Quadratic Coefficients of Goulden–Rattan Character Polynomials

Abstract

Goulden–Rattan polynomials give the exact value of the subdominant part of the normalized characters of the symmetric groups in terms of certain quantities $(C_i)$ which describe the macroscopic shape of the Young diagram. The Goulden–Rattan positivity conjecture states that the coefficients of these polynomials are positive rational numbers with small denominators. We prove a special case of this conjecture for the coefficient of the quadratic term $C_2^2$ by applying certain bijections involving maps (i.e., graphs drawn on surfaces).

1 Introduction

1.1 Normalized Characters

Characters are a basic tool of representation theory. After normalization, they are also useful in asymptotic problems.

If $k\le n$ are natural numbers, then any permutation $\pi \in S_k$ can also be treated as an element of the larger symmetric group $S_n$ by adding $n-k$ additional fixpoints. For any permutation $\pi \in S_k$ and any irreducible representation $\rho ^{\lambda }$ of the symmetric group $S_n$ which corresponds to the Young diagram $\lambda$, we define the normalized character

$$\begin{aligned} \Sigma _{\pi }(\lambda ) = {\left\{ \begin{array}{ll} n(n-1)\cdots (n-k+1)\frac{{\text {Tr}}\rho ^{\lambda }(\pi )}{\text {dimension of }\rho ^{\lambda }} &{}\quad \text {for } k \le n,\\ 0 &{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

Of particular interest are the character values on the cycles, therefore we will use the shorthand notation

$$\begin{aligned} \Sigma _{k}(\lambda )=\Sigma _{(1,2,\ldots ,k)}(\lambda ). \end{aligned}$$

1.2 Free Cumulants

Free cumulants are an important tool of free probability theory [9] and random matrix theory [10]. In the context of the representation theory of the symmetric groups, they can be defined as follows, see [1]. For a Young diagram $\lambda$, we define its free cumulants $R_2(\lambda ), R_3(\lambda ), \ldots$ as

$$\begin{aligned} R_k(\lambda )=\lim _{s\rightarrow \infty } \frac{1}{s^k}\Sigma _{k-1}(s\lambda ), \end{aligned}$$

where the diagram $s\lambda$ is created from the diagram $\lambda$ by dividing each box of $\lambda$ into an $s\times s$ square.

The free cumulants have been defined in such a way as to be very helpful for studying asymptotic behaviour of the characters on a cycle of length k when the size of the Young diagram tends to infinity [2].

1.3 Kerov Character Polynomials

Kerov [6] formulated the following result: for each permutation $\pi$ and any Young diagram $\lambda$, the normalized character $\Sigma _{\pi }(\lambda )$ is equal to the value of some polynomial $K_{\pi }(R_2(\lambda ), R_3(\lambda ), \ldots )$ (now called the Kerov character polynomial) with integer coefficients. The first published proof of this fact was provided by Biane [1]. The Kerov character polynomial is universal because it does not depend on the choice of $\lambda$. We are interested in the values of the characters on cycles; therefore, for $\pi =(1, 2, \ldots , k)$, we use the simplified notation

$$\begin{aligned} \Sigma _k=K_k(R_2, R_3, \ldots ) \end{aligned}$$

(1.1)

for such Kerov polynomials. The first few Kerov polynomials $K_k$ are as follows:

$$\begin{aligned} K_1&=R_2,\\ K_2&=R_3,\\ K_3&=R_4+R_2,\\ K_4&=R_5+5R_3,\\ K_5&=R_6+15R_4+5R_2^2+8R_2,\\ K_6&=R_7+35R_5+35R_3R_2+84R_3, \\ K_7&=R_8+180R_2+224R_2^2+14R_2^3+56R_3^2+469R_4+84R_2R_4+70R_6. \end{aligned}$$

Kerov conjectured that the coefficients of the polynomial $K_k$ are non-negative integers. Goulden and Rattan [5] found an explicit formula for the coefficients of the Kerov polynomial $K_k$; unfortunately, their formula was complicated and did not give any combinatorial interpretation to the coefficients. Later, Féray proved positivity [4] and together with Dołęga and Śniady found a combinatorial interpretation of the coefficients [3]. In this paper, we will use the combinatorial interpretation given by them in the special case of linear and square coefficients.

1.4 Goulden–Rattan Conjecture

Goulden and Rattan [5] introduced a family of functions $C_2,C_3,\dots$ on the set of Young diagrams given by $C_0=1$, $C_1=0$ and

$$\begin{aligned} C_k^{\lambda }=\frac{24}{k(k+1)(k+2)}\lim _{s\rightarrow \infty }\frac{1}{s^k}\Big (\Sigma _{k+1}(s\lambda )-R_{k+2}(s\lambda )\Big ) \end{aligned}$$

for $k\ge 2$.

Śniady [7] proved the explicit form of $C_k$ (conjectured by Biane [1]) as a polynomial in the free cumulants $R_2, R_3, \ldots$ given by

$$\begin{aligned} C_k=\sum _{\begin{array}{c} j_2,j_3,\ldots \ge 0 \\ 2j_2+3j_3+\cdots =k \end{array}} (j_2+j_3+\cdots )!\prod _{i\ge 2} \frac{\big ( (i-1)R_i\big )^{j_i}}{j_i!} \end{aligned}$$

(1.2)

for $k\ge 2$. The aforementioned formula of Goulden and Rattan for the Kerov polynomials was naturally expressed in terms of these quantities $C_2,C_3,\dots$ [5]. More specifically, they constructed an explicit polynomial $L_k$ with rational coefficients such that

$$\begin{aligned} K_k-R_{k+1}=L_k(C_2, C_3, \ldots ). \end{aligned}$$

(1.3)

These polynomials are called the Goulden–Rattan polynomials. They formulated the following conjecture:

Goulden–Rattan conjecture. The coefficients of the Goulden–Rattan polynomials are non-negative numbers with small denominators.

