CLASSIFICATION OF WEIGHTED DUAL GRAPHS WITH ONLY … · 2018-11-16 · CLASSIFICATION OF WEIGHTED DUAL GRAPHS WITH ONLY COMPLETE INTERSECTION SINGULARITIES STRUCTURES FAN CHUNG, YI-JING

TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 361, Number 7, July 2009, Pages 3535–3596S 0002-9947(09)04524-3Article electronically published on March 4, 2009

CLASSIFICATION OF WEIGHTED DUAL GRAPHSWITH ONLY COMPLETE INTERSECTION

SINGULARITIES STRUCTURES

FAN CHUNG, YI-JING XU, AND STEPHEN S.-T. YAU

Dedicated to Henry Laufer on the occasion of his 65th birthday

Abstract. Let p be normal singularity of the 2-dimensional Stein space V .Let π : M → V be a minimal good resolution of V , such that the irreduciblecomponents Ai of A = π−1(p) are nonsingular and have only normal crossings.Associated to A is weighted dual graph Γ which, along with the genera ofthe Ai, fully describes the topology and differentiable structure of A and thetopological and differentiable nature of the embedding of A in M . In this paper

we give the complete classification of weighted dual graphs which have onlycomplete intersection singularities but no hypersurface singularities associatedto them. We also give the complete classification of weighted dual graphswhich have only complete intersection singularities associated with them.

1. Introduction

Let p be a normal singularity of the 2-dimensional Stein space V . Let π : M →V be a resolution of V such that the irreducible components Ai, 1 ≤ i ≤ n,of A = π−1(p) are nonsingular and have only normal crossings. Associated toA is a weighted dual graph Γ (e.g., see [HNK] or [La1]) which, along with thegenera of the Ai, fully describes the topology and differentiable structure of Aand the topological and differentiable nature of the embedding of A in M . Oneof the famous important questions in normal two-dimensional singularities asks:What conditions are imposed on the abstract topology of (V, p) by the completeintersection hypothesis? Recall a theorem of Milnor [Mi, Theorem 2, p. 18] thatessentially says that any isolated singularity is a cone over its link L which isthe intersection of V with a small sphere centered at p. L is a compact real 3-manifold whose oriented homeomorphism type determines and is determined bythe weighted dual graph Γ of a canonically determined resolution (cf. [Ne]). So, wemay equivalently ask: What conditions will the existence of a complete intersectionrepresentative (V, p) put on a weighted dual graph Γ? A complete intersectionsingularity (V, p) is Gorenstein [Ba], [Gr-Ri]. So there exists an integral cycle K onΓ which satisfies the adjunction formula [Se]. The purpose of this paper is to givea complete classification of those weighted dual graphs which have only completeintersection singularities associated to them.

Received by the editors March 2, 2007.2000 Mathematics Subject Classification. Primary 32S25, 58K65, 14B05.The third author’s research was partially supported by an NSF grant.

c©2009 American Mathematical Society

3535

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3536 FAN CHUNG, YI-JING XU, AND STEPHEN S.-T. YAU

M. Artin has studied the rational singularities (those for whichR1π∗(O) = 0). It is well known that rational complete intersection singularitiesare hypersurfaces (cf. Theorem 4.3 below). Artin has shown that all hypersurfacerational singlarities have multiplicities two and the graphs associated with thosesingularities are one of the graphs Ak, k ≥ 1; Dk, k ≥ 4; E6, E7 and E8 whicharise in the classification of simple Lie groups. In [La4], Laufer examines a class ofelliptic singularities which satisfy a minimality condition. These minimally ellipticsingularities have a theory much like the theory for rational singularities. Laufer[La4] proved that p is minimally elliptic if and only if H1(M,O) = C and OV,p

is a Gorenstein ring. Let Z be the fundamental cycle [Ar, p. 132] of the minimalresolution of a minimally elliptic singularity. If Z2 = −1 or −2, then p is a doublepoint [La4]. Laufer [La4] proved that if Z2 = −3, then p is a hypersurface singu-larity with multiplicity 3. In fact he shows that for a minimally elliptic singularityZ2 ≥ −4 if and only if p is a complete intersection singularity.

Now let p be an arbitrary singularity in the Stein normal2-dimensional space V having p as its only singularity. Let Γ denote the weighteddual graph of the exceptional set of the minimal good resolution π : M → V . In[La3], Laufer developed a deformation theory preserving Γ. This theory allowshim to introduce the notion of a property of the associated singularity holdinggenerically for Γ. Now suppose that Γ is a weighted dual graph which does notcorrespond to a rational double point or to a minimally elliptic singlarity. Thena deep theorm of Laufer [La4] asserts that the corresponding singularity is gener-ically non-Gorenstein. In particular, it is generically not a complete intersection.As a consequence we can characterize those weighted dual graphs which have onlycomplete intersection singularities associated to them. These are precisely ratio-nal double point graphs and minimally elliptic graphs with Z2 = −1, −2, −3 or−4. Notice that rational double point graphs and minimally elliptic graphs withZ2 = −1,−2 or −3 are precisely those graphs which have only hypersurface sin-gularities associated with them. Laufer [La4] has completely classified minimallyelliptic graphs with Z2 = −1,−2, or −3. Therefore in order to classify thoseweighted dual graphs which have only complete intersection singularities associ-ated with them, we only need to classify all the minimally elliptic graphs withZ2 = −4. This will be done in section 6. Incidentally, these graphs are preciselythe graphs with complete intersection singularities associated with them but nohypersurface singularities associated with them. We summarize our results in thefollowing theorems.

Theorem A. The complete classification of weighted dual graphs which have onlycomplete intersection singularities but no hypersurface singularities associated tothem consists of the minimally elliptic singularity graphs with Z2 = −4 which arelisted in section 7.

Theorem B. The complete classification of weighted dual graphs which have onlycomplete intersection singularities associated with them consists of rational doublepoint graphs listed in [Ar], minimal elliptic hypersurface singularity graphs listedin [La4] and the minimal elliptic complete intersection singularity graphs listed insection 7.

Our strategy of classification of all minimally elliptic singularity graphs withZ2 = −4 is quite simple. We first introduce the concept of an effective component,


CLASSIFICATION OF WEIGHTED DUAL GRAPHS 3537

which is an irreducible component A∗ of the exceptional set such that A∗ · Z < 0.It turns out that there are at most 4 effective components with known fundamentalcoefficients (Proposition 6.2). Let Γ′ be the subgraph of Γ obtained by removing allthe effective components of Γ. Suppose A∗ is an effective component of Γ. Let Γ1

be any connected component of Γ′ which intersect with A∗. Then Γ1 is necessarilyone of the rational double point graphs appearing in Theorem 4.2. Let Z1 be thefundamental cycle of Γ1. Then A∗ · Z1 ≤ 2. If A∗ · Z1 = 2, then Γ = A∗ ∪ Γ1 andZ = A∗ + Z1; moreover for any Aj ∈ Γ1, Aj · Ak > 0 if and only if Aj · Z1 < 0(Proposition 6.3). In order to find out how one can add A∗ to the rational doublepoint graphs, we use Theorem 3.5 and the adjunction formula (2.3).

2. Preliminaries

Let π : M → V be a resolution of the normal two-dimensional Stein space V .We assume that p is the only singularity of V . Let π−1(p) = A =

⋃Ai, 1 ≤ i ≤ n,

be the decomposition of the exceptional set A into irreducible components.A cycle D =

∑diAi, 1 ≤ i ≤ n is an integral combination of the Ai, with di an

integer. There is a natural partial ordering, denoted by <, between cycles definedby comparing the coefficients. We let supp D =

⋃Ai, di = 0, denote the support

of D.Let O be the sheaf of germs of holomorphic functions on M . Let O(−D) be the

sheaf of germs of holomorphic functions on M which vanish to order di on Ai. LetOD denote O/O(−D). Define

(2.1) χ(D) := dimH0(M,OD) − dimH1(M,OD).

The Riemann-Roch theorem [Se, Proposition IV.4, p. 75] says that

(2.2) χ(D) = −12(D2 + D · K),

where K is the canonical divisor on M . D ·K may be defined as follows. Let ω be ameromorphic 2-form on M . Let (ω) be the divisor of ω. Then D ·K = D · (ω) andthis number is independent of the choice of ω. In fact, let gi be the geometric genusof Ai, i.e., the genus of the desingularization of Ai. Then the adjunction formula[Se, Proposition IV, 5, p. 75] says that

(2.3) Ai · K = −A2i + 2gi − 2 + 2δi,

where δi is the “number” of nodes and cusps on Ai. Each singular point on Ai

other than a node or cusp counts as at least two nodes. It follows immediatelyfrom (2.2) that if B and C are cycles, then

(2.4) χ(B + C) = χ(B) + χ(C) − B · C.

Definition 2.1. Associated to π is a unique fundamental cycle Z [Ar, pp. 131-132]such that Z > 0, Ai · Z ≤ 0 for all Ai and such that Z is minimal with respect tothose two properties. Z may be computed from the intersection as follows via acomputation sequence for Z in the sense of Laufer [La2, Proposition 4.1, p. 607]:

Z0 = 0, Z1 = Ai1 , Z2 = Z1 + Ai2 , . . . , Zj = Zj−1 + Aij, . . . ,

Z = Z−1 + Ai= Z,

where Ai1 is arbitrary and Aij· Zj−1 > 0, 1 < j ≤ .



O(−Zj−1)/(O(−Zj) represents the sheaf of germs of sections of a line bundle

over Aijof Chern class −Aij

· Zj−1. So

H0(M,O(−Zj−1)/O(−Zj)) = 0

for j > 1 and

(2.5) 0 → O(−Zj−1)/O(−Zj) → OZj

→ OZj−1 → 0

is an exact sheaf sequence. From the long exact cohomology sequence for (2.5), itfollows by induction that

H0(M,OZk) = C, 1 ≤ k ≤ ,(2.6)

dim H1(M,OZk) =

∑dimH1

(M,O)(−Zj−1)

/O(−Zj)

),(2.7)

1 ≤ j ≤ k.

Lemma 2.2 ([La4]). Let Zk be part of a computation sequence for Z and such thatχ(Zk) = 0. Then dimH1(M,OD) ≤ 1 for all cycles D such that 0 ≤ D ≤ Zk. Alsoχ(D) ≥ 0.

3. Minimally elliptic singularities

In this section we shall recall some of the properties of minimally elliptic singu-larities which we need for our classification problem.

Definition 3.1. A cycle E > 0 is minimally elliptic if χ(E) = 0 and χ(D) > 0 forall cycles D such that 0 < D < E.

Wagreich [Wa] defined the singularity p to be elliptic if χ(D) ≥ 0 for all cyclesD ≥ 0 and χ(F ) = 0 for some cycles F > 0. He proved that this definition isindependent of the resolution. It is easy to see that under this hypothesis, χ(Z) = 0.The converse is also true [La4]. Henceforth, we shall adopt the following definition.

Definition 3.2. p is said to be weakly elliptic if χ(Z) = 0.

The following proposition and lemma hold for a weakly elliptic singularity.

Proposition 3.3 ([La4]). Suppose that χ(D) ≥ 0 for all cycles D > 0. Let B =∑biAi and C =

∑ciAi, 1 ≤ i ≤ n, be any cycles such that 0 < B, C and

χ(B) = χ(C) = 0. Let F =∑

min(bi, ci)Ai, 1 ≤ i ≤ n. Then F > 0 andχ(F ) = 0. In particular, there exists a unique minimally elliptic cycle E.

Lemma 3.4 ([La4]). Let E be a minimally elliptic cycle. Then for Ai ⊂ supp E,Ai · E = −Ai · K. Suppose additionally that π is the minimal resolution. Then Eis the fundamental cycle for the singularity having supp E as its exceptional set.Also, if Ek is part of a computation sequence for E as a fundamental cycle andAj ⊂ supp (E − Ek), then the computation sequence may be continued past Ek soas to terminate at E = E with Ai

= Aj.

Theorem 3.5 ([La4]). Let π : M → V be the minimal resolution of the normaltwo-dimensional variety V with one singular point p. Let Z be the fundamentalcycle on the exceptional set A = π−1(p). Then the following are equivalent:

(1) Z is a minimally elliptic cycle,(2) Ai · Z = −Ai · K for all irreducible components Ai in A,(3) χ(Z) = 0 and any connected proper subvariety of A is the exceptional set

for a rational singularity.



In [La4], Laufer introduced the notion of minimally elliptic singularity.

Definition 3.6. Let p be a normal two-dimensional singularity. p is minimallyelliptic if the minimal resolution π : M → V of a neighborhood of p satisfies one ofthe conditions of Theorem 3.5.

Proposition 3.7 ([La4]). Let π : M → V and π′ : M ′ → V be the minimal resolu-tion and minimal good resolution respectively for a minimally elliptic singularity p.Then π = π′ and all the Ai are rational curves except for the following cases:

(1) A is an elliptic curve. π is a minimal good resolution.(2) A is a rational curve with a node singularity.(3) A is a rational curve with a cusp singularity.(4) A is two nonsingular rational curves which have first order tangential con-

tact at one point.(5) A is three nonsingular rational curves all meeting transversely at the same

point.

In case (2), the weighted dual graph of the minimal good resolution is

−w1 −1 with w1 ≥ 5.

In cases (3)–(5), π′ has the following weighted dual graph:

−w2

−w1 −w3

with wi ≥ 2, 1 ≤ i ≤ 3.

Minimally elliptic singularities can be characterized without explicit use of theresolution as follows because H1(M,O) can be described in terms of V [La2, The-orem 3.4, p. 604].

Theorem 3.8 ([La4]). Let V be a Stein normal two-dimensional space with p asits only singularity. Let π : M → V be a resolution of V . Then p is a minimallyelliptic singularity if and only if H1(M,O) = C and OV,p is a Gorenstein ring.

4. Weighted dual graphs admitting no complete intersection

singularities structures

In this section, we shall show that there is a large class of weighted dual graphsnot admitting any complete intersection singularity structure. Let (V, p) be a nor-mal 2-dimensional singularity. Let π : M → V be the minimal resolution. Let Z bethe fundamental cycle.

Definition 4.1. p is a rational singularity if χ(Z) = 1.

If p is a rational singularity, then π is also a minimal good resolution, i.e.,exceptional set with nonsingular Ai and normal crossings. Moreover each Ai is arational curve [Ar].



Theorem 4.2 ([Ar]). If p is a hypersurface rational singularity, then p is a rationaldouble point. Moreover the set of weighted dual graphs of hypersurface rationalsingularities consists of the following graphs:

(1) An, n ≥ 1−2 −2 −2 Z = 1 1 . . . 1

(2) Dn, n ≥ 4−2 −2 −2 −2

−2 Z = 1

1

2 1 . . . 1

(3) E6

−2 −2 −2 −2 −2

−2 Z = 1 2

2

3 2 1

(4) E7

−2 −2 −2 −2 −2 −2

−2 Z = 2 3

2

4 3 2 1

(5) E8

−2 −2 −2 −2 −2

−2 −2 −2

Z = 2 4

3

6 5 4 3 2

Theorem 4.3. Let Γ be a weighted dual graph of a rational singularity. If Γ isnot one of the five types in Theorem 4.2, then Γ does not admit any Gorensteinsingularity structure; in particular, Γ does not admit any complete intersectionsingularity structure.

Proof. Since in the definition of a rational singularity, χ(Z) can be computed fromthe weighted dual graph, any singularity associated to Γ is a rational singularity.To prove the theorem, we only need to prove that if p is a Gorenstein rationalsingularity, then its graph is one of the five types in Theorem 4.2. Suppose (V, p)is a Gorenstein rational singularity. Then dimH1(M,O) = 0 [Ar]. By a resultof Laufer [La2], dimH1(M,O) = dimH0(M − A, Ω2)

/H0(M, Ω2) where Ω2 is the

sheaf of germs of holomorphic 2-forms on M . Therefore there exists an effectivecanonical divisor K =

∑kiAi, ki a nonnegative integer, on M . Since M is a

minimal resolution, by the adjunction formula, we have

(4.1) Ai · K ≥ 0 for all Ai ⊆ A.

It follows that

(4.2) K2 =∑

ki(Ai · K) ≥ 0.

On the other hand, the intersection matrix is a negative definition [Gr]. ThereforeK2 ≤ 0. This together with (4.2) implies K2 = 0. The negative definiteness of theintersection matrix implies K = 0. The adjunction formula tells us that A2

i = −2for all Ai. Then as an easy exercise, one can show that the weighted dual graph ofthe exceptional set is one of the five types listed in Theorem 4.2.

5. Characterization of weighted dual graphs admitting only

complete intersection singularities structures

In this section we shall give a characterization of weighted dual graphs admittingonly hypersurface singularities structures. We shall also give a characterization ofweighted dual graphs admitting only complete intersection singularities structures.



