2 Main result 1

In this section, we always consider $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ counted with multiplicity and deal with improvements to the algorithm for finding $\#(\Phi _d^{-1}(\bar {\unicode{x3bb} }))$ . We first fix our notation.

For $d \ge 2$ , we put

$$ \begin{align*} \mathrm{Poly}_d := \{f \in \mathbb{C}[z] \mid \deg f = d \} \quad \textrm{and} \quad \mathrm{Aut}(\mathbb{C}) := \{\gamma (z) = az+b \mid a,b \in \mathbb{C},\ a \ne 0 \}. \end{align*} $$

Since $\gamma \in \mathrm {Aut}(\mathbb {C})$ naturally acts on $f \in \mathrm {Poly}_d$ by $\gamma \cdot f := \gamma \circ f \circ \gamma ^{-1}$ , we can define its quotient $\mathrm {MP}_d := \mathrm {Poly}_d / \mathrm {Aut}(\mathbb {C})$ , which we usually call the moduli space of polynomial maps of degree d. We put $\mathrm {Fix}(f) := \{ z \in \mathbb {C} \mid f(z)=z\}$ for $f \in \mathrm {Poly}_d$ , where $\mathrm {Fix}(f)$ is considered counted with multiplicity. Hence, we always have $\# (\mathrm {Fix}(f)) = d$ . Since the unordered collection of fixed-point multipliers $(f'(\zeta ))_{\zeta \in \mathrm {Fix}(f)}$ of $f \in \mathrm {Poly}_d$ is invariant under the action of $\mathrm {Aut}(\mathbb {C})$ , we can naturally define the map $\Phi _d:\mathrm {MP}_d \to \mathbb {C}^{d}/\mathfrak {S}_{d}$ by $\Phi _d(f):=(f'(\zeta ))_{\zeta \in \mathrm {Fix}(f)}$ . Here, $\mathfrak {S}_{d}$ denotes the dth symmetric group which acts on $\mathbb {C}^{d}$ by the permutation of coordinates. Note that a fixed point $\zeta \in \mathrm {Fix}(f)$ is multiple if and only if $f'(\zeta )=1$ .

By the fixed point theorem for polynomial maps, we always have $\sum _{\zeta \in \mathrm {Fix}(f)} {1}/({1-f'(\zeta )}) = 0$ for $f \in \mathrm {Poly}_d$ if f has no multiple fixed point. (See §12 in [Reference Milnor11] or Proposition 1.1 in [Reference Sugiyama14] for more details.) Hence, putting $\Lambda _d :=\{(\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _d)\in \mathbb {C}^d \mid \sum _{i=1}^d \prod _{j\ne i} ( 1- \unicode{x3bb} _j ) = 0 \}$ and $\widetilde {\Lambda }_d := \Lambda _d / \mathfrak {S}_d$ , we have the inclusion relation $\Phi _d(\mathrm {MP}_d) \subseteq \widetilde {\Lambda }_d \subseteq \mathbb {C}^{d}/\mathfrak {S}_{d}$ . We therefore have the map

$$ \begin{align*} \Phi_d : \mathrm{MP}_d \to \widetilde{\Lambda}_d \end{align*} $$

by $f \mapsto (f'(\zeta ) )_{\zeta \in \mathrm {Fix}(f)}$ , which is the main object of our study.

In this paper, we again restrict our attention to the map $\Phi _d$ on the domain where polynomial maps have no multiple fixed points, that is, on the domains

$$ \begin{align*} V_d := \{ (\unicode{x3bb}_1,\ldots ,\unicode{x3bb}_d) \in \Lambda_d \mid \unicode{x3bb}_i \ne 1 \textrm{ for every } 1\le i\le d \} \quad \textrm{and}\quad \widetilde{V}_d := V_d/ \mathfrak{S}_d, \end{align*} $$

which are Zariski open subsets of $\Lambda _d$ and $\widetilde {\Lambda }_d$ , respectively. Here, note that we also have

$$ \begin{align*} V_d = \bigg\{ (\unicode{x3bb}_1,\ldots ,\unicode{x3bb}_d) \in \mathbb{C}^d\, \bigg|\, \unicode{x3bb}_i \ne 1 \textrm{ for every } 1\le i\le d, \sum_{i=1}^d \frac{1}{1-\unicode{x3bb}_i} = 0 \bigg\}. \end{align*} $$

Throughout this paper, we always denote by $\bar {\unicode{x3bb} }$ the equivalence class of $\unicode{x3bb} \in \Lambda _d$ in $\widetilde {\Lambda }_d$ , that is, $\bar {\unicode{x3bb} }=\textit {pr}(\unicode{x3bb} )$ , where $\textit {pr}: \Lambda _d \to \widetilde {\Lambda }_d$ denotes the canonical projection. Hence, for $\unicode{x3bb} = (\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _d) \in \Lambda _d$ , we sometimes express $\bar {\unicode{x3bb} } = \{ \unicode{x3bb} _1,\ldots , \unicode{x3bb} _d \} \in \widetilde {\Lambda }_d$ . We never denote by $\bar {\unicode{x3bb} }$ the complex conjugate of $\unicode{x3bb} $ in this paper.

The objects defined in the following definition play a central roll in [Reference Sugiyama14] and also in this paper.

Definition 2.1. For $\unicode{x3bb} = (\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _d) \in V_d$ , we put

$$ \begin{align*} \mathfrak{I}(\unicode{x3bb}) := \left\{ \{I_1,\ldots,I_l\}\ \left| \ \begin{matrix} l \ge 2,\ \ I_1 \amalg \cdots \amalg I_l = \{1,\ldots,d \},\\ I_u \ne \emptyset \textrm{ for every } 1\le u\le l,\\ \sum_{i \in I_u} 1/(1-\unicode{x3bb}_i)=0 \textrm{ for every } 1\le u\le l \end{matrix} \right.\right\}\hspace{-2.5pt}, \end{align*} $$

where $I_1 \amalg \cdots \amalg I_l$ denotes the disjoint union of $I_1,\ldots ,I_l$ . By definition, each element of $\mathfrak {I}(\unicode{x3bb} )$ is considered to be a partition of $\{1,\ldots ,d \}$ . The partial order $\prec $ in $\mathfrak {I}(\unicode{x3bb} )$ is defined by the refinement of partitions, namely, for $\mathbb {I}, \mathbb {I}'\in \mathfrak {I}(\unicode{x3bb} )$ , the relation $\mathbb {I} \prec \mathbb {I}'$ holds if and only if $\mathbb {I}'$ is a refinement of $\mathbb {I}$ as partitions of $\{1,\ldots ,d \}$ .

For $\unicode{x3bb} \in V_d$ and for $I \in \mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ , we put $\unicode{x3bb} _I := (\unicode{x3bb} _i)_{i \in I}$ .

In the above definition, note that the condition $I \in \mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ for I is equivalent to the conditions $\emptyset \subsetneq I \subsetneq \{1,\ldots ,d \}$ and $\sum _{i \in I} 1/(1-\unicode{x3bb} _i)=0$ . Hence, we always have $\unicode{x3bb} _I \in V_{\# I}$ for $\unicode{x3bb} \in V_d$ and $I \in \mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ by definition. Also note that $\# I \geq 2$ holds for every $I \in \mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ .

The following object is also very important in this paper.

Definition 2.2. For $\unicode{x3bb} \in V_d$ , we put

$$ \begin{align*} \mathfrak{I}'(\unicode{x3bb}) := \mathfrak{I}(\unicode{x3bb}) \cup \{\{\{ 1, \ldots, d\}\}\}. \end{align*} $$

The partial order $\prec $ in $\mathfrak {I}(\unicode{x3bb} )$ is naturally extended to the partial order $\prec $ in $\mathfrak {I}'(\unicode{x3bb} )$ .

By definition, $\mathfrak {I}'(\unicode{x3bb} )$ is obtained from $\mathfrak {I}(\unicode{x3bb} )$ by adding exactly one element $\mathbb {I}_0 := \{\{ 1, \ldots , d\}\}$ . Here, $\mathbb {I}_0$ is the unique minimum element of $\mathfrak {I}'(\unicode{x3bb} )$ with respect to the partial order $\prec $ . Moreover, $\mathbb {I}_0$ is considered to be a partition of $\{ 1, \ldots , d \}$ which, in practice, does not partition $\{ 1, \ldots , d \}$ . We also have the equality

$$ \begin{align*} \mathfrak{I}'(\unicode{x3bb}) = \left\{ \{I_1,\ldots,I_l\}\ \left| \ \begin{matrix} l \ge 1, I_1 \amalg \cdots \amalg I_l = \{1,\ldots,d \},\\ I_u \ne \emptyset \textrm{ for every } 1\le u\le l,\\ \sum_{i \in I_u} 1/(1-\unicode{x3bb}_i)=0 \textrm{ for every } 1\le u\le l \end{matrix} \right.\right\}\hspace{-2.5pt}. \end{align*} $$

We already have the following theorem by Main Theorem III and Remark 1.8 in [Reference Sugiyama14] and by Theorem B and Proposition C in §6 in [Reference Sugiyama14].

