RAMIFIED COVERINGS OF S^{2} WITH TWO

DEGENERATE BRANCHING POINTS

S. Kuleshov

Abstract. In this work some Hurwitz numbers are calculated. In other words, here one compute numbers of nonequivalent ramified coverings of the Riemann sphere by another Riemann surface with two degenerate branch points for several cases.

1. Introduction

Let C_{g} be a connected Riemann surface of genus g, S^{2} be a Riemann sphere with a fixed infinity and zero, f : C_{g} → S^{2} be a meromorphic function of degree d with k poles of orders

p_{1}, p_{2}, …, p_{k} and m zeros of orders n_{1}, n_{2}, …, n_{m} ( ∑_{i=1}^{k} p_{i} = ∑_{i=1}^{m} n_{i} = d). Such a function

determines a ramified covering of the Riemann sphere S^{2} by the surface C_{g}. Two these functions f and f′ (or coverings) are called equivalent if there exists a homeomorphism γ of C_{g} such that f′ = f ∘ γ.

Our aim is a calculation of numbers of nonequivalent coverings with the following condition.

A covering has only two degenerated ramification points (zero and infinity).

In other words, we assume that ramification points, except zero and infinity are simple, i.e., they have order 2. We shall say that a function (covering) is primitive if it satisfies this condition. The number s of simple ramification points of a primitive covering can be obtained by the Riemann–Hurwitz formula

s = s(g, m, k) = 2g − 2 + m + k.  (1-1)

This work extends research of several mathematicians in this branch. Our starting point it the paper [1], where the authors deal with ramified coverings of S^{2} with one degenerate branching point.

Sometimes we use a notation (1^{i_{1}} 2^{i_{2}} … d^{i_{d}}) for a partition of a number d. It means that in the partition (1^{i_{1}} 2^{i_{2}} … d^{i_{d}}) there are i_{1} numbers equal 1, i_{2} numbers equal 2, and so on. For partitions α and β of a number d we denote by CN_{d}(α; β; g) the number of nonequivalent meromorphic functions on C_{g} with zeros and poles of orders α and β respectively. In the article we calculate all concrete numbers CN_{d}(α; β; g) for d = 3 (see p. 14), d = 4 (p. 19), and d = 5 (p. 31). Besides, some general formulae were written in the case when meromorphic functions have one zero of the maximal order. Namely,

CN_{d}((d); (d); g) = \{ \begin{matrix}\frac{d^{2g−2}(B_{2g+1}(d+1)−2^{2g}B_{2g+1}( \frac{d+2}{2} )−B_{2g+1}(0)(2^{2g}−1)+ \frac{d(2g+1)}{2} )}{2^{2g−1}(2g+1)}, & for even d, \\ \frac{2·d^{2g−2}}{2g+1} (B_{2g+1} (\frac{d+3}{2}) − B_{2g+1}(0)) & for odd d. \end{matrix}

If d is odd, then

CN_{d}((d); (l, d − l); g) = \frac{2d^{2g}}{l(d − l)(2g + 2)} (B_{2g+2} (\frac{d + 1}{2}) − B_{2g+2} (\frac{d + 1 − 2l}{2} )) .

If d is even and l < \frac{d}{2}, then

CN_{d}((d); (l, d − l); g) = \frac{d^{2g}}{(2g + 2)2^{2g}l(d − l)}×

× (B_{2g+2}(d + 1) − B_{2g+2}(d − 2l + 1) + 2^{2g+1} (B_{2g+2} (\frac{d − 2l + 2}{2}) − B_{2g+2} (\frac{d + 2}{2}))) .

If d is even and l = \frac{d}{2}, then CN_{d}((d); (l, d − l); g) =

= \frac{d^{2g−2}}{(2g + 2)2^{2g−1}} (B_{2g+2}(d + 1) − B_{2g+2}(1) + 2^{2g+1} (B_{2g+2}(1) − B_{2g+2} (\frac{d + 2}{2}))) + \frac{d^{2g}}{2^{2g+1}} .

CN_{d}((d); (1^{d−3} 3)) = \frac{d^{(}s − 1)}{(d − 2)!} (σ(d − 1, s + 2, \frac{d − 1}{2}) + \frac{(d − 1)(d − 5)}{12}σ(d − 1, s, \frac{d − 1}{2})) ,

where s = 2g + d − 3.

In formulae above B_{p}(x) denote the Bernoulli polynomial and the numbers σ(n, s, a) have

the property: ∑_{s=0}^{∞} σ(n, s, a)\frac{x^{s}}{s!} = e^{ax}(1 − e^{−x})^{n}.

The author is grateful to M. Shapiro, D. Zvonkin, and S. Lando for useful discussions. I am thankful to Max-Plank-Institute für Mathenatik for hospitality.

2. Method of calculation

2.1. Consider a primitive covering with fixed orders of zeros: α = (n_{1}, …, n_{m}) and of poles: β = (p_{1}, …, π_{n}). Denote by R_{1}, R_{2}, …, R_{s} simple ramification points. A small circle, drowning around a ramification point gives a permutation of covering sheets. If we enumerate the sheets, we can associate to each ramification point an element σ of the symmetric group S_{d}. In particular, each simple ramification point R_{i} corresponds to a transposition π_{i}, zero corresponds to a permutation σ_{α}^{1} of a cycle type α = (n_{1}, …, n_{m}), and infinity corresponds to a permutation σ_{β} of a cycle type β = (p_{1}, …, p_{k}). Obviously,

σ_{α}^{1}π_{1} ⋯ π_{s} = σ_{β}^{−1}.  (2-1)

Note that permutations σ and σ^{−1} have the same cycle type.

It is known that any collection of permutation, satisfying (2-1) can be realized by some primitive covering. Moreover two such collections corresponds to equivalent coverings iff there exists a permutation σ ∈ S_{d} such that

σ′_{α}^{1} = σσ_{α}^{1}σ^{−1}, π′_{i} = σπ_{i}σ^{−1}, i = 1, 2, …, s.

Let Π_{d}^{s}(σ_{α}^{1}, σ_{β}) be the set of ordered collections of transpositions {π_{i}} = (π_{1}, …, π_{s}), where s = s(g, m, k), such that π_{i}’s are permutation of the symmetric group S_{d}, σ_{α}^{1} ∈ S_{d} is a fixed permutation and the product σ_{α}^{1}π_{1} ⋯ π_{s} has the same cycle type as σ_{β} is. By definition, put

x_{d,s}(σ_{α}^{1}, σ_{β}) = |Π_{d}^{s}(σ_{α}^{1}, σ_{β})| (the number of the set′s elemets).

Denote by St_{σ_{α}^{1}} the subgroup of S_{d} consisting of permutations commuting with σ_{α}^{1}. This subgroup acts on the set Π_{d}^{s}(σ_{α}^{1}, σ_{β}) by the rule:

σ : {π_{i}} ↦ {σπ_{i}σ^{−1}}.

Obviously, an orbits set of this action one to one corresponds to equivalent classes of coverings with 2 degenerated ramification points of orders α (a cycle type of σ_{α}^{1}) and β (a cycle type of σ_{β}). We’ll say that collections {π_{i}}, {π′_{i}} ∈ Π_{d}^{s}(σ_{α}^{1}, σ_{β}) are equivalent, if they belong to one orbit of the St_{σ_{α}^{1}} action. The number of all equivalent classes of the collections we denote by N_{d}((α); (β); g).

On the other hand, this number equals a quantity of equivalent classes of coverings f : C_{g} → S^{2} with two degenerated ramification points of orders (n_{1}, …, n_{m}) and (p_{1}, …, p_{k}), but may be disconnected. Therefore, we have the following identity:

N_{d}(α; β; g) = N_{d}(β; α; g).

Let {π_{i}} be a collection of Π_{d}^{s}(σ_{α}^{1}, σ_{β}). Denote by M the subgroup of S_{d} generated by σ_{α}^{1} and π_{i}’s. This subgroup is called a monodromy group of a corresponding covering. We see that the covering is connected iff the group M transitive acts on the set {1, 2, …, d}. Let us call a collection of Π_{d}^{s}(σ_{α}^{1}, σ_{β}) connected if a corresponding covering is connected too and disconnected in the converse case.

For a given set Π_{d}^{s}(σ_{α}^{1}, σ_{β}) we denote by D the subset of all disconnected collections. Then the number CN_{d}(α; β; g) of all connected nonequivalent coverings is coincides with quantity of St_{σ_{α}^{1}} orbits in Π_{d}^{s}(σ_{α}^{1}, σ_{β}) \ D. 2.2. We see that the first calculation step of Hurwitz numbers is computation of numbers x_{d,s}(σ_{α}^{1}, σ_{β}). Let A_{βα} be the number x_{d,1}(σ_{α}^{1}, σ_{β}). In other words, A_{βα} is the number of transpositions π such that σ_{α}^{1}π = σ_{β}.

In this notation we have

x_{d,s}(σ_{α}^{1}, σ_{β}) = ∑_{γ⊢d} x_{d,s−1}(σ_{α}^{1}, σ_{γ})x_{d,1}(σ_{γ}^{1}, σ_{β}) = ∑_{γ⊢d} A_{βγ}x_{d,s−1}(σ_{α}^{1}, σ_{γ}),  (2-2)

or in a matrix form,

X_{d,s} = AX_{d,s−1}, where A = (A_{βα}), X_{d,s} = [x_{d,s}(σ_{α}^{1}, σ_{β})].

We can consider the set of the numbers A_{βα} as a linear operator A on the vector space Zℂ[S_{d}] (the center of group algebra). This operator acts by the rule

A : w_{α} ↦ ∑_{β} A_{βα}w_{β},

where w_{α} is an element of Zℂ[S_{d}] equal to sum of all permutations of cycle type α. 2.3. To solve the recursive relations (2-2) it is convenient to know eigenvectors and eigenvalues of the operator A. Suppose that C_{α}’s are the eigenvectors and μ_{α}’s are the eigenvalues. Additionally, assume that

C_{α} = ∑_{β} c_{βα}w_{β}.

Then recursive relations (2-2) has the form X′_{s} = A′X′_{s−1}, where

X′_{s} = C^{−1}X_{ds}, C = (c_{βα}), A′ = diag (μ_{α}).

Therefore, X′_{s} = (A′)^{s} X′_{0}, or x′_{s,α} = μ_{α}^{s}x′_{0,α} in the element form. Note that the initial data {x′_{0,α}} are determined by the initial data

x_{d,0}(σ_{α}^{1}, σ_{β}) = \{ \begin{matrix}1, & if α = β, \\ 0, & if α ≠ β \end{matrix} :

X′_{0} = C^{−1}X_{d,0}, i.e.,

x′_{0,γ} = ∑_{β⊢d} c′_{γβ}x_{d,0}(σ_{α}^{1}, σ_{β}) = c′_{γα},

where C^{−1} = (c′_{γβ}). Moreover, returning to the old basis, we get

x_{d,s}(σ_{α}^{1}, σ_{β}) = ∑_{γ} c_{βγ}μ_{γ}^{s}c′_{γα}.

Consider another linear operator B on the same vector space Z(ℂS_{d}).

B : w_{α} ↦ w_{α} ⋆ w_{2} = ∑_{β⊢d} B_{βα}w_{β},

where ‘⋆’ denotes the product of ℂS_{d} and w_{2} is the sum of all transpositions. 2.4. Lemma. For any two partitions α and β of d we have

B_{βα} = A_{αβ}.

Proof. Let us calculate a product w_{α} ⋆ w_{2}.

w_{α} = ∑_{i} σ_{α}^{i}, w_{2} = ∑_{j} π_{j}, w_{α}w_{2} = ∑_{ij} σ_{α}^{i}π_{j} = ∑_{β⊢d} B_{βα}w_{β}.

Here B_{βα} is a number of pairs (σ_{α}^{i}, π_{j}) such that σ_{α}^{i}π_{j} = σ_{β}^{1}. But the last equality is equivalent to σ_{α}^{i} = σ_{β}^{1}π_{j}. Therefore, the number B_{βα} is a quantity of transpositions π’s such that σ_{β}^{1}π = σ_{α}, i.e. B_{βα} = A_{αβ}. 2.5. Known facts about representations of S_{d}. In this subsection we calculate eigenvectors and eigenvalues of the operator B (see 2.4), using standard facts of the group representation theory. The set of complex irreducible representations of the symmetric group S_{d} one to one corresponds to the set of all partitions of the number d. Denote by χ_{α} the character of an irreducible representation corresponding to a partition α.

The set of all characters χ_{α}’s farm a basis of the vector space Z(ℂS_{d})^{∗}, dual to the center of group algebra ℂS_{d}. By the known theorem characters of irreducible representations satisfy the following orthogonal property

⟨χ_{α}|χ_{β}⟩ = \frac{1}{d!} ∑_{σ∈S_{n}} χ_{α}(σ)\overline{χ_{β}(σ)} = δ_{αβ} ∀α, β ⊢ d.  (2-3)

This fact is true for irreducible representations of any finite group. In our case values of characters are real and we will omit the symbol of the complex conjugate.

By definition, put χ̃_{α} = \frac{1}{d!} ∑_{σ∈S_{d}} χ_{α}(σ)σ ∈ Z(ℂS_{d}). It follows from (2-3) that

χ_{α}(χ̃_{β}) = δ_{αβ},  (2-4)

i.e., the set of all χ̃_{α}’s form a basis of Z(ℂS_{d}), dual to characters.