It turns out that the latter list minus the former list corresponds to the list ofweighted dual graphs which admit only complete intersection singularities struc-tures but not hypersurface singularities structures.

Theorem 5.1 ([La4]). Let p be a minimally elliptic singularity. Let π : M → V bea resolution of a Stein neighborhood V of p with p as its only singular point. Letm be the maximal ideal in OV,p. Let Z be the fundamental cycle on A = π−1(p).

(1) If Z2 ≤ −2, then O(−Z) = mO on A.(2) If Z2 = −1, and π is the minimal resolution or the minimal resolution with

nonsingular Ai and normal crossings, then O(−Z)/mO is the structure

sheaf for an embedded point.(3) If Z2 = −1 or −2, then p is a double point.(4) If Z2 = −3, then for all integers n ≥ 1, mn ≈ H0

(A,O(−nZ)

)and

dimmn/mn+1 = −nZ2.(5) If −3 ≤ Z2 ≤ −1, then p is a hypersurface singularity.(6) If Z2 = −4, then p is a complete intersection and in fact a tangential

complete intersection.(7) If Z2 ≤ −5, then p is not a complete intersection.

Let p be a normal two-dimensional singularity. Choose the minimal resolutionof p having nonsingular Ai and normal crossings. Let Γ denote the weighted dualgraph along with the genera. See [HNK] or [La1] for a more detailed descriptionof Γ. Γ may be described abstractly. Given Γ, we say that p is a singularityassociated to Γ. As in [La1, Theorem 6.20, p. 132] we may choose a suitably largeinfinitesimal neighborhood B of the exceptional set such that B depends only on Γand determines p. We can deform B in such a way that Γ is preserved. See [La3]for the general theory in this situation.

Definition 5.2. Let Γ be a weighted dual graph, including genera for the vertices.A property is generically true for an associated singularity of Γ if given any normaltwo-dimensional singularity p having Γ as the weighted dual graph of its minimalresolution with nonsingular Ai and normal crossings, then the property is true for allsingularities near p and off a proper subvariety of the parameter space of a completedeformation of a suitable large infinitesimal neighborhood B of the exceptional setfor P .

The following deep theorem is due to Laufer.

Theorem 5.3 ([La4]). All rational double points and all minimally elliptic singu-larities are Gorenstein. Let Γ be a weighted dual graph, including genera for thevertices, associated to a minimal resolution with nonsingular Ai and normal cross-ings of a singularity p. Suppose that p is not a rational double point or minimallyelliptic. Then an associated singularity of Γ is generically non-Gorenstein.

Now we are ready to give a characterization of weighted dual graphs admittingonly complete intersection singularities structures (respectively hypersurface singu-larities structures). Recall that rational and minimally elliptic singularities havetopological definitions; i.e., they can be defined in terms of their weighted dualgraphs.



Theorem 5.4.

(1) The weighted dual graphs which have only hypersurface singularities associ-ated to them are precisely those graphs coming from rational double points,minimally elliptic double points (Z2 = −1, or −2), or minimally elliptictriple points (Z2 = −3).

(2) The weighted dual graphs which have only complete intersection singular-ities associated to them are precisely those graphs coming from rationaldouble points, minimally elliptic double points (Z2 = −1, or −2), mini-mally elliptic triple points (Z2 = −3), or minimally elliptic quadruple points(Z2 = −4).

(3) The weighted dual graphs which have only complete intersection but not hy-persurface singularities associated to them are precisely those graphs comingfrom minimally elliptic quadruple points (Z2 = −4).

Proof. We only need to observe that hypersurface or complete intersection sin-gularities are Gorenstein. Theorem 5.4 follows directly from Theorem 5.1 andTheorem 5.3.

6. Classification of weighted dual graphs with only complete

intersection but not hypersurface singularities structures

By Theorem 5.4, the classification of weighted dual graphs with only completeintersection but not hypersurface singularity structure is equal to the classificationof minimally elliptic singularity weighted dual graphs with Z2 = −4.

Definition 6.1. Let (V, p) be a germ of weakly elliptic singularity. Let π : M → Vbe the minimal resolution with π−1(p) = A =

⋃Ai, 1 ≤ i ≤ n the irreducible

decomposition of the exceptional set. Let Z be the fundamental cycle. The set ofeffective components A∗1, . . . , A∗n is the set Ai : Ai · Z < 0.

Proposition 6.2. Let (V, p) be a germ of minimally elliptic singularity. Let π :M → V be the minimal resolution of p. If π is also a minimal good resolution andZ2 = −4, then the set of effective components A∗1, . . . , A∗m must be one of thefollowing:

(1) A∗1, A2∗1 = −3, z1 = 4

(2) A∗1, A2∗1 = −4, z1 = 2

(3) A∗1, A2∗1 = −6, z1 = 1

(4) A∗1, A∗2, A2∗1 = A2

∗2 = −3, z1 = z2 = 2(5) A∗1, A∗2, A2

∗1 = A2∗2 = −3, z1 = 3, z2 = 1

(6) A∗1, A∗2, A2∗1 = −3, A2

∗2 = −4, z1 = 2, z2 = 1(7) A∗1, A∗2, A2

∗1 = A2∗2 = −4, z1 = z2 = 1

(8) A∗1, A∗2, A2∗1 = −3, A2

∗2 = −5, z1 = z2 = 1(9) A∗1, A∗2, A∗3, A2

∗1 = A2∗2 = A2

∗3 = −3, z1 = z2 = 1, z3 = 2(10) A∗1, A∗2, A∗3, A2

∗1 = A2∗2 = −3, A2

∗3 = −4, z1 = z2 = z3 = 1(11) A∗1, A∗2, A∗3, A∗4, A2

∗1 = −3, zi = 1, i = 1, 2, 3, 4,

where A∗i = A∗j if i = j and zi is the coefficient of A∗i in Z.



Proof. Let A∗1, . . . , A∗m be the set of effective components. Then, by Theo-rem 3.5, we have

−Z2 = −n∑

i=1

zi(Ai · Z) = −m∑

i=1

zi(A∗i · Z)

=m∑

i=1

zi(A∗i · K).

This implies that 4 =∑m

i=1 zi(−A2∗i − 2). By the definition of the effective com-

ponent, we have −A2∗i − 2 = A∗i · K = −A∗i · Z > 0. Hence we have 1 ≤

m ≤ 4. If m = 1, then −4 = z1(A2∗1 + 2) and we are in case (1), case (2)

or case (3). If m = 2, then −4 = z1(A2∗1 + 2) + z2(A2

∗2 + 2). It follows easilythat we are in case (4), case (5), case (6), case (7), or case (8). If m = 3, then−4 = z1(A2

∗1+2)+z2(A2∗2+2)+z3(A2

∗3+2). It is easy to see that we are in case (9)or case (10). If m = 4, then −4 = (A2

∗1 + 2) + (A2∗2 + 2) + (A2

∗3 + 2) + (A2∗4 + 2).

So we are in case (11).

Proposition 6.3. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained by re-moving all the effective components of Γ. Suppose that A∗ is an effective componentof Γ, and let Γ1, . . . , Γn be the set of connected components of Γ′ which intersectwith A∗. Then Γ1, . . . , Γn are necessarily one of the rational double point graphsappearing in Theorem 4.2. Let Z1, . . . , Zn be the fundamental cycles of Γ1, . . . , Γn

respectively. Then A∗ ·Z1 ≤ 2. If A∗ ·Z1 = 2, then Γ = A∗ ∪ Γ1 and Z = A∗ + Z1;moreover for any A1 ∈ Γ1, A1 · A∗ > 0 if and only if A1 · Z1 < 0.

Proof. For any Aj ∈ Γi, 0 = Aj · Z = Aj · (−K) = A2j + 2. Hence A2

j = −2. Itfollows that Γi are rational double point graphs.

Since Γ is the graph of a minimally elliptic singularity, we have

0 ≤ χ(A∗ + Z1)(6.1)

= χ(A∗) + χ(Z1) − A∗ · Z1,

which implies

(6.2) A∗ · Z1 ≤ χ(A∗) + χ(Z1) = 2.

Observe that if Γ = A∗∪Γ1 or Z > A∗ +Z1, then the inequalities in (6.1) and (6.2)are strict inequalities. Hence A∗ ·Z1 = 1. We have proved that if A∗ ·Z1 = 2, thenΓ = A∗ ∪ Γ1 and Z = A∗ + Z1.

We shall assume from now on that A∗ ·Z1 = 2. Let A1 ∈ Γ1 such that A1 ·A∗ > 0.A1 ·Z1 = 0 would imply A1 · (Z1 + A∗) > 0 and hence A1 ·Z > 0, which is absurd.It follows that A1 · Z1 < 0.

Conversely, if A1 ∈ Γ1 and A1 ·Z1 < 0, but A∗ ·A1 = 0, then there is an A2 ∈ Γ1

such that A2 ·A∗ > 0 and A2 ·Z1 < 0. Since Z21 = −2, we have A2 ·Z1 = A1 ·Z1 = −1

and the coefficient of A2 in Z1 is one. It follows that A2 is the only component inΓ1 which intersects with A∗ and A2 · A∗ = 2. Observe that χ(A∗ + A2) = 0 andA∗ + A2 < Z. This contradicts the fact that Z is the minimally elliptic cycle. Sowe have shown that A∗ · A1 > 0 if and only if A1 · Z1 < 0.



Notation. From now on, we shall denote a nonsingular rational curve with −2weight.

Corollary 6.4. Let Γ be the minimal resolution graph of a minimally elliptic sin-gularity with fundamental cycle Z. Let Γ1 be a rational double point subgraph ofΓ with fundamental cycle Z1 in Proposition 6.3. Let A∗ be an effective componentattaching on Γ1. Suppose that A∗ · Z1 = 2. Then one of the following cases holds.

(1) Γ is of the following form:

A∗ ∗ r ≥ 1...............

............ ........ ........ ........ ........ ........ ....... ....... ....... ........ ......... ........ ........ ........ ......... ...........................................................................................................................................................

....... Z = A∗ + A1 + · · · + Ar,

where r denotes with r vertices and r +1 edges.is a nonsingular rational curve with weight −2.(2) Γ1 is either Dn, E6, E7 or E8. There exists a unique A1 in Γ1 such that

A1 ·A∗ = 1 and A1 ·Z1 < 0. The coefficient of A1 in Z1 is 2. Γ = A∗ ∪ Γ1

and Z = A∗ + Z1. Γ is one of the following forms.

(a) (i) r

r ≥ 0

∗ A∗

A1

Z = 112 2 · · ·

12 1

(ii) ∗

A∗

A1 Z = 1

1

2

1

1

(b)

∗ A∗

A1

Z = 1 2

1

2

3 2 1

(c)A∗ A1

∗ Z = 1 2 3

2

4 3 2 1

(d)A∗A1

∗ Z = 2 4

3

6 5 4 3 2 1

Proof. This follows from Proposition 6.3 and Theorem 4.2.

Definition 6.5. Let A1 be an irreducible component in a weighted dual graph Γ.The degree of A1 is defined to be the number of distinct irreducible components inΓ intersecting with A1 positively.

Lemma 6.6. Let Γ be the minimal resolution graph of a minimally elliptic singu-larity with fundamental cycle Z. Let Γ1 be a subgraph of Γ in Proposition 6.3 withfundamental cycle Z1. Let A∗ be an effective component attaching on Γ1. Suppose



that the coefficient z∗ of A∗ in Z is one. Then either A∗ has degree one or Γ is ofthe following form:

∗A∗2A∗∗

rn

r1 r2∗

A∗1

A∗n∗.....................

........... ........... . .......

........ ......... ......... .......... .......... ......... ......... ........ ....... .......... .......... ...... ...................................................................................................................................................

.....................

where n ≥ 1 and Γ1 is r which denotes with r1 ver-tices and r1 + 1 edges.

Proof. By Proposition 6.3, A∗ · Z1 is either 1 or 2. If A∗ · Z1 = 2, then the lemmafollows from Corollary 6.4.

From now on, we shall assume that A∗ · Z1 = 1. To prove the lemma, we onlyneed to prove that if deg A∗ > 1, then Γ must be of the circular form shown asabove. If deg A∗ > 1, then there exists A2 not in Γ1 such that A2 ·A∗ > 0. ClearlyA2 · A∗ = 1 by the minimal ellipticity of Γ. We claim that A2 is connected to Γ1

via a path in Γ which is disjoint from A∗.By Theorem 3.4, we can choose a computation sequence of the fundamental cycle

Z starting from A∗ continuing to Γ1 and ending at A2. Now z∗ = 1, A2∗+2 = A∗ ·Z,

and deg A∗ > 1 implies that the computation sequence contains A∗ only once andthe coefficient of A2 in Z must also be one. Hence the computation sequence mustcontain A2 only once. Moreover A2

2+2 = A2 ·Z implies that deg A2 = 2. Repeatingthe same argument, we see that for every component in that computation sequenceits coefficient in Z is one, its degree is 2 and the computation sequence passes itonly once. Therefore Γ must be the form shown in the lemma.

Remark 6.7. With the same assumption and notation as in Lemma 6.6, so longas the interseciton matrix remains negative definite, A2

∗ can be given any value atmost −2 and Z remains unchanged and Γ still corresponds to a minimally ellipticsingularity.

Proposition 6.8. Let Γ be the minimal resolution graph of minimally elliptic sin-gularity with fundamental cycle Z. Suppose that there is no effective componentwith coefficient in Z strictly greater than 1. Set all A2

∗ of effective components of Γbut one to −2 and the remaining weight to −3. Then the new weighted dual graphΓ, which coincides with Γ except for the weights, is obtained from a rational doublepoint weighted dual graph by the addition of one additional vertex A∗. In fact Γcorresponds to a minimally elliptic double point with Z2 = −1.

Proof. Since A∗ ·Z = −A∗ ·K = A2∗+2, after setting all A2

∗ of effective componentsof Γ but one to −2 and the remaining weight to −3, it is still true that Ai · Z ≤ 0for all i and that A∗Z < 0 for one A∗. Therefore Z is also the fundamental cyclefor Γ and the intersection matrix of Γ is still negative definite [Ar, Proposition 2,pp. 130–131]. By Lemma 6.6, Γ is obtained from a rational double point weighteddual graph by the addition of one additional vertex A∗. Clearly Z2

Γ= −1.



Proposition 6.9. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component of Γ.Suppose that Γ1 is a connected component of Γ′ which corresponds to the An graphin case (1) of Theorem 4.2. Suppose also that Γ1 intersects with A∗ but is disjointfrom other effective components. Let Z1 be the fundamental cycle on Γ1. SupposeA∗ ·Z1 = 1. If the coefficient z∗ of A∗ in Z is four and A2

∗ = −3, then A∗ ∪Γ1 andthe restriction of Z on A∗ ∪ Γ1 must be one of the following forms.

(1) ∗ A∗

Z

∣∣∣∣A∗∪Γ1

=4 2

(2) ∗A∗

Z

∣∣∣∣A∗∪Γ1

=4 3 2 1

(3) ∗A∗

Z

∣∣∣∣A∗∪Γ1

=436 5 4 3 2 1

(4) ∗A∗

Z

∣∣∣∣A∗∪Γ1

=4

369 8 7 6 5 4 3 2 1

(5) ∗A∗

Z

∣∣∣∣A∗∪Γ1

=244 2

(6)

A∗

∗ Z

∣∣∣∣A∗∪Γ1

=2 446 4 2

(7)

A∗

∗ Z

∣∣∣∣A∗∪Γ1

=2 4 648 6 4 2

(8)

A∗

∗ Z

∣∣∣∣A∗∪Γ1

=2 4 6 8410 8 6 4 2

Proof. Consider A∗ attaching on Γ1 in the following form:

∗ A∗ A1 A2 Am

Z

∣∣∣∣A∗∪Γ1

= 4 n1 n2 . . . nm.



Since Ai ·Z = Ai · (−K) = A2i + 2 = 0, 1 ≤ i ≤ m, we have the following system of

equations:⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2n1 + 4 + n2 = 0−2n2 + n1 + n3 = 0

...−2nm−1 + nm−2 + nm = 0−2nm + nm−1 = 0

⇒

ni = (m − i + 1)nm 1 ≤ i ≤ m

nm = 4m+1

.

Therefore m = 1 or m = 3 and we are in case (1) or case (2) respectively.Consider A∗ attaching on Γ1 in the following form:

∗ A∗

A′m2

A′2 A1 A2 Am

Z

∣∣∣∣A∗∪Γ1

= n′m2

. . . n′2

4n1 n2 . . . nm1 .

