rectangle and . We fix points , , , , , ; , , , , , on , so that their projection onto is as in Figure 1.3. Let denote the of with the points fixed. Consider tangle -lattice crossing obtained by placing vertical parallel lines above horizontal parallel lines as in Figure 1.3.
shall denote the set of all crossingless connections between points on . We can place an order on so that it becomes a basis of (see ), and now we have
for some Laurent polynomials . In this paper we aim to determine , for all Catalan states with no returns on a fixed side of .
Let be the set of all Kauffman states of , i.e. the set of all choices of positive and negative markers for crossings as in Figure 1.3. For in , denote by its corresponding diagram obtained by smoothing all crossings of according to . Furthermore, let be the number of closed components of and be the Catalan state resulting from removal of said components. Define the function by . For , the Kauffman state sum is given by:
where and stand for the number of positive and negative markers determined by , respectively. Hence, for , its coefficient is given by:
if ( is a realizable Catalan state), otherwise ( is non-realizable. In  we gave necessary and sufficient conditions for and found a closed form formula for the number of realizable Catalan states. Furthermore, for with no arcs starting and ending on the same side of , a formula for was also obtained333In fact, it was derived by J. Hoste and J. H. Przytycki in while writing  although it was not included in the final version of their paper. Also, as we were told by M. Hajij, a related formula was noted by S. Yamada , . (see ). Let denote the subset of all realizable Catalan states of that have no returns on the floor of . In this paper, we consider a submodule of generated by along with the projection of onto the aforementioned submodule. In particular, we show the following:
For with returns allowed only on the ceiling of , the coefficient is a product of Gaussian polynomials and its coefficients form a unimodal sequence.
The paper is organized as follows. In the second section, we define a poset
associated with Kauffman states and
use it to compute . Next, in section three, for a Catalan
, we define a rooted tree with a delay function and the plucking polynomial . We show that can be computed using , the highest and the lowest coefficients of are equal to one, and all its coefficients in between are positive. In section four, we discuss several important properties of . In particular, using results of , , and , for with returns on its ceiling only, we obtain as corollaries that has unimodal coefficients, and we give criterion for a Laurent polynomial to be . Finally, in the last section, we give a closed formula for , where .
2 Poset of Kauffman States
In this section, for Catalan states with no returns on the floor, we construct a poset of Kauffman states which we later use to compute . Given a Catalan state , its boundary is , where , , , and (see Figure 2.2). Arcs joining and and joining and as well as all arcs that join and ) will be called the innermost upper cups. More generally, for a Catalan state , we will refer to arcs with both ends in or with one end in and the other in either or as upper cups.
Let be a Catalan state of . We say that has no returns on the floor, if none of its arcs have both ends in . For such a state, there are precisely arcs with none of its endpoints in . Consider the set of Kauffman states of which realize . Each Kauffman state can be identified with an matrix , where . Denote by the subset of consisting of all states with rows in the form:
where (see Figure 2.2). In the next proposition, we show that in can be computed using only Kauffman states with . In particular, this significantly reduces the number of Kauffman states that one needs to consider555To compute we need to consider Kauffman states instead of ..
The projection of onto the submodule of generated by is given by
Proof. We compute starting from the top row of crossings to the bottom (row by row) using the Kauffman bracket skein relation and omitting states with returns on the floor (observe that we will never get trivial components). When doing so, we should not take into the account Kauffman states with change of markers from to in a row (see Figure 2.4) as they result in lower caps after the regular isotopy of diagrams (see Figure 2.4).
Therefore, all Catalan states with no returns on the floor can be obtained using only Kauffman states . In particular, we have
Hence, the coefficient of the Catalan state can be computed using the formula:
where consists of all Kauffman states representing (i.e. ).
Let be the set of all sequences , where , and . The sets and are in bijection, so denote by the sequence corresponding to ; analogously, by , we denote the Kauffman state corresponding to . Given , we denote by the Catalan state obtained from (i.e. ).
For every , there is a sequence such that .
Proof. Every realizable Catalan state (not necessarily in ) has
at least one innermost upper cup (see Figure 2.5).
