Blow up formulae for Euler characteristic of moduli space of rank 3 bundles over a surface.^{1}

S. Kuleshov.

1. Introduction.

In this preprint we partially solve the following problem. Let X be a smooth algebraic surface over ℂ. We fix some ample divisor H on X and denote by 𝔐_{H}(r, C, k) the moduli space of Mumford – Takemoto μ_{H}-stable rank r vector bundles on X with Chern classes C and k such that H · C and r are coprime. Denote by Z_{r}(q) the following generating function:

Z_{r}(q) = ∑_{k∈ℤ} e(𝔐_{H}(r, C, k)) · q^{k},

where e(Y) for a complex scheme Y is the topological Euler characteristic of Y_{red}, and e(𝔐_{H}(r, C, k))= 0 whenever 𝔐_{H}(r, C, k) is empty.

Further, let σ : X̃ ⟶ X be the blow up of the surface X at a point x_{0} and E = σ^{−1}(x_{0}) be the exceptional curve. For a positive integer n by definition put H_{n} = n(σ^{∗}H) − E. Denote by 𝔐_{H_{n}}(r, σ^{∗}C − iE, k) the moduli space of Mumford – Takemoto μ_{H_{n}}-stable rank r vector bundles V on X̃ with c_{1}(V) = σ^{∗}C − iE, c_{2}(V) = k. It is clear that 𝔐_{H_{n}}(r, σ^{∗}C − iE, k) ≅ 𝔐_{H_{n+1}}(r, σ^{∗}C−iE, k) if n ≫ 0. We denote by 𝔐_{H_{∞}}(r, σ^{∗}C−iE, k) this moduli space. Consider the generating functions

Z_{r,i}(q) = ∑_{k∈ℤ} e(𝔐_{H_{∞}}(r, σ^{∗}C − iE, k))q^{k}.

Now we can formulate the question. How do the functions Z_{r,i}(q) depend on Z_{r}(q)?

In [7], Vafa and Witten formulated hypothetical answer for this question in the case r = 2 and i = 0 derived from the S-duality conjecture in physics. They presuppose that there exists a universal function u(q) such that

Z_{2,0}(q) = u(q) · Z_{2}(q).

Moreover, the function u(q) is a modular form.

This conjecture was proved by Li and Qin in [3]. The authors obtained the following result. For i = 0, 1 there exist universal functions

u_{2,i}(q) = ∑_{n∈ℤ} q^{n(n+i)}

such that Z_{2,i}(q) = u_{2,i}(q) · Z_{2}(q).

In this preprint the problem is solved in the case r = 3. The main result of the preprint is the following (theorem 6.1 and corollary 6.4)

For i = 0, 1, 2 there exist universal functions u_{3,i}(q) such that Z_{3,i}(q) = u_{3,i}(q) · Z_{3}(q), and

u_{3,i}(q) = q^{δ_{2}^{i}} ∑_{(x,y)∈ℤ^{2}} q^{x^{2}+xy+y^{2}+i(x+y)},

1This work was done during the author‘s stay at the Max Plank Institute für Mathematik in 1998.

where δ_{2}^{i} is Kronecker symbol, i.e. δ_{2}^{i} = \{ \begin{matrix}1 & for & i = 2, \\ 0 & otheraise \end{matrix} .

Further we fix r = 3, C, and H and use the notations

𝔐(k) = 𝔐_{H}(3, C, k), 𝔐(i, k) = 𝔐_{H_{∞}}(r, σ^{∗}C − iE, k)

The idea of proof is the same as in [3]. It is the following.

For integer d_{1} ≥ −\frac{i}{3} and d_{1} ≥ d_{2} ≥ −\frac{d_{1} + i}{2} denote by 𝔐^{(d_{1},d_{2})}(i, k) the subset of 𝔐(i, k) consisting of all bundles V such that

V |_{E} = O_{E}(−d_{1}) ⊕ O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i),

then

e(𝔐(i, k)) = ∑_{(d_{1},d_{2})} e(𝔐^{(d_{1},d_{2})}(i, k)).

Consider the following commutative diagram

\begin{aligned}& & 0 & & 0 \\ & & ↑ & & ↑ \\ 0 & ⟶ & O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i) & ⟶ & V |_{E} & ⟶ & O_{E}(−d_{1}) & ⟶ & 0 \\ & & β ↑ & & res ↑ & & id ↑ \\ 0 & ⟶ & V′ & ⟶ & V & ⟶^{α} & O_{E}(−d_{1}) & ⟶ & 0 \\ & & ↑ & & ↑ \\ & & V (−E) & = & V (−E) \\ & & ↑ & & ↑ \\ & & 0 & & 0 \end{aligned} ,

where V ∈ 𝔐^{(d_{1},d_{2})}(i, k). Note that V′ ∈ 𝔐^{(m_{1},m_{2})}(i + 1, k − d_{1} − i) for some pair of integers (m_{1}, m_{2}) depending on (d_{1}, d_{2}). In other words, by means of the pair (V, α) we construct the pair (V′, β). Later we will show that there exists a bijective morphism from a locally trivial bundle over 𝔐^{(d_{1},d_{2})}(i, k) with fiber ℙ(Hom (V, O_{E}(−d_{1}))) to the join of locally trivial bundles over 𝔐^{(m_{1},m_{2})}(i + 1, k − d_{1} − i) with fiber

{β}/Aut (O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i)) = EP _{−d_{2},d_{1}+d_{2}+i}^{−m_{1},−m_{2},m_{1}+m_{2}+i+1}.

Thus if we know e(EP _{−d_{2},d_{1}+d_{2}+i}^{−m_{1},−m_{2},m_{1}+m_{2}+i+1}), we obtain recursion relations for e(𝔐^{(d_{1},d_{2})}(i, k)). Finally, the isomorphism 𝔐(3, k) ≅ 𝔐(0, k + 3) lets obtain the main result.

This preprint is organized as follows. In section 1 we do some calculations known to experts, which we use to prove the main result. In section 2 we define the set EP _{e,d}^{a,b,c} and find Euler characteristic of it. Sections 3 and 4 deal with recursion relations for e(𝔐^{(d_{1},d_{2})}(i, k)). Finally, in the last section we deduce the blowup formulae.

The author is grateful to A. Rudakov, M. Leenson, A. Zorich, S. Orevkov, Don Zagier and A. Tikhomirov for useful discussions. I am thankful to Max-Plank-Institut für Mathematik for hospitality.

2. Some calculations.

In this section we do some calculations, which we use in the text.

Let X be a smooth algebraic surface and σ : X̃ ⟶ X be a blowup at a point x_{0} ∈ X. Denote by E the corresponding exceptional curve.

2.1. Claim. For an exceptional curve E

c_{2}(O_{E}(−d)) = d − 1.

Proof. Using multiplicativity of the full Chern character and the restriction exact sequence

0 ⟶ OX̃((d − 1)E) ⟶ OX̃(dE) ⟶ O_{E}(−d) ⟶ 0,

we have

(1 + (d − 1)E · t) · (1 + E · t + c_{2}(O_{E}(−d)) · t^{2}) = (1 + dE · t).

Whence,

(d − 1)E^{2} + c_{2}(O_{E}(−d)) = 0 ⟺ c_{2}(O_{E}(−d)) = d − 1.

2.2. Claim. For an exceptional curve E

Tor_{1}(O_{E}, O_{E}(d)) = O_{E}(d + 1).

Proof. Tensoring the restriction sequence

0 ⟶ OX̃(−(d + 1)E) ⟶ OX̃(−dE) ⟶ O_{E}(d) ⟶ 0

by O_{E}, we obtain

0 ⟶ Tor_{1}(O_{E}, O_{E}(d)) ⟶ O_{E}(d + 1) ⟶ O_{E}(d) ⟶ O_{E}(d) ⟶ 0.

Thus,

Tor_{1}(O_{E}, O_{E}(d)) = O_{E}(d + 1).

2.3. Claim. For an exceptional curve E and a locally free sheaf V on X̃

Ext 2 ; X̃(O_{E}(d), V) ≅ Ext _{E}^{0}(V |_{E}, O_{E}(d − 1))^{∗},

Ext 1 ; X̃(O_{E}(d), V) ≅ Hom_{E}(O_{E}(d), V |_{E} ⊗ O_{E}(−1)).

Proof. By Serre’s duality on the surface X̃, we get

Ext q ; X̃(O_{E}(d), V) ≅ Ext 2−q ; X̃(V, O_{E}(d) ⊗ KX̃)^{∗}.

But KX̃ = σ^{∗}(K_{X}) + E, i.e., E · KX̃ = −1. Therefore, O_{E}(d) ⊗ KX̃ = O_{E}(d − 1).

Further, Ext 2−q ; X̃(V, O_{E}(d − 1)) ≅ Ext _{E}^{2−q}(V |_{E}, O_{E}(d − 1)).

In the case q = 1 we use Serre’s duality for the rational curve E:

Ext _{E}^{1}(V |_{E}, O_{E}(d − 1))^{∗} ≅ Hom_{E}(O_{E}(d − 1), V |_{E} ⊗ O_{E}(−2)).

2.4. Claim. For an exceptional divisor E

Ext^{1}(O_{E}(−d), OX̃) ≅ O_{E}(d − 1).

Proof. Let us apply the functor Hom ( · , OX̃) to the restriction sequence

0 ⟶ OX̃((d − 1)E) ⟶ OX̃(dE) ⟶ O_{E}(−d) ⟶ 0.

0 ⟶ OX̃(−dE) ⟶ OX̃((1 − d)E) ⟶ Ext^{1}(O_{E}(−d), OX̃) ⟶ 0.

This implies the claim.

3. The set EP _{e,d}^{a,b,c}.

In this section we calculate Euler characteristic of the set EP _{e,d}^{a,b,c} of all epimorphisms

β : O(a) ⊕ O(b) ⊕ O(c) ⟶ O(e) ⊕ O(d) ⟶ 0

up to the group Aut (O(e) ⊕ O(d)) action. Here we deal with bundles over projective line and we will denote by O the structure sheaf of ℙ^{1}.

For two sheaves A and B on ℙ^{1} we denote by U(A, B) the subset in Hom (A, B) consisting of all epimorphisms.

3.1. Claim. Let a, b, c, d and e be integers such that the set

U_{d,e}^{a,b,c} = U(O(a) ⊕ O(b) ⊕ O(c), O(d) ⊕ O(e))

is nonempty. Then the group G_{d,e} = Aut (O(d) ⊕ O(e)) natural action^{2} on the set U_{d,e}^{a,b,c} is free, i.e., there exists a geometrical quotient

π : U_{d,e}^{a,b,c} ⟶ EP _{d,e}^{a,b,c} = U_{d,e}^{a,b,c}/G_{d,e}

with π^{−1}([β]) ≅ G_{d,e}.

Moreover, there is an open (in Zariski sense) covering ⋃ V_{i} = EP _{d,e}^{a,b,c} such that for each V_{i} there exists a section s_{i} : V_{i} ⟶ U_{d,e}^{a,b,c} with π∘s_{i} = id_{V_{i}}.

Proof. Each element

β ∈ Hom (O(a) ⊕ O(b) ⊕ O(c), O(d) ⊕ O(e))

can be wrote as a matrix

B = ( \begin{matrix}p_{d−a} & p_{d−b} & p_{d−c} \\ p_{e−a} & p_{e−b} & p_{e−c} \end{matrix} ) ,

where p_{k} is a homogeneous polynomial of two variables (x : y) and degree k. By definition put,

m_{12}(x : y) = det B_{12}(x : y) = | \begin{matrix}p_{d−a} & p_{d−b} \\ p_{e−a} & p_{e−b} \end{matrix} | , m_{13}(x : y) = det B_{13}(x : y) = | \begin{matrix}p_{d−a} & p_{d−c} \\ p_{e−a} & p_{e−c} \end{matrix} | ,

m_{23}(x : y) = det B_{23}(x : y) = | \begin{matrix}p_{d−b} & p_{d−c} \\ p_{e−b} & p_{e−c} \end{matrix} | .

We see, that β with the matrix B is epimorphism iff for each point (x : y) ∈ ℙ^{1} the vector

(m_{12}(x : y), m_{13}(x : y), m_{23}(x : y))

is nonzero.

For (ij) ∈ {(12), (13), (23)} denote by V_{ij} the open subset in U_{d,e}^{a,b,c} consisting of all epimorphisms β such that m_{ij} is a nonzero polynomial.

Suppose that there is an epimorphism β ∈ U_{d,e}^{a,b,c} such that φ∘β = β for all φ ∈ H ⊂ G_{d,e}, where H is some subgroup of G_{d,e}. Without loss of generality we can assume that β ∈ V_{11}, i.e., m_{11} ≢ 0. Therefore the matrix B_{11}(x_{0} : y_{0}) belongs to GL(2, ℂ) for some point (x_{0} : y_{0}) ∈ ℙ^{1}. Since φ∘β = β, we see that Φ(x_{0} : y_{0}) · B_{11}(x_{0} : y_{0}) = B_{11}(x_{0} : y_{0}), where Φ(x_{0} : y_{0}) is a matrix of φ(x_{0} : y_{0}). On the other hand, Φ(x_{0} : y_{0}) ∈ GL(2, ℂ). Hence, Φ(x_{0} : y_{0}) = E is identity matrix. It is sufficient to prove that G_{d,e} acts freely in the case d = e. If d < e, we note that there exists an open subset U(β) ⊂ ℙ^{1} such that B_{11}(x : y) ∈ GL(2, ℂ) for each (x : y) ∈ U(β).