Since Ai · Z = Ai · (−K) = A2i + 2 = 0, 1 ≤ i ≤ m1 and similarly A′

j · Z = 0 for2 ≤ j ≤ m′

2, we have the following system of equations:⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2nm1 +nm1−1 =0−2nm1−1+nm1−2+nm1 =0

...−2n3+n2+n4 =0−2n2+n1+n3 =0

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2n′m2

+n′m2−1 =0

−2n′m2−1+n′

m2−2+n′m2

=0...

−2n′3+n′

2+n′4 =0

−2n′2+n′

1+n′3 =0

(6.3)

4 − 2n1 + n2 + n′2 = 0(6.4)

(6.3) implies

ni = (m1 − i + 1)nm1 1 ≤ i ≤ m1,(6.5)

n′j = (m2 − j + 1)n′

m22 ≤ j ≤ m2,(6.6)

m1nm1 = m2n′m2

.(6.7)

Putting (6.5) and (6.6) into (6.4), we get

0 = 4 − 2m1nm1 + (m1 − 1)nm1 + (m2 − 1)n′m2

= 4 − (m1 + 1)nm1 + (m2 − 1)n′m2

= 0.(6.8)

(6.7) and (6.8) imply

(6.9) nm1 + n′m2

= 4.

(6.9) implies that either nm1 = 3, n′m2

= 1 or nm1 = 2 = n′m2

.

Case I. nm1 = 3 and n′m2

= 1. By (6.7), we have 3m1 = m2. Observe that

− 1 = A2∗ + 2 = A∗ · (−K) = A∗ · Z ≥ 4(−3) + n1 = −12 + 3m1

⇒ 3m1 ≤ 11⇒ m1 ≤ 3.



If m1 = 1, or 2, or 3, then we are in case (2), case (3) or case (4) respectively inthe statement of the proposition.

Case II. nm1 = 2 = n′m2

. By (6.7), we have m1 = m2. Observe that

− 1 = A∗ · (−K) = A∗ · Z ≥ 4(−3) + n1 = −12 + 2m1

⇒ 2m1 ≤ 11⇒ m1 ≤ 5.

If m1 = 2, or 3, or 4, or 5, then we are in case (5), case (6), case (7) or case (8)respectively in the statement of the proposition. Proposition 6.10. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component of Γ.Suppose that Γ1 is a connected component of Γ′ which corresponds to the Dn graphin case (2) of Theorem 4.2. Suppose also that Γ1 intersects with A∗ but is disjointfrom other effective components. Let Z1 be the fundamental cycle on Γ1. SupposeA∗ ·Z1 = 1. If the coefficient z∗ of A∗ in Z is four and A2

∗ = −3, then A∗ ∪Γ1 andthe restriction of Z on A∗ ∪ Γ1 must be one of the following forms.

(1) ∗A∗

ZA∗∪Γ1 = 4 424 2

(2) ∗A∗

ZA∗∪Γ1 = 4 536 4 2

(3) ∗A∗

ZA∗∪Γ1 = 4 648 6 4 2

(4) ∗A∗

ZA∗∪Γ1 = 4 7510 8 6 4 2

(5) ∗A∗

ZA∗∪Γ1 = 4 8612 10 8 6 4 2

(6) ∗A∗

ZA∗∪Γ1 = 4 9714 12 10 8 6 4 2

(7) ∗A∗

ZA∗∪Γ1 = 4 10816 14 12 10 8 6 4 2

(8) ∗A∗

ZA∗∪Γ1 = 4 11918 16 14 12 10 8 6 4 2

(9) ← m − 3 →

m ≥ 4

∗

A∗

ZA∗∪Γ1 = 224 4 . . . 4 4



Proof. Consider A∗ attaching on Γ1 in the following form:

∗−3

A∗ A1 A3 A4 Am

A2

ZA∗∪Γ1 = 4 n1

n2n3 n4 . . . nm.

Since Ai ·Z = Ai · (−K) = A2i + 2 = 0, 1 ≤ i ≤ m, we have the following system of

equations:

−2n1 + 4 + n3 = 0(6.10) ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n2 + n3 = 0−2n3 + n1 + n2 + n4 = 0−2n4 + n3 + n5 = 0

...−2nm−1 + nm−2 + nm = 0−2nm + nm−1 = 0

(6.11)

(6.11) implies

(6.12) n1 =m

2nm, n2 =

m − 22

nm, nj = (m − j + 1)nm, 3 ≤ j ≤ m.

(6.10) and (6.12) imply nm = 2 and hence n1 = m. Recall that

−1 = A∗ · (−K) = A∗ · Z ≥ 4(−3) + n1 = −12 + m

⇒ m ≤ 11.

Since 4 ≤ m ≤ 11, we are in case (1)–case (8) of the proposition.We next consider A∗ attaching on Γ1 in the following form:

A1 A3 A4 Am

A2

∗−3

A∗

Z

∣∣∣∣A∗∪Γ1

= n1n2n3 n4 . . . nm 4.

Since Ai · Z = 0, 1 ≤ i ≤ m, we have the following system of equations:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + n3 = 0−2n2 + n3 = 0−2n3 + n1 + n2 + n4 = 0−2n4 + n3 + n5 = 0

...−2nm−1 + nm−2 + nm = 0−2nm + nm−1 + 4 = 0

(6.13)

−2nm + nm−1 + 4 = 0(6.14)

(6.13) implies n3 = n4 = · · · = nm = 2n1 = 2n2. By (6.14), we know that n1 = 2.Therefore we are in case (9) of the proposition.



Proposition 6.11. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to either theE6, E7 or E8 graph in case (3)–case (5) of Theorem 4.2. Suppose also that Γ1

intersects with A∗ but is disjoint from other effective components. Let Z1 be thefundamental cycle of Γ1. Suppose A∗ · Z1 = 1. If the coefficient z∗ of A∗ in Z isfour and A2

∗ ≤ −3, then A∗ ∪Γ1 and the restriction of Z on A∗ ∪Γ1 must be of theform

A1 A2 A3 A5 A6 A7 A∗

A4

∗ Z

∣∣∣∣A∗∪Γ1

= 4 8612 10 8 6 4

Proof. By Theorem 4.2, A∗ attaching on E6 must be of the following form:

∗A∗ A1 A2 A3 A5 A6

A4

Z

∣∣∣∣A∗∪Γ1

= 4 n1 n2n4n3 n5 n6

Since Ai ·Z = Ai(−K) = 0 for 1 ≤ i ≤ 6, we have the following system of equations:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + 4 + n2 = 0−2n2 + n1 + n3 = 0−2n3 + n2 + n4 + n5 = 0−2n4 + n3 = 0−2n5 + n3 + n6 = 0−2n6 + n5 = 0,

which imply n6 = 163 . This contradicts the fact that n6 is an integer.

By Theorem 4.2, A∗ attaching on E7 must be of the following form:

A1 A2 A3 A5 A6 A7

A4

∗A∗

Z

∣∣∣∣A∗∪Γ1

= n1 n2n4n3 n5 n6 n7 4

Since Ai · Z = Ai · (−K) = 0 for 1 ≤ i ≤ 7, we have the following system ofequations: ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + n2 = 0−2n2 + n1 + n3 = 0−2n3 + n2 + n4 + n5 = 0−2n4 + n3 = 0−2n5 + n3 + n6 = 0−2n6 + n5 + n7 = 0−2n7 + n6 + 4 = 0.

An easy exercise shows that we are in the form of the proposition.By Theorem 4.2, A∗ cannot attach on E8 because A∗ · Z1 ≥ 2.



Theorem 6.12. Let (V, p) be a germ of minimally elliptic singularity. Let π :M −→ V be the minimal resolution of p. If case (1) of Proposition 6.2 holds,i.e., there exists only one effective component A∗, and A2

∗ = −3, z∗ = 4, then theweighted dual graph Γ of the exceptional set is one of the following forms.

(1)A∗ ∗

Z =1 2 3

1

2

3

4

2

3 2 1

(2) A∗∗

Z =2

1

2

3

4

3

6 5 4 3 2 1

(3) A∗∗ Z =2 4

3

6

9 8 7 6 5 4 3 2 1

(4)A∗

∗

Z =1 2 3

1

2

3

4 5

3

6 4 2

(5) A∗∗ Z =2 4

3

6 5 4

3

6 5 4 3 2 1

Proof. Let Γ′ be the graph obtained by deleting A∗ from Γ. Let Γ1, . . . , Γm bethe connected components of Γ′ with fundamental cycles Z1, . . . , Zm respectively.Since z∗ = 4, in view of Proposition 6.3, we have A∗ · Zi = 1 for 1 ≤ i ≤ m. ByProposition 6.9, Proposition 6.10 and Proposition 6.11, we have

A∗ · Z/Γ1, . . . , A∗ · Z/Γm ⊆ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.

Since the singularity is minimally elliptic, we have

A∗ · (Z − 4A∗) = −A∗ · (K + 4A∗) = A2∗ + 2 − 4A2

∗ = 11.



Observe that we can write

11 = 2 + 2 + 2 + 2 + 3= 2 + 2 + 2 + 5= 2 + 2 + 3 + 4= 2 + 2 + 7= 2 + 3 + 3 + 3= 2 + 3 + 6= 2 + 4 + 5= 2 + 9= 3 + 3 + 5= 3 + 4 + 4= 3 + 8= 4 + 7= 5 + 6= 11.

In case of 11 = 2 + 2 + 2 + 2 + 3, by Proposition 6.9 (1) and (2) we obtain agraph with the proposed fundamental cycle

A∗∗

Z =

22 2

2 4 3 2 1

On the other hand one may find a positive cycle on the graph

A∗∗

Z ′ =

11 1

1 2 2 2 1

with Z ′ < Z that also satisfies Definition 2.1. Therefore the proposed fundamentalcycle Z does not satisfy the minimum condition stated in Definition 2.1. Hencethere is no dual graph produced from this case.

Similarly, by Propositions 6.9, 6.10 and 6.11 together with Definition 2.1, incases 11 = 2 + 2 + 2 + 5, 11 = 2 + 2 + 3 + 4 and 11 = 2 + 2 + 7, there is no dualgraph produced. In the case of 11 = 1 + 3 + 3 + 3, we have case (1). In the caseof 11 = 2 + 3 + 6, we only have case (2). In the case of 11 = 2 + 4 + 5, there is nodual graph produced. In the case of 11 = 2 + 9, we only have case (3). In the caseof 11 = 3 + 3 + 5, we have case (4). In the cases 11 = 3 + 4 + 4, 11 = 3 + 8 and11 = 4 + 7, there is no dual graph produced. In the case of 11 = 5 + 6, we onlyhave case (5). Finally case 11 = 11 does not produce any dual graph.

Proposition 6.13. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to a rational



double point graph in Theorem 4.2. Suppose also that Γ1 intersects with A∗ but isdisjoint from other effective components. Let Z1 be the fundamental cycle on Γ1.Suppose A∗ ·Z1 = 1. If the coefficient z∗ of A∗ in Z is 2 and A2

∗ = −4, then A∗∪Γ1

and the restriction of Z on A∗ ∪ Γ1 must be one of the following forms.

(1) ∗−4

A∗

Z

∣∣∣∣A∗∪Γ1

=2 1

(2) −4 A∗ ∗ Z

∣∣∣∣A∗∪Γ1

=1 2

2

3 2 1

(3) −4 A∗ ∗ Z

∣∣∣∣A∗∪Γ1

=1 2 3

2

4 3 2 1

(4) −4 A∗ ∗ Z

∣∣∣∣A∗∪Γ1

=1 2 3 4

2

5 4 3 2 1

(5) −4 A∗ ∗ Z

∣∣∣∣A∗∪Γ1

=1 2 3 4 5

2

6 5 4 3 2 1

(6) ∗−4

A∗

Z

∣∣∣∣A∗∪Γ1

=2 3

2

4 3 2 1

(7) ∗−4

A∗

Z

∣∣∣∣A∗∪Γ1

=2 4

3

6 5 4 3 2 1

(8) ∗−4

A∗

Z

∣∣∣∣A∗∪Γ1

=2 5

4

8 7 6 5 4 3 2 1

(9) ∗−4

A∗

Z

∣∣∣∣A∗∪Γ1

=2 6

5

10 9 8 7 6 5 4 3 2 1

(10) rr ≥ 0

∗−4

A∗

Z

∣∣∣∣A∗∪Γ1

=1

1

2 2 . . . . . . 2︸︷︷︸r≥0

2

(11) −4

A∗

∗ Z

∣∣∣∣A∗∪Γ1

=2 4

3

6 5 4 3 2

Proof. The proof is similar to those of Propositions 6.9, 6.10 and 6.11. Theorem 6.14. Let (V, p) be a germ of minimally elliptic singularity. Let π :M → V be the minimal resolution of p. If case (2) of Proposition 6.2 holds, i.e.,if there exists only one effective component A∗, and A2

∗ = −4, z∗ = 2, then theweighted dual graph Γ of the exceptional set is one of the following forms.



(1) A∗ −4

∗

Z = 1

1

1

2

1

1

1

(2)A∗

−4∗ r

r ≥ 0

Z =

1

1

1

2

1

. . . . . .

1

2 1

(3) A∗

−4∗

Z = 1

1

1

2

1

2

3 2 1

(4) A∗

−4∗

Z = 1

1

1

2 3

2

4 3 2 1

(5) A∗

−4∗

Z = 1

1

1

2 3 4 5

3

6 4 2

(6) ∗

A∗

−4r

r ≥ 0s

s ≥ 0Z = 1

1

2 . . . . . .

1

2

1

. . . . . .

1

2 1

(7) ∗

A∗

−4Z = 1

1

2 4

3

6 5 4 3 2 1

(8) ∗

A∗

−4Z = 1

1

2

1

2

3

4 3 2 1

(9) ∗

A∗

−4r

r ≥ 0Z = 1

1

2 . . . . . .

1

2

1

2

3 2 1

(10) ∗

A∗

−4r

r ≥ 0Z = 1

1

2 . . . . . .

1

2 3

2

4 3 2 1



(11) ∗

A∗

−4r

r ≥ 0Z = 1

1

2 . . . . . .

1

2 3 4 5

3

6 4 2

(12) ∗ A∗−4

Z = 1 2 3 4 5

2

1

4 3 2 1

(13) ∗A∗

−4 Z = 1 2 5

4

8 7 6 5 4 3 2 1

(14) ∗

r s−4

A∗

t

Z = 1

1

2 · · · · · · 2

.

..

1

2

1

· · · · · ·1

2 1

(15) ∗r

A∗

−4Z = 1

1

2 · · · · · · 2

1

2

3

4 3 2 1

(16) ∗r

A∗

−4Z = 1

1

2 · · · · · · 2 4

3

6 5 4 3 2 1

(17) ∗

A∗

−4Z = 1 2

1

2

3 2

1

2

3 2 1

(18) ∗

A∗

−4Z = 1 2

1

2

3 2 3

2

4 3 2 1



(19) ∗

A∗

−4Z = 1 2

1

2

3 2 3 4 5

3

6 4 2

(20) ∗A∗

−4 Z = 1 2 3

2

4 3 2 3

2

4 3 2 1

(21) ∗A∗

−4 Z = 1 2 3

2

4 3 2 3 4 5

3

6 4 2

(22) ∗A∗

−4 Z = 2 4

3

6 5 4 3 2 3 4 5

3

6 4 2

(23)

A∗−4 ∗ Z = 1 2 3 4 5

2

6 5 4 3 2 1

(24) ∗A∗

−4 Z = 2 6

5

10 9 8 7 6 5 4 3 2 1

Proof. The proof is the same as those given in Theorem 6.12. By Proposition 6.13,we have

A∗ · Z/Γm, . . . , Z∗ · Z/Γm ⊆ 1, 2, 3, 4, 5, 6.Since the singularity is minimally elliptic, we have

A∗ · (Z − 2A∗) = −A∗(K + 2A∗) = −A2∗ + 2 = 6.

Observe that we can write

6 = 1 + 1 + 1 + 1 + 1 + 1 (case (1))

= 1 + 1 + 1 + 1 + 2 (case (2))

= 1 + 1 + 1 + 3 (case (3), case (4), case (5))

= 1 + 1 + 2 + 2 (case (6))

= 1 + 1 + 4 (case (7), case (8))

= 1 + 2 + 3 (case (9), case (10), case (11))

= 1 + 5 (case (12), case (13))

= 2 + 2 + 2 (case (14))

= 2 + 4 (case (15), case (16))



= 3 + 3 (case (17)–case (22))

= 6 (case (23), case (24)).

Proposition 6.15. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to a rationaldouble point graph in Theorem 4.2. Suppose also that Γ1 intersects with A∗ but isdisjoint from other effective components. Let Z1 be the fundamental cycle on Γ1.Suppose A∗ · Z1 = 1. If the coefficient z∗ of A∗ in Z is 1 and A2

∗ = −6, then sucha graph does not exist.