The first few Goulden–Rattan polynomials are as follows [5]:

$$\begin{aligned} K_1&-R_2=0\\ K_2&-R_3=0\\ K_3&-R_4=C_2,\\ K_4&-R_5=\frac{5}{2}C_3,\\ K_5&-R_6=5C_4+8C_2,\\ K_6&-R_7=\frac{35}{4}C_5+42C_3,\\ K_7&-R_8=14C_6+\frac{469}{3}C_4+\frac{203}{3}C_2^2+180C_2. \end{aligned}$$

Linear coefficients of the Goulden–Rattan polynomials are non-negative, because they are equal to certain scaled coefficients of the Kerov polynomial:

$$\begin{aligned}{}[ C_j ] L_k=\frac{1}{j-1} [ R_j ] K_k. \end{aligned}$$

In this paper, we will prove that the coefficient of $C_2^2$ is non-negative. We hope that edge sliding we will define in this article will also be a useful tool in proving non-negativity of the square coefficients $[ C_i C_j ] L_k$. The next step towards the proof of Goulden–Rattan conjecture would be to understand the cubic coefficients $[ C_i C_j C_u ] L_k$; we hope that our methods will still be applicable there, nevertheless, there seem to be some difficulties related to the inclusion–exclusion principle.

1.5 Graphs on Surfaces, Maps and Expanders

We will consider graphs drawn on an oriented surface. Each face of such a graph has some number of edges ordered cyclically by going along the boundary of the face and touching it with the right hand. We will call it the clockwise boundary direction. If we use the left hand and visit the edges in the opposite order, we will call it the counterclockwise boundary direction.

By a map we mean a bipartite graph drawn without intersections on an oriented and connected surface with minimal genus. The maps which we consider have a fixed choice of colouring of the vertices, i.e., each vertex is coloured black or white, with the edges connecting the vertices of the opposite colours. An example of a map is shown in Fig. 1.

Fig. 1

a An example of a map with 4 vertices and 5 edges drawn on a torus. b The same map drawn for simplicity on the plane (Color figure online)

An expander [8, Appendix A.1] is a map with the following properties.

• It has a distinguished edge (known as the root) and one face.

• Each black vertex is assigned a natural number, known as a weight, such that each non-empty proper subset of the set of black vertices has more white neighbours than the sum of its weights.

• The sum of all weights is equal to the number of white vertices.

The map from Fig. 1 is an expander if each black vertex has weight 1 (any choice of the root is valid).

Using the Euler characteristic we get

$$\begin{aligned} 2-2g=\chi =V-k+1, \end{aligned}$$

(1.4)

where g denotes the genus of the surface, V denotes the number of vertices and k denotes the number of edges.

1.6 Combinatorial Interpretation of the Kerov Polynomial Coefficients

The following two theorems Theorem 1 and Theorem 3 give a combinatorial interpretation to the linear and square coefficients of the Kerov character polynomials [3, Theorem 1.2, Theorem 1.3]. The first is as follows.

Theorem 1

For all integers $l\ge 2$ and $k\ge 1$ the coefficient $[R_l] K_k$ is equal to the number of pairs $(\sigma _1,\sigma _2)$ of permutations $\sigma _1, \sigma _2 \in S(k)$ such that $\sigma _1\sigma _2 =(1 \; 2 \; \cdots \;k)$ and such that $\sigma _2$ consists of one cycle and $\sigma _1$ consists of $l-1$ cycles. (We use the convention $\sigma _1 \sigma _2 = \sigma _2 ( \sigma _1 ) = \sigma _2 \circ \sigma _1.)$

The expanders are a graphical interpretation of these pairs of permutations. There is a natural bijection between such pairs of permutations $(\sigma _1, \sigma _2)$ and the expanders with one face, one black vertex, $l-1$ white vertices and k edges. Additionally, one edge is selected as the root and the unique black vertex has a weight $l-1$. More precisely:

• The edges are numbered $1, 2, \dots , k$. The edge with number 1 is selected as the root.

• The counterclockwise angular cyclic order of the edges on a given vertex (i.e., the order of edges ending at this vertex around it) corresponds to a cycle of a permutation depending on the colour of this vertex, i.e., $\sigma _1$ for white and $\sigma _2$ for black (in this case we have a unique cycle of the permutation $\sigma _2$).

• The unique face corresponds to the unique counterclockwise boundary cycle of the permutation $(1 \; 2 \; \cdots \; k)$.

Since there is only one face, the root determines the numbering of all edges. We can reformulate Theorem 1 as follows. (See Fig. 2a for an example.)

Theorem 2

For all integers $l\ge 2$ and $k\ge 1$ the coefficient $[R_l] K_k$ is equal to the number of expanders with k edges, $l-1$ white vertices and 1 black vertex with the weight $l-1$.

Fig. 2

Examples of expanders with 5 edges and their corresponding pair of permutations $\sigma _1, \sigma _2$ such that $\sigma _1 \sigma _2 = (1 \; 2\; 3 \; 4 \; 5)$. The root is assigned the number 1. a The expander with one black vertex and three white vertices. b The expander with two black vertices and two white vertices (Color figure online)

Similarly, we use the second theorem [3, Theorem 1.3] for square coefficients.

Theorem 3

For all integers $l_1, l_2\ge 2$ and $k\ge 1$ the coefficient $[R_{l_1} R_{l_2}] K_k$ is equal to the number of triples $(\sigma _1, \sigma _2, q)$ with the following properties.

• The permutations $\sigma _1, \sigma _2 \in S_k$ fulfill the equality $\sigma _1 \sigma _2=(1 \; 2 \; \cdots \; k)$.