Let V_{α} be a vector space of an irreducible representation R_{α} : S_{d} → Aut V_{α} corresponding to a partition α, then any element χ̃_{β} can be considering as a linear operator on the space V_{α}. Moreover, χ̃_{β} ∈ Z(ℂS_{d}) and for any representation R_{α}, and σ ∈ S_{d} we have,

R_{α}(σ)R_{α}(χ̃_{β}) = R_{α}(χ̃_{β})R_{α}(σ).

But if A and B are commuting linear operators, then an eigensubspase of the operator a is invariant w.r.t. the operator B. Therefore, an eigensubspase of R_{α}(χ̃_{β}) is invariant w.r.t. R_{α}(σ) for any σ. This contradicts to irreducibility of R_{α}. Hence, R_{α}(χ̃_{β}) = λ_{αβ}Id_{V_{α}}.

On the other hand, tr R_{α}(χ̃_{β}) = dim V_{α}·λ_{αβ}. In addition, be definition tr R_{α}(χ̃_{β}) = χ_{α}(χ̃_{β}). Therefore,

R_{α}(χ̃_{β}) = \{ \begin{matrix}\frac{1}{dim V_{α}} Id_{V_{α}}, & if α = β, \\ 0 & if α ≠ β. \end{matrix}

With help of this fact we can compute a product χ̃_{α}⋆χ̃_{β} as elements of the center. To do this we search its coordinates w.r.t. the basis {χ̃_{α}}. We have R_{γ}(χ̃_{α} ⋆ χ̃_{β}) = R_{γ}(χ̃_{α}) ∘ R_{γ}(χ̃_{β}) = 0 unless γ = α = β, i.e., χ_{γ}(χ̃_{α} ⋆ χ̃_{β}) = 0 in these cases. Besides, R_{α}(χ̃_{α} ⋆ χ̃_{α}) = (R_{α}(χ̃_{α}))^{2} = \frac{1}{(dim V_{α})^{2}} Id_{V_{α}}. Whence, χ_{α}(χ̃_{α} ⋆ χ̃_{α}) = \frac{1}{dim V_{α}}. Thus,

χ̃_{α} ⋆ χ̃_{β} = \{ \begin{matrix}\frac{_{χ_{α}}}{dim V_{α}} , & if α = β, \\ 0 & if α ≠ β. \end{matrix}  (2-5)

Using dual basises, we obtain w_{2} = ∑_{α⊢d} χ_{α}(w_{2})χ̃_{α}. Note that, if we denote the set of all cycle type α permutations by W_{α} (W_{2} is the set of all transpositions), then

χ_{α}(w_{2}) = χ_{α} ( ∑_{π∈W_{2}} π) = χ_{α}(π) · |W_{2}|.

Finally,

χ̃_{α} ⋆ w_{2} = \frac{χ_{α}(w_{2})}{dim V_{α}} χ̃_{α}.

In other words, μ_{α} = \frac{χ_{α}(w_{2})}{dim V_{α}} = |W_{2}|\frac{χ_{α}(π)}{χ_{α}(w_{1})} is an eigenvalue and χ̃_{α} is an eigenvector of the linear operator B. 2.6. Let us calculate the conversion matrix F = (f_{αβ}) from the basis {w_{α}} to the eigenbasis

{χ̃_{β}}, i.e., χ̃_{β} = ∑_{α} f_{αβ}w_{α}.

χ̃_{β} = \frac{1}{d!} ∑_{σ∈S_{d}} χ_{β}(σ)σ = \frac{1}{d!} ∑_{α⊢d} χ_{β}(σ_{α})|w_{α}.

Hence,

f_{αβ} = \frac{χ_{β}(σ_{α})}{d!},  and C^{−1} = (c′_{αβ}) = (\frac{χ_{α}(σ_{β})}{d!}) .  (2-6)

Taking into account 2.3, lemma 2.4, and a standard fact of linear algebra, we need calculate F^{−1} = (g_{αγ}). We use the orthogonal relations of characters (2-3).

δ_{αβ} = \frac{1}{d!} ∑_{σ∈S_{d}} χ_{α}(σ)χ_{β}(σ) = \frac{1}{d!} ∑_{γ⊢d} χ_{α}(σ_{γ})χ_{β}(σ_{γ})|W_{γ}| = ∑_{γ⊢d}(d!|W_{γ}|f_{γα})f_{γβ}.

Whence,

g_{αγ} = d!|W_{γ}|f_{γα}, and C = (F^{−1})^{t} = (c_{αβ}) = (d!|W_{α}|f_{αβ}).  (2-7)

Combining together our calculation, we have x′_{0,γ} = \frac{1}{d!}χ_{γ}(σ_{α}) and

x_{d,s}(σ_{α}^{1}, σ_{β}) = \frac{|W_{β}|}{d!} ∑_{γ⊢d}(χ_{γ}(σ_{β})χ_{γ}(σ_{α}) (\frac{χ_{γ}(w_{2})}{χ_{γ}(w_{1})})^{s}).  (2-8)

It is convenient for concrete calculations to rewrite the last formula in the matrix form. Assume that we ordered partitions of the number d and denote by χ the matrix (χ_{α}(β)) such that in α-th row and β-th column is a value of the character χ_{α} on a permutation σ_{β} of the cycle type β.

Let W and μ be diagonal matrixes such that W_{αα} = |W_{α}| and μ_{αα} = \frac{χ_{γ}(w_{2})}{χ_{γ}(w_{1})}. In this notation we get

(x_{d,s}(σ_{α}^{1}, σ_{β})) = \frac{1}{d!}χ^{t}μ^{s}χW.  (2-9)

3. The case of one zero.

In this section we deal with a covering of the sphere by a Riemann surface of genus g with two degenerate ramification points, one of them has the maximal order. In notation of the introduction, we calculate numbers CN_{d}((d), β; g). We fix symbol τ for a cycle of maximal length, belonging to S_{d}. 3.1. It is known, that if χ_{λ} is a character of irreducible representation of S_{d} corresponding to a partition λ = (1^{i_{1}} 2^{i_{2}} … d^{i_{d}}), then

χ_{λ}(τ) = \{ \begin{matrix}(−1)^{k+1}, & if λ = (1^{k−1} (d − k + 1)), \\ 0, & in other cases. \end{matrix}

Therefore, the general formula (2-8) can be rewritten as follows

x_{d,s}(τ, σ_{β}) = \frac{|W_{β}|}{d!} ∑_{γ⊢d}(χ_{γ}(σ_{β})χ_{γ}(τ) (\frac{χ_{γ}(w_{2})}{χ_{γ}(w_{1})})^{s}) = \frac{|W_{β}|}{d!} ∑_{k=1}^{d} χ_{k}^{d}(σ_{β})(−1)^{k+1}μ_{d,k}^{s},  (3-1)

where χ_{k}^{d} = χ_{(1^{k−1} (d−k+1))} and μ_{d,k} = \frac{χ_{k}^{d}(w_{2})}{χ_{k}^{d}(w_{1})}.

Let us calculate χ_{k}^{d}(12), i.e. the value of the character χ_{k}^{d} on the transposition (12). It is known, that the representation, corresponding to the partition λ_{k} = (1^{k−1} (d − k + 1)) is the (k − 1)th exterior power of the simplicial representation, if k ≥ 2, and the trivial onedimensional representation, if k = 1. Therefore, χ_{1}^{d}(12) = 1.

Let us recall the construction of simplicial representation of the symmetric group S_{d}. Let V be a (d − 1)-dimensional vector space with a basis e_{1}, … e_{d−1}. By definition put powere_{d} =

− ∑_{i=1}^{d−1} e_{i}. Then the simplicial representation of S_{d} on the space V can be defined by the rule:

σ(e_{i}) = e(σ(i)).

Therefore, to calculate χ_{2}^{d}(12) we need to fined basis vectors such that (12)e_{i} = e_{i}. This condition is true for i = 3, 4, …, (d − 1). Hence,

χ_{2}^{d}(12) = d − 3.

The (k − 1)th exterior of the simplicial representation can be defined as

σ : e_{i_{1}} ∧ e_{i_{2}} ∧ ⋯ ∧ e_{i_{k−1}} ↦ e_{σ(i_{1})} ∧ e_{σ(i_{2})} ∧ ⋯ ∧ e_{σ(i_{k−1})}.

We have

(12) : e_{i_{1}} ∧ e_{i_{2}} ∧ ⋯ ∧ e_{i_{k−1}} ↦ \{ \begin{matrix}−e_{i_{1}} ∧ e_{i_{2}} ∧ ⋯ ∧ e_{i_{k−1}}, & if i_{1} = 1, i_{2} = 2, \\ e_{i_{1}} ∧ e_{i_{2}} ∧ ⋯ ∧ e_{i_{k−1}}, & if i_{3} ≥ 3, \\ ≠ ±e_{i_{1}} ∧ e_{i_{2}} ∧ ⋯ ∧ e_{i_{k−1}}, & in other cases. \end{matrix}

This implies that for k ∈ [3, d − 2] the following formulae takes place

χ_{k}^{d}(12) = \binom{d − 3}{k − 1} − \binom{d − 3}{k − 3} = \frac{(d − 2)!(d − 2k + 1)}{(k − 1)!(d − k)!}.  (3-2)

But assuming that \binom{b}{a} = 0 whenever a > b or a < 0, the last formula will true for any k.

Now we recall that χ_{k}^{d}(w_{2}) equals |W_{2}|χ_{k}^{d}(12) = \frac{d(d−1)}{2}χ_{k}^{d}(12) and χ_{k}^{d}(w_{1}) is dimension of the space of the corresponding representation, i.e., χ_{k}^{d}(w_{1}) = \binom{d−1}{k−1}. Thus,

μ_{d,k} = \frac{χ_{k}^{d}(w_{2})}{χ_{k}^{d}(w_{1})} = \frac{d(d − 2k + 1)}{2},  (3-3)

and formula (3-1) can be simplified.

x_{d,s}(τ, σ_{β}) = \frac{|W_{β}|d^{s}}{2^{s}d!} ∑_{k=1}^{d} χ_{k}^{d}(σ_{β})(−1)^{k+1}ν_{k}^{s},  (3-4)

where ν_{k} = d − 2k + 1. 3.2. Note that each a collection {π_{i}} ∈ Π_{d}^{s}(τ, σ_{β}) is connected. Therefore, to calculate a number CN_{d}((d); β; g) it is sufficient to compute the number of St_{τ} orbits in the set Π_{d}^{s}(τ, σ_{β}).

At first we suppose that d is odd. In this case the subgroup St_{τ} does not contain order 2 elements, since order of element should divide order of the subgroup. Moreover, each element of this subgroup acts on the set {1, 2, …, d} without fixed points. Hence, any element of St_{τ} does not commute with a transposition and the subgroup acts free on the set Π_{d}^{s}(τ, σ_{β}) in this case. Therefore,

CN_{d}(τ; β; g) = \frac{1}{d}x_{d,s}(τ, σ_{β})

for some s depending of g, β.

Next we consider the case of even d, i.e. d = 2r. We have, τ^{r} = π_{1}π_{2} ⋯ π_{r}, where π_{i}’s are transpositions. Denote be M the set of these transpositions. It is clear, that an element of St_{τ} commutes with such a transposition iff it is τ^{r}. Therefore,

if a collection from Π_{d}^{s}(τ, σ_{β}) consists of π_{i} ∈ M only, then the orbit containing this collection has r points exactly (such a collection we call a collection of the first type);

if a collection from Π_{d}^{s}(τ, σ_{β}) contains at least one a transposition which does not belong to M, then its orbit has d = 2r points (such a collection we call a collection of the second type).

Let us calculate the number of the first type collections in the set Π_{d}^{s}(τ, σ_{β}) in accordance with β.

Lemma. We can enumerate elements π_{i}’s of the set M such that τπ_{i}τ^{−1} = π_{i ˙+1}, where ˙+ denotes the sum in the ring ℤ_{r}.

Proof. Note that the product τπτ^{−1} belongs to the set M whenever π ∈ M. In fact, this product equals a transposition and commutes with any power of τ. Suppose now, that τ^{p}πτ^{−p} = τ^{q}πτ^{−q} for some integers 0 ≤ p < q < r. Then τ^{p−q}πτ^{q−p} = π, i.e. the number (p − q) is divided by r. This contradiction proves the lemma.

Starting at this moment we use indexing of transpositions from M as in the previous lemma. Lemma. τπ_{i}π_{i ˙+1} = π_{i ˙+1}τπ_{i ˙+1}. Proof. The lemma is proved by direct calculation:

τπ_{i}τ^{−1} = π_{i ˙+1} ⇔ τπ_{i} = π_{i ˙+1}τ ⇔ τπ_{i}π_{i ˙+1} = π_{i ˙+1}τπ_{i ˙+1}.

Let us denote by G the subgroup of S_{d} which is generated by transpositions of the set M. Obviously, the subgroup G is commutative.

Lemma. For any pair of transpositions π_{i} and π_{i ˙+p} there exists an element g ∈ G such that

τπ_{i}π_{i ˙+p} = gτg.

Proof. The proof is by induction on p. The case p = 1 follows from the previous lemma. Let us do the inductive step.

τπ_{i}π_{i ˙+p} = (τπ_{i}π_{i ˙+(p−1)})π_{i ˙+(p−1)}π_{i ˙+p} = (g_{1}τg_{1})π_{i ˙+(p−1)}π_{i ˙+p} =

= g_{1}(τπ_{i ˙+(p−1)}π_{i ˙+p})g_{1} = g_{1}π_{i ˙+p}τπ_{i ˙+p}g_{1} = gτg.

