arXiv:2107.08375v1 [math.NT] 18 Jul 2021

Automorphic forms for PGL(3) over elliptic function fields

Part 1: Graphs of Hecke operators

Roberto Alvarenga, Oliver Lorscheid, and Valdir Pereira Júnior

Abstract. This is a first part of a series of papers in which we develop explicit compu-tational methods for automorphic forms for GL3 and PGL3 over elliptic function fields.In this first part, we determine explicit formulas for the action of the Hecke operatorson automorphic forms on GL2 and GL3 in terms of their graphs. Our primary resultconsists in a complete description of the graphs of degree 1 Hecke operators for GL3.As complementary results, we describe the ‘even component’ of the graphs of degree 2Hecke operators for GL2 and the ‘neighborhood of the identity’ of the graphs of degree2 Hecke operators for GL3. In addition, we establish two dualities for Hecke operatorsfor GLn and PGLn, which hold for all n, all degrees and all function fields.

Dedicated to Gunther Cornelissen on the occasion of his 50th birthday.

Contents

Introduction 11. Graphs of Hecke operators 102. Duality 133. Hecke operators for elliptic curves 174. Graphs of rank 3 and degree 1 235. Proof of the main theorem 336. Degree 2 graphs 57References 71

Introduction

This is the first part of a series of papers, in which we aim for an explicit understandingof the vanishing of periods of automorphic forms for PGL3 over an elliptic functionfield. Before we explain the results of the present paper, we outline the overall scope ofthis series of papers and put it into context with the literature.

Periodical automorphic forms for PGL2. Period integrals for PGL2 are well-under-stood and, in essence, vanishing periods yield (conjecturally) tempered representations.Let us survey this in more detail.

We fix a global field F as our base field and let A be its adele ring. Let H be analgebraic subgroup of PGL2 and f : PGL2(F)\PGL2(A)→C an automorphic form. Wecall f H-periodical if its H-period

ΩH( f )(g) =∫

H(F)\H(A)

f (hg)dh

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2 Roberto Alvarenga, Oliver Lorscheid and Valdir Pereira Júnior

vanishes for all g ∈ PGL2(A). We say that an automorphic representation π (which wealways assume to be irreducible) is H-periodical if all of its elements are H-periodical.Note that this condition is invariant under conjugation of H.

Since PGL2 has dimension 3, the only interesting periods appear for 1 or 2-dimen-sional H, which are of the following types (up to conjugation): the Borel subgroup Band its unipotent radical N, the diagonal torus T , for every separable quadratic fieldextension E of F a non-split torus TE and, in characteristic 2, the Weil-restrictions ofGm,E for non-separable quadratic extensions. Note that B is of secondary interest as thesemidirect product of N with T . The following interpretations of period integrals areknown.

N-periods. N-periodical automorphic forms are, by definition, cusp forms. As a by-product of Drinfeld’s proof of the Langlands conjectures for GL2 over function fields(cf. [Dri83], [Dri88]), N-periodical, or cuspidal, representations are tempered in thefunction field case. The Ramanujan-Petersson conjecture claims the analogous propertyfor number fields.

TE-periods. TE-periods are linked to special values of L-functions, and TE-periodicalautomorphic forms are called E-toroidal. In the case of a cuspidal representation π overa number field and f in π, Waldspurger shows in [Wal85] that∣∣∣∣ ∫

TE(F)\TE(A)

f (tg)dt∣∣∣∣2 = c f ,TE (g) ·L(π, 1

2) ·L(π⊗χE ,12)

for some factor c f ,TE (g) that is nonzero as a function in g, where L(π,s) is the L-series ofπ and χE is the quadratic Hecke character associated with E/F . The analogous formulafor function fields is established by Chuang and Wei in [CW19]; also cf. [Lys08, Lys20].

In the case of an Eisenstein series E(g,χ), where g ∈ PGL2(A) and χ is a Heckecharacter, Zagier shows in [Zag81] that∫

TE(F)\TE(A)

E(tg,χ)dt = cχ,TE (g) ·L(χ, 12) ·L(χχE ,

12)

for some factor cχ,TE (g) that is nonzero as a function in g, where L(χ,s) and L(χχE ,s)are Hecke L-series. In the function field case, the Hasse-Weil theorem (cf. [Wei49])asserts that the real part of χ is contained in [−2γ

√q, 2γ

√q] if E(g,χ) is TE-periodical,

where γ is the genus of F . This condition is equivalent with the temperedness of theautomorphic representation generated by E(g,χ). The analogous claim for numberfields is equivalent to the generalized Riemann hypothesis.

T -periods. Waldspurger’s formula holds also for T -periods: the squared absolute valueof the T -period of a cusp form f in π is a non-zero multiple of L(π, 1

2)2. The T -

period of an Eisenstein series diverges and therefore does not make sense if takenliterally. Cornelissen and the second author show in [CL12] and [Lor13b] that a suitablerenormalization leads to a link with L-series, namely∫T (F)\T (A)

E(tg,χ)− 12

[ΩN(E( · ,χ))(tg)+ΩNt (E( · ,χ))(tg)

]dt = cχ,T (g) ·L(χ, 1

2)2

Automorphic forms for PGL(3) over elliptic function fields - Part 1: Graphs of Hecke operators 3

for some factor cχ,T (g) that is nonzero as a function in g, where Nt is the group of lowertriangular unipotent matrices. Thus we get the same relation as for TE between the tem-peredness for T -periodical Eisenstein series and the generalized Riemann hypothesis.

Generalizations to higher rank groups. Periodical automorphic forms for higherrank groups are far less understood than for PGL2 and an active area of research. We donot attempt to give an overview of this rich field, but we would like to mention a fewnotable results:

• Lafforgue generalizes Drinfeld’s proof of the Langlands conjecture and thetemperedness of cuspidal representations to PGLn in the function field case (cf.[Laf97], [Laf02]).• GLn×GLn′-periods of cusp forms for GLn+n′ (for n′ = n or n+1) are intensely

studied (e.g., see [FMW18]).• Zagier’s computation of TE-periods of Eisenstein series is generalized by

Wielonsky in [Wie85] to a formula for PGLn, which links a certain 1-parameterfamily of residual Eisenstein series with an L-function of a degree n extensionE of F .• Beineke and Bump ([BB03]; also cf. [BB02, BG84, Woo94]) investigate certain

renormalizations of Eisenstein series for PGL3 and the relation between T -periods and L-series. The situation is, however, much more complicated forhigher rank than for PGL2.• Periods of automorphic forms appear in the context of the relative trace formula.

Namely, a comparision of relative trace formulas for two pairs of algebraicgroups yields typically a relation between associated period integrals, and thenon-vanishing of such periods is typically related to the lifting of automorphicrepresentations. As first references, we guide the reader to [Jac05] and [Lap06].

In particular, the understanding of periodical automorphic forms for PGL3 is far fromcomplete. It is largely open which periods are linked to tempered representations. Inthis paper and its sequels, we investigate this question in the case of elliptic functionfields in terms of explicit parametrizations of automorphic forms.

Automorphic representations for PGL3. The theory of automorphic forms for PGL3is richer than that for PGL2, which features only cusp forms and a 1-parameter familyof Eisenstein series. The reason is that PGL3 has more parabolic subgroups that lead toinduced representation: the whole group PGL3, two maximal parabolic subgroups oftypes (2,1) and (1,2) (which are connected by duality) and the Borel subgroup. Thenormalized induction of cuspidal representations from a Levi subgroup of a parabolicsubgroup to PGL3 yields the following types of automorphic forms: cusp form (stem-ming from PGL3), a 1-parameter family of ‘parabolic’ Eisenstein series (stemmingfrom either of the maximal parabolic subgroup) and a 2-parameter family of ‘Borel’Eisenstein series; for more details, we refer to the sequel paper, as well as to [Lau97,App. G] and [PJ20, sections 1.6, 1.7] for general background.

In fact, Langlands shows in [BC79, pp. 203–208] that every automorphic representa-tion is a subquotient of a representation that is induced from a cuspidal representationon a Levi subgroup; also cf. Franke ([Fra98]), and Moeglin-Waldspurger ([MW95]) forthe function field case.


Automorphic forms in the function field case. In the function field case, the domainof an automorphic form with fixed ramification is discrete modulo the right-action ofa compact open subgroup K, which allows for explicit descriptions of automorphicforms. Even though a large part of our methods extends to ramified automorphic forms,we restrict ourselves to unramified automorphic forms for the sake of limiting thecomplexity.

Unramified automorphic forms for function fields behave particularly nice: everyunramified automorphic representation contains a unique unramified automorphic form,or spherical vector, up to scalar multiples. Therefore the unramified automorphic repre-sentations correspond to certain 1-dimensional representations of the (commutative)spherical Hecke algebra HK . In other words, unramified automorphic representationscorrespond to eigenforms of HK and are therefore determined by their eigenvaluesunder the action of Hecke operators.

Geometric Langlands: Weil’s theorem and Hecke operators. The geometric Lang-lands program (e.g., cf. [Fre20, FGV02, Gai15, Lau87]) is rooted in the followingobservation that is usually attributed to Weil (e.g., cf. [Fre04, Lemma 3.1]): the domainof unramified automorphic forms, modulo the right-action of the standard maximalcompact subgroup K of PGLn(A), is naturally identified with the set PBunn X of isomor-phism classes of Pn−1-bundles1 on the smooth, projective and geometrically irreduciblecurve X with function field F .

Let |X | be the set of closed points of X . Satake shows in [Sat63] that the sphericalHecke algebra HK of PGLn(A) is freely generated over C by the pairwise commutingHecke operators Φx,r = Φ

(n)x,r where x ∈ |X | and r = 1, . . . ,n−1. The Hecke operator

Φx,r acts on functions f : PBunn X → C in terms of the formula

(Φx,r. f )(E) = ∑E/F=K⊕r

x

f (F)

where E is a rank n bundle over X representing the class E ∈ PBunn X , where Kx is thesimple torsion sheaf with support at x and where F ranges over all subsheaves of E withquotient K⊕r

x .

Graphs of Hecke operators. We can subsume the information of how Φx,r acts on

automorphic forms in an edge weighted graph G(n)x,r with vertex set PBunn X as follows.

We define the weights mx,r(E,F) of G(n)x,r as the number of subsheaves F′ of E with

quotient K⊕rx such that F′ = F. Then G

(n)x,r has an edge from E to F of weight mx,r(E,F)

if the weight is nonzero. These graphs were introduced by the first and second author in[Alv19] and [Lor13a], respectively.

We draw edges together with their weights m = mx,r(E,F) as

E F

m , or asE F

m m′

1Note that an isomorphism class of a Pn−1-bundle can be identified with an PicX-orbit of isomorphismclasses of rank n bundles on X ; cf. [Har77, Ex. 7.10].


Ny1,2

Nyq,2

Mx,1

Mx,0

O⊕O

O⊕Lx O⊕L2x O⊕L3x

... · · ·q+1

1

q+1

1

1

q 1

q−1

q+1

11 q 1 q

Figure 1. The graph G(2)x,1 for X = Eq with unique degree 1 place x

if m′ = mx,r(F,E) is also nonzero, where we label the nodes with representative rank nbundles (without bars) for better readability. Note that the latter drawing conventionapplies in particular in the case 2r = n due to the first duality theorem (Theorem 2.3and Corollary 2.5):

Theorem A (first duality for Hecke operators). Let E and F be rank n bundles on X.Then mx,r(E,F) 6= 0 if and only if mx,n−r(F,E) 6= 0.

The virtue of the graph G x,r is that we can read off the Φx,r-eigenvalue equations ofan automorphic eigenform f : PBunn X → C as

λx,r · f (E) = ∑

E F

mx,r(E,F) · f (F)

where λx,r is the eigenvalue of f for Φx,r and F ranges over all neighbors of E in G x,r.

An example. The theory of unramified automorphic forms for rational function fieldsis not very exciting from a certain point of view: there are no unramified cusp forms andall vector bundles decompose into a sum of line bundles. Elliptic function fields are ofa more intriguing nature. For example, let x0 be chosen base point of the elliptic curveX , x a degree one place and D = x+(d−1)x0 for d ∈ 0,1. Then there is a nontrivialextension of the twisted line bundle LD = O(D) by the structure sheaf O= OX and thisdefines an indecomposable rank 2 bundle Mx,d . Another example of an indecomposablerank 2 bundle is the trace Ny,2 of the line bundle OX2(y) over constant field extensionX2 = X⊗Fq Fq2 where Fq is the constant field of X and y is a place of X2 that lies overthe degree 2 place y of X .

If X = Eq is an elliptic curve over Fq with a unique degree 1 place x = x0 (there isa unique such curve for q = 2,3,4), then X has q degree 2 places y1, . . . ,yq and the

graph G(2)x,1 of Φx,1 for PGL2 is as illustrated in Figure 1; cf. [CL09], [Lor12], [Ser03]

or [Tak93].Given an eigenform f : PBun2 X → C with eigenvalue λ= λx,1 for Φx,1, the graph

determines the eigenvalue equations for f , such as

λ f (O⊕O) = (q+1) f (O⊕Lx), λ f (Mx,0) = q f (Mx,1)+ f (O⊕Lx),

λ f (Nyi) = (q+1) f (Mx,1), λ f (Mx,1) = f (Mx,0)+q

∑j=1

f (Ny j),


for i = 1, . . . ,q. Note that this system is underdetermined since its q+ 3 equationsinvolve q+4 values of f . Indeed, we know that every eigenvalue λ occurs for someEisenstein series.

The vanishing of a period imposes an additional condition, which typically limitsλ to a finite number of values if we aim for a (nonzero) eigenform f in the solutionspace. Since f lies in a tempered representation if and only if λ ∈ [−2

√q, 2√

q] (cf.[Lor13b, Lemma 9.3]), this yields a method to verify the temperedness for H-periodicaleigenforms, which we inspect in more detail for the following two types of periods.

Cusp forms. Let N be the unipotent radical of the Borel subgroup of PGL2. Then anunramified cusp form f satisfies

0 = ΩN( f )(O⊕O) = f (O⊕O)+(q−1) f (Mx,0).

Solving a system of linear equations shows that we find a nonzero solution only if λ= 0.Since λ ∈ [2

√q, 2√

q], this shows that every unramified cuspidal representation forPGL2(A) is tempered.

Toroidal automorphic forms. Let E = Fq2F . An unramified TE-periodical eigenform fsatisfies

0 = ΩTE ( f )(O⊕O) = f (O⊕O)+2 f (Ny1,2)+ · · ·+2 f (Nyq,2).

Solving a system of system of linear equations shows that it contains nonzero solutionsonly if λ=±q (cf. [CL09]). Thus λ ∈ [−2

√q, 2√

q] for q = 2,3,4, which shows thatevery unramified TE-periodical automorphic representation for PGL2(A) is tempered.

The aim of our study. In the light of the previous examples, we can phrase the overallobjective of this series of papers as follows:

We intend to extend the methods from the previous example to PGL3. Inparticular, we aim to probe into the question of which (combinations of)period integrals lead to tempered automorphic representations.

As a pioneering example, the third author has executed in [PJ20] this program forthe elliptic curve E2 over F2 with one rational point. To name an example, all TFq2F -periodical Eisenstein series as considered by Wielonsky in [Wie85] are tempered.

The main results of this paper. The scope of this paper is the computation of theweights mx,r(E,F) as far as we will use them in our sequel papers. In fact, we willcompute refined weights2 mx,r(E,F), which are defined as the number of subsheavesF′ of E with quotient isomorphic to K⊕r

x such that F′ ' F (as abstract sheaves). This

determines the weights of G(n)x,r as

mx,r(E,F) = ∑F′∈F

mx,r(E,F′).

2In fact, mx,r(E,F) appears as a multiplicity for the action of Φx,r on automorphic forms for GLn.


The primary result that forms the basis for most of our applications is Theorem 4.6whose content is as follows.3 We fix an arbitrary elliptic curve X over Fq.

Result B (main theorem). A description of the weights mx,1(E,F) for a degree 1 placex and all rank 3 bundles E and F on X.

Note that the weights mx,2(E,F) are fully determined by the weights mx,1(E′,F′)

thanks to our second duality result (see Theorem 2.3), which holds, in fact, for everysmooth projective curve.

Theorem C (second duality for Hecke operators). Let E1 and E2 be rank n bundles onX and let E∨1 and E∨2 be their respective duals and x a place of X. Then mx,r(E1,E2) =mx,n−r(E

∨1 ,E

∨2 ⊗OX(x)).

Our first side result, which is used to determine the cusp eigenforms of PGL(2) andthe induced parabolic Eisenstein series for PGL(3), is Theorem 6.6 whose content is asfollows.

Result D (first complementary theorem). A description of the weights my,1(E,F) fordegree 2 places y of X and all rank 2 bundles E and F of even degree.

Our second side result is Theorem 6.15 whose content is as follows.

Result E (second complementary theorem). A description of the weights my,1(O⊕3,F)

for all degree 2 places y and all rank 3 bundles F.

Method of proof: comparison with Hall algebras. The Hall algebra of a curvedefined over a finite field was introduced by Kapranov in [Kap97] and posteriorly studiedby Kapranov, Schiffmann and Vasserot in [KSV17]. Based on structure theorems for theHall algebra of an elliptic curve by Burban-Schiffmann ([BS12]) and Fratila ([Fra13]),the first author has developed in [Alv20] a method to compute the weights mx,r(E,F).In the following, we describe how this method works.

The Hall algebra. The Hall algebra of X is defined as the complex vector space

HX =⊕

F∈Coh(X)

C ·F

that is freely generated by the isomorphism classes F of coherent sheaves on X . Theproduct of two classes is defined as

G∗F = q12 〈G,F〉 · ∑

E

hEG,F ·E

where 〈G,F〉= dimHom(G,F)−dimExt1(G,F) is the Euler form and

hEG,F = #

subsheaves F′ ' F of E such that E/F′ ' G.In particular, the weight mx,r(E,F) equals the structure constant hE

K⊕rx ,F

in the Hallalgebra. Thus the product K⊕r

x ∗F determines the weights mx,r(E,F) simultaneouslyfor all rank n bundles E.

3Since the descriptions of the following results extend over several pages, we omit the full statementsin the introduction.


Structure results. The elliptic Hall algebra HX decomposes as a restricted tensorproduct of the subalgebras H(µ)

X linearly spanned by isomorphism classes of semi-stablesheaves of slope µ ∈ Q∪∞, cf. [BS12, Lemma 2.6]. By Atiyah’s classification[Ati57, Thm. 7] of vector bundles on an elliptic curve, the category Cµ of semi-stablecoherent sheaves of slope µ is equivalent with C∞, the category of torsion sheaves. As aconsequence, H(µ)

X is isomorphic to H(∞)X for every slope µ.

The subalgebra H(∞)X is generated by certain elements T(0,d),x, which average over

torsion sheaves of degree d with support in x. Transporting these generators via theisomorphisms H(µ)

X 'H(∞)X to H(µ)

X yields elements Tv,x in H(µ)X where v = (n,d) ranges

over all pairs with n> 0, d ∈ Z and slope µ(v) = dn equal to µ.

Let ρ be a primitive character of Pic0(Xn), i.e. it is not induced from a character onPic0(Xm) with m< n, and let ρ be its Galois orbit. Taking a ρ-weighted average of thoseTv,x for which the degree of x divides gcd(n,d) yields elements T ρv , which generate thetwisted spherical Hall algebra UρX . The Hall algebra decomposes into a restricted tensorproduct HX =

⊗′UρX where ρ varies over all Galois orbits of primitive characters.This means, in particular, that the elements of UρX and UσX commute for distinct ρ 6= σ.

The commutators of elements in UρX have an explicit description in terms of coefficientsof a generating series in the T ρv ; cf. [Alv20], [BS12], [Fra13] for more details.

The algorithm. In broad strokes, the algorithmic method to compute hEK⊕r

x ,Ffrom the

first author’s paper [Alv20] is as follows:

(1) Base change: Express K⊕rx and F as linear combinations of products of the T ρv .

This uses a reduction to H(∞)X via H(µ)

X ' H(∞)X and calculus in the Macdonald

algebra to express Hall-Littlewood symmetric functions, which correspond totorsion sheaves, as sums of products of power-sum functions, which correspondto the T(0,d),x, followed by a base change to the T ρv .

Output: K⊕rx ∗F = ∑aiT

ρi,1vi,1 · · ·T

ρi,kivi,ki

.

(2) Order by slopes: Exchange factors in the products, so that slopes increase.Exchanging the order of T ρv ·T ρv′ brings the commutator [T ρv ,T

ρv′ ] into play, which

leads to the subdivision of certain triangles in the lattice Z2 and the computationof coefficients of a certain generating series, which describe the commutator ifthe triangle with vertices (0,0), v and v′ has no inner lattice points.

Output: K⊕rx ∗F = ∑biT

ρi,1wi,1 · · ·T

ρi,liwi,li

such that µ(wi,1)6 . . .6 µ(wi,li) for all i.

(3) Reverse base changes and multiplication: Replace the T ρw by the Tv,x, multi-ply and replace by the F.Both reverse base changes as well as the multiplication of two factors are imme-diate, with the exception of the products of the form Tv,xTv′,x′ with µ(v) = µ(v′)and x = x′. These latter products can be computed by expressing products ofpower-sums in the Macdonald algebra in terms of Hall-Littlewood polynomials.Output: K⊕r

x ∗F = ∑hEiK⊕r

x ,FEi.


Note that this description is slightly inaccurate since, in fact, we compute the commu-tator [K⊕r

x ,F] of K⊕rx and F in HX , which agrees with K⊕r

x ∗F up to coherent sheavesthat are not vector bundles. Instead of digging into the details, we explain the algorithmin a concrete case. We refer the interested reader to the introduction and section 4 in[Alv20] for a comprehensive account.

An example. Let q ∈ 2,3,4 and let X = Eq be the elliptic curve over Fq with 1rational point x. Let Ny,2 be the trace of a line bundle L(y) on Eq,2 = Eq⊗Fq Fq2 wherey is a place of Eq,2 over y. We compute Kx ∗F for F =Ny,2⊗OX(−x) in the following.

We begin with the observation that we are only interested in the extensions of Kx byF that are vector bundles, i.e. we want to compute

πvec(Kx ∗F) = q12 〈Kx,F〉 · ∑

vectorbundles E

hEKx,FE.

Note that 〈Kx,F〉=−2 and thus q12 〈Kx,F〉 = q−1. Since the only extension of Kx by F

that is not a vector bundle is F⊕Kx, which is equal to q−1F ∗Kx, we gain an equality

πvec(Kx ∗F) = Kx ∗F−F ∗Kx = [Kx,F].

In order to compute the commutator [Kx,F], we express the involved bundles in termsof the elements T ρ

(n,d):4

Kx = T 1(0,1) and F = c · ∑

ρ∈P2

12

(ρ(y)+ρ(y′)

)T ρ(2,0)

where 1 denotes the trivial character of Pic(Eq), P2 is the group of characters of Pic(Eq,2)

that are trivial on OX(x), the place y′ is the Galois conjugate of y and c = 2√

q(q+1)#Eq(Fq2)

.

Using that [T 1(0,1),T

ρ(2,0)

]=

c−1q−1 ·T 1

(2,1) if ρ= 1,

0 if not,

and T 1(2,1) =Mx,1, we compute

[Kx,F] = c · ∑ρ∈P2

12

(ρ(y)+ρ(y′)

)[T 1(0,1),T

ρ(2,0)

]= q−1 ·T 1

(2,1) = q12 〈Kx,F〉 ·Mx,1,

which determines mx(E,F) as 1 for E'Mx,1 and as 0 for E 6'Mx,1.

Content overview. The text is structured as follows. In section 1, we introduce graphsof Hecke operators for Hecke operators of GLn and PGLn over a global function field,and we explain the relation to Hall algebras. In section 2, we establish the dualitytheorems (Theorems A and C) for Hecke operators. In section 3, we recall and establishseveral facts around Hecke operators for elliptic function fields. In section 4, we statethe main theorem (Result B) of the text, which we prove in section 5. In section 6, westate and proof the two complimentary theorems (Results D and E).

Acknowledgements. The first author was supported by FAPESP [grant number 2017/21259-3]. The second author was supported by a Marie Skłodowska-Curie IndividualFellowship.

4In order to avoid a digression into definition of the T ρ(n,d), we omit the details of this part.


1. Graphs of Hecke operators

In this section, we set up the notations that are used throughout the paper. We reviewthe notion of a graph of a Hecke operator and its connection with a Hall algebra.

Let X be a smooth projective and geometrically irreducible curve over a finite fieldFq, where q is a prime power. In later sections, we specify X to elliptic curves, but thecontent of this section holds for all curves X .

Let F be the function field of X . We denote by |X | the set of closed points of X or,equivalently, the set of places of F . We denote the residue field of X at x ∈ |X | by Fq(x)and by |x|= [Fq(x) : Fq] the degree of x. Let A be the adele ring of F and OA the ringof the adelic integers.

We denote by G the general linear algebraic group GLn. In particular, we useG(F) = GLn(F) and G(A) = GLn(A). We denote the center of G by Z; thus Z(A) is thecenter of G(A). With respect to its adelic topology, G(A) is a locally compact group.Hence G(A) carries a (left or right) Haar measure that is unique up to constant. Notethat G(A) is unimodular i.e. the left and right Haar measure coincide. Let K =GLn(OA)be the standard maximal compact open subgroup of G(A). We fix the Haar measure onG(A) for which vol(K) = 1.

The complex vector space H of all smooth compactly supported functions Φ :G(A)→ C together with the convolution product

Φ1 ∗Φ2 : g 7−→∫

G(A)

Φ1(gh−1)Φ2(h)dh

for Φ1,Φ2 ∈H is called the Hecke algebra for G(A). Its elements are called Heckeoperators. The zero element of H is the zero function, but there is no multiplicativeunit. We define HK to be the subalgebra of all bi-K-invariant elements. This subalgebrahas a multiplicative unit, namely, the characteristic function εK = charK of K acts as theidentity on HK by convolution. We call HK the unramified part of H and its elementsare called unramified Hecke operators. The algebra HK is also known as the sphericalHecke algebra; also cf. the introduction.