Otherwise, as shown in Figure 2.6, has no upper cups
below the dotted line which also cuts in points. Hence, it follows that is non-realizable666As we have shown in  (see Lemma and Theorem ), a Catalan
state is realizable if and only if
every vertical line cuts at most times and every horizontal line
cuts at most times.. The statement of Proposition 2.2 follows by induction on .
If there are Catalan states with no returns on the floor and a single innermost upper cup (see Figure 2.7), so Proposition 2.2 holds with .
Let and . Assume that the statement holds for all numbers smaller than . As we noted before, has an innermost upper cup and we can deform the diagram of to that in Figure 2.5. Consider the Catalan state shown in the bottom of Figure 2.5. Since is in , then by the inductive hypothesis, there is a sequence that realizes it. We conclude that realizes which finishes our proof.
It is clear (from our proof of Proposition 2.2) that for , there might be several sequences777The Catalan state in has exactely one representative iff with , for some , or , , for for some . with . Therefore, for , we let
We note that, for , there is a bijection between the set of all Kauffman states representing and the set .
Let be a sequence with for some (). Let an operation defined by
and be its inverse. Sequences , are called -equivalent if and differ by a finite number of operations.
If and is -equivalent to then
Proof. It suffices to show that if and , then . This is evident from Figure 2.8
since operations correspond (geometrically) to a change of order in which arcs of (with no ends on the floor) are realized.
Using operations, we define a poset structure on as follows: covers (i.e. ) iff there is , such that . Let be the transitive closure of . Clearly, is a poset. Hence is also a poset.
Let and be the directed graph with vertices and directed edges
Let be the graph obtained from by ignoring the edge orientations. Then is a connected graph.
Proof. Connectedness of the graph follows by induction on . For , the statement is obvious since has only one element. Assume that for all , the graph is connected and let , . If then sequences , , where is obtained from by removing its top corresponding to (). By inductive assumption, the graph is connected, so and are -equivalent and consequently, and are also -equivalent. Therefore, there is a path in joining vertices and . WLOG, we can assume that (in fact because and are the innermost cups). In the sequence that represents there is corresponding to the innermost upper cup of . By inductive assumption one changes to using operations, where represents the innermost upper cup in . Now, operation changes the sequence to . Using the same argument as in the first case (i.e. ), we see that and are in the same connected component of . It follows that the graph is connected.
We observe that, if then , hence the directed graph has no directed cycles, i.e. is a Hasse diagram of the poset .
Let denote the weight of sequence . Define , to be a minimal and a maximal sequence representing in the lexicographic order on 888Recall, iff there is , such that for and . respectively.
Let and , be defined as the above. Then , are unique elements having the smallest and the largest weight, respectively.
Proof. Let and be another sequence representing (if it exists). We show that by induction on . For , the statement holds since has exactly one element. Assume that the statement is valid for all numbers smaller than . If we can compare shorter sequences and . Using induction assumption we conclude that the first sequence has the larger weight, i.e. .
Suppose is smaller than . There exists
() which represents the
innermost upper cup in . We consider the following cases:
. Changing order of cups and in , where , results in a sequence , with the weight larger by .
There is different than , for some . Let be the smallest such index (so for ). Notice that the arc giving the innermost upper cup in on the level has index .
There are two possibilities:
If , then the sequence represents the Catalan state with two innermost cups and , where is to the right of . By inductive assumption, the sequence does not have the maximal weight, as it is not maximal in the lexicographical order (i.e. ).
If then represents the innermost upper cup in which is to the right of . This, however, contradicts the maximality of in the lexicographic order.
Therefore, is a unique element with the maximal weight in . A proof for is similar.
Figure 2.9 diagram on the right shows the Hasse diagram associated to the Catalan state see the diagram on the left, and the maximal sequence that realizes see the middle.
From the definition of it directly follows that
For the Catalan state in Figure 2.9, in particular, we have:
3 Coefficient and Plucking Polynomial
In this section we explore the relationship between coefficients of Catalan states and the plucking polynomial of associated rooted trees. We would like to stress the fact that the definition of the plucking polynomial was strongly motivated by .999In fact, the plucking polynomial for rooted trees was discovered just after the paper  was finished.
