Expansions of a chord diagram and alternating permutations

A chord diagram is a set of chords of a circle such that no pair of chords has a common endvertex. A chord diagram E with n chords is called an n-crossing if all chords of E are mutually crossing. A chord diagram E is called nonintersecting if E contains no 2-crossing. For a chord diagram E having a 2-crossing S = {x1x3, x2x4}, the expansion of E with respect to S is to replace E with E1 = (E\S)∪{x2x3, x4x1} or E2 = (E\S)∪{x1x2, x3x4}. It is shown that there is a one-to-one correspondence between the multiset of all nonintersecting chord diagrams generated from an ncrossing with a finite sequence of expansions and the set of alternating permutations of order n + 1.


Introduction
Let us consider a set of chords of a circle.A set of chords is called a chord diagram, if they have no common endvertex.If a chord diagram consists of a set of n mutually crossing chords, it is called an n-crossing.A 2-crossing is simply called a crossing as well.If a chord diagram contains no crossing, it is called nonintersecting.
Let V be a set of 2n vertices on a circle, and let E be a chord diagram of order n, where each chord has endvertices of V .We denote the family of all such chord diagrams by CD(V ).Let x 1 , x 2 , x 3 , x 4 ∈ V be placed on a circle in clockwise order.Let E ∈ CD(V ).For a crossing S = {x 1 x 3 , x 2 x 4 } ⊂ E, let S 1 = {x 2 x 3 , x 4 x 1 }, and S 2 = {x 1 x 2 , x 3 x 4 }.The expansion of E with respect to S is defined as a replacement of E with E 1 = (E \ S) ∪ S 1 or E 2 = (E \ S) ∪ S 2 (see Figure 1).In this procedure, E is called the predecessor of E 1 the electronic journal of combinatorics 23(1) (2016), #P1.7 For E ∈ CD(V ), let us denote the number of 2-crossings of E by c(E).Let E be a successor of E such that E = (E \ S) ∪ S , where S is an original 2-crossing and S is a pair of additional chords.
We claim that c(E ) < c(E).Indeed, for e ∈ E ∩ E , let t (resp.t ) be the number of chords of S (resp.S ) intersecting e.
It is not difficult to see that if t 1 then we have t = t, and if t = 2 then we have t = 2 or t = 0. Hence, we have t t.Since S is a crossing of E which is removed in E , we have c(E ) < c(E).
Lemma 1.Let E ∈ CD(V ) be a chord diagram.Then beginning from E, the resulting mutiset of nonintersecting chord diagrams generated by a maximal set of expansions is uniquely determined.
Proof.We proceed by induction on the number of crossings c of a chord diagram E.
If c = 0 or 1, there is nothing to prove.Let c 2 and let c(E) = c.By inductive hypothesis, for a chord diagram E with c(E ) c−1, we define N CD(E ) as the resulting multisets of nonintersecting chord diagrams generated by E .Moreover, for a set of chord diagrams Let S 1 and S 2 be two 2-crossings of E, and let E i1 and E i2 be two successors of E by an expansion with respect to For E 11 and E 12 , by an expansion with respect to S 2 , we have a set E of four chord diagrams.Then we have N CD(E 1 ) = N CD(E ).In the same way, for E 21 and E 22 , by an expansion with respect to S 1 , we have E , and we have We may assume S 1 = {e 0 , e 1 } and S 2 = {e 0 , e 2 }, where e i = x i y i for 0 i 2. Let V 0 = {x 0 , x 1 , x 2 , y 0 , y 1 , y 2 } and let E = E \ {e 0 , e 1 , e 2 }.Beginning from E i with i = 1, 2, let us consider expansions with respect to a crossing induced by V 0 .Case 2.1.e 1 and e 2 are not crossing.
We may assume x 0 , x 1 , x 2 , y 0 , y 2 , y 1 are placed on a circle in clockwise order.By iterating possible expansions, not depending on the order of the expansions, we always have a set of four chord diagrams Case 2.2.e 1 and e 2 are crossing.
We may assume x 0 , x 1 , x 2 , y 0 , y 1 , y 2 are placed on a circle in clockwise order.By iterating possible expansions, not depending on the order of the expansions, we always have a set of five chord diagrams In any case, we have N CD(E i ) = N CD(E ) for i = 1, 2, as required.
Let us denote the multiset of nonintersecting chord diagrams generated by E ∈ CD(V ) by N CD(E).For E ∈ CD(V ), let us define f (E) as the cardinality of N CD(E) as a multiset.
Example 2. Let C n be an n-crossing.Then we have f A background of expansions of a chord diagram is Ptolemy's theorem and its generalization.For two points x, y on a circle, let xy be the length of a chord xy.Ptolemy's theorem states that if E = {x 1 x 3 , x 2 x 4 } itself is a 2-crossing, then we have In other words, we have where E 1 and E 2 are two successors of E. In general, for a given E ∈ CD(V ), by iterating expansions with applications of Ptolemy's theorem, we have If E is a 3-crossing, the equation ( 2) is known as Fuhrmann's Theorem ( [2]).