A Hecke operator Φ ∈H acts on the space V := C0(G(G) \G(A)) of continuousfunctions f : G(G)\G(A)→ C by the formula

Φ( f )(g) :=∫

G(A)

Φ(h) f (hg)dh.

The above action restricts to an action of HK on VK , the space of right K-invariantfunctions.

Proposition 1.1 ([Lor13a, Proposition 1.3]). For any unramified Hecke operator Φ,there are unique complex numbers m1, . . . ,mr ∈ C∗ and pairwise distinct classesg1, . . . ,gr ∈ G(F)\G(A)/K such that for all f ∈ VK

Φ( f )(g) =r

∑i=1

mi f (gi).

Remark 1.2. The previous proposition (and hence the upcoming definitions) remaintrue if we consider G(F)Z(A) \G(A)/K instead G(F) \G(A)/K. In this case, wedenote the classes g1, . . . ,gr by g1, . . . ,gr. This reflects the situation when G equals


PGLn instead of GLn in our previous discussion. In this text, we consider both casesGLn and PGLn. In order to avoid confusion, we stress whenever G stands for PGLn.

Remark 1.3. The space A of automorphic forms on G(A) is a subspace of V that isinvariant by the action of H (see eg. [BJ79, Sec. 5]). In the parts 2 and 3 of this project,we will use Proposition 1.1 mostly in the case that f is an automorphic form.

Definition 1.4. For g,g1, . . . ,gr ∈ G(F)\G(A)/K and Φ ∈HK , we define

VΦ,K(g) :=(g,gi,mi)

i=1,...,r

where the gi’s and mi’s are as in Proposition 1.1. The graph G(n)Φ,K of Φ (relative to K) is

the edge weighted graph with vertex set

Vert G(n)Φ,K = G(F)\G(A)/K

and with edge set

Edge G(n)Φ,K =

⋃g∈VertGΦ,K

VΦ,K(g)

where a triple (g,gi,mi) denotes an oriented edge from vertex g to vertex gi with weightmi. The classes gi are called the Φ-neighbors of g (relative to K).

1.1. Drawing convention. If VΦ,K(g) :=(g,g1,m1), . . . ,(g,gr,mr)

, we make the

following drawing conventions to illustrate the graph G(n)Φ,K: vertices are represented

by labeled dots, and an edge (g,gi,mi) together with its origin g and its terminus gi isdrawn as

g gi

m

Hence the Φ-neighborhood of g is thus illustrated as

g

g1

gr

mr

m1

If there is an edge from g to g′ as well as an edge from g′ to g, then we draw

g g′m m′ in place of

g g′

m

m′ andg

min place of

g

m


1.2. Graphs of unramified Hecke operators. Fix an integer n > 1. Let Ik be thek× k-identity matrix and Φx,r the characteristic function of

K(πxIr

In−r

)K

for a place x ∈ |X |. The unramified Hecke algebra HK is isomorphic to

C[Φx,1, . . . ,Φx,n,Φ−1x,n]x∈|X |;

cf. see [BCdS+03, Chapter 12, 1.6] for details. In the following, we use the shorternotations Vx,r(g) := VΦx,r,K(g) and G

(n)x,r := G

(n)Φx,r,K . From the preceding discussion, we

conclude that descriptions of the graphs G(n)x,r , for x ∈ |X | and 0 6 r 6 n−1, yields a

complete knowledge of the action of HK on AK ; also see [Alv19, Proposition 1.8]. Thisallows us to restrict our attention to the graphs G

(n)x,r .

Theorem ([Alv19, Thm. 2.6]). Let G(n)x,r denote the graph of the Hecke operator Φx,r

over GLn. If g ∈ Vert G(n)x,r , then the multiplicities of the edges originating in g sum up

to #Gr(n− r,n)(Fq(x)), the number of points in the Grassmannian over the residualfield of x.

The previous theorem is also true if we replace GLn by PGLn. In more detail, wehave a natural action of Z(A) on G

(n)x,r , which preserves the weight of the edges. Locally

around a vertex, the quotient graph looks like G(n)x,r – with the exception of loops, which

appear if two adjacent vertices are identified – and the weights of the outgoing edgesat a vertex still sum up to the same value. We shall denote G

(n)x,r when the graph under

consideration is taken over PGLn.

1.3. Algebraic geometry perspective of graphs of Hecke operators. Our goal is todescribe the graphs of Hecke operators for GL3 (and for PGL3) for any elliptic curve.Our methods use tools from algebraic geometry. In this subsection, we derive analternative description of the graphs G

(n)x,r in terms of algebraic geometry.

By a theorem of André Weil, there is a bijection between G(F) \G(A)/K withthe set Bunn X of isomorphism classes of rank n vector bundles on X . Similarly, abijection between G(F)Z(A)\G(A)/K with the set PBunn X of isomorphism classesof projective rank n vector bundles on X ; cf. section 5 in [Lor13a] for details. Thistheorem allows us to determine the action of an unramified Hecke operator Φx,r in termsof the equivalence classes of short exact sequences of coherent sheaves on X . Namely,we consider exact sequences of the form

0−→ E′ −→ E−→K⊕rx −→ 0

where E′ and E are rank n vector bundles, x is a closed point of X , and K⊕rx is the

skyscraper sheaf on x whose stalk is Fq(x)⊕r. Let mx,r(E,E′) be the number of isomor-

phism classes of exact sequences

0−→ E′′ −→ E−→K⊕rx −→ 0

with fixed E such that E′′ ∼= E′. Equivalently, mx,r(E,E′) is the number of subsheaves

E′′ of E that are isomorphic to E′ and which the quotient E/E′′ is isomorphic to K⊕rx .


We denote by Vx,r(E) the set of triples(E,E′,mx,r(E,E

′))

such that there exists an exactsequence of the above type, i.e. mx,r(E,E

′) 6= 0. Then we have

Vert G(n)x,r = Bunn X and Edge G

(n)x,r =

∐E∈Bunn X

Vx,r(E);

see [Alv19, Theorem 3.4 ].For G = PGLn, i.e. if we consider the graphs over G(F)Z(A)\G(A)/K, we obtain

in consequence

Vert G(n)x,r = PBunn X and Edge G

(n)x,r =

∐E∈PBunn X

Vx,r(E),

where E denotes the class of the rank n vector bundle E on PBunn X .

1.4. Hall algebras. We denote the category of coherent sheaves on X by Coh(X). TheHall algebra HX of coherent sheaves on a smooth projective curve X , as introducedby Kapranov in [Kap97], encodes the extensions of coherent sheaves on X . Let v be asquare root of q−1. The Hall algebra of X is defined as the C-vector space

HX =⊕F

C ·F

where F runs through the set of isomorphism classes of objects in Coh(X), togetherwith the product

F ∗G = v−〈F,G〉 ∑E

hEF,G E

where 〈F,G〉= dimFq Ext0(F,G)−dimFq Ext1(F,G) and

hEF,G =

∣∣0−→ G−→ E−→ F −→ 0∣∣

|Aut(F)| |Aut(G)|The link between Hall algebras and graphs of Hecke operators is as follows. By

[Alv20, Lemma 2.1], we have

mx,r(E,E′) = hE

K⊕rx ,E′

for E,E′ ∈ Bunn X . Thus we can recover the multiplicities mx,r(E,E′) from the products

K⊕rx ∗E′ in the Hall algebra of X ; more precisely, for fixed x and r the graph G

(n)x,r is

determined by the products K⊕rx ∗E′ where E′ runs through the set of rank n vector

bundles on X . In order to prove the main theorem of this article, we apply a method de-veloped in [Alv20], which is based on results of Fratila [Fra13] and Burban-Schiffmann[BS12] about the Hall algebra of an elliptic curve, to calculate the former products forall E′ ∈ Bun3 X , r = 1 and x a place of degree 1. Note that we can derive a descriptionof G

(n)x,r from G

(n)x,r .

2. Duality

Hecke operators and their graphs satisfy two dualities, which we will explore in thissection.


Lemma 2.1. Given E,E′ ∈ Bunn X, a closed point x of X and 16 r < n. Then for everyshort exact sequence of the form

0−→ E′ϕ−→ E−→K⊕r

x −→ 0,

there exists a canonical short exact sequence

0−→ Eϕ∗−→ E′(x)−→K⊕n−r

x −→ 0

where E′(x) := E′⊗OX OX(x).

Proof. Let V = X −x be an open set in X , then localizing the given short exactsequence in V yields

0 // E′|Vϕ|V∼ //

∼i|V

E|V //

ϕ∗|Vzz

0

0 // E′(x)|Vwhich defines ϕ∗|V as i|V (ϕ|V )−1.

Let U be a sufficiently small affine neighborhood of x, such that OX(U) is a principalideal domain. Then localizing the given short exact sequence yields the exact sequence

0−→ E′(U)ϕ(U)−→ E(U)−→ Fq(x)⊕r −→ 0

of OX(U)-modules.By the elementary division theorem (see [Lan02, Thm. 7.8]), there are bases for

E′(U) and E(U) such that ϕ(U) is represented by a matrixa1 0 · · · 00 a2 · · · 0...

... . . . ...0 0 · · · an

∈Matn(OX(U))

with 〈an〉 ⊆ 〈an−1〉 ⊆ · · · ⊆ 〈a1〉.Moreover, since cokerϕ(U) =Fq(x)⊕r, we can assume

〈ai〉=〈1〉 for i = 1, . . . ,n− r〈πx〉 for i = n− r+1, . . . ,n,

where πx is a uniformizer for OX ,x in OX(U).Therefore the diagonal maps

ψx :=

a−1

1 πx 0 · · · 00 a−1

2 πx · · · 0...

... . . . ...0 0 · · · a−1

n πx

,

φx :=

πx 0 · · · 00 πx · · · 0...

... . . . ...0 0 · · · πx

and γx :=

a1 0 · · · 00 a2 · · · 0...

... . . . ...0 0 · · · an


between a dimension-n vector space Fn over the function field F of X yields thefollowing diagram

E′(x)(U)_

E′(U)

22

ϕ(U)

(( _

E(U) _

∃!

;;

Fn

Fnφx

22

γx )) Fnψx

::

And that diagram induces a map from E(U) to E′(x)(U). This shows that there is aunique ψ : E(U)→ E′(x)(U) such that

E′(U) //

ϕ(U) !!

E′(x)(U)

E(U)ψ

;;

commutes. To conclude, there is a unique morphism ϕ∗ : E→ E′(x) such that

E′ //

ϕ

E′(x)

Eϕ∗AA

commutes. Moreover, it follows from the local considerations

ϕ∗(U)∼

a−1

1 πx 0 · · · 00 a−1

2 πx · · · 0...

... . . . ...0 0 · · · a−1

n πx

,that ϕ∗ is a monomorphism with cokernel Fq(x)n−r.

Remark 2.2. In fact, there is a choice involved when considering E′→ E′(x), which isa nonzero constant in the base field Fq. Up to this choice, the matrix

πx 0 · · · 00 πx · · · 0...

... . . . ...0 0 · · · πx

is well defined as a representative of the (restriction of the) morphism E′→ E′(x) (tothe open neighborhood U of x). This choice does not have consequences for the graphssince it does not change the subbundles of E′(x).

We gain the following dualities for G(n)x,r and G

(n)x,n−r over GLn .

Theorem 2.3. Let x ∈ |X |, 16 r < n and E,E′ ∈ Bunn X . Then we have:

First duality: mx,r(E,E′) 6= 0 if and only if mx,n−r(E

′(x),E) 6= 0;

Second duality: mx,r(E,E′) = mx,n−r(E

∨(x),E′∨) = mx,n−r(E∨,E′∨(−x));

Derived duality: mx,r(E,E′) 6= 0 if and only if mx,r(E

′∨,E∨) 6= 0.


Proof. The first duality is a consequence of Lemma 2.1. The second duality can beproven as follows. Consider a short exact sequence

0−→ E′ −→ E−→K⊕rx −→ 0.

Applying (−)∨ =Hom(−,OX) yields the long exact sequence

0−→Hom(K⊕rx ,OX)−→ E∨

ϕ∨−→ E′∨ δ−→ Ext(K⊕rx ,OX)−→ ·· ·

But Hom(K⊕rx ,OX) vanish and Ext(K⊕r

x ,OX) ∼= K⊕rx . Moreover (ϕ∨)∨ = ϕ and

Ext(E,OX) = 0, thus δ must be surjective. This yields the short exact sequence

0−→ E∨ −→ E′∨ −→K⊕rx −→ 0.

Applying the functor (−)∗ from Lemma 2.1 to this latter sequence yields

0−→ E′∨ −→ E∨(x)−→K⊕n−rx −→ 0.

This establishes a bijectionrank n locally free subsheaves

of E with cokernel K⊕rx

((−)∨

)∗←→

rank n locally free subsheavesof E∨(x) with cokernel K⊕n−r

x

which proves the second duality. The dervived duality results from the composition ofthe first with the second duality.

Remark 2.4. In general, mx,r(E,E′) 6= mx,n−r(E

′(x),E), which are the quantities thatappear in the first duality. See for example [Alv20, Thm. 6.2] for E = L⊕L andE′ = L(−x)⊕L in G

(2)x,1 , where X is an elliptic curve, x is a closed point of degree one

and L is any line bundle.This is because mx,r(E,E

′) is the number of isomorphism classes of short exactsequences

0−→ E′ −→ E−→K⊕rx −→ 0,

with fixed E. Which is the same as the number of rank n locally free subsheaves of Ewith cokernel isomorphic to K⊕r

x . While mx,r(E′∨,E∨) is the number of isomorphism

classes of short exact sequences

0−→ E∨ −→ E′∨ −→K⊕rx −→ 0,

with fixed E′∨ i.e. the number of rank n locally free subsheaves of E′∨ with cokernelisomorphic to K⊕r

x .

We can deduce from Theorem 2.3 analogous dualities for the graphs G(n)x,r .

Corollary 2.5. Let x ∈ |X |, 16 r < n and E,E′ ∈ Bunn X. Then we have:

First duality: mx,r(E,E′) 6= 0 if and only if mx,n−r(E′,E) 6= 0;

Second duality: mx,r(E,E′) = mx,n−r(E∨(x),E′∨) = mx,n−r(E

∨,E′∨);

Derived duality: mx,r(E,E′) 6= 0 if and only if mx,r(E′∨,E∨) 6= 0.

Proof. Since mx,r(E,E′) = ∑E′′∈E′mx,r(E,E′′) and mx,r(E,E

′)> 0, the first and deriveddualities follow at once from Theorem 2.3. To establish the second duality, note that

rank n locally free subsheavesof E with cokernel K⊕r

x

((−)∨

)∗←→

rank n locally free subsheavesof E∨(x) with cokernel K⊕n−r

x


⊗OX (−x)←→

rank n locally free subsheavesof E∨ with cokernel K⊕n−r

x

Remark 2.6. Note that (−)∗ in the Lemma 2.1 is indeed a duality in the sense that(0−→ E′ −→ E−→K⊕r

x −→ 0)∗∗

is equals to0−→ E′(x)−→ E(x)−→K⊕r

x −→ 0.Hence (−)∗∗ is isomorphic to −⊗OX OX(x) as a functor on short exact sequences ofthe form 0→ E′→ E→K⊕r

x → 0.

3. Hecke operators for elliptic curves

From this point on, we restrict our attention to elliptic curves. In particular, X willdenote an elliptic curve over the finite field Fq for the rest of the paper. In the followingwe summarize some well known facts about coherent sheaves on X .

3.1. Coherent sheaves on elliptic curves. We denote the rank and the degree of acoherent sheaf F by rk(F) and degF, respectively. We denote the slope of a nonzerosheaf F in Coh(X) by µ(F) = deg(F)

/rk(F). A sheaf F is semistable (resp. stable) if

for any proper nontrivial subsheaf G⊂ F we have µ(G)6 µ(F) (resp. µ(G)< µ(F)).The full subcategory Cµ of Coh(X) consisting of all semistable sheaves of a fixed slopeµ ∈ Q∪∞ is abelian, artinian and closed under extension. Moreover, if F,G aresemistable with µ(F)< µ(G), then Hom(G,F) = Ext(F,G) = 0. Any sheaf F possessesa unique filtration, called the Harder-Narasimhan filtration or HN-filtration,

0 = Fr+1 ⊂ Fr ⊂ ·· · ⊂ F1 = F

for which Fi/Fi+1 is semistable of slope µi and µ1 < · · ·< µr, (cf. [HN75]). Observethat C∞ is the category of torsion sheaves and hence equivalent to the product category∏x Torx where x runs through the set of closed points of X and Torx denotes the categoryof torsion sheaves supported at x. Since Torx is equivalent to the category of finite lengthmodules over the local ring OX ,x of the point x, the torsion sheaf Kx is the unique simplesheaf in Torx. Moreover, for each l ∈ Z>0 there exists a unique (up to isomorphism)indecomposable torsion sheaf K(l)

x of length l and for each partition λ= (l1, . . . , lr) wedenote by K

(λ)x the unique (up to permutation) torsion sheaf K(l1)

x ⊕·· ·⊕K(lr)x supported

at x associated to λ.

Theorem 3.1 ([Ati57]). The following holds.(i) The HN-filtration of any coherent sheaf splits (noncanonically). In particular,

any indecomposable coherent sheaf is semistable;(ii) The set of stable sheaves of slope µ is the class of simple objects of Cµ.

Let E ∈ Bunn X . We call a subsheaf E′ of E a subbundle if the quotient bundle E/E′

is locally free. Note that every subsheaf E′ is contained in a unique minimal subbundleE′ of E, and this subbundle has the same rank as E′, though the degree might increase,i.e. deg(E′)> deg(E′). For brevity, we call a rank k (sub)bundle also an k-(sub)bundle.Following [Alv19] we define

δ(E′,E) := rk(E)deg(E′)− rk(E′)deg(E)


for a subbundle E′ of E,δk(E) := sup

E′→Ek-subbundle

δ(E′,E).

for k = 1, . . . ,n−1, and

δ(E) := maxδ1(E), . . . ,δn−1(E)

.

The definition of these δ-invariants is motivated by [Ser03].

Remark 3.2. A vector bundle E is semistable if and only if δ(E)6 0, and it is stable ifand only if δ(E)< 0.

3.2. Edges in rank 3 coming from rank 2. Next we present our first results on thegraphs G

(3)x,1 which we derive from results on G

(2)x,1 . Namely, we shall describe the edges

Vx,1(E) of G(3)x,1 for all vertices of the form E =M⊕L ∈ Bun3 X where M ∈ Bun2 X

and L ∈ PicX with |degM− 2degL| 0, where the precise bound depends on M

and L, as explained in the following statement.

Lemma 3.3. Let x ∈ X be a closed point of degree |x| and L ∈ PicX. Let M,M′ ∈Bun2 X be such that mx,1(M,M′) 6= 0.

(i) If either(a) M is indecomposable with 2degL−degM> 0 or(b) M = L1⊕L2 is decomposable with Li ∈ PicX and degL > degLi for

i = 1,2,then Ext(M,L) = 0 and Ext(M′,L) = 0.

(ii) If either(a) M is indecomposable with 2degL−degM> 2|x| or(b) M= L1⊕L2 is decomposable with Li ∈ PicX and degL> degLi + |x|

for i = 1,2,then Ext(M,L) = 0 and Ext(M,L(−x)) = 0.

(iii) If either(a) M is indecomposable with 2degL−degM< 0 or(b) M = L1⊕L2 is decomposable with Li ∈ PicX and degL < degLi for

i = 1,2,then Ext(L,M) = 0 and Ext(L(−x),M) = 0.

(iv) If either(a) M,M′ is indecomposable with 2degL−degM<−|x| or(b) M is indecomposable with M′ decomposable and 2degL−degM<−2|x|

or(c) M= L1⊕L2 is decomposable with Li ∈ PicX and degL< degLi−|x|

for i = 1,2,then Ext(L,M) = 0 and Ext(L,M′) = 0.

Proof. (i) We first suppose M is indecomposable. Then M is semistable and by hypoth-esis µ(M) = degM/2< degL= µ(L), which implies Ext(M,L) = Hom(L,M) = 0.If M′ is indecomposable, then M′ is semistable and by hypothesis

µ(M′) = (degM−|x|)/2< degL= µ(L),

which implies Ext(M′,L) = 0.


If M′ is decomposable, we write M′ = L′1⊕L′2 with degL′2−degL′1 = δ1(M′) > 0

(see [Lor13a, Prop. 7.7]). Since degL′2 +degL′1 = degM−|x| it follows that

degL′2 =degM−|x|+ δ1(M

′)2

From [Lor13a, Lemma 8.2] and M semi-stability (i.e. δ1(M) 6 0), it follows thatδ1(M

′)6 |x|. Therefore,

degL′1 6 degL′2 =degM−|x|+ δ1(M

′)2

< degL,

which implies that Ext(M′,L) = H0(X ,M′⊗L∨) = 0.If M is decomposable, then Ext(M,L) = Hom(L,M) = H0(X ,M⊗L∨) = 0. IfM′ is indecomposable we proceed as above. If M′ is decomposable and M is de-composable, we can write M′ = L′1⊕L′2 with degL′i 6 degLi, i = 1,2. Therefore,Ext(M′,L) = H0(X ,M′⊗L∨) = 0.

(ii) If M is indecomposable we have

µ(M) = degM/2< degL−|x|= µ(L(−x))< µ(L),

which implies Ext(M,L) = 0 and Ext(M,L(−x)) = 0.If M is decomposable, by Serre duality Ext(M,L) = H0(X ,M⊗L∨) = 0. SimilarlyExt(M,L(−x)) = 0.

(iii) If M is indecomposable,

degL−|x|= µ(L(−x))< µ(L)< µ(M) = degM/2,

which implies Ext(L,M) = Ext(L(−x),M) = 0.If M is decomposable, by Serre duality we obtain Ext(L,M) = H0(X ,M∨⊗L) = 0.Similarly Ext(L(−x),M) = 0.

(iv) If M and M′ are indecomposable, then they are semistable. By hypothesis

µ(L)< (degM−|x|)/2< µ(M) and µ(L)< (degM−|x|)/2 = µ(M′),

which implies Ext(L,M) = Ext(L,M′) = 0.If M is indecomposable and M′ is decomposable, we write M′=L′1⊕L′2 with degL′2−degL′1 = δ1(M

′)> 0. Moreover degL′2+degL′1 = degM−|x| and δ1(M′)6 |x| since

δ1(M)6 0.Thus

degL′2 > degL′1 =degM−|x|− δ1(M

′)2

> degL,

which implies Ext(L,M′) = 0. That Ext(L,M) = 0 follows from µ(L) < µ(M) andthe semi-stability of M.If M is decomposable, then Ext(L,M) = Hom(M,L) = H0(X ,L⊗M∨) = 0. More-over, if M′ is indecomposable,

µ(L) = degL< (degL1 +degL2−2|x|)/2< µ(M′),

which implies Ext(L,M′) = 0. Otherwise, if M′ is decomposable, we can write M′ =L′1⊕L′2 with degL′i6 degLi, i= 1,2. Since degM= degL1+degL2−|x|, thereforedegL′i > degLi−|x|> degL, which implies Ext(L,M′) = H0(X ,M′∨⊗L) = 0.


Theorem 3.4. Let x ∈ X be a closed point of degree |x| and qx := q|x|. Let L ∈ PicXand M,M′ ∈ Bun2 X be such that mx,1(M,M′) = m 6= 0. In (i)− (iv) below we considerthe corresponding hypotheses of Lemma 3.3. Then

(i) mx,1(M⊕L,M′⊕L)> m;

(ii) mx,1(M⊕L,M⊕L(−x)) = q2x and equality holds in part (i);

(iii) mx,1(M⊕L,M⊕L(−x))> 1;(iv) mx,1(M⊕L,M′⊕L)> m ·qx.

Moreover, equality holds in (iv) if M and L satisfy Lemma 3.3 (iv) for every M′ suchthat mx,1(M,M′) 6= 0. In this case, we also have an equality in part (iii).

Proof. Let Kx be the skyscraper sheaf at x. If M1 = M, . . . ,Mr ∈ Coh(X) are theextensions of M′ by Kx and mi := hMi

Kx,M′for i = 1, . . .r. Then m = m1,

Kx ∗M′ = v2|x|(m1M1 + · · ·+mrMr), Kx ∗L= v|x|

(L(x)+L⊕Kx

)and

Kx ∗L(−x) = v|x|(L+L(−x)⊕Kx

).

We denote by πvec(−) the vector bundle part in the Hall algebra products. For E ∈Bunn X , we have πvec(Kx ∗E) = [Kx,E] , where the commutator is taken in the Hallalgebra HX (cf. [Fra13, Proposition 2.9]).(i) Since Ext(M′,L) = 0, we have in HX that

M′⊕L= v2degL−degM+|x|M′ ∗L.Hence

Kx ∗ (M′⊕L) = v2degL−degM+|x|(Kx ∗M′)∗L

= v2degL−degM+|x|v2|x|(m1M1 + · · ·+mrMr)∗L

= mv2degL−degM+3|x|M∗L+ v2degL−degM+3|x|( r

∑i=2

miMi

)∗L

= mv2degL−degM+3|x|v−(2degL−degM)M⊕L

+ v2degL−degM+3|x|( r

∑i=2

miMi

)∗L

= mv3|x|M⊕L+ v2degL−degM+3|x|(m2M2 + · · ·+mrMr)∗L

Therefore mx,1(M⊕L,M′⊕L

)> v−3|x|mv3|x| = m.

(ii) Since Ext(M,L(−x)) = 0, thus in HX

M⊕L(−x) = v2(degL−|x|)−degMM∗L(−x).