Lemma. Let π_{i_{1}}, π_{i_{2}}, …, π_{i_{2k−1}}, π_{i_{2k}} be transpositions of the set M. The there exists an element g ∈ G such that

τπ_{i_{1}}π_{i_{2}} ⋯ π_{i_{2k−1}}π_{i_{2k}} = gτg.

Proof. The proof is by induction on k. The case k = 1 follows from the previous lemma. Let us do the inductive step.

τπ_{i_{1}}π_{i_{2}} ⋯ π_{i_{2k−1}}π_{i_{2k}} = (τπ_{i_{1}}π_{i_{2}} ⋯ π_{i_{2k−3}}π_{i_{2k−2}})π_{i_{2k−1}}π_{i_{2k}} =
= (g_{1}τg_{1})π_{i_{2k−1}}π_{i_{2k}} = g_{1}(τπ_{i_{2k−1}}π_{i_{2k}})g_{1} = g_{1}(g_{2}τg_{2})g_{1} = gτg.

Lemma. If π ∈ M, then cycle type of the product τπ is (r, r).

Proof. Sins τ = (1, 2, …, 2r) = (2, 3, …, 2r, 1) = ⋯ = (2r, 1, …, 2r − 1) and the transposition π has the form (i, i + r), we can assume, that π = (1, r + 1). Now the lemma follows from calculation:

(1, 2, …, 2r)(1, r + 1) = (1, r + 2, r + 3, …, 2r)(r + 1, 2, 3 …, r).

Corollary. Let π_{1}, π_{2}, …, π_{k} be transpositions from M, then cycle type of the product τπ_{1}π_{2} ⋯ π_{k} is either

(2r), if k is even, or

(r, r), if k is odd. 3.3. Proposition.

CN_{2r+1}(τ, σ_{β}) = \frac{1}{2r + 1}x_{2r+1,s}(τ, σ_{β}).

CN_{2r}(τ, σ_{β}) = \{ \begin{matrix}\frac{1}{2r}x_{2r,s}(τ, σ_{β}) + \frac{r^{s−1}}{2} , & if β = (2r) or β = (r, r) \\ \frac{1}{2r}x_{2r,s}(τ, σ_{β}), & in other cases. \end{matrix}

Proof. As we show in the beginning of the section, the subgroup St_{τ} acts on the set Π_{d}^{s}(τ, σ_{β}) free whenever d = 2r + 1. This implies the first formula.

Assume that d = 2r. Then the set Π_{d}^{s}(τ, σ_{β}) has a collection of the first type only if cycle type σ_{β} is either (2r) or (r, r). This fact follows from the corollary. Therefore, the subgroup St_{τ} acts on the set Π_{d}^{s}(τ, σ_{β}) free unless either β = (2r) or β = (r, r).

Suppose now, that β is one of the singular partitions. Then the set Π_{d}^{s}(τ, σ_{β}) has collections of the first type. Moreover, quantity of such collections is r^{s} (in both cases).

These arguments prove the proposition. 3.4. Here we use formula (3-4) to calculate CN_{d}((d); (d); g). At first, |W_{(d)}| = (d − 1)! and

x_{d,s}(τ, τ′) = \frac{d^{s−1}}{2^{s}} ∑_{k=1}^{d}(d − 2k + 1)^{s},

where s is the number of simple ramification point and it equals 2g in our situation. In particular, s is even.

By definition put ∑_{k=1}^{d} ν_{k}^{s} = D_{s}(d) and consider two cases. The first case is d = 2n + 1. Then

D_{s}(d) = 2^{s} ∑_{k=1}^{2n+1}(n − k + 1)^{s} = 2^{s+1}S_{s}(n + 1),

where S_{s}(m) is the standard notation of a sum ∑_{k=1}^{m−1} k^{s}. This sum can be rewritten with help of the Bernoulli polynomials (see [2]):

S_{s}(m) = \frac{1}{s + 1}(B_{s+1}(m) − B_{s+1}(0)),  (3-5)

where B_{p}(x) = ∑_{k=0}^{p} \binom{p}{k}b_{k}x^{m−k} is the Bernoulli polynomial and b_{k}’s are Bernoulli numbers.

Thus,

x_{d,s}(τ, τ′) = \frac{2 · d^{s−1}}{s + 1} (B_{s+1} (\frac{d + 3}{2}) − B_{s+1}(0)).

Further, taking into account the proposition on page 9, we have

CN_{d}((d); (d); g) = \frac{2 · d^{2g−2}}{2g + 1} (B_{s+1} (\frac{d + 3}{2}) − B_{s+1}(0)) (if d is odd).

The second case is d = 2n. In this case D_{s}(d) = ∑_{k=1}^{2n}(2n−2k+1)^{s} = 2 ∑_{k=1}^{2n}(2k−1)^{s}. Denote by

O_{s}(n) the sum ∑_{k=1}^{n}(2k − 1)^{s} and by E_{s}(n) the sum ∑_{k=1}^{n}(2k)^{s}. Then O_{s}(n) + E_{s}(n) = S_{s}(2n + 1). On the other hand, E_{s}(n) = 2^{s}S_{s}(n + 1). Hence,

O_{s}(n) = S_{s}(2n + 1) − 2^{s}S_{s}(n + 1) = \frac{B_{s+1}(2n + 1) − 2^{s}B_{s+1}(n + 1) + B_{s+1}(0)(2^{s} − 1)}{s + 1}. (3-6)

Therefore,

x_{d,2g}(τ, τ′) = \frac{d^{2g−1}(B_{2g+1}(d + 1) − 2^{2g}B_{2g+1}(\frac{d+2}{2}) − B_{2g+1}(0)(2^{2g} − 1))}{2^{2g−1}(2g + 1)}.

Using the proposition on page 9, we get

CN_{d}((d); (d); g) = \frac{d^{2g−2}(B_{2g+1}(d + 1) − 2^{2g}B_{2g+1}(\frac{d+2}{2}) − B_{2g+1}(0)(2^{2g} − 1) + \frac{d(2g+1)}{2})}{2^{2g−1}(2g + 1)}.

Combining together our calculations, we can formulate a proposition. Proposition.

CN_{d}((d); (d); g) = \{ \begin{matrix}\frac{d^{2g−2}(B_{2g+1}(d+1)−2^{2g}B_{2g+1}( \frac{d+2}{2} )−B_{2g+1}(0)(2^{2g}−1)+ \frac{d(2g+1)}{2} )}{2^{2g−1}(2g+1)} & for even d, \\ \frac{2·d^{2g−2}}{2g+1} (B_{2g+1} (\frac{d+3}{2}) − B_{2g+1}(0)) & for odd d. \end{matrix}

3.5. Denote by σ_{l} a permutation from the group S_{d} of the cycle type (l, d − l) with l ≤ d − l. Then

χ_{k}^{d}(σ_{l}) = \{ \begin{matrix}(−1)^{k+1}, & if k ≤ l, \\ 0, & if l < k ≤ d − l, \\ (−1)^{k}, & if d − l < k. \end{matrix}  (3-7)

Lemma. In the case σ_{β} = σ_{l} formula (3-4) has the form

x_{d,s}(τ, σ_{l}) = \frac{|W_{l}|d^{s}}{d!2^{s−1}} ∑_{k=1}^{l}(d − 2k + 1)^{s}.  (3-8)

Proof. By definition put ν_{k} = (d − 2k + 1) and consider two cases.

The first. If d = 2n − 1, then the permutation τ is even and for any l the permutation σ_{l} is odd. Therefore, s is odd and we have the following relations

(−1)^{k+1} = χ_{k}^{d}(τ) = χ_{d+1−k}^{d}(τ) = (−1)^{k+1}, χ_{k}^{d}(σ_{l}) = −χ_{d+1−k}^{d}(σ_{l}), ν_{k}^{s} = −ν_{d+1−k}^{s}.

This implies that χ_{k}^{d}(τ)χ_{k}^{d}(σ_{l})ν_{k}^{s} = χ_{d+1−k}^{d}(τ)χ_{d+1−k}^{d}(σ_{l})ν_{d+1−k}^{s}.

The second. If d = 2n, then the permutation τ is odd and for any l the permutation σ_{l} is even. Therefore, s is odd and we have

(−1)^{k+1} = χ_{k}^{d}(τ) = −χ_{d+1−k}^{d}(τ) = −(−1)^{d+−k}, χ_{k}^{d}(σ_{l}) = χ_{d+1−k}^{d}(σ_{l}), ν_{k}^{s} = −ν_{d+1−k}^{s}.

In particular, the previous relation is true too. Now the proof follows from (3-7). Proposition. If d is odd, then

CN_{d}((d); (l, d − l); g) = \frac{2d^{2g}}{l(d − l)(2g + 2)} (B_{2g+2} (\frac{d + 1}{2}) − B_{2g+2} (\frac{d + 1 − 2l}{2} )) .

If d is even and l < \frac{d}{2}, then

CN_{d}((d); (l, d − l); g) = \frac{d^{2g}}{(2g + 2)2^{2g}l(d − l)}×

× (B_{2g+2}(d + 1) − B_{2g+2}(d − 2l + 1) + 2^{2g+1} (B_{2g+2} (\frac{d − 2l + 2}{2}) − B_{2g+2} (\frac{d + 2}{2}))) .

If d is even and l = \frac{d}{2}, then CN_{d}((d); (l, d − l); g) =

= \frac{d^{2g−2}}{(2g + 2)2^{2g−1}} (B_{2g+2}(d + 1) − B_{2g+2}(1) + 2^{2g+1} (B_{2g+2}(1) − B_{2g+2} (\frac{d + 2}{2}))) + \frac{d^{2g}}{2^{2g+1}} .

Proof. We calculate the numbers s with help of the Riemann–Hurwitz formulae: s = 2g+1.

Suppose that d = 2n−1 is odd. Then ν_{k} = d−2k+1 = 2(n−k) and ∑_{k=1}^{l} ν_{k}^{s} = 2^{s} ∑_{k=1}^{l}(n−k)^{s} =

2^{s}(S_{s}(n) − S_{s}(n − l)). In addition, |W_{(l,d−l)}| = \frac{d!}{l(d−l)}. Finally, by the proposition on page 9 we have

CN_{d}((d); (d, d − l); g) = \frac{1}{d}x_{d,2g+1}(τ, σ_{l}).

Now the first formula of the proposition follows from (3-5) and the previous lemma.

Suppose that d = 2n is even. In this case ν_{k} = d − 2k + 1 = 2(n − k) + 1 and ∑_{k=1}^{l} ν_{k}^{s} =

O_{s}(n) − O_{s}(n − l), where O_{s}(n) = ∑_{k=1}^{n}(2k − 1)^{s}. Besides,

|W_{(l,d−l)}| = \{ \begin{matrix}\frac{d!}{l(d−l)} & if l < d/2 \\ \frac{d!}{2l^{2}} & if l = d/2. \end{matrix}

Further, taking into account the proposition on the page 9, formulae (3-5), (3-6) and the previous lemma we obtain the last formulae of the proposition. 3.6. Let us research the case when σ_{β} is a 3-cycle, i.e., σ_{β} = (abc). Then

χ_{k}^{d} = \binom{d − 4}{k − 4} + \binom{d − 4}{k − 1}.

It can be checked that

\binom{d − 4}{k − 4} + \binom{d − 4}{k − 1} = \frac{1}{d − 2}\binom{d − 1}{k − 1} (\frac{3}{4}ν_{k}^{2} + \frac{(d − 1)(d − 5)}{4}) ,

where ν_{k} = d + 1 − 2k as usual. On the other hand, |W_{3}| = \frac{d(d−1)(d−2)}{3}. Therefore,

x_{d,s}(τ, (abc)) = \frac{d^{s}}{3 · 2^{s}(d − 2)!} ∑_{k=1}^{d}(−1)^{k+1}\binom{d − 1}{k − 1} (\frac{3}{4}ν_{k}^{2} + \frac{(d − 1)(d − 5)}{4}) ν_{k}^{s}.

To write the number NC_{d}((d), (1^{d−3} 3)); g, we use the notation:

σ(n, s, α) = ∑_{k=0}^{n} \binom{n}{k}(−1)^{k}(α − k)^{s}.

These numbers have the following property.

Lemma ([1]). ∑_{s=0}^{∞} σ(n, s, α)\frac{x^{s}}{s!} = e^{αx}(1 − e^{−x})^{n}.

Taking into account the lemma, we get

∑_{k=1}^{d}(−1)^{k+1}\binom{d − 1}{k − 1}ν_{k}^{s} = 2^{s}σ(d − 1, s, \frac{d − 1}{2}).

Hence,

x_{d,s}(τ, (abc)) = \frac{d^{s}}{(d − 2)!} (σ(d − 1, s + 2, \frac{d − 1}{2}) + \frac{(d − 1)(d − 5)}{12}σ(d − 1, s, \frac{d − 1}{2})) .

Combining this calculation with the proposition on the page 9 and the fact that s = 2g + d − 3, we obtain Proposition.

CN_{d}((d); (1^{d−3} 3)) = \frac{d^{(}s − 1)}{(d − 2)!} (σ(d − 1, s + 2, \frac{d − 1}{2}) + \frac{(d − 1)(d − 5)}{12}σ(d − 1, s, \frac{d − 1}{2})) ,

where s = 2g + d − 3.

4. Coverings of degree 3.

4.1. Here we deal with primitive coverings of degree 3. There exists only one possibility for orders of degenerate ramification points: (3). In the following table we give number s(g,m,k) of simple ramification points for degree 3 primitive coverings.