2φ : β ↦ φ∘β

To prove existence of local sections s_{i} : V_{i} ⟶ U_{d,e}^{a,b,c} for open sets V_{i} ⊂ EP _{d,e}^{a,b,c} it is sufficient to show that there exists closed subsets Z_{i} ⊂ π^{−1}(V_{i}) ⊂ U_{d,e}^{a,b,c} such that intersection of each G_{d,e}-orbit in U_{d,e}^{a,b,c} with Z_{i} equals a point.

Note that each automorphism φ(x : y) ∈ G_{d,e} is determined by its values at e−d+1 different points. Let d_{ij} be degree of the polynomial m_{ij}(x : y) and d′_{ij} = max(d_{ij} + 2, e − d + 1). For s = 1, 2, ..., d′_{ij} we fix different points (x_{s} : y_{s}) ∈ ℙ^{1} and define an open set W_{ij}^{s} ⊂ V_{ij} as the set of matrices B ∈ V_{ij} such that m_{ij}(x_{s} : y_{s}) ≠ 0. By definition put, V_{ij}^{s} = π(W_{ij}^{s}). We see that

⋃_{s,(ij)} V_{ij}^{s} = EP _{d,e}^{a,b,c}.

The description of closed subsets Z_{ij}^{s} ⊂ W_{ij}^{s} is easy to understanding and difficult to explanation. For example, we consider two cases: d = e and d = e − 1.

In the first case Z_{ij}^{s} is defined as the subset of matrices B’s in W_{ij}^{s} such that B_{ij}(x_{s} : y_{s}) = E. In the second one we introduce some more open subsets in W_{ij}^{s}: W_{ij}^{sk} = W_{ij}^{s} ∩ W_{ij}^{k} and if

B(x_{s} : y_{s}) = ( \begin{matrix}a_{11}^{s} & a_{12}^{s} \\ a_{21}^{s} & a_{22}^{s} \end{matrix} ), then

W_{ij}^{sk}(p, q) = {B ∈ W_{ij}^{sk} : a_{1p}^{s} · a_{1q}^{k} ≠ 0}.

Now we can define closed subsets Z_{ij}^{sk}(p, q) as

Z_{ij}^{sk}(p, q) = {B ∈ W_{ij}^{sk}(p, q) : a_{2p}^{s} = 0, a_{1p}^{s} = a_{2(p±1)}^{s} = 1, a_{2q}^{k} = 0}.

These examples conclude the proof.

3.2. Claim. Let X, Y be topological spaces and a group G acts on X and Y such that the action on X is free and there exists a geometrical quotient π : X ⟶ (X/G) with a section s : (X/G) ⟶ X (π∘s = id). Assume that G acts on the direct product X × Y by the rule g(x, y) ↦ (gx, gy). Then there exists geometrical quotient (X × Y)/G and (X × Y)/G ≅ (X/G) × Y.

Proof. We prove that (X × Y)/G ≅ (X/G) × Y, i.e., there are projections

(X/G) ⟵^{p_{X}} (X × Y)/G ⟶^{p_{Y}} Y

such that for any topological space Z with maps q_{X} : Z ⟶ X/G, q_{Y} : Z ⟶ Y there exists a map q : Z ⟶ (X × Y)/G with q∘p_{X} = q_{X}, q∘p_{Y} = q_{Y}.

The projection p_{X} takes each orbit G(x, y) to orbit Gx.

Since the action on X is free, for each x ∈ X we can determine an element g_{x} ∈ G such that g_{x}x = s(Gx), where Gx is the corresponding element of (X/G).

By definition put, p_{y} : G(x, y) ⟶ g_{x}y. Since g_{g_{0}x} = g_{x}g_{0}^{−1}, we see that p_{y} is well defined. Now, determining the map q : Z ⟶ (X × Y)/G by the rule q(z) = G(s(q_{X}(z)), q_{Y} (z)), we

prove the claim.

To calculate Euler characteristic of the set EP _{e,d}^{a,b,c} we need to know Euler characteristic of the set U(m_{1}, m_{2}, m_{3}), which is defined in the following way.

For several nonnegative integers m_{1}, m_{2}, …, m_{s}, let U(m_{1}, m_{2}, …, m_{s}) be the subset of ℙ(H^{0}(O_{ℙ^{1}}(m_{1}) ⊕ O_{ℙ^{1}}(m_{2}) ⊕ ⋯ ⊕ O_{ℙ^{1}}(m_{s}))) parameterizing all collections (f_{1} : f_{2} : …: f_{s}) of coprime homogeneous polynomials with deg f_{i} = m_{i}, i = 1, 2, ..., s.

It is clear, that U(m_{1}, m_{2}, …, m_{s}) parameterizes all surjective maps

⨁_{i=1}^{s} O_{ℙ^{1}}(−m_{i}) ⟶ O_{ℙ^{1}} ⟶ 0.

In the case s = 2 Euler characteristic was calculated by Li and Qin in [3]:

3.3. Lemma. (Li and Qin) For nonnegative integers m_{1} ≤ m_{2}

e(U(m_{1}, m_{2})) = δ_{m_{1}}^{0} + δ_{m_{2}}^{0}.

Here we prove the following

3.4. Lemma. Let m_{1}, m_{2}, m_{3} be integers such that 0 ≤ m_{1} ≤ m_{2} ≤ m_{3}, then

e(U(m_{1}, m_{2}, m_{3})) = δ_{m_{1}}^{0} + δ_{m_{2}}^{0} + δ_{m_{3}}^{0}.

Proof. Let us start with the case: \underline{0 = m_{1} = m_{2} = m_{3}}, i.e., U(0, 0, 0) is the set of all nonzero maps O_{ℙ^{1}} ⊕O_{ℙ^{1}} ⊕O_{ℙ^{1}} ⟶ O_{ℙ^{1}} up to a scalar multiplier. We see, that U(0, 0, 0) = ℙ^{2} and the equality follows from the formulae

e(ℙ^{n}) = n + 1.

In the case of \underline{0 = m_{1} = m_{2} < m_{3}} we have maps

u : O_{ℙ^{1}} ⊕ O_{ℙ^{1}} ⊕ O_{ℙ^{1}}(−m_{3}) ⟶ O_{ℙ^{1}}.

The morphism u is not surjective iff u|_{O_{ℙ^{1}}⊕O_{ℙ^{1}}} = 0. Hence,

U(0, 0, m_{3}) = ℙ(H^{0}(O_{ℙ^{1}} ⊕ O_{ℙ^{1}} ⊕ O_{ℙ^{1}}(m_{3}))) \ ℙ({0} ∪ {0} ∪ H^{0}(O_{ℙ^{1}}(m_{3}))) = ℙ^{m_{3}+2} \ ℙ^{m_{3}}

and e(U(0, 0, m_{3})) = 2 whenever m_{3} > 0.

The set U(0, m_{2}, m_{3}), when \underline{0 < m_{2} ≤ m_{3}}, can be produced as

U(0, m_{2}, m_{3}) = U_{0}(0, m_{2}, m_{3}) ∐ U_{1}(0, m_{2}, m_{3}),

where U_{1}(0, m_{2}, m_{3}) = {(λ, f_{2}, f_{3}) with λ ≠ 0}, and U_{0}(0, m_{2}, m_{3}) = {(0, f_{2}, f_{3})} with coprime f_{2}, f_{3}. It is clear, that

U_{1}(0, m_{2}, m_{3}) = ℙ(H^{0}(O_{ℙ^{1}} ⊕ O_{ℙ^{1}}(m_{2}) ⊕ O_{ℙ^{1}}(m_{3}))) \ ℙ({0} ∪ H^{0}(O_{ℙ^{1}}(m_{2}) ⊕ O_{ℙ^{1}}(m_{3}))) =

= ℙ^{m_{2}+m_{3}+2} \ ℙ^{m_{2}+m_{3}+1},

and U_{0}(0, m_{2}, m_{3}) = U(m_{2}, m_{3}) — the set of all coprime homogeneous polynomials of degree m_{2} and m_{3} respectively. By lemma 3.3, e(U(m_{2}, m_{3})) = 0 whenever 0 < m_{2} ≤ m_{3}.

We have,

e(U(0, m_{2}, m_{3})) = e(U_{1}(0, m_{2}, m_{3})) = m_{2} + m_{3} + 3 − (m_{2} + m_{3} + 2) = 1.

Now, let \underline{0 < m_{1} ≤ m_{2} ≤ m_{3}}. Let us represent U(m_{1}, m_{2}, m_{3}) in the form

U(m_{1}, m_{2}, m_{3}) = U_{0}(0, m_{2}, m_{3}) ∐ U_{1}(m_{1}, m_{2}, m_{3}),

where U_{1}(m_{1}, m_{2}, m_{3}) is the all triples (f_{1}, f_{2}, f_{3}) ∈ U(m_{1}, m_{2}, m_{3}) with f_{1} ≢ 0 and the set U_{0}(0, m_{2}, m_{3}) as before.

Since U_{0}(0, m_{2}, m_{3}) = U(m_{2}, m_{3}), we get

e(U(m_{1}, m_{2}, m_{3})) = e(U_{1}(m_{1}, m_{2}, m_{3})).

Note that

U_{1}(m_{1}, m_{2}, m_{3}) ⊂ ℙ(H^{0}(O_{ℙ^{1}}(m_{1})⊕O_{ℙ^{1}}(m_{2})⊕O_{ℙ^{1}}(m_{3})))\ℙ({0}∪H^{0}(O_{ℙ^{1}}(m_{2})⊕O_{ℙ^{1}}(m_{3}))).

The last set we denote by W = ℙ^{m_{1}+m_{2}+m_{3}+2} \ ℙ^{m_{2}+m_{3}+1}. On the other hand, any triples of polynomials (f_{1}, f_{2}, f_{3}) we can rewrite in the form (hg_{1}, hg_{2}, hg_{3}) such that g_{1}, g_{2}, g_{3} are coprime and h is a polynomial of some degree d. Let us denote by Y_{d} the set of all such triples. We see that U_{1}(m_{1}, m_{2}, m_{3}) = Y_{0}.

We’ll show that e(U_{1}(m_{1}, m_{2}, m_{3})) = 0 for m_{1} > 0 by induction on m_{1}. The set Y_{d} is a product ℙ^{d} and U_{1}(m_{1} −d, m_{2} −d, m_{3} −d) whenever 1 ≤ d ≤ m_{1}, i.e., e(Y_{d}) = e(ℙ^{d})·e(U_{1}(m_{1} − d, m_{2} − d, m_{3} − d)) = 0 for 1 ≤ d < m_{1} (by inductive hypothesis).

If d = m_{1}, then

U_{1}(0, m_{2} − m_{1}, m_{3} − m_{1}) =

= ℙ(H^{0}(O_{ℙ^{1}} ⊕O_{ℙ^{1}}(m_{2} −m_{1})⊕O_{ℙ^{1}}(m_{3} −m_{1})))\ℙ({0}∪H^{0}(O_{ℙ^{1}}(m_{2} −m_{1})⊕O_{ℙ^{1}}(m_{3} −m_{1}))),

therefore e(U_{1}(0, m_{2} −m_{1}, m_{3} −m_{1})) = (m_{2} −m_{1} +m_{3} −m_{1} +3)−(m_{2} −m_{1} +m_{3} −m_{1} +2) = 1, and e(Y_{m_{1}}) = m_{1} + 1.

Finally, we obtain

e(W) = m_{1} + 1 = ∑_{d=0}^{m_{1}} e(Y_{d}) = e(Y_{0}) + e(Y_{m_{1}}) = e(Y_{0}) + m_{1} + 1.

Hence,

e(Y_{0}) = e(U_{1}(m_{1}, m_{2}, m_{3})) = e(U(m_{1}, m_{2}, m_{3})) = 0.

This complates the proof.

Since EP _{e,f}^{a,b,c} ≅ EP _{e−1,f−1}^{a−1,b−1,c−1}, it can be assumed that a = 0 and e = k, f = d − k, i.e., we deal with EP _{k,d−k}^{0,b,c}. By definition, put

e_{k,d−k}^{0,b,c} = e(EP _{k,d−k}^{0,b,c}).

Let us determine possible values of the integers b, c, k and d in the case U_{k,d−k}^{0,b,c} ≠ ∅. Without loss of generality we may assume that

b ≥ c ≥ 0, d − k ≥ k.

On the other hand, there exists an epimorphism

β : O ⊕ O(c) ⊕ O(b) ⟶ O(k) ⊕ O(d − k) ⟶ 0.

This yields that either k = 0, or k ≥ c.

