Catalan States of Lattice Crossing
CCatalan States of Lattice Crossing
Mieczyslaw K. Dabkowski Changsong Li Jozef H. Przytycki
Abstract
For a Lattice crossing L ( m, n ) we show which Catalan connection between 2 ( m + n ) points onboundary of m × n rectangle P can be realized as a Kauffman state and we give an explicit formula forthe number of such Catalan connections. For the case of a Catalan connection with no arc startingand ending on the same side of the tangle, we find a closed formula for its coefficient in the RelativeKauffman Bracket Skein Module of P × I . Keywords:
Knot, Link, Kauffman Bracket, Catalan TanglesMathematics Subject Classification 2000: Primary 57M99; Secondary 55N, 20D
We consider Relative Kauffman Bracket Skein Module ( RKBSM ) of P × I, where P is m × n parallel-ogram with 2 ( m + n ) points on the boundary arranged as it is shown in Figure 1 . , and ( m + n )-tangle L ( m, n ) shown in Figure 1 . m × n -lattice crossing . If Cat m,n denotes the set ofall Catalan states for P (crossingless connections between boundary points) then L ( m, n ) in RKBSM of P × I can be uniquely written in the form L ( m, n ) = (cid:80) C ∈ Cat m,n r ( C ) C, where r ( C ) ∈ Z (cid:2) A ± (cid:3) . n pointsn points m po i n t s m po i n t s Figure 1.1 n-1 red lines m - r e d li n e s l l l m-2h l m-1h l l l n-2v l n-1v Figure 1.2 +1 marker -1 marker
Figure 1.3Let S m,n be the set of all Kauffman states (choices of positive or negative markers for all crossings asshown in Figure 1.3), after applying skein relations we have in RKBSM of P × I : L ( m, n ) = (cid:88) s ∈S m.n A p ( s ) − n ( s ) ( − A − A − ) | s | K ( s ) RKBSM was defined in [6] and it was noted there that,
RKBSM for D × I with 2 k points on the boundary is a freemodule with the basis consisting of Catalan connections. For an extensive discussions of theory of Kauffman bracket skeinmodules see [7]. a r X i v : . [ m a t h . G T ] S e p here, following standard conventions, for each Kauffman state s, p ( s ) (resp. n ( s )) denotes the numberof crossings with positive (resp. negative) marker, | s | the number of trivial components of the diagramobtained from L ( m, n ) by smoothing crossings according to markers defined by s, and K ( s ) denotes thediagram corresponding to state s after removing trivial components. In RKBSM of P × I, we have (cid:88) C ∈ Cat m,n r ( C ) C = (cid:88) s ∈S m.n A p ( s ) − n ( s ) ( − A − A − ) | s | K ( s ) , so, for each Catalan state C ∈ Cat m,n define R ( C ) = { s ∈ S m.n | K ( s ) ≈ C } , where ≈ denotes isotopy of diagrams modulo boundary, thus r ( C ) = (cid:26) if R ( C ) = ∅ (cid:80) s ∈ R ( C ) A p ( s ) − n ( s ) ( − A − A − ) | s | if R ( C ) (cid:54) = ∅ . We call C a realizable Catalan state if R ( C ) (cid:54) = ∅ and denote by Cat
Rm,n the set of all realizable Cata-lan states. Analogously, we call C a forbidden Catalan state if R ( C ) = ∅ and denote by F m,n = Cat m,n \ Cat
Rm,n the set of all forbidden Catalan states . In this paper we will consider the following prob-lems: Which Catalan states C are realizable as Kauffman states of L ( m, n )? What is an explicit formula for r ( C )? What is the number of elements of
Cat
Rm,n ?In the second section we show which Catalan states C can be realized by Kauffman states of L ( m, n ).In the third section, for a Catalan state C with no arc that starts and ends on the same “ side” of L ( m, n ),we give an explicit formula for the corresponding coefficient r ( C ). Finally, in the forth section, we obtainan explicit formula for the number of realizable Catalan states. L ( m, n ) Enumerating horizontal and vertical lines as in Figure 1 . , let l h , l h , ..., l hm − be horizontal lines and l v ,l v , ..., l vn − be vertical lines . For a Catalan state C ∈ Cat m,n , we let d h ( C ) = max (cid:8) (cid:0) l hi ∩ C (cid:1) | i = 1 , , ..., m − (cid:9) and d v ( C ) = max { l vi ∩ C ) | i = 1 , , ..., n − } where l αi ∩ C ) is the minimal number of intersections of l αi and C for α = h or α = v. Lemma 2.1
Let C ∈ Cat m,n and assume that d h ( C ) > n or d v ( C ) > m then C ∈ F m,n . Proof.