(3) (1, 2)
(3)  2g 2g + 1
(1, 2) 2g + 1 2g + 2

We start our calculations with numbers x_{3,s}(σ_{α}^{1}, σ_{β}) with help of formula (2-9).

Order partitions of 3: 1^{3}; 1 2; 3 and write the matrixes χ, μ and W (for character values we use the book [3]).

χ = ( \begin{matrix}1 & 1 & 1 \\ 2 & 0 & −1 \\ 1 & −1 & 1 \end{matrix} ) μ = ( \begin{matrix}3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & −3 \end{matrix} ) W = ( \begin{matrix}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{matrix} ) .

Hence,

\begin{matrix}x_{3,s}((12), (ab)) = 3^{s}, & x_{3,s}((12), (abc)) = 2 · 3^{s−1} \\ x_{3,s}((123), (ab)) = 3^{s}, & x_{3,s}((123), (abc)) = 2 · 3^{s−1} \end{matrix}  (4-1)

4.2. At first we calculate the number N_{3}((1 2); (1 2); g). Fix the permutation σ_{α}^{1} = (12) and consider the set Π_{3}^{2g+2}((12); (ab)). By formula (4-1) and the definition of x_{3,s}((12); (ab)) we obtain

|Π_{3}^{2g+2}((12); (ab))| = 3^{2g+2}.

The subgroup St_{(12)} acts on the set Π_{3}^{2g+2}((12); (ab)) with a unique fixed point {π_{i}} = {(12)}. In addition, order of St_{(12)} is 2. Therefore,

N_{3}((1 2); (1 2); g) = \frac{1}{2}(x_{2,2g+2}((12); (ab)) − 1) + 1 = \frac{1}{2}(3^{2g+2} − 1) + 1.

Further, the subgroup M generated by the transpositions (12) and π_{i}’s acts on the set {1, 2, 3} transitive whenever at least one of the π_{i}’s does not equal (12). Therefore,

CN_{3}((1 2); (1 2); g) = \frac{1}{2}(x_{3,2g+2}((12); (ab)) − 1) = \frac{1}{2}(3^{2g+2} − 1).

4.3. The next number we calculate is N_{3}((1 2); (3); g). To do this we hold fixed a transposition (12) and consider the set Π_{3}^{2g+1}((12); (abc)). According to (4-1),

|Π_{3}^{2g+1}((12); (abc))| = 2 · 3^{2g}.

The subgroup St_{(12)} acts on the set Π_{3}^{2g+1}((12); (abc)) without fixed points. Its order is 2. Hence,

N_{3}((1 2); (3); g) = \frac{1}{2} |Π_{3}^{2g+1}((12); (abc))| = 3^{2g}.

Moreover, since the monodromy group of a corresponding covering has a cycle of the maximal length, all the coverings are connected, i.e.,

CN_{3}((1 2); (3); g) = N_{3}((1 2); (3); g) = 3^{2g}.

4.4. The last number we need to compute is N_{3}((3); (3); g). It follows from (4-1) that the number |Π_{3}^{2g}((123); (abc))| equals 2·3^{2g−1}. The subgroup St_{(123)} acts free on the set Π_{3}^{2g}((123); (abc))

and |St_{(123)}| = 3. Moreover, any collection of this set is connected. Thus,

CN_{3}((3); (3); g) = N_{3}((3); (3); g) = 2 · 3^{2g−2}.

These calculations can be summarized in the following tables.

\begin{aligned}& N_{3}(α; β; g) \\ (3) & (1, 2) \\ (3) & 2 · 3^{2g−2} & 3^{2g} \\ (1, 2) & 3^{2g} & \frac{3^{2g+2}−1}{2} + 1 \end{aligned}  \begin{aligned}& CN_{3}(α; β; g) \\ (3) & (1, 2) \\ (3) & 2 · 3^{2g−2} & 3^{2g} \\ (1, 2) & 3^{2g} & \frac{3^{2g+2}−1}{2} \end{aligned}

5. Coverings of degree 4.

5.1. For degree 4 coverings there exist the following possible orders of degenerated ramification points:

(1, 3), (2, 2), (4).

Using the Riemann-Hurwitz formula (1-1), find quantities s = s(m, k, g) of simple ramification points.

s = s(m, k, g)
(1, 1, 2) (1, 3) (2, 2) (4)
(1, 1, 2) 2g + 4 2g + 3 2g + 3 2g + 2
(1, 3)  2g + 2 2g + 2 2g + 1
(2, 2)  2g + 2 2g + 1
(4)  2g

As in the case of degree 3 we order partitions of 4: 1^{4}; 1^{2} 2; 2^{2}; 1 3; 4 and take character values from the book [3]. We have

χ = ( \begin{matrix}1 & 1 & 1 & 1 & 1 \\ 3 & 1 & −1 & 0 & −1 \\ 2 & 0 & 2 & −1 & 0 \\ 3 & −1 & −1 & 0 & 1 \\ 1 & −1 & 1 & 1 & −1 \end{matrix} ) W = ( \begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 0 & 6 \end{matrix} ) μ = ( \begin{matrix}6 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 0 & −6 \end{matrix} )

and

\begin{matrix}x_{4,s}((12), (ab)) = 2^{s−1}(3^{s} + 1), & x_{4,s}((12), (ab)(cd)) = 2^{s−2}(3^{s} − 1), \\ x_{4,s}((12), (abc)) = 2^{s+1}3^{s−1}, & x_{4,s}((12), (abcd)) = 2^{s−1}(3^{s} − 1), \\ x_{4,s}((12)(34), (ab)) = 2^{s−1}(3^{s} − 1), & x_{4,s}((12)(34), (ab)(cd)) = 2^{s−2}(3^{s} + 1), \\ x_{4,s}((12)(34), (abc)) = 2^{s+1}3^{s−1}, & x_{4,s}((12)(34), (abcd)) = 2^{s−1}(3^{s} + 1), \\ x_{4,s}((123), (ab)) = 2^{s−1}3^{s}, & x_{4,s}((123), (ab)(cd)) = 2^{s−2}3^{s}, \\ x_{4,s}((123), (abc)) = 2^{s+1}3^{s−1}, & x_{4,s}((123), (abcd)) = 2^{s−1}3^{s}, \\ x_{4,s}((1234), (ab)) = 2^{s−1}(3^{s} − 1), & x_{4,s}((1234), (ab)(cd)) = 2^{s−2}(3^{s} + 1), \\ x_{4,s}((1234), (abc)) = 2^{s+1}3^{s−1}, & x_{4,s}((1234), (abcd)) = 2^{s−1}(3^{s} + 1), \end{matrix}  (5-1)

5.2. Now we are ready to calculate the Hurwitz numbers for coverings of degree 4. Let us start with N_{4}((1^{2} 2); (1^{2} 2); g). Recall that this number is the quantity of all the topological nonequivalent coverings f : C_{g} → S^{2} with 2g + 6 simple ramification points. Taking into account (5-1), we have

|Π_{4}^{2g+4}((12), (ab))| = 2^{2g+3}(3^{2g+4} + 1).

In the set Π_{4}^{2g+4}((12), (ab)) there exist fixed collections w.r.t. the St_{(12)} action. The fixed collections consist of the transpositions (12) and (34). Therefore, the number of the fixed collections is 2^{2g+4}. All other orbits have 4 different points. Hence,

N_{4}((1^{2} 2); (1^{2} 2); g) = 2^{2g+4} + \frac{1}{4} (|Π_{4}^{2g+4}((12), (ab))| − 2^{2g+4}) = 2^{2g+1}(3^{2g+4} + 7).

To calculate the number CN_{4}((1^{2} 2); (1^{2} 2); g) consider subsets of Π_{4}^{2g+4}((12), (ab)) consisting of disconnected collections.

D_{1} = {(π_{1}, …, π_{2g+4}) ∈ Π_{4}^{2g+4}((12), (ab))| π_{i} = (12) ∀i.} ,
D_{2} = {(π_{1}, …, π_{2g+4}) ∈ Π_{4}^{2g+4}((12), (ab))| π_{i} ∈ {(12), (34)} ∀i} ,
D_{3} = {(π_{1}, …, π_{2g+4}) ∈ Π_{4}^{2g+4}((12), (ab))| π_{i}(4) = 4 ∀i} ,
D_{4} = {(π_{1}, …, π_{2g+4}) ∈ Π_{4}^{2g+4}((12), (ab))| π_{i}(3) = 3 ∀i, } .

We see that a collection {π_{i}} ∈ Π_{4}^{2g+4}((12), (ab)) corresponds to a connected covering whenever

{π_{i}} ∉ D = ⋃_{α=1}^{4} D_{α}. Besides,

|D_{1}| = 1, |D_{2}| = 2^{2g+4}, |D_{3}| = |D_{4}| = |Π_{3}^{2g+4}((12), (ab))| = 3^{2g+4} (see (4 − 1));
D_{1} ⊂ D_{α} ∀α, D_{α} ∩ D_{β} = D_{1} for α ≠ β.

Thus,

\begin{aligned}CN_{4}((1^{2} 2); (1^{2} 2); g) & = \frac{1}{4} (|Π_{4}^{2g+4}((12), (ab))| − 2|D_{3}| − |D_{2}| + 2) = \\ & = \frac{1}{2}(3^{2g+4} − 1)(2^{2g+2} − 1). \end{aligned}

5.3. The next number is N_{4}((1^{2} 2); (1 3); g). It follows from (5-1) that

|Π_{4}^{2g+3}((12), (abc))| = 2^{2g+4} · 3^{2g+2}.

On the set Π_{4}^{2g+3}((12), (abc)) the group St_{(12)} acts free. Since all orbits of this action have 4 collections exactly, we obtain

N_{4}((1^{2} 2); (1 3); g) = 2^{2g+2} · 3^{2g+2}.

Now we compute the number CN_{4}((1^{2} 2); (1 3); g) of connected coverings. Consider the subsets of Π_{4}^{2g+3}((12), (abc)) consisting of disconnected collections.

D_{3} = {(π_{1}, …, π_{2g+3} ∈ Π_{4}^{2g+3}((12), (abc))| π_{i}(3) = 3 ∀i} ,
D_{4} = {(π_{1}, …, π_{2g+3} ∈ Π_{4}^{2g+3}((12), (abc))| π_{i}(4) = 4 ∀i} .

It is clear that |D_{3}| = |D_{4}|, D_{3} ∩ D_{4} = Ø and |D_{4}| = |Π_{3}^{2g+3}((12), (abc))| = 2 · 3^{2g+2} (see 4-1). Therefore,

CN_{4}((1^{2} 2); (1 3); g) = \frac{1}{4} (|Π_{3}^{2g+3}((12), (abc))| − 2|D_{4}|) = 3^{2g+2}(2^{2g+2} − 1).

5.4. Let us calculate the number N_{4}((1^{2} 2); (2^{2}); g). By (5-1) the quantity of elements of the set Π_{4}^{2g+3}((12); (ab)(cd)) is 2^{2g+1}(3^{2g+3} − 1). As before, the group St_{(12)} acts on the set Π_{4}^{2g+3}((12), (ab)(cd)) by conjugations and we need to determine number of the action orbits. A collection {π_{i}} is fixed w.r.t. this action iff π_{i} ∈ {(12), (34)}. On the other hand such a collection belongs to Π_{4}^{2g+3}((12), (ab)(cd)) only if quantity of the transpositions (12) into it is even and, consequently, quantity of (34) is odd. Thus the subset of fixed collections (i.e., one point orbits) has 2^{2g+2} elements. Another orbit has 4 elements. Hence,

N_{4}((1^{2} 2); (2^{2}); g) = 2^{2g+2} + 2^{2g−1}(3^{2g+3} − 3) = 2^{2g−1}(3^{2g+3} + 5).

Further, to compute CN_{4}((1^{2} 2); (2^{2}); g) we remark that only fixed collections w.r.t. St_{(12)} action are disconnected. Therefore,

CN_{4}((1^{2} 2); (2^{2}); g) = 2^{2g−1}(3^{2g+3} − 3).

5.5. The next number is N_{4}((1^{2} 2); (4); g). Recall that Π_{4}^{2g+2}((12), (abcd)) is the set of all collections of transpositions (π_{1}, …, π_{2g+2}) satisfying (2-1) with σ_{α}^{1} = (12) and σ_{β} = (abcd), a 4-cycle. By formula (5-1) we have

|Π_{4}^{2g+2}((12), (abcd))| = 2^{2g+1}(3^{2g+2} − 1).

The subgroup St_{(12)} acts on the set Π_{4}^{2g+2}((12), (abcd)) free and each collection of this set is connected. Thus,

CN_{4}((1^{2} 2); (4); g) = N_{4}((1^{2} 2); (4); g) = 2^{2g−1}(3^{2g+2} − 1).

5.6. Let us compute the number N_{4}((1 3); (1 3); g). By relations (5-1),

|Π_{4}^{2g+2}((123), (abc))| = x_{4,2g+2}((123), (abc)) = 2^{2g+3} · 3^{2g+1}.

The subgroup St_{(123)} acts free on the set Π_{4}^{2g+2}((123), (abc)). Its order equals 3. Whence,

N_{4}((1 3); (1 3); g) = 2^{2g+3} · 3^{2g}.