Proof. The proof is similar to those of Propositions 6.9, 6.10 and 6.11. Theorem 6.16. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (3) of Proposition 6.2 holds, i.e., if thereexists one effective component A∗, and A2

∗ = −6, z∗ = 1, then the weighted dualgraph Γ of the exceptional set is one of the following forms.

(1) ∗A∗ −6 r r ≥ 1..............

................. ....... ........ ......... ......... ........ ....... ....... .......... ..............

.......................................................................................................

.......r ≥ 1Z = 1 .......

.................

....... ....... ........ ......... ......... ........ ....... ....... .......... ..............................................................................................................

..............

(2) (i) ∗

A∗−6

Z = 1

1

2

1

1

(ii) ∗

A∗−6 Z = 1

1

2 . . . . . .

1

2 1

(3) ∗ A∗−6

Z = 1 2

1

2

3 2 1

(4) ∗ A∗

−6Z = 1 2 3

2

4 3 2 1

(5) ∗ A∗

−6Z = 2 4

3

6 5 4 3 2 1

Proof. This follows easily from Proposition 6.3, Corollary 6.4 and Proposition 6.15.

Proposition 6.17. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to a rational



double point graph in Theorem 4.2. Suppose also that Γ1 intersects with A∗ but isdisjoint from other effective components. Let Z1 be the fundamental cycle on Γ1.Suppose A∗ ·Z1 = 1. If the coefficient z∗ of Z∗ in Z is 2 and A2

∗ = −3, then A∗∪Γ1


(1) A∗

−3∗ Z

∣∣∣∣A∗∪Γ1

= 2 1

(2) ∗

A∗−3

Z

∣∣∣∣A∗∪Γ1

= 1

2

2 1

(3) ∗

A∗−3

Z

∣∣∣∣A∗∪Γ1

= 1 2

2

3 2 1

(4) ∗

A∗−3 Z

∣∣∣∣A∗∪Γ1

= 1 2 3

2

4 3 2 1

(5) ∗

A∗−3 Z

∣∣∣∣A∗∪Γ1

= 1 2 3 4

2

5 4 3 2 1

(6) ∗A∗

−3 Z

∣∣∣∣A∗∪Γ1

= 2 2

1

2 1

(7) ∗A∗

−3 Z

∣∣∣∣A∗∪Γ1

= 2 3

2

4 3 2 1

(8) ∗A∗

−3 Z

∣∣∣∣A∗∪Γ1

= 2 4

3

6 5 4 3 2 1

(9) ∗A∗

−3 Z

∣∣∣∣A∗∪Γ1

= 2 5

4

8 7 6 5 4 3 2 1

(10) r

∗

−3

A∗

Z

∣∣∣∣A∗∪Γ1

= 1

1

2 2 . . . . . . 2

(11) ∗A∗

−3Z

∣∣∣∣A∗∪Γ1

= 2 4

3

6 5 4 3 2

Proof. The proof is similar to those of Propositions 6.9, 6.10 and 6.11.

Proposition 6.18. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1 and A∗2 be two effective com-ponents of Γ. Suppose that Γ1 is a connected component of Γ′ which corresponds toa rational double point graph in Theorem 4.2. Suppose also that Γ1 intersects withboth A∗1 and A∗2, but no other effective component. Let Z1 be the fundamentalcycle on Γ1. Suppose A∗1 · Z1 = A∗2 · Z1 = 1. If A∗1 · A∗2 = 0 and the coefficients



z∗1 of A∗1 and z∗2 of A∗2 in Z are 2 and A2∗1 = A2

∗2 = −3, then A∗1 ∪ A∗2 ∪ Γ1

and the restriction of Z on A∗1 ∪ A∗2 ∪ Γ1 must be one of the following forms.

(1) ∗

A∗

A∗−3

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2

2

3 2 1

(2) A∗1−3 −3

A∗2

∗ ∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2

2

4

2

2

(3) ∗ ∗−3 −3

A∗1 A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 2 . . . . . . 2︸︷︷︸r≥1

2

(4) ∗

A∗1−3

∗

−3

A∗2Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3

2

4 3 2

(5) ∗

A∗1−3 ∗

−3

A∗2Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3 4

2

5 4 3 2

(6) ∗ ∗

r

r ≥ 0

−3 A∗1 A∗2−3

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2

2

4 4 . . . . . . 4︸︷︷︸r≥0

2

4 2

(7) ∗A∗2

−3

A∗1−3 ∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2

2

5

3

6 4 2

(8) ∗ ∗

∗(m − 4)

m − 4 ≥ 0

−3 A∗1 A∗2−3

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2

2

4 4 . . . . . . 4︸︷︷︸m−4≥0

2

4 2

(9) ∗A∗1

−3 A∗1−3 ∗

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 3

2

3

4 2

(10) ∗A∗1

−3 ∗ A∗1−3

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4

2

4

6 4 2



(11) ∗A∗1

−3 ∗A∗1−3

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 5

2

5

8 4 2

(12) ∗A∗1

−3 A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4

3

6 5 4 3 2

(13) ∗A∗1

−3 A∗2

−3 ∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 5

4

8 7 6 5 4 3 2

(14) ∗A∗1

−3 A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4 6

4

8 6 4 2

Proof. (I) Assume that Γ1 is of the form of case (1) in Theorem 4.2.Consider A∗1 and A∗2 attaching on Γ1 in the following form:

A∗1 A1 A2 Am

A∗2−3

−3∗ ∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 22n1 n2 . . . . . . nm.

As in the proof of Proposition 6.9, we have m = 1 or m = 3. If m = 1, then we arein case (3). If m = 3, then we are in case (1).

Consider A∗1 and A∗2 attaching on Γ1 in the following form:

A′

m1A′

2 A1 A2 Am1

A∗1−3−3

A∗2

∗ ∗

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= n′m2

. . . n′2

2

n1

2

n2 . . . nm1 .

As in the proof of Proposition 6.9, we have either nm1 = 3, n′m2

= 1 or nm1 =2 = n′

m2.

If nm1 = 3, n′m2

= 1, then m2 = 3m1 and n1 = 3m1. Since −1 = A2∗1 + 2 =

A∗1 · (−K) = A∗1 · Z ≥ 2(−3) + n1 ⇒ n1 = 3m1 ≤ 5, therefore m1 = 1, m2 = 3and we are in case (1).

If nm1 = 2 = n′m2

, then m1 = m2 and n1 = 2m1. The same argument as aboveshows that 2m1 ≤ 5, i.e., m1 ≤ 2. If m1 = 1, then we are in case (3). If m1 = 2,then we are in case (2).

Consider A∗1 and A∗2 attaching a Γ1 in the following form:

∗ ∗A∗1 A1 A2 Am A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 n1 n2 . . . nm 2.



Since Ai · Z = Ai · (−K) = A2i + 2 = 0, 1 ≤ i ≤ m, we have⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

−2n1 + 2 + n2 = 0−2n2 + n1 + n3 = 0

...−2nm−1 + nm−2 + nm = 0,

(6.15)

−2nm + nm−1 + 2 = 0.(6.16)

(6.15) implies

(6.17) nj = jn1 − 2(j − 1), 2 ≤ j ≤ m.

(6.16) and (6.17) imply n1 = 2 = n2 = · · · = nm. We are in case (3).Consider A∗1 and A∗2 attaching on Γ1 in the following form:

∗−3

A∗2

∗ A∗1−3

A′m2

A′2 A1 A2 Am1 Z

∣∣∣∣A∗1∪A∗2∪Γ1

= n′m2

. . . n′2

2n1 n2 . . . nm1 2.

Since Ai · Z = Ai · (−K) = A2i + 2 = 0, 1 ≤ i ≤ m and A′

j · Z = A′j · (−K) =

A′2j + 2 = 0, 2 ≤ j ≤ m2, we have⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2nm1 + nm1−1 + 2 = 0−2nm1−1 + nm1−2 + nm1 = 0

...−2n3 + n2 + n4 = 0−2n2 + n1 + n3 = 0,

(6.18)

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2n′m2

+ n′m2−1 = 0

−2n′m2−1 + n′

m2−2 + n′m2

= 0...

−2n′3 + n′

2 + n′4 = 0

−2n′2 + n1 + n′

3 = 0,

(6.19)

2 − 2n1 + n2 + n′2 = 0.(6.20)

(6.18) implies

(6.21) nj = (m1 − j + 1)nm1 − 2(m1 − j), 1 ≤ j ≤ m1 − 1.

(6.19) implies

(6.22) n′j = (m2 − j + 1)n′

m2, 1 ≤ j ≤ m2 − 1.

(6.21) and (6.22) imply

(6.23) m1nm1 − 2(m1 − 1) = m2n′m2

= n1.

(6.20), (6.21) and (6.22) imply nm1 + n′m2

= 4. We have either nm1 = 3, n′m2

= 1or nm1 = 2 = n′

m2.

If nm1 = 3 and n′m2

= 1, then (6.23) implies m2 = m1 +2 = n1. −1 = A2∗1 +2 =

A∗1 · (−K) = A∗1 · Z ≥ 2(−3) + n1 implies m1 + 2 = n1 ≤ 5, i.e., m1 ≤ 3. If



m1 = 1, then m2 = 3 and we are in case (1). If m1 = 2, then m2 = 4 and we arein case (4). If m1 = 3, then m2 = 5 and we are in case (5).

If nm1 = 2 = n′m2

, then (6.23) implies m2 = 1 and we are in case (3).Consider A∗1 and A∗2 attaching on Γ1 in the following form:

∗−3

Am1+1 Am1+m3

∗ A∗1 A∗2−3

A′m2

A′2 A1 A2 Am1

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= n′m2

. . . n′2

2n1 n2 · · ·

2nm1 nm1+1 . . . nm1+m3 .

By the same argument as before, we have the following equations:⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2n′m2

+ n′m2−1 = 0

−2n′m2−1 + n′

m2−2 + n′m2

= 0...

−2n′3 + n′

2 + n′4 = 0

−2n′2 + n1 + n′

3 = 0,

(6.24)

− 2n1 + n′2 + n2 + 2 = 0,(6.25) ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2n2 + n1 + n3 = 0−2n3 + n2 + n4 = 0

...−2nm1−2 + nm1−3 + nm1−1 = 0−2nm1−1 + nm1−2 + nm1 = 0,

(6.26)

− 2nm1 + nm1−1 + nm1+1 + 2 = 0,(6.27) ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

−2nm1+1 + nm1 + nm1+2 = 0−2nm1+2 + nm1+1 + nm1+3 = 0

...−2nm1+m3−1 + nm1+m3−2 + nm1+m3 = 0−2nm1+m3 + nm1+m3−1 = 0.

(6.28)

(6.24) implies

(6.29) n′j = (m2 − j + 1)n′

m2, 1 ≤ j ≤ m2 − 1.

(6.25) and (6.29) imply

(6.30) n2 = (m2 + 1)n′m2

− 2.

(6.30) and (6.26) imply

(6.31) nj = (m2 + j − 1)n′m2

− 2(j − 1), 3 ≤ j ≤ m1.

(6.28) implies

(6.32) nm1+j = (m3 − j + 1)nm1+m3 , 0 ≤ j ≤ m3 − 1.

(6.31) and (6.32) imply

(6.33) nm1 = (m2 + m1 − 1)n′m2

− 2(m1 − 1) = (m3 + 1)nm1+m2.



(6.31), (6.32) and (6.27) imply

(6.34) (m2 + m1)n′m2

− 2m1 = m3nm1+m3 + 2.

(6.33) and (6.34) imply n′m2

+ nm1+m3 = 4. Therefore we have either n′m2

= 1,nm1+m3 = 3 or n′

m2= 2 = nm1+m3 .

If n′m2

= 1 and nm1+m3 = 3, then m2 = m1 +3m3 +2 by (6.34) and n1 = m2 by(6.29). Since −1 = A2

∗1 + 2 = A∗1(−K) = A∗1 · Z ≥ 2(−3) + n1, we have m2 ≤ 5.Hence m1 + 3m3 ≤ 3. Since m1 ≥ 2, we have either m3 = 0, m1 = 2, m2 = 4, orm3 = 0, m1 = 3, m2 = 5. If m3 = 0, m1 = 2, m2 = 4, then we are in case (4). Ifm3 = 0, m1 = 3, m2 = 5, then we are in case (5).

If n′m2

= 2 = nm1+m3 , then n1 = 2m2 and m2 = m3 + 1 by (6.33). Since−1 = A2

∗1 + 2 = A∗1 · (−K) = A∗1 · Z ≥ 2(−3) + n1, we have 2m2 ≤ 5, whichimplies m2 ≤ 2 and m3 ≤ 1. If m3 = 0 and m2 = 1, then we are in case (3). Ifm3 = 1 and m2 = 2, then nj = 4, 1 ≤ j ≤ m1 and we are in case (6).

(II) Assume that Γ1 is of the form Dm of case (2) in Theorem 4.2.


−3

A∗1∗

−3

A1

∗A∗2

A3

A4

A2

Am

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 22n1

n2

n3 n4 . . . nm.

As in the proof of Proposition 6.10, we have nm = 2, n1 = m, n2 = m− 2, nj =2(m− j +1), 3 ≤ j ≤ m. Since −1 = A2

∗1 +2 = A∗1 · (−K) = A∗1 ·Z ≥ 2(−3)+n1,we have m ≤ 5. If m = 4, then we are in case (8). If m = 5, then we are in case (7).


∗ ∗

A2 A∗2−3

A1 A3 A4 Am A∗1

−3Z

∣∣∣∣A∗1∪A∗2∪Γ1

= n1

n2

n3 n4 . . .

2nm 2.

As in the proof of Proposition 6.10, we have n1 = n2 = 2, n3 = · · · = nm = 4and we are in case (8).


−3

A∗1∗

−3

A1

∗

A∗2

A3

A4

A2

Am

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 n1

2n2

n3 n4 . . . nm.

By the same argument as before, we have the following equations:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + 2 + n3 = 0−2n2 + 2 + n3 = 0−2n4 + n3 + n5 = 0

...−2nm−1 + nm−2 + nm = 0−2nm + nm−1 = 0,

(6.35)

− 2n3 + n1 + n2 + n4 = 0.(6.36)



(6.35) implies

(6.37) n1 = n2 = 1 +m − 2

2nm, nj = (m − j + 1)nm, 3 ≤ j ≤ m.

(6.36) and (6.37) imply nm = 2 and n1 = n2 = m − 1. Since −1 = A2∗1 + 1 =

A∗1 · (−K) = A∗1 · Z ≥ 2(−3) + n1, we have m ≤ 6. Hence we are in case (9) andcase (10) and case (11).


−3

A∗1∗

A1

A3

A4

A2

Am

−3∗A∗2

Z

∣∣∣∣A1∗∪A2∗∪Γ1

= 2 n1

n2

n3 n4 . . . nm 2.

By the same argument as before, we have the following equations:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + 2 + n3 = 0−2n2 + n3 = 0−2n3 + n1 + n2 + n4 = 0−2n4 + n3 + n5 = 0

...−2nm−1 + nm−2 + nm = 0,

(6.38)

− 2nm + nm−1 + 2 = 0.(6.39)

(6.38) implies

(6.40) n1 = 1 + n2, nj = 2n2 − (j − 3), 3 ≤ j ≤ m.

(6.39) and (6.40) imply n2 = m2 . In particular m is even. Since −1 = A2

∗1 + 2 =A∗1 · (−K) = A∗1 · Z ≥ 2(−3) + n1, we have n1 ≤ 5, which implies 1 + m

2 ≤ 5and hence m ≤ 8. If m = 4, 6, 8, then we are in case (9), case (12) and case (13)respectively.

(III) Assume that Γ1 is of the form E6 of case (3) in Theorem 4.2.

Consider A∗1 and A∗2 attaching on E6 in the following form:

∗A∗1

−3

A∗2−3 ∗ A4 A6A5A3A2A1

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 22n1 n2

n4

n3 n5 n6.

As in the proof of Proposition 6.11, we find out that this case is not possible.Consider A∗1 and A∗2 attaching on E6 in the following form:

∗−3

A∗1

−3

A∗2

A4 ∗A6A5A3A2A1

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 n1 n2

n4

n3 n5 n6 2.



By the same argument as before, we have the following equations:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + 2 + n2 = 0−2n2 + n1 + n3 = 0−2n3 + n2 + n4 + n5 = 0−2n4 + n3 = 0−2n5 + n3 + n6 = 0−2n6 + n5 + 2 = 0.

An easy exercise shows that we are in case (14).

(IV) Assume that Γ1 is of the form E7 of case (4) in Theorem 4.2.

Consider A∗1 and A∗2 attaching on E7 in the following form:

A1

A2

A3

A5

A4 A6

A∗1A7

A∗2

∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= n1 n2

n4

n3 n5

2n6 n7 2.

