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Available at: http://www.ictp.trieste.it/~pub-off IC/99/186
United Nations Educational Scientific and Cultural Organizationand
International Atomic Energy Agency
THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS
COMBINATORICS OF RATIONAL SURFACE SINGULARITIES
Le Dung Trang1
Universite de Provence, Centre de Mathematiques et d'Informatique,39 rue F. Joliot-Curie, F-13 453 Marseille Cedex 13, France
and
Meral Tosun2
The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy.
Abstract
A normal surface singularity is rational if and only if the dual graph of a desingularization
satisfies some combinatorial properties. In fact, these graphs are trees. In this paper we give
geometric features of these trees. In particular, we prove that the number of vertices of valency
> 3 in the dual tree of the minimal desingularization of a rational singularity of multiplicity
m > 3 is at most m — 2.
MIRAMARE - TRIESTE
December 1999
1E-mail: [email protected]: [email protected]
1 Introduction
Rational surface singularities are the singularities of normal surfaces whose geometric genus does
not change by a desingularization. These singularities had been studied, for the first time, by
P. Du Val in [4].
Following the work of M. Artin [1], M. Spivakovsky (see [15], p. 421) has emphasized the
fact that a normal surface singularity is rational if and only if the dual graph associated with a
desingularization of the singularity satisfies combinatorial properties.
On the other hand, in the complex case, W. Neumann proved in [13] that the dual graph
associated with the minimal good desingularization of a normal surface singularity (algebraic
or analytic) is determined by the topology of the surface in a neighborhood of the singularity.
So, to obtain a topological classification of rational complex surface singularities, it is important
to study the graphs which are the dual graphs associated with a desingularization of these
singularities.
In this work, in place of getting a classification, we give several combinatorial properties of the
dual graphs associated with desingularizations of rational singularities. In particular, we obtain
a bound for the complexity of the dual graphs of minimal desingularizations by means of the
multiplicity of the corresponding rational singularity. In this case, the complexity is measured
by the number of vertices of valency > 3. The properties given in this paper lead to a complete
list of the dual graphs associated with the minimal desingularizations of rational singularities of
multiplicity 5 (see [19]). The lists for the cases of multiplicity 2, 3, and 4 were already given in
[4], [1] and [16] respectively.
2 Rational surface singularities
In this paragraph, we recall basic properties of rational surface singularities.
Let O be a normal local ring of Krull dimension 2. Denote S := Spec(O).
Definition 2.1 We say that the local ring O has a rational singularity (or that O is rational)
if there is a desingularization Π : X —> S such that H1(X, OX) = 0.
This definition is known to be independent of the desingularization π (see e.g. [2], theorem 2.3).
Recall that, if π : X —> S is a desingularization of S, by Zariski Main Theorem, the excep-
tional divisor vr" 1^) = E is connected and has dimension 1 (see [7], Chap V, theorem 5.2). We
call positive cycle with support in E a formal sum of the irreducible components Ei of E with
non-negative integral coefficients and with at least one positive coefficient. The set of positive
cycle is naturally ordered by the product order. So the positive cycle J2i aiEi is bigger than
ibiEi if and only if, for all i, ai > bi. The support of a positive cycle J2iai^i is the union of
the components Ei for which ai is = 0. From the proof of proposition 1 in [1], we deduce the
following :
2
Theorem 2.2 Let O be a local ring of dimension 2 and let S := Spec(O). Suppose that O
has a rational singularity. Let S1 be a normal surface and ρ : S' —> S be a proper birational
map which is not the identity. Let D be a positive cycle with support in the exceptional divisor
P " 1 ^ ) - Then we have H1(|D|,OD) = 0 where |D| is the reduced curve associated with D.
Proof : We just follow the proof of proposition 1 in [1]. By the Main Theorem of Zariski, the
fibre p~l(0 is connected and of dimension 1. Let E1, • • ,Ek be the irreducible components of
p - 1 ({) . Denote by Y(r) = J2 rjEj, with (r) = (r1, • • •, rk), a positive cycle supported on p~1(£).
By ([7], III.11.1 or [3], III.4.2.1), we have :
(R'p.Os')^ = limH1(|Y(r)|,OY(r))
where the projective limit is taken as (r) tends to +oo and (Rlp*Os')P is the completion of the
module R1p*Os> of finite type on OS,ξ for the m-adic topology, where m is the maximal ideal of
OS,ξ. The fact that ξ is a rational singularity implies ( i ^ p * ^ ' ) ^ = 0. Because the spaces Y (
have dimension 1, the canonical map
Hl(\Y{r)\,OY(r)) ^ HWY^IOY^)
is surjective when rj > r'j for all j . This gives H1(|Y(r)|, OY(r)) = 0 for all (r).