In the case k = 0, we have the following commutative diagram

\begin{aligned}& & & & 0 & & 0 \\ & & & & ↑ & & ↑ \\ & & 0 & ⟶ & O & ⟶^{id} & O & ⟶ & 0 \\ & & ↑ & & ↑ & & ↑ \\ 0 & ⟶ & O(c + b − d) & ⟶^{α^{∗}} & O ⊕ O(c) ⊕ O(b) & ⟶^{β} & O ⊕ O(d) & ⟶ & 0 \\ & & ↑ & & ↑ & & ↑ \\ 0 & ⟶ & O(c + b − d) & ⟶ & O(c) ⊕ O(b) & ⟶^{β′} & O(d) & ⟶ & 0 \\ & & ↑ & & ↑ & & ↑ \\ & & 0 & & 0 & & 0 \end{aligned} .

Therefore, we obtain: either d = c or d ≥ b.

Suppose that d − k < b, then the following commutative diagram

0  0
↑  ↑
0  ⟶  O ⊕ O(c)  ⟶ O(k) ⊕ O(d − k) ⟶ 0
↑  ↑  ↑
0 ⟶ O(c + b − d) ⟶ O ⊕ O(c) ⊕ O(b) ⟶ O(k) ⊕ O(d − k) ⟶ 0
↑  ↑  ↑
0 ⟶  O(b)  ⟶  O(b)  ⟶  0
↑  ↑
0  0

implies, k = 0, d = c. Thus we obtain. If b ≥ c ≥ 0 and d − k ≥ k ≥ 0, then

U_{k,d−k}^{0,b,c} ≠ ∅ ⟺ \{ \begin{matrix}either d/2 ≥ k ≥ c, d ≥ b + k, \\ or k = 0, and d = c, or d ≥ b. \end{matrix}  (3.1)

\underline{Suppose that d < c + b} (in particular, d can be equaled to c if b > 0), then each epimorphism α : O(−b) ⊕ O(−c) ⊕ O ⟶ O(d − b − c) ⟶ 0 has the kernel ker α = O ⊕ O(−d). Therefore we have bijective correspondence between EP _{0,d}^{0,b,c} and the set of all epimorphisms α : O(−b) ⊕ O(−c) ⊕ O ⟶ O(d − b − c) ⟶ 0 up to a scalar multiplier. The last set either is isomorphic to U(d − b, d − c) whenever d ≥ b or consists of one point only (when d − b) < 0.

Thus we get the following statement

3.5. Statement. For integers b ≥ c ≥ 0 and 0 ≤ d < b + c the following equalities

e(EP _{k,d−k}^{0,b,c}) = e_{k,d−k}^{0,b,c} = \{ \begin{matrix}0, & when k > 0, \\ δ_{b}^{d} + δ_{c}^{d}, & when d ≥ b, k = 0 \\ 1, & when b > d = c, k = 0. \end{matrix}

hold.

\underline{Suppose that d = c = b = 0}, then we see that EP _{0,0}^{0,0,0} ≅ U(0, 0, 0) and

e_{0,0}^{0,0,0} = 3.  (3.2)

\underline{Now we assume that d ≥ b + c > c}. Hence d ≥ b + k.

Denote by U^{k}(d − b − c, d − b, d − c) the subset of U(d − b − c, d − b, d − c), consisting of all epimorphism classes

α : O(−b) ⊕ O(−c) ⊕ O ⟶ O(d − c − b) ⟶ 0

such that ker α = O(−k) ⊕ O(k − d).

3.6. Statement. For integers b, c, k and d, satisfying conditions (3.1) the following relations

1. U^{k}(d − b − c, d − b, d − c) ≅ EP _{k,d−k}^{0,b,c};

2. e_{0,d}^{0,b,c} + ∑_{{c≤k≤d/2 \atop 0<k≤d−b}} e_{k,d−k}^{0,b,c} = e(U(d − b − c, d − b, d − c))

hold whenever d ≥ b + c.

Proof. Each class [β] ∈ EP _{k,d−k}^{0,b,c} unequally determines an epimorphism

α : O(−b) ⊕ O(−c) ⊕ O ⟶ O(d − c − b) ⟶ 0

up to a scalar multiplier. On the other hand, ker α is a rank 2 bundle of degree d, i.e., ker α = O(−k) ⊕ O(k − d) with 0 ≤ k ≤ d/2. This proves the first relations.

Consider the following commutative diagram:

\begin{aligned}& & 0 & & 0 & & 0 \\ & & ↑ & & ↑ & & ↑ \\ 0 & → & O(−r) & ⟶ & O(−b) & ⟶ & O_{ξ} & → & 0 \\ & & ↑ & & ↑ & & ↑ \\ 0 & → & O(−k) ⊕ O(k − d) & ⟶^{β^{∗}} & O(−b) ⊕ O(−c) ⊕ O & ⟶^{α} & O(d − b − c) & → & 0 \\ & & ↑ & & ↑ & & ↑ \\ 0 & → & O(r − d) & ⟶ & O(−c) ⊕ O & ⟶ & O(d − c − r) & → & 0 \\ & & ↑ & & ↑ & & ↑ \\ & & 0 & & 0 & & 0 \end{aligned} ,

where O(−r) is the image of the map from O(−k) ⊕ O(k − d) to O(−b). We see that −r ≤ −b and (since −k ≥ k−d) either −k ≤ −r, or −r = k−d. If −k ≤ −r, we get, that k ≥ r ≥ b ≥ c, i.e., k ≥ c. If −r = k − d, then the last row of the diagram has the form

0 ⟶ O(−k) ⟶ O(−c) ⊕ O ⟶ O(k − c) ⟶ 0.

Therefore, either k = 0, or k ≥ c. This concludes the proof.

3.7. Lemma. Suppose that integers b, c, d and k satisfy the conditions: d ≥ b + c, b ≥ c ≥ 0 and either c ≤ k ≤ min(d/2, d − b), k < d − k, or k = 0, c > 0. Then

e_{k,d−k}^{0,b,c} = \{ \begin{matrix}∑_{{m=−b \atop −c≤m≤(k−b−c)/2} ; m≤k−b−c} e(U(d − b − c − m, d − k + m)) · e_{m,k−b−c−m}^{−b,−c,0}, for k ≥ c ≥ 0, \\ e(U(d − c, d − b)) · e_{0,b−c}^{0,b−c,b} for k = 0, c > 0. \end{matrix}

Proof. Consider the following commutative diagram:

0  0  0
↑  ↑  ↑
0 ⟶  O(k − d)  ⟶  F  ⟶^{η} O(d − b − c) ⟶ 0
↑  ε ↑  id ↑
0 ⟶ O(−k) ⊕ O(k − d) ⟶^{β^{∗}} O(−b) ⊕ O(−c) ⊕ O ⟶^{α} O(d − b − c) ⟶ 0
γ ↑  δ ↑  ↑
0 ⟶  O(−k)  ⟶^{id}  O(−k)  ⟶  0
↑  ↑
0  0

(3.3)

Recall that by statement 3.6 a class [β] = {φ∘β}_{φ∈G_{k}} ∈ EP _{k,d−k}^{0,b,c} unequally determines a class [α] = {λ · α}_{λ∈ℂ^{∗}} ∈ U^{k}(d − b − c, d − c, d − b), where G_{k} = Aut (O(−k) ⊕ O(k − d)).

If we hold fixed a nonzero map

γ ∈ Hom (O(−k), O(−k) ⊕ O(k − d)) = Hom (O(−k), O(−k)),

we have a subset

[δ] = {β^{∗}∘φ∘γ}_{φ∈G_{k}} ⊂ Hom (O(−k), O(−b) ⊕ O(−c) ⊕ O).

Since the group G_{k} acts transitive on the set Hom (O(−k), O(−k)) \ {0}, the subset [δ] does not depend on a choice of γ. Therefore we can assume that γ has the matrix Γ = (1, 0)^{t}. On the other hand, the map β^{∗} can be wrote as a matrix

B = ( \begin{matrix}p_{k−b} & p_{d−b−k} \\ p_{k−c} & p_{d−c−k} \\ p_{k} & p_{d−k} \end{matrix} ) ,

where p_{i} is a homogeneous polynomial of two variables and degree i. An automorphism φ has a matrix

Φ = ( \begin{matrix}λ & p_{d−2k} \\ 0 & μ \end{matrix} )

with λ, μ ∈ ℂ^{∗}. Therefore a map δ ∈ [δ] has a matrix

D = B · Φ · Γ = (λ · p_{k−b}, λ · p_{k−c}, λ · p_{k})^{t},

i.e., each class [β] unequally determines a map δ up to a scalar multiplier. Hence having a class [β] via [δ] we unequally construct a bundle F = coker δ and a class of epimorphisms [ε]

ε ∈ Hom (O(−b) ⊕ O(−c) ⊕ O, F)

up to the group Aut F action: [ε] = {ψ∘ε}_{ψ∈Aut F}.

Furthermore, for fixed epimorphisms α and ε there exists a unique epimorphism η ∈ Hom (F, O(d − b − c)) such that α = η∘ε. Moreover, if we are twisting the map ε by an automorphism ψ ∈ Aut F, the map η is changing to η∘ψ^{−1}. Besides, α is determined up to a scalar multiplier, consequently by means of class [β] ∈ EP _{k,d−k}^{0,b,c} we unequally construct a set

[ε, η] = {ψ∘ε, λ · η∘ψ^{−1}}_{λ∈ℂ^{∗}, ψ∈Aut F} .

We see that the pair [ε, η] is an element of quotient

(U(O(−b) ⊕ O(−c) ⊕ O, F) × U(F, O(d − b − c)))/(Aut F × ℂ^{∗}),

where the group Aut F × ℂ^{∗} acts by the rule

(ψ, λ) : (ε, η) ↦ (ψ∘ε, λ · η∘ψ^{−1}),

and U(A, B) is the set of epimorphisms from A to B. Let us denote this quotient by CO(F). The following lemma is consequence of claims 3.1 and 3.2.

3.8. Lemma. There exists a locally trivial bundle

CO(F) ⟶ U(O(−b) ⊕ O(−c) ⊕ O, F)/Aut F

with fiber U(F, O(d − b − c))/ℂ^{∗}.

Moreover, it is easy to see that the class [β] is unequally reconstructed by means of a pair [ε, η] ∈ CO(F). Thus there exists one-to-one correspondence between EP _{k,d−k}^{0,b,c} and ∐_{F} CO(F).

Let us determine possible F ’s. F is a rank 2 bundle with degree (k − b − c). Therefore it has the form F = O(m) ⊕ O(k − b − c − m), where \frac{k − b − c}{2} ≥ m. The bundle F is included in commutative diagram (3.3). In particular, there exists an epimorphism

ε : O(−b) ⊕ O(−c) ⊕ O ⟶ O(m) ⊕ O(k − b − c − m) ⟶ 0.

This yields that the following possibilities take place:

k − b − c − m ≥ 0, m = −b,
k − b − c − m ≥ 0, m ≥ −c ≥ −b,
k − b − c − m < 0 ⟹ k = 0, m = −b.

Besides, F is included in the exact sequence

0 ⟶ O(k − d) ⟶ O(m) ⊕ O(k − b − c − m) ⟶ O(d − c − b) ⟶ 0,

whence,

k − d ≤ m ≤ k − b − c − m ≤ d − b − c ⟺ m ≥ k − d,

or

m = d − b − c.

By our assumption, −b ≥ k − d. Therefore, inequality m ≥ k − d is true.

If k − b − c ≥ m and k < d − k, the equality m = d − b − c is impossible.

If k = 0 and c > 0, then we see that there exists only one possibilities for F F = O(−b) ⊕ O(−c). This concludes the proof.

3.9. Corollary. Under the conditions of lemma 3.7 we have

e_{k,d−k}^{0,b,c} = {e_{d−b,b}^{0,b,c} = e_{0,d−c}^{0,b,b−c}, when k = d − b, d < 2b, 0, otherwise.

Proof. By lemma 3.7,

e_{k,d−k}^{0,b,c} = ∑_{{m=−b \atop −c≤m≤(k−b−c)/2} ; m≤k−b−c} e(U(d − b − c − m, d − k + m)) · e_{m,k−b−c−m}^{−b,−c,0}, for k ≥ c ≥ 0,

or e(U(d − c, d − b)) · e_{0,b−c}^{0,b−c,b} for k = 0, c > 0.

In the last case e(U(d − c, d − b)) = 0, since d ≥ b + c > b ≥ c.

It follows from the assumptions that

2m ≤ k − b − c ⟺ d − k + m ≤ d − b − c − m.

Besides, d − k + m ≥ 0 and e(U(p, q)) = δ_{p}^{0} + δ_{q}^{0}. Therefore, e(U(d − b − c − m, d − k + m)) = 0 iff m = k − d.

\underline{Suppose that −b = m = k − d}, i.e., k = d − b. The inequality k < d − k (by assumptions) implies that d < 2b. In this case e(U(d − b − c − m, d − k + m)) = 1 + δ_{d−c}^{0}.