Hence

Kx ∗ (M⊕L(−x)) = v2(degL−|x|)−degM Kx ∗M∗L(−x)

= v2(degL−|x|)−degM(M∗Kx +[Kx,M])∗L(−x)

= v2(degL−|x|)−degM(M∗Kx ∗L(−x)+ [Kx,M]∗L(−x))


= v2(degL−|x|)−degM(M∗ v|x|L+M∗ v|x|

(Kx⊕L(−x)

)+πvec(Kx ∗M)∗L(−x)

)= v−|x|M⊕L+ v2(degL−|x|)−degM

(M∗ v|x|

(Kx⊕L(−x)

)+πvec(Kx ∗M)∗L(−x)

)Therefore, mx,1

(M⊕L,M⊕L(−x)

)> v−3|x|v−|x| = q2|x| = q2

x . Moreover, if M and L

are such that satisfies Lemma 3.3(ii), then it satisfies Lemma 3.3(i) and hence (i) holds.Summing up over all multiplicities

∑E′∈Bun3 X

mx,1(M⊕L,E′

)> q2

x + ∑M′∈Bun2 X

mx,1(M,M′

)= q2

x +qx +1

where E′ runs over the neighbors of M⊕L in G(3)x,1 obtained on (i) and (ii). The identi-

ties follow from Theorem 1.2.

(iii) Since Ext(L(−x),M) = 0, we have in HX that

L(−x)⊕M= vdegM−2(degL−|x|)L(−x)∗M.

Hence,

Kx ∗(L(−x)⊕M

)= vdegM−2(degL−|x|)Kx ∗L(−x)∗M= vdegM−2(degL−|x|)v|x|v−(degM−2degL)L⊕M

+ vdegM−2(degL−|x|)v|x|(Kx⊕L(−x)

)∗M.

Therefore we conclude that mx,1(L⊕M,L(−x)⊕M

)> v−3|x|v3|x| = 1.

(iv) Since Ext(L,M′) = 0, we have M′⊕L= vdegM′−2degLL∗M′. Hence,

Kx ∗(M′⊕L

)= vdegM′−2degLKx ∗L∗M′

= vdegM−|x|−2degL(L∗Kx ∗M′+[Kx,L]∗M′)

= vdegM−|x|−2degL(L∗ v2|x|

( r

∑i=1

miMi

)+πvec(Kx ∗L)∗M′

)= mv|x|L⊕M+ vdegM+|x|−2degLL∗

( r

∑i=2

miMi

)+ vdegM−|x|−2degLπvec(Kx ∗L)∗M′.

Therefore, mx,1(M⊕L,M′⊕L

)> v−3|x|v|x|m = mqx. Moreover, if M and L are such

that satisfies Lemma 3.3(iv), then it satisfies Lemma 3.3(iii) and hence (iii) above holds.Summing up over all multiplicities

∑E′∈Bun3 X

mx,1(M⊕L,E′

)> ∑

M′∈Bun2 Xmx,1

(M,M′

)qx +1 = q2

x +qx +1

where E′ runs over the neighbors of M⊕L in G(3)x,1 obtained on (iii) and (iv). The

identities follow from Theorem 1.2.


Corollary 3.5. Let M ∈ Bun2 X and L ∈ PicX. We denote by E the rank 3 vectorbundle M⊕L.

(i) Suppose that M and L verify the hypothesis in Lemma 3.3 (ii). Then

Vx,1(E) =(

E,M⊕L(−x),q2x),(E,M′⊕L,m

) ∣∣ (M,M′,m)∈ Vx,1(M)

.

(ii) Suppose that M and L verify the hypothesis in Lemma 3.3 (iv) for every M′ ∈Bun2 X such that

(M,M′,m

)∈ Vx,1(M). Then

Vx,1(E) =(

E,M⊕L(−x),1),(E,M′⊕L,mqx

) ∣∣ (M,M′,m)∈ Vx,1(M)

.

Proof. It follows immediately from the proofs of Theorem 3.4 (ii) and (iv) and fromthe fact

∑E′∈Bun3 X

mx,1(E,E′) = q2

x +qx +1,

that is proved in Theorem 1.2.

Corollary 3.6. Let M ∈ Bun2 X, L ∈ PicX and set E the rank 3 vector bundle M⊕L.

(i) Suppose that either M is indecomposable and 2degL−degM> 2|x| or M=L1⊕L2 with degL1 6 degL2 6 degL and degL−degL2 > |x|. Then

Vx,2(E)=(

E,M(−x)⊕L,1),(E,(M′∨⊕L)(−x),mqx

) ∣∣ (M∨,M′,m)∈Vx,1(M∨).

(ii) Suppose that either M is indecomposable and 2degL− degM < −2|x| orM=L1⊕L2 with degL6 degL1 6 degL2 and degL1−degL> |x| . Then

Vx,2(E) =(

E,M(−x)⊕L,q2x),(E,(M′∨⊕L)(−x),m

) ∣∣ (M∨,M′,m) ∈ Vx,1(M∨).

Proof. If M,L satisfy the hypothesis of (i) (resp. (ii)), thus E∨ satisfies the conditionsof Lemma 3.3 (iv) (resp. (ii)). Hence item (i) (resp. (ii)) follows from Corollary 3.5and Theorem 2.3.

We conclude this section with an improvement of [Alv19, Theorem 4.15] in rank 3case.

Proposition 3.7. Let E,E′,E′′ ∈ Bun3 X such that E′ (resp. E′′) is a neighbor of E inG

(3)x,1 (resp. G

(3)x,2 ). Then:

(i) δ1(E′) ∈

δ1(E)−2|x|, . . . ,δ1(E)+ |x|

and δ1(E

′)− δ1(E)≡ |x| (mod3);

(ii) δ2(E′) ∈

δ2(E)−|x|, . . . ,δ2(E)+2|x|

and δ2(E

′)− δ2(E)≡ 2|x| (mod3);

(iii) δ1(E′′) ∈

δ1(E)−|x|, . . . ,δ1(E)+2|x|

and δ1(E

′′)− δ1(E)≡ 2|x| (mod3);

(iv) δ2(E′′) ∈

δ2(E)−2|x|, . . . ,δ2(E)+ |x|

and δ2(E

′′)− δ2(E)≡ |x| (mod3);

where x ∈ |X | is a closed point of degree |x|.Proof. Let M be a 2-subbundle of E and L := E/M. Hence, L is a line bundle and, byduality, L∨ is a line subbundle of E∨. Moreover, E∨/L∨ is isomorphic to M∨. We seethat to maximize degM is equivalent to maximize degL∨. Let M as above such thatdegL∨ attains the maximum, then

degL∨ =degE∨+ δ1(E

∨)3

and degM= degE+degL∨.


Therefore,δ2(E) = 3degM−2degE= δ1(E

∨).

If E′′ is a neighbor of E in G(3)x,2 , by [Alv19, Theorem 4.15] and Theorem 2.3 we

obtain

δ2(E′′) = δ1(E

′′∨) = δ1(E′′∨(−x)) ∈

δ2(E)−|x|, . . . ,δ2(E)+2|x|

.

Analogously we prove that δ2(E′) ∈

δ2(E)−|x|, . . . ,δ2(E)+2|x|

if E′ is a neighbor

of E in G(3)x,1 . This together with [Alv19, Theorem 4.15] finishes the proof.

4. Graphs of rank 3 and degree 1

In this section we state the main theorem of this work, which consists of a description ofthe graphs G

(3)x,1 of the Hecke operators Φx,1 (over GL3) for a closed point x of degree one

over an elliptic curve. We begin recalling Atiyah’s famous theorem about classificationon vector bundles on elliptic curves and then we provide a description of all rank 3vector bundles on an elliptic curve.

Theorem 4.1 (Atiyah). Let X be an elliptic curve defined over Fq. The choice ofany rational point x0 ∈ X(Fq) induces an exact equivalence of abelian categoriesεν,µ : Cµ→ Cν for any µ,ν ∈Q∪∞.

Atiyah’s proof of Theorem 4.1 also provides an algorithm to compute the equiva-lences, cf. [Ati57, Thm. 7]. His proof holds for an algebraically closed field of anycharacteristic. Burban and Schiffmann in [BS12, Thm. 1.1] revamp the statement andextend Atiyah’s theorem to finite fields.

4.1. Classification of rank 3 bundles. We introduce the following notation that willbe used throughout the paper. From Atiyah’s classification, an indecomposable vectorbundle E of rank n and degree d corresponds to a torsion sheaf K(l)

y , as introduced insubsection 3.1, for some y ∈ |X | and l ∈ Z>0 such that l|y|= gcd(n,d). Therefore E iscompletely determined by its rank n, degree d, a positive integer l and a closed point y.We shall denote E by E

(n,d)(y,l) .

Remark 4.2. An extension of scalars FqmF/F , or geometrically π : Xm → X whereXm := X⊗Fqm, defines the constant extension of vector bundles

π∗ : Bunn X −→ Bunn Xm

as pull backs along π. As a consequence of Lang’s theorem [Lan56, Cor. to Thm. 1],π∗ is injective. On the other hand, π : Xm→ X defines the trace of vector bundles

π∗ : Bunn Xm −→ Bunmn X

as the direct image. As shown in [AEJ92, Thm. 1.8], every indecomposable vectorbundle over X is the trace of a geometrically indecomposable bundle over some constantextension Xm of X .

Example 4.3. The trace of a rank n bundle over Xm is the direct image sheaf with respectto π : Xm→ X , which is a rank mn bundle. If y ∈ |X | of degree m and y is a degree 1place of Xm over y, then E

(n′,d′)(y,l) is the trace of the geometrically indecomposable bundle

E(n,d)(y,l) , for n′ = mn and d′ = md.


Example 4.4. The (indecomposable) line bundles over X are the bundles E(1,d)(x,1) with

x ∈ |X | of degree 1 and d ∈ Z. For n = 2,3, the indecomposable rank n bundles over Xare the bundles E(n,d)

(x,l) with x ∈ |X | of degree 1, d ∈ Z and l = gcd(n,d) (geometrically

indecomposable bundles) and E(n,nd)(y,1) with y ∈ |X | of degree n and d ∈ Z (traces of line

bundles over Xn).

In the following, we give a complete list of isomorphism classes of rank 3 bundleson X . To begin with, the Krull-Schmidt theorem [Ati56, Thm. 2] holds for the categoryof vector bundles over X , i.e. every vector bundle on X has a unique decomposition intoa direct sum of indecomposable subbundles, up to permutation of factors.

By our previous discussion, every indecomposable bundle on X is the trace π∗E ofa bundle E on a finite extension Xm of X , which satisfies rk(π∗E) = m · rk(E). Thusthe indecomposable bundles of rank at most 3 on X are geometrically indecomposablebundles of rank at most 3 and traces of line bundles over X2 and X3.

Taking direct sums of all possible combinations of indecomposable bundles yieldsthe following list of rank 3 bundles on X :

E(3,d)(y,l) for |y|= 1 or 3 and d ∈ Z,

E(1,d1)(x1,1)

⊕E(2,d2)(x2,l2)

for |x1|= 1, |x2|= 1 or 2 and d1,d2 ∈ Z,

E(1,d1)(x1,1)

⊕E(1,d2)(x2,1)

⊕E(1,d3)(x3,1)

for |x1|= |x2|= |x3|= 1 and d1,d2,d3 ∈ Z.

Note that this justifies that Theorem 4.6 provides a full description of the graphs G(3)x,1

where x is a degree one place of X .

Remark 4.5. Note that we can derive readily a list of representatives for PBun3 X fromthe above description, which is as follows:

E(3,d)(y,l) for |y|= 1 or 3 and d ∈ 0,1,2

E(1,d1)(x1,1)

⊕E(2,d2)(x2,l2)

for |x1|= 1, |x2|= 1 or 2, d1,∈ Z and d2 ∈ 0,1

E(1,0)(x1,1)

⊕E(1,d2)(x2,1)

⊕E(1,d3)(x3,1)

for |x1|= |x2|= |x3|= 1 and d2,d3 ∈ Z.

This allows us to deduce a description of G(3)x,1 from Theorem 4.6.

4.2. The main theorem. We are prepared to state the central result of this paper.Identities of divisors in the theorem are taken modulo the subgroup 〈x0〉 of PicX wherex0 is the base point of our elliptic curve X ; e.g. if we write z = x+ y, then this has to beread as the equality z−|z| · x0 = (x−|x| · x0)+(y−|y| · x0) in Pic0 X .

Theorem 4.6. Let X be an elliptic curve defined over a finite field Fq and x be a degreeone closed point at X. Then the graph G

(3)x,1 is given as follows.

(i) Let E := E(3,0)(y,1) , then

Vx,1(E) =(

E,E(3,−1)(x′,1) ,q

2 +q+1),

where x′ = y− x.


(ii) Let E := E(3,0)(y,3) , then

Vx,1(E) =(

E,E(3,−1)(x′,1) ,q

2),(E,E(2,−1)(x′′,1) ⊕E

(1,0)(y,1) ,q

),(E,E

(1,−1)(y−x,1)⊕E

(2,0)(y,2) ,1

),

where x′ = 3y− x and x′′ = 2y− x.

(iii) Let E := E(3,1)(y,1) , then

Vx,1(E) =(

E,E(3,0)(z,3) ,1

),(E,E

(3,0)(w,1),1

),(E,E

(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,0)(x3,1)

,1),(

E,E(1,0)(z1,1)⊕E

(2,0)(z2,1)

,1),(E,E

(1,0)(y′,1)⊕E

(2,0)(y′′,2),1

)where 3z = y− x, w = y− x, x1,x2,x3 are pairwise distinct and x1 + x2 + x3 = y− x,z2 + z1 = y− x, and y′ 6= y′′ with y′+2y′′ = y− x.

(iv) Let E := E(3,2)(y,1) , then

Vx,1(E) =(

E,E(3,1)(y−x,1),q

2 +q+1−N1),(E,E

(1,0)(x′,1)⊕E

(2,1)(x′′,1),1

),

where x′,x′′ ∈ |X | are such that x′+ x′′ = y− x.

(v) Let E := E(2,0)(y,2) ⊕E

(1,0)(y′,1) with y 6= y′, then,

Vx,1(E) =(

E,E(3,−1)(z,1) ,q2−q

),(E,E

(1,−1)(y′−x,1)⊕E

(2,0)(y,2) ,1

),(E,E

(2,−1)(y−x,1)⊕E

(1,0)(y,1) ,q

),(

E,E(2,−1)(x′,1) ⊕E

(1,0)(x′′,1),q−1

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y′,1),1

),

where z = y′+2y− x, x′ = y+ y′− x and x′′ = x+ y− y′.

(vi) Let E := E(2,0)(y,2) ⊕E

(1,0)(y,1) , then,

Vx,1(E) =(

E,E(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,1

),(E,E

(2,−1)(y−x,1)⊕E

(1,0)(y,1) ,q

2),(E,E

(1,−1)(y−x,1)⊕E

(2,0)(y,2) ,q

).

(vii) Let E := E(2,0)(y,2) ⊕E

(1,1)(y′,1) with y 6= y′− x, then

Vx,1(E) =(

E,E(2,−1)(2y−x,1)⊕E

(1,1)(y′,1),q

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y′,1),1

),(

E,E(2,0)(y,2) ⊕E

(1,0)(y′−x,1),q

2).(viii) Let E := E

(2,0)(y,2) ⊕E

(1,1)(y′,1), with y = y′− x, then

Vx,1(E) =(

E,E(2,−1)(2y−x,1)⊕E

(1,1)(y′,1),q

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y′,1),1

),(

E,E(3,0)(y,3) ,q

2−q),(E,E

(2,0)(y,2) ⊕E

(1,0)(y,1) ,q

).


(ix) Let E := E(2,0)(y,2) ⊕E

(1,−1)(y′,1) , with y 6= y′+ x, then

Vx,1(E) =(

E,E(1,−1)(y′,1) ⊕E

(2,−1)(2y−x,1),q

2),(E,E(1,−2)(y′−x,1)⊕E

(2,0)(y,2) ,1

),(

E,E(1,−1)(y′,1) ⊕E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ,q

).

(x) Let E := E(2,0)(y,2) ⊕E

(1,−1)(y′,1) , with y = y′+ x, then

Vx,1(E) =(

E,E(1,−1)(y′,1) ⊕E

(2,−1)(2y−x,1),q

2),(E,E(2,−2)(y′,2) ⊕E

(1,0)(y,1) ,q−1

),(

E,E(1,−2)(y′−x,1)⊕E

(2,0)(y,2) ,1

),(E,E

(1,−1)(y′,1) ⊕E

(1,−1)(y′,1) ⊕E

(1,0)(y,1) ,1

).

(xi) Let E := E(2,0)(y,2) ⊕E

(1,d)(y′,1), with |d|> 2, then

Vx,1(E) =(

E,E(2,0)(y,2) ⊕E

(1,d−1)(y′−x,1),q

2),(E,E

(2,−1)(2y−x,1)⊕E

(1,d)(y′,1),q

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,d)(y′,1),1

)if d > 2, and

Vx,1(E) =(

E,E(1,d−1)(y′−x,1)⊕E

(2,0)(y,2) ,1

)(E,E

(1,d)(y′,1)⊕E

(2,−1)(2y−x,1),q

2),(E,E

(1,d)(y′,1)⊕E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ,q

)if d 6−2.

(xii) Let E := E(2,0)(y,1) ⊕E

(1,0)(y′,1), then

Vx,1(E) =(

E,E(3,−1)(x′,1) ,q

2−1),(E,E

(2,0)(y,1) ⊕E

(1,−1)(y′−x,1),1

),(E,E

(2,−1)(z,1) ⊕E

(1,0)(y′,1),q+1

)where x′ = y′+ y− x and z = y− x.

(xiii) Let E := E(2,0)(y,1) ⊕E

(1,d)(y′,1), with |d|> 1, then

Vx,1(E) =(

E,E(2,0)(y,1) ⊕E

(1,d−1)(y′−x,1),q

2),(E,E(2,−1)(z,1) ⊕E

(1,d)(y′,1),q+1

)if d > 1, and

Vx,1(E) =(

E,E(2,0)(y,1) ⊕E

(1,d−1)(y′−x,1),1

),(E,E

(2,−1)(z,1) ⊕E

(1,d)(y′,1),q

2 +q)

if d 6−1, where z = y− x.

(xiv) Let E := E(2,1)(y,1) ⊕E

(1,0)(y′,1), then

Vx,1(E) =(

E,E(1,0)(y′,1)⊕E

(1,0)(x1,1)

⊕E(1,0)(x2,1)

,q),(E,E

(1,0)(y′,1)⊕E

(2,0)(z,1) ,q

),(

E,E(1,0)(y′,1)⊕E

(2,0)(w,2),q

),(E,E

(1,0)(x′,1)⊕E

(2,0)(y′,2),q−1

),(E,E

(2,1)(y,1) ⊕E

(1,−1)(y′−x,1),1

)⋃(

E,E(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,0)(y−y′−x,1),1

) ∣∣ if y 6= y′


⋃(E,E

(1,0)(y′,1)⊕E

(2,0)(y′,2),1

),(E,E

(3,0)(y′,3),q−1

) ∣∣ if y = 2y′+ x

where y′,x1,x2 are pairwise distinct and x1 + x2 = y− x, z is such that z = y− x,2w = y− x with w 6= y′ and x′+ x = y− y′ with x′ 6= y′.

(xv) Let E := E(2,1)(y,1) ⊕E

(1,1)(y′,1), then

Vx,1(E) =(

E,E(2,0)(z,1) ⊕E

(1,1)(y′,1),1

),(E,E

(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,1)(y′,1),1

),(

E,E(2,0)(w,2)⊕E

(1,1)(y′,1),1

),(E,E

(2,1)(y,1) ⊕E

(1,0)(y′−x,1),q

),(E,E

(3,1)(z′,1),q

2−q)

where z ∈ |X | of degree two is such that z = y− x, x1 6= x2 are such that x1 + x2 = y− x,2w = y− x and z′ = y+ y′− x.

(xvi) Let E := E(2,1)(y,1) ⊕E

(1,d)(y′,1), then

Vx,1(E) =(

E,E(2,0)(z,1) ⊕E

(1,d)(y′,1),1

),(E,E

(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,d)(y′,1),1

),(

E,E(2,0)(w,2)⊕E

(1,d)(y′,1),1

),(E,E

(2,1)(y,1) ⊕E

(1,d−1)(y′−x,1),q

2)if d > 2, and

Vx,1(E) =(

E,E(2,0)(z,1) ⊕E

(1,d)(y′,1),q

),(E,E

(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,d)(y′,1),q

),(

E,E(2,0)(w,2)⊕E

(1,d)(y′,1),q

),(E,E

(2,1)(y,1) ⊕E

(1,d−1)(y′−x,1),1

)if d 6−1, where in each case z ∈ |X | of degree two is such that z = y− x, x1 6= x2 aresuch that x1 + x2 = y− x and 2w = y− x.

(xvii) Let E := E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) , then

Vx,1(E) =(

E,E(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,q

2 +q+1).

(xviii) Let E= E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y′,1) with y 6= y′, then

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,−1)(y′−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y′,1),q+1

),(

E,E(2,−1)(2y−x,1)⊕E

(1,0)(y,1) ,q

2−1).

(xix) Let E := E(1,0)(y1,1)

⊕E(1,0)(y2,1)

⊕E(1,0)(y3,1)

, with y1,y2,y3 pairwise distinct, then

Vx,1(E) =(

E,E(3,−1)(z,1) ,q2−2q+1

),(E,E

(1,−1)(yi−x,1)⊕E

(1,0)(y j,1)⊕E

(1,0)(yk,1)

,1),(

E,E(2,−1)(zi,1)

⊕E(1,0)(yi,1)

,q−1)

where z = y1+y2+y3−x and zi = y j +yk−x for i, j,k = 1,2,3 are such that i, j,k=1,2,3.


(xx) Let E := E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,1)(y2,1)

, then

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,0)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,0)(y1,1)

⊕E(1,1)(y2,1)

,1),(

E,E(1,−1)(y1−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y2,1)

,1),(E,E

(2,−1)(y+y1−x,1)⊕E

(1,1)(y2,1)

,q−1)

if y,y1,y2− x are pairwise distinct,

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(2,0)(y2,2)

,q2−q),(E,E

(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,0)(y1,1)

,q)

(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y1,1)

⊕E(1,1)(y2,1)

,1),(E,E

(1,−1)(y1−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y2,1)

,1),(

E,E(2,−1)(y+y1−x,1)⊕E

(1,1)(y2,1)

,q−1)

if y 6= y1 and y2 = y1 + x,

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(2,0)(y,2) ,q

2−1),(E,E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,1

)(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y2,1)

,q+1)

if y = y1 and y2 = y+ x, and

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y2,1)

,q+1)

if y = y1 and y2 6= y+ x.

(xxi) Let E := E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,−1)(y2,1)

, then

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,−1)(y−x,1)⊕E

(1,−1)(y2,1)

,q2 +q),(E,E

(1,−2)(y2−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,1

)if y = y1 and y 6= y2 + x;

Vx,1(E) =(

E,E(2,−2)(y2,2)

⊕E(1,0)(y,1) ,q

2−1),(E,E

(1,−1)(y2,1)

⊕E(1,−1)(y2,1)

⊕E(1,0)(y,1) ,q+1

)(E,E

(1,−2)(y2−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,1

)if y = y1 and y = y2 + x;

Vx,1(E) =(

E,E(1,−1)(y2,1)

⊕E(2,−1)(y+y1−x,1),q

2−q),(E,E

(1,−1)(y2,1)

⊕E(1,−1)(y1−x,1)⊕E

(1,0)(y,1) ,q

)(E,E

(1,−1)(y2,1)

⊕E(1,−1)(y−x,1)⊕E

(1,0)(y1,1)

,q)(E,E

(1,−2)(y2−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y1,1)

,1)

if y2,y1− x,y− x are pairwise distinct;

Vx,1(E) =(

E,E(1,−1)(y2,1)

⊕E(1,−1)(y1−x,1)⊕E

(1,0)(y,1) ,q

),(E,E

(2,−2)(y2,2)

⊕E(1,0)(y1,1)

,q−1),(

E,E(1,−1)(y2,1)

⊕E(2,−1)(y+y1−x,1),q

2−q),(E,E

(1,−1)(y2,1)

⊕E(1,−1)(y2,1)

⊕E(1,0)(y1,1)

,1),(

E,E(1,−2)(y2−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y1,1)

,1)

if y 6= y1 and y1 6= y2 + x;

Vx,1(E) =(

E,E(1,−1)(y2,1)

⊕E(1,−1)(y−x,1)⊕E

(1,0)(y,1) ,q

),(E,E

(2,−2)(y2,2)

⊕E(1,0)(y,1) ,q−1

),

Automorphic forms for PGL(3) over elliptic function fields - Part 1: Graphs of Hecke operators 29(E,E

(1,−1)(y2,1)

⊕E(2,−1)(y+y1−x,1),q

2−q),(E,E

(1,−1)(y2,1)

⊕E(1,−1)(y2,1)

⊕E(1,0)(y,1) ,1

),(

E,E(1,−2)(y2−x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y1,1)

,1)

if y 6= y1 and y 6= y2 + x.

(xxii) Let E := E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,d)(y2,1)

for |d|> 2, then

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,d−1)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,0)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,−1)(y1−x,1)⊕E

(1,d)(y2,1)

,1),(E,E

(2,−1)(y+y1−x,1)⊕E

(1,d)(y2,1)

,q−1)

if d > 2 and y 6= y1;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,d−1)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,d)(y2,1)

,q+1)

if d > 2 and y = y1;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y1,1)

⊕E(1,d)(y2,1)

,q),(

E,E(1,0)(y,1) ⊕E

(1,−1)(y1−x,1)⊕E

(1,d)(y2,1)

,q),(E,E

(2,−1)(y+y1−x,1)⊕E

(1,d)(y2,1)

,q2−q)

if d 6−2 and y 6= y1;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,0)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,d)(y2,1)

,q2 +q)

if d 6−2 and y = y1.