Since, for all positive cycles Y supported on the fibre p~1(£), there exists (r) such that Y C Y(r),
we have H1(Y,OY) = 0. •
Corollary 2.3 The irreducible components of the fibre p~x(0 are rational nonsingular curves.
Proof: Let E1,..., En be the irreducible components of p~l(0- We have 1 —p(Ei) = χ(OEi) =
dimH0(Ei,OEi) - dimH1(Ei,OEi). We have dimC(H°(Ei, OEi)) = 1 ([7], theorem I.3.4), and
by taking D = Ei in theorem 2.2, we have H1(Ei, OEi) = 0. This implies p(Ei) = 0.
Let n : Ei i> Ei be the normalisation of Ei. Since Ei is nonsingular, we have the following
exact sequence of coherent sheaves on Ei ([7], ex. IV.1.8) :
0 - ^ OEi -^ n*OEi -^ J2peEi Op/Op -^ 0
where Op is the integral closure sheaf of Op, so Op/Op is a coherent sheaf concentrated at the
singular points of Ei. This gives %(n*Ogi) = χ(OE i) + x(J2PeEi Op/Op). Moreover since n is a
finite map we have x(n*Oe i) = xi^sj ( s e e [7], exercise III.4.1), and
χ ( ]T Op/Op) = dimCH°(Ei, ]T Op/Op) = ^ δp
where δp = length(Op/Op). Since dimH0(Ei, OE) = 1, we obtain
X(OEi) = l-dimHl(Et,OEi) = l-p(Et).
3
So, we have p(Ei) = p{Ei) + J2PeEi <V Since Ei is a nonsingular curve, we have p{Ei) > 0 ([7],
p.181) and δp > 0 for all p G Ei. Therefore p(Ei) = 0 implies p(Ei) = 0 and J2peEt δp = 0; then
δp = 0 for every point p of Ei. This shows that Ei is a nonsingular rational curve. •
Now, let π be a desingularization of Spec(O). The set of the positive cycles Y with support
equal to E := |vr 1 (ξ)| such that (Y • Ei) < 0 for all i, (1 < i < n), has a smallest element for
the product order considered above. This smallest element is called the fundamental cycle of the
desingularization Π. It is easy to prove that the support of the fundamental cycle is the whole
curve E. M. Artin [1] proved that the singularity ξ of S is rational if and only if the arithmetic
genus of the fundamental cycle of any desingularization of Spec(O) vanishes.
Recall that π : X —• S is a good desingularization of S if the exceptional divisor has normal
crossings and the irreducible components Ei, (i = 1, • • • ,n), of E are all nonsingular curves. If,
in addition, Ei and Ej, for i / j , intersects transversally at one point at most, we say that π is
a very good desingularization. Another important result of M. Artin is (consequence of theorem
??, see e.g. [14] Proposition of §5) :
Theorem 2.4 If the 2-dimensional local ring O is rational, any desingularization of Spec(O)
is very good.
3 Rational Trees
Let Γ be a graph without loops, with vertices Ei,---,En, weighted by pairs (wi,gi) at each
vertex Ei, (1 < i < n), where wi is a positive integer called the weight of Ei, and gi is a
non-negative integer, called the genus of Ei.
We associate a symmetric matrix M(Γ) = (%-)i<tj<n, with the weighted graph Γ in the
following way: an = —Wi and ay is the number of edges linking the vertices Ei and Ej whenever
i / j . We call M(Γ) the incidence matrix of Γ. Now, a weighted graph Γ is called a singular
graph if the associated incidence matrix is negative definite. Therefore, if Γ is a singular graph,
in the free abelian group G generated by the vertices Ei of Γ, the incidence matrix M(Γ) defines
a symmetric bilinear form. We shall denote (Y • Z) the value of this bilinear form on a pair
(Y, Z) of elements in the free abelian group G.
The elements of the free abelian group generated by the Ei's will be called cycles of the graph
Γ. A positive cycle is a cycle in which all the coefficients are non-negative and at least one is
positive. Let Y = J2i aiEi be a positive cycle. The support of Y is the set of vertices such that
a i = 0 .