By the same argument as before, we have the following equations:

(6.41)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−2n1 + n2 = 0−2n2 + n1 + n3 = 0−2n3 + n2 + n4 + n5 = 0−2n4 + n3 = 0−2n5 + n3 + n6 = 0−2n6 + n5 + n7 = 0−2n7 + n6 + 4 = 0.

(6.11) implies

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 4 8612 10 8

26 2.

Since −1 = A2∗1 + 2 = A∗1 · (−K) = A∗1 · Z ≥ 2(−3) + 6 = 0, this is absurd. This

case is impossible.

(V) Assume that Γ1 is of the form E8 of case (5) in Theorem 4.2. This casecannot happen because A∗1 · Z1 ≥ 2. Theorem 6.19. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (4) of Proposition 6.2 holds, i.e., if thereexist two effective components A∗1 and A∗2 with A2

∗1 = −3 = A2∗2 and z∗1 = 2 =

z∗2, then the weighted dual graph Γ of the exceptional set is one of the followingforms.

(1a1) ∗

−3

A∗1

A∗2

∗

Z = 1

1

2

1

2

3

1

2 1



(1a2) ∗

A∗1−3

∗−3

A∗2

r

r ≥ 0

Z = 1

1

2

1

2

3 2 . . .

1

2 1

(1b) r ≥ 0

r

A∗1∗

−3

A∗2∗

−3

s ≥ 0

s Z = 1

1

2 . . . 2

1

2

3 2 . . .

1

2 1

(3a1) ∗

r ≥ 0

r

A∗1

−3

A∗2∗ −3

Z = 1

1

2

1

. . .

1

2

1

1

(3a2) ∗

r ≥ 0

r

A∗1

−3

A∗2∗ −3

s ≥ 0

s Z = 1

1

2

1

. . .

1

2 . . .

1

2 1

(3a3) ∗

r ≥ 0

r

A∗1

−3

A∗2∗−3 Z = 1

1

2

1

. . . 2

1

2

3 2 1

(3a4) ∗

r ≥ 0

r

A∗1

−3

A∗2∗−3 Z = 1

1

2

1

. . . 2 3

2

4 3 2 1

(3a5)−3

A∗1∗

r ≥ 0

r

A∗2

−3∗ Z = 1

1

2

1

. . . 2 3 4 5

3

6 4 2

(3b1)

s ≥ 0

s

r ≥ 0

r

t ≥ 0

t−3∗A∗1

−3

A∗2∗

Z = 1

1

2 . . .

1

2 . . .

1

2 . . .

1

2 1

(3b2)

s ≥ 0

s

A∗1∗−3

r ≥ 0

r

A∗2∗−3

Z = 1

1

2 . . .

1

2 . . . 2

1

2

3 2 1

(3b3)

s ≥ 0

s

A∗1∗−3

r ≥ 0

r

A∗2∗−3 Z = 1

1

2 . . .

1

2 . . . 2 3

2

4 3 2 1

(3b4)

s ≥ 0

s

A∗1∗−3

r ≥ 0

r

A∗2∗−3 Z = 1

1

2 . . .

1

2 . . . 2 3 4 5

3

6 4 2



(3c1)

A∗1∗

−3

r ≥ 0

r

A∗2∗

−3

Z = 1 2

1

2

3 2 . . . 2

1

2

3 2 1

(3c2)

A∗1∗

−3

r ≥ 0

r

A∗2∗

−3 Z = 1 2

1

2

3 2 . . . 2 3

2

4 3 2 1

(3c3)

A∗1∗

−3

r ≥ 0

r

A∗2∗

−3 Z = 1 2

1

2

3 2 . . . 2 3 4 5

3

6 4 2

(3d1) ∗

−3

A∗1

−3

r ≥ 0

r

A∗2∗ Z = 1 2 3

2

4 3 2 . . . 2 3

2

4 3 2 1

(3d2) ∗

−3

A∗1

−3

r ≥ 0

r

A∗2∗ Z = 1 2 3

2

4 3 2 . . . 2 3 4 5

3

6 4 2

(3e) ∗−3

A∗1

−3

r ≥ 0

r

A∗2∗ Z =2 4

3

6 5 4 3 2 . . . 2 3 4 5

3

6 4 2

(4a1) ∗ −3

∗

A∗1 −3

A∗2Z = 1 2 3

1

2

4 3

1

2 1

(4a2) A∗1 ∗

−3

A∗2

−3∗

r ≥ 0

r Z = 1 2 3

1

2

4 3 2 . . .

1

2 1

(5a1) A∗1 ∗

∗ −3A∗2

−3 ∗

Z = 1 2 3 4

2

5 4 3

1

2 1

(5a2) ∗A∗1 −3 ∗

r ≥ 0

r

A∗2

−3 Z = 1 2 3 4

2

5 4 3 2 . . .

1

2 1

(9a1) ∗−3

A∗1

∗

−3

A∗2

Z = 1

1

2 3

2

4 3

1

2 1

(9a2) ∗−3

A∗1

−3

A∗2∗

r ≥ 0

r Z = 1

1

2 3

2

4 3 2 . . .

1

2 1



(9b) r ≥ 0

r −3

A∗1∗ ∗−3

A∗2

s Z = 1

1

2 . . . 2 3

2

4 3 2 . . .

1

2 1

(12a1) ∗ −3

A∗2

∗

−3

A∗2

Z = 1 2 4

3

6 5 4 3

1

2 1

(12a2) −3∗

A∗1

∗ r ≥ 0

r−3

A∗1Z = 1 2 4

3

6 5 4 3 2 . . .

1

2 1

(13a1) ∗ −3

A∗2

−3

A∗2∗

Z = 2 5

4

8 7 6 5 4 3

1

2 1

(13a2) ∗−3

A∗1

∗A∗2

−3∗ r

r ≥ 0

Z = 2 5

4

8 7 6 5 4 3 2 . . .

1

2 1

Proof. Since the singularity is minimally elliptic, A2∗i = −3, z∗i = 2 for i = 1, 2, we

have

(6.42) A∗i · (Z − 2A∗i) = −A∗i · (K + 2A∗i) = A2∗i + 2 − 2A2

∗i = 5.

Let Γ′ be the graph obtained by deleting A∗1 and A∗2 from Γ. Let Γ1, . . . , Γm bethe connected components of Γ′ with fundamental cycles Z1, . . . , Zm respectively.(6.42) implies that

(6.43)m∑

j=1

A∗i · Z/Γj = 5, for i = 1, 2.

Since we have two effective components, by Corrollay 6.4 we have

(6.44) A∗i · Zj = 1 for i = 1, 2 and 1 ≤ j ≤ m.

Consider first that A∗1 and A∗2 do not meet. Then Proposition 6.18 applies.For case (1) of Proposition 6.18 if the decomposition (6.43) at A∗1 is 5 = 1 + 1 + 3and the decomposition (6.43) of A∗2 is 5 = 1 + 1 + 3, then we are in case (1a1). Ifthe decomposition (6.43) at A∗1 is 5 = 1 + 1 + 3 and the decomposition of (6.43)at A∗2 is 2 + 3, then we are in case 1(a2). If the decomposition (6.43) at A∗1 andA∗2 are 2 + 3, then we are in case (1b).

For case (2) of Proposition 6.18, the decomposition of (6.43) at A∗1 and A∗2 mustbe 5 = 4+1. From Proposition 6.17, we obtain a possible dual graph together witha proposed fundamental cycle. One may check that the proposed fundamental cycledoes not meet the minimum condition required by Definition 2.1. Therefore thereis no dual graph produced from this case.

For case (3) of Proposition 6.18, if the decomposition of (6.43) at A∗1 is 5 =2+1+1+1 and the decomposition at A∗2 is either 5 = 2+1+1+1, or 5 = 2+2+1, or5 = 2+3, according to Proposition 6.17, we are in case (3a1), . . . , (3a5) respectively.



If the decomposition of (6.43) at A∗1 is 5 = 2+2+1 and the decomposition at A∗2is 5 = 2+2+1, or 5 = 2+3, we are in case (3b1), . . . , (3b4). If the decompositionsof (6.43) at A∗1 and at A∗2 are both 5 = 2 + 3, we are in case (3c1), (3c2), (3c3),(3d1), (3d2), and (3e).

For case (4) of Proposition 6.18, if the decomposition of (6.43) at A∗1 is 5 = 4+1and the decomposition of (6.43) at A∗2 is 5 = 3 + 1 + 1, we have case 4(a1). If thedecompositions of (6.43) at A∗1 and A∗2 are 5 = 4 + 1 and 5 = 3 + 2 respectively,we have case 4(a2).

For case (5) of Proposition 6.18, the decomposition of (6.43) at A∗1 must be5 = 5+0. If the decomposition of (6.43) at A∗2 is 5 = 3+1+1, we have case 5(a1).If the decomposition of (6.43) at A∗2 is 5 = 3 + 2, we have case 5(a2).

For case (6) of Proposition 6.18, the decompositions of (6.43) at A∗1 and A∗2must be 5 = 4 + 1. For case (7) of Proposition 6.18, the decomposition of (6.43) atA∗1 and A∗2 must be 5 = 5+0. For case (8) of Proposition 6.18, the decompositionsof (6.43) at A∗1 and A∗2 must be 5 = 4 + 1. In all these cases, the proposedfundamental cycles on the possible dual graphs obtained via Proposition 6.17 donot meet the minimum condition required in Definition 2.1. Therefore there is nodual graph produced from these cases.

For case (9) of Proposition 6.18, if the decomposition of (6.43) at A∗1 is 5 =3 + 1 + 1 and the decomposition of (6.43) at A∗2 is 5 = 3 + 1 + 1 or 5 = 3 + 2, wehave cases 9(a1) and 9(a2) respectively. If the decomposition of (6.43) at A∗1 andA∗2 is 5 = 3 + 2, we have case 9(b).

For case (10) of Proposition 6.18, the decomposition of (6.43) at A∗1 and A∗2must be 5 = 4+1. For case (11) of Proposition 6.18, the decomposition of (6.43) atA∗1 and A∗2 must be 5 = 5 + 0. The proposed fundamental cycles on the possibledual graphs obtained via Proposition 6.17 do not meet the minimum condition ofDefinition 2.1. Hence there is no dual graph produced from these two cases.

For case (12) of Proposition 6.18, the decomposition of (6.43) at A∗1 must be5 = 4 + 1. The decomposition of (6.43) at A∗2 must be 5 = 3 + 1 + 1 or 5 = 3 + 2.We have cases 12(a1) and 12(a2) respectively.

For case (13) of Proposition 6.18, the decomposition of (6.43) at A∗1 must be5 = 5 + 0. The decomposition of (6.43) at A∗2 must be 5 = 3 + 1 + 1 or 5 = 3 + 2.We have cases 13(a1) and 13(a2) respectively.

For case (14) of Proposition 6.18, the decomposition of (6.43) at A∗1 and A∗2must be 5 = 4 + 1. Again the proposed fundamental cycle does not meet theminimum condition of Definition 2.1. There is no dual graph produced from thiscase.

We next consider the case A∗1 · A∗2 > 0. Since the singularity is minimallyelliptic and z∗1 = z∗2 = 2, it follows that A∗1 · A∗2 = 1. Then we are in cases(3a1)–(3e) by an argument similar to the above.

Proposition 6.20. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to a rationaldouble point graph in Theorem 4.2. Suppose also that Γ1 intersects with A∗ but isdisjoint from other effective components. Let Z1 be the fundamental cycle on Γ1.Suppose A∗ ·Z1 = 1. If the coefficient z∗ of A∗ in Z is 3 and A2

∗ = −3, then A∗∪Γ1




(1)−3

A∗∗ Z

∣∣∣∣A∗∪Γ1

= 3 2 1

(2) ∗

A∗−3

Z

∣∣∣∣A8∪Γ1

= 1 2 3

3

4 2

(3) ∗

A∗−3 Z

∣∣∣∣A∗∪Γ1

= 1 2 3 4 5

3

6 4 2

(4) ∗A∗−3 Z

∣∣∣∣A∗∪Γ1

= 1 2 3 4 5 6 7

3

8 6 4 2

(5)A∗

−3∗ Z

∣∣∣∣A∗∪Γ1

= 3 4 5

3

6 4 2

Proof. The proof is similar to those of Propositions 6.9, 6.10 and 6.11. Proposition 6.21. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to a rationaldouble point graph in Theorem 4.2. Suppose also that Γ1 intersects with A∗, but isdisjoint from other effective components. Let Z1 be the fundamental cycle on Γ1.Suppose A∗ · Z1 = 1. If the coefficient z∗ of A∗ in Z is 1 and A2

∗ ≤ −3, then sucha graph does not exist.

Proof. The proof is similar to those of Propositions 6.9, 6.10, and 6.11. Proposition 6.22. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1 and A∗2 be two effective com-ponents of Γ. Suppose that Γ1 is a connected component of Γ′ which correspondsto a rational double point graph in Theorem 4.2. Suppose also that Γ1 intersectswith A∗1 and A∗2, but is disjoint from other effective components. Let Z1 be thefundamental cycle on Γ1. Suppose A∗1 ·Z1 = 1 = A∗2 ·Z1. If A∗1 ·A∗2 = 0, the coef-ficients z∗1 and z∗2 of A∗1 and A∗2 are 3 and 1 respectively, and A2

∗1 = A2∗2 = −3,

then A∗1 ∪ A∗2 ∪ Γ1 and the restriction of Z on A∗1 ∪ A2 ∪ Γ1 must be one of thefollowing forms.

(1)−3

A∗1∗ −3

A∗2∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 3 2 1

(2) −3 ∗A∗1 A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2

3

4 3 2 1



(3) −3 ∗ A∗1 A∗2

−3 ∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4

3

6 5 4 3 2 1

(4) −3 ∗ A∗1 A∗2

−3 ∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4 6

3

8 7 6 5 4 3 2 1

Proof. The proof is the same as those in Proposition 6.18. Theorem 6.23. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (5) of Proposition 6.2 holds, i.e., if thereexist two effective components A∗1 and A∗2 with A2

∗1 = −3 = A2∗2 and z∗1 = 3,

z∗2 = 1, then the weighted dual graph Γ of the exceptional set is one of the followingforms.

(1a) −3∗A∗1

A∗2

−3∗ Z = 1 2

1

2

3

2

1

2 1

(1b1)

A∗1∗−3

−3∗A∗2

Z = 1 2 3

2

4

1

2

3 2 1

(1b2) ∗

A∗1

−3 −3∗A∗2

Z = 2 4

3

6 5 4

1

2

3 2 1

(1c) A∗1∗−3

−3∗A∗2

Z = 1 2 3 4 5

2

4

6 3 2 1

(2a)

A∗1∗

A∗2∗ Z = 1 2

1

2

3

2

4 3 2 1

(2b1)

∗−3

A∗1

−3

A∗2∗ Z = 1 2 3

2

4 3

2

4 3 2 1

(2b2) ∗−3

A∗1

−3

A∗2∗ Z = 2 4

3

6 5 4 3

2

4 3 2 1



(3) −3

A∗1∗

−3

A∗2∗ Z = 1 2 3

2

4

6 5 4 3 2 1

(4) −3A∗1 ∗ −3

A∗2∗ Z = 2 4 6

3

8 7 6 5 4 3 2 1

Proof. Since z∗2 = 1, z∗1 = 3, A2∗2 = −3 and A∗2 · Z = −1, we have A∗1 · A∗2 = 0.

The proof is similar to those of Theorem 6.19 by using Propositions 6.20, 6.21,and 6.22. Proposition 6.24. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtaeind byremoving all the effective components of Γ. Let A∗1 and A∗2 be two effective com-ponents of Γ. Suppose that Γ1 is a connected component of Γ′ which correspondsto a rational double point graph in Theorem 4.2. Suppose also that Γ1 intersectswith A∗1 and A∗2, but is disjoint from other effective components. Let Z1 be thefundamental cycle on Γ1. Suppose that A∗1 · Z1 = 1 = A∗2 · Z1. If A∗1 · A∗2 = 0,z∗1 = 2, z∗2 = 1 (coefficients of A∗1 and A∗2 in Z respectively), and A2

∗1 = −3,A2

∗2 = −4 or −3, then A∗1 ∪ A∗2 ∪ Γ1 and the restriction of Z on A∗1 ∪ A∗2 ∪ Γ1

must be one of the following forms. (In case A2∗2 = −3, replace −4 by −3, in the

following graphs.)

(1) −3 ∗A∗1

∗−4 A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1

2

2 1

(2) −3 ∗A∗1

∗−4 A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2

2

3 2 1

(3) −3 ∗A∗1 −4

∗A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3

2

4 3 2 1

(4) −3 ∗A∗1 −4

∗A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3 4

2

5 4 3 2 1

(5) ∗A∗1

−3 r

r ≥ 0

∗A∗2

−4Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 . . . . . .