Observe that for any Kauffman state s of L ( m, n ) , corresponding diagram L s ( m, n ) obtained bysmoothing crossings according to choice of markers defined by s cuts horizontal and vertical separatinglines precisely n times and m times. Hence for the Catalan state C ( s ) obtained from L s ( m, n ) we musthave d h ( C ) ≤ n and d v ( C ) ≤ m .We say that a Catalan state C satisfies horizontal-line condition, H ( m, n ) if d h ( C ) ≤ n and, corre-spondingly, C satisfies vertical-line condition, V ( m, n ) if d v ( C ) ≤ m. We show that if C ∈ Cat m,n satisfiesboth conditions H ( m, n ) and V ( m, n ) then C is a realizable Catalan state. The following lemma givessufficient conditions for a Catalan state C ∈ Cat m,n to meet condition V ( m, n ) or/and H ( m, n ). Lemma 2.2
For Catalan state C, (1) if C has no returns on the top horizontal boundary then C satisfies condition V ( m, n ) ;22) if C, in addition, has no returns on the right side, then C satisfies both H ( m, n ) and V ( m, n ) . Proof.
For (1) , we argue by case distinction. Let x , ..., x n be boundary points on top of C , and (cid:96) be aseparating line between x k and x k +1 (see Figure 2 . x k + i th connection cuts (cid:96) for some 1 ≤ i ≤ n − k, then all x , ..., x k connections do not cut (cid:96) . Sincethere are m + 2 k points on the right of (cid:96) and among them at least 2 k points yield connections in C without cutting (cid:96) , it follows that there are at most m points on the right of (cid:96) which can form connectionswith points on the left of (cid:96) . Using similar reasoning, if 1 ≤ i ≤ k then at most m points on the left of (cid:96) can form connections with points on the right of (cid:96) . Therefore, statement (1) follows from basic counting.If each x i connection does not cut (cid:96) for all i = 1 , , ..., n, we observe that, in particular, x , ..., x k connections do not cut (cid:96). Thus, as in the previous case, we argue that at most m points on the left of (cid:96) can be connected with points on the right of (cid:96) .Statement (2) , follows directly from (1) , since (1) can be applied to any side of the rectangle. x n x k+i x k+1 x k x Figure 2.1 y y m y m +1 y m Figure 2.2For two Catalan states C ∈ Cat m ,n and C ∈ Cat m ,n with no returns on the bottom of the firstand the top for the second one defines an operation ∗ v of vertical composition of Catalan states (seeFigure 2 . ∗ v allows us to define Catalan state C = C ∗ v C , for which the number ofintersections with its vertical lines can easily be counted provided that the number of intersections withvertical lines of the factor states is known. Analogously, one defines horizontal composition ∗ h of Catalanstates. Lemma 2.3
Let C ∈ Cat m ,n and C ∈ Cat m ,n and assume that bottom of C and top of C are noreturn sides respectively ( see Figure . . Then the vertical composition C = C ∗ v C of states C and C is a Catalan state with the following properties :(1) Every vertical line (cid:96) intersects C at most m = m + m times and every horizontal line (cid:96) (cid:48) either in C or C cuts C the same number of times as in C or C respectively . (2) Catalan states C and C satisfy conditions H ( m i , n ) and V ( m i , n ) , for i = 1 , , respectively if andonly if C satisfies conditions H ( m, n ) and V ( m, n ) . Proof.