By a proof analogous to the one of O. Zariski given in ([20], theorem 7.1), for the case of curve
configurations, we can prove that, for any singular graph Γ with vertices Ei, there are non-zero
cycles Y of the graph Γ,Y = J27=i miEi, such that (Y • Ei) < 0 for any i, (1 < i < n) ; by using
the connexity of the graph, these elements satisfy mi > 1. As in ([12], §18), let E +(Γ) be the
sets of these elements. It is an additive monoid.
Let A be a set of positive cycles with support equal to the set of all the vertices of the singular
graph Γ. We define inf A as inf A = Z0 = J27=i aiEi where
where multYEi is the coefficient mi of Ei in the positive cycle Y. The cycle Z0 is a positive
cycle since mi G N* for any i.
Using ([1]-[12]), we can prove :
Theorem 3.1 Let Γ be a singular graph. For any subset A of E+(Γ), we have inf A e + ( Γ ) .
Therefore, we define :
Definition 3.2 We call Z = inf E + (Γ) the fundamental cycle of the singular graph Γ.
By proposition 4.1 of [10], we can find Z by constructing a sequence of positive cycles in the
following way : Let Z1 = Y^l=\Ei- Now, suppose that j > 1, we are given the term Zj of
the sequence, either (Zj • Ei) < 0 for any i, then we put Z = Zj, or there is an Eij such that
(Zj • Eij) > 0, then we put Zj+1 = Zj + Eij. Then, the fundamental cycle of Γ is the first
cycle Zk of this sequence, such that (Zk • Ei) < 0, for any i. This construction is called Laufer
algorithm.
By plumbing (see [11]), a weighted graph Γ, as above, defines a (non-unique) complex curve
configuration embedded in a non-singular complex analytic surface. By a result of H. Grauert
(see [6] p. 367), a weighted graph (as above) is singular if it is the intersection graph of the
exceptional divisor of a desingularization of a normal complex analytic surface singularity. So,
we have :
Theorem 3.3 IfΓ is a singular graph, there is a normal complex analytic surface singularity
and a desingularization of this singularity, such that the intersection graph of its exceptional
divisor is Γ.
Following M. Spivakovsky ([15], chap.II, def.1.9), we define :
Definition 3.4 (Rationality Conditions) The graph Γ is rational if:
(i) it is a tree,
(ii) the set F of non-zero cycles D = J27=i kiEi, where the ki 's are positive integers, such that
(D • Ei) < 0 for all i = 1, • • •, n, is not empty and there is D0 e F such that D02 < 0,
(iii) The genera gi of all the vertices of Γ are trivial,
(iv) Let inf F = Z = £ aiEi. Then
1 n
p(Z) := 1(Z • Z + J2<wi - 2)) + 1 = 0
Notice that the condition (ii) is equivalent to say that Γ is a singular graph (see [1] Proposition
2) and that the cycle Z of (iv) is the fundamental cycle of Γ.
In what follows, we shall only consider trees with vertices of trivial genus and R will denote a
rational tree with n vertices.
Now, we shall say that a singularity of a complex analytic surface is rational, if its analytic
local ring is rational. As in [15], (chap. II, proposition 1.11), we have :
Theorem 3.5 If R is a rational tree, there is a rational complex analytic surface singularity
and a desingularization of this singularity, such that the intersection graph of its exceptional
divisor is R.
In fact, for rational trees, in [1] M. Artin proves a general statement :
Theorem 3.6 If R is a rational tree, there is a rational local ring O of dimension 2 and a
desingularization of Spec(O), such that the intersection graph of its exceptional divisor is R.
Conversely, if a local ring O of dimension 2 is rational, the dual graph of the exceptional
divisor of any desingularization of Spec(O) is a rational tree (see [15]). Now, we have :
Proposition 3.7 Let R be a rational tree and let Z be its fundamental cycle. Let Y = J2aiEi
be any positive cycle ofR. Then
where wi := —Ei.Ei.
Proof. In fact, this proposition is consequence of theorem ?? (see [1]). •
It will be convenient to call p(Y) the arithmetic genus of the positive cycle Y in the sequence
of this paper.
Proposition 3.8 Any subtree of a rational tree is rational.
Proof : Let TZ' be a subtree of R. The tree TZ' is a singular graph ([9], lemma 5.11). Let
Z' = J2ieAa'i-Ei with A = {1, • • • ,n} be the fundamental cycle of TZ'. Proposition ?? implies
p(Z') < 0. However, p(Z') > 0 for the fundamental cycle of any desingularization of a normal
surface singularity (see [1], theorem 3). Therefore TZ' is a rational tree. •
6
4 Properties of Rational Trees
In this section we shall give some remarkable features of rational trees. In particular cases, these
properties of rational trees will lead to a classification of these trees.