1

2 1

(6) ∗A∗1

−3 ∗A∗2

−4Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 3

2

4 3 2 1



(7) ∗A∗1

−3 ∗A∗2

−4Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4

3

6 5 4 3 2 1

(8) ∗A∗1

−3 ∗A∗2

−4Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 5

4

8 7 6 5 4 3 2 1

Proof. The proof is the same as that in Proposition 6.22. Theorem 6.25. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (6) of Proposition 6.2 holds, i.e., if thereexist two effective components A∗1 and A∗2 with A2

∗1 = −3, A2∗2 = −4 and z∗1 = 2,

z∗2 = 1, then the weighted dual graph Γ of the exceptional set is one of the followingforms.

(1a) ∗−3 −4

A∗2

A∗1

∗ Z = 1

1

2

1

1

2 1

(1b) r

r ≥ 0∗

−3

A∗1

−4

A∗2

∗ Z = 1

1

2 . . .

1

2

1

2 1

(1c1)

∗ ∗−3

A∗1

−4

A∗2Z = 1 2

1

2

3 2

1

2 1

(1c2) ∗ ∗ −4

A∗2

−3

A∗1Z = 1 2 3

2

4 3 2

1

2 1

(1c3) −3

A∗1∗ −4

A∗2∗ Z = 2 4

3

6 5 4 3 2

1

2 1

(2a) ∗−3

A∗1

∗−4

A∗2Z = 1

1

2

1

2

3 2 1

(2b) r ≥ 0

r ∗−3

A∗1∗

∗−4

A∗2Z = 1

1

2 . . . 2

1

2

3 2 1



(3a) ∗−3

A∗1

∗−4

A∗2Z = 1 2

1

2

3

4 3 2 1

(4) ∗−3 A∗1 ∗−4

A∗2Z = 1 2 3 4

2

5 4 3 2 1

(5a) ∗

r ≥ 0

r ∗−3

A∗1

∗−4

A∗2Z = 1

1

2

1

. . .

1

2 1

(5b) s ≥ 0

s ∗

−3

A∗1 r ≥ 0

r

∗−4

A∗2Z = 1

1

2 . . .

1

2 . . .

1

2 1

(5c1)

A∗1∗−3

r ≥ 0

r

∗−4

A∗2Z = 1 2

1

2

3 2 . . .

1

2 1

(5c2) −3

A∗1∗

r ≥ 0

r ∗−4

A∗2Z = 1 2 3

2

4 3 2 . . .

1

2 1

(5c3) −3

A∗1∗

r ≥ 0

r ∗−4

A∗2Z = 2 4

3

6 5 4 3 2 . . .

1

2 1

(6a) ∗−3

A∗1

∗−4

A∗2Z = 1

1

2 3

2

4 3 2 1

(6b) r ≥ 0

r ∗−3

A∗1∗ ∗−4

A∗2Z = 1

1

2 . . . 2 3

2

4 3 2 1

(7a) ∗−3

A∗1

∗−4

A∗2Z = 1 2 4

3

6 5 4 3 2 1

(8) ∗−3

A∗1

∗−4

A∗2Z = 2 5

4

8 7 6 5 4 3 2 1

(9) −3

∗

A∗1

A∗2

−4∗ Z = 1

1 1

2

1

1



(10) −3

A∗1

r ∗

∗−4

A∗2Z = 1

1

2 . . .

1

2

1

1

(11)

∗

−3

A∗1

−4

A∗2∗ Z = 1 2

1

2

3 . . .

1

2 1

(12) ∗

−3

A∗1

−4

A∗2∗ Z = 1 2 3

2

4 3

1

2 1

(13) A∗1∗

−3 −4

A∗2∗ Z = 2 4

3

6 5 4 3

1

2 1

(14) r ≥ 0

r∗

∗−4

−3

A∗2

A∗1s ≥ 0

s

Z = 2

1

2 . . .

1

2 . . .

1

2 1

(15)

A∗1∗∗−3

A∗2∗−4

Z = 1 2 3

1

2

3

4 2 1

(16) A∗1

−3∗∗A∗2∗−4

Z = 1 2 3 4 5

3

6 4 2 1

Proof. If A∗1 · A∗2 = 0, then the proof is similar to that of Theorem 6.23 by usingProposition 6.21, Proposition 6.24 and Proposition 6.17. We have case (1a) tocase (8).

If A∗1 ·A∗2 = 0, then A∗1 ·A∗2 = 1. It follows that A∗1 · (Z − 2A∗1−A∗2) =−A∗1 · (K + 2A∗1 + A∗2) = −A2

∗1 + 1 = 4. For 4 = 1 + 1 + 1 + 1, we are in case (9).For 4 = 1 + 1 + 2, we are in case (10). For 4 = 1 + 3, we are in case (11), case (12)and case (13). For 4 = 2 + 2, we are in case (15). For 4 = 4, we are in case (16)and case (17).

Proposition 6.26. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1 and A∗2 be two effective com-ponents of Γ. Suppose that Γ1 is a connected component of Γ′ which correspondsto a rational double point graph in Theorem 4.2. Suppose also that Γ1 intersects



with both A∗1 and A∗2, but with no other effective component. Let Z1 be the fun-damental cycle on Γ1. Suppose A∗1 · Z1 = A∗2 · Z1 = 1. If A∗1 · A∗2 = 0 andthe coefficients z∗1 of A∗1 and z∗2 of A∗2 in Z are one and A2

∗1 ≤ −3, A2∗2 ≤ −3,

then A∗1 ∪ A∗2 ∪ Γ1 and the restriction of Z on A∗1 ∪A∗2 ∪ Γ1 must be one of thefollowing forms.

(1) ∗A∗1

rr ≥ 0 A∗2

∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 1 . . . . . . 1

(2) s ∗A∗1 ∗A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1

1

2 . . . . . .

1

2 1

(3) ∗A∗1 A∗2∗

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1

1

2

1

1

(4) A∗1

(m − 4)(m − 4) ≥ 0

∗ ∗A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1

1

2 . . .12 1

(5) ∗A∗1

∗A∗2

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2

1

2

3 2 1

(6) ∗A∗1

A∗2∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3

2

4 3 2 1

Proof. The proof is the same as that in Proposition 6.24. Theorem 6.27. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (7) of Proposition 6.2 holds, i.e., if thereexist two effective components A∗1 and A∗2 with A2

∗1 = −4 = A2∗2 and z∗1 = 1 =

z∗2, then the weighted dual graph Γ of the exceptional set is one of the followingforms.

(1) ∗ ∗A∗1

−4 −4

A∗2

s ≥ 0

r ≥ 0

.............. ....... ....... ....... ........ ......... ........ ......... ......... ........ ........ ........ ........ ....... ........ ..... ...............................................................................................................................................

....Z = .....

......... ....... ....... ....... ........ ......... ........ ......... ......... ........ ........ ........ ........ ....... ........ ..... ...............................................................................................................................................

....1 1

1

1

(2) s ∗A∗1−4 ∗A∗2−4

Z = 1

1

2 2 . . . 2

1

2 1

(3) A∗1

−4∗ ∗−4

A∗2 Z = 1

1

2

1

1



(4) (m − 4) ≥ 0

(m − 4) −4 A∗1∗

∗A∗2

−4 Z = 1

1

2 . . .

1

2 1

(5) −4

A∗1∗

∗A∗2−4

Z = 1 2

1

2

3 2 1

(6) −4

A∗1∗

A∗2

−4∗ Z = 1 2 3

2

4 3 2 1

Proof. This follows from Proposition 6.3, Proposition 6.21 and Proposition 6.26.

Proposition 6.28. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1 and A∗2 be two effective com-ponents of Γ. Suppose that Γ1 is a connected component of Γ′ which correspondsto a rational double point graph in Theorem 4.2. Suppose also that Γ1 intersectswith A∗1 and A∗2, but is disjoint from other effective components. Let Z1 be thefundamental cycle on Γ1. Suppose that A∗1 · Z1 = 1 = A∗2 · Z1. If A∗1 · A∗2 = 0,z∗1 = 1 = z∗2, and A2

∗1 = −3, A2∗2 = −5, then A∗1 ∪ A∗2 ∪ Γ1 and the restriction

of Z on A∗1 ∪ A∗2 ∪ Γ1 must be one of the following forms.

(1) ∗A∗1

−3 r ∗A∗2

−5 Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 1 . . . 1

(2) s

s ≥ 0 ∗A∗1−3 ∗A∗2−5

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1

1

2 . . .

1

2 1

(3) mm ≥ 0

∗

A∗1−3 ∗

A∗2

−5Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1

1

2 . . .

1

2 1

(4) −3

A∗1∗

∗A∗2−5

Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2

1

2

3 2 1

(5) −3

A∗1∗

A∗2

−5∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3

2

4 3 2 1



Proof. The proof is the same as that in Proposition 6.26. Theorem 6.29. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (8) of Proposition 6.2 holds, i.e., ifthere exist two effective components A∗1 and A∗2 with A2

∗1 = −3, A2∗2 = −5 and

z∗1 = z∗2 = 1, then the weighted dual graph Γ of the exceptional set is one of thefollowing forms.

(1) ∗ ∗A∗1

−3 −5

A∗2

s ≥ 0

r ≥ 0

.............. ....... ....... ....... ........ ......... ........ ......... ......... ........ ........ ........ ........ ....... ........ ..... ...............................................................................................................................................

....

.................

..... ........... ..... ..... .......................................

................................................Z = 1

1

1

1

(2) ss ≥ 0

∗A∗1−3 ∗A∗2−5

Z = 1

1

2 . . .

1

2 1

(3) m

m ≥ 0 ∗

A∗1−3 ∗

A∗2

−5Z = 1

1

2 . . .

1

2 1

(4) −3

A∗1∗

∗A∗2−5

Z = 1 2

1

2

3 2 1

(5) −3

A∗1∗

A∗2

−5∗ Z = 1 2 3

2

4 3 2 1

Proof. This follows from Proposition 6.3, Proposition 6.21 and Proposition 6.28.

Proposition 6.30. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1, A∗2 and A∗3 be three effectivecomponents of Γ. Suppose that Γ1 is a connected component of Γ′ which correspondsto a rational double point graph in Theorem 4.2. Suppose also that Γ1 intersects withA∗1, A∗2 and A∗3, but is disjoint from other effective components. Let Z1 be thefundamental cycle on Γ1. Suppose that A∗1·Z1 = 1 = A∗2·Z1 = A∗3·Z1. If A∗1, A∗2and A∗3 are mutually disjoint, z∗1 = 1 = z∗2, z∗3 = 2, and A2

∗1 = −3 = A2∗2 = A2

∗3,then A∗1 ∪ A∗2 ∪ A∗3 ∪ Γ1 and the restriction of Z on A∗1 ∪ A∗2 ∪ A∗3 ∪ Γ1 mustbe one of the following forms.

(1) −3

A∗1

−3

A∗2∗ ∗ ∗ A∗3−3 Z

∣∣∣∣A∗1∪A∗2∪A3∪Γ1

= 1 2

2

3 2 1



(2) −3

A∗1∗ −3 ∗A∗3

A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪A3∪Γ1

= 1 2 3

2

4 3 2 1

(3) −3

A∗1∗ −3 A∗3∗

A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪A3∪Γ1

= 1 2 3 4

2

5 4 3 2 1

(4) ∗−3

A∗1

−3 ∗A∗2

m ≥ 0m −3

A∗3∗ Z

∣∣∣∣A∗1∪A∗2∪A3∪Γ1

= 1

1

2 . . . 2 2

Proof. The proof is the same as that in Proposition 6.28. Proposition 6.31. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗ be an effective component ofΓ. Suppose that Γ1 is a connected component of Γ′ which corresponds to a rationaldouble point graph in Theorem 4.2. Suppose also that Γ1 intersects with A∗ but isdisjoint from other effective components. Let Z1 be the fundamental cycle on Γ1.Suppose A∗ ·Z1 = 1. If the coefficient z∗ of A∗ in Z is 2 and A2

∗ = −3, then A∗∪Γ1


(1)A∗1

−3∗ A

∣∣∣∣A∗∪Γ1

= 2 1

(2) −3∗

A∗ A

∣∣∣∣A∗∪Γ1

= 1

2

2 1

(3) ∗

A∗−3

A

∣∣∣∣A∗∪Γ1

= 1 2

2

3 2 1

(4) ∗

A∗−3 A

∣∣∣∣A∗∪Γ1

= 1 2 3

2

4 3 2 1

(5) ∗

A∗−3 A

∣∣∣∣A∗∪Γ1

= 1 2 3 4

2

5 4 3 2 1

(6) ∗A∗

−3 A

∣∣∣∣A∗∪Γ1

= 2 3

2

4 3 2 1

(7) ∗A∗

−3 A

∣∣∣∣A∗∪Γ1

= 2 4

3

6 5 4 3 2 1

(8) ∗A∗

−3 A

∣∣∣∣A∗∪Γ1

= 2 5

4

8 7 6 5 4 3 2 1



(9) m ≥ 0

m ∗

−3

A∗

A

∣∣∣∣A∗∪Γ1

= 1

1

2 . . . . . . 2

(10) ∗A∗

−3A

∣∣∣∣A∗∪Γ1

= 2 4

3

6 5 4 3 2

Proof. The proof is the same as that in Proposition 6.13. Proposition 6.32. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1 and A∗2 be two effective com-ponents of Γ. Suppose that Γ1 is a connected component of Γ′ which corresponds toa rational double point graph in Theorem 4.2. Suppose also that Γ1 intersects withA∗1 and A∗2, but is disjoint from other effective components. Let Z1 be the funda-mental cycle on Γ1. Suppose that A∗1 ·Z1 = 1 = A∗2 ·Z1. If A∗1 ·A∗2 = 0, z∗1 = 2,z∗2 = 1 (coefficient of A∗1 and A∗2 in Z respectively), and A2

∗1 = −3 = A3∗2, then

A∗1∪A∗2∪Γ1 and the restriction of Z on A∗1∪A∗2∪Γ1 must be one of the followingforms.

(1)A∗1

−3∗

r ≥ 0

r ∗A∗2

−3Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 . . . . . .

1

2 1

(2) A∗1−3 ∗ ∗A∗2

−3Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2

2

3 2 1

(3) −3 ∗A∗1 ∗A∗2

−3Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3

2

4 3 2 1

(4) −3 ∗A∗1 ∗A∗2

−3Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 1 2 3 4

2

5 4 3 2 1

(5)−3

A∗1∗

A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 3

2

4 3 2 1

(6)−3

A∗1∗

A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 4

3

6 5 4 3 2 1

(7)−3

A∗1∗

A∗2

−3∗ Z

∣∣∣∣A∗1∪A∗2∪Γ1

= 2 5

4

8 7 6 5 4 3 2 1

Proof. The proof is the same as that in Proposition 6.24.



Theorem 6.33. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (9) of Proposition 6.2 holds, i.e., if thereexist three effective components A∗1, A∗2 and A∗3 with A2

∗1 = A2∗2 = A2

∗3 = −3 andz∗1 = z∗2 = 1, z∗3 = 2, then the weighted dual graph Γ of the exceptional set is oneof the following forms.