Theorem 2.3. We can define the non-negative integer $e_{\mathbb {I}}(\unicode{x3bb} )$ for each $d \geq 4$ , $\unicode{x3bb} \in V_d$ , and $\mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ , and can also define the non-negative integer $s_d(\unicode{x3bb} )$ for each $d \geq 2$ and $\unicode{x3bb} \in V_d$ inductively by the equalities

(2.1) $$ \begin{align} s_d(\unicode{x3bb}) = (d-2)! - \sum_{\mathbb{I} \in \mathfrak{I}(\unicode{x3bb})} \bigg( e_{\mathbb{I}}(\unicode{x3bb})\cdot \prod_{k=d-\# \mathbb{I} +1}^{d-2}k \bigg) \end{align} $$

for $d \geq 2$ and $\unicode{x3bb} \in V_d$ , and

(2.2) $$ \begin{align} e_{\mathbb{I}}(\unicode{x3bb}) = \prod_{I \in \mathbb{I}} ( ( \# I -1 ) \cdot s_{\# I}(\unicode{x3bb}_{I}) ) \end{align} $$

for $d \geq 4$ , $\unicode{x3bb} \in V_d$ , and $\mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ . Here, in the case $\# \mathbb {I} = 2$ , we put $\prod _{k=d-\# \mathbb {I} +1}^{d-2}k = \prod _{k=d-1}^{d-2}k = 1$ .

If we consider $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ ‘counted with multiplicity’ for $d \geq 2$ and $\unicode{x3bb} \in V_d$ , then we have

$$ \begin{align*} \#(\Phi_d^{-1}(\bar{\unicode{x3bb}})) = s_d(\unicode{x3bb}). \end{align*} $$

In the rest of this paper, we always assume that $e_{\mathbb {I}}(\unicode{x3bb} )$ and $s_d(\unicode{x3bb} )$ are the non-negative integers defined in Theorem 2.3.

We already made a minor improvement to the above algorithm by Main Theorem III in [Reference Sugiyama14] and by Proposition D in §6 in [Reference Sugiyama14], as in the following.

Theorem 2.5. The non-negative integer $e_{\mathbb {I}}(\unicode{x3bb} )$ for $\unicode{x3bb} \in V_d$ and $\mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ defined in Theorem 2.3 also satisfies the equality

(2.3) $$ \begin{align} e_{\mathbb{I}}(\unicode{x3bb}) = \bigg(\prod_{I \in \mathbb{I}} ( \# I -1 )! \bigg) - \sum_{{{\begin{matrix} \mathbb{I}' \in \mathfrak{I}(\unicode{x3bb}) \\ \mathbb{I}' \succ \mathbb{I}, \; \mathbb{I}' \ne \mathbb{I} \end{matrix}}}} \bigg( e_{\mathbb{I}'}(\unicode{x3bb}) \cdot \prod_{I \in \mathbb{I}} \bigg( \prod_{k=\# I - \chi_I(\mathbb{I}')+1 }^{\# I -1}k \bigg) \bigg), \end{align} $$

where we put $\chi _I(\mathbb {I}'):=\#(\{ I'\in \mathbb {I}' \mid I' \subseteq I \})$ for $\mathbb {I}' \succ \mathbb {I}$ and $I \in \mathbb {I}$ . Here, in the case $\chi _I(\mathbb {I}') = 1$ , we put $\prod _{k=\# I - \chi _I(\mathbb {I}')+1 }^{\# I -1}k = \prod _{k=\# I}^{\# I -1}k = 1$ .

Remark 2.8. We present a rough outline of the proof of Theorem 2.5 in this remark, since the proof can be an easy exercise for the proof of Theorem I in this paper. (See ‘Proof of Proposition D’ on pp. 175–177 in [Reference Sugiyama14] for details.) In the case where $d = \# I$ and $\unicode{x3bb} = \unicode{x3bb} _I$ , equation (2.1) is equivalent to the following:

(2.4) $$ \begin{align} ( \# I - 1 )! = ( \# I - 1 ) s_{\# I}(\unicode{x3bb}_I) + \sum_{\mathbb{I} \in \mathfrak{I}(\unicode{x3bb}_I)} \bigg( e_{\mathbb{I}}(\unicode{x3bb}_I)\cdot \prod_{k=\#I -\# \mathbb{I}+1}^{\#I - 1}k \bigg). \end{align} $$

Plugging equation (2.4) into $\prod _{I \in \mathbb {I}} ( \# I -1 )!$ and using equation (2.2) carefully, we have equation (2.3).

In this paper, we make a drastic improvement to the above algorithm as in the following.

Theorem I. The non-negative integer $s_d(\unicode{x3bb} )$ for $d \geq 2$ and $\unicode{x3bb} \in V_d$ defined in Theorem 2.3 is expressed in the form

(2.5) $$ \begin{align} (d-1)s_d(\unicode{x3bb}) = \sum_{\mathbb{I}\in \mathfrak{I}'(\unicode{x3bb})} \bigg( \{ -(d-1) \}^{\#\mathbb{I} - 1} \cdot \prod_{I \in \mathbb{I}}( \#I - 1 )!\!\bigg). \end{align} $$

Hence, if we consider $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ ‘counted with multiplicity’ for $d \geq 2$ and $\unicode{x3bb} \in V_d$ , then we have

(2.6) $$ \begin{align} \#(\Phi_d^{-1}(\bar{\unicode{x3bb}})) = -\sum_{\mathbb{I}\in \mathfrak{I}'(\unicode{x3bb})} \bigg( \{ -(d-1) \}^{\#\mathbb{I} - 2} \cdot \prod_{I \in \mathbb{I}}( \#I - 1 )!\!\bigg). \end{align} $$

Theorem I is proved in §4.

3 Main result 2

In this section, we proceed to the next step, in which we discuss the possibility of improving the algorithm for counting the number of discrete elements of $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ . Therefore, in this section, $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ is not considered counted with multiplicity; $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ is considered to be a set. In this setting, we have already obtained an algorithm for counting the number of discrete elements of $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ by using $\{s_{d'}(\unicode{x3bb} ') \mid 2 \leq d' \leq d,\ \unicode{x3bb} ' \in V_{d'} \}$ in the third and fourth steps in Main Theorem III in [Reference Sugiyama14]. To review the result more precisely and to discuss further properties, we first fix our notation.

The following objects are important in this section.

Definition 3.1. For $\unicode{x3bb} = (\unicode{x3bb} _1, \ldots , \unicode{x3bb} _d) \in V_d$ , we put

$$ \begin{align*} \mathfrak{K}(\unicode{x3bb}) := \left\{K \ \left| \ \begin{matrix} \emptyset \subsetneq K \subseteq \{1,\ldots,d \},\\ i,j \in K \Rightarrow \unicode{x3bb}_i = \unicode{x3bb}_j,\\ i\in K,\ j\in\{1,\ldots,d\}\setminus K\Longrightarrow\unicode{x3bb}_i\ne\unicode{x3bb}_j \end{matrix} \right.\right\}\hspace{-2.5pt}. \end{align*} $$

Note that if we put $\mathfrak {K}(\unicode{x3bb} ) =: \{ K_1,\ldots , K_q \}$ , then $K_1,\ldots , K_q$ are mutually disjoint, and the equality $K_1 \amalg \cdots \amalg K_q = \{1,\ldots ,d\}$ holds by definition; and hence $\mathfrak {K}(\unicode{x3bb} )$ is a partition of $\{1,\ldots ,d\}$ .

Definition 3.2. We denote the family of monic centered polynomials of degree d by

$$ \begin{align*} \mathrm{MC}_d := \bigg\{ f(z) = z^d + \sum_{k=0}^{d-2}a_kz^k \ \bigg| \ a_k \in \mathbb{C} \text{ for } 0 \leq k \leq d-2 \bigg\}, \end{align*} $$

denote the composite mapping of $\mathrm {MC}_d \subset \mathrm {Poly}_d \twoheadrightarrow \mathrm {Poly}_d / \mathrm {Aut}(\mathbb {C}) = \mathrm {MP}_d$ by $p : \mathrm {MC}_d \to \mathrm {MP}_d$ , and also denote the composite mapping of $p: \mathrm {MC}_d \to \mathrm {MP}_d$ and $\Phi _d : \mathrm {MP}_d \to \widetilde {\Lambda }_d$ by $\widehat {\Phi }_d : \mathrm {MC}_d \to \widetilde {\Lambda }_d$ , that is, $\widehat {\Phi }_d := \Phi _d \circ p$ .