We call valency of the vertex Ei in R the number of the vertices adjacent to Ei in R. We
denote it by υR(Ei). We prove the following property of a rational tree, given by M. Spivakovsky
([15], remark 2.3) :
Proposition 4.1 IfRisa rational tree, for any vertex Ei of R, we have
Proof : First, assume that R corresponds to the minimal desingularization π : X —> S of a
rational singularity of S i.e. wi > 2 for any Ei. Assume that in the tree R there is a vertex
E with valency ΥR(E) > WE + 2. Consider the subtree TZ' of R which contains the vertices
E1,... Ev ,(E) adjacent to E and such that V-R'(E) = WE + 2. Since the tree R is rational, the
subtree TZ' has to be rational. Since WE > 2, we have 2WE > WE + 2 and by Laufer algorithm,
we obtain the fundamental cycle of TZ' as :
vn,(E)
Z' = 2E+ ]T Ei.i=1
We have Z'2 = 8 — Yl7=i^2 wEi, and so p(Z') = 1. But TZ' is assumed to be rational, so we get a
contradiction. Then, WE + 1 > υR(E).
Now assume that R corresponds to a desingularization of a rational sigularity which is not
a minimal desingularization. It is well known that every desingularization vr' : X —> S of S
is obtained from the minimal desingularization X by a finite sequence of point blowing-ups
a : X —>• X. We consider σi : Xi —> -X"i-i, (i = 1, • • •, n + 1), with X0 := X, Xn+1 := X'. We
have σ = σ1 o σ2 o • • σ n + 1 and vr' = π o σ. If we denote by Γ0 the rational tree of the minimal
desingularization π : X —> S then Γk is the rational tree of the desingularization π o σ 1 • • • σk.
We prove our result by induction on k. Our former result shows that it is true for k = 0. Assume
that it is true for £ > 0. There are two cases depending if a^+1 blows up a non-singular point of
the exceptional divisor of Xi lying on the component D or if it blows-up the intersection point
of two components D and D' of the exceptional divisor of Xg. In the first case, the weights and
the valencies of the strict transforms by ae of all other components of the exceptional divisor of
Xi but D remain the same. So the inequality to be proved is true for all vertices except the ones
corresponding to the strict transform of D by ag and the exceptional divisor of erg. Since the
weight of the exceptional divisor of ae is 1 and its valency is 1, and the weight and the valency
of the strict transform of D by ag both increase by 1, the inequalities are true for all the vertices
of the dual tree of the desingularization vr σ <n σ ag. In the other case, a similar argument proves
the desired inequalities. This proves our result for R by induction on k and ends our proof. •
Definition 4.2 We call bad (resp. good) vertex a vertex Ei ofR such thatwi + 1 = υR(i) (resp.
wi > υR(i)) ; in particular, we call very good vertex a vertex Ei ofR such that wi > υR(i).
Theorem 4.3 Let R be a rational tree where the weights are > 2 and containing two bad
vertices and let C be the smallest path (i.e. the geodesic) in the tree R linking these two bad
vertices without containing them. Then, at least one of the vertices of the subtree C is very
good.
Proof: Let E and F be the two bad vertices of R. Assume that C is not empty. Let A1, • • •, An
be the vertices of C. We have (A1 •E) = (An-F) = 1 et (Aj • E) = 0 for j = 1 and (A,- • F) = 0
for j / n. Consider the subtree R1 of R which contains E, F, C and the vertices adjacent to
the vertices E, F and adjacent to C. Assume that the vertices of C are all good vertices, but
not very good. Since R is rational, the subtree R1 has to be rational. By Laufer algorithm, we
obtain its fundamental cycle Z1 as :
WE n wF k wni—'2
Z1 = YjEJ+2E + YJiAl + 2F+YJFm + YJ E Blj=1 i=1 m=1 i=1 l=1
where Ej, Fm et Blni are vertices adjacent to E, F and to Ani respectively.
WE wF k wni—'2
j=1 m=1 i=1 l=1
We have :k k Wnj-2
p(Z1) = i = 1 2
i=1 l=1 + 1 = 1
This contradicts the hypothesis. Therefore the subtree C contains at least one very good vertex.
