aa r X i v : . [ m a t h . G T ] O c t GRAM DETERMINANT OF PLANAR CURVES
J ´OZEF H. PRZYTYCKI AND XIAOQI ZHU
Abstract.
We investigate the Gram determinant of the bilinear form based on curves ina planar surface, with a focus on the disk with two holes. We prove that the determinantbased on n − n curves. Motivated by the workon Gram determinants based on curves in a disk and curves in an annulus (Temperley-Lieb algebra of type A and B , respectively), we calculate several examples of the Gramdeterminant based on curves in a disk with two holes and advance conjectures on thecomplete factorization of Gram determinants. Introduction
Let F n , be a unit disk with 2 n points on its boundary. Let B n, be the set of all pos-sible diagrams, up to deformation, in F n , with n non-crossing chords connecting these 2 n points. It is well-known that | B n, | is equal to the n th Catalan number C n = n +1 (cid:0) nn (cid:1) [10].Accordingly, we will call B n, the set of Catalan states .We will now generalize this setup. Let F ,k ⊂ D be a plane surface with k + 1 boundarycomponents. F , = D , and for k ≥ F ,k is equal to D with k holes. Let F n ,k be F ,k with 2 n points, a , . . . , a n − , arranged counter-clockwise along the outer boundary,cf. Figure 1. Figure 1.
Throughout the paper, we number the points counter-clockwisebeginning at the top of outer boundary. We label and differentiate betweenthe holes.Let B n,k be the set of all possible diagrams, up to equivalence, in F n ,k with n non-crossing chords connecting these 2 n points. We define equivalence as follows: for each diagram b ∈ B n,k , there is a corresponding diagram γ ( b ) ∈ B n, obtained by filling the k holes in b . We call γ ( b ) the underlying Catalan state of b (cf. Figure 2). In addition, a Figure 2. b γ ( b )given diagram in F n ,k partitions F ,k into n + 1 regions. Two diagrams are equivalent if andonly if they have the same underlying Catalan state and the labeled holes are distributedin the same manner across regions. Accordingly, | B n,k | = ( n + 1) k − (cid:0) nn (cid:1) . We remark thatin the cases k = 0 and k = 1, two diagrams are equivalent if they are homotopic, but for k >
2, this is not the case (for an example, see Figure 3).In this paper, we define a pairing over B n,k and investigate the Gram matrix of the pair-ing. This concept is a generalization of a problem posed by W. B. R. Lickorish for type A (based on a disk, i.e. k = 0) Gram determinants, and Rodica Simion for type B (based onan annulus, i.e. k = 1) Gram determinants, cf. [4, 5], [7, 8]. Significant research has beencompleted for the Gram determinants for type A and B . In particular, P. Di Francescoand B. W. Westbury gave a closed formula for the type A Gram determinant [3], [11]; acomplete factorization of the type B Gram determinant was conjectured by Gefry Baradand a closed formula was proven by Q. Chen and J. H. Przytycki [1] (see also [6]). Thetype A Gram determinant was used by Lickorish to find an elementary construction ofReshetikhin-Turaev-Witten invariants of oriented closed 3-manifolds.We specifically investigate the Gram determinant G n of the bilinear form defined over B n, and prove that det G n − divides det G n for n >
1. Furthermore, we investigate thediagonal entries of G n and give a method for computing terms of maximal degree in det G n .We conclude the paper by briefly discussing generalizations of the Gram determinant andpresenting some open questions.2. Definitions for B n, Consider F n , , a unit disk with two holes, along with 2 n points along the outer boundary.Denote the holes in F n , by ∂ X and ∂ Y , or more simply, just X and Y . To differentiatebetween the two holes, we will always place X to the left and Y to the right if labels arenot present.Let B n := B n, := { b n , . . . , b n ( n +1) ( nn ) } , the set of all possible diagrams, up to equivalencein F n , with n non-crossing chords connecting these 2 n points. For simplicity, we will oftenuse b i instead of b ni , when the number of points along the outer boundary can be inferredfrom context. RAM DETERMINANT OF PLANAR CURVES 3
Figure 3.
Two non-isotopic but equivalent diagrams in F , . They corre-spond to the same state in B . A complete specification of B can be foundin the Appendix. Figure 4.