Since C and C have no return sides on the bottom and the top respectively, vertical composition C = C ∗ v C of states C and C (as it is shown in Figure 2 .
2) is a Catalan state. Furthermore, if (cid:96) is avertical line in C, then from Lemma 2.2(1) it follows that the number of times that (cid:96) cuts C is boundedabove by the number m + m = m . Finally, if (cid:96) (cid:48) is a horizontal line either in C or C then (cid:96) (cid:48) is also aseparating line in the composed state C and, since C and C are composed along sides with no returns,the number of times (cid:96) (cid:48) cuts C is identical as the number of times (cid:96) (cid:48) cuts either C or C . This finishesour proof of (1) . Statement (2) follows from (1). 3or Catalan states C satisfying conditions H ( m, n ) and V ( m, n ) , we apply induction on m + n toprove that such states could be obtained as Kauffman states of L ( m, n ) . An important step in ourinductive proof is an observation that Catalan states with d h ( C ) = n (or d v ( C ) = m ) could be obtainedas a vertical (horizontal) composition of Catalan states. Lemma 2.4 (Split Lemma)
Let C ∈ Cat m,n and assume that C satisfies conditions H ( m, n ) and V ( m, n ) , and d h ( C ) = n . Then (1) C = C ∗ v C , where C ∈ Cat m ,n , C ∈ Cat m ,n , m + m = m, and (2) Catalan states C and C satisfies conditions H ( m i , n ) and V ( m i , n ) , for i = 1 , , respectively. Proof.
Since d h ( C ) = n , there is a horizontal separating line (cid:96) that cuts C precisely n times. Splitting C along (cid:96) yields a pair of Catalan states C ∈ Cat m ,n and C ∈ Cat m ,n , where m + m = m. Since n is the number of intersections of (cid:96) with Catalan state C, it follows that C has the bottom side with noreturns and C has the top side with no returns. Therefore, state C can be expressed as C = C ∗ v C .Since C satisfies conditions H ( m, n ) and V ( m, n ) , it follows from Lemma 2.3(2) that each C i satisfiesconditions H ( m i , n ) and V ( m i , n ), for i = 1 , , respectively.Obviously, the statement of Split Lemma holds when d v ( C ) = m . In such a case, C = C ∗ h C , where C ∈ Cat m,n , C ∈ Cat m,n , n + n = n, and each C i satisfies conditions H ( m, n i ) and V ( m, n i ),for i = 1 , , respectively. Now, we state and prove the main result for this section. Theorem 2.5
Assume that C satisfies conditions H ( m, n ) and V ( m, n ) , then C can be obtained as aKauffman state of L ( m, n ) . Proof.
We proceed by induction on m + n . Clearly, statement is vacuously true for m = n = 0, and it isobvious when m = n = 1. We assume that statement is true for all pairs ( m, n ) such that m, n > m + n ≤ k, where k ≥ . We show that statement is true for m, n > m + n = k + 1 and WLOGwe can assume m ≥ C ∈ Cat m,n and C satisfies conditions H ( m, n ) and V ( m, n ). We have the following cases d h ( C ) = n or d v ( C ) = m If d h ( C ) = n, then applying Split Lemma, we have C = C ∗ v C , where C ∈ Cat m ,n , C ∈ Cat m ,n ( m + m = m ) , and each C i satisfies conditions H ( m i , n ) and V ( m i , n ), for i = 1 ,
2. Since m i + n ≤ k ,applying induction hypothesis, we obtain corresponding Kauffman states s and s for L ( m , n ) and L ( m , n ) respectively such that C ≈ K ( s ) and C ≈ K ( s ). Let s be Kauffman state obtained byputting markers according to s in the first m rows of L ( m, n ) and according to s in the remaining m − m = m rows of L ( m, n ). Clearly, for Kauffman state s, we have C ≈ K ( s ).Proof in the case when d v ( C ) = m is analogous. d h ( C ) < n and d v ( C ) < m, that is, each horizontal line cuts C less than n times and each verticalline cuts C less than m times.In this case we will construct a Kauffman state s such that C ≈ K ( s ) using an algorithm given inthe following lemma. Lemma 2.6
Assume that d h ( C ) < n and d v ( C ) < m , then C can be obtained as a Kauffman state of L ( m, n ) . Proof.