To calculate the number of connected coverings, we need reject from Π_{4}^{2g+2}((123), (abc)) collections {π_{i}} such that π_{i}(4) = 4 ∀i. Denote the set of all such collections by D_{4}. It is clear, that D_{4} = Π_{3}^{2g+2}((123), (abc)). Hence, |D_{4}| = 2 · 3^{2g+1} (see (4-1)). Thus,

CN_{4}((1 3); (1 3); g) = 3^{2g}(2^{2g+3} − 2).

5.7. Now we wont to search the number N_{4}((1 3); (2^{2}); g). The set Π_{4}^{2g+2}((123), (ab)(cd)) has x_{4,2g+2}((123), (ab)(cd)) = 3^{2g+2} · 2^{2g} elements (see (5-1)). The subgroup St_{(123)} acts on this set free. Hence,

N_{4}((1 3); (2^{2}); g) = 3^{2g+1} · 2^{2g}.

Moreover, any covering with degenerate ramification points of such a type is connected. Therefore,

CN_{4}((1 3); (2^{2}); g) = N_{4}((1 3); (2^{2}); g) = 3^{2g+1} · 2^{2g}.

5.8. N_{4}((1 3); (4); g). It follows from (5-1) that the set Π_{4}^{2g+1}((123), (abcd)) has 3^{2g+1} · 2^{2g} elements. The group St_{(123)} acts on this set free and any collection of Π_{4}^{2g+1}((123), (abcd)) is connected. We have

CN_{4}((1 3); (4); g) = N_{4}((1 3); (4); g) = 3^{2g} · 2^{2g}.

5.9. Each a collection {π_{i}} ∈ Π_{4}^{2g+2}((12)(34), (ab)(cd)) corresponds to a meromorphic function f : C_{g} → S^{2} with 2 zeros of multiplicities 2 and 2, and 2 poles with the same orders. The subgroup S_{(12)(34)} consists of the following permutations:

St_{(12)(34)} = {id, (12), (34), (12)(34), (1324), (4231), (13)(24), (14)(23)}.

In particular, |St_{(12)(34)}| = 8. This group acts on the set Π_{4}^{2g+2}((12)(34), (ab)(cd)) by conjugations. Suppose that a collection {π_{i}} ∈ Π_{4}^{2g+2}((12)(34), (ab)(cd)) is fixed w.r.t this action. Then there exists a transposition π such that σπσ^{−1} = π for any σ of the subgroup St_{(12)(34)}. But |St_{π}| = 4. This contradiction proves that the action of the subgroup is free and

N_{4}((2^{2}); (2^{2}); g) = \frac{1}{8}x_{4,2g+2}((12)(34), (ab)(cd)) = 2^{2g−3}(3^{2g+2} + 1).

As usually, D is the subset of Π_{4}^{2g+2}((12)(34), (ab)(cd)) consisting of disconnected collections. It is clear that

D = {{π_{i}} ∈ Π_{4}^{2g+2}((12)(34), (ab)(cd))| π_{i} ∈ {(12), (34)} ∀i} .

A collection {π} consisting of (12) and (34) belongs to D iff the number of transpositions (12)’s into it is even. Hence, |D| = 2^{2g+1} and

CN_{4}((2^{2}); (2^{2}); g) = \frac{1}{8}(2^{2g}(3^{2g+2} + 1) − 2^{2g+1}) = 2^{2g−3}(3^{2g+2} − 1).

5.10. Finally, the group St_{(12)(34)} acts free on the set Π_{4}^{2g+1}((12)(34), (abcd)) and each a collection of this set corresponds to a connected covering. We get

CN_{4}((2^{2}); (4); g) = N_{4}((2^{2}); (4); g) = 2^{2g−3}(3^{2g+1} + 1).

5.11. To finish our considerations of degree 4 coverings it remains only to calculate

CN_{4}((4); (4); g) = N_{4}((4); (4); g).

It follows from (5-1) that |Π_{4}^{2g}((1234), (abcd))| = 2^{2g−1}(3^{2g} + 1). Consider the subgroup St_{(1234)} of S_{4} consisting of permutations commuting with (1234). If we denote (1234) by σ, then St_{(1234)} = {id, σ, σ^{2}, σ^{−1}}. It is clear, that for any transpositions π the inequality σπσ^{−1} ≠ π is true. But in the case π ∈ {(13), (24)} we have σ^{2}πσ^{−2} = π.

This observation and the definition of the set D imply

D = {{π_{i}} ∈ Π_{4}^{2g}((1234), (abcd))| π_{i} ∈ {(13), (24)} ∀i} .

Obviously, |D| = 2^{2g} and each an orbit of the St_{(1234)} actions belonging to D has two points exactly. Besides, the subgroup St_{(1234)} acts on the set Π_{4}^{2g}((1234), (abcd)) \ D free. Therefore,

CN_{4}((4); (4); g) = N_{4}((4); (4); g) = 2^{2g−1} + \frac{1}{4}(2^{2g−1}(3^{2g} + 1) − 2^{2g}) = 2^{2g−3}(3^{2g} + 3).

Rewrite the Hurwitz numbers for degree 4 primitive coverings of S^{2} in the following tables.

N_{4}(α; β; g)
(1^{2} 2)  (1 3)  (2^{2})  (4)
(1^{2} 2) 2^{2g+1}(3^{2g+4} + 7) 2^{2g+2}3^{2g+2} 2^{2g−1}(3^{2g+3} + 5) 2^{2g−1}(3^{2g+2} − 1)
(1 3)  2^{2g+3}3^{2g}  2^{2g}3^{2g+1}  2^{2g}3^{2g}
(2^{2})  2^{2g−3}(3^{2g+2} + 1) 2^{2g−3}(3^{2g+1} + 1)
(4)  2^{2g−3}(3^{2g} + 3)

CN_{4}(α; β; g)
(1^{2} 2) (1 3) (2^{2}) (4)
(1^{2} 2) \frac{(3^{2g+4}−1)(2^{2g+2}−1)}{2} 3^{2g+2}(2^{2g+2} − 1) 2^{2g−1}(3^{2g+3} − 3) 2^{2g−1}(3^{2g+2} − 1)
(1 3) 3^{2g}(2^{2g+3} − 2) 2^{2g}3^{2g+1} 2^{2g}3^{2g}
(2^{2})  2^{2g−3}(3^{2g+2} − 1) 2^{2g−3}(3^{2g+1} + 1)
(4)  2^{2g−3}(3^{2g} + 3)

6. The case of degree 5

6.1. Let us put in order partitions of integer 5 in the following way:

α_{1} = 1^{5}, α_{2} = 1^{3} 2, α_{3} = 1 2^{2}, α_{4} = 1^{2} 3, α_{5} = 2 3, α_{6} = 1 4, α_{7} = 5;

We use characters values of the symmetric group S_{5} from the book [3] to write the matrixes:

χ = ( \begin{matrix}1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 4 & 2 & 0 & 1 & −1 & 0 & −1 \\ 5 & 1 & 1 & −1 & 1 & −1 & 0 \\ 6 & 0 & −2 & 0 & 0 & 0 & 1 \\ 5 & −1 & 1 & −1 & −1 & 1 & 0 \\ 4 & −2 & 0 & 1 & 1 & 0 & −1 \\ 1 & −1 & 1 & 1 & −1 & −1 & 1 \end{matrix} ) , W = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 10 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 15 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 20 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 20 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 30 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 24 \end{matrix} ) ,

μ = ( \begin{matrix}10 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −5 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & −10 \end{matrix} ) .

Then, taking into account (2-9), we have

\begin{matrix}x_{5,s}((12345), (ab)) = \frac{1}{3}5^{s}(2^{s−1} − 1), & x_{5,s}((12345), (ab)(cd)) = 5^{s}2^{s−2}, \\ x_{5,s}((12345), (abc)) = \frac{1}{3}5^{s}(2^{s} − 1), & x_{5,s}((12345), (ab)(cde)) = \frac{1}{3}5^{s}(2^{s} + 1), \\ x_{5,s}((12345), (abcd)) = 5^{s}2^{s−1}, & x_{5,s}((12345), (abcde)) = 5^{s−1}(2^{s+1} + 2). \end{matrix}  (6-1)

\begin{matrix}x_{5,s}((1234), (abcd)) = 2^{s−1}(5^{s} + 1), & x_{5,s}((1234), (ab)(cde)) = \frac{1}{3}2^{s}(5^{s} − 1), \\ x_{5,s}((1234), (abc)) = \frac{1}{3}2^{s}(5^{s} + 1), & x_{5,s}((1234), (ab)(cd)) = 2^{s−2}(5^{s} − 1), \\ x_{5,s}((1234), (ab)) = \frac{1}{3}2^{s−1}(5^{s} − 1), \end{matrix}  (6-2)

\begin{matrix}x_{5,s}((12)(345), (ab)(cde)) = \frac{1}{3}(10^{s} + 5^{s} + 2^{s}), & x_{5,s}((12)(345), (abc)) = \frac{1}{3}(10^{s} − 5^{s} − 2^{s}), \\ x_{5,s}((12)(345), (ab)(cd)) = 2^{s−2}(5^{s} + 1), & x_{5,s}((12)(345), (ab)) = \frac{1}{3}(2^{s−1}(5^{s} + 1) − 5^{s}); \end{matrix}

(6-3)

\begin{matrix}x_{5,s}((123), (abc)) = \frac{1}{3}(10^{s} + 5^{s} + 2^{s}), & x_{5,s}((123), (ab)(cd)) = 2^{s−2}(5^{s} − 1), \\ x_{5,s}((123), (ab)) = \frac{1}{3}(2^{s−1}(5^{s} − 1) + 5^{s}); \end{matrix}  (6-4)

x_{5,s}((12)(34), (ab)(cd)) = 2^{s−2}(5^{s} + 1), x_{5,s}((12)(34), (ab)) = \frac{1}{3}2^{s−1}(5^{s} + 1);  (6-5)

x_{5,s}((12), (ab)) = \frac{1}{3}(2^{s−1}(5^{s} + 1) + 2 · 5^{s}).  (6-6)

6.2. Number of simple ramification points. We use Riemann–Hurwitz formulae (1-1) to fill the table of numbers of simple ramification points for corresponding coverings.

1^{3} 2 1 2^{2}  1^{2} 3  2 3  1 4 5
1^{3} 2 2g + 6
1 2^{2} 2g + 5 2g + 4
1^{2} 3 2g + 5 2g + 4 2g + 4
2 3 2g + 4 2g + 3 2g + 3 2g + 2
1 4 2g + 4 2g + 3 2g + 3 2g + 2 2g + 2
5 2g + 3 2g + 2 2g + 2 2g + 1 2g + 1 2g

6.3. Now we calculate numbers N_{5}((5); β; g). It is clear, that

N_{5}((5); β; g) = CN_{5}((5); β; g).

The subgroup St_{(12345)} acts on sets Π_{5}^{s}((12345), σ_{β}) free and each an orbit of this action has 5 points. Therefore,

CN_{5}((5); (5); g) = N_{5}((5); (5); g) = \frac{1}{5}x_{5,2g}((12345), (abcde)) = 5^{2g−2}(2^{2g+1} + 2);

CN_{5}((5); (1 4); g) = N_{5}((5); (1 4); g) = \frac{1}{5}x_{5,2g+1}((12345), (abcd)) = 5^{2g}2^{2g};

CN_{5}((5); (2 3); g) = N_{5}((5); (2 3); g) = \frac{1}{5}x_{5,2g+1}((12345), (ab)(cde)) = 5^{2g} \frac{2^{2g+1}+1}{3};

CN_{5}((5); (1^{2} 3); g) = N_{5}((5); (1^{2} 3); g) = \frac{1}{5}x_{5,2g+2}((12345), (abc)) = 5^{2g+1} \frac{2^{2g+2}−1}{3};

CN_{5}((5); (1 2^{2}); g) = N_{5}((5); (1 2^{2}); g) = \frac{1}{5}x_{5,2g+2}((12345), (ab)(cd)) = 5^{2g+1}2^{2g};

CN_{5}((5); (1^{3} 2); g) = N_{5}((5); (1^{3} 2); g) = \frac{1}{5}x_{5,2g+3}((12345), (ab)) = 5^{2g+2} \frac{2^{2g+2}−1}{3}.

6.4. Applying the proposition on page 9, we compute the numbers N_{5}((1 4); β; g).

N_{5}((1 4); (1 4); g) = \frac{1}{4}(x_{5,2g+2}((1234), (abcd)) − 2^{2g+2}) + 2^{2g+1} = 2^{2g−1}(5^{2g+2} + 3).

N_{5}((1 4); (1 2^{2}); g) = \frac{1}{4}(x_{5,2g+3}((1234), (ab)(cd)) − 2^{2g+3}) + 2^{2g+2} = 2^{2g−1}(5^{2g+3} + 3).

CN_{5}((1 4); (2 3); g) = N_{5}((1 4); (2 3); g) = \frac{1}{4}x_{5,2g+2}((1234), (ab)(cde)) = 2^{2g} \frac{5^{2g+2} − 1}{3}.

N_{5}((1 4); (1^{2} 3); g) = \frac{1}{4}x_{5,2g+3}((1234), (abc)) = 2^{2g+1} \frac{5^{2g+3} + 1}{3}.

N_{5}((1 4); (1^{3} 2); g) = \frac{1}{4}x_{5,2g+4}((1234), (ab)) = 2^{2g+1} \frac{5^{2g+4} − 1}{3}.

Note that if a transposition τ ∈ S_{5} satisfy the condition: τ(5) = 5, then

Π_{5}^{2g+3}((1234), τ) ⊃ Π_{4}^{2g+3}((1234), τ).