Pictorial representations of six states { b , b , b , b , b , b } ⊂ B .We stress that this is not a natural ordering of states in B .Let X and Y be the inversions of X and Y , respectively, with respect to the unitdisk, and let S = { X , X , Y , Y } . Given b i ∈ B n , let b i ∗ denote the inversion of b i .Given b i , b j ∈ B n , we glue b i with b j ∗ along the outer boundary, respecting the labels ofthe marked points. b i and b j each contains n non-crossing chords, so b i ◦ b j ∗ can have atmost n closed curves. The resulting diagram, denoted by b i ◦ b j ∗ , is then a set of up to n closed curves in the 2-dimensional sphere ( D ∪ ( D ) ∗ ) with four holes: X , X , Y , Y (wedisregard the outer boundary, ∂D ). Each closed curve partitions the set S into two sets.Two closed curves are of the same type if they partition S the same way. For each b i ◦ b j ∗ ,there are then up to eight types of disjoint closed curves, whose multiplicities we index bythe following variables: n d = the number of curves with { X , X , Y , Y } on the same side n x = the number of curves that separate { X } from { X , Y , Y } n x = the number of curves that separate { X } from { X , Y , Y } n y = the number of curves that separate { Y } from { X , X , Y } n y = the number of curves that separate { Y } from { X , X , Y } n z = the number of curves that separate { X , X } from { Y , Y } n z = the number of curves that separate { X , Y } from { X , Y } n z = the number of curves that separate { X , Y } from { X , Y } Inversion is an involution defined on the sphere C ∪ ∞ by z ↔ z | z | . J ´OZEF H. PRZYTYCKI AND XIAOQI ZHU
Let R := Z [ d, x , x , y , y , z , z , z ], and R B n be the free module over the ring R withbasis B n . We define a bilinear form h , i : R B n × R B n → R by: h b i , b j i = d n d x n x x n x y n y y n y z n z z n z z n z h b i , b j i is a monomial of degree at most n . Some examples of paired diagrams and theircorresponding monomials, using examples from Figure 4, are given in Figure 5. Figure 5.
From left to right: h b , b i = x h b , b i = x x h b , b i = dz h b , b i = x Let G n = (cid:0) g ij (cid:1) = (cid:0) h b i , b j i (cid:1) ≤ i,j ≤ ( n +1) ( nn )be the Gram matrix of the pairing on B n . For example, G = d y x z y z z x x z z y z x y d up to ordering of B anddet G = (( d + z )( z + z ) − ( x + y )( x + y ))(( d − z )( z − z ) − ( x − y )( x − y )) . We remark that for b i , b j ∈ B n , h b j , b i i can be obtained by taking b i ◦ b j ∗ and interchang-ing the roles of X and Y with X and Y , respectively. Let h t be an involution on theentries of G n which interchanges the variables x with x and y with y . It follows that h b i , b j i = h t ( h b j , b i i ). The transpose matrix is then given by: t G n = (cid:0) h t ( h b i , b j i ) (cid:1) We note that the variables d, z , z , z are preserved by h t (cf. Theorem 3.2(4)).We can define more generally: given A = { b n , b n , . . . , b n p } ⊆ B n and B = { b m , b m , . . . , b m q } ⊆ B n , let h A, B i be an p × q submatrix of G n given by: h A, B i = (cid:0) h b n i , b m j i (cid:1) ≤ i ≤ p, ≤ j ≤ q For example, we can express the matrix G n as h B n , B n i . The i th row of G n can be writtenas h b i , B n i . RAM DETERMINANT OF PLANAR CURVES 5
Figure 6.
A pictorial representation of curves used to define G This paper is mostly devoted to exploring possible factorizations of det G n , and is thefirst step toward computing det G n in full generality, which we conjecture to have a nicedecomposition.Let i : B n → B n +1 be the embedding map defined as follows: for b i ∈ B n , i ( b i ) ∈ B n +1 is given by adjoining to b i a non-crossing chord close to the outer boundary thatintersects the outer circle at two points between a and a n − , cf. upper part of Figure 8.We will also use a generalization of i , for which we need first the following definition. Forany real number α , consider the homeomorphism r α : C → C on the annulus R ′ ≤ | z | ≤ α -Dehn Twist , defined by: r α ( z ) = ze iα (1 − (1 −| z | ) / (1 − R ′ )) Note that r α ( z ) = z as | z | = R ′ . Therefore, we can extend the domain of r α to D bydefining r α ( z ) = z for 0 ≤ | z | ≤ R ′ . Fix R ′ such that a circle of radius R ′ encloses X and Y . Then r α acts on b i ∈ B n as a clockwise rotation of a diagram close to the outerboundary. Figure 7. A π/ r π ( b i ) = b i (cf. Figure 3). J ´OZEF H. PRZYTYCKI AND XIAOQI ZHU
Consider the k -conjugated embedding i k : B n → B n +1 defined by: i k ( b i ) = r π/n +1 k i r π/n − k ( b i )Intuitively, if for b i ∈ B n +1 there exists b j ∈ B n such that i k ( b j ) = b i , then b i is composedof b j and a non-crossing chord close to the outer boundary connecting a k and a k − , Figure8. Figure 8.
An embedding b i i ( b i ), top; a 1-conjugated embedding b i i ( b i ), bottom; b i ∈ B .For every b i ∈ B n , let p k ( b i ) be the diagram obtained by gluing to b i a non-crossing chordconnecting a k and a k − outside the circle, and pushing the chord inside the circle. Theproperties of p k will be explored in greater detail in Section 4. We conclude this sectionwith a basic identity linking i and p : Proposition 2.1.
For any b i ∈ B n , b j ∈ B n − , b i ◦ i ( b j ) ∗ = p ( b i ) ◦ b j ∗ . Basic Properties of Gram Determinant
In this section, we prove basic properties of det G n . In particular, we show that thedeterminant of G n is nonzero. Lemma 3.1. h b i , b j i is a monomial of maximal degree if and only if γ ( b i ) = γ ( b j ) .Proof. b i ◦ b j ∗ has n closed curves if and only if each closed curve is formed by exactly twoarcs, one in b i and one in b j ∗ . Hence, any two points connected by a chord in b i must alsobe connected by a chord in b j , so γ ( b i ) = γ ( b j ). (cid:3) Theorem 3.1. det G n = 0 for all integers n ≥ . Throughout this paper, we use a k and a k − to denote two adjacent points along the outer boundary,where k is taken modulo 2 n . RAM DETERMINANT OF PLANAR CURVES 7
Proof.