(xxiii) Let E := E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

for d 6−1 or d > 2, then

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q)

if d > 2 and y 6= y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,1),(E,E

(2,0)(y,2) ⊕E

(1,d)(y2,1)

,q−1)

if d > 2 and y = y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,q),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q2)if d <−1 and y 6= y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,q),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q),(E,E

(2,0)(y,2) ⊕E

(1,d)(y2,1)

,q2−q)


if d <−1 and y = y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q)

if d = 2, y1 6= y+ x and y1 6= y2− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),q

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q),(E,E

(1,0)(y,1) ⊕E

(2,2)(y1,2)

,q2−q)

if d = 2, y1 6= y+ x and y1 = y2− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),q

2),(E,E(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,1),(E,E

(2,0)(y,2) ⊕E

(1,d)(y2,1)

,q−1)

if d = 2, y1 = y+ x and y1 6= y2− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),q

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,1),(E,E

(2,0)(y,2) ⊕E

(1,d)(y2,1)

,q−1),(

E,E(1,0)(y,1) ⊕E

(2,2)(y1,2)

,q2−q)

if d = 2, y1 = y+ x and y1 = y2− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,q),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q2)if d =−1, y 6= y2 + x and y 6= y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,q),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q),(E,E

(1,d)(y2,1)

⊕E(2,0)(y,2) ,q

2−q)

if d =−1, y 6= y2 + x and y = y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q2),(E,E(2,−2)(y2,2)

⊕E(1,1)(y1,1)

,q−1)

if d =−1, y = y2 + x and y 6= y1− x;

Vx,1(E) =(

E,E(1,0)(y,1) ⊕E

(1,1)(y1,1)

⊕E(1,d−1)(y2−x,1),1

),(E,E

(1,−1)(y−x,1)⊕E

(1,1)(y1,1)

⊕E(1,d)(y2,1)

,1),(

E,E(1,0)(y,1) ⊕E

(1,0)(y1−x,1)⊕E

(1,d)(y2,1)

,q),(E,E

(2,−2)(y2,2)

⊕E(1,1)(y1,1)

,q−1),(

E,E(1,d)(y2,1)

⊕E(2,0)(y,2) ,q

2−q)

if d =−1, y = y2 + x and y = y1− x.


(xxiv) Let E := E(1,d)(y,1) ⊕E

(1,d1)(y1,1)

⊕E(1,d2)(y2,1)

with d2−d1 > 1 and d1−d > 1, then

Vx,1(E) =(

E,E(1,d−1)(y−x,1) ⊕E

(1,d1)(y1,1)

⊕E(1,d2)(y2,1)

,1),(E,E

(1,d)(y,1) ⊕E

(1,d1−1)(y1−x,1) ⊕E

(1,d2)(y2,1)

,q),(

E,E(1,d)(y,1) ⊕E

(1,d1)(y1,1)

⊕E(1,d2−1)(y2−x,1) ,q

2).Proof. The proof is based on the algorithm developed in [Alv20]. For better readabilitywe separated the proof in several lemmas and present it in the next section.

Item (i) follows from [Alv20, Thm. 5.3]. Item (ii) follows from [Alv20, Thm. 5.3]and Lemmas 5.6 and 5.7. Item (iii) follows from Lemmas 5.4, and 5.17, [Alv20, Thm.5.1], and Lemmas 5.8 and 5.9. Item (iv) follows from Lemmas 5.5 and 5.10. Item(v) follows from [Alv20, Thm. 5.3] and Lemmas 5.6, 5.7 and 5.18. Item (vi) followsfrom Lemmas 5.19, 5.6 and 5.7. Item (vii) follows from [Alv20, Lemma 6.7] joint withTheorem 3.4, [Alv20, Lemma 6.8] joint with Theorem 3.4 and Lemma 5.9. Item (viii)follows from [Alv20, Lemma 6.7] joint with Theorem 3.4, [Alv20, Lemma 6.8] jointwith Theorem 3.4, Lemmas 5.4 and 5.11. Item (ix) follows from Theorem 3.4 (iv) jointwith [Alv20, Lemma 6.7], Theorem 3.4 (iii) and Lemma 5.20. Item (x) follows fromTheorem 3.4 (iv) joint with [Alv20, Lemma 6.7], Lemmas 5.12 and 5.21, and Theorem3.4 (iii).

For (xi) observes that if d > 2 (resp. d 6 2), then E satisfies the hypothesis ofCorollary 3.5 (i) (resp. (ii)) and thus [Alv20, Lemmas 6.7, 6.8] completes the proof.

For (xii) observe that since y 6= y′, Atiyah’s classification 4.1 implies

Ext(E(1,0)(y′,1),E

(2,0)(y,1)

)= Ext

(E(1,−1)(y′−x,1),E

(2,0)(y,1)

)= 0.

Hence E satisfy the hypothesis of Lemma 3.3 (iii) and thus Theorem 3.4 (iii) with[Alv20, Thm. 5.3] and Lemma 5.6 complete the proof.

Item (xiii) follows from Lemma 5.8 and Theorem 3.4 (i) joint with [Alv20, Lemma6.7] for d = 1 and Corollary 3.5 for d > 2. Since E

(2,0)(y,1) has only one neighbor in the

rank 2 graph Gx,1 and it is indecomposable, Corollary 3.5 implies the case d 6−1.Item (xiv) follows from Lemmas 5.4, 5.17, 5.8, 5.9. 5.11, 5.22, 5.13, 5.23, 5.16 and

[Alv20, Thm. 5.1]. Item (xv) follows from Lemmas 5.15, 5.24, 5.14, 5.10, 5.5 and5.25. Item (xvi) follows from Corollary 3.5 (i) if d > 2 and Corollary 3.5 (ii) if d 6−1.Item (xvii) follows from Lemma 5.19(ii). Item (xviii) follows from Lemmas 5.19(i),5.18(iv) and 5.6(vi). Item (xix) follows from [Alv20, Thm. 5.3] and Lemmas 5.18(v)and 5.6. Item (xx) follows from Lemmas 5.17, 5.9, 5.11, 5.22, 5.23 and Theorem3.4 joint with [Alv20, Thm. 6.2]. Item (xxi) follows from Lemmas 5.26, 5.20, 5.12,5.21, 5.10, 5.27, 5.28 and by observing that mx,r(E,E

′) = mx,r(E⊗L,E′⊗L) for everyL ∈ PicX . Item (xxii) follows from Corollary 3.5 joint with [Alv20, Thm. 6.2]. Item(xxiii) for d > 2 or d <−1 follows from Corollary 3.5 coupled with [Alv20, Thm. 6.2].For d =−1,2 item (xxiii) follows from Theorem 3.4 joint with [Alv20, Thm. 6.2] andLemmas 5.7, 5.18, 5.19. Item (xxiv) follows from [Alv20, Thm. 5.2].

Before we state and prove the lemmas that compose to the proof of our main theorem,we give an example that illustrates the graph of a Hecke operator for an elliptic curvewith only one rational point over PGL(3).

Example 4.7. In this example we describe the projective rank 3 and degree 1 graphG

(3)x,1 for an elliptic curve with only one rational point x. There are up to an isomorphism


T1

Tq

2

S2

S1

S0

...T

1 [−2]

Tq [−

2]

T1 [−

1]

Tq [−

1]

T1 [0]

Tq [0]

T1 [1]

Tq [1]

T1 [2]

Tq [2]

...

...

S1 [−

2]

S1 [−

1]

S1 [0]

S1 [1]

S1 [2]

...

...

S0 [−

2]

S0 [−

1]

S0 [0]

S0 [1]

S0 [2]

...

O(1

:1)

O(0

:0)

O(0

:1)

O(1

:2)

O(2

:3)

O(0

:2)

···...

···...

q2+

q+

1

q2+

q+

1 1

11

11

q2

q

1

q2+

q

1

q2−

1

q2−

1

11

q+

1

q+

1

q2+

qq

2+

q

11

q2+

qq

2+

q

q2

q2

q+

1

q+

1

q2

q2

q

q

q

1

q

q

q

1

qq

1q−

1

q

11

1

q2−

q

1

qq

q q2

q ···

1

q2

q−1

1

q

q2

1

q

q

q2−

q

1

q2

··· q2−

1

q+

1

q2+

q+

1

1

q2−

1

q+

1

q

q2−

q

1 1

q−1

Figure 2. The graph G(3)3,1 for the elliptic curve Eq over Fq for q = 2,3,4


three such elliptic curves (see for example [Ser03, 2.4.4 and Ex. 3 of 2.4] and [CL09]):E2 over F2 defined by the Weierstrass equation y2 + y = x3 + x+1, E3 over F3 definedby the Weierstrass equation y2 = x3 +2x+2 and E4 over F4 defined by the Weierstrassequation y2 + y = x3 +α, where F4 = F2(α).

The number of traces of line bundles of constant field extensions of Eq depends on thenumber Fqn-rational points of Eq. A comparison of the coefficients of the zeta functionsof Eq and Eq,n = Eq×Fq Fqn shows that the cardinality of Eq(Fqn) is determined by thecardinality of Eq(Fq). Since the latter number is 1, we obtain #Eq(Fq2) = 2q+1 and#Eq(Fq3) = 3q2 +1 in our situation; cf. [Sil09, Exerc. 5.13] for details. We denote byy1, . . . ,yq the q closed points of Eq of degree 2 and by z1, . . . ,zq2 the q2 closed points ofEq of degree 3.

Since PicX has trivial torsion, it is particularly easy to describe a system of represen-tatives for PBun3 X . Namely, E= E′ and degE= degE′ imply that E' E′ as vectorbundles, which allows us to choose the following set of representatives. If E is inde-composable, then E has a unique representative E′ with 06 degE′ 6 2. If E=M⊕L

for an indecomposable rank 2 bundle M, then E has a unique representative of the formM′⊕L′ with 06 degM6 1. If E is the sum of three line bundles, then E has a uniquerepresentative of the form O⊕O(kx)⊕O(lx) with 06 k 6 l.

For better readability, we adopt the following notation for the vertices of G(3)x,1:

Ti :=[E(3,0)(zi,1)

], i = 1, . . . ,q2, S0 :=

[E(3,0)(x,3)

], S1 :=

[E(3,1)(x,1)

], S2 :=

[E(3,2)(x,1)

],

Ti[k] :=[E(2,0)(yi,1)⊕OEq(kx)

], i = 1, . . . ,q, S0[k] :=

[E(2,0)(x,2)⊕OEq(kx)

],

S1[k] :=[E(2,1)(x,1)⊕OEq(kx)

]and O(k : `) :=

[OEq⊕OEq(kx)⊕OEq(`x)

],

where k, ` ∈ Z.Finally we remark that mx,1(E,E′) = mx,1(E,E

′) since two subbundles E′ and E′′ ofE with quotient E/E′ 'Kx ' E/E′′ have the same degree and therefore E= E′ impliesthat E ' E′ as vector bundles. Thanks to these deductions, the graph G

(3)x,1 for Eq is

determined by Theorem 4.6 and as illustrated in Figure 2.

5. Proof of the main theorem

This section is devoted to the proof of Theorem 4.6. Throughout the section, X is anelliptic curve with a fixed base point x0 and x stands for a second (possibly equal) degree1 place of X .

5.1. Norm maps on the Picard group of X . Let Xn be the base field extension of Xto Fqn . By the Riemann-Roch theorem, the map

X(Fqn) = Xn(Fqn)1:1−→ Pic0(Xn)

x 7−→ OXn(x− x0)

is bijective. This allows us to transport the group structure from Pic0(Xn) to Xn(Fqn)such that x0 is the neutral element. Moreover, the inclusions X(Fqm)⊆ X(Fqn) for m|nare compatible with the respective group structures. To shorten notation, we let Nn standfor |X(Fqn)|, the number of rational points of X over Fqn .


We consider the (geometric) Frobenius endomorphism FrX on the scheme (X ,OX),which is the identity at the level of (topological) points and the raising at the q-th powerat the level of functions. The Frobenius endomorphism FrX acts on the set X(Fqn) ofFqn-rational points of X by composition FrX x : Spec(Fqn)→ X , where x ∈ X(Fqn).This action is compatible with the group structure of X(Fqn). If m|n are positive integers,then we have the identification X(Fqm) = X(Fqn)Frm

X . In particular X(Fqn) = X(Fq)Frn

X .

We can also identify the set |X | with X(Fq)/

FrX , where the quotient means that twoFq-rational points of X are identified if they have the same orbit under the Frobeniusaction. Similarly, we can identify |Xn| with X(Fq)

/Frn

X . These two equalities allow usto define an action of FrX on the closed points of Xn.

The above isomorphism X(Fqn)∼= Pic0(Xn) yields the commutative diagram

Pic0(Xn)

o

Fr∗X ,n // Pic0(Xn)

o

X(Fqn)FrX // X(Fqn).

Let m,n be positive integers such that m|n. We define the relative norm map

Normnm : Pic0(Xn)−→ Pic0(Xm)

by

Normnm(L) :=

n/m⊗i=1

(Fr∗X ,n)mi(L).

The fact that this map is well defined follows from Galois descent [AEJ92, Proposition3.4].

5.2. Characters on the Picard group of X . For a finite abelian group G, we denoteby G its characters group. The Frobenius Fr∗X ,n acts by duality on each group Pic0(Xn)

and we will denote this action simply by FrX ,n .

We denote Pic0(Xn)(resp. Pic0(Xn)

/FrX ,n

)shortly by Pn (resp. Pn). For a character

ρ ∈ Pn, we denote by ρ its class on Pn. Moreover, X stands for⊔

n Pn and we shall abusethe language and call the elements on X by characters even though they are actuallyorbits of characters.

By dualizing the norm maps we also obtain the relative norm maps between thecharacter groups for which we will use the same notation

Normnm : Pic0(Xm)−→ Pic0(Xn).

Since we have fixed a rational point x0 on X , for any integers n> 1 and d ∈ Z we canidentify Picd(Xn) with Pic0(Xn) by subtract dx0. This allows us to extend any characterρ of Pic0(Xn) to a character of Pic(Xn) by putting ρ(OXn(x0)) = 1. Unless otherwisespecified we will view the characters of Pic0(Xn) as characters of Pic(Xn).


5.3. A grading on the elliptic Hall algebra. The Grothendieck group K0(X) ofCoh(X) is isomorphic to Z⊕ Pic(X) via the map F 7→

(rk(F),det(F)

). Moreover,

if we compose this morphism with the one sending a line bundle to its degree, then weget a map

ϕ : Coh(X) −→ K′0(X) := Z2

F 7−→(rk(F),deg(F)

).

into the numerical Grothendieck group K′0(X). Its image is

Z :=(n,d) ∈ K′0(X)

∣∣n> 0 or n = 0 and d > 0.

The Hall algebra HX is naturally graded over Z with homogeneous parts HX [v] :=⊕[F]=vC ·F for v ∈ Z. If v = (n,d) ∈ Z, we denote by γ(v) the integer gcd(n,d).We shall denote by Htor

X :=⊕

d>0 HX [0,d] the subalgebra of torsion sheaves and byHvec

X :=⊕

(n,d)∈Z,n>0 HX [n,d] the subalgebra of vector bundles. We denote by πvec theprojection map HX → Hvec

X .

5.4. A special set of generators for the elliptic Hall algebra. Next we introducea special element of the Hall algebra of X that will be essential to the proof of thefollowing lemmas. Let y be a closed point of X and m a positive integer. We set

[m] :=vm− v−m

v− v−1 and ci :=vi[i]Ni

i,

where v2 = q−1 and Ni := |X(Fqi)|. Let λ= (λ1, . . . ,λl) be a partition of length l(λ) = land weight |λ|= λ1 + · · ·+λl . We define

T(0,m),y :=

[m]|y|

m ∑|λ|=m/|y|

nuy(l(λ)−1)K(λ)y if |y| divides m

0 otherwise,

where nuy(l) = ∏li=1(1−u−2i

y ), uy is a square root of q−1y and K

(λ)y the unique torsion

sheaf supported at y associated to λ.For any µ ∈Q∪∞, we consider the subspace H(µ)

X ⊂HX that is spanned by classesF | F ∈ Cµ

. Since the category Cµ is stable under extensions, H(µ)

X is a subalgebra ofHX . The Hall algebra HX is isomorphic to the restrict tensor product of the subalgebrasH(µ)

X ordered from left to right in increasing order for µ ∈Q∪∞, see [BS12, Lemma2.6].

The exact equivalence εµ,ν from Theorem 4.1 gives rise to an algebra isomorphismεµ,ν : H(ν)

X∼→ H(µ)

X . Note that T(0,m),y ∈ H(∞)X . For every v ∈ Z with slope µ, we define

the element Tv,y by εµ,∞(T(0,m),y). Namely, there is f ∈ SL2(Z) such that f (0,m) = vand thus Tv,y := f ·T(0,m),y, see also [BB07, pag. 16].

Definition 5.1. For a character ρ ∈ Pn and a closed point y ∈ X , we define

ρ(y) :=1n

n−1

∑i=0

ρ((FrX ,n)i(OXn(y

′))) =1d

d−1

∑i=0

ρ((FrX ,n)i(OXn(y

′)))

where y′ ∈ Xn is a closed point that sits above y and d = gcd(|y|,n).


Definition 5.2. For a character ρ ∈ X and for a point v ∈ Z we define the element

T ρv := ∑y∈|X |

ρ(y)Tv,y.

From [Fra13, Prop. 3.4] the elements T ρv in the above definition generates the ellipticHall algebra HX for v ∈ Z and ρ ∈ Pd with d = γ(v).

5.5. Edges terminating in geometrically indecomposable bundles. For the rest ofthis section, we fix a degree one place x of X . In order to keep the exposition compact,we refer the reader to [Alv20, Sec. 4] (or more specific [Alv20, Thms. 4.2, 4.6 and4.8]) for further details on the algorithm that we use in the following proofs.

Remark 5.3. Note that the following lemmas cover all edges of G(3)x,1 between rank 3

bundles E and E′. In particular, this means that if E does not appear in the lemma listingthe arrows of G(3)

x,1 with target E′, then mx,1(E,E′) = 0.

Lemma 5.4. Let E′ := E(3,0)(z,3) be a rank 3 vector bundle on X. Then

(i) mx,1(E(2,0)(z,2) ⊕E

(1,1)(x+z,1),E

′)= q2−q;

(ii) mx,1(E(1,0)(z,1) ⊕E

(2,1)(x+2z,1),E

′)= q−1;

(iii) mx,1(E(3,1)(x+3z,1),E

′)= 1;

Proof. Via Atiyah’s classification 4.1, the vector bundle E′ corresponds to the torsionsheaf K(3)

x . Hence

E′ = v−6(v4−2v2 +1)(T(1,0),z

)3+ v−6(v4−1)T(2,0),zT(1,0),z + v−6(v4 + v2 +1)T(3,0),z.

Base changing and calculating the product Kx ∗E′ in the Hall algebra HX yields

πvec(Kx ∗E′) = N−41 (v−2−2v−4 + v−6)

· ∑σ,ρi∈P1i=1,2,3

σ(−x)ρ1(−z)ρ2(−z)ρ3(−z)[T σ(0,1),T

ρ1(1,0)T

ρ2(1,0)T

ρ3(1,0)

]+N−2

1 N−12 (v−2− v−6) ∑

σ,ρ1∈P1ρ2∈P2

σ(−x)ρ1(−z)ρ2(−z)[T σ(0,1),T

ρ2(2,0)T

ρ1(1,0)

]+N−1

1 N−13 (v−6 + v−4 + v−2) ∑

σ∈P1,ρ3∈P3

σ(−x)ρ3(−z)[T σ(0,1),T

ρ3(3,0)

].

Denote the first, second and third term on the above identity by A,B and C, respectively.Then

A = v5+v3

2 (v−2−2v−4 + v−6)E(1,0)(z,1) ⊕E

(1,0)(z,1) ⊕E

(1,1)(x+z,2)

+ 3v−1(v−2−2v−4 + v−6)E(2,0)(z,2) ⊕E

(1,1)(x+z,1)

+ 3v(v−2−2v−4 + v−6)E(1,0)(z,2) ⊕E

(2,1)(x+2z,1)+ v3(v−2−2v−4 + v−6)E

(3,1)(x+3z,1),

B = (v−2− v−6)v5(v−2−1)

2 E(1,0)(z,1) ⊕E

(1,0)(z,1) ⊕E

(1,1)(x+z,2)+ (v−2− v−6)−v5

2 E(2,0)(z,2) ⊕E

(1,1)(x+z,1)


+ (v−2− v−6)−v7

2 E(2,1)(x+2z,2)⊕E

(1,0)(z,1) + (v−2− v−6)−v9

2 E(3,1)(x+3z,1)

and

C = (v4 + v2 +1) v3

3 E(3,1)(x+3z,1).

Putting all together we have precisely the assertion of the lemma.

Lemma 5.5. Let E′ := E(3,1)(z,1) be a rank 3 vector bundle on X . Then

(i) mx,1(E(3,2)(x+z,1),E

′)= q2 +q+1−N1;

(ii) mx,1(E(2,1)(x′,1)⊕E

(1,1)(x′′,1),E

′)= q2−q;

where x′,x′′ ∈ |X | are such that x′+ x′′ = z+ x.

Proof. We first observe that E′ = N−11 ∑ρ∈P1 ρ(−z)T ρ

(3,1). Thus

πvec(Kx ∗E′) = N−21 ∑

σ,ρ∈P1

σ(−x)ρ(−z)[T σ(0,1),T

ρ(3,1)

].

From [Fra13, Thm. 5.2],[T σ(0,1),T

ρ(3,1)

]vanishes unless σ = ρ. In this case we have[

T σ(0,1),Tσ(3,1)

]= (c2− c1 + c2

2(v−1− v)2−1)T σ(3,2)+ c1(v−1− v)T σ(2,1)T

σ(1,1).

Therefore

πvec(Kx ∗E′) = v([3]− vc1)E(3,2)(x+z,1)+(v−1− v)E(2,1)

(x′,1)⊕E(1,1)(x′′,1).

Multiplying the above equation by v−3 yields the desired statement.

5.6. Edges terminating in a sum of rank 2 bundle with a line bundle.

Lemma 5.6. Let E′ := E(1,0)(y,1) ⊕E

(2,−1)(y′,1) be a rank 3 vector bundle on X. Then

(i) mx,1(E(1,0)(y,1) ⊕E

(2,0)(z,1) ,E

′)= q+1, where z = x+ y′;

(ii) mx,1(E(1,0)(y,1) ⊕E

(2,0)(z,2) ,E

′)= q, where 2z = x+ y′ and z 6= y;

(iii) mx,1(E(1,0)(y,1) ⊕E

(2,0)(y,2) ,E

′)= q2, if 2y = x+ y′;

(iv) mx,1(E(3,0)(y,3) ,E

′)= q, if 2y = x+ y′;

(v) mx,1(E(1,0)(y,1) ⊕E

(1,0)(z,1) ⊕E

(1,0)(z′,1),E

′)= q−1, where y,z,z′ are pairwise distinct andz+ z′ = x+ y′;

(vi) mx,1(E(1,0)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= q2−1, where z 6= y and z+ y = x+ y′;

(vii) mx,1(E(1,0)(z,1) ⊕E

(2,0)(y,2) ,E

′)= q−1, where z 6= y and z+ y = x+ y′;

(viii) mx,1(E(1,1)(x+y,1)⊕E

(2,−1)(y′,1) ,E

′)= q2.


Proof. By [Alv20, Thm. 4.6], we may write

E′ = vN−21 ∑

ρ,ρ1∈P1

ρ1(−y′)ρ(−y)T ρ1(2,−1)T

ρ(1,0)

andKx = N−1

1 ∑σ∈P1

σ(−x)T σ(0,1).

See [Alv20, Section 4] for further details on the following algorithm. Then

πvec(Kx ∗E′) = vN−31 ∑

σ,ρ1,ρ∈P1

σ(−x)ρ1(−y′)ρ(−y)[T σ(0,1),T

ρ1(2,−1)T

ρ(1,0)

].

Since [T σ(0,1),T

ρ1(2,−1)T

ρ(1,0)

]= c1

Θ(2,0)

(v−1− v)T ρ(1,0)+ c1T ρ1

(2,−1)Tσ(1,1)

andΘ(2,0) = (v−1− v)T Norm2

1 σ

(2,0) + (v−1−v)2

2 T σ(1,0)Tσ(1,0),

we obtain

πvec(Kx ∗E′) =(

v2T(2,0),z +v2(v−1−v)

2 T(1,0),zT(1,0),z′)

T(1,0),y + v2T(2,−1),y′T(1,1),x+y,

where z+ z′ = x+ y′, and z ∈ |X | is such that z = x+ y. We calculated each termseparated. By the definition,

T(2,0),z =

[2]E(2,0)(z,1) if |z|= 2

[2]2

((1− v−2)E

(1,0)(z,1) ⊕E

(1,0)(z,1) +E

(2,0)(z,2)

)if |z|= 1.

Thus,T(2,0),zT(1,0),y = [2] E(1,0)

(y,1) ⊕E(2,0)(z,1)

if |z|= 2,

T(2,0),zT(1,0),y =[2](1−v−2)

2 E(1,0)(z,1) ⊕E

(1,0)(z,1) ⊕E

(1,0)(y,1) +

v2[2]2 E

(1,0)(y,1) ⊕E

(2,0)(z,2)

if |z|= 1 and z 6= y, and

T(2,0),zT(1,0),y =[2]2

((1− v−6)E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +E

(1,0)(y,1) ⊕E

(2,0)(y,2) +E

(3,0)(y,3)

)if |z|= 1 and z = y. On the other hand,

T(1,0),zT(1,0),z′T(1,0),y = E(1,0)(z,1) ⊕E

(1,0)(z′,1)⊕E

(1,0)(y,1)

if z,z′,y are pairwise distinct,

T(1,0),zT(1,0),z′T(1,0),y =((1+ v−2)E

(1,0)(z,1) ⊕E

(1,0)(z,1) ⊕E

(1,0)(y,1) +E

(1,0)(y,1) ⊕E

(2,0)(z,2)

)if z = z′ and z 6= y,

T(1,0),zT(1,0),z′T(1,0),y = 2((1+ v−2)E

(1,0)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +E

(1,0)(z,1) ⊕E

(2,0)(y,2)

)if z 6= z′ and z′ = y (or equivalently if z 6= z′ and z = y, interchanging z by z′ in the aboveequation), and finally

T(1,0),zT(1,0),z′T(1,0),y =((1+2v−2 +2v−4 + v−6)E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1)


+(1+2v−2)E(1,0)(y,1) ⊕E

(2,0)(y,2) +E

(3,0)(y,3)

)if z = z′ = y. Taking care of summing up the multiplicities of a same vertex, this provesthe lemma.