The equality d = c contradicts the assumptions d ≥ b + c, b ≥ c ≥ 0 of the lemma with d < 2b. Hence e(U(d − b − c − m, d − k + m)) = 1.

\underline{Consider the situation −c ≤ m = k − d ≤ \frac{k − b − c}{2}.} The left side of this inequalities with the assumption of the lemma k ≤ d − b, b ≥ c, d ≥ b + c give:

d − k = b = c,  k = d − b,  d ≥ b + c = 2b.

On the other hand, by assumptions of the lemma 3.7 k < d − k. This with k = d − b gives: d < 2b. This contradiction proves that e_{k,d−k}^{0,b,b} = 0 whenever 0 ≤ b ≤ k ≤ d − b, k < d − k and d ≥ 2b.

The above reasons yield that the sum of lemma 3.7 has one nonzero summand only: e_{m,k−b−c−m}^{−b,−c,0} with m = k − d = −b in the case d < 2b.

We see that

e_{m,k−b−c−m}^{−b,−c,0} = e_{−b,d−c−b}^{−b,−c,0} = e_{0,d−c}^{0,b,b−c}.

This completes the proof.

3.10. Proposition. For integers b ≥ c ≥ 0, d/2 ≥ k ≥ 0 the numbers e_{k,d−k}^{0,b,c} are nonzero iff either k = 0 or k = c > 0. Moreover,

e_{0,d}^{0,b,c} = \{ \begin{matrix}3, & when & d = b = c = 0; \\ 2, & when & d = b = c > 0, or d = b > 0, c = 0; \\ 1, & when & d = b > c > 0, or b > d = c; \\ 0, & otherwise. \end{matrix}

e_{c,d−c}^{0,b,c} = \{ \begin{matrix}1, & when & d − c = b, c > 0; \\ 0, & otherwise. \end{matrix}

Proof. We assume that e_{k,d−k}^{0,b,c} = 0 whenever U_{k,d−k}^{0,b,c} = ∅. Suppose that 0 ≤ d < b + c. Using statement 3.5 and equality (3.2), we see that e_{k,d−k}^{0,b,c} = 0

if k > 0 and

e_{0,d}^{0,b,c} = \{ \begin{matrix}3, & when & d = b = c = 0; \\ 2, & when & d = b = c > 0; \\ 1, & when & d = b > c > 0, or b > d = c; \\ 0, & otherwise. \end{matrix}  (3.4)

Suppose that d ≥ b + c and k < d − k, then by corollary 3.9, e_{k,d−k}^{0,b,c} = 0 unless k = d − b. Besides, e_{d−b,b}^{0,b,c} = e_{0,d−c}^{0,b,b−c} and d < 2b, i.e., d − c < b + (b − c). Now we use equality (3.4) and obtain that

e_{0,d−c}^{0,b,b−c} = \{ \begin{matrix}3, & when & d − c = b = b − c = 0; \\ 2, & when & d − c = b = b − c > 0; \\ 1, & when & d − c = b > b − c > 0, or b > d − c = b − c; \\ 0, & otherwise. \end{matrix}

The equalities d − c = b = b − c = 0 are equivalent to d = b = c = 0, what contradicts the inequality d < 2b.

The equalities d − c = b = b − c > 0 imply d = b > 0, c = 0. The relations d − c = b > b − c > 0 yield that d = b + c, c > 0. In this situation the

assumption d < 2b gives b > c.

The relations b > d − c = b − c imply d = b, c > 0, what contradicts the assumption d ≥ b + c.

Therefore,

e_{d−b,d}^{0,b,c} = \{ \begin{matrix}e_{0,b}^{0,b,0} = 2, & when & d = b > 0, c = 0; \\ e_{c,b}^{0,b,c} = 1, & when & d = b + c, b > c > 0; \\ 0, & otherwise. \end{matrix}

It is remains to consider the case k = d − k and d ≥ b + c. Let us denote d = 2n and calculate the numbers e_{n,n}^{0,b,c}.

Taking into account statement 3.6, we get

e_{0,2n}^{0,b,c} + ∑_{{c≤k≤n \atop 0<k≤2n−b}} e_{k,2n−k}^{0,b,c} = δ_{b+c}^{2n} + δ_{b}^{2n} + δ_{c}^{2n}.  (3.5)

Since

e_{0,2n}^{0,b,c} = \{ \begin{matrix}2, & when & b = 2n, c = 0; \\ 0, & otherwise. \end{matrix} ,

and for integer k such that c ≤ k ≤ 2n − b, 0 < k < n the following relations hold

e_{k,2n−k}^{0,b,c} = \{ \begin{matrix}1, & when & k = c, b + c = 2n, b > c > 0; \\ 0, & otherwise. \end{matrix}

we see that the left side of equality (3.5) has the form

e_{n,n}^{0,b,c} + \{ \begin{matrix}2, & when & b = 2n, c = 0; \\ 1, & when & b + c = 2n, b > c > 0; \\ 0, & otherwise. \end{matrix} .

On the other hand, since 2n ≥ b + c, the right side of equality (3.5) is equal to

\{ \begin{matrix}2, & when & b = 2n, c = 0; \\ 1, & when & b + c = 2n, c > 0; \\ 0, & otherwise. \end{matrix} .

Therefore,

e_{n,n}^{0,b,c} = \{ \begin{matrix}1, & when & b = c = n > 0; \\ 0, & otherwise. \end{matrix} .

4. Lemma about stability.

Recall that σ : X̃ ⟶ X is a blow up of a smooth projective surface X at a point x_{0} ∈ X and E = σ^{−1}(x_{0}) is an exceptional curve on X̃. We denote by H̃ the divisor σ^{∗}(H) and H_{n} = nH̃ − E, where H is an ample divisor on X.

Besides, we recall the definition of stability in Mumford–Takemoto sense and some properties of stable sheaves.

A torsion free sheaf F on a surface X is called μ_{H}-(semi)stable w.r.t. an ample (or numerically effective) divisor H, if for any its subsheaf G with rk G < rk F the following takes place

μ_{H}(G) = \frac{c_{1}(G) · H}{rk G} < \frac{c_{1}(F) · H}{rk F} = μ_{H}(F) (≤ for semistable F).

It is known (see, for example [6], [2]), that μ_{H}-stable sheaf F is simple, i.e., End(F) = ℂ, and if F is locally free then the dual sheaf F^{∗} is μ_{H}-stable too.

4.1. Lemma. Assume, V is a bundle over X̃ such that c_{1}(V) · H̃ and rk (V) = r are coprime. Denote by V′ the kernel of an epimorphism V ⟶ O_{E}(d) ⟶ 0:

0 ⟶ V′ ⟶ V ⟶ O_{E}(d) ⟶ 0.  (4.1)

Then V′ is a locally free sheaf and it is μ_{H_{n}}-stable for n ≫ 0 iff V is μ_{H_{n}}-stable for n ≫ 0 too.

Proof. At first we note that V′ has no torsion, because of it is a subsheaf of torsion free sheaf V. If V′ is not locally free, we have a commutative diagram:

\begin{aligned}& & & & & & 0 \\ & & & & & & ↓ \\ & & 0 & & & & Q \\ & & ↓ & & & & ↓ \\ 0 & ⟶ & V′ & ⟶ & V & ⟶ & O_{E}(d) & ⟶ & 0 \\ & & ↓ & & || & & ↓ \\ 0 & ⟶ & (V′)^{∗∗} & ⟶ & (V)^{∗∗} & ⟶ & Q′ & ⟶ & 0 \\ & & ↓ & & & & ↓ \\ & & Q & & & & 0 \\ & & ↓ \\ & & 0 \end{aligned} ,

where dim Supp Q = 0. But Q is a subsheaf of a bundle over the curve, i.e., Q = 0. The second statement of the lemma we prove in several steps. Step 1. If a sheaf F is μ_{H_{n}}-stable for n ≫ 0, then it is μ_{H}-semistable. For any subsheaf

G of F we have

μ_{H_{n}}(G) < μ_{H_{n}}(F) ⟺ \frac{nc_{1}(G) · H̃ − c_{1}(G) · E}{r(G)} < \frac{nc_{1}(F) · H̃ − c_{1}(F) · E}{r(F)} ,

i.e.,

n (\frac{c_{1}(G) · H̃}{r(G)} − \frac{c_{1}(F) · H̃}{r(F)} ) < \frac{c_{1}(G) · E}{r(G)} − \frac{c_{1}(F) · E}{r(F)}

for n ≫ 0. Therefore,

\frac{c_{1}(G) · H̃}{r(G)} ≤ \frac{c_{1}(F) · H̃}{r(F)} .

Step 2. Assume that a sheaf F is μ_{H_{n}}-stable for some n. Then it is μ_{H_{k}}-stable for all k > n. The proof follows from the previous step.

Step 3. If V is μ_{H_{n}}-stable for n ≫ 0, Then V′ is μ_{H_{n}}-stable for n ≫ 0. Suppose, that V′ is not μ_{H_{n}}-stable, then there exists a subsheaf G ⊂ V′ of rank r(G) < r such that

μ_{H_{n}}(V′) ≤ μ_{H_{n}}(G).  (4.2)

But G is a subsheaf of V too. Hence,

μ_{H_{n}}(G) < μ_{H_{n}}(V).  (4.3)

Since c_{1}(V′) = c_{1}(V) − E and rank of V′ equals r, we have,

μ_{H_{n}}(V′) = \frac{1}{r}(c_{1}(V′)(nH̃ − E)) = \frac{1}{r}(c_{1}(V)(nH̃ − E) − 1),

μ_{H_{n}}(V) = \frac{1}{r}(c_{1}(V)(nH̃ − E)), μ_{H_{n}}(G) = \frac{1}{r(G)}(c_{1}(G)(nH̃ − E)),

Combining (4.2) and (4.3), we have

\frac{r(G)}{r}(c_{1}(V)(nH̃ − E)) − \frac{r(G)}{r} ≤ c_{1}(G)(nH̃ − E) < \frac{r(G)}{r}(nc_{1}(V) · H̃ − c_{1}(V) · E)) (4.4)

By assumption of the lemma c_{1}(V) · H̃ and r are coprime, therefore the number

\frac{r(G)}{r}(nc_{1}(V) · H̃ − c_{1}(V) · E))

is integer for n ≡ (c_{1}(V) · E)(c_{1}(V) · H̃)^{−1} (mod r). The number c_{1}(G)(nH̃ − E) is integer

too, and 0 < \frac{r(G)}{r} < 1. Therefore, inequality (4.4) is not possible. Now the step 3 follows from the second one.

Step 4. If V′ is μ_{H_{n}}-stable for n ≫ 0, Then V is μ_{H_{n}}-stable for n ≫ 0. Let us apply the functor Hom ( · , OX̃) to the sequence (4.1):

0 ⟶ V^{∗} ⟶ (V′)^{∗} ⟶ Ext^{1}(O_{E}(d), OX̃) ⟶ 0.

It follows from claim 2.4 that Ext^{1}(O_{E}(d), OX̃) ≅ O_{E}(1 − d). Since the μ_{H_{n}}-stability of V and V^{∗} is equivalent, our lemma follows from the step 3.

4.2. Lemma. For a μ_{H_{n}}-stable bundle V on X̃ the space Hom (V (−E), V) has dimension 1. Proof. Suppose, that the restriction of V to E has the form V |_{E} = ⨁_{i=1}^{r} O_{E}(d_{i}) such that

d_{1} ≤ d_{2} ≤ … ≤ d_{r}. We prove the lemma by induction on Δ(V) = d_{r} − d_{1}.

In the case of Δ(V) = 0 we can assume, that d_{1} = d_{2} = … = d_{r} = 0, since

Hom (V (−E + dE), V (dE)) ≅ Hom (V (−E), V).

Applying the functor Hom ( · , V) to the restriction sequence

0 ⟶ V (−E) ⟶ V ⟶ V |_{E} ⟶ 0,

we obtain

0 ⟶ Hom (V, V) ⟶ Hom (V (−E), V) ⟶ Ext^{1}(rO_{E}, V),

where rO_{E} = O_{E} ⊕ … ⊕ O_{E}. Since V is μ_{H_{n}}-stable, it is simple, i.e., Hom (V, V) = ℂ. On the other hand, by claim 2.3 it follows, that Ext^{1}(rO_{E}, V) ≅ Ext _{E}^{1}(V |_{E}, rO_{E}(−1))^{∗} = 0. Therefore, the base of induction is proved.

Suppose, Δ(V) > 0 and d_{1} = … = d_{s} < d_{s+1}. Consider an exact sequence

0 ⟶ V′ ⟶ V ⟶ ⨁_{i=1}^{s} O_{E}(d_{i}) ⟶ 0  (4.5)

and restrict it to the curve E:

0 ⟶ Tor_{1}(⨁_{i=1}^{s} O_{E}(d_{i}), O_{E}) ⟶ V′|_{E} ⟶ V |_{E} ⟶ ⨁_{i=1}^{s} O_{E}(d_{i}) ⟶ 0.