Assume h b i , b j i is a monomial of maximal degree consisting only of the variables d and z . Because γ ( b i ) = γ ( b j ) by Lemma 3.1, it follows that any two points connected in b i are also connected in b j . Each connection in b i can be drawn in four different ways withrespect to X and Y , since there are two ways to position the chord relative to each hole.Because h b i , b j i is assumed to consist only of the variables d and z , it follows that eachpair of arcs that form a closed curve in b i ◦ b j ∗ either separates { X , X } from { Y , Y } orhas { X , X , Y , Y } on the same side of the curve. One can check each of the four casesto see that this condition implies that any two arcs that form a closed curve in b i ◦ b j ∗ must be equal, so b i = b j . Using Laplacian expansion, this implies that the product of thediagonal of G n is the unique summand of degree n ( n + 1) (cid:0) nn (cid:1) in det G n consisting only ofthe variables d and z . (cid:3) We need the following notation for the next theorem: let f : α ↔ α denote a function f which acts on the entries of G n by interchanging variables α with α . We can extendthe domain of f to G n . Let f ( G n ) denote the matrix formed by applying f to all theindividual entries of G n .Let h , h , h be involutions acting on the entries of G n with the following definitions:(1) h : x ↔ y z ↔ z (2) h : x ↔ y z ↔ z (3) h = h h : x ↔ y x ↔ y (4) h t : x ↔ x y ↔ y Theorem 3.2. (1) det h ( G ) = − det G , and for n > , det h ( G n ) = det G n . (2) det h ( G ) = − det G , and for n > , det h ( G n ) = det G n . (3) det h ( G n ) = det G n . (4) det h t ( G n ) = det G n .Proof. For (1), note that h ( G n ) corresponds to exchanging the positions of the holes X and Y for all b i ∈ B n . b j ∗ is unchanged, so h can be realized by a permutation of rows.For states where X and Y lie in the same region, their corresponding rows are unchangedby h . The number of such states is given by n +1 | B n | . Thus, the total number of rowtranspositions is equal to12 (cid:18) | B n | − (cid:18) n + 1 (cid:19) | B n | (cid:19) = n (cid:18) nn (cid:19) = (cid:18) n ( n + 1)2 (cid:19) C n where C n = n +1 (cid:0) nn (cid:1) . It is a known combinatorial fact that C n is odd if and only if n = 2 m − m , [2]. Hence, C n is odd implies that n ( n + 1)2 = 2 m (2 m − m − (2 m − m >
1. Thus, h ( G n ) can be obtained from G n by an even permuta-tion of rows for n >
1, so det h ( G n ) = det G n . h ( G ) is given by an odd number of row J ´OZEF H. PRZYTYCKI AND XIAOQI ZHU transpositions on G , so det h ( G ) = − det G .(2) can be shown using the same method of proof as before. h ( G n ) corresponds toexchanging the positions of the holes X and Y for all b i ∈ B n . h can thus be realizedby a permutation of columns, and the rest of the proof follows in a similar fashion as theprevious one. Since h ( G n ) can be obtained from G n by an even permutation of columnsfor n >
1, det h ( G n ) = det G n . h ( G ) is given by an odd number of column transposi-tions on G , so det h ( G ) = − det G , which proves (2).Since h = h h , it follows immediately that det h ( G n ) = det G n for n >
1. The sum oftwo odd permutations is even, so the equality also holds for n = 1, which proves (3). (4)follows because det h t ( G n ) = det t G n = det G n . (cid:3) Theorem 3.3. det G n is preserved under the following involutions on variables: (1) g : x ↔ − x , x ↔ − x , z ↔ − z , z ↔ − z (2) g : y ↔ − y , y ↔ − y , z ↔ − z , z ↔ − z (3) g : x ↔ − x , y ↔ − y , z ↔ − z , z ↔ − z (4) g g : x ↔ − x , x ↔ − x , y ↔ − y , y ↔ − y (5) g g : x ↔ − x , y ↔ − y , z ↔ − z , z ↔ − z (6) g g : x ↔ − x , y ↔ − y , z ↔ − z , z ↔ − z (7) g g g : x ↔ − x , y ↔ − y , z ↔ − z , z ↔ − z Proof.
To prove (1), we show that g can be realized by conjugating the matrix G n bya diagonal matrix P n of all diagonal entries equal to ±
1. The diagonal entries of P n aredefined as p ii = ( − q ( b i ,F x ) where q ( b i , F x ) is the number of times b i intersects F x modulo 2, cf. Figure 9. The theoremfollows because curves corresponding to the variables x , x , z and z intersect F x ∪ F x ∗ in an odd number of points, whereas curves corresponding to the variables d , z , y and y cut it an even number of times. Figure 9.