By assumption, each horizontal line cuts C less than n times and each vertical line cuts C lessthan m times, so the top boundary side of C must have at least one return (see Figure 2 . . C has no corner connection, we start constructionby defining Kauffman state s (first row of desired Kauffman state s ) for the returning connection of C r Figure 2.3 or Figure 2.4between pair of points x i and x i +1 on the top of C with minimal index i , i = 1 , , ..., n − . s (first row of Kauffmanstate s ) the state shown in Figure 2 . C into an equivalent state (cid:101) C with top as shown in Figure 2 . m −
1) + n )-Catalan state . Since every vertical line cuts C lessthan m times, we observe that it must cut each vertical line no more then m − m −
1) + n )-Catalan state C (cid:48) ∈ Cat m − ,n obtained from (cid:101) C by removing its top (see Figure 2 .
4) cuts allvertical separating lines no more than m − C (cid:48) satisfies conditions H ( m − , n ) and V ( m − , n ) and one may apply induction hypothesis for the state C (cid:48) . Therefore, thereis a Kauffman state s (cid:48) of L ( m − , n ) such that K ( s (cid:48) ) = C (cid:48) . In order to complete our construction, wetake for s the state obtained by putting markers according to state s in the first row of L ( m, n ) andputting markers according to s (cid:48) in its remaining ( m −
1) rows. For the choice of markers determined by s , we clearly have K ( s ) ≈ C. Proof of Lemma 2.6 also completes our proof of Theorem 2.5.
RK BSM
In this section we prove result concerning coefficients of realizable Catalan states for L ( m, n ) for the caseof Catalan states with no returns. This result was proved in 1992 by J. Hoste and J. H. Przytycki whilethey were working on the Kauffman bracket skein module of the Whitehead type manifolds. Theorem 3.1 In RKBSM of D × I with m + n ) points we have L ( m, n ) = m (cid:88) k =0 P k,m − k,h T k, m − k, h + Remainder, where P k,m − k,h = A nm − k + h ) k (cid:2) mk (cid:3) A − ; T k,m − k,h is the Catalan state with k positive diagonal arcs, m − k ) negative diagonal arcs and h = n − m ≥ vertical arcs ( see Figure . and the remainder is a polynomial ( over Z (cid:2) A ± (cid:3) ) in variables consisting of realizable Catalan states that have at least onereturn. Authors of [3] didn’t publish this result. We were informed by Mustafa Hajij that a related formula was noted by S.Yamada [9],[2]. It is worth to observe that if we decorate lattice crossing L ( m, n ) by Jones-Wenzl idempotents f m and f n (respectively)then reminder is 0 since f k e i = 0 = e i f k for any i and k . points m po i n t s km-k n-m m-kk Figure 3.1Before we give a proof of Theorem 3.1, we need to recall the notation q -analogue of an integer m and q -analogue of binomial coefficient . A q -analogue of an integer m is defined as follows [ m ] q = 1 + q + q + ... + q m − , and if [ m ] q ! = [1] q · [2] q · ... · [ m ] q then q -analogue of binomial coefficient is defined by (cid:20) mk (cid:21) q = [ m ] q ![ k ] q ! · [ m − k ] q One shows using the following identity for q -analogue of binomial coefficient (cid:20) mk (cid:21) q = (cid:20) m − k (cid:21) q + q m − k (cid:20) m − k − (cid:21) q and induction that (cid:2) mk (cid:3) q ∈ Z [ q ]. Proof.