In the above definition, the map p is surjective since every affine conjugacy class of polynomial maps contains monic centered polynomials. Moreover, two monic centered polynomials $f, g \in \mathrm {MC}_d$ are affinely conjugate if and only if there exists a $(d-1)$ th radical root a of $1$ such that the equality $g(z) = a f(a^{-1}z)$ holds. Hence, the group $\{ a \in \mathbb {C} \mid a^{d-1} = 1 \} \cong \mathbb {Z} /(d-1)\mathbb {Z}$ naturally acts on $\mathrm {MC}_d$ , and the induced mapping $\overline {p} : \mathrm {MC}_d / ( \mathbb {Z} /(d-1)\mathbb {Z} ) \to \mathrm {MP}_d$ is an isomorphism. Since $\mathrm {MC}_d \cong \mathbb {C}^{d-1}$ , we also have $\mathrm {MP}_d \cong \mathbb {C}^{d-1} / ( \mathbb {Z} /(d-1)\mathbb {Z} )$ . Here, the action of $\mathbb {Z} /(d-1)\mathbb {Z}$ on $\mathrm {MC}_d$ is not free for $d \geq 3$ , and $\mathrm {MP}_d$ has the set of singular points $\mathrm {Sing}(\mathrm {MP}_d)$ for $d \geq 4$ . Hence, in some sense, the map $p : \mathrm {MC}_d \to \mathrm {MP}_d$ can be considered to be a ‘desingularization’ of $\mathrm {MP}_d$ for $d \geq 4$ .

We already have the following theorem by Remark 1.9 in [Reference Sugiyama14].

Theorem 3.3. For $d \geq 2$ and $\unicode{x3bb} \in V_d$ , we put $\mathfrak {K}(\unicode{x3bb} ) =: \{ K_1,\ldots , K_q \}$ and denote by $g_w$ the greatest common divisor of $\#K_1,\ldots ,\#K_{(w-1)},(\#K_w)-1,\#K_{(w+1)},\ldots ,\#K_q$ for each $1\le w\le q$ . If $g_w=1$ holds for every $1 \leq w \leq q$ , then we have

(3.1) $$ \begin{align} \#(\Phi_d^{-1}(\bar{\unicode{x3bb}})) = \frac{s_d(\unicode{x3bb})}{(\#K_1)!\cdots(\#K_q)!} = \frac{s_d(\unicode{x3bb})}{\prod_{K \in \mathfrak{K}(\unicode{x3bb})}(\#K)!}, \end{align} $$

where $s_d(\unicode{x3bb} )$ is the non-negative integer defined in Theorem 2.3 and rewritten in Theorem I. Here, note that $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ is not considered counted with multiplicity, and hence $\#(\Phi _d^{-1}(\bar {\unicode{x3bb} }))$ denotes the number of discrete elements of $\Phi _d^{-1}(\bar {\unicode{x3bb} })$ .

In the case of $g_w \geq 2$ for some w, we also have an algorithm for finding $\#(\Phi _d^{-1}(\bar {\unicode{x3bb} }))$ in the third and fourth steps in Main Theorem III in [Reference Sugiyama14]. However, it contains induction processes and is much more complicated than equation (3.1); and hence we omit to describe it again in this paper.

As we already mentioned in Remark 1.9 in [Reference Sugiyama14], we find that for $d \geq 4$ and for $\unicode{x3bb} \in V_d$ , the inequality $g_w \geq 2$ holds for some w only if $\bar {\unicode{x3bb} } \in \Phi _d(\mathrm {Sing}(\mathrm {MP}_d))$ . Since $\mathrm {MC}_d$ is a ‘desingularization’ of $\mathrm {MP}_d$ , it is natural to expect that the map $\widehat {\Phi }_d = \Phi _d\circ p : \mathrm {MC}_d \to \widetilde {\Lambda }_d$ is simpler than the map $\Phi _d : \mathrm {MP}_d \to \widetilde {\Lambda }_d$ itself. In the following, we consider $\mathrm {MC}_d$ instead of $\mathrm {MP}_d$ , and also consider $\widehat {\Phi }_d : \mathrm {MC}_d \to \widetilde {\Lambda }_d$ instead of $\Phi _d : \mathrm {MP}_d \to \widetilde {\Lambda }_d$ .

We now state the second main theorem in this paper.

Theorem II. For $d \geq 2$ , $\unicode{x3bb} \in V_d$ , and $\widehat {\Phi }_d : \mathrm {MC}_d \to \widetilde {\Lambda }_d$ , we have

(3.2) $$ \begin{align} \#( \widehat{\Phi}_d^{-1}(\bar{\unicode{x3bb}}) ) = \frac{(d-1)s_d(\unicode{x3bb})}{\prod_{K \in \mathfrak{K}(\unicode{x3bb})}(\#K)!}, \end{align} $$

where $s_d(\unicode{x3bb} )$ is the non-negative integer defined in Theorem 2.3 and rewritten in Theorem I. Here, note that $\widehat {\Phi }_d^{-1}(\bar {\unicode{x3bb} })$ is not considered counted with multiplicity, and hence $\#(\widehat {\Phi }_d^{-1}(\bar {\unicode{x3bb} }))$ denotes the number of discrete elements of $\widehat {\Phi }_d^{-1}(\bar {\unicode{x3bb} })$ .

Theorem II is proved in §5.

Combining Theorems I and II, we have the following.

Corollary III. For $d \geq 2$ , $\unicode{x3bb} \in V_d$ , and $\widehat {\Phi }_d : \mathrm {MC}_d \to \widetilde {\Lambda }_d$ , we have

$$ \begin{align*} \#( \widehat{\Phi}_d^{-1}(\bar{\unicode{x3bb}}) ) = \frac{\sum_{\mathbb{I}\in \mathfrak{I}'(\unicode{x3bb})} ( \{ -(d-1) \}^{\#\mathbb{I} - 1} \cdot \prod_{I \in \mathbb{I}}( \#I - 1 )! )}{\prod_{K \in \mathfrak{K}(\unicode{x3bb})}(\#K)!}. \end{align*} $$

4 Proof of Theorem I

In this section, we prove Theorem I. We assume $d \geq 2$ and $\unicode{x3bb} =(\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _d) \in V_d$ , and denote by $\mathbb {I}_0 = \{\{ 1,\ldots ,d \}\}$ the minimum element of $\mathfrak {I}'(\unicode{x3bb} )$ , which are fixed throughout this section.

First we put

$$ \begin{align*} e_{\mathbb{I}_0}(\unicode{x3bb}) := (d-1)s_d(\unicode{x3bb}) \end{align*} $$

for $\mathbb {I}_0 = \{\{ 1,\ldots ,d \}\} \in \mathfrak {I}'(\unicode{x3bb} )$ . Then, equation (2.2) for $\mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ is rewritten in the form

(4.1) $$ \begin{align} e_{\mathbb{I}}(\unicode{x3bb}) = \prod_{I \in \mathbb{I}} e_{\{I\}}(\unicode{x3bb}_I). \end{align} $$

Here, $\{I\}$ denotes the minimum element of $\mathfrak {I}'(\unicode{x3bb} _I)$ . Moreover, equation (2.1) is rewritten in the form

(4.2) $$ \begin{align} e_{\mathbb{I}_0}(\unicode{x3bb}) = (d-1)! - \sum_{\mathbb{I} \in \mathfrak{I}(\unicode{x3bb})} \bigg( e_{\mathbb{I}}(\unicode{x3bb})\cdot \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \bigg), \end{align} $$

which is also equivalent to the equality

$$ \begin{align*} (d-1)! = \sum_{\mathbb{I} \in \mathfrak{I}'(\unicode{x3bb})} \bigg( e_{\mathbb{I}}(\unicode{x3bb})\cdot \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \bigg) \end{align*} $$

since for $\mathbb {I}_0 \in \mathfrak {I}'(\unicode{x3bb} )$ , we have $e_{\mathbb {I}_0}(\unicode{x3bb} )\cdot \prod _{k=d-\# \mathbb {I}_0 +1}^{d-1}k = e_{\mathbb {I}_0}(\unicode{x3bb} )\cdot \prod _{k=d}^{d-1}k = e_{\mathbb {I}_0}(\unicode{x3bb} )$ . Equation (2.5), which we would like to prove in this section, is also rewritten in the form

(4.3) _d $$\begin{align} e_{\mathbb{I}_0}(\unicode{x3bb}) = \sum_{\mathbb{I} \in \mathfrak{I}'(\unicode{x3bb}) } \bigg( \{ -(d-1) \}^{\#\mathbb{I} - 1} \cdot \prod_{I \in \mathbb{I}}( \#I - 1 )!\!\bigg). \end{align}$$

Hence, to prove Theorem I, it suffices to derive equation (4.3) _d from equations (4.1) and (4.2).

In the following, we show equation (4.3) _d by induction on d.