3.1 Rooted Tree with Delay Function Associated to
Let be a Catalan state in . We define, in a standard way, a planar tree by taking the dual graph to . That is, the set of arcs of splits the rectangle into bounded regions . For each region we have a vertex and, two vertices are adjacent in iff their corresponding regions share boundary in common (see Figure 3.1).
For a Catalan state with no returns on the floor (), we use a modified version of the tree described above. Let be the set of arcs of with an end on the floor of , and be the set of arcs of . Denote by the dual graph to and observe that is an embedded planar tree with vertices (corresponding to regions of ) and edges (corresponding to arcs101010Each edge is dual to a unique arc . of ). There is an obvious choice for the root of , i.e. is the vertex assigned to the regions containing (as a part of its boundary) the floor of (see right of Figure 3.2).
For a vertex , let be the number of vertices adjacent to (degree of ). A vertex of degree one () is called a leaf. Denote by the set of all leaves of . Let be defined by setting to be , if has both ends in ; and to be the maximal index of the end point or of the arc , otherwise. Define the delay function by putting , where corresponds to the edge incident to . Let be the rooted tree with a delay function associated to (see Figure 3.4). We note that there might be several different Catalan states with the same (compare Figure 3.4 and Figure 3.4).
3.2 Plucking Polynomial of Rooted Trees with Delay Function
We recall, after [13, 14], the definition of the plucking polynomial of a plane rooted tree with a delay function from leaves to positive integers. Let be a plane rooted tree (we assume that our trees are growing upwards, see Figure 3.5). For consider the unique path from to , and let be the number of vertices of to the right of the path.
Let be a plane rooted tree with the root and delay function . Denote by the set of all leaves of with . The plucking polynomial of is a polynomial in variable defined as follows: If has no edges, we put ; otherwise
where if is a leaf of , and if is a new leaf of .
Clearly, if and, we note that this is never the case when is a tree with the delay function associated to a Catalan state.
Our next result establishes a relationship between the coefficient () and the plucking polynomial of a rooted tree associated to .
Let be a Catalan state with no returns on the floor, be the minimum degree of in and its evaluation at . Then the coefficient of in is given by
In particular, if has returns only on its ceiling or the left side, then .
Proof. The above statement holds since, in the definition of , we just follow computations of outlined in Proposition 2.1 (using the ”row by row” approach). In particular, we observe that deleting from the sequence (i.e. assigning markers according to in the first row of ) results in removing a leaf which corresponds to the innermost upper cup of . Coefficients of the Laurant polynomial can then be found by comparing it with . That is, we observe that if a move is applied on , the new sequence adds a new monomial to obtained by multiplying the monomial by a factor of . Since each sequence yields a unique order of removing vertices from , each contributes a monomial to . Therefore, a move on results in adding a monomial multiplied by to . We describe this in detail for a move on . By definition, the move changes to , hence the monomial corresponding to changes to . Now we observe that, a move induces a change of order of leaves which results in decreasing by (i.e. the corresponding monomial is multiplied by ). Since the graph is connected, we conclude that
for some that depends only on . Now, to find , it suffices to compare the maximal power of with the minimal power of the variable in ( i.e. ). The formula given in the theorem follows and, in particular, if has only returns on its top or the left side then , so .
Let be a Catalan state with no returns on the floor and
be its coefficient in , where . Then and , for all .
4 Coefficients of Catalan States with Returns on One Side
The recursion for given in
Definition 3.1 can be used to prove many important
results about . In particular, we can find a
closed form formula for coefficients of Catalan
states with returns on the ceiling only. Furthermore, for such states
the polynomial depends only on
the tree rather than its
particular planar embedding (see Theorem 4.1). To simplify
our notations, let stand for the rooted tree . We recall also the notation of the -analogue of an integer and
-analogue of a multinomial coefficient. The -analogue of an
integer is defined by
and let . Therefore, the -analogue of the multinomial coefficient is given by
In particular, the binomial coefficient (Gauss polynomial) is defined by
Let be the wedge product of rooted trees and having the common root . Then,
Wedge Product Formula If is the wedge product of rooted trees which have the common root , then
Product Formula Let and denote by the rooted subtree of with the root . Assume that , where are rooted trees with the common root , and let