Lemma 5.7. Let E′ := E(2,0)(y,2) ⊕E

(1,−1)(y′,1) be a rank 3 vector bundle on X.

(i) If y 6= x+ y′, then

(a) mx,1(E(1,0)(x+y′,1)⊕E

(2,0)(y,2) ,E

′)= 1;

(b) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1),E

′)= q2−q;

(c) mx,1(E(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1),E

′)= q.

(ii) If y = x+ y′, then(a) mx,1

(E(1,0)(y,1) ⊕E

(2,0)(y,2) ,E

′)= q;

(b) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1),E

′)= q2−q;

(c) mx,1(E(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1),E

′)= q;

(d) mx,1(E(3,0)(y,3) ,E

′)= 1.

Proof. Writing E′ is terms of elements T ρv (see Definition 5.2) we get

E′ = v2(

N−31 (1−v2)

2 ∑ρ,ρ1,χ∈P1

χ(−y′)ρ(−y)ρ1(−y)T χ(1,−1)T

ρ(1,0)T

ρ1(1,0)

+N−1

1 N−12 (v2+1)[2] ∑

χ∈P1,ρ2∈P2

χ(−y′)ρ2(−y)T χ(1,−1)T

ρ2(2,0)

).

Hence

πvec(Kx ∗E′) = N−41 (v2−v4)

2 ∑σ,ρ1,ρ,χ∈P1

σ(−x)χ(−y′)ρ1(−y)ρ(−y)[T σ(0,1),T

χ(1,−1)T

ρ(1,0)T

ρ1(1,0)

]+

N−21 N−1

2 (v2+v4)[2] ∑

σ,χ∈P1,ρ2∈P2

σ(−x)χ(−y′)ρ2(−y)[T σ(0,1),T

χ(1,−1)T

ρ2(2,0)

].

Denote the first term in the above identity by A and the second term by B. Using thealgorithm in [Alv20] we obtain

A = (v3)−v5

2 T(1,0),x+y′T(1,0),yT(1,0),y + (v3− v5)T(1,−1),y′T(1,0),yT(1,1),x+y +v4−v6

2 T(1,−1),y′T(2,1),x+2y.

We have two cases to analyze. If y 6= x+ y′,

A = (v−v5)2 E

(1,0)(x+y′,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +

(v3−v5)2 E

(1,0)(x+y′,1)⊕E

(2,0)(y,2)

+(v−1− v)E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1)+

(v−v3)2 E

(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1).

If y = x+ y′,

A = (−v5−v3+v−1+v−3)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +

(−v3−v5+2v)2 E

(1,0)(y,1) ⊕E

(2,0)(y,2)

+ (v3−v5)2 E

(3,0)(y,3) +(v−1− v)E(1,−1)

(y′,1) ⊕E(1,0)(y,1) ⊕E

(1,1)(x+y,1)+

(v−v3)2 E

(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1).


The second term of πvec(Kx ∗E′) yields

B =N−2

1 N−12 (v4+v2)[2]

(c1N1N2T(1,0),x+yT(2,0),y + c2N2

1 T(1,−1),yT(2,1),x+y).

Hence

B= (v5−v)2 E

(1,0)(x+y′,1)⊕E

(1,0)(y,1)⊕E

(1,0)(y,1)+

(v3+v5)2 E

(1,0)(x+y′,1)⊕E

(2,0)(y,2)+

(v+v3)2 E

(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1),

for y 6= x+ y′. The case where y = x+ y′ yields

B = (v5+v3−v−1−v−3)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +

(v3+v5)2 E

(1,0)(y,1) ⊕E

(2,0)(y,2)

+ (v3+v5)2 E

(3,0)(y,3) +

(v+v3)2 E

(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1).

Thereforeπvec(Kx ∗E′) = v3E

(1,0)(x+y′,1)⊕E

(2,0)(y,2) +(v−1− v)E(1,−1)

(y′,1) ⊕E(1,0)(y,1) ⊕E

(1,1)(x+y,1)+ vE(1,−1)

(y′,1) ⊕E(2,1)(x+2y,1)

for y 6= x+ y′, and

πvec(Kx ∗E′) = v E(1,0)(y,1) ⊕E

(2,0)(y,2) + v3E

(3,0)(y,3) + vE(1,−1)

(y′,1) ⊕E(2,1)(x+2y,1)

+(v−1− v)E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1)

for y = x+ y′. Multiplying above identities by v−3 yields the desired assertion.

Lemma 5.8. Let E′ := E(1,0)(y′,1)⊕E

(2,0)(y,1) be a rank 3 vector bundle on X. Then

(i) mx,1(E(2,0)(y,1) ⊕E

(1,1)(x+y′,1),E

′)= q2;

(ii) mx,1(E(1,0)(y′,1)⊕E

(2,1)(z,1) ,E

′)= q, with z = x+ y;

(iii) mx,1(E(3,1)(w,1),E

′)= 1, with w = x+ y′+ y.

Proof. Our proof starts with the observation that

E′ := E(1,0)(y′,1)⊕E

(2,0)(y,1) = E

(1,0)(y′,1) ∗E

(2,0)(y,1) .

Since

E(1,0)(y′,1) = N−1

1 ∑ρ1∈P1

ρ1(−y′)T ρ1(1,0) and E

(2,0)(y,1) = 2[2]−1N−1

2 ∑ρ2∈P2

ρ2(−y)T ρ2(2,0),

we obtain

E′ = 2[2]−1N−12 N−1

1 ∑ρ1∈P1,ρ2∈P2

ρ1(−y′)ρ2(−y)T ρ1(1,0)T

ρ2(2,0).

From the algorithm in [Alv20], it follows that

πvec(Kx ∗E′) = 2[2]−1N−12 N−2

1 ∑σ,ρ1∈P1ρ2∈P2

σ(−x)ρ1(−y′)ρ2(−y)[T σ(0,1),T

ρ1(1,0)T

ρ2(2,0)

]= 2[2]−1N−1

2 N−21

(c1 ∑

σ∈P1ρ2∈P2

σ(−x− y′)ρ2(−y)T ρ2(2,0)T

σ(1,1)

+ c2 ∑σ,ρ1∈P1

σ(−x− y)ρ1(−y′)T ρ1(1,0)T

σ(2,1)+ c1c2 ∑

σ∈P1

σ(−x− y′− y)T σ(3,1))


Thus

πvec(Kx ∗E′) = v[2]T(2,0),yT(1,1),x+y′+ v2T(1,0),y′T(2,1),x+y + v3T(3,1),x+y′+y

= v−1E(2,0)(y,1) ⊕E

(1,1)(x+y′,1)+ vE(1,0)

(y′,1)⊕E(2,1)(x+y,1)+ v3E

(3,1)(x+y′+y,1).

Multiplying the above equation by v−3 completes the proof.

Lemma 5.9. Let E′ := E(1,0)(y,1) ⊕E

(2,0)(y′,2) be a rank 3 vector bundle on X with y 6= y′. Then

(i) mx,1(E(2,0)(y′,2)⊕E

(1,1)(x+y,1),E

′)= q2;

(ii) mx,1(E(1,0)(y,1) ⊕E

(1,0)(y′,1)⊕E

(1,1)(x+y′,1),E

′)= q2−q;

(iii) mx,1(E(1,0)(y′,1)⊕E

(2,1)(z,1) ,E

′)= q−1, where z = x+ y+ y′;

(iv) mx,1(E(1,0)(y,1) ⊕E

(2,1)(z′,1),E

′)= q, where z′ = x+2y′;

(v) mx,1(E(3,1)(w,1),E

′)= 1, where w = x+ y+2y′.

Proof. Since y 6= y′, we can write E′ = E(1,0)(y,1) ∗E

(2,0)(y′,2) in HX . Moreover,

E(1,0)(y,1) = N−1

1 ∑ρ1∈P1

ρ1(−y)T ρ1(1,0)

and

E(2,0)(y′,2) =

(1−v2)N−21

2 ∑ρ,χ∈P1

ρ(−y′)χ(−y′)T ρ(1,0)T

χ(1,0)

+ [2]−1(v2 +1)N−12 ∑

ρ2∈P2

ρ2(−y′) T ρ2(2,0).

Hence

πvec(Kx ∗E′) = (1−v2)

2N41

∑σ,ρ1,ρ,χ∈P1

σ(−x)ρ1(−y)ρ(−y′)χ(−y′)[T σ(0,1),T

ρ1(1,0)T

ρ(1,0)T

χ(1,0)

]+ (1+v2)

[2]N21 N2

∑σ,ρ1∈P1ρ2∈P2

σ(−x)ρ1(−y)ρ2(−y′)[T σ(0,1),T

ρ1(1,0)T

ρ1(1,0)T

ρ2(2,0)

].

We denote by A (resp. B) the first (resp. second) term in the above summation. Againusing the algorithm in [Alv20], we obtain

A = (v−3−v)2 E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,1)(x+y,1)+

(v−1−v)2 E

(2,0)(y′,2)⊕E

(1,1)(x+y,1)+

(v3−v5)2 E

(3,1)(w,1)

+(v− v3)E(1,0)(y′,1)⊕E

(2,1)(z,1) +

(v−v3)2 E

(1,0)(y,1) ⊕E

(2,1)(z′,1)+(v−1− v)E(1,0)

(y,1) ⊕E(1,0)(y′,1)⊕E

(1,1)(x+y′,1)

and

B = (v−v−3)2 E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,1)(x+y,1)+

(v3+v5)2 E

(3,1)(w,1)

+ (v+v3)2 E

(1,0)(y,1) ⊕E

(2,1)(z′,1)+

(v−1+v)2 E

(2,0)(y′,2)⊕E

(1,1)(x+y,1)


where z = x+ y+ y′, z′ = x+2y′ and w = x+ y+2y′. Adding A to B yields preciselythe assertion of the lemma.

Lemma 5.10. Let E′ := E(1,0)(x′,1)⊕E

(2,1)(x′′,1) be a rank 3 vector bundle on X . Then

(i) mx,1(E(1,1)(x+x′,1)⊕E

(2,1)(x′′,1),E

′)= q;

(ii) mx,1(E(3,2)(z,1) ,E

′)= 1, for z = x+ x′+ x′′;

(iii) mx,1(E(1,0)(x′,1)⊕E

(2,2)(w,2),E

′)= q2, if |w|= 1;

(iv) mx,1(E(1,0)(x′,1)⊕E

(2,2)(w,1),E

′)= q2 +q, if |w|= 2;

(v) mx,1(E(1,0)(x′,1)⊕E

(1,1)(y1,1)

⊕E(1,1)(y2,1)

,E′)= q2−q,

where w ∈ |X | is such that w = x+ x′′ and y1 6= y2 are such that y1 + y2 = x+ y.

Proof. In the Hall algebra of X , we write

E′ = v E(1,0)(x′,1) ·E

(2,1)(x′′,1) = vN−2

1 ∑ρ1,ρ2∈P1

ρ1(−x′)ρ2(−x′′)T ρ1(1,0)T

ρ2(2,1).

Hence, using [Alv20, Section 4],

πvec(Kx ∗E′) = vN−31 ∑

σ,ρ1,ρ2∈P1

σ(−x)ρ1(−x′)ρ2(−x′′)[T σ(0,1),T

ρ1(1,0)T

ρ2(2,1)

]= vN−3

1 ∑σ,ρ1,ρ2∈P1σ=ρ1

σ(−x− x′)ρ2(−x′′)c1T ρ2(2,1)T

σ(1,1)

+ vN−31 ∑

σ,ρ1,ρ2∈P1σ=ρ1=ρ2

σ(−x− x′− x′′)c21T σ(3,2)

+ vN−31 ∑

σ,ρ1,ρ2∈P1σ=ρ2

σ(−x− x′′)ρ1(−x′)c1T ρ1(1,0)

Θ(2,2)(v−1−v)

where Θ(2,2) := (v−1− v)T Norm21(σ)

(2,2) + (v−1−v)2

2 T σ(1,1)Tσ(1,1). Thus

πvec(Kx ∗E′) = vN−31

(c1N2

1 T(2,1),x′′T(1,1),x+x′+ c21N1T(3,2),z + c1N2

1 T(1,0),x′T(2,2),w

+ c1N21(v−1−v)

2 T(1,0),x′T(1,1),y1T(1,1),y2

),

where z = x+ x′+ x′′, w = x+ x′′, and y1 + y2 = x+ x′′. Since

T(2,2),w =

[2]E(2,2)(w,1) if |w|= 2

[2]2

((1− v−2)E

(1,1)(w,1)⊕E

(1,1)(w,1)+E

(2,2)(w,2)

)if |w|= 1,

and

T(1,1),y1T(1,1),y2 =

E(1,1)(y,1) ⊕E

(1,1)(y,1) if y1 6= y2

(1+ v−2)E(1,1)(y,1) ⊕E

(1,1)(y,1) +E

(2,2)(y,2) if y = y1 = y2

we obtain

πvec(Kx ∗E′) = vE(1,1)(x+x′,1)⊕E

(2,1)(x′′,1)+ v3E

(3,2)(z,1) +A+B


where

A =

(v+ v−1)E(1,0)(x′,1)⊕E

(2,2)(w,1) if |w|= 2;

(v−v−3)2 E

(1,0)(x′,1)⊕E

(1,1)(w,1)⊕E

(1,1)(w,1)+

(v+v−1)2 E

(1,0)(x′,1)⊕E

(2,2)(w,2) if |w|= 1

and

B =

(v−1− v)E(1,0)(x′,1)⊕E

(1,1)(y1,1)

⊕E(1,1)(y2,1)

if y1 6= y2;(v−3−v)

2 E(1,0)(x′,1)⊕E

(1,1)(y,1) ⊕E

(1,1)(y,1) +

(v−1−v)2 E

(1,0)(x′,1)⊕E

(2,2)(y,2) if y = y1 = y2.

The lemma follows from putting all together and multiplying by v−3.

Lemma 5.11. Let E′ := E(1,0)(y,1) ⊕E

(2,0)(y,2) be a rank 3 vector bundle on X. Then,

(i) mx,1(E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1),E

′)= q2−1;

(ii) mx,1(E(2,0)(y,2) ⊕E

(1,1)(x+y,1),E

′)= q;

(iii) mx,1(E(1,0)(y,1) ⊕E

(2,1)(x+2y,1),E

′)= 1.

Proof. Following the algorithm in [Alv20, Section 4], we write E′ in terms of elementsin Definition 5.2:

E′ = N−31 (2v2−v6−v4)

6 ∑σi∈P1

i=1,2,3

σ1(−y)σ2(−y)σ3(−y)T σ1(1,0)T

σ2(1,0)T

σ3(1,0)

+N−1

1 N−12 (v6+v4)[2] ∑

ρ1∈P1ρ2∈P2

ρ1(−y)ρ2(−y)T ρ2(2,0)T

ρ1(1,0)+

N−13 (−v6−v4−v2)

[3] ∑ρ3∈P3

ρ3(−y)T ρ3(3,0).

Thus

πvec(Kx ∗E′) = N−41 (2v2−v6−v4)

6 ∑σ,σi∈P1i=1,2,3

σ(−x)σ1(−y)σ2(−y)σ3(−y)[T σ(0,1),T

σ1(1,0)T

σ2(1,0)T

σ3(1,0)

]

+N−2

1 N−12 (v6+v4)[2] ∑

σ,ρ1∈P1ρ2∈P2

σ(−x)ρ1(−y)ρ2(−y)[T σ(0,1),T

ρ2(2,0)T

ρ1(1,0)

]

+N−1

3 N−11 (−v6−v4−v2)

[3] ∑σ∈P1,ρ3∈P3

σ(−x)ρ3(−y)[T σ(0,1),T

ρ3(3,0)

].

Calculating the Lie brackets in HX and base changing back, the first term of the abovesum is given by

(v−v5−2v3+2v−1)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1)+

(2v−v5−v3)2 E

(2,0)(y,2) ⊕E

(1,1)(x+y,1)

+ (2v3−v5−v7)2 E

(1,0)(y,1) ⊕E

(2,1)(x+2y,1)+

(2v5−v9−v7)6 E

(3,1)(x+3y,1).

The second term in πvec(Kx ∗E′) becomes

(v5−v)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1)+

(v9+v7)2 E

(3,1)(x+3y,1)+

(v5+v3)2 E

(2,0)(y,2) ⊕E

(1,1)(x+y,1)

+ (v5+v7)2 E

(1,0)(y,1) ⊕E

(2,1)(x+2y,1),


and the third term in πvec(Kx ∗E′) is given by (−v9−v7−v5)3 E

(3,1)(x+3y,1). Putting all together

concludes the proof.

Lemma 5.12. Let E′ := E(2,−2)(y′,2) ⊕E


(i) mx,1(E(2,−2)(y′,2) ⊕E

(1,1)(x+y,1),E

′)= q2;

(ii) mx,1(E(2,−1)(x+2y′,1)⊕E

(1,0)(y,1) ,E

′)= 1;

(iii) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y′+x,1)⊕E

(1,0)(y,1) ,E

′)= q−1 if x+ y′ 6= y;

(iv) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= q2−1 if x+ y′ = y;

(v) mx,1(E(1,−1)(y′,1) ⊕E

(2,0)(y,2) ,E

′)= q−1 if x+ y′ = y.

Proof. Since µ(E(2,−2)(y′,2) )< µ(E

(1,0)(y,1) ) and E

(2,−2)(y′,2) corresponds to K

(2)y′ via Atiyah’s clas-

sification of vector bundles on elliptic curves, we obtain

E′ = N−31 v2(1−v2)

2 ∑ρ,ρi∈P1i=1,2,

ρ1(−y′)ρ2(−y′)ρ(−y)T ρ1(1,−1)T

ρ2(1,−1)T

ρ(1,0)

+N−1

2 N−11 v2(1+v2)[2] ∑

ρ∈P1χ∈P2

χ(−y′)ρ(−y)T χ(2,−2)T

ρ(1,0).

Hence

πvec(Kx ∗E′) = N−41 v2(1−v2)

2 ∑σ,ρ,ρi∈P1

i=1,2,

σ(−x)ρ1(−y′)ρ2(−y′)ρ(−y)[T σ(0,1),T

ρ1(1,−1)T

ρ2(1,−1)T

ρ(1,0)

]

+N−1

2 N−21 v2(1+v2)[2] ∑

σ,ρ∈P1χ∈P2

σ(−x)χ(−y′)ρ(−y)[T σ(0,1),T

χ(2,−2)T

ρ(1,0)

].

We denote by A (resp. B) the first (resp. second) term in the above sum. Proceeding asthe algorithm in [Alv20, Section 4], we obtain

A = (v3− v5)T(1,−1),y′T(1,0),x+y′T(1,0),y +(v3−v5)

2 T(1,−1),y′T(1,−1),y′T(1,1),x+y

+ (v4−v6)2 T(2,−1),2y′+xT(1,0),y

= (v3−v5)2

(v−6(v2 +1)E(1,−1)

(y′,1) ⊕E(1,−1)(y′,1) ⊕E

(1,1)(x+y,1)+ v−4E

(2,−2)(y′,2) ⊕E

(1,1)(x+y,1)

)+ (v3−v5)

2 E(2,−1)(x+2y′,1)⊕E

(1,0)(y,1) +A′

where

A′ =

(v− v3)E(1,−1)(y′,1) ⊕E

(1,0)(y′+x,1)⊕E

(1,0)(y,1) if x+ y′ 6= y

(v− v3)((1+ v−2)E

(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +E

(1,−1)(y′,1) ⊕E

(2,0)(y,2)

)if x+ y′ = y.

We further obtain

B = v4+v2

[2]

(v2[2]

2 T(2,−1),2y′+xT(1,0),y + v T(2,−2),2y′T(1,1),x+y

)


= (v+v−1)2

((1− v−2)E

(1,−1)(y′,1) ⊕E

(1,−1)(y′,1) ⊕E

(1,1)(x+y,1)+E

(2,−2)(y′,2) ⊕E

(1,1)(x+y,1)

)+ (v3+v5)

2 E(2,−1)(x+2y′,1)⊕E

(1,0)(y,1) .

Adding the multiplicities of a same vector bundle that shows up in A and B andmultiplying by v−3 completes the proof.

Lemma 5.13. Let E′ := E(1,−1)(y′,1) ⊕E


(i) mx,1(E(1,0)(y′+x,1)⊕E

(2,1)(y,1) ,E

′)= 1;

(ii) mx,1(E(1,−1)(y′,1) ⊕E

(2,2)(z,2) ,E

′)= q2;

(iii) mx,1(E(1,−1)(y′,1) ⊕E

(2,2)(w,2),E

′)= q2 +q;

where |z|= 1 with z = x+ y and |w|= 2 with w = x+ y.

Proof. In HX we may write

E′ = v3N−21 ∑

ρi∈P1i=1,2,

ρ1(−y′)ρ2(−y)T ρ1(1,−1)T

ρ2(2,1).

Thus

πvec(Kx ∗E′) = v3N−31 ∑

σ,ρi∈P1i=1,2

σ(−x)ρ1(−y′)ρ2(−y)[T σ(0,1),T

ρ1(1,−1)T

ρ2(2,1)

]

= v4(

T(1,0),x+y′T(2,1),y + ∑x′∈|X |

x′=x+y

T(1,−1),y′(T(2,2),x′+

(v−1−v)2 T(1,1),x′T(1,1),x′

))

= v3E(1,0)(y′+x,1)⊕E

(2,1)(y,1) + v−1E

(1,−1)(y′,1) ⊕E

(2,2)(z,2) +(v+ v−1)E

(1,−1)(y′,1) ⊕E

(2,2)(w,2),

where |z|= 1 with z = x+ y and |w|= 2 with w = x+ y. Note that the above sum on x′

vanishes if |x′|> 2. The proof follows after multiplying by v−3.

Lemma 5.14. Let E′ := E(2,0)(y,2) ⊕E

(1,1)(y′,1) be a rank 3 vector bundle on X. Then

(i) mx,1(E(2,0)(y,2) ⊕E

(1,2)(y′+x,1),E

′)= q2;

(ii) mx,1(E(2,1)(x+2y,1)⊕E

(1,1)(y′,1),E

′)= 1;

(iii) mx,1(E(1,0)(y,1) ⊕E

(1,1)(y+x,1)⊕E

(1,1)(y′,1),E

′)= q−1, if y+ x 6= y′;

(iv) mx,1(E(1,0)(y,1) ⊕E

(1,1)(y′,1)⊕E

(1,1)(y′,1),E

′)= q2−1, if y+ x = y′;

(v) mx,1(E(1,0)(y,1) ⊕E

(2,2)(y′,2),E

′)= q−1, if y+ x = y′.

Proof. In the Hall algebra of X , E′ = v2E(2,0)(y,2) ∗E

(1,1)(y′,1),

E(2,0)(y,2) =

N−21 (1−v2)

2 ∑ρ,ρ1∈P1

ρ(−y)ρ1(−y)T ρ(1,0)T

ρ1(1,0)+

N−12 (1+v2)

[2] ∑ρ2∈P2

ρ2(−y)T ρ2(2,0)


andE(1,1)(y′,1) = N−1

1 ∑ρ′∈P1

ρ′(−y′)T ρ′

(1,1).

Therefore

πvec(Kx ∗E′) = N−41 (v2−v4)

2 ∑σ,ρ,ρ1,ρ′∈P1

σ(−x)ρ(−y)ρ1(−y)ρ′(−y′)[T σ(0,1),T

ρ(1,0)T

ρ1(1,0)T

ρ′

(1,1)

]+

N−21 N−1

2 (v4+v2)[2] ∑

σ.ρ′∈P1ρ2∈P2

σ(−x)ρ2(−y)ρ′(−y′)[T σ(0,1),T

ρ2(2,0)T

ρ′

(1,1)

].

Let us set A (resp. B) the first (resp. second) term in the previous sum. By [Alv20,Section 4], we obtain

A = (v3−v5)2

((v−4 + v−6)E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,2)(y′+x,1)+ v−4E

(2,0)(y,2) ⊕E

(1,2)(y′+x,1)

)+ (v3−v5)

2 E(2,1)(x+2y,1)⊕E

(1,1)(y′,1)+A′

where

A′ =

(v− v3)

((1+ v−2)E

(1,0)(y,1) ⊕E

(1,1)(y′,1)⊕E

(1,1)(y′,1)+E

(1,0)(y,1) ⊕E

(2,2)(y′,2)

)if x+ y = y′,

(v− v3)(E(1,0)(y,1) ⊕E

(1,1)(y+x,1)⊕E

(1,1)(y′,1)

)if x+ y 6= y′.

Analogously,

B = (v3+v5)2 E

(2,1)(x+2y,1)⊕E

(1,1)(y′,1)+

(v+v−1)2

((1− v−2)E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,2)(y′+x,1)+E

(2,0)(y,2) ⊕E

(1,2)(y′+x,1)

).

Multiplying the sum A+B by v−3 yields the desired result.

Lemma 5.15. Let E′ := E(2,0)(y,1) ⊕E


(i) mx,1(E(2,1)(y+x,1)⊕E

(1,1)(y′,1),E

′)= 1;

(ii) mx,1(E(2,0)(y,1) ⊕E

(1,2)(y′+x,1),E

′)= q2.