For $d=2$ or $3$ , we have $s_d(\unicode{x3bb} )=1$ and $\mathfrak {I}'(\unicode{x3bb} ) = \{ \mathbb {I}_0 \}$ for every $\unicode{x3bb} \in V_d$ . Hence, for $\unicode{x3bb} \in V_d$ , we always have

$$ \begin{align*} e_{\mathbb{I}_0}(\unicode{x3bb}) = (d-1)s_d(\unicode{x3bb}) = d-1 \end{align*} $$

and also have

$$ \begin{align*} \sum\limits_{{\mathbb{I}\in \mathfrak{I}'}(\unicode{x3bb}) } \bigg( \{ -(d-1) \}^{\#\mathbb{I} - 1} \cdot \prod_{I \in \mathbb{I}}( \#I - 1 )!\!\bigg) &= \{ -(d-1) \}^{\#\mathbb{I}_0 - 1} \cdot \prod_{I \in \mathbb{I}_0}( \#I - 1 )!\\ &= \{ -(d-1) \}^{1 - 1} \cdot ( d - 1 )! = (d-1)!. \end{align*} $$

Since $d-1 = (d-1)!$ for $d=2$ or $3$ , we have equations $(4.3)_2$ and $(4.3)_3$ .

In the following, we assume $d \geq 4$ and show equation (4.3) _d by the assumption of equations $(4.3)_2, (4.3)_3, \ldots , (4.3)_{d-1}$ , (4.1), and (4.2).

For each $\mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ with $\unicode{x3bb} \in V_d$ , we put $\mathbb {I}=:\{ I_1,\ldots ,I_l \}$ . Then, by using equations (4.1) and $(4.3)_{d'}$ for $2 \leq d' < d$ , we have the following equalities:

(4.4) $$ \begin{align}\begin{aligned} e_{\mathbb{I}}(\unicode{x3bb}) &= \prod_{I \in \mathbb{I}} e_{\{I\}}(\unicode{x3bb}_I) = \prod_{u=1}^l e_{\{I_u\}}(\unicode{x3bb}_{I_u}) \\ &= \prod_{u=1}^l \bigg( \sum_{\mathbb{I}^{\prime}_u\in \mathfrak{I}'(\unicode{x3bb}_{I_u})} \bigg[ \{ -(\#I_u-1) \}^{\#\mathbb{I}^{\prime}_u - 1} \cdot \prod_{I^{\prime}_u \in \mathbb{I}^{\prime}_u}( \#I^{\prime}_u - 1 )! \bigg] \bigg) \\ &= \sum_{\mathbb{I}^{\prime}_1\in \mathfrak{I}'(\unicode{x3bb}_{I_1})} \cdots \sum_{\mathbb{I}^{\prime}_l\in \mathfrak{I}'(\unicode{x3bb}_{I_l})} \prod_{u=1}^l \bigg[ \{ -(\#I_u-1) \}^{\#\mathbb{I}^{\prime}_u - 1} \cdot \prod_{I^{\prime}_u \in \mathbb{I}^{\prime}_u}( \#I^{\prime}_u - 1 )! \bigg] \\ &= \sum_{\mathrm{\scriptsize {\begin{matrix} \mathbb{I}' \in \mathfrak{I}(\unicode{x3bb}) \\ \mathbb{I}' \succ \mathbb{I} \end{matrix}}}} \bigg[ \bigg( \prod_{I' \in \mathbb{I}'}( \#I' - 1 )! \bigg) \cdot \bigg( \prod_{u=1}^l \{ -(\#I_u-1) \}^{\chi_{I_u}( \mathbb{I}' ) - 1} \bigg) \bigg] \end{aligned}\end{align} $$

since we have the equality

$$ \begin{align*} \{ \mathbb{I}^{\prime}_1 \amalg \cdots \amalg \mathbb{I}^{\prime}_l \mid \mathbb{I}^{\prime}_1 \in \mathfrak{I}'( \unicode{x3bb}_{I_1} ), \ldots, \mathbb{I}^{\prime}_l \in \mathfrak{I}'( \unicode{x3bb}_{I_l} ) \} = \{ \mathbb{I}' \in \mathfrak{I}(\unicode{x3bb}) \mid \mathbb{I}' \succ \mathbb{I} \} \end{align*} $$

by definition. Here, since $\mathbb {I} \succ \mathbb {I}$ holds for $\mathbb {I} \in \mathfrak {I}(\unicode{x3bb} )$ , we have $\mathbb {I} \in \{ \mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} ) \mid \mathbb {I}' \succ \mathbb {I} \}$ . Note that in equation (4.4), $\chi _{I_u}( \mathbb {I}' ) = \#( \{ I' \in \mathbb {I}' \mid I' \subseteq I_u \} )$ is the function defined in Theorem 2.5.

Substituting equation (4.4) into equation (4.2), we have

(4.5) $$ \begin{align}\begin{aligned} &e_{\mathbb{I}_0}(\unicode{x3bb}) \\ &= (d-1)! - \sum_{\mathbb{I} \in \mathfrak{I}(\unicode{x3bb})} \bigg\{ \sum_{\mathrm{\scriptsize {\begin{matrix} \mathbb{I}' \in \mathfrak{I}(\unicode{x3bb}) \\ \mathbb{I}' \succ \mathbb{I} \end{matrix}}}} \bigg(\prod_{I' \in \mathbb{I}'}( \#I' - 1 )! \bigg) \cdot \bigg( \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} \bigg) \bigg\} \cdot \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \\ &= (d-1)! - \sum_{\mathbb{I}' \in \mathfrak{I}(\unicode{x3bb})} \bigg\{ \prod_{I' \in \mathbb{I}'}( \#I' - 1 )! \bigg\} \cdot \bigg\{ \sum_{\mathrm{\scriptsize {\begin{matrix} \mathbb{I} \in \mathfrak{I}(\unicode{x3bb}) \\ \mathbb{I} \prec \mathbb{I}' \end{matrix}}}} \bigg( \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} \bigg) \cdot \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \bigg\}. \end{aligned}\end{align} $$

Here, equation (4.3) _d , which we would like to prove in this section, is equivalent to the equality

$$ \begin{align*} e_{\mathbb{I}_0}(\unicode{x3bb}) = (d-1)! + \sum_{\mathbb{I}\in \mathfrak{I}(\unicode{x3bb})} \bigg( \{ -(d-1) \}^{\#\mathbb{I} - 1} \cdot \prod_{I \in \mathbb{I}}( \#I - 1 )!\!\bigg), \end{align*} $$

which is also equivalent to

(4.6) $$ \begin{align} e_{\mathbb{I}_0}(\unicode{x3bb}) = (d-1)! + \sum_{\mathbb{I}'\in \mathfrak{I}(\unicode{x3bb})} \bigg[ \bigg\{\prod_{I' \in \mathbb{I}'}( \#I' - 1 )!\bigg\} \cdot \{ -(d-1) \}^{\#\mathbb{I}' - 1} \bigg]. \end{align} $$

Hence, comparing equations (4.5) and (4.6), we find that to prove equation (4.3) _d , we only need to show the following equality for each $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ :

(4.7) $$ \begin{align} \{ -(d-1) \}^{\#\mathbb{I}' - 1} = - \sum_{\mathrm{\scriptsize {\begin{matrix} \mathbb{I} \in \mathfrak{I}(\unicode{x3bb}) \\ \mathbb{I} \prec \mathbb{I}' \end{matrix}}}} \bigg( \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} \bigg) \cdot \prod_{k=d-\# \mathbb{I} +1}^{d-1}k. \end{align} $$

Here, equation (4.7) is equivalent to the equality

(4.8) $$ \begin{align} \sum_{\mathrm{\scriptsize {\begin{matrix} \mathbb{I} \in \mathfrak{I}'(\unicode{x3bb}) \\ \mathbb{I} \prec \mathbb{I}' \end{matrix}}}} \bigg( \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \bigg) \cdot \bigg( \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} \bigg) = 0 \end{align} $$

since for $\mathbb {I}_0 \in \mathfrak {I}'(\unicode{x3bb} )$ and $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ , we have $\mathbb {I}_0 \prec \mathbb {I}'$ and

$$ \begin{align*} \bigg( \prod_{k=d-\# \mathbb{I}_0 +1}^{d-1}k \bigg) \cdot \prod_{I \in \mathbb{I}_0} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} = \bigg( \prod_{k=d}^{d-1}k \bigg) \cdot \{ -(d-1) \}^{\#\mathbb{I}' - 1} = \{ -(d-1) \}^{\#\mathbb{I}' - 1}. \end{align*} $$

Hence, to prove Theorem I, we only need to show equation (4.8) for every $d \geq 4$ , $\unicode{x3bb} \in V_d$ , and $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ . In the following, instead of expressing $\sum _{\mathbb {I} \in \mathfrak {I}'(\unicode{x3bb} ),\, \mathbb {I} \prec \mathbb {I}'}$ for $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ , we simply express $\sum _{\mathbb {I} \prec \mathbb {I}'}$ , because if $\mathbb {I}$ is a partition of $\{1,\ldots ,d\}$ and $\mathbb {I} \prec \mathbb {I}'$ for $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ , then we automatically have $\mathbb {I} \in \mathfrak {I}'(\unicode{x3bb} )$ .

To prove equation (4.8), we make use of the following.

Definition 4.1. For $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ with $\#\mathbb {I}'=l$ and for $k \in \mathbb {Z}$ , we put

$$ \begin{align*} f_{l, k} := \sum_{\mathbb{I} \prec \mathbb{I}',\, \#\mathbb{I}=k} \, \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1}. \end{align*} $$

Example 4.3. Let us find $f_{l,l}$ and $f_{l,1}$ for $l \geq 2$ in this example.