Since Tor_{1}(O_{E}(d), O_{E}) = O_{E}(d + 1) (see claim 2.2), we get

0 ⟶ ⨁_{i=1}^{s} O_{E}(d_{i} + 1) ⟶ V′|_{E} ⟶ ⨁_{i=s+1}^{r} O_{E}(d_{i}) ⟶ 0.

Whence, V′|_{E} = ⨁_{i=1}^{r} O_{E}(d′_{i}) with d_{1} < d′_{1} ≤ d′_{2} ≤ … ≤ d′_{r} ≤ d_{r}, i.e., Δ(V′) < Δ(V). Besides,

V′ is μ_{H_{n}}-stable (see lemma 4.1). By the inductive hypothesis, Hom (V′(−E), V′) = ℂ. Let us apply the functor Hom (V (−E), · ) to the sequence (4.5):

0 ⟶ Hom (V (−E), V′) ⟶ Hom (V (−E), V) ⟶ Hom (V (−E), ⨁_{i=1}^{s} O_{E}(d_{i})).

Since Hom (V (−E), ⨁_{i=1}^{s} O_{E}(d_{i})) = Hom_{E}( ⨁_{i=1}^{r} O_{E}(d_{i} + 1), ⨁_{i=1}^{s} O_{E}(d_{i})) = 0,

Hom (V (−E), V′) ≅ Hom (V (−E), V).

Further, tensoring the sequence (4.5) by O(−E) and applying the functor Hom ( · , V′) to it, we get:

0 ⟶ Hom (V (−E), V′) ⟶ Hom (V′(−E), V′).

Because of the equality Hom (V′(−E), V′) = ℂ, dimension of the space Hom (V (−E), V′) is less or equal to 1. Finally,

1 ≤ dim Hom (V (−E), V) = dim Hom (V (−E), V′) ≤ 1.

This concludes the proof.

5. Mutations.

Let 𝔐(i, k) be a moduli space of μ_{H_{n}}-stable (for n ≫ 0) rank 3 bundles V on X̃ with c_{1}(V) = σ^{∗}(C) − iE, c_{2}(V) = k (see Introduction). Then the restriction of a bundle V ∈ 𝔐(i, k) to the rational curve E has the form

V |_{E} = O_{E}(−d_{1}) ⊕ O_{E}(−d_{2}) ⊕ O_{E}(i + d_{1} + d_{2})  (5.1)

We may assume that i = 0, 1, 2 or 3 and −\frac{i}{3} ≤ d_{1}, −d_{1} ≤ −d_{2} ≤ i + d_{1} + d_{2}, i.e., −\frac{d_{1} + i}{2} ≤ d_{2} ≤ d_{1}.

Denote by 𝔐^{(d_{1},d_{2})}(i, k) the subset of 𝔐(i, k) consisting of bundles V with restriction to E in the form (5.1). Then

𝔐(i, k) = ∐_{{− \frac{i}{3} ≤d_{1} \atop − \frac{d_{1}+i}{2} ≤d_{2}≤d_{1}}} 𝔐^{(d_{1},d_{2})}(i, k).

5.1. Lemma. For integers d ≥ 0, and i = 0, 1, 2 the following formula takes place

2 · e(𝔐^{(d,d)}(i, k)) = ∑_{− \frac{d+i+1}{2} ≤l≤d−1} e_{0,3d+i}^{0,2d+l+i+1,d−l} · e(𝔐^{(d,l)}(i + 1, k − i − d))

unless i = d = 0, where e(Y) is Euler characteristic of Y_{red} and

e_{0,3d+i}^{0,2d+l+i+1,d−l} = e(EP _{0,3d+i}^{0,2d+l+i+1,d−l})

is Euler characteristic of the quotient of all epimorphism set

U(O_{E} ⊕ O_{E}(2d + l + i + 1) ⊕ O_{E}(d − l), O_{E} ⊕ O_{E}(3d + i))

by the group Aut (O_{E} ⊕ O_{E}(3d + i)) action.

Proof. Let V be a bundle from 𝔐^{(d,d)}(i, k), i.e.,

V |_{E} = O_{E}(−d) ⊕ O_{E}(−d) ⊕ O_{E}(2d + i).

Consider an arbitrary epimorphism α : V ⟶ O_{E}(−d). Note that

α ∈ Hom (V, O_{E}(−d)) ≅ Hom_{E}(O_{E}(−d) ⊕ O_{E}(−d), O_{E}(−d)) ≅ ℂ^{2}.

Therefore, α up to a scalar multiplier is element of ℙ^{1}. Denote by V′_{α} the kernel of α:

0 ⟶ V′_{α} ⟶ V ⟶^{α} O_{E}(−d) ⟶ 0.  (5.2)

It follows from lemma 4.1, that V′_{α} is a μ_{H_{n}}-stable bundle. We see that

rk (V′_{α}) = 3, c_{1}(V′_{α}) = c_{1}(V) − E = σ^{∗}(C) − (i + 1)E

and c_{2}(V′_{α}) can be calculated by the standard way, using multiplicativity of the full Chern character and exact triple (5.2):

(1 + c_{1}(V′_{α})t + c_{2}(V′_{α})t^{2})(1 + Et + c_{2}(O_{E}(−d))t^{2}) = 1 + c_{1}(V)t + k t^{2}.

Since c_{2}(O_{E}(−d)) = d − 1 (see claim 2.1),

c_{2}(V′_{α}) = k − d + 1 − c_{1}(V) · E + E^{2} = k − d − i.

Whence, V′_{α} ∈ 𝔐(i + 1, k − d − i).

Exact sequence (5.2) induces the following commutative diagram

\begin{aligned}& & 0 & & 0 \\ & & ↑ & & ↑ \\ 0 & ⟶ & O_{E}(−d) ⊕ O_{E}(2d + i) & ⟶ & V |_{E} & ⟶ & O_{E}(−d) & ⟶ & 0 \\ & & β ↑ & & res ↑ & & id ↑ \\ 0 & ⟶ & V′_{α} & ⟶ & V & ⟶^{α} & O_{E}(−d) & ⟶ & 0 \\ & & ↑ & & ↑ \\ & & V (−E) & = & V (−E) \\ & & ↑ & & ↑ \\ & & 0 & & 0 \end{aligned}  (5.3)

Thus we constructed a bundle V′_{α} ∈ 𝔐(i + 1, k − d − i) and an epimorphism β : V′_{α} ⟶ O_{E}(−d)⊕O_{E}(2d+i) ⟶ 0 by means of the bundle V and the epimorphism α : V ⟶ O_{E}(−d). Let us show, that different α’s give various bundles V′_{α}.

If we apply the functor Hom ( · , V) to the left column of the diagram, we get

0 ⟶ Hom (V′_{α}, V) ⟶ Hom (V (−E), V).

Taking into account lemma 4.2, we obtain Hom (V′_{α}, V) ≅ ℂ. Moreover, this equality holds for the kernel of any epimorphism α ∈ Hom (V, O_{E}(−d)).

Assume, that V ′ ; ᾱ is the kernel of another epimorphism ᾱ ∈ Hom (V, O_{E}(−d)), and there exists an isomorphism φ : V′_{α} ⟶ V ′ ; ᾱ. Because of the bundle V is simple, the map of composition 0 ⟶ Hom (V, V) ⟶ Hom (V′_{α}, V) is an isomorphism, and we have the commutative diagram

0 ⟶ V′_{α} ⟶ V ⟶^{α} O_{E}(−d) ⟶ 0
φ ↓ φ_{V} ↓ φ_{O} ↓
0 ⟶ V ′ ; ᾱ ⟶ V ⟶^{ᾱ} O_{E}(−d) ⟶ 0

with nonzero φ_{V} and φ_{O}. The sheaves V and O_{E}(−d) are simple, therefore, φ_{V} = λ·id_{V} and φ_{O} = μ·id_{O_{E}(−d)}. Thus bundles V′_{α} and V ′ ; ᾱ are isomorphic iff α = νᾱ for some ν ∈ ℂ^{∗}.

5.2. Lemma. Let β and β′ be epimorphisms from Hom (V′_{α}, O_{E}(−d) ⊕ O_{E}(2d + i)), then ker β ≅ ker β′ iff there exists an element φ ∈ Aut(O_{E}(−d) ⊕ O_{E}(2d + i)) such that β′ = φ ∘ β.

Proof. It is clear, that if β′ = φ ∘ β, then ker β ≅ ker β′. Now suppose that V (−E) = ker β ≅ ker β′ = V′(−E) and ψ : V (−E) ⟶ V′(−E) is isomorphism:

\begin{aligned}0 & ⟶ & V (−E) & ⟶^{γ} & V′_{α} & ⟶^{β} & O_{E}(−d) ⊕ O_{E}(2d + i) & ⟶ & 0 \\ & & ψ ↓ \\ 0 & ⟶ & V′(−E) & ⟶^{γ′} & V′_{α} & ⟶^{β′} & O_{E}(−d) ⊕ O_{E}(2d + i) & ⟶ & 0 \end{aligned} .

Therefore, there exists a map γ′ ∘ ψ : V (−E) ⟶ V′_{α}. Let us apply the functor Hom ( · , V′_{α}) to the first sequence of the diagram

0 ⟶ Hom (V′_{α}, V′_{α}) ⟶^{p} Hom (V (−E), V′_{α}).

On the other hand, we know, that

0 ⟶ V′_{α} ⟶ V ⟶^{α} O_{E}(−d) ⟶ 0,

and Hom (V (−E), V) ≅ ℂ. Hence, Hom (V (−E), V′_{α}) ≅ ℂ, i.e., the map p is isomorphism. This implies that there exists the following commutative diagram

\begin{aligned}0 & ⟶ & V (−E) & ⟶^{γ} & V′_{α} & ⟶^{β} & O_{E}(−d) ⊕ O_{E}(2d + i) & ⟶ & 0 \\ & & ψ ↓ & & ψ′ ↓ & & φ ↓ \\ 0 & ⟶ & V′(−E) & ⟶^{γ′} & V′_{α} & ⟶^{β′} & O_{E}(−d) ⊕ O_{E}(2d + i) & ⟶ & 0 \end{aligned} ,

where ψ′ and φ are isomorphisms. This concludes the proof.

Prove that there exists one-to-one correspondence between

{(V, α)} and {(V′_{α}, β/mod Aut(O_{E}(−d) ⊕ O_{E}(2d + i))}.

At first, the bundle V is uniquely determined by (V′_{α}, β) as (ker β) ⊗ O(E). Let us reconstruct commutative diagram (5.3).

Applying the functor Hom ( · , V (−E)) to the first row of the diagram, we obtain

0 ⟶ Ext^{1}(O_{E}(−d), V (−E)) ⟶^{p} Ext^{1}(V |_{E}, V (−E)) ⟶^{q} Ext^{1}(O_{E}(−d)⊕O_{E}(2d+i), V (−E))

(5.4)

By claim 2.3, we get

Ext^{1}(O_{E}(−d), V (−E)) ≅ Hom_{E}(O_{E}(−d), V |_{E}), Ext^{1}(V |_{E}, V (−E)) ≅ End_{E}(V |_{E}),

Ext^{1}(O_{E}(−d) ⊕ O_{E}(2d + i), V (−E)) ≅ Hom_{E}(O_{E}(−d) ⊕ O_{E}(2d + i), V |_{E}).

Consequently triple (5.4) has the form

0 ⟶ Hom_{E}(O_{E}(−d), V |_{E}) ⟶^{p} End_{E}(V |_{E}) ⟶^{q} Hom_{E}(O_{E}(−d) ⊕ O_{E}(2d + i), V |_{E}).

By dimension consideration we get, that q is epimorphism. If the left column of the diagram is determined by an element e ∈ Ext^{1}(O_{E}(−d) ⊕ O_{E}(2d + i), V (−E)), then taking an exact sequence corresponding to q^{−1}(e) ∈ Ext^{1}(V |_{E}, V (−E)), we obtain the following commutative diagram

0  0
↑  ↑
0 ⟶ O_{E}(−d) ⊕ O_{E}(2d + i) ⟶ V |_{E} ⟶ O_{E}(−d) ⟶ 0
β ↑  res ↑  id ↑
0 ⟶  V′_{α}  ⟶ V̄  ⟶^{ᾱ} O_{E}(−d) ⟶ 0
↑  ↑
V (−E)  = V (−E)
↑  ↑
0  0

Now we need to check, that V̄ ≅ V and ᾱ = να. It is sufficient to prove the first equality since the second one follows from the first equality and the fact that was proved above.

First we note that the group Aut(V |_{E}) acts on the space Ext^{1}(V |_{E}, V (−E)) ≅ End_{E}(V |_{E}) and extensions determined by elements from one orbit are isomorphic.

Let us show that elements belonging to the open orbit determines locally free extensions. Apply the functor Hom ( · , OX̃) to the middle column of the last diagram

0 ⟶ V̄^{∗} ⟶ V^{∗}(E) ⟶^{f} Ext^{1}(V |_{E}, OX̃) ⟶ Ext^{1}(V̄ , OX̃).