Proof. In the Hall algebra HX , we can write E′ as follows

E′ = v2(E(2,0)(y,1) ∗E

(1,1)(y′,1)

)=

2v2N−12 N−1

1[2] ∑

ρi∈Pii=1,2,

ρ1(−y′)ρ2(−y)T ρ2(2,0)T

ρ1(1,1).

Hence

πvec(Kx ∗E′) = 2v2N−12 N−2

1[2] ∑

σ,ρi∈Pii=1,2,

σ(−x)ρ1(−y′)ρ2(−y)[T σ(0,1),T

ρ2(2,0)T

ρ1(1,1)

].

Since [T σ(0,1)T

ρ2(2,0)

]=

c2T σ(2,1) if ρ2 = Norm2

1(σ)

0 otherwiseand [

T σ(0,1)Tρ1(2,0)

]=

c1T σ(1,2) if ρ1 = σ

0 otherwise,


we haveπvec(Kx ∗E′) = v3E

(2,1)(y+x,1)⊕E

(1,1)(y′,1)+ v−1E

(2,0)(y,1) ⊕E

(1,2)(y′+x,1).

Therefore the lemma follows from multiplying πvec(Kx ∗E′) by v−3.

Lemma 5.16. Let E′ := E(2,−1)(y,1) ⊕E


(i) mx,1(E(2,0)(y+x,2)⊕E

(1,1)(y′,1),E

′)= q;

(ii) mx,1(E(2,−1)(y,1) ⊕E

(1,2)(x+y′,1),E

′)= q2.

Proof. We can write E′ in HX as follows:

E′ = v3(E(2,−1)(y,1) ∗E

(1,1)(y′,1)

)= v3N−2

1 ∑ρi∈P1i=1,2,

ρ1(−y)ρ2(−y′)T ρ1(2,−1)T

ρ2(1,1).

Thus

πvec(Kx ∗E′) = v3N−31 ∑

σ,ρi∈P1i=1,2,

σ(−x)ρ1(−y)ρ2(−y′)[T σ(0,1),T

ρ1(2,−1)T

ρ2(1,1)

]

= v4((

T(2,0),w + (v−1−v)2 T(1,0),wT(1,0),w

)T(1,1),y′+T(2,−1),yT(1,2),w′

)= v E(2,0)

(w,2)⊕E(1,1)(y′,1)+ v−1E

(2,−1)(y,1) ⊕E

(1,2)(w′,1),

where w = x+ y and w′ = x+ y′. The proof follows from multiplying above equationby v−3.

5.7. Edges terminating in a sum of three line bundles.

Lemma 5.17. Let E′ := E(1,0)(x1,1)⊕E

(1,0)(x2,1)⊕E

(1,0)(x3,1)

be a rank 3 vector bundle and x1,x2,x3

be pairwise distinct closed points on X. Then

(i) mx,1(E(1,0)(xi,1)⊕E

(1,0)(x j,1)⊕E

(1,1)(x+xk,1)

,E′)= q2, where i, j,k= 1,2,3;

(ii) mx,1(E(1,0)(xi,1)⊕E

(2,1)(y,1) ,E

′)= q, where y = x+ x j + xk and i, j,k= 1,2,3;

(iii) mx,1(E(3,1)(z,1) ,E

′)= 1, where z = x+ x1 + x2 + x3.

Proof. Applying [Alv20, Thm. 4.6] we can write

E′ = N−31 ∑

σ1,σ2,σ3∈P1

σ1(−x1)σ2(−x2)σ3(−x3)Tσ1(1,0)T

σ2(1,0)T

σ3(1,0).

Hence

πvec(Kx ∗E′) = N−41 ∑σ,σ1,σ2,σ3∈P1

σ(−x)σ1(−x1)σ2(−x2)σ3(−x3)[T σ(0,1),T

σ1(1,0)T

σ2(1,0)T

σ3(1,0)

].

By the algorithm in [Alv20, Section 4] we obtain that

πvec(Kx ∗E′) = v(E(1,0)(x1,1)

∗E(1,0)(x2,1)

∗E(1,1)(x+x3,1)

+E(1,0)(x1,1)

∗E(1,0)(x3,1)

∗E(1,1)(x+x2,1)

+E(1,0)(x2,1)

∗E(1,0)(x3,1)

∗E(1,1)(x+x1,1)

)+ v2

(E(1,0)(x1,1)

∗E(2,1)(x+x2+x3,1)

+E(1,0)(x2,1)

∗E(2,1)(x+x1+x3,1)

+E(1,0)(x3,1)

∗E(2,1)(x+x1+x2,1)

)+ v3E

(3,1)(z,1)


where z = x+x1 +x2 +x3. By definition of the product of the Hall algebra, multiplyingthe above identity by v−3 yields the desired assertion.

Lemma 5.18. Let E′ := E(1,−1)(y1,1)

⊕E(1,0)(y2,1)

⊕E(1,0)(y3,1)

be a rank 3 vector bundle on X withy2 6= y3. Then,

(i) mx,1(E(1,−1)(y1,1)

⊕E(1,0)(y2,1)

⊕E(1,1)(x+y3,1)

,E′)= q2;

(ii) mx,1(E(1,−1)(y1,1)

⊕E(1,0)(y3,1)

⊕E(1,1)(x+y2,1)

,E′)= q2;

(iii) mx,1(E(1,−1)(y1,1)

⊕E(2,1)(x+y2+y3,1)

,E′)= q.

Moreover, depending on y1,y2,y3 we have the following additional edges:(iv) If x+ y1 = yi, then

(a) mx,1(E(1,0)(yi,1)⊕E

(1,0)(yi,1)⊕E

(1,0)(y j,1)

,E′)= q+1;

(b) mx,1(E(2,0)(yi,2)⊕E

(1,0)(y j,1)

,E′)= 1,

for i, j= 2,3.(v) If x+ y1 6= y2 and x+ y1 6= y3, then

mx,1(E(1,0)(y1+x,1)⊕E

(1,0)(y2,1)

⊕E(1,0)(y3,1)

,E′)= 1.

Proof. Since y2 6= y3, we might write E′ in the Hall algebra HX as follows

E′ = v2 E(1,−1)(y1,1)

∗E(1,0)(y2,1)

∗E(1,0)(y3,1)

= v2N−31 ∑

ρi∈P1i=1,2,3

ρ1(−y1)ρ2(−y2)ρ3(−y3)Tρ1(1,−1)T

ρ2(1,0)T

ρ3(1,0).

By the algorithm in [Alv20], we conclude that

πvec(Kx ∗E′) = v2N−41

(c1N3

1 T(1,0),x+y1T(1,0),y2T(1,0),y3

+ c1N31 T(1,−1),y1T(1,0),y2T(1,1),x+y3

+ c1N31 T(1,−1),y1T(1,0),y3T(1,1),x+y2 + c2

1N21 T(1,−1),y1T(2,1),x+y2+y3

).

We denote the four terms in the above identity by

A := v3T(1,0),x+y1T(1,0),y2T(1,0),y3 , B := v3T(1,−1),y1T(1,0),y2T(1,1),x+y3,

C := v3T(1,−1),y1T(1,0),y3T(1,1),x+y2 and D := v4T(1,−1),y1T(2,1),x+y2+y3.

Since the elements in B,C and D have slopes strictly increasing, using the definition ofHall algebras product and multiplying by v−3 yields the first three assertions.

If x+ y1 = y2 in A, thus

A = (v3 + v)E(1,0)(y2,1)

⊕E(1,0)(y2,1)

⊕E(1,0)(y3,1)

+ v3E(2,0)(y2,2)

⊕E(1,0)(y3,1)

.

Analogous for x+ y1 = y3. Multiplying by v−3 we have the desired last two assertions.


Lemma 5.19. Let E′ := E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) be a rank 3 vector bundle on X.

(i) If x+ y′ 6= y, then

(a) mx,1(E(1,0)(y′+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= 1;

(b) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1),E

′)= q.

(ii) If x+ y′ = y, then

(a) mx,1(E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= q2 +q+1;

(b) mx,1(E(2,0)(y,2) ⊕E

(1,0)(y,1) ,E

′)= 1;

(c) mx,1(E(1,−1)(y−x,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1),E

′)= q.

Proof. We first write E′ in the terms of elements T ρv in the Definition 5.2. Proceedingas in the proof of [Alv20, Thms. 4.2 and 4.6], we obtain

E′ = v2E(1,−1)(y′,1) ∗

(E(1,0)(y,1) ∗E

(1,0)(y,1)

)= v2

(N−1

1 ∑χ∈P1

χ(−y′)T χ(1,−1)

)∗(

v2N−21

2 ∑ρ,ρ1∈P1

ρ(−y)ρ1(−y)T ρ(1,0)Tρ1(1,0)−

v2N−12

[2] ∑ρ2∈P2

ρ2(−y)T ρ2(2,0)

)= v4

2

(N−3

1 ∑χ,ρ,ρ1∈P1

χ(−y′)ρ(−y)ρ1(−y)T χ(1,−1)Tρ(1,0)T

ρ1(1,0)−

2N−11 N−1

2[2] ∑

χ∈P1ρ2∈P2

χ(−y′)ρ2(−y)T χ(1,−1)Tρ2(2,0)

).

Hence the product Kx ∗E′ in HvecX is given by


2 ∑σ,χ,ρ,ρ1∈P1

σ(−x)χ(−y′)ρ(−y)ρ1(−y)[T σ(0,1),T

χ(1,−1)T

ρ(1,0)T

ρ1(1,0)

]− v4N−2

1 N−12

[2] ∑σ,χ∈P1ρ2∈P2

σ(−x)χ(−y′)ρ2(−y)[T σ(0,1),T

χ(1,−1)T

ρ2(2,0)

].

We calculate each term of the above sum separatedly. We denote by A the first term andby B the second one.

Suppose x+ y′ 6= y. Applying the algorithm developed in [Alv20, Section 4] yields

A = (v3+v5)2 E

(1,0)(y′+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +

v5

2 E(1,0)(y′+x,1)⊕E

(2,0)(y,2) +

v3

2 E(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1)

+ vE(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(x+y,1)

and

B = (v3−v5)2 E

(1,0)(y′+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) − v5

2 E(1,0)(y′+x,1)⊕E

(2,0)(y,2) − v3

2 E(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1).

Suppose x+ y′ = y. Applying the algorithm in [Alv20, Section 4] once again yields

A = 2−1(v5 +2v3 +2v+ v−1)E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +

(v5+2v3)2 E

(2,0)(y,2) ⊕E

(1,0)(y,1)

+ v5

2 E(3,0)(y,3) + vE(1,−1)

(y−x,1)⊕E(1,0)(y,1) ⊕E

(1,1)(x+y,1)+

v3

2 E(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1)

and

B = (v−1−v5)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) − v5

2 E(2,0)(y,2) ⊕E

(1,0)(y,1) − v5

2 E(3,0)(y,3)


− v3

2 E(1,−1)(y′,1) ⊕E

(2,1)(x+2y,1).

Adding the multiplicities for each vector bundle proves the lemma.

Lemma 5.20. Let E′ := E(1,−1)(y′,1) ⊕E

(1,−1)(z,1) ⊕E

(1,0)(y,1) be a rank 3 vector bundle on X with

z 6= y′. Then,

(i) mx,1(E(1,−1)(y′,1) ⊕E

(1,−1)(z,1) ⊕E

(1,1)(y+x,1),E

′)= q2;

(ii) mx,1(E(2,−1)(w,1) ⊕E

(1,0)(y,1) ,E

′)= 1, with w = y′+ z+ x;

(iii) mx,1(E(1,−1)(z,1) ⊕E

(1,0)(x+y′,1)⊕E

(1,0)(y,1) ,E

′)= q, if y′+ x 6= y;

(iv) mx,1(E(1,−1)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= q2 +q, if y′+ x = y;

(v) mx,1(E(1,−1)(z,1) ⊕E

(2,0)(y,2) ,E

′)= q, if y′+ x = y;

(vi) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(x+z,1)⊕E

(1,0)(y,1) ,E

′)= q, if z+ x 6= y;

(vii) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= q2 +q, if z+ x = y;

(viii) mx,1(E(1,−1)(y′,1) ⊕E

(2,0)(y,2) ,E

′)= q, if z+ x = y.

Proof. Since z 6= y′, in the Hall algebra of X we can write

E′ = N−31 v2

∑ρi∈P1

i=1,2,3

ρ1(−y′)ρ2(−z)ρ3(−y)T ρ1(1,−1)T

ρ2(1,−1)T

ρ3(1,0).

Thus

πvec(Kx ∗E′) = N−41 v2

∑σ,ρi∈P1i=1,2,3

σ(−x)ρ1(−y′)ρ2(−z)ρ3(−y)[T σ(0,1),T

ρ1(1,−1)T

ρ2(1,−1)T

ρ3(1,0)

]= v3

(T(1,−1),zT(1,0),x+y′T(1,0),y +T(1,−1),y′T(1,0),x+zT(1,0),y

+ v T(2,−1),x+y′+zT(1,0),y +T(1,−1),y′T(1,−1),zT(1,1),x+y

)= v−1E

(1,−1)(y′,1) ⊕E

(1,−1)(z,1) ⊕E

(1,1)(y+x,1)+ v3E

(2,−1)(x+y′+z,1)⊕E

(1,0)(y,1)

+ v3T(1,−1),zT(1,0),x+y′T(1,0),y + v3T(1,−1),y′T(1,0),x+zT(1,0),y.

We define A := v3T(1,−1),zT(1,0),x+y′T(1,0),y and B := v3T(1,−1),y′T(1,0),x+zT(1,0),y. Analyz-ing the cases in which x+ y′ and x+ z are equal to and different from y, we obtain

A =

v E(1,−1)(z,1) ⊕E

(1,0)(x+y′,1)⊕E

(1,0)(y,1) if y′+ x 6= y;

(v+ v−1)E(1,−1)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) + v E(1,−1)

(z,1) ⊕E(2,0)(y,2) if y′+ x = y

and

B =

v E(1,−1)(y′,1) ⊕E

(1,0)(x+z,1)⊕E

(1,0)(y,1) if z+ x 6= y;

(v+ v−1)E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) + v E(1,−1)

(y′,1) ⊕E(2,0)(y,2) if z+ x = y.

Putting all together and multiplying by v−3 yields the desired assertion.


Lemma 5.21. Let E′ := E(1,−1)(y′,1) ⊕E

(1,−1)(y′,1) ⊕E

(1,0)(y,1) be a rank 3 vector bundle on X. If

y = y′+ x, then

(i) mx,1(E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= q+1;

(ii) mx,1(E(1,−1)(y′,1) ⊕E

(2,0)(y,2) ,E

′)= 1;

(iii) mx,1(E(1,−1)(y′,1) ⊕E

(1,−1)(y′,1) ⊕E

(1,1)(y+x,1),E

′)= q2.

Proof. In HX we may write

E′ = v4

2

(N−3

1 ∑ρ,ρi∈P1i=1,2,

ρ1(−y′)ρ2(−y′)ρ(−y)T ρ1(1,−1)T

ρ2(1,−1)T

ρ(1,0)

− 2N−12 N−1

1[2] ∑

ρ∈P1χ∈P2

χ(−y′)ρ(−y)T χ(2,−2)T

ρ(1,0)

).

Since Kx = N−11 ∑σ∈P1 σ(−x)T σ(0,1),


2 ∑σ,ρ,ρi∈P1

i=1,2,

σ(−x)ρ1(−y′)ρ2(−y′)ρ(−y)[T σ(0,1),T

ρ1(1,−1)T

ρ2(1,−1)T

ρ(1,0)

]

− N−12 N−2

1 v4

[2] ∑σ,ρ∈P1χ∈P2

σ(−x)χ(−y′)ρ(−y)[T σ(0,1),T

χ(2,−2)T

ρ(1,0)

].

Let A (resp. B) be the first (resp. second) term on the right side of this identity. Usingthe algorithm of [Alv20, Section 4] as the previous lemma, we obtain

A = (v3 + v)E(1,−1)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) + v3E

(1,−1)(y′,1) ⊕E

(2,0)(y,2)

+ (v+v−1)2 E

(1,−1)(y′,1) ⊕E

(1,−1)(y′,1) ⊕E

(1,1)(y+x,1)+

v2E

(2,−2)(y′,2) ⊕E

(1,1)(y+x,1)+

v5

2 E(2,−1)(2y′+x,1)⊕E

(1,0)(y,1)

and

B = v5

2 E(2,−1)(2y′+x,1)⊕E

(1,0)(y,1) +

v2 E

(2,−2)(y′,2) ⊕E

(1,1)(y+x,1)+

(v−v−1)2 E

(1,−1)(y′,1) ⊕E

(1,−1)(y′,1) ⊕E

(1,1)(y+x,1).

Subtracting B from A and multiplying by v−3 concludes the proof.

Lemma 5.22. Let E′ := E(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E


y 6= y′. Then

(i) mx,1(E(1,0)(y′,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y′+x,1),E

′)= q;

(ii) mx,1(E(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,1)(y+x,1),E

′)= q2;

(iii) mx,1(E(1,0)(y′,1)⊕E

(2,1)(x+y+y′,1),E

′)= 1.

Proof. Since y′ 6= y, we can write in the Hall algebra of X

E′ =(E(1,0)(y′,1)⊕E

(1,0)(y′,1)

)∗E(1,0)

(y,1) and E(1,0)(y,1) = N−1

1 ∑ρ∈P1

ρ(−y)T ρ(1,0).


By Atiyah’s classification of vector bundles on an elliptic curve (Theorem 4.1), thevector bundle E

(1,0)(y′,1)⊕E

(1,0)(y′,1) corresponds to K⊕2

y′ . The isomorphism between the Hallalgebra of torsion sheaves supported at a closed point and the Macdonald ring ofsymmetric functions (see [BS12, Prop. 4.1] and [Mac15, Chapter III]), coupled withthe definition of Tv,y′ in [BS12, Paragraph 4.2], yields

E(1,0)(y′,1)⊕E

(1,0)(y′,1) =

v2

2

(T(1,0),y′T(1,0,y′)− 2

[2]T(2,0),y′).

Hence

E′ = N−31 v2

2 ∑ρ,ρi∈P1i=1,2,

ρ1(−y′)ρ2(−y′)ρ(−y)T ρ1(1,0)T

ρ2(1,0)T

ρ(1,0)

− N−12 N−1

1 v2

[2] ∑ρ∈P1χ∈P2

χ(−y′)ρ(−y)T χ(2,0)T

ρ(1,0).

Thus


2 ∑σ,ρ,ρi∈P1

i=1,2,

σ(−x)ρ1(−y′)ρ2(−y′)ρ(−y)[T σ(0,1),T

ρ1(1,0)T

ρ2(1,0)T

ρ(1,0)

]

− N−12 N−2

1 v2

[2] ∑σ,ρ∈P1χ∈P2

σ(−x)χ(−y′)ρ(−y)[T σ(0,1),T

χ(2,0)T

ρ(1,0)

]

= v3T(1,0),y′T(1,0),yT(1,1),x+y′+v3

2 T(1,0),y′T(1,0),y′T(1,1),x+y

+ v5

2 T(3,1),x+2y′+y + v4T(1,0),y′T(2,1),x+y+y′+v4

2 T(1,0),yT(2,1),x+2y′

− v3

[2]T(2,0),y′T(1,1),x+y− v4

2 T(1,0),yT(2,1),x+2y′− v5

2 T(3,1),x+2y′+y

= v E(1,0)(y′,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y′+x,1)+

(v+v−1)2 E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,1)(y+x,1)

+ v3E(1,0)(y′,1)⊕E

(2,1)(x+y+y′,1)+

v2

2 E(2,0)(y′,2)⊕E

(1,1)(x+y,1)− v2

2 E(2,0)(y′,2)⊕E

(1,1)(x+y,1)

− (v−v−1)2 E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,1)(y+x,1)

= v E(1,0)(y′,1)⊕E

(1,0)(y,1) ⊕E

(1,1)(y′+x,1)+ v−1E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,1)(y+x,1)

+ v3E(1,0)(y′,1)⊕E

(2,1)(x+y+y′,1).

We conclude the proof by multiplying the above identity by v−3.

Lemma 5.23. Let E′ := E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E


mx,1(E(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y+x,1),E

′)= 1.

Proof. We first observe from Lemma 5.19 (ii) and Theorem 2.3 that there exists only oneE ∈ Bun3 X such that mx,1

(E,E′

)6= 0, and E must be equal to E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y+x,1).

Hence we are left with computing its multiplicity.


Atiyah’s theorem (Theorem 4.1) and the structure of the Macdonald ring of symmetricfunctions yields

E′ = v6(

16

(T(1,0),y

)3− 1[2]T(1,0),yT(2,0),y +

1[3]T(3,0),y

).

Hence


6 ∑σ,ρi∈P1i=1,2,3

σ(−x)ρ1(−y)ρ2(−y)ρ3(−y)[T σ(0,1),T

ρ1(1,0)T

ρ2(1,0)T

ρ3(1,0)

]

− v6N−21 N−1

2[2] ∑

σ,ρ∈P1χ∈P2

σ(−x)ρ(−y)χ(−y)[T σ(0,1),T

ρ(1,0)T

χ(2,0)

]

+v6N−1

1 N−13

[2] ∑σ∈P1λ∈P3

σ(−x)λ(−y)[T σ(0,1),T

λ(3,0)

].

We denote by A,B and C the first, second and third term in the above sum, respectively.Proceeding as before and following the algorithm in [Alv20], we obtain

A = (v5+v3)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y+x,1)

+ v5

2 E(2,0)(y,2) ⊕E

(1,1)(y+x,1)+

v7

2 E(1,0)(y,1) ⊕E

(2,1)(2y+x,1)+

v9

6 E(3,1)(3y+x,1),

B = (v3−v5)2 E

(1,0)(y,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y+x,1)

− v5

2 E(2,0)(y,2) ⊕E

(1,1)(y+x,1)− v7

2 E(1,0)(y,1) ⊕E

(2,1)(2y+x,1)− v9

2 E(3,1)(3y+x,1)

and C = v9

3 E(3,1)(3y+x,1). Summing A, B and C and multiplying the outcome by v−3 yields

the desired assertion.

Lemma 5.24. Let E′ := E(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,1)(y′,1) be a rank 3 vector bundle on X with

x1 6= x2. Then

(i) mx,1(E(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,2)(x+y′,1),E

′)= q2;

(ii) mx,1(E(2,1)(x+x1+x2,1)

⊕E(1,1)(y′,1),E

′)= 1;

(iii) mx,1(E(1,0)(x2,1)

⊕E(1,1)(x1+x,1)⊕E

(1,1)(y′,1),E

′)= q, if x1 + x 6= y′;

(iv) mx,1(E(1,0)(x1,1)

⊕E(1,1)(x2+x,1)⊕E

(1,1)(y′,1),E

′)= q, if x2 + x 6= y′;

(v) mx,1(E(1,0)(x2,1)

⊕E(1,1)(y′,1)⊕E

(1,1)(y′,1),E

′)= q2 +q, if x1 + x = y′;

(vi) mx,1(E(1,0)(x2,1)

⊕E(2,2)(y′,1),E

′)= q, if x1 + x = y′;

(vii) mx,1(E(1,0)(x1,1)

⊕E(1,1)(y′,1)⊕E

(1,1)(y′,1),E

′)= q2 +q, if x2 + x = y′;

(viii) mx,1(E(1,0)(x1,1)

⊕E(2,2)(y′,1),E

′)= q, if x2 + x = y′.


Proof. Since x1 6= x2 and µ(E(1,0)(x2,1)

)< µ

(E(1,1)(y′,1)

), we have

E′ = v2N−31 ∑

ρ,ρi∈P1i=1,2,

ρ1(−x1)ρ2(−x2)ρ(−y′)T ρ1(1,0)T

ρ2(1,0)T

ρ(1,1).

Hence

πvec(Kx ∗E′) = v2N−41 ∑

σ,ρ,ρi∈P1i=1,2,

σ(−x)ρ1(−x1)ρ2(−x2)ρ(−y′)[T σ(0,1),T

ρ1(1,0)T

ρ2(1,0)T

ρ(1,1)

],

where the commutator is taken in the Hall algebra of X . By [Fra13, Thm. 5.2], we cancompute above commutator in HX . Thus proceeding with the algorithm in [Alv20]yields

πvec(Kx ∗E′) = v3T(1,0),x2T(1,1),x1+xT(1,1),y′+ v3T(1,0),x1T(1,1),x2+xT(1,1),y′

+ v−1E(1,0)(x1,1)

⊕E(1,0)(x2,1)

⊕E(1,2)(x+y′,1)+ v3E

(2,1)(x+x1+x2,1)

⊕E(1,1)(y′,1).

To conclude the proof, set A := T(1,0),x2T(1,1),x1+xT(1,1),y′ , B := T(1,0),x1T(1,1),x2+xT(1,1),y′and observe that

A =

v−2E(1,0)(x2,1)

⊕E(1,1)(x1+x,1)⊕E

(1,1)(y′,1) if x1 + x 6= y′,

(v−2 + v−4)E(1,0)(x2,1)

⊕E(1,1)(y′,1)⊕E

(1,1)(y′,1)+ v−2E

(1,0)(x2,1)

⊕E(2,2)(y′,1) if x1 + x = y′.

Changing x1 by x2 above, we obtain an analogous expression for B. Putting all togetherand multiplying by v−3 completes the proof.

Lemma 5.25. Let E′ := E(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E


(i) mx,1(E(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,2)(y+x,1),E

′)= q2;

(ii) mx,1(E(1,0)(y′,1)⊕E

(1,1)(y′+x,1)⊕E

(1,1)(y,1) ,E

′)= 1, if y′+ x 6= y;

(iii) mx,1(E(1,0)(y′,1)⊕E

(1,1)(y,1) ⊕E

(1,1)(y,1) ,E

′)= q+1, if y′+ x = y;

(iv) mx,1(E(1,0)(y′,1)⊕E

(2,2)(y,2) ,E

′)= 1, if y′+ x = y.