Applying Kauffman bracket skein relation (see Figure 1 .
3) for the crossing in top left corner of L ( m, n ) (see Figure 3 .
2) we obtain the following recursive formula P k,m − k,h = A ( m + n ) − P k,m − k − ,h + A − ( m + n ) P k − ,m − k,h : L(m,n) = m-1 n-1 = A + A -1 = A + A -1 + A = A m+n-1 + Remainder m-1 n-1 m-1 n-1n-1m-1m-1 n-1m-1 n-1 m-1 n-1 Figure 3.2Now, we let P m, ,h = A − mn and P ,m,h = A mn . One verifies that P k,m − k,h = A mn − k + h ) k (cid:2) mk (cid:3) A − satisfies the above recurrence . Namely, we have by the property of q -analogue of binomial coefficient For the initial conditions we check: P m, ,h = A mn − m + h ) m = A mn − m + n − m ) m = A − mn , and P ,m,h = A mn − = A mn . q = A − : P k,m − k,h = A mn − k + h ) k (cid:20) mk (cid:21) A − = A mn − k + h ) k (cid:20) m − k (cid:21) A − + A mn − k + h ) k A − m − k ) (cid:20) m − k − (cid:21) A − = A ( m + n ) − A − m − n + mn − k + h ) k (cid:20) m − k (cid:21) A − + A − ( m + n ) A m + n − mn − k + h ) k − m − k ) (cid:20) m − k − (cid:21) A − = A ( m + n ) − A ( m − n − − k + h ) k (cid:20) m − k (cid:21) A − + A − ( m + n ) A ( m − n − k − h )( k − (cid:20) m − k − (cid:21) A − = A ( m + n ) − P k,m − k − ,h + A − ( m + n ) P k − ,m − k,h as needed. Remark 3.2
One of the goals for future research is to find a ” relatively good ” formula for coefficientsof all realizable Catalan states of L ( m, n ) in RKBSM of D × I with m + n ) points. In particular,Theorem 3.1 gives an explicit formula for all coefficients of Catalan states with no returns. A naturalgeneralization of this family Catalan states would be families of Catalan states with three, two, or just oneside that has no return. As we proved in Lemma 2.2 even if only one side ( left or right ) of the Catalanstate has no return then any horizontal separating line can cut such a Catalan state in at most n points. L ( m, n ) In this section, we find a closed formula for (cid:12)(cid:12)(cid:12)
Cat
Rm,n (cid:12)(cid:12)(cid:12) – the number of Catalan states that could beobtained as Kauffman states of L ( m, n ) , or alternatively, we find number of elements of the set F m,n = Cat m,n \ Cat
Rm,n of all
Forbidden Catalan states.
As we showed in Theorem 2.5, a Catalan state C ∈ Cat m,n is realizable if and only if C satisfies conditions H ( m, n ) and V ( m, n ) . Therefore, every forbidden state C must intersects at least one of the horizontal or vertical lines more than n or m times respectively.Let T h ( m, n ) and T v ( m, n ) denote sets of Catalan states that are excluded by horizontal and verticalcondition respectively. More precisely, we have for the lines in Figure 1 . T h ( m, n ) = (cid:8) C ∈ Cat m,n | d h ( C ) > n (cid:9) and T v ( m, n ) = { C ∈ Cat m,n | d v ( C ) > m } We call them respectively horizontally excluded and vertically excluded Catalan states . Clearly, we seethat T h ( m, n ) ∪ T v ( m, n ) = F m,n . However, more is true, i.e. T h ( m, n ) ∩ T v ( m, n ) = ∅ , which is a consequence of the following proposition. Proposition 4.1
Let S be a circle with an even number of points a + a + a + a and consider a pairof lines intersecting at one point and splitting S into sectors A i with a i points each sectors, i = 1 , , , see Figure . . Let C be a Catalan state and m ( C ) be the number of intersections of C with the pairof lines. (i) If points from sectors A and A are not connected in C, then m ( C ) ≤ a + a . (ii) If points from sectors A and A are not connected in C, then m ( C ) ≤ a + a . (iii) For any
C, m ( C ) ≤ min { a + a , a + a } . Proof.