• The permutation $\sigma _1$ consists of two cycles and the permutation $\sigma _2$ consists of $l_1+l_2-2$ cycles.

• The function q associates the numbers $l_1$ and $l_2$ to the two cycles of $\sigma _1$. Furthermore, for each cycle c of $\sigma _1$ there exist at least q(c) cycles of $\sigma _2$ which nontrivially intersect c.

Analogously, we can also reformulate Theorem 3. (See Fig. 2b for an example.)

Theorem 4

For all integers $l_1, l_2\ge 2$ and $k\ge 1$ the coefficient $[R_{l_1} R_{l_2}] K_k$ is equal to the number of expanders with k edges, $l_1+l_2-2$ white vertices and 2 black vertices with weights $l_1-1, l_2-1$.

1.7 Relationship Between Coefficients of Goulden–Rattan Polynomials and Coefficients of Kerov Polynomials

The formula (1.2) allows us to express $(C_i)$ in terms of free cumulants; we see that the coefficients of the terms $R_i R_j$, $R_{i+j}$, $R_j^2$ and $R_{2j}$ in the expressions $C_{i} C_{j}$, $C_{i+j}$, $C^2_{j}$ and $C_{2j}$ are given for $i \ne j$ by

$$\begin{aligned} C_i C_j&= (i-1)(j-1) R_i R_j + 0 R_{i+j} + \text {(sum of other terms)},\\ C_{i+j}&=2(i-1)(j-1)R_iR_j+(i+j-1)R_{i+j}+\text {(sum of other terms)},\\ C_{j}^{2}&=(j-1)^2R_j^2+0 R_{2j}+\text {(sum of other terms)},\\ C_{2j}&=(j-1)^2R_j^2+(2j-1)R_{2j}+\text {(sum of other terms)}. \end{aligned}$$

Moreover, any product $C_{i_1} C_{i_2}\cdots C_{i_t}$ of at least $t\ge 3$ factors does not contain any of the terms $C_i C_j$, $C_{i+j}, C^2_{j}$ and $C_{2j}$. It follows that the square coefficients of the Goulden–Rattan polynomial are related to the coefficients of the Kerov polynomial via

$$\begin{aligned} \frac{\partial ^2L_k}{\partial {C_i}\partial {C_j}}\Bigg |_{0=C_1=C_2=\cdots }&=\frac{1}{(i-1)(j-1)}\frac{\partial ^2K_k}{\partial {R_i}\partial {R_j}}\Bigg |_{0=R_1=R_2=\cdots }\\&\quad -2\frac{\partial L_k}{\partial {C_{i+j}}}\Bigg |_{0=C_1=C_2=\cdots }\\&=\frac{1}{(i-1)(j-1)}\frac{\partial ^2K_k}{\partial {R_i}\partial {R_j}}\Bigg |_{0=R_1=R_2=\cdots }\\&\quad -\frac{2}{(i+j-1)}\frac{\partial K_k}{\partial {R_{i+j}}}\Bigg |_{0=R_1=R_2=\cdots } \end{aligned}$$

for $i \ne j$. Whereas the quadratic coefficients are related via

$$\begin{aligned} \frac{\partial ^2L_k}{\partial {C_j^2}}\Bigg |_{0=C_1=C_2=\cdots }&=\frac{1}{(j-1)^2}\frac{\partial ^2K_k}{\partial {R_j^2}}\Bigg |_{0=R_1=R_2=\cdots } -2 \frac{\partial L_k}{\partial {C_{2j}}}\Bigg |_{0=C_1=C_2=\cdots }\\&=\frac{1}{(j-1)^2}\frac{\partial K_k^2}{\partial {R_j^2}}\Bigg |_{0=R_1=R_2=\cdots } -\frac{2}{(2j-1)}\frac{\partial K_k}{\partial {R_{2j}}}\Bigg |_{0=R_1=R_2=\cdots } \end{aligned}$$

for any natural number j. Thus, we obtain the explicit formula for the square coefficients of the Goulden–Rattan polynomial:

$$\begin{aligned}{}[C_j^2]L_k&=\frac{1}{(j-1)^2}[R_j^2]K_k-\frac{1}{2j-1}[R_{2j}]K_k \end{aligned}$$

(1.5)

and

$$\begin{aligned}{}[C_iC_j]L_k&=\frac{1}{(i-1)(j-1)}[R_iR_j]K_k-\frac{2}{i+j-1}[R_{i+j}]K_k \qquad \text {for }i\ne j.\nonumber \\ \end{aligned}$$

(1.6)

2 The Main Result

Let $Y_k(u)$ denote the set of expanders with k edges, $u-1$ white vertices and one black vertex. Let $X_k(i, j)$ denote the set of expanders with k edges, $i+j-2$ white vertices and two black vertices with weights $i-1$ and $j-1$. Using Theorems 2 and 4 we can also reformulate the Goulden–Rattan conjecture for the square coefficients in terms of expanders, as follows.

Conjecture 1

Let $i \ne j$ be natural numbers. Then

$$\begin{aligned} (2j-1)\ \left\| X_k(j, j) \right\|&\ge (j-1)^2\ \left\| Y_k(2j) \right\| \end{aligned}$$

and

$$\begin{aligned} (i+j-1)\ \left\| X_k(i, j) \right\|&\ge 2(i-1)(j-1)\ \left\| Y_k(i+j) \right\| \end{aligned}$$

for any natural number k.

These inequalities are equivalent to the positivity of the coefficients $[C_j^2] L_k$ and $[C_i C_j] L_k$ respectively. In this text we prove only the first inequality in the special case $j=2$. We hope to present a proof of Conjecture 1 in its general form in a future paper.