Proof. As the first step of the algorithm in [Alv20], we must write E′ in terms of theelements T ρv in the Definition 5.2. Atiyah’s theorem (Theorem 4.1) implies that E(1,1)

(y,1)

corresponds to Ky and E(1,0)(y′,1)⊕E

(1,0)(y′,1) corresponds to K⊕2

y′ . Hence by the definition ofTv,y′ (and Tv,y) in terms of elements in the Macdonald ring of symmetric functions (cf.[BS12, Paragraph 4.2]) and [Alv20, Prop. 3.5], we can write

E(1,0)(y′,1)⊕E

(1,0)(y′,1) =

v2

2

(N−2

1 ∑ρ,ρ1∈P1

ρ(−y′)ρ1(−y′)T ρ(1,0)T

ρ1(1,0)−

2N−12

[2] ∑ρ2∈P2

ρ2(−y′)T ρ2(2,0)

)and

E(1,1)(y,1) = N−1

1 ∑χ∈P1

χ(−y)Tχ(1,1)

in the Hall algebra HX . Thus E′ = v2(E(1,0)(y′,1)⊕E

(1,0)(y′,1)

)∗E(1,1)

(y,1) is equal to


v2(

v2N−31

2 ∑ρ,ρ1,χ∈P1

ρ(−y′)ρ1(−y′)χ(−y)T ρ(1,0)T

ρ1(1,0)T

χ(1,1)

− v2N−11 N−1

2[2] ∑

χ∈P1ρ2∈P2

ρ2(−y′)χ(−y)T ρ2(2,0)T

χ(1,1)

)and then


2 ∑σ,ρ,ρ1,χ∈P1

σ(−x)ρ(−y′)ρ1(−y′)χ(−y)[T σ(0,1),T

ρ(1,0)T

ρ1(1,0)T

χ(1,1)

]− v4N−2

1 N−12

[2] ∑σ,χ∈P1ρ2∈P2

σ(−x)ρ2(−y′)χ(−y)[T σ(0,1),T

ρ2(2,0)T

χ(1,1)

].

We denote by A and B the first and second term of the above sum, respectively. Next weapply [Fra13, Thm. 5.2] to calculate the previous commutators in HX . Hence[

T σ(0,1),Tρ(1,0)T

ρ1(1,0)T

χ(1,1)

]= c1 T ρ

(1,0)Tσ(1,1)T

χ(1,1)+ c1 T ρ1

(1,0)Tσ(1,1)T

χ(1,1)

+ c1 T ρ(1,0)T

ρ1(1,0)T

σ(1,2)+ c2

1 T σ(2,1)Tχ(1,1)

and [T σ(0,1),T

ρ2(2,0)T

χ(1,1)

]= c2T σ(2,1)T

χ(1,1)+ c1T ρ2

(2,0)Tσ(1,2).

Replacing the above calculations in the expression of πvec(Kx ∗E′) and base changingback yields

A = v5T(1,0),y′T(1,1),y′+xT(1,1),y +v5

2 T(1,0),y′T(1,0),y′T(1,2),y+x +v6

2 T(2,1),x+2y′T(1,1),y

= v5

2 E(2,1)(x+2y′,1)⊕E

(1,1)(y,1) +

v2

((1+ v−2)E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,2)(y+x,1)+E

(2,0)(y′,2)⊕E

(1,2)(y+x,1)

)+ v5T(1,0),y′T(1,1),y′+xT(1,1),y.

Let A′ denote T(1,0),y′T(1,1),y′+xT(1,1),y, then

A′ =

v−2E(1,0)(y′,1)⊕E

(1,1)(y′+x,1)⊕E

(1,1)(y,1) if y′+ x 6= y,

v−2((1+ v−2)E

(1,0)(y′,1)⊕E

(1,1)(y,1) ⊕E

(1,1)(y,1) +E

(1,0)(y′,1)⊕E

(2,2)(y,2)

)if y′+ x = y.

Analogous for B, we obtain

B = v6

2 T(2,1),x+2y′T(1,1),y +v5

[2]T(2,0),y′T(1,2),y+x

= v5

2 E(2,1)(x+2y′,1)⊕E

(1,1)(y,1) +

v2

((1− v−2)E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,2)(y+x,1)+E

(2,0)(y′,2)⊕E

(1,2)(y+x,1)

).

Thereforev−3 πvec(Kx ∗E′) = v−4E

(1,0)(y′,1)⊕E

(1,0)(y′,1)⊕E

(1,2)(y+x,1)+ v2A′,

which proves our statement since v−2 = q.

Lemma 5.26. Let E′ := E(1,−2)(y′,1) ⊕E

(1,0)(y,1) ⊕E


(i) mx,1(E(1,−1)(y′+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) ,E

′)= 1;

(ii) mx,1(E(1,−2)(y′,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y+x,1),E

′)= q.


Proof. Similar to the proof of Lemma 5.25, we can write E′ in HX as follows:

E′ = v4 E(1,−2)(y′,1) ∗

(E(1,0)(y,1) ⊕E

(1,0)(y,1)

)= v6

2

(N−3

1 ∑ρ,ρ1,χ∈P1

ρ1(−y′)ρ(−y)χ(−y)T ρ1(1,−2)T

ρ(1,0)T

χ(1,0)

− 2N−11 N−1

2[2] ∑

ρ1∈P1ρ2∈P2

ρ1(−y)ρ2(−y)T ρ1(1,−2)T

ρ2(2,0)

).

Hence


2 ∑σ,ρ,ρ1,χ∈P1

σ(−x)ρ1(−y′)ρ(−y)χ(−y)[T σ(0,1),T

ρ1(1,−2)T

ρ(1,0)T

χ(1,0)

]− v6N−2

1 N−12

[2] ∑σ,ρ1∈P1ρ2∈P2

σ(−x)ρ1(−y)ρ2(−y)[T σ(0,1),T

ρ1(1,−2)T

ρ2(2,0)

].

Let us denote by A (resp. B) the first (resp. second) term in the above subtraction.Proceeding with the algorithm in [Alv20, Section 4] yields

A = v5

2

((1+ v−2)E

(1,−1)(y′+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +E

(1,−1)(y′+x,1)⊕E

(2,0)(y,2)

)+ v E(1,−2)

(y′,1) ⊕E(1,0)(y,1) ⊕E

(1,1)(y+x,1)+

v3

2 E(1,−2)(y′,1) ⊕E

(2,1)(2y+x,1)

and

B = v5

2 E(1,−1)(y′+x,1)⊕E

(2,0)(y,2) +

(v5−v3)2 E

(1,−1)(y′+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y,1) +

v3

2 E(1,−2)(y′,1) ⊕E

(2,1)(2y+x,1).

The proof follows from multiplying A−B by v−3.

Lemma 5.27. Let E′ := E(1,−1)(z,1) ⊕E

(1,−1)(z,1) ⊕E


y 6= z+ x. Then

(i) mx,1(E(1,−1)(z,1) ⊕E

(1,0)(z+x,1)⊕E

(1,0)(y,1) ,E

′)= 1;

(ii) mx,1(E(1,−1)(z,1) ⊕E

(1,−1)(z,1) ⊕E

(1,1)(y+x,1),E

′)= q2.

Proof. As in the proof of Lemma 5.21,


2 ∑σ,ρ,ρi∈P1

i=1,2

σ(−x)ρ1(−z)ρ2(−z)ρ(−y)[T σ(0,1),T

ρ1(1,−1)T

ρ2(1,−1)T

ρ(1,0)

]

− N−12 N−2

1 v4

[2] ∑σ,ρ∈P1χ∈P2

σ(−x)χ(−z)ρ(−y)[T σ(0,1),T

χ(2,−2)T

ρ(1,0)

].

Let A (resp. B) be the first (resp. second) term in the right side of above identity. SinceT(1,d),zT(1,d),z = (1+ v−2)E

(1,d)(z,1) ⊕E

(1,d)(z,1) +E

(2,2d)(z,2) in HX for any integer d,

A = v3E(1,−1)(z,1) ⊕E

(1,0)(z+x,1)⊕E

(1,0)(y,1) +

v5

2 E(2,−1)(2z+x,1)⊕E

(1,0)(y,1)

+ (v−1+v)2 E

(1,−1)(z,1) ⊕E

(1,−1)(z,1) ⊕E

(1,1)(y+x,1)+

v2E

(2,−2)(z,2) ⊕E

(1,1)(y+x,1)


by the algorithm in [Alv20, Section 4]. Moreover,

B = v5

2 E(2,−1)(2z+x,1)⊕E

(1,0)(y,1)

+ v2

((1− v−2)E

(1,−1)(z,1) ⊕E

(1,−1)(z,1) ⊕E

(1,1)(y+x,1)+

v2E

(2,−2)(z,2) ⊕E

(1,1)(y+x,1)

)since

T(2,−2),z := [2]2

((1− v−2)E

(1,−1)(z,1) ⊕E

(1,−1)(z,1) +E

(2,−2)(z,2)

)for |z|= 1.

Therefore we have

πvec(Kx ∗E′) = v3E(1,−1)(z,1) ⊕E

(1,0)(z+x,1)⊕E

(1,0)(y,1) + v−1E

(1,−1)(z,1) ⊕E

(1,−1)(z,1) ⊕E

(1,1)(y+x,1)

and the lemma follows from multiplying the above identity by v−3.

Lemma 5.28. Let E′ := E(1,−2)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,0)(y′,1) be a rank 3 vector bundle on X with

y 6= y′. Then

(i) mx,1(E(1,−1)(z+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y′,1),E

′)= 1;

(ii) mx,1(E(1,−2)(z,1) ⊕E

(1,0)(y′,1)⊕E

(1,1)(y+x,1),E

′)= q2;

(iii) mx,1(E(1,−2)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y′+x,1),E

′)= q2;

(iv) mx,1(E(1,−2)(z,1) ⊕E

(2,1)(y+y′+x,1),E

′)= q2.

Proof. Since y 6= y′, we can write in HX

E′ = v4N−31 ∑

ρi∈P1i=1,2,3

ρ1(−z)ρ2(−y)ρ3(−y′)T ρ1(1,−2)T

ρ2(1,0)T

ρ3(1,0).

Thus

πvec(Kx ∗E′) = v4N−41 ∑

σ,ρi∈P1i=1,2,3

σ(−x)ρ1(−z)ρ2(−y)ρ3(−y′)[T σ(0,1),T

ρ1(1,−2)T

ρ2(1,0)T

ρ3(1,0)

]= v5

(T(1,−1),x+zT(1,0),yT(1,0),y′+T(1,−2),zT(1,1),y+xT(1,0),y′

+T(1,−2),zT(1,0),yT(1,1),y′+x + vT(1,−2),zT(2,1),y+y′+x

)= v3E

(1,−1)(z+x,1)⊕E

(1,0)(y,1) ⊕E

(1,0)(y′,1)+ v−1E

(1,−2)(z,1) ⊕E

(1,0)(y′,1)⊕E

(1,1)(y+x,1)

+ v−1E(1,−2)(z,1) ⊕E

(1,0)(y,1) ⊕E

(1,1)(y′+x,1)+ v−1E

(1,−2)(z,1) ⊕E

(2,1)(y+y′+x,1).

We complete the proof multiplying above identity by v−3.

6. Degree 2 graphs

In this last section, we describe the ‘even component’ of G(2)y,1 and the neighborhood of

O⊕3X in G

(3)y,1 for a degree two place y of X . Our results are based on the computation

of the commutators of certain elements in twisted spherical Hall algebras, which weintroduce in the following.


6.1. Twisted spherical Hall algebras. Recall from section 5 that Pn stands for Pic0Xn

and for ρ ∈ Pn, ρ denotes its class on Pn := Pic0Xn/FrX ,n.

Definition 6.1. We call a character ρ ∈ Pn primitive of degree n if its orbit underthe Frobenius FrX ,n is of maximal cardinality, which is n. Equivalently, by [Fra13,Lemma 3.8], ρ is primitive if there does not exist a character χ ∈ Pm for m|n such thatρ= Normn

m(χ).

Note that ρ ∈ Pn is either primitive or that there is a primitive character χ ∈ Pm form|n such that ρ= Normn

m(χ), see also [Fra13, Cor. 3.9].

Definition 6.2. Let n> 1 and ρ ∈ Pn be a primitive character. One defines the algebraUρ

X as the subalgebra of HX that is generated byT Normnγ(v)

n (ρ)nv

∣∣ v ∈ Z,

where Z :=(m,n) ∈ Z2 | m > 0 or m = 0 and n > 0

and γ(v) := gcd(m,n) for v =

(m,n) ∈ Z. This subalgebra is called the twisted spherical Hall algebra of X and ρ.

For non-collinear v,w ∈ Z, we define εv,w := sign(det(v,w)) ∈ ±1. We denote by∆v,w the triangle formed by the vectors o,v,v+w where o = (0,0).

Definition 6.3. Fix α,β ∈ C∗ with α,β 6∈ ±1. We set ν := (αβ)−12 and

ci(α,β) := (αi/2−α−i/2)(βi/2−β−i/2)[i]ν/

i where [i]ν :=ν i−ν−i

ν−ν−1 .

Let Enα,β be the C-algebra generated by tv | v ∈ Z modulo the following relations:

(1) If o,v,v′ are collinear then [tv, tv′] = 0.(2) If v,w are such that γ(v) = 1 and that ∆v,w has no interior lattice point, then

[tw, tv] = εv,wcnγ(w)(α,β)θv+w

n(ν−1−ν)where the elements θz, with z∈Z are defined by the following generating series:

∑i>0

θiz0si = exp

(n(ν−1−ν)∑

i>1tiz0si

)for any z0 ∈ Z such that γ(z0) = 1.

Note that θz = n(ν−1−ν)tz whenever γ(z) = 1.

Given an elliptic curve X defined over Fq, by a theorem of Hasse (see [Sil09, Thm.2.3.1]) there exists conjugate algebraic numbers α,α satisfying αα= q, such that

Nm := #X(Fqm) = qm +1− (αm +αm)

for any m> 1. The numbers α,α are the eigenvalues of the Frobenius automorphismacting on H1

et(X ,Q`) for some prime number ` distinct from the characteristic of Fq andX := X×Spec(Fq) Spec(Fq) (cf. [Har77, App C] for further details). We observe that inthis case, ci(α,α) = ci as defined in subsection 5.1. The connection between the abovedefinition and the Hall algebra HX is given by the following theorem.


Theorem 6.4 ([Fra13, Thm. 5.2]). With the previous choice of α,α, for any two integersn,m> 1 and any two different primitive characters ρ ∈ Pn and σ ∈ Pm, we have:

(1) The twisted spherical Hall algebra UρX is isomorphic to En

α,α. This isomorphismis given by

tv 7−→ T Normnγ(v)n (ρ)

nv .

(2) The algebras UρX and Uσ

X commute with each other.(3) The Hall algebra HX decomposes into a commutative restricted tensor product

HX ∼=⊗ρ∈P

′ UρX

where P is the set of equivalence classes of primitive characters on⊔

n>1 Pn.

Finally we can state and prove a result that is fundamental to the computation of theedges for a degree 2 graph.

Lemma 6.5. Let ρ,σ ∈ P2 and ρ, σ its classes in P2. Then

[T σ(0,2),T

ρ(2,0)

]=

c2T ρ

(2,2) if σ is primitive and ρ= σ;

c22(v−1−v)2c2

T χ(1,1)T

χ(1,1)+

c22−2c1c2

c1T σ(2,2) if σ = ρ= Norm2

1(χ);

0 otherwise;

where χ ∈ P1 and the commutator is taken in the elliptic Hall algebra HX .

Proof. For the case that σ is primitive, we refer [Fra13, Section 8.4 - Step 2n]. If σis not primitive, then σ = Norm2

1(χ) for some χ ∈ P1. If ρ 6= σ, then these charactersbelong to different twisted spherical Hall subalgebras of HX and

[T σ(0,2),T

ρ(2,0)

]vanishes

by Theorem 6.4, part (2).If σ = ρ = Norm2

1(χ), then[T σ(0,2),T

ρ(2,0)

]can be calculated in Uσ

X . By the isomor-phism given in Theorem 6.4, part (1), we are left with computing

[t(0,2), t(2,0)

]in En

α,α.Since [

t(0,1), t(2,−1)]= c1

(t(2,0)+

(v−1−v)2 t(1,0)t(1,0)

),

we have [t(0,2), t(2,0)

]=[t(0,2),c

−11[t(0,1), t(2,−1)

]]− (v−1−v)

2

[t(0,2), t(1,0)t(1,0)

]= c−1

1[t(0,1),

[t(0,2), t(2,−1)

]]− c2(v−1− v)t(1,0)t(1,2)

− c1c2(v−1−v)

2

(t(2,2)+

(v−1−v)2 t(1,1)t(1,1)

).

Next we observe that[t(0,2), t(2,−1)

]= c2(v−1− v)t(1,0)t(1,1)+ c2([3]− vc1− 1)t(2,1).

Therefore [t(0,2), t(2,0)

]= c2

(c2c1−2)

t(2,2)+ c22(v−1−v)

2c1t(1,1)t(1,1).

Going back to HX via the isomorphism from Theorem 6.4 concludes the proof.


6.2. The even component of graphs of rank 2 and degree 2. We continue with adescription of the even component of the graph G

(2)y,1 for a closed point y of degree two.

Theorem 6.6. Let y ∈ |X | a degree two closed point. Then the even component of G(2)y,1

is given as follows.

(i) Let E := L1⊕L2 with Li ∈ PicX , i = 1,2. Suppose either degL2 > degL1 +2 ordegL2 = degL1 +2 and L1⊗OX(y) 6∼= L2, then

Vy,1(E) =(

E,L1(−y)⊕L2,1),(E,L1⊕L2(−y),q2).

(ii) Let E := L1⊕L2 with Li ∈ PicX , i = 1,2. Suppose degL2 = degL1 + 2 andL1⊗OX(y)∼= L2, then

Vy,1(E) =(

E,L1(−y)⊕L2,1),(E,L1⊕L2(−y),q

),(E,E

(2,2d1)(x,2) ,q2−q

),

where the Atiyah representation of L1 is E(1,d1)(x,1) .

(iii) Let E := L⊕L with L ∈ PicX, then

Vy,1(E) =(

E,L(−y)⊕L,q+1),(E,E

(2,2d−2)(y,1) ,q2−q

),

where d = degL.

(iv) Let E := L1⊕L2 with Li ∈ PicX , i = 1,2. Suppose degL2 = degL1, then

Vy,1(E) =(

E,L1(−y)⊕L2,1),(E,L1⊕L2(−y),1

),(E,E

(2,2d−2)(x,2) ,q−1

),(

E,E(2,2d−2)(y′,1) ,q−1

),

where x = x1 +x2−y, y′+y = x1 +x2 and the Atiyah representations of L1 and L2 areE(1,d)(x1,1)

and E(1,d)(x2,1)

, respectively.

(v) Let E := E(2,0)(x,2) , then

Vy,1(E) =(

E,E(1,−2)(x−y,1)⊕E

(1,0)(x,1),1

),(E,E

(1,−1)(x1,1)

⊕E(1,−1)(x2,1)

,q),(E,E

(2,−2)(x,2) ,q

),(

E,E(2,−2)(y′,1) ,q

),

where x1 6= x2 with x1 + x2 = x+ y and y′ 6= y with y′ = 2x− y.

(vi) Let E := E(2,0)(y,1) , then

Vy,1(E) =(

E,E(1,−1)(x0,1)

⊕E(1,−1)(x0,1)

,1),(E,E

(2,−2)(y,1) ,m1

),(E,E

(2,−2)(y′,1) ,m2

),(

E,E(1,−1)(x1,1)

⊕E(1,−1)(x2,1)

,q+1),

where y′ 6= y and x1 6= x2 with x1 + x2 = 0. Let z,z′ ∈ |X2| above y and w,w′ ∈ |X2|above y′, the multiplicities m1, m2 are given as follows

m1 =

q+1 if 3z 6= 01 if 3z = 0

and m2 =

q+1 if w 6= 2z and w 6= 2z′

1 if w = 2z or w = 2z′.


(vii) Let E := E(2,0)(y′,1) with y′ 6= y, then

Vy,1(E) =(

E,E(2,−2)(y′′,1) ,1

),(E,E

(2,−2)(y′,1) ,m1

),(E,E

(2,−2)(y,1) ,m2

),(E,E

(2,−2)(x,2) ,q+1

),(

E,E(1,−1)(x1,1)

⊕E(1,−1)(x2,1)

,q+1),

where y,y′,y′′ are pairwise distinct, x = y− y′ and x1 6= x2 with x1 + x2 = y′− y. More-over if z,z′ ∈ |X2| sit above y, w,w′ ∈ |X2| sit above y′ and v,v′ ∈ |X2| sit above y′′, theny′′ is such that either v = z+w or v = z+w′ and

m1 =

1 if z = 2w or z = 2w′

q+1 if z 6= 2w and z 6= 2w′and m2 =

1 if w = 2z or w = 2z′

q+1 if w 6= 2z and w 6= 2z′.

Proof. Item (i) follows from [Alv20, Thm. 5.2] and Lemma 6.9. Item (ii) follows fromLemma 6.10, [Lor13a, Thm. 8.5] and Lemma 6.11. Applying either Lemma 6.8 or[Lor13a, Thm. 8.5] coupled with either Lemma 6.12 or [Lor12, Cor. 6.5] give item(iii). Item (iv) follows from Lemmas 6.8, 6.11, 6.12 and 6.13. Item (v) follows fromLemmas 6.8, 6.9, 6.11 and 6.13. Item (vi) follows from Lemmas 6.10, 6.12, 6.13 and6.9. Finally, item (vii) follows from Lemmas 6.13. 6.12, 6.11 and 6.9.

Before we state and prove the lemmas implying this theorem, we provide twoexamples that clarify the difference in the statement of the neighbors of traces bundles.

Example 6.7. Let E2 and E3 be the elliptic curves as in the Example 4.7 with only onerational point i.e. Eq(Fq) = x for q = 2,3. We denote by y1, . . . ,yq the q closed pointsof Eq of degree 2 and

Eq(Fq2) =

x,z1,z′1,z2,z′2, . . . ,zq,z′q

the 2q+ 1 rational points of Eq(Fq2) prescribed such that zi,z′i sit above yi for i =1, . . . ,q. Moreover, since Cl0(Eq⊗Fq2) ' Z/(2q+ 1)Z, we can suppose nz1 = zn inCl0(Eq⊗Fq2), for n = 1, . . . ,q and q = 2,3.

We fix y = y1 and adopt the following notation for the vertices of G(2)y,1

cnx := OEq⊕OEq(nx), tyi := E(2,0)(yi,1)

, s0 := E(2,0)(x,2) and sx := E

(2,1)(x,1)

for i = 1, . . . ,q. Hence

PBundec2 Eq =

cnx

n∈N, PBungi2 Eq =

s0,sx

, PBuntr

2 Eq =

ty, ty2, . . . , tyq

.

Therefore the even component of the graph G(2)y,1 over E2 is given by

ty

ty2

c0

s0

c2 c4· · ·

3

1

1

1

3

1

2

1

2

23

2

21 4 1


and the even component of the graph G(2)y,1 over E3 is

ty2

ty3

ty

s0

c0 c2 c4· · ·

11

4

4

1

1

4

4

1 4

4

1

3

3

1

3

64 3

6

1 9 1

We conclude the example with the following observation. The fact that there existsan edge from E′ to E whenever there exists an edge from E to E′ in above graphsis not a coincidence. Indeed, let X be any elliptic curve, for every place x ∈ |X | ifthere exists an edge (E,E′,m) ∈ Edge G

(2)x,1 , then there is a positive m′ ∈ Z× such that

(E′,E,m′) ∈ Edge G(2)x,1 . This follows from Theorem 2.3. See also [Lor13a, section 3]

for a different approach and a connection with Bruhat-Tits trees.

6.3. Proof of Theorem 6.6. The proof of this section’s main theorem is based on thealgorithm developed in [Alv20, Sec. 4] and is given by the following lemmas.

Lemma 6.8. Let E′ := E(1,−2)(x1,1)

⊕E(1,0)(x2,1)

be a rank 2 vector bundle on X. Then,

(i) my,1(E(1,0)(x1+y,1)⊕E

(1,0)(x2,1)

,E′)= 1, if x2 6= x1 + y;

(ii) my,1(E(1,0)(x2,1)

⊕E(1,0)(x2,1)

,E′)= q+1, if x2 = x1 + y;

(iii) my,1(E(2,0)(x2,2)

,E′)= 1, if x2 = x1 + y;

(iv) my,1(E(1,−2)(x1,1)

⊕E(1,2)(x2+y,1),E

′)= q2.

Proof. We first observe that

E′ = v2E(1,−2)(x1,1)

∗E(1,0)(x2,1)

= v2N−21 ∑

ρ1,ρ2∈P1

ρ1(−x1)ρ2(−x2)Tρ1(1,−2)T

ρ2(1,0)

andKy = 2N−2

2 [2]−1∑σ∈P2

σ(−y) T σ(0,2).

Thus,

πvec(Ky ∗E′) = 2v2N−21

[2]N2 ∑ρ1,ρ2∈P1σ∈P2

σ(−y)ρ1(−x1)ρ2(−x2)[T σ(0,2),T

ρ1(1,−2)T

ρ2(1,0)

]= v4T(1,0),x1+yT(1,0),x2 + v4T(1,−2),x1T(1,2),x2+y.

Moreover,

T(1,0),x1+yT(1,0),x2 =

E(1,0)(x1+y,1)⊕E

(1,0)(x2,1)

if x2 6= x1 + y

(1+ v−2)E(1,0)(x2,1)

⊕E(1,0)(x2,1)

+E(2,0)(x2,2)

if x2 = x1 + y,


andT(1,−2),x1T(1,2),x2+y = v−4E

(1,−2)(x1,1)

⊕E(1,2)(x2+y,1).