The sheaf V̄ is locally free iff Ext^{1}(V̄ , OX̃) = 0, i.e., the map f is an epimorphism. Taking into account assumptions and claim 2.4, we get

Ext^{1}(V |_{E}, OX̃) ≅ O_{E}(d − 1) ⊕ O_{E}(d − 1) ⊕ O_{E}(−2d − 1).

On the other hand, the map f is an epimorphism iff its restriction f|_{E} to the curve E is one. In other words, f|_{E} is automorphism of V^{∗}(E)|_{E}. Therefore elements of Ext^{1}(V |_{E}, V (−E)) ≅ End_{E}(V |_{E}), determining locally free extensions are automorphisms of V |_{E}.

For the same reason, the element e ∈ Ext^{1}(O_{E}(−d) ⊕ O_{E}(2d + i), V (−E)), determining the left column of the diagram is a monomorphism e : O_{E}(−d) ⊕ O_{E}(2d + i) ⟶ V |_{E}, since V′_{α} is locally free. Hence, to reconstruct the diagram we can choose q^{−1}(e) in the such way that it be an automorphism of V |_{E}. Therefore, V̄ is locally free and is isomorphic to V.

Let us calculate the restriction of V′_{α} to the curve E. The left column of diagram (5.3) tensoring by O_{E} has the form:

0 ⟶ Tor_{1}(O_{E}(−d)⊕O_{E}(2d+i), O_{E}) ⟶ V (−E)|_{E} ⟶ V′_{α}|_{E} ⟶^{β|_{E}} O_{E}(−d)⊕O_{E}(2d+i) ⟶ 0.

It follows from claim 2.2, that

Tor_{1}(O_{E}(−d) ⊕ O_{E}(2d + i), O_{E}) ≅ O_{E}(1 − d) ⊕ O_{E}(2d + i + 1).

Hence V′_{α}|_{E} is included in the following exact triple:

0 ⟶ O_{E}(1 − d) ⟶^{γ} V′_{α}|_{E} ⟶^{β|_{E}} O_{E}(−d) ⊕ O_{E}(2d + i) ⟶ 0,  (5.5)

and

V′_{α}|_{E} = O_{E}(−d) ⊕ O_{E}(−l) ⊕ O_{E}(d + l + i + 1),

with 1 − d ≤ −l ≤ d + l + i + 1 ≤ 2d + i ⟺ −\frac{d + 1 + i}{2} ≤ l ≤ d − 1.

5.3. Remark. The set 𝔐̃^{(d,d)}(i, k) of all pairs (V, α), where V ∈ 𝔐^{(d,d)}(i, k) and α ∈ ℙHom (V, O_{E}(−d)) is a locally trivial bundle over 𝔐^{(d,d)}(i, k) with fiber ℙ^{1}, i.e.,

e(𝔐̃^{(d,d)}(i, k)) = 2e(𝔐^{(d,d)}(i, k)).

Proof. Since n ≫ 0 and C · H ≢ 0 (mod 3), there are no μ_{H_{n}}-semistable bundles V on X̃ with c_{1}(V) = C − iE, rk (V) = 3 and c_{2}(V) = k. Therefore, by remark A.7 in [4] there exists a universal bundle V over X̃ × 𝔐^{(d,d)}(i, k). Denote by p_{i} the projection of the direct product to the i-th factor. Consider the following sheaf on 𝔐^{(d,d)}(i, k)

F = R_{p_{2}_{∗}}^{0} (V^{∗} ⊗ p_{1}^{∗}O_{E}(−d)) .

We see that for each bundle V ∈ 𝔐^{(d,d)}(i, k) the fiber of F over V is isomorphic to

Hom X̃(V, O_{E}(−d)) ≅ Hom_{E}(O_{E}(−d) ⊕ O_{E}(−d) ⊕ O_{E}(2d + i), O_{E}(−d)) ≅ ℂ^{2}.

Thus, F is the locally free sheaf (see [5], corollary 2 on p. 50), and ℙ(F) is the set of the pairs (V, α). This concludes the proof of the remark.

Let W_{l}^{(d,d)} be the subset of 𝔐̃^{(d,d)}(i, k) consisting of all pairs (V, α) such that the above defined V′_{α} belongs to 𝔐^{d,l}(i + 1, k − i − d). Then

𝔐̃^{(d,d)}(i, k) = ∐_{− \frac{d+i+1}{2} ≤l≤d−1} W_{l}^{(d,d)}

and the following statement takes place.

5.4. Remark. The set W_{l}^{(d,d)} is a locally trivial bundle over 𝔐^{(d,l)}(i + 1, k − i − d) with the fibers isomorphic to EP _{−d,2d+i}^{−d,d+l+i+1,−l} ≅ EP _{0,3d+i}^{0,2d+l+i+1,d−l}, i.e.,

e(W_{l}^{(d,d)}) = e_{0,3d+i}^{0,2d+l+i+1,d−1} · e(𝔐^{(d,l)}(i + 1, k − i − d)).

Proof. It follows from the previous proof that 𝔐̃^{(d,d)}(i, k) ≅ ℙ(F), where F is the vector bundle over 𝔐^{(d,d)}(i, k) parameterizing morphisms from V ∈ 𝔐^{(d,d)}(i, k) to O_{E}(−d). Denote by π : ℙ(F) ⟶ 𝔐^{(d,d)}(i, k) the natural projection, by q the projection from the direct product X̃ × ℙ(F) to the second factor, and by O_{F}(−1) the tautological line bundle over ℙ(F). Then on the direct product X̃ × ℙ(F) there exists an exact sequence of sheaves ([1])

0 ⟶ V′_{d} ⟶ (idX̃ × π)^{∗}V ⊗ q^{∗}(O_{F}(−1)) ⟶ (idX̃ × π)^{∗}p_{1}^{∗}O_{E}(−d) ⟶ 0,

which restriction to X̃ × (V, α) has the form

0 ⟶ V′_{α} ⟶ V ⟶^{α} O_{E}(−d) ⟶ 0.

If we restrict the sheaf V′_{d} to X̃ × W_{l}^{(d,d)}, we obtain a family of μ_{H_{n}}-stable bundles V′_{α} from 𝔐^{(d,l)}(i + 1, k − i − d). But the moduli space 𝔐(i + 1, k − i − d) is fine, i.e., there exists a universal bundle V′ over X̃ × 𝔐(i + 1, k − i − d) ([4]). Moreover, there exists a morphism f from the base of the family V′_{d} W_{l}^{(d,d)} to 𝔐(i + 1, k − i − d) such that V′_{d} = (idX̃ × f)^{∗}V′. It follows from the above construction that the image of f is 𝔐^{(d,l)}(i + 1, k − i − d) and its fiber over V′_{α} has the form

f^{−1}(V_{α}) = U(V′_{α}, O_{E}(−d) ⊕ O_{E}(2d + i))/Aut (O_{E}(−d) ⊕ O_{E}(2d + i)) = EP _{0,3d+i}^{0,2d+l+i+1,d−1}.

This completes the proof of the remark.

Combining remarks 5.3 and 5.4, we prove lemma 5.1.

5.5. Corollary.

e(𝔐^{(d,d)}(0, k)) = \frac{1}{2}e(𝔐^{(d,d−1)}(1, k − d)) for d ≥ 1;

e(𝔐^{(d,d)}(1, k)) = \{ \begin{matrix}e(𝔐^{(0,−1)}(2, k − d − 1)) for d = 0, \\ \frac{1}{2}e(𝔐^{(d,d−1)}(2, k − d − 1)) for d ≥ 1; \end{matrix}

e(𝔐^{(d,d)}(2, k)) = \frac{1}{2}e(𝔐^{(d+1,d)}(0, k − d + 1)) for d ≥ 0.

Proof. Let us calculate nonzero values of e_{0,3d+i}^{0,2d+l+i+1,d−l} for i = 0, 1, 2, and −\frac{d + i + 1}{2} ≤ l ≤ d − 1.

Since d − l ≤ 2d + l + i + 1, taking into account proposition 3.10 we get

e_{0,3d+i}^{0,2d+l+i+1,d−l} = \{ \begin{matrix}3, & if & 3d + i = 2d + l + i + 1 = d − l = 0; \\ 2, & if & 3d + i = 2d + l + i + 1 = d − l > 0; \\ 2, & if & 3d + i = 2d + l + i + 1 > 0, d − l = 0; \\ 1, & if & 3d + i = 2d + l + i + 1 > d − l > 0; \\ 1, & if & 2d + l + i + 1 > 3d + i = d − l; \\ 0, & & otherwise. \end{matrix}

It can be proved by the direct calculation that under our assumption about l, d, i we have,

e_{0,3d+i}^{0,2d+l+i+1,d−l} = \{ \begin{matrix}2, & when & d = 0, i = 1, l = −1; \\ 1, & when & d > \frac{1 − i}{3}, l = d − 1; \\ 0, & & otherwise. \end{matrix}

Therefore, the sum of lemma 5.1 has only one nonzero summand. To prove the corollary it remains to note that by the next lemma there exists an isomorphism of moduli spaces

𝔐^{(d_{1},d_{2})}(3, k) ≅ 𝔐^{(d_{1}+1,d_{2}+1)}(0, k + 3).

5.6. Lemma.

e(𝔐^{(d_{1},d_{2})}(3, k)) = e(𝔐^{(d_{1}+1,d_{2}+1)}(0, k + 3)).

Proof. It is clear, that the map τ : 𝔐(3, k) ⟶ 𝔐(0, k′), taking each bundle V to the tensor product V (E) = V ⊗ OX̃(E) is isomorphism. Recall, that c_{2}(V (E)) = c_{2}(V) + 3E^{2} + 2E · c_{1}(V) = c_{2}(V) + 3, i.e. k′ = k + 3. Moreover, if V ∈ 𝔐^{(d_{1},d_{2})}(3, k), then

V |_{E} = O_{E}(−d_{1}) ⊕ O_{E}(−d_{2}) ⊕ O_{E}(3 + d_{1} + d_{2}).

Therefore,

V (E)|_{E} = O_{E}(−d_{1} − 1) ⊕ O_{E}(−d_{2} − 1) ⊕ O_{E}(2 + d_{1} + d_{2})

This proves that the restriction of τ to the subset 𝔐^{(d_{1},d_{2})}(3, k) is isomorphism with the subset 𝔐^{(d_{1}+1,d_{2}+1)}(0, k + 3).

5.7. Lemma. If integers d_{1}, d_{2} satisfy the conditions: d_{1} ≥ 0, −\frac{d_{1} + i}{2} ≤ d_{2} < d_{1}, with i = 0, 1, 2, then

e(𝔐^{(d_{1},d_{2})}(i, k)) = ∑_{(m_{1},m_{2})∈D_{i}(d_{1},d_{2})} e_{m_{1}−d_{2},d_{1}+d_{2}+i+1+m_{1}}^{0,2m_{1}+m_{2}+i+1,m_{1}−m_{2}} · e(𝔐^{(m_{1},m_{2})}(i + 1, k − d_{1} − i)),

where

D_{i}(d_{1}, d_{2}) = \{(m_{1}, m_{2}) ∈ ℤ^{2} | (−\frac{i + 1}{3} ≤ m_{1} ≤ d_{1} − 1, ; max(d_{2}, −\frac{m_{1} + i + 1}{2}) ≤ m_{2} ; m_{2} ≤ min(m_{1}, d_{1} + d_{2} − m_{1} − 1) ) when d_{2} < −\frac{i + 1}{3} (d_{2} < m_{1} ≤ d_{1} − 1, ; max(d_{2}, −\frac{m_{1} + i + 1}{2}) ≤ m_{2} ; m_{2} ≤ min(m_{1}, d_{1} + d_{2} − m_{1} − 1) ) when d_{2} ≥ −\frac{i + 1}{3} \binom{m_{1} = d_{2},}{−\frac{m_{1} + i + 1}{2} ≤ m_{2} ≤ m_{1}} when d_{2} ≥ −\frac{i + 1}{3} \}

Proof. Let V be a bundle from 𝔐^{(d_{1},d_{2})}(i, k) with d_{1} > d_{2}, i.e.,

V |_{E} = O_{E}(−d_{1}) ⊕ O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i).

Hence there exists a unique up to a scalar multiplier epimorphism V ⟶ O_{E}(−d_{1}) ⟶ 0. Denoting by V′ its kernel we can construct the commutative diagram

\begin{aligned}& & 0 & & 0 \\ & & ↑ & & ↑ \\ 0 & ⟶ & O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i) & ⟶ & V |_{E} & ⟶ & O_{E}(−d_{1}) & ⟶ & 0 \\ & & β ↑ & & res ↑ & & id ↑ \\ 0 & ⟶ & V′ & ⟶ & V & ⟶ & O_{E}(−d_{1}) & ⟶ & 0 \\ & & ↑ & & ↑ \\ & & V (−E) & = & V (−E) \\ & & ↑ & & ↑ \\ & & 0 & & 0 \end{aligned}  (5.6)

Arguing as above, we see that V′ ∈ 𝔐(i + 1, k − i − d_{1}). Furthermore, there exists one-to-one correspondence between 𝔐^{(d_{1},d_{2})}(i, k) and the set of pairs

{(V′, β/mod Aut(O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i)))}.