We observe that in a Catalan state C we cannot have points from A connected with points from A , and points from A connected with points from A at the same time. For ( i ) , we notice that sincepoints from sectors A and A are not connected in C, if u is the number of connections in C betweenpoints from sectors A and A then these connections intersect pair of lines at 2 u points and involve 2 u A and A in total. Consequently, there is a total of a + a − u points left in sectors A and A that may form connections intersecting the pair of lines at most once each. It follows thattotal number of intersections of Catalan state C with the pair of lines is bounded above by a + a .Proof of ( ii ) follows by changing the roles of A , A and A , A , and statement ( iii ) is a directconsequence of ( i ) and ( ii ). a a a a Figure 4.1Proposition 4.1 has the following important consequence.
Corollary 4.2
Horizontally and vertically excluded Catalan states are disjoint.
Proof.
For a pair of lines consisting of a vertical and a horizontal line as in Figure 1 . , we see that a + a = a + a = m + n. Applying result of Proposition 4.1 for such a pair of lines, for any Catalanstate C, we have m ( C ) ≤ m + n. Therefore, if the horizontal line intersects C more then n times thenthe vertical line must intersect C less than m times. Analogously, if the vertical line intersects C morethen m times then the horizontal line must intersect C less than n times.As a consequence of Corollary 4.2 and Theorem 2.5, we obtain the following result. Corollary 4.3
Let T h ( m, n ) and T v ( m, n ) be the number of horizontally and vertically excluded Catalanstates. The number of Forbidden Catalan states equals T h ( m, n ) + T v ( m, n ) . It is worth to note that following relation between numbers T h ( m, n ) and T v ( m, n ) holds T h ( m, n ) = T v ( n, m ) . Consequently, from now on, we will consider only T h ( m, n ).In order to find T h ( m, n ) , we start by constructing bijection between the set Cat m,n and the set ofDyck paths D m,n of semilength m + n from (0 ,
0) to ( m + n, m + n ) that never pass below the line y = x and consists only the vertical steps V = (0 ,
1) ( ascents ) and horizontal steps H = (1 ,
0) ( descents ) as itis shown in Figure 4 . m + n )+1 lines l , l , ..., l m + n ) as in Figure 4 .
3. Any Catalan state C ∈ Cat m,n uniquelydetermines the minimal number of intersections n j ( C ) = l j ∩ C of C with each line l j , ≤ j ≤ m + n )and let r j ( C ) = n j ( C ) − n j − ( C ) , j = 1 , , ..., m + n ) . Lemma 4.4
For all j = 1 , , ..., m + n ) , we have r j ( C ) = ± . x(0,0) (m+n,m+n)x=y+1 Figure 4.2 l l l j l j+1 l j+2 l l k l k+1 l k+2 Figure 4.3 kk-2
Figure 4.4a) kk Figure 4.4b) kk Figure 4.4c) kk+2
Figure 4.4d)
Proof.