Fig. 3

a The graph with the multiset of cycles $\big \{ (2) (1) (5 \; 6 \; 4) (2 \; 3 \; 4 \; 5 \; 6)(3 \; 1)\big \}$. b The graph with the multiset of cycles $\big \{ (2) (1) (5 \; 6 \; 4) (2 \; 3 ) ( 4 \; 5 \; 6 \; 3 \; 1)\big \}$ (Color figure online)

Using Eq. (1.5), we can calculate several examples of the coefficient of $C_2^2$ of the Goulden–Rattan polynomials

$$\begin{aligned}{}[C_2^2] L_4&=0-0=0,\\ [C_2^2] L_5&=5-\frac{1}{3} \cdot 15=0,\\ [C_2^2] L_6&=0-0=0, \\ [C_2^2] L_7&=224-\frac{1}{3} \cdot 469=\frac{203}{3}, \\ [C_2^2] L_8&=0-0=0. \end{aligned}$$

Note that if k is even then $[C_2^2] L_k=0$ because there does not exist an expander with 4 vertices and an even number of edges, since $2-2g=2j-k+1$ by Eq. (1.4). Additionally, $[C_2^2]L_1=0$ and $[C_2^2]L_3=0$. Thus, we can assume that the number of edges k is odd and $k \ge 5$.

Let

$$\begin{aligned} X_k&= X_k(2, 2), \end{aligned}$$

(2.1)

$$\begin{aligned} Y_k&= Y_k(4). \end{aligned}$$

(2.2)

The set $X_k$ consists of expanders with 2 black vertices and 2 white vertices such that each black vertex is connected with both white vertices; each black vertex necessarily has weight equal to 1. The set $Y_k$ consists of expanders with one black vertex (which necessarily has weight 3) connected with all 3 white vertices. From now on we will omit the weights of the black vertices.

The main goal of this paper is to prove the following:

Theorem 5

The inequality

$$\begin{aligned} 3\big \Vert X_k\big \Vert \ge \big \Vert Y_k\big \Vert \end{aligned}$$

is true for any natural number k.

Fig. 4

An example of the single edge sliding for the number ${a=4}$ in the clockwise boundary direction from the cycle ${c=(2 \; 3 \; 4)}$ to the cycle ${c'=(1 \; 3)}$ along the number ${h=3}$. a The initial graph G. The edge labelled a is dashed and coloured red. b Visualization of the single edge sliding. c The resulting graph with two updated cycles: ${s(c)= \left[ (2 \; 3 \; 4) \circ \left( 3 \; 4\right) \right] {\setminus } \left( 4\right) = \left( 2 \; 3 \right) }$ and ${s'(c')= (3 \; 4) \circ \left[ \left( 1 \; 3\right) \; \left( 4 \right) \right] = (1 \; 4 \; 3)}$ (Color figure online)

3 Edge Sliding

In this section, we provide some necessary background details for the proof of Theorem 5. We will denote a transposition exchanging a and h by $(a \; h)$. For any set $H \subset \{1, 2, \dots \}$ such that $a, h \in H$, we can treat the transposition $(a \; h)$ as a permutation of the set H. (By adding fixed points to the transposition.) Therefore, for any permutation $\pi$ of the set H, the products $\pi \circ (a \; h)$ and $(a \; h) \circ \pi$ are also permutations of the set H.

Let G be a graph with edges numbered $1, \dots , k$ drawn without intersections on an oriented and connected surface with minimal genus. There is a natural bijection between the graph G and a multiset of cycles $M_G$ (cyclic permutations of a subset of $\{1, \dots , k\}$) such that each number j belongs to exactly two cycles. More precisely, the counterclockwise angular cyclic order of edges at a given vertex of the graph G (i.e., the order of edges ending at this vertex around it) corresponds to a single cycle of $M_G$. Two examples of a graph with its multiset of cycles are shown in Fig. 3.

In this section, each end of every edge (or equivalently, each element of every cycle) will have one of three values assigned to it: clockwise direction, counterclockwise direction, or no direction. Note that the two ends of the same edge can be assigned different values. By default, we assume that a cycle element to which we have not assigned a direction has an assigned the value no direction.

3.1 Edge Sliding for a Single Number

Let $M_G$ be the multiset of cycles of a graph G. Let a belonging to a cycle $c \in M_G$ be a number with assigned the clockwise direction. (The number “a” from a cycle different from “c” and any number different from “a” from any cycle do not have a direction assigned to them.) Let $c' \ne c$ be the second cycle of the multiset $M_G$ containing the number $h=c^{-1}(a)$. We assume that the number a does not belong to the cycle $c'$. We define the single edge sliding for the number a in the clockwise boundary direction from the cycle c to the cycle $c'$ along the number h as the replacement of the cycles $c, c'$ in the multiset $M_G$ by two new cycles $s(c), s'(c')$ given by

$$\begin{aligned} s(c)&= \left[ c \circ \left( a \; h\right) \right] {\setminus } \left( a\right) ,\\ s'(c')&= (a \; h) \circ \left[ \left( a \right) \; c' \right] . \end{aligned}$$

Note that the product $c \circ \left( a \; h\right)$ consists of two cycles: the cycle (a) and the cycle formed from the cycle c by removing the number a. Therefore, the above transformation removes the number a from the cycle c and adds it to the cycle $c'$ before the number h. Finally, we change the direction of the number a to the counterclockwise direction. We can naturally think of such operations as sliding of edges in a graph. An example of the single edge sliding is shown in Fig. 4.