Since $\{ \mathbb {I} \mid \mathbb {I} \prec \mathbb {I}',\, \#\mathbb {I}=l \} = \{ \mathbb {I}' \}$ , we have

$$ \begin{align*} f_{l, l} = \prod_{I \in \mathbb{I}'} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} = \prod_{I \in \mathbb{I}'} \{ -(\#I-1) \}^{1 - 1} = 1. \end{align*} $$

Let us consider $f_{l,1}$ next. Since $\{ \mathbb {I} \mid \mathbb {I} \prec \mathbb {I}',\, \#\mathbb {I}=1 \} = \{ \mathbb {I}_0 \}$ , we have

$$ \begin{align*} f_{l,1} = \prod_{I \in \mathbb{I}_0} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} = \{ -(d-1) \}^{l - 1}. \end{align*} $$

Example 4.4. Let us also find $f_{4,2}$ in this example. For $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ with $\#\mathbb {I}'=4$ , we can put $\mathbb {I}'=\{ I_1, I_2, I_3, I_4 \}$ , and in this expression, we have $\{ \mathbb {I} \mid \mathbb {I} \prec \mathbb {I}',\, \#\mathbb {I}=2 \} = \{ \mathbb {I}_1,\ldots , \mathbb {I}_7 \}$ , where

$$ \begin{align*} &\mathbb{I}_1=\{ I_1,\ I_2\amalg I_3\amalg I_4 \},\quad \mathbb{I}_2=\{ I_2,\ I_1\amalg I_3\amalg I_4 \},\\ &\mathbb{I}_3=\{ I_3,\ I_1\amalg I_2\amalg I_4 \},\quad \mathbb{I}_4=\{ I_4,\ I_1\amalg I_2\amalg I_3 \},\\ &\mathbb{I}_5=\{ I_1\amalg I_2,\ I_3\amalg I_4 \},\quad \mathbb{I}_6=\{ I_1\amalg I_3,\ I_2\amalg I_4 \},\quad \text{and} \quad \mathbb{I}_7=\{ I_1\amalg I_4,\ I_2\amalg I_3 \}. \end{align*} $$

We put $\#I_u=:i_u$ for $1 \leq u \leq 4$ . Note that the equality $i_1 + i_2 + i_3 + i_4 = d$ holds. We have

$$ \begin{align*} &\prod_{I \in \mathbb{I}_1} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} = \{ -(i_1-1) \}^{1-1} \cdot \{ -(i_2+i_3+i_4-1) \}^{3-1},\\ &\prod_{I \in \mathbb{I}_5} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} = \{ -(i_1+i_2-1) \}^{2-1} \cdot \{ -(i_3+i_4-1) \}^{2-1}, \end{align*} $$

for instance, which implies

$$ \begin{align*} \sum_{u=1}^4 \prod_{I \in \mathbb{I}_u} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} &= \sum_{u=1}^4 (i_1+i_2+i_3+i_4 - i_u-1)^2 = \sum_{u=1}^4 (d - i_u-1)^2\\ &= 4(d-1)^2 - 2(d-1)d + \sum_{u=1}^4 i_u^2,\\ \sum_{u=5}^7 \prod_{I \in \mathbb{I}_u} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} &= (i_1+i_2-1)(i_3+i_4-1) + (i_1+i_3-1)(i_2+i_4-1) \\ & \quad + (i_1+i_4-1)(i_2+i_3-1) = 2\sum_{1 \leq u <v \leq 4}i_ui_v - 3d + 3. \end{align*} $$

Hence, we have

$$ \begin{align*} f_{4, 2} &= \sum_{u=1}^7 \prod_{I \in \mathbb{I}_u} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1}\\ &= 4(d-1)^2 - 2(d-1)d + \sum_{u=1}^4 i_u^2 + 2\sum_{1 \leq u <v \leq 4}i_ui_v - 3d + 3\\ &= 2d^2 - 9d + 7 + \bigg(\sum_{u=1}^4 i_u \bigg)^2 = 3d^2 - 9d + 7. \end{align*} $$

Example 4.5. By a similar computation to Example 4.4, we have the following for $l \leq 5$ :

$$ \begin{align*} f_{2,1} &= -d+1, & f_{3,1} &= (d-1)^2, & f_{4,1} &= \{-(d-1)\}^3, & f_{5,1} &= \{-(d-1)\}^4,\\f_{2,2} &= 1, & f_{3,2} &= -2d+3, & f_{4,2} &= 3d^2-9d+7, & f_{5,2} &= -4d^3+18d^2-28d+15,\\& & f_{3,3} &= 1, & f_{4,3} &= -3d+6, & f_{5,3} &= 6d^2-24d+25,\\& & & & f_{4,4} &= 1, & f_{5,4} &= -4d+10,\\& & & & & & f_{5,5} &= 1. \end{align*} $$

The following is the key proposition to prove equation (4.8).

Proposition 4.6. The number $f_{l, k}$ defined in Definition 4.1 is a function of $l, k$ , and d, and does not depend on the choice of $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ with $\#\mathbb {I}'=l$ . Moreover, for $l, k \in \mathbb {Z}$ with $l \geq 2$ , we have the equality

$$ \begin{align*} f_{l+1, k} = f_{l,k-1} - (d-k)f_{l,k}. \end{align*} $$

Proof of Proposition 4.7

If $\#\mathbb {I}'=2$ , then we can put $\mathbb {I}'=\{I_1, I_2\}$ and have $\{ \mathbb {I} \mid \mathbb {I} \prec \mathbb {I}' \} = \{\mathbb {I}_0, \mathbb {I}'\}$ . Hence, we have

$$ \begin{align*} \sum_{\mathbb{I} \prec \mathbb{I}'}& \bigg( \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \bigg) \cdot \bigg( \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} \bigg)\\[3pt] &= 1\cdot \{ -(d-1) \}^{2-1} + (d-1) \cdot \{ -(\#I_1-1) \}^{1-1} \cdot \{ -(\#I_2-1) \}^{1-1}\\[3pt] &= -(d-1) + (d-1) = 0. \end{align*} $$

In the case where $\#\mathbb {I}' \geq 3$ , we put $\#\mathbb {I}' =:l+1$ . Then we have $l \geq 2$ and have the following equalities by Proposition 4.6:

$$ \begin{align*} \sum_{\mathbb{I} \prec \mathbb{I}'}& \bigg( \prod_{k=d-\# \mathbb{I} +1}^{d-1}k \bigg) \cdot \bigg( \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} \bigg)\\ &= \sum_{k=1}^{l+1} \bigg( \prod_{k'=d-k +1}^{d-1}k' \bigg) \cdot f_{l+1, k}\\ &= \sum_{k=1}^{l+1} \bigg( \prod_{k'=d-k +1}^{d-1}k' \bigg) \cdot ( f_{l, k-1} - (d-k)f_{l,k} )\\ &= \sum_{k=1}^{l+1} \bigg( \prod_{k'=d-k +1}^{d-1}k' \bigg) \cdot f_{l, k-1} - \sum_{k=1}^{l+1} \bigg( \prod_{k'=d-k +1}^{d-1}k' \bigg) \cdot (d-k)f_{l,k}\\ &= \sum_{k=0}^{l} \bigg( \prod_{k'=d-k}^{d-1}k' \bigg) \cdot f_{l, k} - \sum_{k=1}^{l+1} \bigg( \prod_{k'=d-k}^{d-1}k' \bigg) \cdot f_{l,k}\\ &= \bigg( \prod_{k'=d}^{d-1}k' \bigg) \cdot f_{l, 0} - \bigg( \prod_{k'=d-(l+1)}^{d-1}k' \bigg) \cdot f_{l,l+1} = 0, \end{align*} $$

which completes the proof of Proposition 4.7.

In the rest of this section, we shall prove Proposition 4.6. We make use of the following polynomial to prove Proposition 4.6.

Definition 4.8. For $l, k \in \mathbb {Z}$ with $l \geq 2$ , we define $\mathfrak {J}_l(k)$ as follows: if $k \leq 0$ or $k \geq l+1$ , then we put $\mathfrak {J}_l(k) = \emptyset $ ; if $1 \leq k \leq l$ , then we put

$$ \begin{align*} \mathfrak{J}_l(k) := \left\{ \{J_1,\ldots,J_k\}\ \left| \ \begin{matrix} J_1 \amalg \cdots \amalg J_k = \{1,\ldots,l \},\\ J_v \ne \emptyset \textrm{ for every } 1\le v\le k \end{matrix} \right.\right\}, \end{align*} $$

where $J_1 \amalg \cdots \amalg J_k$ denotes the disjoint union of $J_1,\ldots , J_k$ . Moreover, for $l, k \in \mathbb {Z}$ with $l \geq 2$ , we put

$$ \begin{align*} g_{l,k}(X_1,\ldots,X_l) := \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \prod_{J\in \mathbb{J}} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1}. \end{align*} $$

By definition, $\mathfrak {J}_l(k)$ is the set of all the partitions of $\{1, \ldots ,l\}$ into k pieces. Note that the equality $g_{l,k}(X_1,\ldots ,X_l)=0$ trivially holds for $k \leq 0$ or $k \geq l+1$ .