We observe that for any three successive lines l j , l j +1 , and l j +2 , where j = 1 , , ..., ( m + n ) − . a ) − d )) for the intersections of C with lines l j , l j +1 , and l j +2 :For instance, in the second case shown in Figure 4 . b ), we have n j ( C ) = k, n j +1 ( C ) = k + 1 , and n j +2 ( C ) = k, thus r j +1 ( C ) = 1 , and r j +2 ( C ) = − . Analogously, for the remaining cases shown in Figure 4 . a ) , Figure 4 . c ) , and Figure 4 . d ) . Therefore,for all j = 3 , , ..., m + n ) − , we have r j ( C ) = ± l , l , l , and l m + n − , l m + n ) − , l m + n ) , we see that n ( C ) = n m + n ) ( C ) = 0 , n ( C ) = n m + n ) − ( C ) = 1 , and n ( C ) = 0 , n m + n − ( C ) = 0 , . Therefore, we have r ( C ) = 1 , r ( C ) = − r ( C ) = 1 and analogously r m + n ) − ( C ) = 1 or r m + n ) − ( C ) = − , r m + n ) ( C ) = −
1. Hence, we have r j ( C ) = ± , for all j = 1 , , ..., m + n ) as weclaimed.Now, let C ∈ Cat m,n and let a j ( C ) = (cid:26) H if r j ( C ) = − V if r j ( C ) = 1 , j = 1 , , ..., m + n )We define a lattice path p ( C ) by putting p ( C ) = a ( C ) a ( C ) ...a m + n ) ( C ) . C ∈ Cat m,n , lattice path p ( C ) of semilength m + n from (0 ,
0) to ( m + n, m + n ) neverpass below the line y = x, thus p ( C ) is a Dyck path. Furthermore, we have the following result. Proposition 4.5
The map ψ : Cat m,n → D m,n given by ψ ( C ) = p ( C ) is a bijection . Proof.
It suffices to show that ψ : Cat m,n → D m,n is injective. Let C , C ∈ Cat m + n and suppose that ψ ( C ) = p ( C ) = p ( C ) = ψ ( C ) . therefore, for every j = 1 , , ..., m + n ) , a j ( C ) = a j ( C ) , that is r j ( C ) = n j ( C ) − n j − ( C ) = n j ( C ) − n j − ( C ) = r j ( C )Since n ( C ) = n ( C ) , and r ( C ) = r ( C ) , we have n ( C ) = n ( C ) . By induction, we have n j ( C ) = n j ( C ) , j = 1 , , ..., m + n ) . It follows that C and C have identical minimal number of intersections with all lines l , l , ..., l m + n ) , hence C = C .Following the notations used in [5] (see pp. 9 and Lemma 4 A, pp. 12), let L ( m, n ; t ) denote the set ofpaths from (0 ,
0) to ( m, n ) not touching the line y = x − t, where t > , and let L ( m, n ; t, s ) denotethe set of paths from (0 ,
0) to ( m, n ) not touching the lines y = x − t and y = x + s, where t, s > . As it was shown in [5] , we have | L ( m, n ; t ) | = (cid:18) m + nn (cid:19) − (cid:18) m + nm − t (cid:19) , and | L ( m, n ; t, s ) | = (cid:88) k ∈ Z (cid:34)(cid:18) m + nm − k ( t + s ) (cid:19) + − (cid:18) m + nm + k ( t + s ) + t (cid:19) + (cid:35) , where (cid:18) yz (cid:19) + = (cid:26) (cid:0) yz (cid:1) if y ≥ z ≥ if y < y < z Theorem 4.6
The number T h ( m, n ) of horizontally excluded Catalan states ( see also Figure . isgiven by T h ( m, n ) = ∞ (cid:88) i =0 (cid:20)(cid:18) m + 2 nm − i ( n + 3) − (cid:19) − (cid:18) m + 2 nm − i ( n + 3) − (cid:19) + (cid:18) m + 2 nm − i ( n + 3) − (cid:19)(cid:21) Before we prove Theorem 4.6, we note that if i = 2 j, lines l i , l i +1 and l i +2 may intersect a Catalanstate C satisfying horizontal line condition H ( m, n ) in the pattern: n i ( C ) = n, n i +1 ( C ) = n + 1 and n i +2 ( C ) = n (see Figure 4 . . Recall, for a lattice path p = a a ...a k that starts from (0 , H j ( p ) = { a i | a i = H, i = 1 , , ..., j } - number of horizontal steps V j ( p ) = { a i | a i = V, i = 1 , , ..., j } - number of vertical steps Solution to the restricted ballot problem with two boundary conditions, including very interesting historical commentof Kelvin and Maxwell’s method of images, is given in the classic book by W.Feller [1]; see also [4], [8]. n n+1 Figure 4.5 y x(0,0) (m+n,m+n)x=y+1x=y-(n+2)
Figure 4.6where 1 ≤ j ≤ k, and S j ( p ) = V j ( p ) − H j ( p ) , ≤ j ≤ k. Obviously, path p ( C ) , for any j = 1 , , ..., m + n ) , we have: S j ( p ( C )) = V j ( p ) − H j ( p ) = j (cid:88) i =1 r i ( C ) = n j ( C ) ≥ . Proof.