Let us restrict the left column of diagram (5.6) to the curve E:

0 → Tor_{1}(O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i), O_{E}) → V (−E)|_{E} →

→ V′|_{E} ⟶^{β|_{E}} O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i) → 0.

Since

Tor_{1}(O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i), O_{E}) = O_{E}(1 − d_{2}) ⊕ O_{E}(1 + d_{1} + d_{2} + i),

(see claim 2.2), the restriction V′|_{E} is included in the exact sequence

0 ⟶ O_{E}(1 − d_{1}) ⟶^{γ} V′|_{E} ⟶^{β|_{E}} O_{E}(−d_{2}) ⊕ O_{E}(d_{1} + d_{2} + i) ⟶ 0.  (5.7)

Whence,

V′|_{E} = O_{E}(−m_{1}) ⊕ O_{E}(−m_{2}) ⊕ O_{E}(m_{1} + m_{2} + i + 1),

that is V′ belongs to 𝔐^{(m_{1},m_{2})}(i + 1, k − d_{1} − i). We regulate the direct summand of V′|_{E} in the following way: −m_{1} ≤ −m_{2} ≤ m_{1} + m_{2} + i + 1. This implies that

−\frac{i + 1}{3} ≤ m_{1}; −\frac{m_{1} + i + 1}{2} ≤ m_{2} ≤ m_{1}.

Besides, it follows from exact triple (5.7), that 1 − d_{1} ≤ −m_{1}, and β|_{E} is an epimorphism in two cases only:

either  − m_{1} < −d_{2},  −m_{2} ≤ −d_{2},  m_{1} + m_{2} + i + 1 ≤ d_{1} + d_{2} + i,
or  − m_{1} = −d_{2},  −d_{2} ≤ −m_{2} ≤ m_{1} + m_{2} + i + 1 ≤ d_{1} + d_{2} + i.

In other words, for i = 0, 1, 2, d_{1} ≥ 0 and −\frac{d_{1} + i}{2} ≤ d_{2} < d_{1} a pair (m_{1}, m_{2}) can be belonged to the set D_{i}(d_{1}, d_{2}) from the lemma.

If we denote by W_{(m_{1},m_{2})}^{(d_{1},d_{2})} the subset of 𝔐^{(d_{1},d_{2})}(i, k) consisting of all V with

V′|_{E} = O_{E}(−m_{1}) ⊕ O_{E}(−m_{2}) ⊕ O_{E}(m_{1} + m_{2} + i + 1),

then

𝔐^{(d_{1},d_{2})}(i, k) = ∐_{(m_{1},m_{2})∈D_{i}(d_{1},d_{2})} W_{(m_{1},m_{2})}^{(d_{1},d_{2})}.

It can be proved by the same way as in remark 5.4 that W_{(m_{1},m_{2})}^{(d_{1},d_{2})} is a locally trivial bundle over 𝔐^{(m_{1},m_{2})}(i + 1, k − d_{1} − i) with the fibers isomorphic to EP _{m_{1}−d_{2},d_{1}+d_{2}+i+m_{1}}^{0,2m_{1}+m_{2}+i+1,m_{1}−m_{2}}. This concludes the proof.

5.8. Corollary. Under the conditions of the previous lemma

e(𝔐^{(d_{1},d_{2})}(0, k)) = \{ \begin{matrix}2e(𝔐^{(d_{1}−1,d_{2})}(1, k − d_{1})) & for & d_{1} ≥ 1, d_{2} = d_{1} − 1, \\ e(𝔐^{(d_{1}−1,d_{2})}(1, k − d_{1})) & for & d_{1} ≥ 2, d_{1} − 1 > d_{2} ≥ −\frac{d_{1}}{2} ; \end{matrix}

e(𝔐^{(d_{1},d_{2})}(1, k)) = \{ \begin{matrix}2e(𝔐^{(d_{1}−1,d_{2})}(2, k − d_{1} − 1)) & for & d_{1} ≥ 1, d_{2} = d_{1} − 1, \\ e(𝔐^{(d_{1}−1,d_{2})}(2, k − d_{1} − 1)) & for & d_{1} ≥ 1, d_{1} − 1 > d_{2} ≥ −\frac{d_{1} + 1}{2}; \end{matrix}

e(𝔐^{(d_{1},d_{2})}(2, k)) = \{ \begin{matrix}3e(𝔐^{(0,0)}(0, k + 1)) & for & d_{1} = 0, d_{2} = −1, \\ 2e(𝔐^{(d_{1},d_{2}+1)}(0, k − d_{1} + 1)) & for & d_{1} ≥ 1, d_{2} = d_{1} − 1, \\ e(𝔐^{(d_{1},d_{2}+1)}(0, k − d_{1} − 1)) & for & d_{1} ≥ 1, d_{1} − 1 > d_{2} ≥ −\frac{d_{1} + 2}{2}; \end{matrix}

Proof. Since a pair (m_{1}, m_{2}) ∈ D_{i}(d_{1}, d_{2}) satisfies the conditions m_{1}−m_{2} ≤ 2m_{1}+m_{2}+i+1 and m_{1} − d_{2} ≤ d_{1} + d_{2} + i + m_{1}, using proposition 3.10, we get e_{m_{1}−d_{2},d_{1}+d_{2}+i+m_{1}}^{0,2m_{1}+m_{2}+i+1,m_{1}−m_{2}} is nonzero iff either m_{1} − d_{2} = 0 or m_{1} − d_{2} = m_{1} − m_{2} > 0.

Suppose that m_{1} = d_{2}, then

e_{m_{1}−d_{2},d_{1}+d_{2}+i+m_{1}}^{0,2m_{1}+m_{2}+i+1,m_{1}−m_{2}} =

= e_{0,d_{1}+2d_{2}+i}^{0,2d_{2}+m_{2}+i+1,d_{2}−m_{2}} = \{ \begin{matrix}3, & for & d_{1} + 2d_{2} + i = 2d_{2} + m_{2} + i + 1 = d_{2} − m_{2} = 0, \\ 2, & for & d_{1} + 2d_{2} + i = 2d_{2} + m_{2} + i + 1 = d_{2} − m_{2} > 0, \\ 2, & for & d_{1} + 2d_{2} + i = 2d_{2} + m_{2} + i + 1 > 0, d_{2} − m_{2} = 0, \\ 1, & for & d_{1} + 2d_{2} + i = 2d_{2} + m_{2} + i + 1 > d_{2} − m_{2} > 0, \\ 1, & for & 2d_{2} + m_{2} + i + 1 > d_{1} + 2d_{2} + i = d_{2} − m_{2}, \\ 0, & & otherwise. \end{matrix}

In the case m_{1} = d_{2} integers m_{2} and d_{2} satisfy the conditions: d_{1} − 1 ≥ d_{2} ≥ −\frac{i + 1}{3},

d_{2} ≥ m_{2} ≥ −\frac{d_{2} + i + 1}{2}. This gives,

e_{0,d_{1}+2d_{2}+i}^{0,2d_{2}+m_{2}+i+1,d_{2}−m_{2}} = \{ \begin{matrix}3, & when & i = 2, d_{1} = 0, m_{2} = d_{2} = −1, \\ 2, & when & m_{2} = d_{2} = d_{1} − 1, d_{1} > \frac{2 − i}{3}, \\ 0, & & otherwise. \end{matrix}

Suppose that m_{1} − d_{2} = m_{1} − m_{2} > 0, i.e., m_{2} = d_{2}, m_{1} > m_{2}, then

e_{m_{1}−d_{2},d_{1}+d_{2}+i+m_{1}}^{0,2m_{1}+m_{2}+i+1,m_{1}−m_{2}} =

= e_{m_{1}−d_{2},d_{1}+d_{2}+i+m_{1}}^{0,2m_{1}+d_{2}+i+1,m_{1}−d_{2}} = \{ \begin{matrix}1, & for & m_{1} = d_{1} − 1 > m_{2} = d_{2}, \\ 0, & & otherwise. \end{matrix}

Thus we see that the sum of lemma 5.7 has only one nonzero summand, that is equal to

m_{i}(d_{1}, d_{2})e(𝔐^{(d_{1}−1,d_{2})}(i + 1, k − d_{1} − i)),

where

m_{i}(d_{1}, d_{2}) = \{ \begin{matrix}3 & when & i = 2, d_{1} = 0, d_{2} = −1, \\ 2, & when & d_{1} ≥ 1, d_{2} = d_{1} − 1, \\ 1, & & otherwise. \end{matrix}

This with lemma 5.6 completes the proof.

6. Deduction of blow up formulae.

Combining corollaries 5.5 and 5.8 together we obtain.

For d_{1} ≥ 1, d_{1} ≥ d_{2} ≥ −\frac{d_{1}}{2} e(𝔐^{(d_{1},d_{2})}(0, k)) = m_{0}(d_{1}, d_{2})e(𝔐^{(d_{3},d_{4})}(1, k − d_{1})), where

(d_{3}, d_{4}) = \{ \begin{matrix}(d_{1} − 1, d_{2}) & if & d_{2} < d_{1}, \\ (d_{1}, d_{2} − 1) & if & d_{2} = d_{1}; \end{matrix} m_{0}(d_{1}, d_{2}) = \{ \begin{matrix}2 & if & d_{2} = d_{1} − 1, \\ 1 & if & d_{2} < d_{1} − 1 \\ \frac{1}{2} & if & d_{2} = d_{1}. \end{matrix}

For d_{3} ≥ 0, d_{3} ≥ d_{4} ≥ −\frac{d_{3} + 1}{2}

e(𝔐^{(d_{3},d_{4})}(1, k − d_{1})) = m_{1}(d_{3}, d_{4})e(𝔐^{(d_{5},d_{6})}(2, k − d_{1} − d_{3} − 1)), where

(d_{5}, d_{6}) = \{ \begin{matrix}(d_{3} − 1, d_{4}) & if & d_{4} < d_{3}, \\ (d_{3}, d_{4} − 1) & if & d_{4} = d_{3}; \end{matrix} m_{1}(d_{3}, d_{4}) = \{ \begin{matrix}2 & if & d_{4} = d_{3} − 1, \\ 1 & if & d_{4} < d_{3} − 1, \\ 1 & if & d_{4} = d_{3} = 0, \\ \frac{1}{2} & if & d_{4} = d_{3} > 0. \end{matrix}

For d_{5} ≥ 0, d_{5} ≥ d_{6} ≥ −\frac{d_{5} + 2}{2}

e(𝔐^{(d_{5},d_{6})}(2, k − d_{1} − d_{3} − 1)) = m_{2}(d_{5}, d_{6})e(𝔐^{(d_{7},d_{8})}(0, k − d_{1} − d_{3} − d_{5})), where

(d_{7}, d_{8}) = \{ \begin{matrix}(d_{5}, d_{6} + 1) & if & d_{6} < d_{5}, \\ (d_{5} + 1, d_{6}) & if & d_{6} = d_{5}; \end{matrix} m_{2}(d_{5}, d_{6}) = \{ \begin{matrix}3 & if & d_{6} = d_{5} − 1 = −1, \\ 2 & if & d_{6} = d_{5} − 1 ≥ 0, \\ 1 & if & d_{6} < d_{5} − 1, \\ \frac{1}{2} & if & d_{6} = d_{5}. \end{matrix}

Therefore,

e(𝔐(0, k)) = ∑_{{d_{1}≥0 \atop d_{1}≥d_{2}≥− \frac{d_{1}}{2}}} e(𝔐^{(d_{1},d_{2})}(0, k)) =

= ∑_{{d_{1}≥0 \atop d_{1}≥d_{2}≥− \frac{d_{1}}{2}}} m_{0}(d_{1}, d_{2})m_{1}(d_{3}, d_{4})m_{2}(d_{5}, d_{6})e(𝔐^{(d_{7},d_{8})}(0, k − d_{1} − d_{3} − d_{5}))

Note that each pair (d_{1}, d_{2}) suah that d_{1} ≥ 0, d_{1} ≥ d_{2} ≥ −\frac{d_{1}}{2} determains a finite sequence of integers

s(d_{1}, d_{2}) = d_{1}, d_{2}, d_{3}, d_{4}, d_{5}, d_{6}, …, d_{6k−1}, d_{6k}, 0, 0  (6.1)

by the following rule: if m ≡ 0 (mod 3), then

(d_{2m+1}, d_{2m+2}) = \{ \begin{matrix}(d_{2m−1}, d_{2m} + 1) & when & d_{2m−1} > d_{2m}, \\ (d_{2m−1} + 1, d_{2m}) & when & d_{2m−1} = d_{2m}; \end{matrix}

if m ≢ 0 (mod 3), then

(d_{2m+1}, d_{2m+2}) = \{ \begin{matrix}(d_{2m−1} − 1, d_{2m}) & when & d_{2m−1} > d_{2m}, \\ (d_{2m−1}, d_{2m} − 1) & when & d_{2m−1} = d_{2m}. \end{matrix}

Let us define some functions by means of sequence s(d_{1}, d_{2}). S_{0}(d_{1}, d_{2}) = d_{1} + d_{3} + d_{5} + ⋯ + d_{6k−1}, P_{0}(d_{1}, d_{2}) = m_{0}(d_{1}, d_{2})m_{1}(d_{3}, d_{4}) ⋯ m_{2}(d_{6k−1}, d_{6k}), S_{1}(d_{3}, d_{4}) = d_{3} + d_{5} + d_{7} + ⋯ + d_{6k−1}, P_{1}(d_{3}, d_{4}) = m_{1}(d_{3}, d_{4})m_{2}(d_{5}, d_{6}) ⋯ m_{2}(d_{6k−1}, d_{6k}), S_{2}(d_{5}, d_{6}) = d_{5} + d_{7} + d_{9} + ⋯ + d_{6k−1}, P_{2}(d_{5}, d_{6}) = m_{2}(d_{5}, d_{6})m_{0}(d_{7}, d_{8}) ⋯ m_{2}(d_{6k−1}, d_{6k}), where the number k is determined by formula (6.1). In this notation we have

e(𝔐(0, k)) = ∑_{{d_{1}≥0 \atop d_{1}≥d_{2}≥− \frac{d_{1}}{2}}} P_{0}(d_{1}, d_{2})e(𝔐^{(0,0)}(0, k − S_{0}(d_{1}, d_{2}))).

e(𝔐(1, k)) = ∑_{{d_{3}≥0 \atop d_{3}≥d_{4}≥− \frac{d_{3}+1}{2}}} P_{1}(d_{3}, d_{4})e(𝔐^{(0,0)}(0, k − S_{1}(d_{3}, d_{4}))).

e(𝔐(2, k)) = ∑_{{d_{5}≥0 \atop d_{5}≥d_{6}≥− \frac{d_{5}+2}{2}}} P_{2}(d_{5}, d_{6})e(𝔐^{(0,0)}(0, k − S_{2}(d_{5}, d_{6}))).