The proof follows by multiplying πvec(Ky ∗E′) by v−4.

Lemma 6.9. Let E′ := E(1,0)(x1,1)

⊕E(1,0)(x2,1)

be a rank 2 vector bundle on X with x1 6= x2.Then,

(i) my,1(E(1,2)(x1+y,1)⊕E

(1,0)(x2,1)

,E′)= q2;

(ii) my,1(E(1,0)(x1,1)

⊕E(1,2)(x2+y,1),E

′)= q2;

(iii) my,1(E(2,2)(w,2),E

′)= q if w = x1 + x2 + y;

(iv) my,1(E(2,2)(y′,1),E

′)= q+1 if y′ = x1 + x2 + y;

(v) my,1(E(1,1)(x′1,1)

⊕E(1,1)(x′2,1)

,E′)= q−1 if x′1 + x′2 = x1 + x2 + y and x′1 6= x′2.

Proof. Since x1 6= x2, we can write E′ in the Hall algebra HX as follows

E′ = E(1,0)(x1,1)

∗E(1,0)(x2,1)

= N−21 ∑

ρ1,ρ2∈P1

ρ1(−x1)ρ2(−x2)Tρ1(1,0)T

ρ2(1,0).

Hence,

πvec(Ky ∗E′) = 2N−21

[2]N2

(c2 ∑

ρ1,ρ2∈P1σ=Norm2

1(ρ1)

σ(−y)ρ1(−x1)ρ2(−x2)Tρ1(1,2)T

ρ2(1,0)

+ c2 ∑ρ1,ρ2∈P1

σ=Norm21(ρ2)

σ(−y)ρ1(−x1)ρ2(−x2)Tρ1(1,0)T

ρ2(1,2)

)

= v2T(1,0),x2T(1,2),x1+y + v2T(1,0),x1T(1,2),x2+y + v3∑

w∈|X |w=x1+x2+y

T(2,2),w

+ (v2−v4)2 ∑

x′1,x′2∈|X |

x′1+x′2=x1+x2+y

T(1,1),x′1T(1,1),x′2.

We conclude the proof by observing that

T(1,0),x2T(1,2),x1+y = v−2E(1,2)(x1+y,1)⊕E

(1,0)(x2,1)

andT(1,0),x1T(1,2),x2+y = v−2E

(1,0)(x1,1)

⊕E(1,2)(x2+y,1).

Moreover

T(1,1),x′1T(1,1),x′2 =

(1+ v−2)E(1,1)(x′1,1)

⊕E(1,1)(x′1,1)

+E(2,2)(x′1,2)

if x′1 = x′2

E(1,1)(x′1,1)

⊕E(1,1)(x′2,1)

if x′1 6= x′2


and

T(2,2),w =

E(2,2)(w,2)+(1− v−2)E

(1,1)(w,1)⊕E

(1,1)(w,1) if |w|= 1

[2]E(2,2)(w,1) if |w|= 1.

Putting all together and multiplying by v−4 yields the desire statement.

Lemma 6.10. Let E′ := E(1,0)(x,1)⊕E

(1,0)(x,1) be a rank 2 vector bundle on X. Then,

(i) my,1(E(1,0)(x,1)⊕E

(1,2)(x+y,1),E

′)= q;

(ii) my,1(E(2,2)(y,1) ,E

′)= 1.

Proof. As in [Alv20, Lemma 6.6] we can write E′ in terms of elements in the twistedspherical Hall algebras as follows

E′ = v2N−21

2 ∑ρ,ρ1∈P1

ρ(−x)ρ1(−x)T ρ(1,0)T

ρ1(1,0)−

v2N−21

[2] ∑ρ2∈P2

ρ2(−x)T ρ2(2,0).

Thus,

πvec(Ky ∗E′) = v2N−21

[2]N2 ∑ρ,ρ1∈P1σ∈P2

σ(−y)ρ(−x)ρ1(−x)[T σ(0,2),T

ρ(1,0)T

ρ1(1,0)

]− 2v2

[2]2N22

∑σ,ρ2∈P2

σ(−y)ρ2(−x)[T σ(0,2),T

ρ2(2,0)

].

As usual, we denote the first term in the above sum by A and the second term by B. Letus calculate each term separated. Proceeding as previous lemma we conclude that

A = v2E(1,0)(x,1)⊕E

(1,2)(x+y,1)+

(v6+v4)2 E

(2,2)(y′,1)+

v4

2 E(2,2)(x′,2)+

(v4−v6)2 E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

where |y′|= 2 with y′ = y+2x, |x′|= 1 with 2x′ = y+2x and x1 6= x2 with x1 + x2 =y+2x.

From Lemma 6.5,

B = 2v2

[2]2N22

(∑χ∈P1

Norm21(χ)(−y−x)

(c2

2(v−1−v)2c1

T χ(1,1)T

χ(1,1)+

(c22

c1−2c2

)T Norm2

1(χ)

(2,2)

)−2c2 ∑

σ∈P2primitive

σ(−y− x)T σ(2,2)

)

= ∑w∈|X |

T(2,2),w(

2v2

[2]2N22

)(c2

2c1 ∑χ∈P1

Norm21(χ)(w−y−x)−2c2 ∑

σ∈P2

σ(w− y− x)

)+ (v4−v6)

4 ∑x′1,x

′2∈|X |

T(1,1),x′1T(1,1),x′2

where x′1 + x′2 = y+2x. Since

∑σ∈P2

σ(w− y− x) =

N2/2 if w = y0 otherwise


we conclude that

B = (v4−v6)2 E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

+ v4

2 E(2,2)(x′,2)+

(v6+v4)2 E

(2,2)(y′′,1)+

(v6−v4)2 E

(2,2)(y,1)

where y′′ 6= y. Therefore,

πvec(Ky ∗E′) = v2E(1,0)(x,1)⊕E

(1,2)(x+y,1)+ v4E

(2,2)(y,1)

and we have the desired statement multiplying above equation by v−4.

Lemma 6.11. Let E′ := E(2,0)(x,2) be a rank 2 vector bundle on X. Then,

(i) my,1(E(2,2)(x′,2),E

′)= q if 2x′ = y+2x;

(ii) my,1(E(1,1)(x1,1)

⊕E(1,1)(x2,1)

,E′)= q−1 for x1 6= x2 and x1 + x2 = y+2x;

(iii) my,1(E(2,2)(y′,1),E

′)= q+1 for y′ 6= y and y′ = y+2x;

(iv) my,1(E(1,0)(x,1)⊕E

(1,2)(x+y,1),E

′)= q2−q.

Proof. As in [Alv20, Lemma 6.5]

E′ = (v2+1)N2[2] ∑

ρ∈P2

ρ(−x)T ρ(2,0)+

(1−v2)

2N21

∑ρ1,ρ2∈P1

ρ1(−x)ρ2(−x)T ρ1(1,0)T

ρ2(1,0).

Thus,

πvec(Ky ∗E′) = (1−v2)N−21

[2]N2 ∑ρ1,ρ2∈P1σ∈P2

σ(−y)ρ1(−x)ρ2(−x)[T σ(0,2),T

ρ1(1,0)T

ρ2(1,0)

]+ 2(v2+1)

[2]2N22

∑σ,ρ∈P2

σ(−y)ρ(−x)[T σ(0,2),T

ρ(2,0)

].

Let A (resp. B) denote the first (resp. second) term in the above sum. Analogous to theprevious lemma, applying the algorithm in [Alv20] yields

A=(1−v2)E(1,0)(x,1)⊕E

(1,2)(x+y,1)+

(v2−v6)2 E

(2,2)(y′′,1)+

(v2−v4)2 E

(2,2)(x′,2)+

(v2−2v4+v6)2 E

(1,1)(x1,1)⊕E(1,1)

(x2,1)

where |y′′|= 2 with y′′ = y+2x, |x′|= 1 with 2x′ = y+2x and x1 6= x2 with x1 + x2 =y+2x. On the other hand, Lemma 6.5 implies,

B = ∑w∈|X |

T(2,2),w(

2(v2+1)[2]2N2

2

)(c2

2c1 ∑χ∈P1

Norm21(χ)(w−y−x)−2c2 ∑

σ∈P2

σ(w− y− x)

)+ (v2−v6)

4 ∑x′1,x

′2∈|X |

T(1,1),x′1T(1,1),x′2

where x′1 + x′2 = y+2x. As in the proof of previous lemma we observe

∑σ∈P2

σ(w− y− x) =

N2/2 if w = y0 otherwise

in order to conclude that

B = (v2+v4)2 E

(2,2)(x′,2)+

(v2−v6)2 E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

+ (v6−v2)2 E

(2,2)(y,1) +

(v6+2v4+v2)2 E

(2,2)(y′,1)


where x′,x1,x2 are given as before and y′ is such that y′ = y+2x and y′ 6= y. The lemmafollows by multiplying v−4 in the sum A+B.

Lemma 6.12. Let E′ := E(2,0)(y,1) be a rank 2 vector bundle on X and z,z′ ∈ |X2| siting

above y. Then,

(i) my,1(E(1,1)(x,1)⊕E

(1,1)(x,1),E

′)= q2−q, where 2x = 2y;

(ii) my,1(E(1,1)(x1,1)

⊕E(1,1)(x2,1)

,E′)= q−1, where x1 6= x2 and x1 + x2 = 2y;

(iii) my,1(E(2,2)(y,1) ,E

′)= m1, where m1 = q+1 if 3z 6= 0 and m1 = 1 if 3z = 0;

(iv) my,1(E(2,2)(y′,1),E

′)= m2, where m2 = 1 if either w = 2z or w = 2z′ and m2 = q+1otherwise . Whence w ∈ |X2| sits above y′ and y′ 6= y.

Proof. As practice, we start by writing E′ in terms of elements in the twisted sphericalHall algebras,

E′ = 2N2[2] ∑

ρ∈P2

ρ(−y)T ρ(2,0).

Hence,

πvec(Ky ∗E′) = 4[2]2N2

2∑

σ,ρ∈P2

σ(−y)ρ(−y)[T σ(0,2),T

ρ(2,0)

]= 4

[2]2N22

∑σ,ρ∈P2

ρ primitive

σ(−y)ρ(−y)[T σ(0,2),T

ρ(2,0)

]

+ 4[2]2N2

2∑

σ∈P2,χ∈P1ρ=Norm2

1(χ)

σ(−y)ρ(−y)[T σ(0,2),T

ρ(2,0)

].

Lemma 6.5 yields

πvec(Ky ∗E′) = v4(v−1−v)2c1 ∑

χ∈P1

∑x′1,x

′2∈|X |

χ(x′1 + x′2)Norm21(χ)(−2y)T(1,1),x′1T(1,1),x′2

+ ∑x′∈|X |

T(2,2),x′(

4[2]2N2

2

)(c2

2c1 ∑χ∈P1

Norm21(χ)(x

′−2y)−2c2 ∑σ∈P2

σ(x′−2y)

).

Let y′ ∈ |X | be a degree two closed point distinct from y. Let z,z′ ∈ |X2| siting above yand w,w ∈ |X2| siting above y′. A straightforward calculation shows that

∑σ∈P2

σ(x′−2y) =

N2/2 if |x′|= 1N2/4 if x′ = y′ and either w = 2z or w = 2z′

N2/4 if x′ = y and 3z = 00 otherwise.

Therefore,

πvec(Ky ∗E′) = (1− v2)E(1,1)(x,1)⊕E

(1,1)(x,1) +(v2− v4)E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

+m′1E(2,2)(y,1) +m′2E

(2,2)(y′,1)


for 2x = 2y, x1 6= x2 with x1 + x2 = 2y and y′ 6= y. Where

m′1 =

v2 + v4 if 3z 6= 0v4 if 3z = 0

and m′2 =

v4 if either w = 2z or w = 2z′

v4 + v2 otherwise

with z,z′,w,w′ ∈ |X2| are as above. Multiplying πvec(Ky ∗E′) by v−4 yields the desiredstatement.

Lemma 6.13. Let E′ :=E(2,0)(y′,1) be a rank 2 vector bundle on X with y′ 6= y. Let z,z′ ∈ |X2|

siting above y and w,w′ ∈ |X2| siting above y′. Then,

(i) my,1(E(2,2)(x,2),E

′)= q, where 2x = y+ y;

(ii) my,1(E(1,1)(x1,1)

⊕E(1,1)(x2,1)

,E′)= q−1, where x1 6= x2 and x1 + x2 = y+ y′;

(iii) my,1(E(2,2)(y′′,1),E

′)= 1, where y,y′,y′′ are pairwise distinct;

(iv) my,1(E(2,2)(y,1) ,E

′) = m1, where m1 = 1 if w = 2z or w = 2z′ and m1 = q + 1otherwise;

(v) my,1(E(2,2)(y′,1),E

′)= m2, where m2 = 1 if either z = 2w or z = 2w′ and m2 = q+1otherwise.

Proof. We proceed mutatis mutandis last lemma. Hence,

πvec(Ky ∗E′) = 4[2]2N2

2∑

σ,ρ∈P2

σ(−y)ρ(−y′)[T σ(0,2),T

ρ(2,0)

]= 4

[2]2N22

∑σ,ρ∈P2

ρ primitive

σ(−y)ρ(−y′)[T σ(0,2),T

ρ(2,0)

]

+ 4[2]2N2

2∑

σ∈P2,χ∈P1ρ=Norm2

1(χ)

σ(−y)ρ(−y′)[T σ(0,2),T

ρ(2,0)

]

= v4(v−1−v)2c1 ∑

χ∈P1

∑x′1,x

′2∈|X |

χ(x′1 + x′2)Norm21(χ)(−y−y′)T(1,1),x′1T(1,1),x′2

+ ∑x′∈|X |

T(2,2),x′(

4[2]2N2

2

)(c2

2c1 ∑

χ∈P1

Norm21(χ)(x

′−y−y′)−2c2 ∑σ∈P2

σ(x′− y− y′)).

where the last equality is given by Lemma 6.5. Splitting above sums in all possibilitiesfor x′1,x

′2,x′ and observing

∑σ∈P2

σ(x′− y− y′) =

N2/4 if |x′|= 2 and x′,y,y′ are pairwise distinctN2/4 if x′ = y and either w = 2z or w = 2z′

N2/4 if x′ = y′ and either z = 2w or z = 2w′

0 otherwise,

we conclude

πvec(Ky ∗E′) = v2E(2,2)(x,2) +(v2− v4)E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

+ v4E(2,2)(y′′,1)+m′1E

(2,2)(y,1) +m′2E

(2,2)(y′,1).


where 2x = y+ y, x1 6= x2 with x1 + x2 = y+ y′, y′′ 6= y′, y′′ 6= y,

m′1 =

v4 if either w = 2z or w = 2z′

v4 + v2 otherwise

and

m′2 =

v4 if either z = 2w or z = 2w′

v4 + v2 otherwise.

Multiplying πvec(Ky ∗E′) by v−4 yields the proof.

6.4. Some edges for graphs of rank 3 and degree 2. Let y be a place of degree two.In the following, we finish the first part of this project by determining some edges andgive a complete description of the Φy,1-neighborhood of OX ⊕OX ⊕OX in G

(3)y,1 .

Lemma 6.14. Let E′ := E(1,−2)(x,1) ⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1) be a rank 3 vector bundle on X. Then

(i) my,1(E(1,0)(x+y,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1),E

′)= 1, if x+ y 6= x′;

(ii) my,1(E(1,0)(x′,1)⊕E

(2,0)(x′,1),E

′)= 1, if x+ y = x′;

(iii) my,1(E(1,0)(x′,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1),E

′)= q2 +q+1, if x+ y = x′;

(iv) my,1(E(1,−2)(x,1) ⊕E

(1,0)(x′,1)⊕E

(1,2)(x′+y,1),E

′)= q3;

(v) my,1(E(1,−2)(x,1) ⊕E

(2,2)(y,1) ,E

′)= q2.

Moreover, there is no other edge with target to E′ in G(3)y,1 .

Proof. Analogous to the proofs of the lemmas in the former section, the proof is basedon the algorithm in [Alv20] to compute the product Ky ∗E′ in the Hall algebra HX of X .Since y is a closed point of degree two, its skyscraper sheaf is given by

Ky = 2N−22 [2]−1

∑σ∈P2

σ(−y) T σ(0,2)

in HX . On the other hand, we can write E′ = v4 E(1,−2)(x,1) ∗

(E(1,0)(x′,1)⊕E

(1,0)(x′,1)

)as

v6N−31

2 ∑χ,ρ,ρ1∈P1

χ(−x)ρ(−x′)ρ1(−x′)T χ(1,−2)Tρ(1,0)T

ρ1(1,0)−

v6N−11

[2]N2 ∑χ∈P1ρ2∈P2

χ(−x)ρ2(−x′)T χ(1,−2)Tρ2(2,0)

in terms of generators of the Hall algebra of X . Hence

πvec(Ky ∗E′) = v6N−31

N2[2] ∑χ,ρ,ρ1∈P1σ∈P2

σ(−y)χ(−x)ρ(−x′)ρ1(−x′)[T σ(0,2),T

χ(1,−2)T

ρ(1,0)T

ρ1(1,0)

]

− 2v6N−11

N22 [2]

2 ∑χ∈P1σ,ρ2∈P2

σ(−y)χ(−x)ρ2(−x′)[T σ(0,2),T

χ(1,−2)T

ρ2(2,0)

].

We denote by A the first term in the above sum and by B the second one. In order tocalculate A, observe[

T σ(0,2),Tχ(1,−2)T

ρ(1,0)T

ρ1(1,0)

]=[T σ(0,2),T

χ(1,−2)

]T ρ(1,0)T

ρ1(1,0)+T χ

(1,−2)

[T σ(0,2),T

ρ(1,0)

]T ρ1(1,0)


+T χ(1,−2)T

ρ(1,0)

[T σ(0,2),T

ρ1(1,0)

].

Moreover, [T σ(0,2),T

χ(1,−2)

]T ρ(1,0)T

ρ1(1,0) = c2 T χ

(1,0)Tρ(1,0)T

ρ1(1,0)

if σ = Norm21(χ),

T χ(1,−2)

[T σ(0,2),T

ρ(1,0)

]T ρ1(1,0) = c2 T χ

(1,−2)Tρ1(1,0)T

ρ(1,2)

if σ = Norm21(ρ) and ρ 6= ρ1,

T χ(1,−2)

[T σ(0,2),T

ρ(1,0)

]T ρ1(1,0) = c1c2 T χ

(1,−2)

(T ρ(2,2)+

(v−1−v)2 T ρ

(1,1)Tρ(1,1)

)if σ = Norm2

1(ρ) with ρ= ρ′, and

T χ(1,−2)T

ρ(1,0)

[T σ(0,2),T

ρ1(1,0)

]= c2 T χ

(1,−2)Tρ(1,0)T

ρ1(1,2)

if σ = Norm21(ρ1). Hence

A = v8

2 ∑x′′∈|X |

x′′=x+y

T(1,0),x′′T(1,0),x′T(1,0),x′+ v8 T(1,−2),xT(1,0),x′T(1,2),x′+y

+ v9

2 ∑w∈|X |

2w=2x′+y

T(1,−2),xT(2,2),w + (v8−v10)4 ∑

x1,x2∈|X |x1+x2=2x′+y

T(1,−2),xT(1,1),x1T(1,1),x2

Considering all the possibilities for the above sums, we have

A = v8

2 E(1,0)(x′′,1)⊕E

(2,0)(x′,2)+

v8(1+v−2)2 E

(1,0)(x′′,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1)+

(v4+v2)2 E

(1,−2)(x,1) ⊕E

(2,2)(w,1)

+ (v8+2v6+2v4+v2)2 E

(1,0)(x′,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1)+

(v8+2v6)2 E

(1,0)(x′,1)⊕E

(2,0)(x′,2)+

v8

2 E(3,0)(x′,3)

+E(1,−2)(x,1) ⊕E

(1,0)(x′,1)⊕E

(1,2)(x′+y,1)+

v2

2 E(1,−2)(x,1) ⊕E

(2,2)(w′,2)+

(v2−v4)2 E

(1,−2)(x,1) ⊕E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

where x′′ = x+ y 6= x′, 2w = 2x′+ y, 2w′ = 2x′+ y and x1 6= x2 with x1 + x2 = 2x′+ y.Now we turn our attention to calculate B. By Theorem 6.4, we have[

T σ(0,2),Tχ(1,−2)T

ρ2(2,0)

]=[T σ(0,2),T

χ(1,−2)

]T ρ2(2,0)+T χ

(1,−2)

[T σ(0,2),T

ρ2(2,0)

]= c2T χ

(1,0)Tρ2(2,0)+T χ

(1,−2)

[T σ(0,2),T

ρ2(2,0)

]if σ = Norm2

1(χ). We use Lemma 6.5 and the algorithm from [Alv20] to conclude that

B = v8

2 E(1,0)(x′′,1)⊕E

(2,0)(x′,2)+

v8(1−v−2)2 E

(1,0)(x′′,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1)+

v2

2 E(1,−2)(x,1) ⊕E

(2,2)(w′,2)

+ v8

2 E(1,0)(x′,1)⊕E

(2,0)(x′,2)+

v8

2 E(3,0)(x′,3)+

(v2−v4)2 E

(1,−2)(x,1) ⊕E

(1,1)(x1,1)

⊕E(1,1)(x2,1)

+ (v8−v2)2 E

(1,0)(x′,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1)+

(v4+v2)2 E

(1,−2)(x,1) ⊕E

(2,2)(y′,1)+

(v4−v2)2 E

(1,−2)(x,1) ⊕E

(2,2)(y,1)

where y′ 6= y. Subtracting B from A and multiplying the outcome by v−6 completes theproof.


Let x ∈ |X | and x′ ∈ |Xm| be such that x′ lies above x, i.e. π(x′) = x. We rememberthat there are gcd(m, |x|) closed points in |Xm| lying above x and |x′|= |x|/gcd(m, |x|).

Theorem 6.15. Let y ∈ |X | of degree two and E := E(1,0)(x,1) ⊕E

(1,0)(x,1) ⊕E

(1,0)(x,1) for some

closed point x in X of degree one. Then

Vy,1(E) =(

E,E(1,−2)(x−y,1)⊕E

(1,0)(x,1)⊕E

(1,0)(x,1),q

2 +q+1),(E,E

(2,−2)(y,1) ⊕E

(1,0)(x,1),q

4−q)

is the Φy,1-neighborhood of E in G(3)y,1 .

Proof. Let X2 := X⊗Fq2 be the degree two constant extension of X . Let x′ ∈ |X2| lyingabove x, i.e. π(x′) = x, where π : X2→ X is the constant extension map. Note that x′ isunique and |x′|= 1. By [Lor13a, Lemma 6.5] π∗ restricts to an injective map from theset of traces and decomposable bundles on X to the set of decomposable bundles on X2.Moreover,

π∗(E(1,0)(x,1)⊕E

(1,0)(x,1)⊕E

(1,0)(x,1)

)= E

(1,0)(x′,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1).

Suppose E,E′ ∈ Bun3 X are such that my,1(E,E′) 6= 0, i.e. there exists a short exact

sequence0−→ E′ −→ E−→Ky −→ 0.

Since extension of constants is an exact functor, we still have an exact sequence whenwe extend the scalars to Fq2

0−→ π∗(E′)−→ π∗(E)−→Kz⊕Kz′ −→ 0(1)

where z,z′ are the two closed points of X2 lying above y. The above short exact sequence(1) splits into two exact sequences

0→ E′′→ π∗(E)→Kz→ 0 and 0→ π∗(E′)→ E′′→Kz′ → 0

where E′′ ∈ Bun3 X2 is the kernel of π∗(E)→Kz.

Hence for every(E,E′,m

)in Edge G

(3)y,1 , there are a vector bundle E′′ ∈ Bun3 X2 and

edges(E,E′′,m′

)in Edge G

(3)z,1 and

(E′′,E′,m′′

)in Edge G

(3)z′,1 .

From Theorem 4.6 item (xvii) (or directly Lemma 5.20) we have only one option forE′′ in G

(3)z,1 , namely E′′ = E

(1,−1)(x′−z,1)⊕E

(1,0)(x′,1)⊕E

(1,0)(x′,1). Since z 6= z′, item (xxi) of our main

theorem (or directly Lemma 5.26) implies that E′′ has two neighbors in G(3)z′,1 . Therefore

E must have at most two neighbors in G(3)y,1 over X .

Lemma 6.14 shows that

my,1(E,E

(1,−2)(x−y,1)⊕E

(1,0)(x,1)⊕E

(1,0)(x,1)

)= q2 +q+1,

i.e. E(1,−2)(x−y,1)⊕E

(1,0)(x,1)⊕E

(1,0)(x,1) is one of the two neighbors of E in G

(3)y,1 . It follows from

[Lor12, Prop. 6.4] that my,1(E(1,0)(x,1)⊕E

(1,0)(x,1),E

(2,−2)(y,1)

)is non-zero in G

(2)y,1 . Thus,

Vy,1(E) =(

E,E(1,−2)(x−y,1)⊕E

(1,0)(x,1)⊕E

(1,0)(x,1),q

2 +q+1),(E,E

(2,−2)(y,2) ⊕E

(1,0)(x,1),m

)for some positive integer m. By Theorem 1.2, the sum over all the multiplicities of edgesoriginated in a vertex of G

(3)y,1 equals the number of rational points on the Grassmannian


Gr(2,3) over the residual field of y. Therefore m = q4 − q, which completes theproof.

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Roberto Alvarenga, Instituto de Ciências Matemáticas e de Computação - USP, São Carlos, BrazilEmail address: [email protected]

Oliver Lorscheid, University of Groningen, the Netherlands, and IMPA, Rio de Janeiro, BrazilEmail address: [email protected]

Valdir Pereira Júnior, Instituto Nacional de Matemática Pura e Aplicada, Rio de Janeiro, BrazilEmail address: [email protected]

arXiv:2107.08375v1 [math.NT] 18 Jul 2021

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