Fig. 5

An example of the package edge sliding for the numbers ${a_1=4}$, ${a_2=5}$, ${a_3=6}$ in the clockwise boundary direction from the cycle ${c=(2 \; 3 \; 4 \; 5 \; 6)}$ to the cycle ${c'=(1 \; 3)}$ along the number ${h=3}$. a The initial graph G. Edges numbered $a_1, a_2, a_3$ are dashed and coloured red. b Visualization of the package edge sliding. c The resulting graph with two updated cycles: ${s_3(c)= \left[ (2 \; 3 \; 4 \; 5 \; 6) \circ \left( 3 \; 4\right) \circ \left( 3 \; 5\right) \circ \left( 3 \; 6\right) \right] {\setminus } \left[ \left( 4\right) \left( 5\right) \left( 6\right) \right] = \left( 2 \; 3 \right) }$ and ${s'_3(c')= (3 \; 6) \circ (3 \; 5) \circ (3 \; 4) \circ \left[ \left( 1 \; 3\right) \; \left( 4 \right) \right] = (1 \; 4 \; 5 \; 6 \; 3)}$ (Color figure online)

Analogously, we define the single edge sliding for the number a in the counterclockwise boundary direction from the cycle c to the cycle $c'$ along the number $h=c(a)$ as the replacement of the cycles $c, c'$ in the multiset $M_G$ by two new cycles $s(c), s'(c')$ given by

$$\begin{aligned} s(c)&= \left[ \left( a \; h\right) \circ c \right] {\setminus } \left( a\right) ,\\ s'(c')&= \left[ \left( a \right) \; c'\right] \circ (a \; h). \end{aligned}$$

3.2 Edge Sliding for a Sequence of Numbers

Let $M_G$ be the multiset of cycles of a graph G. Let $a_1, \dots , a_l$ be a sequence of successive but not all numbers belonging to a cycle $c\in M_G$, i.e. such that $a_j=c^{j-1}(a_1)$ for each index $j \le l$. Moreover, to each of the numbers $a_1, \dots , a_l$ is assigned the clockwise direction. Let $c' \ne c$ be the second cycle of the multiset of $M_G$ containing the number $h=c^{-1}(a_1)$. We assume that the numbers $a_1, \dots , a_l$ do not belong to the cycle $c'$. We define the package edge sliding for the sequence of the numbers $a_1, \dots , a_l$ in the clockwise boundary direction from the cycle c to the cycle $c'$ along the number h as the replacement of the cycles $c, c'$ in the multiset $M_G$ by two new cycles $s_l(c), s'_l(c')$ obtained by recursion with initial conditions $s_0(c)=c, s'_0(c')=c'$ and a recursive step given for each $1 < j \le l$ by

$$\begin{aligned} s_{j}(c)&= \left[ s_{j-1}(c) \circ \left( a_j \; h\right) \right] {\setminus } \left( a_j \right) ,\\ s'_{j}(c')&= (a_j \; h) \circ \left[ \left( a_j \right) \; s'_{j-1}(c') \right] . \end{aligned}$$

Note that the package edge sliding for a sequence of numbers $a_1, \dots , a_l$ is actually equivalent to the sequential single edge sliding for numbers $a_1, \dots , a_l$. Therefore, the above transformation removes the numbers $a_1, \dots a_l$ from the cycle c and adds them in the same order to the cycle $c'$ before the number h. Finally, we change the directions of the numbers $a_1, \dots , a_l$ to the counterclockwise boundary direction. Note that the single edge sliding is a special case of the package edge sliding. An example of the package edge sliding is shown in Fig. 5.

Analogously, we define the package edge sliding for the numbers $a_1, \dots , a_l$ in the counterclockwise boundary direction from the cycle c to the cycle $c'$ along the number $h=c(a)$ as the replacement of the cycles $c, c'$ in the multiset $M_G$ by two new cycles $s_l(c), s_l(c')$ obtained by recursion with initial conditions $s_0(c)=c, s'_0(c')=c'$ and a recursive step given for each $0 < j \le l$ by

$$\begin{aligned} s_{j}(c)&= \left[ \left( a_j \; h\right) \circ s_{j-1}(c) \right] {\setminus } \left( a_j\right) ,\\ s'_{j}(c')&= \left[ \left( a_j \right) \; s'_{j-1} (c') \right] \circ (a_j \; h). \end{aligned}$$

In other words, if we would like to slide a number a in the clockwise boundary direction from a cycle c along a number h, then $c^{-1}(a)=h$ or before that, a number $c^{-1}(a)$ must be slid in the same direction. (Similarly for the counterclockwise boundary direction and number c(a).) For each number a in a cycle c with a fixed direction, the number h along which it will be slid is uniquely determined. Therefore, for simplicity, we will say that each of the numbers $a_1, \dots , a_l$ from the cycle c is slid in a fixed direction. Of course, we assume that at least one number in the cycle c has no direction assigned.

3.3 Edge Sliding in the General Case

We define the edge sliding on a graph as follows. We start from the graph G in which some ends of certain edges are assigned a clockwise or counterclockwise boundary direction. We assume that:

• At each vertex at least one edge has no direction assigned.

• Two consecutive edges do not have conflicting directions, i.e., there is no situation in which a number a from a cycle c has assigned the counterclockwise boundary direction and the number c(a) has assigned the clockwise boundary direction.

Any such set of directions can be decomposed into an package edge sliding system for sequences of numbers:

Furthermore, we require that:

• The numbers, along which other numbers are sliding, are not themselves sliding, i.e.,

  $$\begin{aligned} \{ a_{1,1}, \dots , a_{t, l_t}\} \cap \{ h_1, \dots , h_t \} = \emptyset . \end{aligned}$$

• Two ends of the same edge will not appear in the same vertex, i.e., if $a_{i_1, j_1}=a_{i_2, j_2}$, then

  $$\begin{aligned} \{ c_{i_1} , {c'}_{i_1} \} \cap \{ c_{i_2} , {c'}_{i_2} \} = \emptyset . \end{aligned}$$

  (This condition can be weakened to ${c'}_{i_1} \ne {c'}_{i_2}$, but it is not necessary in this paper.)