Moreover, the action of the subgroup St_{(1234)} on this sets commutes with the inclusion, and disconnected collections belong to Π_{4}^{2g+3}((1234), τ). Therefore,

CN_{5}((1 4); (1, p_{2}, …, p_{m}); g) = N_{5}((1 4); (1, p_{2}, …, p_{m}); g) − N_{4}((4); (p_{2}, …, p_{m}); g + 1). We have

CN_{5}((1 4); (1 4); g) = 2^{2g−1}(5^{2g+2} − 3^{2g+2});

CN_{5}((1 4); (1 2^{2}); g) = 2^{2g−1}(5^{2g+3} − 3^{2g+3} + 2);

CN_{5}((1 4); (1^{2} 3); g) = 2^{2g+1} \frac{5^{2g+3} − 2 · 3^{2g+3} + 1}{3}.

CN_{5}((1 4); (1^{3} 2); g) = 2^{2g+1} \frac{5^{2g+4} − 3^{2g+5} + 2}{3}.

6.5. Lemma. Let M^{r} be the set of ordered collections (a_{1}, …, a_{r}) such that a_{i} ∈ {−1, 1, 2, 3}. By definition put,

M_{+}^{r} = {(a_{1}, …, a_{r})|a_{1} · a_{2} ⋯ a_{r} > 0}, M_{−}^{r} = {(a_{1}, …, a_{r})|a_{1} · a_{2} ⋯ a_{r} < 0}. Then

|M^{r}| = 4^{r},  |M_{+}^{r}| = \frac{4^{r} + 2^{r}}{2} ,  |M_{−}^{r}| = \frac{4^{r} − 2^{r}}{2} .

Proof. The first equality is trivial. We’ll prove the other equalities for even r only. The case of odd r is proved similarly. For a collection (a_{1}, …, a_{r}) we denote by n(a_{1}, …, a_{r}) a quality of its elements a_{i}’s such that a_{i} = −1. Besides, we denote by M_{+}^{r}(n) the subset of M_{+}^{r}

consisting of collections (a_{1}, …, a_{r}) such that n(a_{1}, …, a_{r}) = 2n. Then M_{+}^{2r} = ⋃_{n=0}^{r} M_{+}^{2r}(n).

It is clear that |M_{+}^{2r}(n)| = 3^{2r−2n}\binom{2r}{2r−2n}, i.e.,

|M_{+}^{2r}| = ∑_{n=0}^{r} 3^{2r−2n}\binom{2r}{2r − 2n}.

By the binomial formula we have

4^{2r} = (1 + 3)^{2r} = ∑_{n=0}^{r} 3^{2r−2n}\binom{2r}{2r − 2n} + ∑_{n=0}^{r−1} 3^{2r−2n−1}\binom{2r}{2r − 2n − 1}.

2^{2r} = (1 − 3)^{2r} = ∑_{n=0}^{r} 3^{2r−2n}\binom{2r}{2r − 2n} − ∑_{n=0}^{r−1} 3^{2r−2n−1}\binom{2r}{2r − 2n − 1}.

Therefore, |M_{+}^{2r}| = \frac{4^{2r}+2^{2r}}{2} . This completes the proof. 6.6. The number N_{5}((2 3); (2 3); g). On the set Π_{5}^{2g+2}((12)(345); (ab)(cde); g) the subgroup St_{(12)(345)} acts by conjugations. To compute number of orbits of this action we consider the set Π consisting of ordered collections (π_{1}, …, π_{2g+2}), where π_{i}’s are transpositions. Let us divide the set Π into 3 subsets:

Π_{6} is a join of all length 6 St_{(12)(345)} orbits;

Π_{3} is a join of all length 3 St_{(12)(345)} orbits;

Π_{1} is a join of all length 1 St_{(12)(345)} orbits. It is clear, that Π_{1} includes one point only. Namely, Π_{1} = {(12), (12), …, (12)}. Collections of transpositions, belonging to the subset Π_{3} consist of transpositions from {(12), (34), (35), (45)}.

We see that Π_{1} ⊂ Π_{5}^{2g+2}((12)(345); (ab)(cde); g), since (12)(345) \underbrace{(12) ⋯ (12)}_{2g+2} = (12)(345).

In addition, if a collection (π_{1}, …, π_{2g+2}) belongs to Π_{3} ∩ Π_{5}^{2g+2}((12)(345); (ab)(cde); g) then it has even quality of the transposition (12). It follows from lemma 6.5, that |Π_{3} ∪ Π_{1}| = \frac{4^{2g+2}+2^{2g+2}}{2} , i.e. |Π_{3}| = \frac{4^{2g+2}+2^{2g+2}}{2} − 1.

By definition put,

Π_{3}(id) = {(π_{1}, …, π_{2g+2}) ∈ Π_{3}| (12)(345)π_{1} ⋯ π_{2g+2} = id},
Π_{3}(345) = {(π_{1}, …, π_{2g+2}) ∈ Π_{3}| (12)(345)π_{1} ⋯ π_{2g+2} = (345)},
Π_{3}(543) = {(π_{1}, …, π_{2g+2}) ∈ Π_{3}| (12)(345)π_{1} ⋯ π_{2g+2} = (543)},

We have

Π_{3} ∩ Π_{5}^{2g+2}((12)(345); (ab)(cde); g) = Π_{3}(345) ∪ Π_{3}(543).

Let φ be the map from Π_{3} to Π_{3} taking a collection (π_{1}, …, π_{2g+1}) to the collection, that can be obtained by the following rule. We change the last transposition different from (12) (π_{j} for example) to the product π_{j}(34)(35). We see, that φ is bijective and

φ : Π_{3}(id) → Π_{3}(543), φ : Π_{3}(543) → Π_{3}(345), φ : Π_{3}(345) → Π_{3}(id).

Therefore, |Π_{3}(id)| = |Π_{3}(345)| = |Π_{3}(543)| = \frac{1}{3}|Π_{3}|. Thus,

|Π_{3} ∩ Π_{5}^{2g+2}((12)(345); (ab)(cde); g)| = \frac{4^{2g+2} + 2^{2g+2} − 2}{3}.

Now we are ready to calculate the number N_{5}((2, 3); (2, 3); g).

N_{5}((2 3); (2 3); g) = \frac{1}{6} (x_{5,2g+2}((12)(345)) − |Π_{3} ∩ Π_{5}^{2g+2}((12)(345); (ab)(cde); g)| − |Π_{1}|) +

+\frac{1}{3}|Π_{3} ∩ Π_{5}^{2g+2}((12)(345); (ab)(cde); g)| + 1 =

= \frac{1}{18}(10^{2g+2} + 5^{2g+2} + 4^{2g+2} + 2^{2g+3} + 13).

Remark now that a connected collection from the set Π_{5}^{2g+2}((12)(345); (ab)(cde)) includes at least one transposition different from (12), (34), (35) and (45). Therefore, the subset of all connected collections is

Π_{5}^{2g+2}((12)(345); (ab)(cde)) \ (Π_{3} ∪ Π_{1}) and

CN_{5}((2 3); (2 3); g) = \frac{1}{18}(10^{2g+2} + 5^{2g+2} − 4^{2g+2} − 1).

6.7. The number N_{5}((2 3); (1^{2} 3); g). Here we denote by the symbol Π the set of ordered collections (π_{1}, …, π_{2g+3}) with π_{i}’s are transpositions of the group S_{5}. As before the subgroup St_{(12)(345)} acts on this set and we divide the set into 3 subsets Π = Π_{6} ∪ Π_{3} ∪ Π_{1}, where Π_{k} is a join of all length k St_{(12)(345)} orbits. Obviously,

Π_{1} = {(12), (12), …, (12)} ∈ Π_{5}^{2g+3}((12)(345); (abc))

and a collection from Π_{3}∩Π_{5}^{2g+3}((12)(345); (abc)) consists of transpositions (12), (34), (35), (45). Moreover, a quantity of the transpositions (12) in it is odd. We denote by Π_{3}^{−} the subset of collections from Π_{3}, satisfying the last condition. By lemma 6.5 we get |Π_{3}^{−} ∪ Π_{1}| = \frac{1}{2}(4^{2g+3} − 2^{2g+3}), i.e., |Π_{3}^{−}| = \frac{4^{2g+3}−2^{2g+3}−2}{2}.

As in the previous section we divide the set Π_{3}^{−} into 3 classes:

Π_{3}^{−}(id) = {(π_{1}, …, π_{2g+3}) ∈ Π_{3}^{−}| (12)(345) · π_{1} ⋯ π_{2g+3} = id};

Π_{3}^{−}(345) = {(π_{1}, …, π_{2g+3}) ∈ Π_{3}^{−}| (12)(345) · π_{1} ⋯ π_{2g+3} = (345)};

Π_{3}^{−}(543) = {(π_{1}, …, π_{2g+3}) ∈ Π_{3}^{−}| (12)(345) · π_{1} ⋯ π_{2g+3} = (543)}. As before, we have |Π_{3}^{−}(id)| = |Π_{3}^{−}(345)| = |Π_{3}^{−}(543)| = \frac{1}{3}|Π_{3}^{−}| and

|Π_{3} ∩ Π_{5}^{2g+3}((12)(345), (abc))| = \frac{2}{3}|Π_{3}^{−}| = \frac{4^{2g+3} − 2^{2g+3} − 2}{3}.

Whence,

N_{5}((2 3); (1^{2} 3); g) =

= \frac{1}{6} |Π_{5}^{2g+3}((12)(345), (abc)) \ (Π_{3} ∪ Π_{1})| + \frac{1}{3} |Π_{5}^{2g+3}((12)(345), (abc)) ∩ Π_{3}| + |Π_{1}| =

= \frac{1}{18}(10^{2g+3} − 5^{2g+3} + 4^{2g+3} − 2^{2g+4} + 13).

It is remains to note that all connected collections belong to Π_{5}^{2g+3}((12)(345), (abc)) \ (Π_{3} ∪ Π_{1}), i.e.

CN_{5}((2 3); (1^{2} 3); g) = \frac{1}{18}(10^{2g+3} − 5^{2g+3} − 4^{2g+3} − 1).

6.8. The number N_{5}((2 3); (1 2^{2}); g). We use notations Π, Π_{1}, Π_{3}, from the previous section. As above, the set Π_{3} includes collections consisting of the transpositions (12), (34), (35), (45). But in this case Π_{1} ∩ Π_{5}^{2g+3}((12)(345), (ab)(cd)) = ∅ and a collection belonging to Π_{3} ∩ Π_{5}^{2g+3}((12)(345), (ab)(cd)) has even quantity of transposition equal to (12), i.e.,

Π_{3} ∩ Π_{5}^{2g+3}((12)(345), (ab)(cd)) ⊂ Π_{3}^{+}.

On the other hand, if (π_{1}, …, π_{2g+3}) ∈ Π_{3}^{+}, then the product (12)(345) · π_{1} ⋯ π_{2g+3} is a permutation of cycle type (2, 2). Therefore, Π_{3} ∩ Π_{5}^{2g+3}((12)(345), (ab)(cd)) = Π_{3}^{+}. Taking into account lemma 6.5, we get |Π_{3}^{+}| = \frac{4^{2g+3}+2^{2g+3}}{2} . Thus,

N_{5}((2 3); (1 2^{2}); g) = \frac{1}{6} |Π_{5}^{2g+3}((12)(345), (ab)(cd)) \ Π_{3}^{+}| + \frac{1}{3} |Π_{3}^{+}| =

= 2^{2g} \frac{5^{2g+3} + 2^{2g+4} + 3}{3}

and

CN_{5}((2 3); (1 2^{2}); g) = \frac{1}{6} |Π_{5}^{2g+3}((12)(345), (ab)(cd)) \ Π_{3}^{+}| = 2^{2g} \frac{5^{2g+3} − 2^{2g+4} − 1}{3}.

6.9. The number N_{5}((2 3); (1^{3} 2); g). Let Π be the set of ordered collections (π_{1}, …, π_{2g+4}). The subgroup St_{(12)(345)} acts on this set and we denote by Π_{k} the subset of Π as usual. At first we note, that Π_{1} ∩ Π_{5}^{2g+4}((12)(345); (ab)) = ∅.

Let Π_{3}^{+} be the subset of Π_{3} consisting of collections such that a quantity of the transpositions equal to (12) in it is even. In addition, we put Π_{3}^{−} = Π_{3} \ Π_{3}^{+}. It is easy to prove, that

Π_{3} ∩ Π_{5}^{2g+4}((12)(345); (ab)) = Π_{3}^{−} ∪ Π_{3}^{+}(id),

where

Π_{3}^{+}(id) = {(π_{1}, …, π_{2g+4}) ∈ Π_{3}^{+}| (345)π_{1} ⋯ π_{2g+4} = id}.

Besides,

|Π_{3}^{+}(id)| = \frac{1}{3} (|Π_{3}^{+}| − 1) = \frac{4^{2g+4} + 2^{2g+4} − 2}{6} and |Π_{3}^{−}| = \frac{4^{2g+4} − 2^{2g+4}}{2} .

Therefore,

|Π_{3} ∩ Π_{5}^{2g+4}((12)(345); (ab))| = \frac{4^{2g+5} − 2^{2g+5} − 2}{6}.

Thus,

N_{5}((2 3); (1^{3} 2)) = \frac{1}{6} |Π_{5}^{2g+4}((12)(345); (ab)) \ Π_{3}| + \frac{1}{3} |Π_{3} ∩ Π_{5}^{2g+4}((12)(345); (ab))| =

= \frac{1}{36}(10^{2g+4} − 2 · 5^{2g+4} + 4^{2g+5} − 2^{2g+4} − 2).