Lemma 4.9. For $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ with $\#\mathbb {I}'=l$ and for every $k \in \mathbb {Z}$ , putting $\mathbb {I}' =: \{ I_1, \ldots , I_l \}$ and $\# I_u =: i_u$ for $1 \leq u \leq l$ , we have

(4.9) $$ \begin{align} f_{l,k} = g_{l,k}(i_1,\ldots,i_l). \end{align} $$

Proof. If $k \leq 0$ or $k \geq l+1$ , then equation (4.9) trivially holds since both sides of equation (4.9) are equal to zero. In the following, we assume $1 \leq k \leq l$ .

By definition, we have

$$ \begin{align*} f_{l,k} =\! \sum_{\mathbb{I} \prec \mathbb{I}',\, \#\mathbb{I}=k} \prod_{I \in \mathbb{I}} \{ -(\#I-1) \}^{\chi_{I}( \mathbb{I}' ) - 1} = \!\sum_{\mathbb{I} \prec \mathbb{I}',\, \#\mathbb{I}=k} \, \prod_{I \in \mathbb{I}} \bigg\{\! -\bigg( \sum_{1 \leq u \leq l,\ I_u \subset I}\! i_u-1 \bigg) \bigg\}^{\chi_{I}( \mathbb{I}' ) - 1}. \end{align*} $$

Hence, putting

$$ \begin{align*} \tilde{g}_{l,k}(X_1,\ldots,X_l) := \sum_{\mathbb{I} \prec \mathbb{I}',\, \#\mathbb{I}=k} \, \prod_{I \in \mathbb{I}} \bigg\{ -\bigg( \sum_{1 \leq u \leq l,\ I_u \subset I} X_u-1 \bigg) \bigg\}^{\chi_{I}( \mathbb{I}' ) - 1}, \end{align*} $$

we obviously have $\tilde {g}_{l,k}(i_1,\ldots ,i_l) = f_{l,k}$ .

Here, we can make a bijection $\mathfrak {J}_l(k) \to \{\mathbb {I} \mid \mathbb {I} \prec \mathbb {I}',\, \#\mathbb {I}=k \}$ by

$$ \begin{align*} \mathbb{J} \mapsto \{ \amalg_{u \in J} I_u \mid \ J \in \mathbb{J} \}, \end{align*} $$

which implies that

$$ \begin{align*} \tilde{g}_{l,k}(X_1,\ldots,X_l) &= \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \, \prod_{I \in \{ \amalg_{u \in J} I_u \mid \ J \in \mathbb{J} \}} \bigg\{ -\bigg( \sum_{1 \leq u \leq l,\ I_u \subset I} X_u-1 \bigg) \bigg\}^{\chi_{I}( \mathbb{I}' ) - 1}\\ &= \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \, \prod_{J \in \mathbb{J}} \bigg\{ -\bigg( \sum_{1 \leq u \leq l,\ I_u \subset \amalg_{u' \in J} I_{u'}} X_u - 1 \bigg) \bigg\}^{\chi_{(\amalg_{u' \in J} I_{u'})}( \mathbb{I}' ) - 1}\\ &= \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \, \prod_{J \in \mathbb{J}} \bigg\{ -\bigg( \sum_{u \in J} X_u - 1 \bigg) \bigg\}^{\#J - 1} = g_{l,k}(X_1,\ldots,X_l). \end{align*} $$

Hence, we have equation (4.9).

Proposition 4.11. For $l, k \in \mathbb {Z}$ with $l \geq 2$ , we have

$$ \begin{align*} g_{l+1, k}(X_1, \ldots, X_l, 0) = g_{l, k-1}(X_1,\ldots, X_l) - (X_1+\cdots+X_l - k) g_{l,k}(X_1,\ldots, X_l). \end{align*} $$

Proof. First, we put

$$ \begin{align*} \mathfrak{J}_{l+1}^1(k) := \{ \mathbb{J} \in \mathfrak{J}_{l+1}(k) \mid \{l+1\} \in \mathbb{J} \} \quad \text{and} \quad \mathfrak{J}_{l+1}^2(k) := \{ \mathbb{J} \in \mathfrak{J}_{l+1}(k) \mid \{l+1\} \notin \mathbb{J} \} \end{align*} $$

for $l \geq 2$ . Then we have $\mathfrak {J}_{l+1}^1(k) \, \amalg \, \mathfrak {J}_{l+1}^2(k) = \mathfrak {J}_{l+1}(k)$ for every k. Moreover, we have $\mathfrak {J}_{l+1}^1(k)= \emptyset $ for $k \leq 1$ or $k \geq l+2$ , and $\mathfrak {J}_{l+1}^2(k)= \emptyset $ for $k \leq 0$ or $k \geq l+1$ .

For $\mathbb {J} \in \mathfrak {J}_{l+1}^1(k)$ , we can express $\mathbb {J} = \{J_1,\ldots ,J_{k-1}, \{l+1\}\}$ , where $J_1 \amalg \cdots \amalg J_{k-1} = \{1,\ldots ,l\}$ . Hence, we can make a bijection $\pi _1: \mathfrak {J}_{l+1}^1(k) \to \mathfrak {J}_{l}(k-1)$ by $\mathbb {J} \mapsto \mathbb {J} \setminus \{\{l+1\}\}$ . Moreover, for $J=\{l+1\} \in \mathbb {J} \in \mathfrak {J}_{l+1}^1(k)$ , we have

$$ \begin{align*} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} = \bigg\{ - \bigg( X_{l+1} - 1 \bigg) \bigg\}^{1-1} = 1. \end{align*} $$

Hence, we have

(4.10) $$ \begin{align}\begin{aligned} \sum_{\mathbb{J} \in \mathfrak{J}_{l+1}^1(k)} \prod_{J\in \mathbb{J}} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} &= \sum_{\mathbb{J} \in \mathfrak{J}_{l+1}^1(k)} \prod_{J\in \pi_1(\mathbb{J})} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1}\\ &= \sum_{\mathbb{J} \in \mathfrak{J}_{l}(k-1)} \prod_{J\in \mathbb{J}} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1}\\ &= g_{l,k-1}(X_1,\ldots,X_l). \end{aligned}\end{align} $$

For $\mathbb {J}' \in \mathfrak {J}_{l+1}^2(k)$ , we can express $\mathbb {J}' = \{J_1,\ldots ,J_k\}$ with $\{l+1\} \subsetneq J_k$ , and in this expression, we have $\{J_1,\ldots ,J_{k-1}, (J_k\setminus \{l+1\}) \} \in \mathfrak {J}_{l}(k)$ . Hence, we can make a surjection $\pi _2: \mathfrak {J}_{l+1}^2(k) \to \mathfrak {J}_{l}(k)$ by $\mathbb {J}' \mapsto \{ J \setminus \{l+1\} \mid J \in \mathbb {J}' \}$ . For each $\mathbb {J} = \{J_1,\ldots ,J_k\} \in \mathfrak {J}_{l}(k)$ , its fiber $\pi _2^{-1}( \mathbb {J} )$ consists of k elements, which are $\{J_v \mid 1 \leq v \leq k,\ v \ne v' \} \cup \{ J_{v'}\amalg \{l+1\} \}$ for $1 \leq v' \leq k$ . Hence, for each $\mathbb {J} = \{J_1,\ldots ,J_k\} \in \mathfrak {J}_{l}(k)$ , we have

$$ \begin{align*} \sum_{\mathbb{J}' \in \pi_2^{-1}( \mathbb{J} )}& \prod_{J\in \mathbb{J}'} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} \bigg|_{X_{l+1}=0}\\ &= \sum_{v'=1}^k \bigg[ \bigg\{ -\bigg( \sum_{u \in J_{v'}\amalg\{l+1\}} X_u -1 \bigg) \bigg\}^{\#( J_{v'}\amalg \{l+1\} ) - 1} \\ &\quad\times \prod_{1 \leq v \leq k,\ v \ne v'} \bigg\{ -\bigg( \sum_{u \in J_v} X_u - 1 \bigg) \bigg\}^{\#J_v - 1} \bigg] \bigg|_{X_{l+1}=0}\\ &= \sum_{v'=1}^k \bigg[ \bigg\{ -\bigg( \sum_{u \in J_{v'}} X_u -1 \bigg) \bigg\}^{\# J_{v'}} \cdot \prod_{1 \leq v \leq k,\ v \ne v'} \bigg\{ -\bigg( \sum_{u \in J_v} X_u - 1 \bigg) \bigg\}^{\#J_v - 1} \bigg]\\ &= \sum_{v'=1}^k \bigg[ \bigg\{ -\bigg( \sum_{u \in J_{v'}} X_u -1 \bigg) \bigg\} \cdot \prod_{v=1}^k \bigg\{ -\bigg( \sum_{u \in J_v} X_u - 1 \bigg) \bigg\}^{\#J_v - 1} \bigg]\\ &= \bigg[ \sum_{v'=1}^k \bigg\{ -\bigg( \sum_{u \in J_{v'}} X_u -1 \bigg) \bigg\} \bigg] \cdot \prod_{v=1}^k \bigg\{ -\bigg( \sum_{u \in J_v} X_u - 1 \bigg) \bigg\}^{\#J_v - 1}\\ &= - \bigg( \sum_{u=1}^l X_u - k \bigg) \cdot \prod_{J \in \mathbb{J}} \bigg\{ -\bigg( \sum_{u \in J} X_u - 1 \bigg) \bigg\}^{\#J - 1}. \end{align*} $$