More precisely, for g ij = h b i , b j i = d n d x n x x n x y n y y n y z n z z n z z n z , RAM DETERMINANT OF PLANAR CURVES 9 the entry g ′ ij of P n G n P n − satisfies: g ′ ij = p ii g ij p jj = p ii p jj g ij = ( − q ( b i ,F x )+ q ( b j ,F x ) g ij = ( − n x + n x + n z + n z g ij = d n d ( − x ) n x ( − x ) n x y n y y n y z n z ( − z ) n z ( − z ) n z For (2) and (3), we use the same method of proof as for (1). In (2), we use F y and F y ∪ F y ∗ . In (3), we use ˜ F x and ˜ F x ∪ F y ∗ . (4) through (7) follow directly from (1), (2) and(3). (cid:3) Terms of Maximal Degree in det G n Theorem 3.1 proves that the product of the diagonal entries of G n is the unique termof maximal degree, n ( n + 1) (cid:0) nn (cid:1) , in det G n consisting only of the variables d and z . Moreprecisely, the product of the diagonal of G n is given by δ ( n ) = Y b i ∈ B n h b i , b i i = d α ( n ) z β ( n )1 with α ( n ) + β ( n ) = n ( n + 1) (cid:0) nn (cid:1) . δ ( n ) for the first few n are given here: δ (1) = d z δ (2) = d z δ (3) = d z δ (4) = d z Computing the general formula for δ ( n ) can be reduced to a purely combinatorial prob-lem. We conjectured that β ( n ) = (2 n )4 n − and this was in fact proven by Louis Shapirousing an involved generating function argument [9]. The result is stated formally below. Theorem 4.1. δ ( n ) = d n ( n +1) ( nn ) − (2 n )4 n − z (2 n )4 n − Let h (det G n ) denote the truncation of det G n to terms of maximal degree, that is, ofdegree n ( n + 1) (cid:0) nn (cid:1) . Each term is a product of ( n + 1) (cid:0) nn (cid:1) entries in G n , each of which isa monomial of degree n . By Lemma 3.1, h b i , b j i has degree n if and only if b i and b j havethe same underlying Catalan state. There are C n = n +1 (cid:0) nn (cid:1) elements in B n, . Divide B n into subsets corresponding to underlying Catalan states, that is, into subsets A , . . . , A C n ,such that for all b i , b j ∈ A k , γ ( b i ) = γ ( b j ). Then from Lemma 3.1 we have Proposition 4.1. h (det G n ) = C n Y k =1 det h A k , A k i Note that h A k , A k i are simply blocks in G n whose determinants can be multiplied to-gether to give the highest terms in det G n . Finding the terms of maximal degree in det G n can give insight into decomposition of det G n for large n . Example 1. B corresponds to the single Catalan state in B , . Thus, det G = h (det G ) ,a homogeneous polynomial of degree 4. Example 2. B can be divided into two subsets, corresponding to the two Catalan statesin B , . We can thus find h (det G ) by computing two × block determinants. Thetwo Catalan states in B , are equivalent up to rotation, so the two block determinants areequal. Specifically, we have: h (det G ) = d ( x x + x y + x y + y y − dz − z z − dz − z z ) ( − x x + x y + x y − y y + dz − z z − dz + z z ) ( − x x z − y y z + dz + x y z + x y z − dz ) ( − x x y y + dx x z + dy y z − d z + dx y z + dx y z − d z ) = d det G ( − x x z − y y z + dz + x y z + x y z − dz ) ( − x x y y + dx x z + dy y z − d z + dx y z + dx y z − d z ) Example 3. B can be divided into five subsets, corresponding to the five Catalan states in B , . We can thus find h (det G ) by computing the determinants of five blocks in B . Thedeterminant of each block gives a homogeneous polynomial of degree / . B , formstwo equivalence classes up to rotation, so there are only two unique block determinants. Forprecise terms, we refer the reader to the Appendix.
5. det G n − Divides det G n In this section, we prove that the Gram determinant for n − n chords. We need several lemmas: Lemma 5.1.
For any b i ∈ B n , p ( b i ) ∈ B n − if and only if b i contains no chord connecting a and a n − .Proof. Suppose a and a n − are not connected by a chord in b i , say, a is connected to a j and a n − is connected to a k . Then p ( b i ) connects a and a n − by a chord outsidethe outer boundary, and this chord does not form a closed curve. Because a j is connectedto a and a k is connected to a n − , p ( b i ) contains single path from a k to a j , which wecan deform through isotopy so that it fits inside the outer circle. Thus, p ( b i ) ∈ B n − , cf.Figure 10.If b i contains an arc connecting a and a n − , then p ( b i ) contains a closed curve enclosingsome subset of { X , Y } , and cannot be in B n . (cid:3) Lemma 5.2.
For any b i ∈ B n , if p ( b i ) / ∈ B n − , there exists b α ( i ) ∈ B n − such that, forall b j ∈ B n − , one of the following is true: (1) h p ( b i ) , b j i = d h b α ( i ) , b j i (2) h p ( b i ) , b j i = x h b α ( i ) , b j i (3) h p ( b i ) , b j i = y h b α ( i ) , b j i (4) h p ( b i ) , b j i = z h b α ( i ) , b j i . RAM DETERMINANT OF PLANAR CURVES 11
Figure 10.