(1)−3

A∗2

−3

A∗1∗

∗

−3

A∗3∗

r ≥ 0

r

Z =

1 2 13 2 2 . . . 2︸︷︷︸

1 2 r ≥ 0 1

(2)−3

A∗2

−3

A∗1∗

∗

−3

A∗3∗ Z =

1 2 34 2 1

1 2 3

(3)−3

A∗2

−3

A∗1∗

∗

−3

A∗3∗ Z =

1 2 3 45 2

1 2 3 4

(4) −3

A∗2

−3

A∗1

∗

∗

∗

r ≥ 0r −3

A∗3Z =

1 12 . . . 2︸︷︷︸ 2 1

1 r ≥ 0 1

(5) −3

A∗2

−3

A∗1∗

∗

r1 ≥ 0r1 −3

A∗3∗

r2 ≥ 0

r2

Z =

1 1 12 . . . 2︸︷︷︸ 2 2 . . . 2︸︷︷︸

1 r1 ≥ 0 r2 ≥ 1 1

(6) −3

A∗2

−3

A∗1∗

∗

r1 ≥ 0r1 −3

A∗3∗

Z =

1 2 12 . . . 2︸︷︷︸ 2 3

1 r ≥ 0 2 1

(7) −3

A∗2

−3

A∗1∗

∗

r1 ≥ 0r1 −3

A∗3∗ Z =

1 22 . . . 2︸︷︷︸ 2 3 4 3 2 1

1 r ≥ 0

(8) −3

A∗2

−3

A∗1∗

∗

r1 ≥ 0r1 −3

A∗3∗ Z =

1 32 . . . 2︸︷︷︸ 2 3 4 5 6 4 2

1 r ≥ 0

(9)

−3

A∗1

−3

A∗2∗

−3

A∗3∗

r ≥ 0

r ∗

Z =

12 1322 . . . 2︸︷︷︸

12 r ≥ 01



(10)

−3

A∗1∗

−3

A∗3∗ ∗−3

A∗2Z =

1 2 34 2 1

1 2 3

(11)

−3

A∗1∗

r1 ≥ 0

r1 −3

A∗3∗r2 ≥ 0

r2

−3

∗A∗2 Z =

1 1 12 . . . 2︸︷︷︸ 2 2 . . . 2︸︷︷︸

1 r1 ≥ 0 r2 ≥ 0 1

(12)

−3

A∗1∗

r1 ≥ 0

r1 −3

A∗3∗∗r2 ≥ 0

r2

−3

A∗2

Z =

1 1 12 . . . 2︸︷︷︸ 2 2 . . . 2︸︷︷︸

1 r1 ≥ 0 r2 ≥ 1 1

(13)

−3

A∗2∗

r1 ≥ 0

r1 −3

A∗3∗ ∗

A∗1

−3Z =

1 22 . . . 2︸︷︷︸ 2 3 4 3 2 1

1 r ≥ 0

(14) ∗∗

∗A∗1

−3

A∗2−3

A∗2−3

Z = 1 2 3

1

2

4 3 2 1

(15) ∗ ∗ ∗A∗1

−3

A∗2

−3

A∗3

−3Z = 1 2 4

3

6 5 4 3 2 1

Proof. Observe that A∗i · (Z − A∗i) = 2 for i = 1, 2 and that A∗3 · (Z − 2A∗3) =2 − A2

∗3 = 5.Also, by Lemma 6.6, A∗1 and A∗2 must be of degree one in Z. Otherwise

Lemma 6.6 says that the Γ must be a circular graph and that will force a∗3 = 1,while we assume that z∗3 = 2. It follows that A∗3 intersects with every connectedcomponent of Γ′ = Γ − A∗1, A∗2, A∗3 and A∗i, i = 1, 2, intersects with at mostone connected component.

If A∗1, A∗2 and A∗3 are mutually disjoint and they are all attached to the sameconnected component of Γ′, by Proposition 6.30 and Proposition 6.31 we are incases (1)–(8).

If A∗1, A∗2 and A∗3 are mutually disjoint and they are not all attached to thesame connected component of Γ′, then A∗1 and A∗3 are both attached to a con-nected component of Γ′, while A∗2 and A∗3 are both attached to another connectedcomponent. By Proposition 6.32 and Proposition 6.31 we are in case (9), (11) and(13).

If A∗1 · A∗3 = 0 and A∗2 · A∗3 = 1, then Γ′ has only one connected componentwhere A∗1 and A∗3 are both attached and A∗2 does not intersect with any connectedcomponent of Γ′. By Proposition 6.32 and Proposition 6.31, we are in case (9) with



r = 0, case (10), case (11) with r2 = 0, case (12), case (13) with r = 0, case (14)and case (15).

If A∗i · A∗3 = 1, i = 1, 2, then every connected component of Γ′ must onlyintersect with A∗3. By Proposition 6.31 we are in case (4) with r = 0, case (5) withr1 = 0, case (6), (7), (8) with r = 0, case (11) with r1 = r2 = 0 and case (12) withr1 = 0. Proposition 6.34. Let Γ be the minimal resolution graph of a minimally ellipticsingularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1, A∗2 and A∗3 be three effectivecomponents of Γ. Suppose that Γ1 is a connected component of Γ′ which correspondsto a rational double point graph in Theorem 4.2. Suppose also that Γ1 intersectswith A∗1, A∗2, A∗3, but is disjoint from other effective components. Let Z1 be thefundamental cycle on Γ1. Suppose that A∗1 · Z1 = 1 = A∗2 · Z1 = A∗3 · Z1. If A∗1,A∗2, and A∗3 are mutually disjoint, z∗1 = 1 = z∗2 = z∗3, and A2

∗1 ≤ −3, A2∗2 ≤ −3,

A2∗3 ≤ −3, then A∗1∪A∗2∪A∗3∪Γ1 and the restriction of Z on A∗1∪A∗2∪A∗3∪Γ1

must be one of the following forms.

(1) ∗A∗1

A∗2

∗m ≥ 1

∗A∗3 Z

∣∣∣∣A∗1∪A∗2∪A∗3∪Γ1

= 1

1

2 . . .

1

2 1

(2) ∗

∗

A∗1

A∗2

∗A∗3

Z

∣∣∣∣A∗1∪A∗2∪A∗3∪Γ1

= 1

1

2

1

1

(3) ∗A∗1

A∗2 ∗ ∗

A∗3

Z

∣∣∣∣A∗1∪A∗2∪A∗3∪Γ1

= 1 2

1

2

3 2 1

Proof. The proof is the same as that in Proposition 6.30. Theorem 6.35. Let (V, p) be a germ of minimally elliptic singularity. Let π : M →V be the minimal resolution of p. If case (10) of Proposition 6.2 holds, i.e., if thereexist three effective components A∗1, A∗2 and A∗3 with A2

∗1 = A2∗2 = −3, A2

∗3 = −4and z∗1 = z∗2 = z∗3 = 1, then the weighted dual graph Γ of the exceptional set isone of the following forms.

(1) ∗A∗1

−3

−3 ∗A∗2

m ≥ 1

m

−4 ∗A∗3 Z = 1

1

2 . . . . . .

1

2 1

(2) ∗A∗1

−3

−4 ∗A∗3

m ≥ 1

m

−3 ∗A∗2 Z = 1

1

2 . . . . . .

1

2 1



(3) ∗∗

∗−4

A∗3

−3 A∗1

−3 A∗2

Z = 1

1

2

1

1

(4) ∗A∗1

−3 A∗2−3 ∗ ∗−4

A∗3

Z = 1 2

1

2

3 2 1

(5) ∗A∗1

−3

∗−3A∗2

A∗3−3

∗t

s

r

.......

............................. ........

......... .......... ........... . ......... . ......... .......... .......... .................. ...............................................................................................................................................................................

.............................

1

1

1

.......

............................. ........

......... .......... ........... . ......... . ......... . ......... .......... .................. ...............................................................................................................................................................................

.............................Z =

Proof. This follows from Proposition 6.3, Proposition 6.21, Proposition 6.26 andProposition 6.32. Proposition 6.36. Let Γ be the minimal resolution graph of a minimally ellip-tic singularity with fundamental cycle Z. Let Γ′ be the subgraph of Γ obtained byremoving all the effective components of Γ. Let A∗1, A∗2, A∗3 and A∗4 be foureffective components of Γ. Suppose that Γ1 is a connected component of Γ′ whichcorresponds to a rational double point graph in Theorem 4.2. Suppose also that Γ1

intersects with A∗1, A∗2, A∗3 and A∗4, but is disjoint from other effective com-ponents. Let Z1 be the fundamental cycle on Γ1. Suppose that A∗1 · Z1 = 1 =A∗2 · Z1 = A∗3 · Z1 = A∗4 · Z1. If A∗1, A∗2, A∗3 and A∗4 are mutually disjoint,z∗1 = z∗2 = z∗3 = z∗4 = 1, and A2

∗1 ≤ −3, A2∗2 ≤ −3, A2

∗3 ≤ −3, A2∗4 ≤ −3, then

A∗1 ∪A∗2 ∪A∗3 ∪A∗4 ∪ Γ1 and the restriction of Z on A∗1 ∪A∗2 ∪A∗3 ∪A∗4 ∪ Γ1

must be one of the following forms.

(1) ∗A∗1

∗A∗2

m ≥ 1

∗A∗3∗A∗4

Z

∣∣∣∣A∗1∪A∗2∪A∗3∪A∗4∪Γ1

= 1

1

2 . . .

1

2 1

(2) ∗A∗1

A∗2

A∗3A∗4

∗

∗

∗ Z

∣∣∣∣A∗1∪A∗2∪A∗3∪A∗4∪Γ1

= 1

1

2

1

1

Proof. The proof is the same as that in Proposition 6.32. Theorem 6.37. Let (V, p) be a germ of minimally elliptic singularity. Let π :M −→ V be the minimal resolution of p. If case (11) of Proposition 6.2 holds, i.e.,if there exist four effective components A∗1, A∗2, A∗3 and A∗4 with A2

∗1 = −3 =A2

∗2 = A2∗3 = A2

∗4 and z∗1 = 1 = z∗2 = z∗3 = z∗4, then the weighted dual graph Γ ofthe exceptional set is one of the following forms.



(1) ∗A∗1

−3

A∗3

−3

r2r1

r3r4

−3 A∗2

A∗4−3∗

∗

∗

.......

............................. ........

......... .......... ........... . ......... . ......... .......... .......... .................. ...............................................................................................................................................................................

.............................

.......

............................. ........

......... .......... ........... . ......... . ......... .......... .......... .................. ...............................................................................................................................................................................

.............................Z =

11 1

1

(2) ∗A∗1

∗A∗2

−3

−3 −3

m

m ≥ 1

∗A∗2

Z = 1

1

2 . . . . . .

1

2 1

(3) ∗ ∗∗

∗−4

A∗4A∗1

−3

−3 A∗2

−3 A∗3

Z = 1

1

2

1

1

Proof. This follows from Proposition 6.3, Proposition 6.21, Proposition 6.26, Propo-sition 6.32 and Proposition 6.34.

7. Complete list of weighted dual graphs of minimally elliptic

singularities with Z2 = −4

In the following, we shall list all the weighted dual graphs of minimally ellipticsingularities with Z2 = −4 according to Proposition 6.2. Before we do this, weshall adopt the following notation, some of which was used by Laufer [La4]. Thespecial cases of Proposition 3.7, where it is not true that the Ai are nonsingularrational curves with normal crossings, are described and named individually.

(1) · · · · · · r · · · · · · denotes −−− −−− · · · · · · −−−with r vertices and r + 1 edges.The case r = 0 is included.is a nonsingular rational curve with weight −2.

(2) E ∗ The vertex A∗ is a nonsingular elliptic curve.(3) N0 (a) ∗ The vertex A∗ is a rational curve with a node singularity.

(b) r

s

u

t

∗

∗

∗

∗r

∗

t

s

∗ ∗

...............

.......... .... ..... ...... ..... .....................

...............................

..........................∗ ∗

r

s

...............

.......... .... ..... ...... ..... .....................

...............................

..........................∗ r≥1

Each A∗ is a nonsingular rational curve.(4) Cu ∗ The vertex A∗ is a rational curve with a cusp

singularity.

(5) Ta ∗

∗ The two A∗ are nonsingular rational curveswhich meet tangentially to first order.

(6) Tr

∗ ∗

∗The three A∗ are nonsingular rational curveswhich meet transversely at the same point.



(7) A∗,m,n ∗ ︸︷︷︸m

︸︷︷︸n(a) A∗,0,1 ∗

A∗A∗,0,2 ∗ A∗

A∗,0,3 ∗ A∗

(b) A∗,1,2 ∗ A∗

A∗,1,4 A∗∗

A∗,1,6 A∗∗

(c) A∗,2,3 A∗∗

A∗,2,6 A∗∗

A∗,2,9 A∗∗

(d) A∗,3,4 A∗∗

A∗,3,8 A∗∗

(e) A∗,4,5 A∗ ∗

(f) A∗,5,6 A∗ ∗

(8) A∗∗,n,m, A∗∗ ∗

A∗

︸︷︷︸n

︸︷︷︸m

︸︷︷︸l



(a) A∗∗,0,1,2

∗A∗1

∗A∗2

A∗∗,0,1,1

∗A∗1

∗A∗2

A∗∗,1,1,1 ∗

A∗1∗

A∗2

(b) A∗∗,3,2,0

A∗2A∗1∗ ∗

A∗∗,2,2,0 A∗1∗

A∗2∗

(c) A∗∗,1,3,0 A∗1∗

A∗2∗

A∗∗,3,3,0 A∗1∗

A∗2∗

A∗∗,4,3,0 A∗1∗

A∗2∗

(d) A∗∗,4,4,0 A∗1∗

A∗2∗

(e) A∗∗,2,5,0 A∗1∗

A∗2∗

(f) A∗∗,3,7,0 A∗1∗

A∗2∗

(g) A∗∗,0,m,0 ︸︷︷︸m ≥ 1

A∗1 A∗2∗ ∗

A∗∗,1,m,0 ︸︷︷︸m ≥ 1

A∗1 A∗2∗ ∗

A∗∗,1,m,1 ︸︷︷︸m ≥ 1

A∗1 A∗2∗ ∗



(9) A∗∗∗,m,n,k ∗ ︸︷︷︸m

A∗1∗︸︷︷︸n

∗A∗2

︸︷︷︸k

(a) A∗∗∗,1,1,0

A∗2

A∗1 A∗3∗ ∗ ∗

(b) A∗∗∗,1,n,0

A∗2

A∗1∗

∗ ︸︷︷︸n

∗A∗3

(c) A∗∗∗,2,2,0

A∗1∗

A∗2

∗ ∗A∗3

(d) A∗∗∗,3,3,0

A∗1∗

A∗2∗ ∗A∗3

(e) A∗∗∗,4,4,0

A∗1∗

A∗2∗ ∗A∗3

(f) A∗∗∗,1,n,1

A∗1∗ ︸︷︷︸

n ≥ 1

A∗2 A∗3∗ ∗

(10) A∗∗∗∗,m

A∗1∗ ︸︷︷︸

m ≥ 1

A∗2 A∗3∗ ∗

∗A∗3

(11) Dm,∗ ︸︷︷︸

m − 2 ≥ 2

∗A∗

(12) D′m,∗ ∗

A∗ ︸︷︷︸m − 3 ≥ 1

(a) D′5,∗

A∗∗

(b) D′6,∗

A∗∗



(c) D′7,∗

A∗∗

(d) D′8,∗

A∗∗

(e) D′9,∗

A∗∗

(f) D′10,∗

A∗∗

(g) D′11,∗

A∗∗

(h) D′12,∗

A∗∗ (13) D′′

m,∗ ︸︷︷︸

m − 3 ≥ 1

∗A∗

(14) Dm,∗∗ ︸︷︷︸

m − 2 ≥ 2

∗A∗1 ∗A∗2

(15) D′5,∗∗ ∗

∗

(16) (a) D′′4,∗∗

A∗1∗ ∗A∗2

(b) D′′5,∗∗

A∗1∗ ∗A∗2

(17) (a) D′′′

5,∗∗A∗1∗ ∗

A∗2



(b) D′′′6,∗∗

A∗1∗ ∗A∗2

(c) D′′′7,∗∗

A∗1∗ ∗A∗2

(d) D′′′8,∗∗

A∗1∗ ∗A∗2

(e) D′′′9,∗∗

A∗1∗ ∗A∗2

(18) D4,∗∗∗A∗1∗

∗A∗2

∗A∗3

(19) E6,∗ ∗A∗

(20) E′6,∗

∗A∗

(21) E6,∗∗ ∗A∗1

∗A∗2

(22) E7,∗ ∗A∗

(23) E′7,∗ ∗

A∗

(24) E8,∗ ∗A∗

In the weighted dual graphs in the following tables, we may use ∗′ or o to replacethe ∗ in graphs (1)–(24) above if it is necessary. Except in part I of the tables, atthe beginning of each part of the tables, we will list values of A∗ · A∗, A∗′ · A∗′ ,Ao · Ao, z∗, z∗′ , and zo when they are used in the dual graphs of that part.

Example 1. At the beginning of Table V, we give the values A∗·A∗ = A∗′ ·A∗′ = −3and z∗ = z∗′ = 2. Therefore the notation A∗,0,1+A∗,0,1+A∗,∗′,0,1,2+A∗′,0,1+A∗′,0,1

denotes the weighted dual graph



∗

−3

∗′ −3

Z = 112

123

12 1

Example 2. At the beginning of Table VII, we give A∗ ·A∗ = −4, A∗′ ·A∗′ = −3,z∗ = 1 and z∗′ = 2. The notation A∗,∗′,m,0 + A∗′,0,1 + A∗′,0,1 + A∗′,0,1 denotes thegraph

∗−4

∗′−3 Z = 1

12 . . . . . .

121

1

Example 3. At the beginning of Table X, we give A∗ ·A∗ = A∗′ ·A∗′ = Ao ·Ao = −3z∗ = z∗′ = 1 and zo = 2. The notation A∗,o,0,0,0 + Ao,0,1 + Ao,∗′,1,1,0 denotes thegraph

∗−3

∗′−3

Z = 11

21 1

2 1

TABLE

The weighted dual graphs for minimally elliptic singularities withZ · Z = −4.