It is easy to see that

e(𝔐(0, k)) = ∑_{n≥0} ( ∑_{{d_{1}≥0 \atop d_{1}≥d_{2}≥− \frac{d_{1}}{2}} ; S_{0}(d_{1},d_{2})=n} P_{0}(d_{1}, d_{2})) e(𝔐^{(0,0)}(0, k − n)).

e(𝔐(1, k)) = ∑_{n≥0} ( ∑_{{d_{3}≥0 \atop d_{3}≥d_{4}≥− \frac{d_{3}+1}{2}} ; S_{1}(d_{3},d_{4})=n} P_{1}(d_{3}, d_{4})) e(𝔐^{(0,0)}(0, k − n)).

e(𝔐(2, k)) = ∑_{n≥0} ( ∑_{{d_{5}≥0 \atop d_{5}≥d_{6}≥− \frac{d_{5}+2}{2}} ; S_{2}(d_{5},d_{6})=n} P_{2}(d_{5}, d_{6})) e(𝔐^{(0,0)}(0, k − n)).

Recall that 𝔐^{(0,0)}(0, k) ≅ 𝔐(k) is the moduli space of rank 3 μ_{H}-stable vector bundles on the surface X with the first Chern class equals C ∈ Pic(X) and the second one is k. If we denote by Z_{3,i}(q) and Z_{3}(q) the generating functions

Z_{3,i}(q) = ∑_{k∈ℤ} e(𝔐(i, k))q^{k}, Z_{3}(q) = ∑_{k∈ℤ} e(𝔐(k))q^{k},

assuming that e(𝔐(i, k)) = 0 whenever 𝔐(i, k) is empty, we obtain the following theorem.

6.1. Theorem. There exist universal functions u_{3,i}(q) = ∑_{n≥0} u_{3,i,n}q^{n} such that

Z_{3,i}(q) = u_{3,i}(q) · Z_{3}(q),

where i = 0, 1, 2 and

u_{3,i,n} = ∑_{{d≥0 \atop \frac{3d+i}{2} ≥k≥0} ; S_{i}(d,d−k)=n} P_{i}(d, d − k).

Now we calculate the universal functions u_{3,i}(q). At first we shall find values of the functions P_{0}(d, d − k) and S_{0}(d, d − k). It is clear, that S_{0}(0, 0) = 0 and P_{0}(0, 0) = 1.

6.2. Lemma. For integers d ≥ 1, \frac{3d}{2} ≥ k ≥ 0 the following formulae

S_{0}(d, d − k) = 3d^{2} − 3dk + k^{2}, P_{0}(d, d − k) = \{ \begin{matrix}3, & when k = 0 or k = \frac{3d}{2}, \\ 6, & otherwise \end{matrix}

hold.

Proof. The proof is by induction on d and k. We start with the case k = 0. Using formulae (6.1) and the definition of the functions P_{0}, S_{0}, m_{i}(d_{1}, d_{2}), we have

s(d, d) = d, d, d, d − 1, d − 1, d − 1, d, d − 1, d − 1, d − 1, d − 1, d − 2, d − 1, d − 1, …,

therefore, S_{0}(d, d) = 6d − 3 + S_{0}(d − 1, d − 1) = 6d − 3 + 3(d − 1)^{2} = 3d^{2}.

P_{0}(d, d) = \{ \begin{matrix}3, & if d = 1, \\ P_{0}(d − 1, d − 1), & otherwise \end{matrix} \} = 3.

If k = 1, then

s(d, d − 1) = d, d − 1, d − 1, d − 1, d − 1, d − 2, d − 1, d − 1, …,

hence, S_{0}(d, d − 1) = 3d − 2 + S(d − 1, d − 1) = 3d − 2 + 3(d − 1)^{2} = 3d^{2} − 3d + 1.

P_{0}(d, d − 1) = \{ \begin{matrix}6, & if d = 1, \\ 2P_{0}(d − 1, d − 1), & otherwise \end{matrix} \} = 6.

In the case k = 2,

s(d, d − 2) = d, d − 2, d − 1, d − 2, d − 2, d − 2, d − 1, d − 2, …,

whence, S_{0}(d, d−2) = 3d−3+S_{0}(d−1, d−1−1) = 3d−3+3(d−1)^{2}−3(d−1)+1 = 3d^{2}−3·2d+2^{2}. P_{0}(d, d − 2) = P_{0}(d − 1, d − 1 − 1) = 6.

Finally suppose, k ≥ 3, then

s(d, d − k) = d, d − k, d − 1, d − k, d − 2, d − k, d − 2, d − k + 1, …, .

It follows that S_{0}(d, d − k) = 3d − 3 + S_{0}(d − 2, d − 2 − (k − 3)). By the induction hypothesis,

S_{0}(d − 2, d − 2 − (k − 3)) = 3(d − 2)^{2} − 3(d − 2)(k − 3) + (k − 3)^{2},

i.e., S_{0}(d, d − k) = 3d^{2} − 3kd + k^{2}. Furthermore,

P_{0}(d, d − k) = \{ \begin{matrix}3, & if d = 2, k = 3, \\ 2P_{0}(d − 2, d − 2), & if d > 2, k = 3, \\ P_{0}(d − 2, d − 2 − (k − 3)), & otherwise \end{matrix} \} = \{ \begin{matrix}3, & if k = \frac{3d}{2}, \\ 6, & otherwise. \end{matrix}

This completes the proof.

By the similar calculations we get 6.3. Lemma. For integers d_{3} ≥ 0, d_{3} ≥ d_{4} ≥ −\frac{d_{3} + 1}{2} the following formulae take place

S_{1}(d_{3}, d_{4}) = d_{3}^{2} + d_{4}^{2} + d_{3}d_{4} + d_{3} + d_{4}, P_{1}(d_{3}, d_{4}) = \{ \begin{matrix}3 & when & d_{4} = d_{3}, or d_{4} = −\frac{d_{3} + 1}{2}, \\ 6 & & otherwise. \end{matrix}

If integers d_{5}, d_{6} satisfy the conditions d_{5} ≥ 0, d_{5} ≥ d_{6} ≥ −\frac{d_{5} + 2}{2}, then the functions S_{2}(d_{5}, d_{6}) and P_{2}(d_{5}, d_{6}) have the form

S_{2}(d_{5}, d_{6}) = d_{5}^{2}+d_{6}^{2}+d_{5}d_{6}+2(d_{5}+d_{6})+1, P_{2}(d_{5}, d_{6}) = \{ \begin{matrix}3 & when & d_{6} = d_{5}, or d_{6} = −\frac{d_{5} + 2}{2}, \\ 6 & & otherwise. \end{matrix}

6.4. Corollary. Under the assumption of theorem 6.1 we have,

u_{3,0}(q) = ∑_{(x,y)∈ℤ^{2}} q^{x^{2}+xy+y^{2}}.

u_{3,1}(q) = ∑_{(x,y)∈ℤ^{2}} q^{x^{2}+xy+y^{2}+x+y}.

u_{3,2}(q) = ∑_{(x,y)∈ℤ^{2}} q^{x^{2}+xy+y^{2}+2(x+y)+1}.

Proof. We prove the first equality of the corollary only, because of other are proved in the similar way. Taking into account the theorem and lemma 6.2, we obtain,

u_{3,0}(q) = 1 + ∑_{{d≥1 \atop \frac{3d}{2} ≥k≥0} ; S_{0}(d,d−k)=n} P_{0}(d, d − k) = 1 + ∑_{d≥1} (3q^{S_{0}(d,d)} + 3q^{S_{0}(d,− \frac{d}{2} )} + 6 ∑_{\frac{3d}{2} >k>0} q^{S_{0}(d,d−k)}) .

Note that,

∑_{d≥1} q^{S(d,− \frac{d}{2} )} = ∑_{s≥1} q^{S_{0}(2s,2s−3s)} = ∑_{s≥1} q^{3s^{2}}.

Therefore,

u_{3,0}(q) = 1 + 6 ∑_{{d≥1 \atop \frac{3d}{2} >k≥0}} q^{3d^{2}−3dk+k^{2}}.

If c = d − k, then

u_{0}(q) = 1 + 6 ( ∑_{{d≥1 \atop d≥c>− \frac{d}{2}}} q^{d^{2}+dc+c^{2}} ) .

The quadratic form Q(d, c) = d^{2} + dc + c^{2} has the following symmetries

t_{0}(d, c) = (d, c),  t_{1}(d, c) = (−d, −c),  t_{2}(d, c) = (c, d),
t_{3}(d, c) = (−c, −d),  t_{4}(d, c) = (−d, d + c),  t_{5}(d, c) = (d, −d − c).

In other words, we have

Q(d, c) = Q(c, d) = Q(−c, −d) = Q(−d, d + c) = Q(d, −d − c) = Q(−d, −c).

Moreover,

⋃_{i=0}^{6} t_{i}(A) = ℤ^{2} \ ℤ · (2, −1), (t_{2}(A) ∐ t_{3}(A)) ∩ (t_{0}(A) ∐ t_{1}(A)) = ℤ · (1, 1),

and other intersections are empty. Thus, for each vector v_{0} = (d, c) ∈ A with d > c there exist 5 different vectors v_{i} such that Q(v_{0}) = Q(v_{i}), (i = 1, …, 6). In the case v_{0} = (d, c) ∈ A with d = c with help of the symmetries we obtain only 3 different vectors: v_{1} = (−d, −d), v_{2} = (−d, 2d), v_{3} = (d, −2d) such that Q(v_{0}) = Q(v_{i}) (i = 1, 2, 3). On the other hand, Q(d, d) = Q(−2d, d) = Q(2d, −d) and the last two vectors belong to the line ℤ · (2, −1). This proves the corollary.

It is known to experts that

u_{3,0}(q) = 1 + 6 ∑_{n≥1} (∑_{d|3} ( \frac{d}{3} )) q^{n}

is Eisenstain series, where ( \frac{d}{3} ) is Legendre symbol:

( \frac{d}{3} ) = \{ \begin{matrix}+1 & if & d ≡ +1 (mod 3) \\ −1 & if & d ≡ −1 (mod 3) \\ 0 & if & d ≡ 0 (mod 3) \end{matrix}

In other words, u_{3,0}(q) is a modular form of weight 1.

References.

[1] W. Fulton, Intersection theory. Springer-Verlag (1983).

[2] S. Kuleshov, Exceptional and rigid sheaves on surfaces with anticanonical class without base components. MPI 1995 – 13.

[3] Wei-Ping Li and Zhenbo Qin, On blowup formulae for the S-duality conjecture of Vafa and Witten. math.AG/9805055

[4] S. Mukai, On the moduli spaces of bundles on K3 surfaces, I, Vector bundles on algebraic varieties (1987), Oxford university press, Oxford.

[5] D. Mumford, Abelian Varieties, Oxford university press, Tata Inst. of Fundamental Research, 1970.

[6] K. Okonek, M. Schneider, H. Spindler, Vector bundles on complex projective spaces. Birkhäuser (1984).

[7] C. Vafa, E. Witten, A strong coupling test of S-duality, Nucl. Phys. B 431 (1995), 3-77.

E-mail address: kuleshov@kuleshov.mccme.rssi.ru