From b i , we obtain p ( b i ) by adjoining a chord outside theouter boundary between a and a n − , and pushing the chord inside theboundary. If b i does not contain a chord connecting a and a n − , then p ( b i ) ∈ B n − . Proof.
By Lemma 5.1, b i contains a chord connecting points a and a n − , so p ( b i ) mustconsist of some diagram in B n − and a closed curve enclosing some subset of { X , Y } .The former is given by h b α ( i ) , b j i for some b α ( i ) ∈ B n − , and the latter curve is given byone of the following variables: d, x , y , z . (cid:3) The previous two lemmas, combined with Proposition 2.1, leads to the following corol-lary.
Corollary 5.1.
Let A = { , d, x , y , z } . For any b i ∈ B n , there exists b α ( i ) ∈ B n − and c ∈ A such that h b i , i ( B n − ) i = c h b α ( i ) , i ( B n − ) i . That is, the rows of h B n , i ( B n − ) i are each either equal to some row of G n − , or tosome row of G n − multiplied by one of the following variables: d , x , y , z . We now haveall the lemmas needed for our main result of this section. Theorem 5.1.
For n > , det G n − | det G n .Proof. We begin by proving that for every row of the matrix G n − , there exists an equiva-lent row in the submatrix h B n , i ( B n − ) i of G n . Fix b i ∈ B n − and take the row of G n − given by h b i , B n − i . We claim that the row in h B n , i ( B n − ) i given by h i ( b i ) , i ( B n − ) i is equal to h b i , B n − i . In other words, h i ( b i ) , i ( B n − ) i is equal to the i th row of G n − , afact which we leave to the reader for the moment, but will demonstrate explicitly in thenext section, cf. Theorem 6.1.Reorder the elements of B n so that h i ( B n − ) , i ( B n − ) i forms an upper-leftmost blockof G n and h i ( B n − ) , i ( B n − ) i forms a block directly underneath h i ( B n − ) , i ( B n − ) i . This is illustrated below: G n = h i ( B n − ) , i ( B n − ) i ∗ ∗ ∗ ∗ ∗h i ( B n − ) , i ( B n − ) i ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗ = h i ( B n − ) , i ( B n − ) i ∗ ∗ ∗ ∗ ∗ G n − ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ ∗ Corollary 5.1 implies that every row of h B n , i ( B n − ) i is a multiple of some row in G n − . Let j , . . . , j k denote the indices of all rows of h B n , i ( B n − ) i other than those in h i ( B n − ) , i ( B n − ) i . Let G n ′ be the matrix obtained by properly subtracting multiples ofrows in h i ( B n − ) , i ( B n − ) i from rows j , . . . , j k of G n so that the submatrix obtained byrestricting G n ′ to rows j , . . . , j k and columns corresponding to states in i ( B n − ) is equalto 0: G n ′ = ∗ ∗ ∗ ∗ ∗ G n − ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Thus, G n ′ restricted to the columns corresponding to states in i ( B n − ) contains precisely n (cid:0) n − n − (cid:1) nonzero rows, each equal to some unique row of G n − . The determinant of this sub-matrix is equal to det G n − . Since det G n − | det G n ′ and det G n ′ = det G n , this completesthe proof. (cid:3) Further Relation Between det G n − and det G n As was first noted in the proof of Theorem 5.1, there exists a submatrix of G n equal to G n − . This section will be focused on identifying multiple nonoverlapping submatrices in G n equal to multiples of G n − . This will prove useful for simplifying the computation ofdet G n . We start with the main lemma for this section and for Theorem 5.1. Lemma 6.1.
For any b i , b j ∈ B n − , h i ( b i ) , i ( b j ) i = h i ( b i ) , i ( b j ) i = h b i , b j i .Proof. We begin with the equality h i ( b i ) , i ( b j ) i = h b i , b j i . By Proposition 2.1, i ( b i ) ◦ i ( b j ) ∗ = p i ( b i ) ◦ b j ∗ , so it suffices to prove that p i ( b i ) = p r π/n i r π/n − − ( b i ) = b i .This is demonstrated pictorially: RAM DETERMINANT OF PLANAR CURVES 13
Figure 11.
Thus, h i ( b i ) , i ( b j ) i = h b i , b j i . Recall that h b i , b j i = h t ( h b j , b i i ). From this and theprevious equality, it follows that h i ( b i ) , i ( b j ) i = h t ( h i ( b j ) , i ( b i ) i ) = h t ( h b j , b i i ) = h t ( h b i , b j i ) = h b i , b j i . (cid:3) Corollary 6.1. h i ( B n − ) , i ( B n − ) i = h i ( B n − ) , i ( B n − ) i = G n − . Lemma 6.2.