I. The following graphs correspond to those exceptional cases in Proposition 3.7.A∗ · A∗1. E −42. N0 −43. Cu −44. Ta −2,−65. Tr −2,−2,−66. Ta −3,−57. Tr −2,−3,−58. Ta −4,−49. Tr −2,−4,−4

10. Tr −3,−3,−4

II. The following graphs correspond to those in Theorem 6.12.A∗ · A∗ = −3, z∗ = 4.1. A∗,0,1 + A∗,0,3 + A∗,0,3 + A∗,0,3

2. A∗,0,1 + A∗,0,3 + A∗,1,6

3. A∗,0,1 + A∗,2,9

4. A∗,0,3 + A∗,0,3 + D′9,∗

5. D′5,∗ + A∗,1,6



III. The following graphs correspond to those in Theorem 6.14.A∗ · A∗ = −4, z∗ = 2.1. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,0,1

2. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,1,2

3. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,0,1 + Dm,∗4. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,2,3

5. A∗,0,1 + A∗,0,1 + A∗,0,1 + D′6,∗

6. A∗,0,1 + A∗,0,1 + A∗,0,1 + E7,∗7. A∗,1,2 + A∗,0,1 + A∗,0,1 + A∗,1,2

8. Dr,∗ + A∗,0,1 + A∗,0,1 + A∗,1,2

9. Dr,∗ + A∗,0,1 + A∗,0,1 + Ds,∗10. A∗,0,1 + A∗,0,1 + D′

8,∗11. A∗,0,1 + A∗,0,1 + A∗,3,4

12. A∗,1,2 + A∗,0,1 + A∗,2,3

13. Dm,∗ + A∗,0,1 + A∗,2,3

14. A∗,1,2 + A∗,0,1 + D′6,∗

15. Dm,∗ + A∗0,1 + D′6,∗

16. A∗,1,2 + A′∗,0,1 + E7,∗

17. Dm,∗ + A∗,0,1 + E7,∗18. A∗,0,1 + A∗,4,5

19. A∗,0,1 + D′10,∗

20. A∗,1,2 + A∗,1,2 + A∗,1,2

21. A∗,1,2 + A∗1,2 + Dm,∗22. A∗,1,2 + Dm,∗ + Dn,∗23. Dm,∗ + Dn,∗ + Dk,∗24. A∗,1,2 + A∗,3,4

25. Dm,∗ + A∗,3,4

26. A∗,1,2 + D′8,∗

27. Dm,∗ + D′8,∗

28. A∗,2,3 + A∗,2,3

29. A∗,2,3 + D′6,∗

30. A∗,2,3 + E7,∗31. D′

6,∗ + D′6,∗

32. D′6,∗ + E7,∗

33. E7,∗ + E7,∗34. A∗,5,6

35. D′12,∗

IV. The following graphs correspond to those in Theorem 6.16.A∗ · A∗ = −6, z∗ = 1.1. No, r ≥ 12. D′′

m,∗3. E′

6,∗4. E′

7,∗5. E8,∗



V. The following graphs correspond to those in Theorem 6.19.A∗ · A∗ = A∗′ · A∗′ = −3, z∗ = z∗′ = 2.1. A∗,0,1 + A∗,0,1 + A∗,∗′,0,1,2 + A∗′,0,1 + A∗′,0,1

2. A∗,0,1 + A∗,0,1 + A∗,∗′,0,1,2 + A∗′,1,2

3. A∗,0,1 + A∗,0,1 + A∗,∗′,0,1,2 + D′m,∗′

4. A∗,1,2 + A∗,∗′,0,1,2 + A∗′,1,2

5. A∗,1,2 + A∗,∗′,0,1,2 + Dm,∗′

6. Dm,∗ + A∗,∗′,0,1,2 + Dn,∗′

7. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,0,1 + A∗′,0,1 + A∗′,0,1

8. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,0,1 + A∗′,1,2

9. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,0,1 + Dm,∗′

10. A∗,0,1 + A∗.0,1 + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,2,3

11. A∗,0,1 + A∗.0,1 + A∗,0,1 + A∗,∗′,0,m,0 + D′6,∗′

12. A∗,0,1 + A∗.0,1 + A∗,0,1 + A∗,∗′,0,m,0 + E7,∗′

13. A∗,1,2 + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,0,1 + A∗′,1,2

14. Dn,∗ + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,0,1 + A∗′,1,2

15. Dn,∗ + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,0,1 + Dk,∗′

16. A∗,1,2 + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,2,3

17. Dn,∗ + A∗,0,1 + A∗,∗′,0,m,0 + A∗′,2,3

18. A∗,1,2 + A∗,0,1 + A∗,∗′,0,m,0 + D′6,∗′

19. Dn,∗ + A∗,0,1 + A∗,∗′,0,m,0 + D′6,∗′

20. A∗,1,2 + A∗,0,1 + A∗,∗′,0,m,0 + E7,∗′

21. Dn,∗ + A∗,0,1 + A∗,∗′,0,m,0 + E7,∗′

22. A∗,2,3 + A∗,∗′,0,m,0 + A∗′,2,3

23. A∗,2,3 + A∗,∗′,0,m,0 + D′6,∗′

24. A∗,2,3 + A∗,∗′,0,m,0 + E7,∗′

25. D′6,∗ + A∗,∗′,0,m,0 + D′

6,∗′

26. D′6,∗ + A∗,∗′,0,m,0 + E7,∗′

27. E7,∗ + A∗,∗′,0,m,0 + E7,∗′

28. A∗,0,1 + A∗,∗′,3,2,0 + A∗′,0,1 + A∗′,0,1

29. A∗,0,1 + A∗,∗′,3,2,0 + A∗′,1,2

30. A∗,0,1 + A∗,∗′,3,2,0 + Dm,∗′

31. A∗,∗′,4,3,0 + A∗′,0,1 + A∗′,0,1

32. A∗,∗′,4,3,0 + A∗′,1,2

33. A∗,∗′,4,3,0 + Dm,∗′

34. A∗,0,1 + A∗,0,1 + D′′4,∗,∗′ + A∗′,0,1 + A∗′,0,1

35. A∗,0,1 + A∗,0,1 + D′′4,∗,∗′ + Dm,∗′

36. A∗,0,1 + A∗,0,1 + D′′4,∗,∗′ + A∗′,1,2

37. Dm,∗ + D′′4,∗,∗′ + A∗′,1,2

38. Dm,∗ + D′′4,∗,∗′ + Dn,∗′

39. A∗′,1,2 + D′′4

40. A∗,0,1 + D′′′6,∗,∗′ + A∗′,0,1 + A∗′,0,1

41. A∗,0,1 + D′′′6,∗,∗′ + A∗′,1,2

42. A∗,0,1 + D′′′6,∗,∗′ + Dn,∗′

43. D′′′8,∗,∗′ + A∗′,0,1 + A∗′,0,1

44. D′′′8,∗,∗′ + A∗′,1,2

45. D′′′8,∗,∗′ + Dm,∗′



VI. The following graphs correspond to those in Theorem 6.23.A∗ · A∗ = A∗′ · A∗′ = −3, z∗ = 3, z∗′ = 1.1. A∗,0,2 + A∗,0,2 + A∗,0,2 + A∗,∗′,0,1,0

2. A∗,1,4 + A∗,0,2 + A∗,∗′,0,1,0

3. E6,∗ + A∗,0,2 + A∗,∗′,0,1,0

4. A∗,2,6 + A∗,∗′,0,1,0

5. A∗,0,2 + A∗,0,2 + A∗,∗′,1,3,0

6. A∗,1,4 + A∗,∗′,1,3,0

7. E6,∗ + A∗,∗′,1,3,0

8. A∗,0,2 + A∗,∗′,2,5,0

9. A∗,∗′,3,7,0

VII. The following graphs correspond to those in Theorem 6.25.A∗ · A∗ = −3, A∗′ · A∗′ = −4, z∗ = 2, z∗′ = 1.1. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,∗′,1,1,0

2. A∗,1,2 + A∗,0,1 + A∗,∗′,1,1,0

3. Dn,∗ + A∗,0,1 + A∗,∗′,1,1,0

4. A∗,2,3 + A∗,∗′,1,1,0

5. D′6,∗ + A∗,∗′,1,1,0

6. E7,∗ + A∗,∗′,1,1,0

7. A∗,0,1 + A∗,0,1 + A∗,∗′,2,2,0

8. A∗,1,2 + A∗,∗′,2,2,0

9. Dn,∗ + A∗,∗′,2,2,0

10. A∗,0,1 + A∗,∗′,3,3,0

11. A∗,∗′,4,4,0

12. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,∗′,0,m,1

13. A∗,1,2 + A∗,0,1 + A∗,∗′,0,m,1

14. Dn,∗ + A∗,0,1 + A∗,∗′,0,m,1

15. A∗,2,3 + A∗,∗′,0,m,1

16. D′6,∗ + A∗,∗′,0,m,1

17. E7,∗ + A∗,∗′,0,m,1

18. A∗,0,1 + A∗,0,1 + D′′′5,∗,∗′

19. A∗,1,2 + D′′′5,∗,∗′

20. Dn,∗ + D′′′5,∗,∗′

21. A∗,0,1 + D′′′7,∗,∗′

22. D′′′9,∗,∗′

23. A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,0,1 + A∗,∗′,0,0,0

24. A∗,1,2 + A∗,0,1 + A∗,0,1 + A∗,∗′,0,0,0

25. Dn,∗ + A∗,0,1 + A∗,0,1 + A∗,∗′,0,0,0

26. A∗,2,3 + A∗,0,1 + A∗,∗′,0,0,0

27. D′6,∗ + A∗,0,1 + A∗,∗′,0,0,0

28. E7,∗ + A∗,0,1 + A∗,∗′,0,0,0

29. A∗,1,2 + A∗,1,2 + A∗,∗′,0,0,0

30. Dn,∗ + A∗,1,2 + A∗,∗′,0,0,0

31. Dn,∗ + Dm,∗ + A∗,∗′,0,0,0

32. A∗,3,4 + A∗,∗′,0,0,0

33. D′8,∗ + A∗,∗′,0,0,0



VIII. The following graphs correspond to those in Theorem 6.27.A∗ · A∗ = A∗′ · A∗′ = −4, z∗ = z∗′ = 1.1. No2. A∗,∗′,1,m,1

3. A∗,∗′,1,1,1

4. Dm,∗,∗′

5. D′′5,∗,∗′

6. E6,∗,∗′

IX. The following graphs correspond to those in Theorem 6.29.A∗ · A∗ = −3, A∗′ · A∗′ = −5, z∗ = z∗′ = 1.1. No2. A∗,∗′,1,m,1

3. Dm,∗,∗′,m≥4

4. D′′5,∗,∗′

5. E6,∗,∗′

X. The following graphs correspond to those in Theorem 6.33.A∗ · A∗ = A∗′ · A∗′ = Ao · Ao = −3, z∗ = z∗′ = 1, zo = 2.1. A∗,o,∗′,2,2,0 + Ao,1,2

2. A∗,o,∗′,2,2,0 + Dn,o

3. A∗,o,∗′,3,3,0 + Ao,0,1

4. A∗,o,∗′,4,4,0

5. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,0,1 + Ao,0,1 + Ao,0,1

6. A∗,∗′,o,1,n,0 + Ao,0,1 + Ao,0,1 + Ao,0,1

7. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,0,1 + Ao,1,2

8. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,0,1 + Dn,o

9. A∗,o,∗′,1,n,0 + Ao,0,1 + Ao,1,2

10. A∗,o,∗′,1,n,0 + Ao,0,1 + Dn,o

11. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,2,3

12. A∗,o,∗′,1,n,0 + Ao,2,3

13. A∗,o,0,0,0 + A∗′,o,0,0,0 + D′6,o

14. A∗,o,∗′,1,n,0 + D′6,o

15. A∗,o,0,0,0 + A∗′,o,0,0,0 + E7,o

16. A∗,o,∗′,1,n,0 + E7,o

17. A∗,o,0,2,2 + A∗′,o,0,0,0 + Ao,0,1

18. A∗,o,0,2,2 + Ao,∗′,0,m,1

19. A∗,o,0,3,3 + Ao,∗′,0,0,0

20. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,0,1 + Ao,0,1 + Ao,0,1

21. A∗,o,0,0,0 + Ao,0,1 + Ao,0,1 + Ao,∗′,0,m,1

22. A∗,o,1,m,0 + Ao,0,1 + Ao,∗′,0,n,1

23. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,0,1 + Ao,1,2

24. A∗,o,0,0,0 + A∗′,o,0,0,0 + Ao,0,1 + Dn,o

25. A∗,o,1,m,0 + A∗′,o,0,0,0 + Ao,1,2

26. A∗,o,1,m,0 + A∗′,o,0,0,0 + Dn,o

27. A∗,o,0,0,0 + Ao,0,1 + D′′′5,o,∗′

28. A∗,o,1,m,0 + D′′′5,o,∗′

29. A∗′,o,0,0,0 + Ao,∗,3,3,0

30. A∗′,o,0,0,0 + D′′′7,o,∗



XI. The following graphs correspond to those in Theorem 6.35.A∗ · A∗ = A∗′ · A∗′ = −3, Ao · Ao = −4, z∗ = z∗′ = zo = 1.1. A∗,∗′,o,1,n,1, n ≥ 12. A∗,o,∗′,1,n,1, n ≥ 13. D4,∗,∗′,o

4. N0XII. The following graphs correspond to those in Theorem 6.37.

A∗ · A∗ = −3, z∗ = 1 for all four effective components.1. No2. A∗,∗,∗,∗,m, m ≥ 1

References

[Ar] M. Artin, On isolated rational singularities of surfaces, Amer. J. Math. 88 (1996), 129–136. MR0199191 (33:7340)

[Ba] H. Bass, On the ubiquity of Gorenstein rings, Math. Z. 82 (1963), 8–28. MR0153708(27:3669)

[Gr] H. Grauert, Uber Modifikationen und exzeptionelle analytische Mengen, Math. Ann. 146(1962), 331–368. MR0137127 (25:583)

[Gr-Ri] H. Grauert and O. Riemenschneider, Verschwindungssatze fur analytische Kohomolo-giegruppen auf komplexen Raumen, Invent. Math. 11 (1970), 263–292. MR0302938(46:2081)

[HNK] F. Hirzebruch, W. Neumann, and S. Koh, Differentiable manifolds and quadratic forms,Lecture Notes in Pure and Applied Mathematics, vol. 4, Marcel Dekker, New York, 1971.MR0341499 (49:6250)

[La1] H. Laufer, Normal two-dimensional singularities, Annals of Math. Studies, no. 71, Prince-ton University Press, Princeton, NJ, 1971. MR0320365 (47:8904)

[La2] , On rational singularities, Amer. J. Math. 94 (1972), 597–608. MR0330500(48:8837)

[La3] , Deformations of resolutions of two-dimensional singularities, Rice UniversityStudies 1 (1973), no. 59, 53–96. MR0367277 (51:3519)

[La4] , On minimally elliptic singularities, Amer. J. Math. 99 (1977), 1257–1295.MR0568898 (58:27961)

[Mi] J. Milnor, Singular points of complex hypersurfaces, Ann. Math. Studies, no. 61, PrincetonUniversity Press, Princeton, NJ, 1968. MR0239612 (39:969)

[Ne] W. Neumann, A calculus for plumbing applied to the topology of complex surface singu-larities and degenerating complex curves, Trans. A.M.S. 268 (1981), 299–344. MR632532(84a:32015)

[Se] J.-P. Serre, Groupes algebriques et corps de classes, Actualites Scientifiques et Indus-trielles, no. 1264, Hermann, Paris, 1959. MR0103191 (21:1973)

[Wa] P. Wagreich, Elliptic singularities of surfaces, Amer. J. Math. 92 (1970), 419–454.MR0291170 (45:264)

[Ya1] S. S.-T. Yau, Normal singularities of surfaces, Proc. Sympos. Pure Math. 72 (1978),195–198. MR520537 (80f:14020)

[Ya2] , On maximally elliptic singularities, Trans. A.M.S. 257 (1980), 269–329.MR552260 (80j:32021)

Department of Mathematics, University of California at San Diego, La Jolla, Cal-

ifornia 92093-0112

Department of Mathematics, John Tyler Community College, 13101 Jefferson Davis

Highway, Chester, Virginia 23831-5316

Department of Mathematics, Statistics, and Computer Science (M/C 249), University

of Illinois at Chicago, 851 South Morgan Street, Chicago, Illinois 60607-7045

E-mail address: [email protected]


http://www.ams.org/mathscinet-getitem?mr=0199191






























CLASSIFICATION OF WEIGHTED DUAL GRAPHS WITH ONLY … · 2018-11-16 · CLASSIFICATION OF WEIGHTED DUAL GRAPHS WITH ONLY COMPLETE INTERSECTION SINGULARITIES STRUCTURES FAN CHUNG, YI-JING

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