• On one side of the edge along which we slide, the numbers do not slide in opposite directions, i.e., for each number $h_j$ belonging to the cycles $c \ne c'$, there is no situation in which a number $c(h_j)$ has assigned the clockwise direction and the number ${c'}^{-1}(h_j)$ has assigned the counterclockwise direction.

We will call the selection of directions that satisfy the above conditions as correct.

We define the edge sliding as applying sequentially the package edge sliding for all sequences in any order. Such an action is well-defined, since permutation multiplication is associative and for any distinct numbers $a_1, a_2, h_1, h_2$ holds

$$\begin{aligned} (a_1 \; h_1) \circ (a_2 \; h_2) = (a_2 \; h_2) \circ (a_1 \; h_1) . \end{aligned}$$

An example of the edge sliding is shown in Fig. 6.

Fig. 6

An example of edge sliding. a A graph with the slid edges dashed and coloured red. The directions of the edge ends are indicated by arrows. b The graph during the edge sliding in the clockwise boundary direction. c The graph after the edge sliding. Directions have already been reversed (Color figure online)

The edge sliding is an involution on the set of graphs drawn on an oriented and connected surface with a correct selection of directions. Edge sliding is an invertible transformation, with the inverse also given by edge sliding.

In addition, it is easy to see that the edge sliding on a graph does not change the number of faces of this graph.

In the rest of this article, we will treat the edge sliding as transformation on a graph.

3.4 The Set $X_k$ of Maps

We consider any map from the set $X_k$. This map has one face and an odd number of edges $k \ge 5$. We denote the black vertices by $b_1, b_2$ and the white vertices by $w_1, w_2$. There is at least one edge between each pair of the vertices of different colours. Of course, ${{\,\textrm{deg}\,}}(b_1)+{{\,\textrm{deg}\,}}(b_2)=k$ is an odd number. Without loss of generality we may assume that ${{\,\textrm{deg}\,}}(b_1)>0$ is an odd number and ${{\,\textrm{deg}\,}}(b_2)>0$ is an even number. Let $k_1, k_2 > 0$ denote the numbers of edges which connect the vertex $b_1$ with the vertices $w_1, w_2$, respectively. As ${{\,\textrm{deg}\,}}(b_1)=k_1+k_2$ is an odd number, without loss of generality we may assume that $k_1$ is even and $k_2$ is odd. For example, the unique (up to choice of the root) map from the set $X_5$ is shown in Fig. 7a.

Fig. 7

Examples of maps from a the set $X_5$, b the set $T_5^{odd}$, c the set $T_5^{even}$, d the set $T_5^{rest}$. Slid edges are dashed and coloured. The directions are indicated by arrows, and the root is not marked (Color figure online)

3.5 The Set $Y_k$ of Maps

Let $\sigma$ be the cycle that encodes the clockwise boundary cyclic order of the corners on the unique face of G. We will say that the vertex $w_j$ is a descendant of the vertex $w_i$ (we denote it by $w_i \rightarrow w_j$) if using the clockwise boundary order of the corners on the unique face of the map we can move (by walking along the edges and holding them with the right hand) in two steps from a certain corner $c_i$ of the vertex $w_i$ to a certain corner $c_j$ of the vertex $w_j$, i.e., $\sigma ^2(c_i)=c_j$.

We consider any map from the set $Y_k$. Any such map has one face and an odd number of edges $k \ge 5$. We denote the black vertex by b and the white vertices by $w_1, w_2, w_3$. We will write the set $Y_k$ as a union of three sets which will be defined below.

Let $Y_{k}^{{{\,\textrm{odd}\,}}}\subseteq Y_k$ be the set of maps for which there exists an odd degree white vertex (let us say it is $w_3$) which has the other two white vertices as descendants, i.e., $w_3\rightarrow w_1$ and $w_3\rightarrow w_2$. Let $T_k^{{{\,\textrm{odd}\,}}}$ be the set of all maps from the set $Y_k^{{{\,\textrm{odd}\,}}}$ with a distinguished vertex $w_3$ with this property. Moreover, each edge between the vertex $w_3$ and the vertex b, has a clockwise boundary direction assigned at the vertex b. The unique (up to choice of the root) map from the set $T_{5}^{{{\,\textrm{odd}\,}}}$ is shown in Fig. 7b. Clearly

$$\begin{aligned} |T_{k}^{{{\,\textrm{odd}\,}}}| \ge |Y_{k}^{{{\,\textrm{odd}\,}}}|. \end{aligned}$$

(3.1)

Let $Y_{k}^{{{\,\textrm{even}\,}}}\subseteq Y_k$ be the set of maps such that there exists an even degree white vertex (let us say it is $w_3$) which has the other two white vertices as descendants, i.e., $w_3\rightarrow w_1$ and $w_3\rightarrow w_2$. Let $T_k^{{{\,\textrm{even}\,}}}$ be the set of all the maps from the set $Y_k^{{{\,\textrm{even}\,}}}$ with a distinguished vertex $w_3$ with this property. Moreover, each edge between the vertex $w_3$ and the vertex b, has a clockwise boundary direction assigned at the vertex b. The unique (up to choice of the root) map from the set $T_{5}^{{{\,\textrm{even}\,}}}$ is shown in Fig. 7c. Clearly

$$\begin{aligned} |T_{k}^{{{\,\textrm{even}\,}}}| \ge |Y_{k}^{{{\,\textrm{even}\,}}}|. \end{aligned}$$

(3.2)

Let $Y_{k}^{{{\,\textrm{rest}\,}}}\subseteq Y_k$ be the set of maps not included in the sets $Y_{k}^{{{\,\textrm{odd}\,}}}$ and $Y_{k}^{{{\,\textrm{even}\,}}}$, i.e.,

$$\begin{aligned} Y_{k}^{{{\,\textrm{rest}\,}}}=Y_k {\setminus } (Y_{k}^{{{\,\textrm{odd}\,}}} \cup Y_{k}^{{{\,\textrm{even}\,}}}). \end{aligned}$$

(3.3)

Consider some map $m\in Y_{k}^{{{\,\textrm{rest}\,}}}$. Obviously $w_1 \rightarrow w_2 \rightarrow w_3 \rightarrow w_1$ or the other way around. Without loss of generality we may assume that $w_1 \rightarrow w_2 \rightarrow w_3 \rightarrow w_1$ and as a consequence $w_1 \nleftarrow w_2 \nleftarrow w_3 \nleftarrow w_1$.