CN_{5}((2 3); (1^{3} 2)) = \frac{1}{6} |Π_{5}^{2g+4}((12)(345); (ab)) \ Π_{3}| =

= \frac{1}{36}(10^{2g+4} − 2 · 5^{2g+4} − 4^{2g+5} + 3 · 2^{2g+4} + 2).

6.10. The number N_{5}((1^{2} 3); (1^{2} 3); g). Let Π be the set of ordered length 2g + 4 collections of transpositions from the group S_{5}. The order 6 subgroup St_{(123)} acts on the set Π_{5}^{2g+4}((123); (abc)) ⊂ Π by conjugations and the number N_{5}((1, 1, 3); (1, 1, 3); g) is the quantity of St_{(123)} orbits in this set. We divide the set Π into 3 subsets: Π = Π_{1} ∪ Π_{3} ∪ Π_{6}, where Π_{k} is a join of length k St_{(123)} orbits in the set Π.

We see that Π_{1} = {(45), (45), …, (45)} ⊂ Π_{5}^{2g+4}((123); (abc)), a collection from Π_{3} consists of the transpositions (12), (13), (23) and (45). Moreover, Π_{3} ∩ Π_{5}^{2g+4}((123); (abc)) ⊂ Π_{3}^{+}, where Π_{3}^{+} is subset of Π_{3}, consisting of collections (π_{1}, …, π_{2g+4}) such that the number of (45) in them is even. By lemma 6.5,

|Π_{3}^{+} ∪ Π_{1}| = \frac{4^{2g+4} + 2^{2g+4}}{2} , i.e., |Π_{3}^{+}| = \frac{4^{2g+4} + 2^{2g+4} − 2}{2}.

By definition put

Π_{3}^{+}(id) = {(π_{1}, …, π_{2g+4}) ∈ Π_{3}^{+}| (123)π_{1} ⋯ π_{2g+4} = id}
Π_{3}^{+}(123) = {(π_{1}, …, π_{2g+4}) ∈ Π_{3}^{+}| (123)π_{1} ⋯ π_{2g+4} = (123)}
Π_{3}^{+}(321) = {(π_{1}, …, π_{2g+4}) ∈ Π_{3}^{+}| (123)π_{1} ⋯ π_{2g+4} = (321)} .

By the same argument, as before, we obtain

|Π_{3}^{+}(321)| = |Π_{3}^{+}(123)| = \frac{1}{3} |Π_{3}^{+}| = \frac{4^{2g+4} + 2^{2g+4} − 2}{6}

and

Π_{3} ∩ Π_{5}^{2g+4}((123), (abc)) = Π_{3}^{+}(123) ∪ Π_{3}^{+}(321),

i.e.,

|Π_{3} ∩ Π_{5}^{2g+4}((123), (abc))| = \frac{4^{2g+4} + 2^{2g+4} − 2}{3}.

Therefore,

N_{5}((1^{2} 3); (1^{2} 3); g) =

= \frac{1}{6} |Π_{5}^{2g+4}((123), (abc)) \ (Π_{3} ∪ Π_{1})| + \frac{1}{3} |Π_{5}^{2g+4}((123), (abc)) ∩ Π_{3})| + 1 =

= \frac{1}{18}(10^{2g+4} + 5^{2g+4} + 4^{2g+4} + 2^{2g+5} + 13).

The subset D of disconnected collections in the set Π_{5}^{2g+4}((123), (abc)) can be represented as a join D = D_{5} ∪ D_{4} ∪ D_{45}, where

D_{5} = {(π_{1}, …, π_{2g+4}) ∈ Π_{5}^{2g+4}((123), (abc))| ∀i π_{i}(5) = 5};
D_{4} = {(π_{1}, …, π_{2g+4}) ∈ Π_{5}^{2g+4}((123), (abc))| ∀i π_{i}(4) = 4};
D_{45} = {(π_{1}, …, π_{2g+4}) ∈ Π_{5}^{2g+4}((123), (abc))| ∀i π_{i}({4.5}) = {4, 5}} .

Since the subgroup St_{(123)} acts on the set Π_{5}^{2g+4}((123), (abc)) \ D free, we have

CN_{5}((1^{2} 3); (1^{2} 3); g) = \frac{1}{6} (x_{5,2g+4}((123), (abc)) − |D|) and

|D| = 2|D_{5}| − |D_{5} ∩ D_{4}| + |D_{45} \ (D_{5} ∪ D_{4})|.

Furthermore,

|D_{4}| = |D_{5}| = x_{4,2g+4}((123), (abc)) = 2^{2g+5} · 3^{2g+3}; (see (5-1))

|D_{5} ∩ D_{4}| = x_{3,2g+4}((123), (abc)) = 2 · 3^{2g+3} (see (4-1));

D_{45} \ (D_{5} ∪ D_{4}) = (Π_{3} ∩ Π_{5}^{2g+4}((123), (abc))) \ (D_{4} ∩ D_{5}), and

|D_{45} \ (D_{5} ∪ D_{4})| = \frac{1}{3}(4^{2g+4} + 2^{2g+4} − 2) − 2 · 3^{2g+3}.

Thus,

CN_{5}((1^{2} 3); (1^{2} 3); g) = \frac{1}{18}(10^{2g+4} − 4 · 6^{2g+4} + 5^{2g+4} − 4^{2g+4} + 4 · 3^{2g+4} + 2).

6.11. The number N_{5}((1^{2} 3); (1 2^{2}); g). We use the notations Π, Π_{1}, Π_{3} of the previous section. It is clear, that Π_{1} ∩ Π_{5}^{2g+4}((123), (ab)(cd)) = Ø and

Π_{3} ∩Π_{5}^{2g+4}((123), (ab)(cd)) = Π_{3}^{−} (collections with odd quantity of the transpositions (45)). Using lemma 6.5 to calculate the number |Π_{3}^{−}|, we have

N_{5}((1^{2} 3); (1 2^{2}); g) = \frac{1}{6} |Π_{5}^{2g+4}((123), (ab)(cd)) \ Π_{3}| + \frac{1}{3} |Π_{3}^{−}| =

= \frac{1}{6}(25 · 10^{2g+2} − 3 · 2^{2g+2} + 2 · 4^{2g+3}).

Moreover, the set disconnected collections coincides with Π_{3} ∩ Π_{5}^{2g+4}((123), (ab)(cd)), i.e.,

CN_{5}((1^{2} 3); (1 2^{2}); g) = \frac{1}{6}(25 · 10^{2g+2} + 2^{2g+2} − 2 · 4^{2g+3}).

6.12. The number N_{5}((1^{2} 3); (1^{3} 2); g). Let Π be the set of ordered collections (π_{1}, …, π_{2g+5}), where π_{i}’s are transpositions, Π_{k} be a subset of Π equal to a join of length k St_{(123)} orbits. We see that Π_{1} ∩ Π_{5}^{2g+5}((123), (ab)) = Ø. If Π_{3}^{+} denotes the subset of Π_{3}, consisting of collections with even quantity of the transpositions (54), then Π_{3}^{+} ⊂ Π_{3} ∩ Π_{5}^{2g+5}((123), (ab)) and

|Π_{3}^{+}| = \frac{4^{2g+5} + 2^{2g+5}}{2} .

By definition put Π_{3}^{−} = Π_{3} \ Π_{3}^{+},

Π_{3}^{−}(id) = {(π_{1}, …, π_{2g+5}) ∈ Π_{3}^{−}| (123)π_{1} ⋯ π_{2g+5} = (45)};
Π_{3}^{−}(123) = {(π_{1}, …, π_{2g+5}) ∈ Π_{3}^{−}| (123)π_{1} ⋯ π_{2g+5} = (45)(123)};
Π_{3}^{−}(321) = {(π_{1}, …, π_{2g+5}) ∈ Π_{3}^{−}| (123)π_{1} ⋯ π_{2g+5} = (45)(321)} .

Then,

|Π_{3}^{−} ∪ Π_{1}| = \frac{4^{2g+5} − 2^{2g+5}}{2} , Π_{3}^{−} ∩ Π_{5}^{2g+5}((123), (ab)) = Π_{3}^{−}(id) and

Π_{3}^{−}(id) = \frac{1}{3} |Π_{3}^{−}| = \frac{4^{2g+5} − 2^{2g+5} − 2}{6}.

We obtain

|Π_{3} ∩ Π_{5}^{2g+5}((123), (ab))| = |Π_{3}^{+}| + |Π_{3}^{−}(id)| = \frac{1}{6}(4^{2g+6} + 2^{2g+6} − 2).

Therefore,

N_{5}((1^{2} 3); (1^{3} 2); g) = \frac{1}{6} |Π_{5}^{2g+5}((123), (ab)) \ Π_{3}| + \frac{1}{3} |Π_{5}^{2g+5}((123), (ab)) ∩ Π_{3}| =

= \frac{1}{36}(10^{2g+5} + 2 · 5^{2g+5} + 4^{2g+6} + 2^{2g+5} − 2).

To calculate the number CN_{5}((1^{2} 3); (1^{3} 2); g) we apply the method of section 6.10. Namely, we consider the subset of Π_{5}^{2g+5}((123), (ab)) consisting of disconnected collections:

D_{5} = {(π_{1}, …, π_{2g+5}) ∈ Π_{5}^{2g+5}((123), (ab))| ∀i π_{i}(5) = 5};
D_{4} = {(π_{1}, …, π_{2g+5}) ∈ Π_{5}^{2g+5}((123), (ab))| ∀i π_{i}(4) = 4};
D_{45} = {(π_{1}, …, π_{2g+5}) ∈ Π_{5}^{2g+5}((123), (ab))| ∀i π_{i}({4.5}) = {4, 5}} .

The subset D of all disconnected collections in Π_{5}^{2g+5}((123), (ab)) is D_{4} ∪ D_{5} ∪ D_{45}. The subgroup St_{(123)} acts on the complement Π_{5}^{2g+5}((123), (ab)) \ (D_{4} ∪ D_{5} ∪ D_{45}) free and the number of orbits of this action is CN_{5}((1, 1, 3); (1, 1, 1, 2); g). Furthermore,

|D_{4}| = |D_{5}| = x_{4,2g+5}((123), (ab) = 2^{2g+4}3^{2g+5} (see (5-1);
|D_{5} ∩ D_{4}| = x_{3,2g+5}((123), (ab) = 3^{2g+5} (see (4-1),
D_{54} \ (D_{5} ∪ D_{4}) = (Π_{3} ∩ Π_{5}^{2g+5}((123), (ab))) \ (D_{5} ∩ D_{4}),

|D_{54} \ (D_{5} ∪ D_{4})| = \frac{4^{2g+6} + 2^{2g+6} − 2 · 3^{2g+6} − 2}{6}.

Therefore,

CN_{5}((1^{2} 3); (1^{3} 2); g) =

= \frac{1}{6}(|Π_{5}^{2g+5}((123), (ab)| − 2|D_{4}| + |D_{4} ∩ D_{5}| − |D_{54} \ (D_{4} ∪ D_{5})|) =

= \frac{1}{36}(10^{2g+5} − 6^{2g+6} + 2 · 5^{2g+5} − 4^{2g+6} + 4 · 3^{2g+6} − 3 · 2^{2g+5} + 2).

6.13. The number N_{5}((1 2^{2}); (1 2^{2}); g). The subgroup St_{(12)(34)} has order 8. It can be proved that

St_{(12)(34)} ∩ St_{(ab)} = \{ \begin{matrix}⟨(12), (34)⟩, & if (ab) = (12) or (ab) = (34), \\ id & in other cases. \end{matrix}

Hence, the set Π of collections (π_{1} …, π_{2g+4}) can be divided into 2 subsets Π = Π_{2} ∪ Π_{8}, where Π_{k} is a join of length k St_{(12)(34)} orbits.

The subset Π_{2} consists of collections of the transpositions (12) and (34). Moreover, if a collection (π_{1}, …, π_{2g+4}) ∈ Π_{2} ∩ Π_{5}^{2g+4}((12)(34), (ab)(cd)), then the quantity of transpositions π_{i}’s equal to (12) is even. Hence,

|Π_{2} ∩ Π_{5}^{2g+4}((12)(34), (ab)(cd)| = 2^{2g+3}.

Thus,

N_{5}((1 2^{2}); (1 2^{2}); g) = \frac{1}{8} |Π_{5}^{2g+4}((12)(34), (ab)(cd) \ Π_{2}| + \frac{1}{2} |Π_{5}^{2g+4}((12)(34), (ab)(cd) ∩ Π_{2}| =

= 2^{2g−1}5^{2g+4}.

To calculate CN_{5}((1 2^{2}); (1 2^{2}); g) by definition put

D_{5} = {(π_{1}, …, π_{2g+4}) ∈ Π_{5}^{2g+4}((12)(34), (ab)(cd))| ∀i π_{i}(5) = 5};
D_{12} = {(π_{1}, …, π_{2g+4}) ∈ Π_{5}^{2g+4}((12)(34), (ab)(cd))| ∀i π_{i}({1, 2}) = {1, 2}};
D_{34} = {(π_{1}, …, π_{2g+4}) ∈ Π_{5}^{2g+4}((12)(34), (ab)(cd))| ∀i π_{i}({3, 4}) = {3, 4}} .

The subset D of all disconnected collections belonging to Π_{5}^{2g+4}((12)(34), (ab)(cd)) is D_{5} ∪ D_{12} ∪ D_{34}.