For any b i , b j ∈ B n − , h i ( b i ) , i ( b j ) i = h i ( b i ) , i ( b j ) i = d h b i , b j i .Proof. i ( b i ) ◦ i ( b j ) ∗ is composed of b i ◦ b j ∗ in addition to a chord close to the boundary gluedwith its inverse. The latter pairing gives a trivial circle. Thus, h i ( b i ) , i ( b j ) i = d h b i , b j i for all b i , b j ∈ B n − .By symmetry, h i ( B n − ) , i ( B n − ) i = dG n − . (cid:3) Corollary 6.2. h i ( B n − ) , i ( B n − ) i = h i ( B n − ) , i ( B n − ) i = dG n − . Using these two facts, we can construct from G n a ( | B n | − | B n − | ) × ( | B n | − | B n − | )matrix whose determinant is equal to det G n / (1 − d ) n ( n − n − )det G n − . This allows us tocompute det G n with greater ease, assuming we know det G n − . This process is shown inthe next theorem. Theorem 6.1.
There exists an integer k ≥ such that, for all integers n > , det G n − | det G n (1 − d ) k . Clearly k is bounded above by ( n + 1) ` nn ´ , or even better, by | B n | − | B n − | . There are obviouslybetter approximations possible, but we do not address them in this paper. Proof.
Order the elements of B n , (or equivalently, the rows and columns of G n ) as shownin Theorem 5.1. We apply the procedure from Theorem 5.1 to construct G n ′ , whose formis given roughly below: G n ′ = − d ) G n − ∗ ∗ ∗ ∗ G n − dG n − ∗ ∗ ∗ ∗ ∗∗∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗∗ ∗ ∗ ∗∗ ∗ ∗ ∗ Consider the block in G n ′ whose columns correspond to states in i ( B n − ) and whose rowscorrespond to states in neither i ( B n − ) nor i ( B n − ) (boxed above). Every row in thissubmatrix is a linear combination of two rows from G n − . More precisely, each row is ofthe form a l − a dl , where l and l are two rows, not necessarily distinct, in G n − , and a , a ∈ A = { , d, x , y , z } . If we assume (1 − d ) is invertible in our ring, then eachrow is a linear combination of two rows from (1 − d ) G n − . We then simplify G n ′ as follows.Let G n ′′ be the matrix obtained by properly subtracting linear combinations of the first n (cid:0) n − n − (cid:1) rows of G n ′ from the rows which correspond to states in neither i ( B n − ) nor i ( B n − ) so that the submatrix obtained by restricting G n ′′ to columns corresponding tostates in i ( B n − ) and rows corresponding to states in neither i ( B n − ) nor i ( B n − ) isequal to 0: G n ′′ = − d ) G n − ∗ ∗ ∗ ∗ G n − dG n − ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗∗ ∗ ∗ ∗∗ ∗ ∗ ∗ The block decomposition at this point proves that det G n ′′ is equal to (1 − d ) n ( n − n − )(det G n − ) times the determinant of the boxed block, which we denote by ¯ G n . The latter containsa power of (1 − d ) − , whose degree is unspecified. Thus, det G n − | det G n ′′ (1 − d ) k forsome integer k ≥
0. We remind the reader that G n ′′ is obtained from G n ′ via determinantpreserving operations, and hence det G n ′ = det G n . (cid:3) Note that if det ¯ G n has fewer than n (cid:0) n − n − (cid:1) powers of (1 − d ) − , then det G n − | det G n .It remains an open problem as to whether this is true. For an example of this decomposi-tion, we refer the reader to the Appendix.7. Future Directions
In this section, we discuss briefly generalizations of the Gram determinant and presenta number of open questions and conjectures.
RAM DETERMINANT OF PLANAR CURVES 15
The case of a disk with k holes. We can generalize our setup by considering F n ,k ,a unit disk with k holes, in addition to 2 n points, a , . . . , a n − , arranged in a similar wayto points in F n , . For b i , b j ∈ B n,k , let b i ◦ b j ∗ be defined in the same way as before. Eachpaired diagram b i ◦ b j ∗ consists of up to n closed curves on the 2-sphere ( D ∪ ( D ) ∗ ) with2 k holes. Let S denote the set of all 2 k holes. We differentiate between the closed curvesbased on how they partition S . We define a bilinear form by counting the multiplicitiesof each type of closed curve in the paired diagram. In the case k = 2, we assigned toeach paired diagram a corresponding element in a polynomial ring of eight variables, eachvariable representing a type of closed curve. In the general case, the number of types ofclosed curves is equal to 2 | S | k k − so we can define the Gram matrix of the bilinear form for a disk with k holes and 2 n pointswith ( n + 1) k − (cid:0) nn (cid:1) × ( n + 1) k − (cid:0) nn (cid:1) entries, each belonging to a polynomial ring of 2 k − variables. We denote this Gram matrix by G F ,k n . For n = 1 and k = 3, we can easily writethis 8 × F n , by ∂ , ∂ and ∂ , and their inversions by ∂ − , ∂ − and ∂ − , respectively. Hence, each closed curvein the surface encloses some subset of S = { ∂ , ∂ − , ∂ , ∂ − , ∂ , ∂ − } . Let x a ,a ,a denote acurve separating the set of holes { ∂ a , ∂ a , ∂ a } from S − { ∂ a , ∂ a , ∂ a } . We can similarlydefine x a ,a and x a . The Gram matrix is then: G F , = d x − x − x − , − x − x − , − x − , − x , , x x , − x − , x , − , x − , x , , − x , , − x , x x , − x , − x , − , x − , x , − , x , , − x , x , x , − , − x , − , − x , − x , − , − x , − x , − x x x , − x , − x , − , − x , − x , − , − x , − , − x , x , x , , − x , − , x − , x , − , x , − x , − x x , x , , − x , , − x , − x , − , x − , x , − x x , , x − , − x − , − x − x − , − x − x − d It would be tempting to conjecture that the determinant of the above matrix has a straight-forward decomposition of the form ( u + v )( u − v ). We found that it is the case for thesubstitution x a = x a ,a = 0 with a , a ∈ {− , − , − , , , } (see Appendix). Howeverin general, the preliminary calculation suggests that det G F , n may be an irreducible poly-nomial.Finally, we observe that many results we have proven for det G F , n holds for generaldet G F ,k n . For example, det G F ,k n = 0 and det G F ,k − n | det G F ,k n . In the specific case ofdet G F , n we conjecture that the diagonal term is of the form δ ( n ) = d α ( n ) ( x , − x , − x , − ) β ( n ) ,where α ( n ) + 3 β ( n ) = n ( n + 1) (cid:0) nn (cid:1) and β ( n ) = n ( n + 1)4 n − .7.2. Speculation on factorization of det G n . Section 5 establishes that det G n − | det G n ,but we conjecture that there are many more powers of det G n − in det G n . Indeed, even in the base case, det G k | det G for k up to 4. Finding the maximal power of det G n − indet G n in the general case is an open problem and can be helpful toward computing thefull decomposition of det G n .Examining the terms of highest degree in det G n , that is, h (det G n ) may also yield helpfulhints toward the full decomposition. In particular, we note that:det G | h (det G ) and h (det G ) det G ! | h (det G )We can conjecture that (cid:18) det G det G (cid:19) | det G so it follows that det G | det G . We therefore offer the following conjecture: Conjecture 1. det G ( nn − ) | det G n for n ≥ . In addition, we also offer the following conjecture, motivated by observations of det G and det G : Conjecture 2.
Let H n denote the factors of det G n not in det G n − , that is, H n | det G n and gcd( H n , det G n − ) = 0 . Then ( H n − ) n | det G n . Conjecture 3.
Let, as before, R = Z [ d, x , x , y , y , z , z , z ] and R be a subgroup of R of elements invariant under h , h , h t , and g , g , g . Similarly, let R be a subgroup of R composed of elements w ∈ R such that h ( w ) = h ( w ) = − w and h t ( w ) = g ( w ) = g ( w ) = g ( w ) . Then (1) det G n = u − v , where u ∈ R and v ∈ R . (2) det G n = Q α ( u α − v α ) , where u α ∈ R and v α ∈ R , and u α − v α and u α + v α areirreducible polynomials. (3) det G n = Q ni =1 ( u i − v i )( nn − i ) , where u i ∈ R and v i ∈ R .Notice that if w = u − v and w = u − v , then w w = ( u u + v v ) − ( u v + u v ) . We have little confidence in Conjecture 3(3). It is closely, maybe too closely, influencedby the Gram determinant of type B (det G Bn = det G F , n ). That is Theorem 7.1. ( [1, 6] ) det G Bn = n Y i =1 (cid:0) T i ( d ) − a (cid:1) ( nn − i ) where T i ( d ) is the Chebyshev polynomial of the first kind: T = 2 , T = d, T i = d T i − − T i − ; d and a in the formula, correspond to the trivial and the nontrivial curves in the annulus F , , respectively. R A M D E T E R M I NAN T O F P L ANA R C U R V E S Appendix B G d dy dx dz z x y d y z x z z z x y y x z z dy dz dz dx x z z y x y x x x z x z z y z x z x x dx dz dz dy y z z x y y x y y z y z z y z x z y y dz dx dy d d y x z dy dx dz d y x x y x y d dz x y d d dy dx dz y x z z x z y z z z x y x z z y dy dz dz dx z z x x z x x x y x x x z y z y z z x dx dz dz dy z z y y z x y y y y y x z y z d y x z dz dx dy d x y d dz dx dy d d x y x y y z x y y y dy y z z x dz dz dx y z z x z y z y y x z x x x y dx x z z y dz