Main Results
For two nonnegative integers k and n with k n, we define A(n, k) as a chord diagram of order n + 1, in which there is an n-crossing E 0 with an extra chord e such that e crosses the electronic journal of combinatorics 23(1) (2016), #P1.7 exactly k chords of E 0 .Note that A(n − 1, n − 1) is simply an n-crossing, and that A(n, 0) is a union of an n-crossing and an isolated chord.Hence, we have f (A(n − 1, n − 1)) = f (A(n, 0)).The values of f (A(n, k)) for small nonnegative integers n and k are shown in Table 1.A permutation σ of [n] = {1, 2, . . ., n} is called an alternating permutation if (σ(i) − σ(i−1))(σ(i+1)−σ(i)) < 0 for 2 i n−1 (see [9] for an excellent survey of alternating permutations).An alternating permutation σ is called an up-down permutation (resp.down-up permutation) if σ(1) < σ(2) (resp.σ(1) > σ(2)).Let UDP(n, k) denote the set of up-down permutations of [n] with the first term at most k.Similarly, let DUP(n, k) denote the set of down-up permutations of [n] with the first term at least n − k + 1.Note that by definition, there is a natural bijection from UDP(n, k) to DUP(n, k).
The main result of the paper is the following theorem.Theorem 3.For 0 k n, there is a bijection from N CD(A(n, k)) to UDP(n + 2, k + 1).
For 0 k n, Entringer number E n,k is defined as the number of down-up permutations of [n + 1] with the first term k + 1 [1], which equals the cardinality of UDP(n, k).Since for n 1, E n+1,1 equals Euler number E n , the number of down-up permutations of [n], we have the following Corollary.Corollary 4. For 0 k n, we have f (A(n, k)) = E n+2,k+1 .In particular, we have Several combinatorial interpretations for Entringer numbers are known ( [4,5,6,7,8]).The generating function for Entringer number is treated in [3] as an exercise, Exer.6.75.According to [3], it follows that By Corollary 4, we have .
the electronic journal of combinatorics 23(1) (2016), #P1.7 3 Proof of Theorem 3 , and the vertices are placed on a circle in clockwise order for each i = 1, 2. Suppose that v 1,α v 1,β ∈ F 1 holds if and only if v 2,α v 2,β ∈ F 2 holds.Then we say that F 1 and F 2 are isomorphic, and furthermore we say that F 1 and F 2 are isomorphic as well.
In order to prove Theorem 3, we will recursively construct a bijection from N CD(A(n, k)) to UDP(n + 2, k + 1) for 0 k n.
Firstly, we will show a recurrence for N CD(A(n, k)), which is a key ingredient for the proof of Theorem 3.
Lemma 5.For 1 k n, we have a bijection between N CD(A(n, k)) and N CD(A(n, k − 1)) ∪ N CD(A(n − 1, n − k)).In particular, we have Proof.Let E be a chord diagram isomorphic to A(n, k).We may assume E contains an n-crossing E 0 and an extra edge e = xz such that e crosses exactly k edges of E 0 .
Let f = yw be an edge of E 0 such that (1) x, y, z, w are placed on a circle in clockwise order and (2) there is no endvertex of E 0 between x and y. (See Figure 3 For the sake of completeness, we recall the well-known recurrence relation for UDP(n, k).Lemma 6.For 1 k n, we have a bijection between UDP(n + 2, k + 1) and UDP(n + 2, k) ∪ UDP(n + 1, n − k + 1).Proof.By the definition, UDP(n + 2, k + 1) is a set of up-down permutations of [n + 2] with the first term at most k + 1. UDP(n + 2, k + 1) is partitioned into UDP(n + 2, k) and T = UDP(n + 2, k + 1) \ UDP(n + 2, k), where T is a set of up-down permutations of [n + 2] with the first term k + 1.
For σ ∈ T , let us remove the first term of σ.The resulting permutation σ is a downup permutation of [n + 2] \ {k + 1} with the first term at least k + 2. Hence, there is a natural bijection from T to DUP(n + 1, n − k + 1), which has a one-to-one correspondence to UDP(n + 1, n − k + 1).Now, we return to the proof of Theorem 3.For n = 0 and k = 0, a set of a single chord of N CD(A(0, 0)) clearly corresponds to a single permutation 12 of UDP(2, 1).
Let n 1 and k 0. By the inductive hypothesis, we have a bijection from N CD(A(n , k )) to UDP(n + 2, k + 1) for n < n or n = n and k < k.
Let k 1.In this case, by Lemma 5 and Lemma 6, we can recursively construct a bijection from N CD(A(n, k)) to UDP(n + 2, k + 1).
This completes the proof.

Figure 1 :
Figure 1: The expansion of a chord diagram.