Lemma

The map m has a white vertex of odd degree, greater than 1.

Proof

By contradiction, suppose this is not the case. The map m has at least one odd degree white vertex, because ${{\,\textrm{deg}\,}}(w_1)+{{\,\textrm{deg}\,}}(w_2)+{{\,\textrm{deg}\,}}(w_3)=k$ is odd. Without loss of generality we may assume that ${{\,\textrm{deg}\,}}(w_1)$ is odd. Since

$$\begin{aligned} {{\,\textrm{deg}\,}}(w_1)+{{\,\textrm{deg}\,}}(w_2)+{{\,\textrm{deg}\,}}(w_3)=k>3= 1+1+1, \end{aligned}$$

it follows that ${{\,\textrm{deg}\,}}(w_1)=1$ and ${{\,\textrm{deg}\,}}(w_2),{{\,\textrm{deg}\,}}(w_3)$ are even, because m does not have a white vertex with odd degree greater than 1. The vertex $w_1$ is a leaf and thus has a unique corner which we denote by $c_1$.

Naturally $\sigma ^2(c_1)$ is a corner of the vertex $w_2$. Note that $\sigma ^2$ is a permutation of the corners of the white vertices which has only one cycle, because the map m has only one face. The corners of the white vertices can be labelled 1, 2, 3 according to the names of the vertices they are in. If a corner c has the label a, its descendant $\sigma ^2(c)$ has either the label a or $1+a {\text {mod}} 3$. There is only one corner which has the label 1, so the corner labels of m (arranged in the cyclic order according to the unique cycle of $\sigma ^2$) are $(1, 2, \ldots , 2, 3, \ldots , 3)$. Since there exists only one corner $c_2$ of the white vertex $w_2$ such that $\sigma ^2(c_2)$ is a corner of the vertex $w_3$, then there exists a unique corner $c_0=\sigma (c_2)$ of the vertex b such that $\sigma (c_0)$ is the corner of the vertex $w_3$ and $\sigma ^{-1}(c_0)$ is the corner of the vertex $w_2$. Thus the clockwise angular cyclic order of the edges around the black vertex b is as follows: one edge connected to the vertex $w_1$, a certain number of edges connected to the vertex $w_2$, a certain number of edges connected to the vertex $w_3$. Figure 8a visualizes this situation.

Let $k'=k-\left( {{\,\textrm{deg}\,}}(w_2)-1\right)$. The number $k'$ is even because k is odd and ${{\,\textrm{deg}\,}}(w_2)$ is even. We remove all edges except one of the white vertex $w_2$ obtaining a new graph $m'$ with $k'$ edges. Obviously, the clockwise angular cyclic order of the edges around the black vertex b is as follows: one edge connected to the vertex $w_1$, one edge connected to the vertex $w_2$, a certain number of edges connected to the vertex $w_3$. Whereas the corner labels of $m'$ are $(1, 2, 3, \ldots , 3)$. Therefore, $m'$ has one face. Figure 8b visualizes this situation. If $m'$ does not have selected a root, we choose any edge of $m'$ as the root. If the genus of the surface on which the map $m'$ is drawn is not minimal, we draw the map $m'$ on a surface with minimal genus. As a result, we obtain an expander from the set $Y_{k'}$ with 4 vertices, one face and an even number of edges $k'$. We get a contradiction because such a map does not exist (see Eq. (1.4)). Therefore, the map m has a white vertex with an odd degree greater than 1. $\square$

Fig. 8

a The map m before replacement including the marked corner $c_0$. b A non-existent hypothetical map with four vertices and an even number of edges (Color figure online)

Now, we fix the directions. To all ends in the vertex b of the edges between the vertex $w_3$ and b, we assign the direction in such way that among them there is an even number with the clockwise boundary direction and an odd number with the counterclockwise boundary direction. This is always possible, e.g. for a single edge with the counterclockwise boundary direction.

Let $T_k^{{{\,\textrm{rest}\,}}}$ be the set of all the maps from the set $Y_k^{{{\,\textrm{rest}\,}}}$ with a distinguished white vertex denoted by $w_3$ with a fixed choice of the set of special edges together with the directions of their ends satisfying the conditions just mentioned above. The unique (up to choice of the root) example of the map from the set $T_{5}^{{{\,\textrm{rest}\,}}}$ is shown in Fig. 7d. Clearly

$$\begin{aligned} |T_{k}^{{{\,\textrm{rest}\,}}}| \ge |Y_{k}^{{{\,\textrm{rest}\,}}}|. \end{aligned}$$

(3.4)

4 Proof of Main Result

In this section, we will construct three bijections which show the cardinalities of the corresponding sets are equal. Using these equalities and the definitions of these sets we will prove Theorem 5.