We therefore have

(4.11) $$ \begin{align}\begin{aligned} \sum_{\mathbb{J}' \in \mathfrak{J}_{l+1}^2(k)}& \prod_{J\in \mathbb{J}'} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} \bigg|_{X_{l+1}=0}\\ &= \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \sum_{\mathbb{J}' \in \pi_2^{-1}( \mathbb{J} )} \prod_{J\in \mathbb{J}'} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} \bigg|_{X_{l+1}=0}\\ &= \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \bigg[ - \bigg( \sum_{u=1}^l X_u - k \bigg) \cdot \prod_{J \in \mathbb{J}} \bigg\{ -\bigg( \sum_{u \in J} X_u - 1 \bigg) \bigg\}^{\#J - 1} \bigg]\\ &= - \bigg( \sum_{u=1}^l X_u - k \bigg) \sum_{\mathbb{J} \in \mathfrak{J}_l(k)} \prod_{J \in \mathbb{J}} \bigg\{ -\bigg( \sum_{u \in J} X_u - 1 \bigg) \bigg\}^{\#J - 1}\\ &= - (X_1+\cdots+X_l-k)g_{l,k}(X_1,\ldots,X_l). \end{aligned}\end{align} $$

By equations (4.10) and (4.11), we have

$$ \begin{align*} g_{l+1,k}(X_1,\ldots,X_l,0) &= \sum_{\mathbb{J} \in \mathfrak{J}_{l+1}(k)} \prod_{J\in \mathbb{J}} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} \bigg|_{X_{l+1}=0}\\&= \sum_{\mathbb{J} \in \mathfrak{J}_{l+1}^1(k)} \prod_{J\in \mathbb{J}} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1}\\& \quad + \sum_{\mathbb{J}' \in \mathfrak{J}_{l+1}^2(k)} \prod_{J\in \mathbb{J}'} \bigg\{ - \bigg( \sum_{u\in J} X_u - 1 \bigg) \bigg\}^{\#J-1} \bigg|_{X_{l+1}=0}\\&= g_{l,k-1}(X_1,\ldots,X_l) - (X_1+\cdots+X_l-k)g_{l,k}(X_1,\ldots,X_l), \end{align*} $$

which completes the proof of Proposition 4.11.

Lemma 4.12. For every $l, k \in \mathbb {Z}$ with $l \geq 2$ , there exists a polynomial $h_{l,k}(Y) \in \mathbb {Z}[Y]$ such that the equality

(4.12) $$ \begin{align} g_{l,k}(X_1,\ldots,X_l)= h_{l,k}(X_1 + \cdots + X_l) \end{align} $$

holds. Moreover, for every $l, k \in \mathbb {Z}$ with $l \geq 2$ , the equality

(4.13) $$ \begin{align} h_{l+1,k}(Y) = h_{l,k-1}(Y) - (Y-k)h_{l,k}(Y) \end{align} $$

holds.

Proof. In the case where $l=2$ , we have $g_{2,1}(X_1, X_2) = -(X_1 + X_2 - 1)$ and $g_{2,2}(X_1, X_2) = 1$ by a direct calculation. Hence, putting $h_{2,1}(Y) \kern1.3pt{=}\kern1.3pt -(Y-1)$ , $h_{2,2}(Y) \kern1.3pt{=}\kern1.3pt 1$ , and $h_{2,k}(Y) = 0$ for $k \ne 1, 2$ , we have $g_{2,k}(X_1, X_2) = h_{2, k}(X_1+X_2)$ for every $k \in \mathbb {Z}$ .

For $l \geq 3$ and for every $k\in \mathbb {Z}$ , we define the polynomials $h_{l,k}(Y)$ inductively by equation (4.13). Then we obviously have $h_{l,k}(Y)=0$ for $k \leq 0$ or $k \geq l+1$ . Hence, equation (4.12) holds for $k \leq 0$ or $k \geq l+1$ . In the following, we show equation (4.12) for $l\geq 3$ and $1 \leq k \leq l$ by induction on l. Hence, we suppose equation (4.12) for every $k \in \mathbb {Z}$ , and show the equality $g_{l+1,k}(X_1,\ldots ,X_{l+1})= h_{l+1,k}(X_1 + \cdots + X_{l+1})$ for $1 \leq k \leq l+1$ .

By the assumption and Proposition 4.11, we have

$$ \begin{align*} g_{l+1, k}(X_1, \ldots, X_l, 0) &= g_{l, k-1}(X_1,\ldots, X_l) - (X_1+\cdots+X_l - k) g_{l,k}(X_1,\ldots, X_l)\\ &= h_{l, k-1}(X_1+\cdots+X_l) - (X_1+\cdots+X_l - k) h_{l,k}(X_1+\cdots+X_l)\\ &= h_{l+1,k}(X_1+\cdots+X_l). \end{align*} $$

Hence, putting $P_{l+1,k}(X_1,\ldots ,X_{l+1}) := g_{l+1, k}(X_1, \ldots , X_{l+1}) - h_{l+1,k}(X_1+\cdots +X_{l+1})$ , we have $P_{l+1,k}(X_1,\ldots ,X_l,0)=0$ . Moreover, by Lemma 4.10, the polynomial $P_{l+1,k}(X_1,\ldots ,X_{l+1})$ is symmetric in $l+1$ variables $X_1,\ldots ,X_{l+1}$ .

We denote by $\sigma _{l+1, m}=\sigma _{l+1,m}(X_1,\ldots ,X_{l+1})$ the elementary symmetric polynomial of degree m in $l+1$ variables $X_1,\ldots ,X_{l+1}$ . Since $P_{l+1,k}(X_1,\ldots ,X_{l+1})$ is a symmetric polynomial with coefficients in $\mathbb {Z}$ , we have $P_{l+1,k}(X_1,\ldots ,X_{l+1}) \in \mathbb {Z}[ \sigma _{l+1,1},\ldots ,\sigma _{l+1,l+1} ]$ . Moreover, since $\deg g_{l+1,k} = \deg h_{l+1,k} = l+1-k \leq l$ , we have $\deg P_{l+1,k} \leq l$ , which implies that $P_{l+1,k}(X_1,\ldots ,X_{l+1}) \in \mathbb {Z}[ \sigma _{l+1,1},\ldots ,\sigma _{l+1,l} ]$ .

Since $\sigma _{l+1,m}(X_1,\ldots ,X_l,0) = \sigma _{l,m}(X_1,\ldots ,X_l)$ for $1 \leq m \leq l$ , we have a ring isomorphism $\varphi : \mathbb {Z}[ \sigma _{l+1,1},\ldots ,\sigma _{l+1,l} ] \to \mathbb {Z}[ \sigma _{l,1},\ldots ,\sigma _{l,l} ]$ by substituting $X_{l+1}=0$ , and under the map $\varphi $ , we have $\varphi ( P_{l+1,k} ) = P_{l+1,k}(X_1,\ldots ,X_l,0)=0$ . Hence, injectivity of $\varphi $ implies $P_{l+1,k}(X_1,\ldots ,X_{l+1})=0$ . We therefore have $g_{l+1,k}(X_1,\ldots ,X_{l+1})= h_{l+1,k}(X_1 + \cdots + X_{l+1})$ , which completes the proof of Lemma 4.12 by induction on l.

Proof of Proposition 4.6

By Definition 4.1, $f_{l,k}$ is originally a function of $d \geq 4$ , $\mathbb {I}' \in \mathfrak {I}(\unicode{x3bb} )$ , and $k \in \mathbb {Z}$ . However, putting $\#\mathbb {I}'=l$ , $\mathbb {I}' =: \{ I_1, \ldots , I_l \}$ , and $\# I_u =: i_u$ for $1 \leq u \leq l$ , we have by Lemmas 4.9 and 4.12 the equality

(4.14) $$ \begin{align} f_{l,k} = g_{l,k}(i_1,\ldots,i_l)=h_{l,k}(i_1+\cdots+i_l) = h_{l,k}(d). \end{align} $$

Hence, $f_{l,k}$ is in practice a function of $l, k$ , and d since the polynomial $h_{l,k}(Y)$ depends only on l and k.