dz dy x z z x z y z x x z x z y z dz z x y d dx dy d z x y x y z z z x y d z x z y z dz y x z d dy dx z z y x x z z y x z x x x y dx z z x dy dz dz x x y z x z y z z x y z x y y y dy z z y dx dz dz y y y z x z y y z y z x y z x y d x z x z x z x y z z z z x x z x z x y z x y d y z y z y z x y z z z z z x y d x x z x z x y y x z y y z y z z z z z z x y d y y z y z x y y x z x x z x z z z z z J ´ O Z E F H . P R Z Y T Y C K I AN D X I A OQ I Z HU G (defined in Theorem 6.1), after simplification x y − dz − ( − d ) d (1 + d ) − d ( − y )(1 + y ) − d ( − z )(1 + z ) − ( − d )(1 + d ) y − ( d − y )( d + y ) − ( y − z )( y + z ) ( x − z )( x + z ) − dy + y z − ( d − z )( d + z ) ( x − y )( x + y ) ( d − z )( d + z ) − dy + x z − dx + x z − y z + y z − − d )(1 + d ) y y − y y − dy z y − dx z − y z − dy y + 2 z − d z − dy + x z − y z + x z dy − x z x x − dz − x − y + z + z − dx y − z ) − x y z + 2 z − y y z − ( − x + y ) z x − dx z − dy z − dy + y z − x z + x z − − d )(1 + d ) x x − x y − dy z x − dy z − x z − dx y + 2 z − d z − dx + y z x z − y z dx − y z x y − dz − d + x + y − z − − d )(1 + d ) z − dx y + 2 z − y z − z ( − d + z ) 2 x − d x − dy z − ( d − x )( d + x ) − ( y − z )( y + z ) ( y − z )( y + z ) − dy + x z y − dx z − ( − d )(1 + d ) x x − dy z − ( − d )(1 + d ) z − d ( − x )(1 + x ) − x z + x z − dx + y z x z − y z x y − dz − ( x − z )( x + z ) dy − x z − dx + y z dx − y z − ( x − y )( x + y ) − dx x + 2 z − x y z − dx y + 2 z − d z − dx y − y y z + 2 z x − d x − dy z y − x y − dx z − x x + dz x y − dz − x y + dz y z − x z − x ( − x + y ) 2 y − dy z − dx z − ( − x + y ) y − d ( − x + y ) − x y z + 2 z − x x z − dx y − x x z + 2 z − dx x + 2 z − d z − dy y + 2 z − x y z y − d y − dx z x − x x − dx z − x y + dz y y − dz − y y + dz − x z + y z − x ( − d + z ) 2 y − d y − dx z − y ( − d + z ) − d ( − d + z ) − dx y + 2 z − x z y z − x z − dx + x z − y z + y z x y − dz − ( x − z )( x + z ) det ¯ G = det G (1 − d ) det G G det G = − d ( − x x + x y + x y − y y + dz − z z − dz + z z ) ( − x x − x y − x y − y y + dz + z z + dz + z z ) ( − x x z − y y z + dz + x y z + x y z − dz ) (8 d − d − x + 2 d x − x + 2 d x + 2 x x + 8 dx y − d x y − dx x y − y + 2 d y + 2 x y +8 dx y − d x y − dx x y + 2 d x x y y − dx y y − y + 2 d y + 2 x y − dx y y + 2 y y +2 dx x z − d x x z − x y z + 2 d x y z − x y z + 2 d x y z + 2 dy y z − d y y z + 8 z − d z + d z − d z + 2 d z + dx x z z − x y z z − x y z z + dy y z z + 4 z z − d z z + 8 z − d z − x x z +2 d x x z + 2 dx y z − d x y z + 2 dx y z − d x y z − y y z + 2 d y y z − x x z z + dx y z z + dx y z z − y y z z + 8 z − d z + d z + 4 z z − d z z )(8 d − d − x + 2 d x − x + 2 d x + 2 x x − dx y + 2 d x y + 2 dx x y − y + 2 d y + 2 x y − dx y + 2 d x y + 2 dx x y + 2 d x x y y + 2 dx y y − y + 2 d y + 2 x y + 2 dx y y + 2 y y +2 dx x z − d x x z + 4 x y z − d x y z + 4 x y z − d x y z + 2 dy y z − d y y z + 8 z − d z + d z +8 d z − d z − dx x z z − x y z z − x y z z − dy y z z − z z + d z z + 8 z − d z + 4 x x z − d x x z + 2 dx y z − d x y z + 2 dx y z − d x y z + 4 y y z − d y y z − x x z z − dx y z z − dx y z z − y y z z + 8 z − d z + d z − z z + d z z ) R A M D E T E R M I NAN T O F P L ANA R C U R V E S G h (det G ) = h (det G ) det G − d w ¯ w = d ( − x x + x y + x y − y y + dz − z z − dz + z z ) ( − x x − x y − x y − y y + dz + z z + dz + z z ) ( − x x z − y y z + dz + x y z + x y z − dz ) (2 x x y y − dx x z − dy y z + d z − dx y z − dx y z + d z ) ( x x y y z − dx x z − dy y z + d z − x x y y z + dx y z + dx y z − d z ) ( x x y y z − dx x z − dy y z + d z + x x y y z − dx y z − dx y z + d z ) G with substitution x = x = y = y = z = 0 det G | x x y y z = ( − d ) ( − d ) d (1 + d ) (2 + d ) ( − d ) ( z − z ) ( z + z ) ( z − z z + z )( z + z z + z )( − d − z + d z − z + d z ) ( − d − z + d z + z z − d z z − z + d z ) ( − d − z + d z − z z + d z z − z + d z ) G F , with substitution x a = x a ,a = 0 for all variables of the form x a and x a ,a det G F , | xa xa ,a = − ( d − x , , )( d + x , , )( x , , − x , − , x , − , − + x , , − x , − , x , − , − − x , , − x , − , x , − , − − x , − , − x , − , − x , − , − x , , − x , , − x , − , − + x , , − x , − , x , − , − ) References [1] Q. 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Department of MathematicsThe George Washington UniversityWashington, DC 20052, USA [email protected]