4.1 Three Bijections

The first bijection between $X_k$ and $T_k^{{{\,\textrm{odd}\,}}}$. We start from a map $m \in X_{k}$. Recall that we have assumed that ${{\,\textrm{deg}\,}}(b_1)$ is odd. All edges connecting a vertex $b_1$ to white vertices, have the counterclockwise boundary direction assigned at the vertices $w_1, w_2$. The choice of directions is correct because m is a bipartite graph. We apply the edge sliding to the map m. Then we change the colour of the black vertex $b_1$ to white and its name to $w_3$, and the name of the vertex $b_2$ to b. Of course, the degree of the vertex $w_3$ does not change and is odd. In addition, $w_3\rightarrow w_1$ and $w_3\rightarrow w_2$, because any map from the set $X_{k}$ has at least one edge between each pair of the vertices of different colours. We obtain a map from the set $T_{k}^{{{\,\textrm{odd}\,}}}$. (At all times one of the edges is selected as the root.) Moreover, each map from the set $T_{k}^{{{\,\textrm{odd}\,}}}$ can be produced in this way. Such a transformation is a bijection between the set $X_k$ and the set $T_k^{{{\,\textrm{odd}\,}}}$, since the edge sliding is reversible. Figure 9 shows an example of this bijection for $k=5$. Thus,

$$\begin{aligned} |X_k|=|T_k^{{{\,\textrm{odd}\,}}}|. \end{aligned}$$

(4.1)

Fig. 9

The example of the first bijection for the 5-edged map. a The map from the set $X_{5}$. b The map during the edge sliding. c The map after the edge sliding. (d) The map from the set $Y_{5}^{{{\,\textrm{odd}\,}}}$ (Color figure online)

The second bijection between $X_k$ and $T_k^{{{\,\textrm{even}\,}}}$. We start from a map $m \in X_{k}$. Recall that we have assumed that ${{\,\textrm{deg}\,}}(b_2)$ is even. All edges connecting a vertex $b_1$ to white vertices, have the counterclockwise boundary direction assigned at the vertices $w_1, w_2$. The choice of directions is correct because m is a bipartite graph. We apply the edge sliding to the map m. Then we change the colour of the black vertex $b_2$ to white and its name to $w_3$, and the name of the vertex $b_1$ to b. Of course, the degree of the vertex $w_3$ does not change and is even. In addition, $w_3\rightarrow w_1$ and $w_3\rightarrow w_2$, because any map from the set $X_{k}$ has at least one edge between each pair of the vertices of different colours. We obtain a map from the set $T_{k}^{{{\,\textrm{even}\,}}}$. (At all times one of the edges is selected as the root.) Moreover, each map from the set $T_{k}^{{{\,\textrm{even}\,}}}$ can be produced. Such a transformation is a bijection between the set $X_k$ and the set $T_k^{{{\,\textrm{even}\,}}}$, since the edge sliding is reversible. Figure 10 shows an example of this bijection for $k=5$. Thus,

$$\begin{aligned} |X_k|=|T_k^{{{\,\textrm{even}\,}}}|. \end{aligned}$$

(4.2)

Fig. 10

a–d The example of the second bijection for the 5-edged map (Color figure online)

The third bijection. We start from a map $m \in X_{k}$. Recall that we have assumed that ${{\,\textrm{deg}\,}}(b_1)$ is odd. All edges connecting a vertex $b_1$ to white vertices, have the counterclockwise boundary direction assigned at the vertex $w_1$ and the clockwise boundary direction assigned at the vertex $w_2$. The choice of directions is correct because m is a bipartite graph. We apply the edge sliding to the map m. Then we change the colour of the black vertex $b_1$ to white and its name to $w_3$, and the name of the vertex $b_2$ to b. Of course, the degree of the vertex $w_3$ does not change and is odd. In addition, $w_3\rightarrow w_1$ and $w_2\rightarrow w_3$ (and $w_1\rightarrow w_2$), because any map from the set $X_{k}$ has at least one edge between each pair of the vertices of different colours. We do not necessarily obtain a map from the set $T_{k}^{{{\,\textrm{rest}\,}}}$ (it may be that we obtain a map from set $T_{k}^{{{\,\textrm{odd}\,}}}$), but it can be seen that each map from the set $T_{k}^{{{\,\textrm{rest}\,}}}$ can be produced. (At all times one of the edges is selected as the root.) Such a transformation is a bijection between the set $X_k$ and some superset of the set $T_k^{{{\,\textrm{odd}\,}}}$, since the edge sliding is reversible. Figure 11 shows an example of this bijection for $k=5$. Thus

$$\begin{aligned} |X_k| \ge |T_k^{{{\,\textrm{rest}\,}}}|. \end{aligned}$$

(4.3)

Fig. 11

a–d The example of the third bijection for the 5-edged map (Color figure online)

4.2 The Conclusion of the Proof

We can now proceed to the proof of Theorem 5, we have:

$$\begin{aligned} 3[C_2^2]L_k&=3[R_2^2]K_k-[R_4]K_k \qquad \quad \qquad \ \ {}{} & {} \text {by }(1.5)\\&=3|X_k|-|Y_k| \qquad \qquad \ {}{} & {} \text {by }(2.1), (2.2)\\&\ge |T_k^{{{\,\textrm{odd}\,}}}|+|T_k^{{{\,\textrm{even}\,}}}|+|T_k^{{{\,\textrm{rest}\,}}}|-|Y_k| \quad{} & {} \text {by }(4.1), (4.2), (4.3)\\&\ge |Y_k^{{{\,\textrm{odd}\,}}}|+|Y_k^{{{\,\textrm{even}\,}}}|+| Y_k^{{{\,\textrm{rest}\,}}}|-|Y_k|\quad{} & {} \text {by }(3.1), (3.2), (3.4)\\&=|Y_k^{{{\,\textrm{odd}\,}}} \cap Y_k^{{{\,\textrm{even}\,}}}| \qquad \qquad \quad \qquad \qquad{} & {} \text {by }(3.3)\\&\ge 0. \end{aligned}$$

$\square$