Using bijection ψ : Cat m,n → D m,n from Proposition 4.5, we have: T h ( m, n ) = (cid:12)(cid:12) T h ( m, n ) (cid:12)(cid:12) = |{ C ∈ Cat m,n | S j ( p ( C )) > n + 1 , for some 0 ≤ j ≤ m + n ) }| = | D m,n | − |{ p ∈ D m,n | ≤ S j ( p ) ≤ n + 1 , for all 0 ≤ j ≤ m + n ) }| Since | D m,n | = | L ( m + n, m + n ; 1) | and |{ p ∈ D m,n | ≤ S j ( p ) ≤ n + 1 , for all 0 ≤ j ≤ m + n ) }| = | L ( m + n, m + n ; 1 , n + 2) | After standard algebraic simplifications, we have T h ( m, n ) = | L ( m + n, m + n ; 1) | − | L ( m + n, m + n ; 1 , n + 2) | = ∞ (cid:88) i =0 (cid:20)(cid:18) m + 2 nm − i ( n + 3) − (cid:19) − (cid:18) m + 2 nm − i ( n + 3) − (cid:19) + (cid:18) m + 2 nm − i ( n + 3) − (cid:19)(cid:21) . This finishes our proof.
Corollary 4.7
The number of all realizable Catalan states by m × n lattice of crossings T ( m,n ) is givenby (cid:12)(cid:12)(cid:12) Cat
Rm,n (cid:12)(cid:12)(cid:12) = 1 m + n + 1 (cid:18) m + n ) m + n (cid:19) − T h ( m, n ) − T h ( n, m ) . In particular, for m = n, (cid:12)(cid:12)(cid:12) Cat
Rn,n (cid:12)(cid:12)(cid:12) = 12 n + 1 (cid:18) n n (cid:19) − (cid:18)(cid:18) nn − (cid:19) − (cid:18) nn − (cid:19) + (cid:18) nn − (cid:19)(cid:19) Acknowledgments
J. H. Przytycki was partially supported by the NSA-AMS 091111 grant, and by the GWU REF grant. Au-thors would like to thank Ivan Dynnikov and Krzysztof Putyra for many useful computations/discussions.
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Bull. Ac. Pol. : Math ., (1-2), 1991, pp. 91-100.E-print: arXiv:math/0611797 [math.GT] [7] J. H. Przytycki, Fundamentals of Kauffman bracket skein modules, Kobe Math. J . , (1), 1999, pp.45-66. E-print: arXiv:math/9809113 [math.GT] [8] L. Tak´acs, Combinatorial Methods in the Theory of Stochastic Processes , Wiley Series in Probabilityand Mathematical Statistics, 1967.[9] S. Yamada, An invariant of spatial graphs,
J. Graph Theory (5), 1989, pp. 537-551.Mieczyslaw K. Dabkowski Changsong LiDepartment of Mathematical Sciences Department of Mathematical SciencesUniversity of Texas at Dallas University of Texas at DallasRichardson TX, 75080 Richardson TX, 75080 e-mail : [email protected] e-mail : [email protected] Jozef H. PrzytyckiDepartment of MathematicsThe George Washington UniversityWashington, DC 20052 e-mail : [email protected]@gwu.edu