We see that

|D_{5}| = x_{4,2g+4}((12)(34), (ab)(cd)) = 2^{2g+2}(3^{2g+4} + 1);

|D_{12}| = |D_{34}|, D_{12} ∩ D_{34} = D_{5} ∩ D_{12} = D_{5} ∩ D_{34} = Π_{2} ∩ Π_{5}^{2g+4}((12)(34), (ab)(cd)).

Therefore, |D_{12} ∩ D_{34}| = 2^{2g+3}.

Let us calculate |D_{12}|. If (π_{1}, …, π_{2g+4}) ∈ D_{12}, then π ∈ {(12), (34), (35), (45)) and the quantity of (12) in it is even. Taking into account lemma 6.5, we get |D_{12}| = \frac{4^{2g+4}+2^{2g+4}}{2} .

Thus

CN_{5}((1 2^{2}); (1 2^{2}); g) = \frac{1}{8} (|Π_{5}^{2g+4}((12)(34); (ab)(cd))| − |D_{5}| − 2|D_{12}| + 2|D_{12} ∩ D_{34}|) =

= 2^{2g−1}(5^{2g+4} − 3^{2g+4}) − 2 · 4^{2g+2}.

6.14. The number N_{5}((1 2^{2}); (1^{3} 2); g). A St_{(12)(34)} orbit in the set Π_{5}^{2g+5}((12)(34), (ab)) has either 2 or 8 points. The subset Π_{2} ⊂ Π_{5}^{2g+5}((12)(34), (ab)) of all length 2 St_{(12)(34)} orbits contains collections (π_{1}, …, π_{2g+5}) such that ∀i π_{i} ∈ {(12), (34)}. Therefore, |Π_{2}| = 2^{2g+5} and

N_{5}((1 2^{2}); (1^{3} 2); g) = \frac{1}{8} |Π_{5}^{2g+5}((12)(34), (ab)) \ Π_{2}| + \frac{1}{2} |Π_{2}| = 2^{2g+1} \frac{5^{2g+5} + 19}{3}.

As before by definition put

D_{5} = {(π_{1}, …, π_{2g+5}) ∈ Π_{5}^{2g+5}((12)(34), (ab))| ∀i π_{i}(5) = 5};
D_{12} = {(π_{1}, …, π_{2g+5}) ∈ Π_{5}^{2g+5}((12)(34), (ab))| ∀i π_{i}({1, 2}) = {1, 2}};
D_{34} = {(π_{1}, …, π_{2g+5}) ∈ Π_{5}^{2g+5}((12)(34), (ab))| ∀i π_{i}({3, 4}) = {3, 4}} .

The subset D of all disconnected collections belonging to Π_{5}^{2g+5}((12)(34), (ab)) is D = D_{5} ∪ D_{12} ∪ D_{34}. Besides,

|D_{5}| = x_{4,2g+5}((12)(34), (ab)) = 2^{2g+4}(3^{2g+5} − 1); |D_{12}| = |D_{34}|;
D_{5} ∩ D_{12} = D_{5} ∩ D_{34} = D_{12} ∩ D_{34} = Π_{2}.

Denote by M the set of collections (π_{1}, …, π_{2g+5}) such that ∀i π_{i} ∈ {(12), (34), (35), (45)}. Then D_{12} ⊂ M. Let M^{−} be the subset of M, consisting of collections with odd quantity of the transpositions (12) and M^{+} = M \ M^{−}. Then M^{−} ⊂ D_{12} and |M^{−}| = \frac{1}{2}(4^{2g+5} − 2^{2g+5}). By definition put

M^{+}(id) = {(π_{1}, …, π_{2g+5}) ∈ M^{+}| (34)π_{1} ⋯ π_{2g+5} = id},
M^{+}(345) = {(π_{1}, …, π_{2g+5}) ∈ M^{+}| (34)π_{1} ⋯ π_{2g+5} = (345)},
M^{+}(543) = {(π_{1}, …, π_{2g+5}) ∈ M^{+}| (34)π_{1} ⋯ π_{2g+5} = (543)}.

Then M^{+} ∩ D_{12} = M^{+}(id), |M^{+}(id)| = \frac{1}{3} |M^{+}| = \frac{4^{2g+5}+2^{2g+5}}{6} . Therefore, |D_{12}| = |M^{+}(id)| + |M^{−}| = \frac{4^{2g+6}−2^{2g+6}}{6} .

Thus,

CN_{5}((1 2^{2}); (1^{3} 2); g) = \frac{1}{8}(|Π_{5}^{2g+5}((12)(34), (ab))| − |D_{5}| − 2|D_{12}| + 2|D_{12} ∩ D_{34}|) =

= \frac{2^{2g+1}}{3} (5^{2g+5} − 3^{2g+6} − 2^{2g+8} + 20).

6.15. The last number N_{5}((1^{3} 2); (1^{3} 2); g). Order of the subgroup St_{(54)} is 12. Obviously,

St_{(54)} ∩ St_{(ab)} = \{ \begin{matrix}⟨(54), (ab)⟩, & if (ab) ∈ St_{(45)}; \\ id, & in the convers case. \end{matrix}

Let Π be the set of ordered collections (π_{1}, …, π_{2g+6}) where π_{i}’s are transpositions. We divide the set Π into 3 subsets: Π = Π_{1} ∪ Π_{3} ∪ Π_{12}, where as usual, Π_{k} is a join of length k St_{(45)} orbits. It is clear, that

Π_{1} = {((45), …, (45))} ⊂ Π_{5}^{2g+6}((45), (ab)).

In particular, |Π_{1}| = 1.

By definition put Π_{3}(ab) = {(π_{1}, …, π_{2g+6}) ∈ Π_{3}| π_{i} ∈ {(45), (ab)}}. Then

Π_{3} = Π_{3}(12) ∪ Π_{3}(13) ∪ Π_{3}(23) = Π_{3} ∩ Π_{5}^{2g+6}((45), (ab)).

We see that

|Π_{3}(12)| = |Π_{3}(13)| = |Π_{3}(23)| = 2^{2g+6} − 1.

Therefore,

N_{5}((1^{3} 2); (1^{3} 2); g) = \frac{1}{12}(x_{5,2g+6}((45), (ab)) − |Π_{3}| − |Π_{1}|) + \frac{1}{3} |Π_{3}| + 1 =

= \frac{1}{12}(25 · 10^{2g+4} + 5^{2g+6} + 31 · 2^{2g+4} + 3).

The subset D of disconnected collections in Π_{5}^{2g+6}((45), (ab)) can be represented as a join D = D_{45} ∪ D_{12} ∪ D_{23} ∪ D_{13}, where

D_{αβ} = {(π_{1}, …, π_{2g+6}) ∈ Π_{5}^{2g+6}((45), (ab))| ∀i π_{i}({α, β}) = {α, β}}.

The subset D_{45} contains collections (π_{1}, …, π_{2g+6}), with π_{i} ∈ {(45), (12), (23), (13)} only. Moreover, such a collection belongs to D_{45} in the two cases:

(i) the number of (45) in it is odd (we denote the corresponding subset by D_{45}^{−}), (ii) the number of (45) is even and the product π_{1} ⋯ π_{2g+6} = id (we denote the correspond-

ing subset by D_{45}^{+}(id)).

By lemma 6.5

|D_{45}^{−}| = \frac{4^{2g+6} − 2^{2g+6}}{2} , |D_{45}^{+}(id)| = 1 + \frac{4^{2g+6} + 2^{2g+6} − 2}{6}.

Therefore,

|D_{45}| = \frac{4^{2g+7} − 2^{2g+7} + 4}{6}.

The subset D_{12} consists of all ordered length 2g + 6 collections of the transpositions with (12), (34), (35), (45) such that either a quantity of the transpositions (12) in them is even and the product of all elements of collection equals id, or

a quantity of the transpositions (12) in them is odd and the product of all elements of collection equals (45).

The first subset we denote by D_{12}^{+}(id) and the second one by D_{12}^{−}(45), i.e., D_{12} = D_{12}^{+}(id) ∪ D_{12}^{−}(45).

Taking into account lemma 6.5, and the standard trick, we get

|D_{12}^{+}(id)| = \frac{4^{2g+6} + 2^{2g+6} + 4}{6}, |D_{12}^{−}(45)| = \frac{4^{2g+6} − 2^{2g+6}}{6} .

Therefore, |D_{12}| = \frac{4^{2g+6}+2}{3}. Similarly,

|D_{23}| = |D_{13}| = \frac{4^{2g+6} + 2}{3}.

The intersection D_{45} ∩D_{12} consists of all length 2g +6 collections of the transpositions (45) and (12). Whence, |D_{45} ∩ D_{12}| = 2^{2g+6}. Likewise,

|D_{45} ∩ D_{13}| = |D_{45} ∩ D_{23}| = 2^{2g+6}.

It is clear, that

D_{12} ∩ D_{13} = D_{12} ∩ D_{23} = D_{23} ∩ D_{13} = {((45), …, (45))}, i.e.,
|D_{12} ∩ D_{13}| = |D_{12} ∩ D_{23}| = |D_{23} ∩ D_{13}| = 1.

In addition, an intersection of any 3 different subsets D_{ij} has one point and the intersection of all D_{ij}’s is one point set too. Finally, the subgroup St_{(45)} acts free on the set Π_{5}^{2g+6}((45); (ab))\ D. Thus,

CN_{5}((1^{3} 2); (1^{3} 2)) = \frac{1}{12}(|Π_{5}^{2g+6}((45), (ab))| − |D_{45}| − 3|D_{13}| + 3|D_{45} ∩ D_{23}| + 3 − 4 + 1) =

= \frac{1}{72}(10^{2g+6} + 4 · 5^{2g+6} − 10 · 4^{2g+6} + 21 · 2^{2g+6} − 16).

6.16. Let us collect all the numbers CN_{5}(α; β; g) together.

CN_{5}((5); (5); g) = 5^{2g−2}(2^{2g+1} + 2);  CN_{5}((5); (1 4); g) = 5^{2g}2^{2g};
CN_{5}((5); (2 3); g) = 5^{2g} \frac{2^{2g+1}+1}{3};  CN_{5}((5); (1^{2} 3); g) = 5^{2g+1} \frac{2^{2g+2}−1}{3};
CN_{5}((5); (1 2^{2}); g) = 5^{2g+1}2^{2g};  CN_{5}((5); (1^{3} 2); g) = 5^{2g+2} \frac{2^{2g+2}−1}{3};
CN_{5}((1 4); (1 4); g) = 2^{2g−1}(5^{2g+2} − 3^{2g+2});  CN_{5}((1 4); (2 3); g) = 2^{2g} \frac{5^{2g+2}−1}{3};
CN_{5}((1 4); (1 2^{2}); g) = 2^{2g−1}(5^{2g+3} − 3^{2g+3} + 2); CN_{5}((1 4); (1^{2} 3); g) = 2^{2g+1} \frac{5^{2g+3}−2·3^{2g+3}+1}{3};

CN_{5}((1 4); (1^{3} 2); g) = 2^{2g+1} \frac{5^{2g+4}−3^{2g+5}+2}{3}; CN_{5}((2 3); (2 3); g) = \frac{1}{18}(10^{2g+2} + 5^{2g+2} − 4^{2g+2} − 1);

CN_{5}((2 3); (1^{2} 3); g) = \frac{1}{18}(10^{2g+3} − 5^{2g+3} − 4^{2g+3} − 1);

CN_{5}((2 3); (1 2^{2}); g) = 2^{2g} \frac{5^{2g+3}−2^{2g+4}−1}{3}

CN_{5}((2 3); (1^{3} 2)) = \frac{1}{36}(10^{2g+4} − 2 · 5^{2g+4} − 4^{2g+5} + 3 · 2^{2g+4} + 2)

CN_{5}((1^{2} 3); (1^{2} 3); g) = \frac{1}{18}(10^{2g+4} − 4 · 6^{2g+4} + 5^{2g+4} − 4^{2g+4} + 4 · 3^{2g+4} + 2);

CN_{5}((1^{2} 3); (1 2^{2}); g) = \frac{1}{6}(25 · 10^{2g+2} + 2^{2g+2} − 2 · 4^{2g+3});

CN_{5}((1^{2} 3); (1^{3} 2); g) = \frac{1}{36}(10^{2g+5} − 6^{2g+6} + 2 · 5^{2g+5} − 4^{2g+6} + 4 · 3^{2g+6} − 3 · 2^{2g+5} + 2)

CN_{5}((1 2^{2}); (1 2^{2}); g) = 2^{2g−1}(5^{2g+4} − 3^{2g+4}) − 2 · 4^{2g+2};

CN_{5}((1 2^{2}); (1^{3} 2); g) = \frac{2^{2g+1}}{3} (5^{2g+5} − 3^{2g+6} − 2^{2g+8} + 20);

CN_{5}((1^{3} 2); (1^{3} 2)) = \frac{1}{72}(10^{2g+6} + 4 · 5^{2g+6} − 10 · 4^{2g+6} + 21 · 2^{2g+6} − 16).

References

[1] B.Z.Shapiro, M.Z.Shapiro and A.D.Vainshtein, Ramified coverings of S^{2} with one degenerate branching point and enumeration of edge-ordered graphs, AMS Translations 180 (1997), 219 – 227.

[2] Graham, Knuth, Patashnik. Concrete mathematics. Addison-Wesley Publ. Comp., 1989.

[3] Littlewood E., The theory of group characters. Oxford, At the Clarendon press, 1950.