Moreover, by equation (4.14) and Lemma 4.12, we have

$$ \begin{align*} f_{l+1,k} = h_{l+1,k}(d) = h_{l,k-1}(d) - (d-k)h_{l,k}(d) = f_{l,k-1} - (d-k)f_{l,k} \end{align*} $$

for every $l,k\in \mathbb {Z}$ with $l \geq 2$ , which completes the proof of Proposition 4.6.

To summarize the above mentioned, we have completed the proof of Theorem I.

5 Proof of Theorem II

In this section, we prove Theorem II. Throughout this section, we always assume $\unicode{x3bb} =(\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _d) \in V_d$ , and moreover assume that $s_d(\unicode{x3bb} )$ is the non-negative integer defined in Theorem 2.3.

First, we consider the case where $d=2$ . If $d=2$ , then the maps $p: \mathrm {MC}_2 \to \mathrm {MP}_2$ and $\Phi _2: \mathrm {MP}_2 \to \widetilde {\Lambda }_2$ are bijective. Hence, we have $\#( \widehat {\Phi }_2^{-1} ( \bar {\unicode{x3bb} } ) ) = 1$ for every $\unicode{x3bb} \in V_2$ . Regarding the right-hand side of equation (3.2), since $s_2(\unicode{x3bb} )= 1$ and $\mathfrak {K}(\unicode{x3bb} ) = \{ \{1\}, \{2\} \}$ for every $\unicode{x3bb} \in V_2$ , we always have

$$ \begin{align*} \frac{(d-1)s_d(\unicode{x3bb})}{ \prod_{K\in\mathfrak{K}(\unicode{x3bb})}(\#K)!} = \frac{(2-1)s_2(\unicode{x3bb})}{ 1!\cdot 1!} = 1. \end{align*} $$

Hence, equation (3.2) holds for every $\unicode{x3bb} \in V_2$ .

In the rest of this section, we consider the case $d \geq 3$ . We denote by $\mathbb {P}^{d-1}$ the complex projective space of dimension $d-1$ , and put

$$ \begin{align*} \Sigma_d(\unicode{x3bb}) := \left\{(\zeta_1:\cdots:\zeta_d) \in \mathbb{P}^{d-1} \ \left|\ \begin{matrix} \sum\limits_{i=1}^d \zeta_i = 0 \\ \sum\limits_{i=1}^d (1/(1-\lambda_i))\zeta_i^{k} = 0 \quad \textrm{for} \quad 1 \le k \le d-2 \\ \zeta_1,\ldots,\zeta_d \textrm{ are mutually distinct} \end{matrix} \right. \right\}. \end{align*} $$

We already have the following proposition by Propositions 4.3 and 9.1 in [Reference Sugiyama14].

Proposition 5.1. The equality $\#( \Sigma _d(\unicode{x3bb} ) ) = s_d(\unicode{x3bb} )$ holds. Moreover, we can define the surjection $\pi (\unicode{x3bb} ) : \Sigma _d(\unicode{x3bb} ) \to \Phi _d^{-1}(\bar {\unicode{x3bb} })$ by

$$ \begin{align*} (\zeta_1:\cdots:\zeta_d) \mapsto f(z)=z + \rho (z - \zeta_1)\cdots(z-\zeta_d), \end{align*} $$

where $-{1}/{\rho }= \sum _{i=1}^d (1/(1-\lambda_i))\zeta_i^{d-1}$ .

We put

$$ \begin{align*} \widetilde{\Sigma}_d(\unicode{x3bb}) := \left\{(\zeta_1,\ldots,\zeta_d) \in \mathbb{C}^{d} \ \left|\ \begin{matrix} \sum\limits_{i=1}^d \zeta_i = 0 \\ \sum\limits_{i=1}^d ({1}/{(1-\unicode{x3bb}_i)})\zeta_i^k = \begin{cases} 0 & \textrm{for } 1 \le k \le d-2 \\ -1 & \textrm{for } k = d-1 \end{cases}\\ \zeta_1,\ldots,\zeta_d \textrm{ are mutually distinct} \end{matrix} \right. \right\}\hspace{-2.5pt}. \end{align*} $$

Then the natural projection $\widetilde {\Sigma }_d(\unicode{x3bb} ) \to \Sigma _d(\unicode{x3bb} )$ defined by $(\zeta _1,\ldots ,\zeta _d) \mapsto (\zeta _1:\cdots :\zeta _d)$ is a $(d-1)$ -to-one map because for every $(\zeta _1:\cdots :\zeta _d) \in \Sigma _d(\unicode{x3bb} )$ , we have $\sum _{i=1}^d ({1}/({1-\unicode{x3bb} _i}))\zeta _i^{d-1} \ne 0$ by Proposition 5.1. Hence, we have

(5.1) $$ \begin{align} \#( \widetilde{\Sigma}_d(\unicode{x3bb}) ) = (d-1) \#( \Sigma_d(\unicode{x3bb}) ) = (d-1)s_d(\unicode{x3bb}). \end{align} $$

We consider next the relation between $\widetilde{\Sigma}_d(\unicode{x3bb})$ and $\widehat {\Phi }_d^{-1}(\bar {\unicode{x3bb} })$ . We can define the surjection $\widehat {\pi }(\unicode{x3bb} ) : \widetilde {\Sigma }_d(\unicode{x3bb} ) \to \widehat {\Phi }_d^{-1}(\bar {\unicode{x3bb} })$ by

$$ \begin{align*} (\zeta_1,\ldots,\zeta_d) \mapsto f(z)=z+(z - \zeta_1)\cdots(z-\zeta_d) \end{align*} $$

by lifting up the map $\pi (\unicode{x3bb} ) : \Sigma _d(\unicode{x3bb} ) \to \Phi _d^{-1}(\bar {\unicode{x3bb} })$ in Proposition 5.1. Here, since $d \geq 3$ , every polynomial $f(z)=z+(z - \zeta _1)\cdots (z-\zeta _d)$ for $(\zeta _1,\ldots ,\zeta _d) \in \widetilde {\Sigma }_d(\unicode{x3bb} )$ is monic and centered.

We put

$$ \begin{align*} \mathfrak{S}( \mathfrak{K}(\unicode{x3bb}) ) : = \{ \sigma \in \mathfrak{S}_d \mid i \in K \in \mathfrak{K}(\unicode{x3bb}) \Longrightarrow \sigma(i) \in K \}. \end{align*} $$

Here, note that we also have $\mathfrak {S}( \mathfrak {K}(\unicode{x3bb} ) ) = \{ \sigma \in \mathfrak {S}_d \mid \unicode{x3bb} _{\sigma (i)} = \unicode{x3bb} _i \text { for every } 1 \leq i \leq d \}$ . Moreover, $\mathfrak {S}( \mathfrak {K}(\unicode{x3bb} ) )$ is a subgroup of $\mathfrak {S}_d$ and is isomorphic to $\prod _{K \in \mathfrak {K}(\unicode{x3bb} )} \mathrm {Aut}(K) \cong \prod _{K \in \mathfrak {K}(\unicode{x3bb} )} \mathfrak {S}_{\#K}$ .

The group $\mathfrak {S}( \mathfrak {K}(\unicode{x3bb} ) )$ naturally acts on $\widetilde {\Sigma }_d(\unicode{x3bb} )$ by the permutation of coordinates, and its action is free. Moreover, for $\zeta , \zeta ' \in \widetilde {\Sigma }_d(\unicode{x3bb} )$ , the equality $\widehat {\pi }(\unicode{x3bb} )(\zeta ) = \widehat {\pi }(\unicode{x3bb} )(\zeta ')$ holds if and only if the equality $\zeta ' = \sigma \cdot \zeta $ holds for some $\sigma \in \mathfrak {S}( \mathfrak {K}(\unicode{x3bb} ) )$ , which can be verified by a similar argument to the proof of Lemma 4.5(6) in [Reference Sugiyama14]. We therefore have the bijection

$$ \begin{align*} \overline{\widehat{\pi}(\unicode{x3bb})} : \widetilde{\Sigma}_d(\unicode{x3bb}) / \mathfrak{S}( \mathfrak{K}(\unicode{x3bb}) ) \cong \widehat{\Phi}_d^{-1}(\bar{\unicode{x3bb}}), \end{align*} $$

which implies the equality

(5.2) $$ \begin{align} \#( \widehat{\Phi}_d^{-1}(\bar{\unicode{x3bb}}) ) = \frac{\#( \widetilde{\Sigma}_d(\unicode{x3bb}) )}{\#( \mathfrak{S}( \mathfrak{K}(\unicode{x3bb}) ) )} = \frac{\#( \widetilde{\Sigma}_d(\unicode{x3bb}) )}{\prod_{K \in \mathfrak{K}(\unicode{x3bb})}(\#K)!}. \end{align} $$

Combining equations (5.1) and (5.2), we have

$$ \begin{align*} \#( \widehat{\Phi}_d^{-1}(\bar{\unicode{x3bb}}) ) = \frac{(d-1)s_d(\unicode{x3bb})}{\prod_{K \in \mathfrak{K}(\unicode{x3bb})}(\#K)!}, \end{align*} $$

which completes the proof of Theorem II.