In the case C is empty, we have a similar proof. We consider the subtree R1 of R which
contains E, F and the vertices adjacent to the vertices E, F. In this case, the fundamental cycle
Z1 of R1 is obtained as :WE wF
j=1 m=1
Then we show that p(Z1) = 1, which is a contradiction. •
Definition 4.4 The vertex Ei is a rupture vertex ofR if υR(i) > 3.
Remark 4.5 Theorem ?? implies that a rational tree where all the weights are equal to 2 has
at most one rupture vertex. Of course, this fact is already known, since in this case the possible
trees are A n , D n , E 6 , E 7 and E 8 .
The following result gives a strong restriction for a tree to be rational :
Theorem 4.6 Let R be a rational tree. The vertices of a subtree TZ' ofR whose valency in R
is different from its valency in TZ', have multiplicity 1 in the fundamental cycle ofTZ'.
8
Proof : Let O be a rational 2-dimensional local ring such that R is the dual tree associated
with the exceptional divisor of the desingularization Π:X —>• Spec(O).
Let F be a vertex of TV whose valency in TV is not the same as in R, and let E be a vertex
mTZ — TV which is adjacent to F.
Theorem ?? says that, by contracting all the components of the exceptional divisor of π which
correspond to the vertices of TV, we obtain a normal surface S' having a rational singularity
and birational morphisms κ : X —> S' and ρ: S —> Spec(O) such that π = ρ o κ. Denote the
singularity of S' by ξ1. Since the morphism ρ is birational and O is a rational ring, Corollary
?? shows that the components of p~1(^), where ξ is the closed point of S := Spec(O), are non-
singular rational curves. In particular, κ(E) which is a component of p~1({) is a non-singular
rational curve.
A result of G. Gonzalez-Sprinberg and M. Lejeune-Jalabert in [5] implies that, since the curve
Κ(E) is non-singular, the strict transform of κ(E) by κ intersects the exceptional divisor of κ
transversally at a component which has multiplicity 1 in the maximal divisor in K~1(£I) of the
singularity ξ1. Since TZ' is a rational tree, ξ1 is a rational singularity and the maximal divisor in
K ~ 1 ( £ I ) of the singularity ξ1 coincides with the fundamental cycle of Κ. This cycle corresponds
to the fundamental cycle of TV. Therefore, the multiplicity of F in this fundamental cycle is 1.
•
Corollary 4.7 The rational tree E8 cannot be a subtree strictly contained in a rational tree.
This comes from the fact that the multiplicities of all the vertices of E8 in its fundamental cycle
are > 2. •
Definition 4.8 Let R1 andR2 be two weighted trees. The weighted tree R obtained by attaching
the vertices E1 of R1 and E2 of R2 by an edge is called the glueing tree of R1 and R2 at E1 and
E2.
Another corollary of Theorem ?? is :
Corollary 4.9 If the glueing tree ofR1 and R2 at E1 and E2 is rational, the multiplicity of E1
(resp. E2) in the fundamental cycle ofR1 (resp. R2) is 1.
Proof: In the glueing tree, the weights of the vertices do not change, but the valencies of E1
and E2 change. Theorem ?? gives the result. •
Remark 4.10 We may always consider a rational tree to be the glueing tree of vertices of
weight > 3 and rational trees of type A n , D n , E6, E7 or E8. The vertices of these rational
trees of multiplicity 2, where the glueing happens, have multiplicity 1 in the fundamental cycle
of the corresponding tree. We saw that E8 cannot be the strict subtree of a rational tree. In
the case of E6, we cannot glue any tree at any vertices of E6, except the ends of the long tails,
9
since the multiplicities of those vertices are > 2 in the fundamental cycle of E6. Similarly, to
obtain rational trees by glueing E7, only one end vertex is available and, for D n only the ends
are available. However, there are rational trees obtained by glueing An at any of its points.
The following theorem shows that the glueing of rational trees gives a rational tree only under
some important necessary conditions :
Theorem 4.11 Let R1 and R2 be rational trees with weights > 2 and let Z1 and Z2 be respec-
tively their fundamental cycles. Suppose that the glueing tree R ofR1 and R2 at the vertices E1
and E2 is rational, then either (Z1.E1) < 0 or (Z2.E2) < 0.
Before giving a proof of this theorem, it will be useful to introduce the following definitions.
A vertex E of a singular graph Γ is called non-Tjurina for an element T of E +(Γ), if it satisfies
(T.E) < 0 (compare with [15] III Definition 3.1). Whenever A is a rational tree, a vertex E is
non-Tjurina for the fundamental cycle of A, if and only if it represents the strict transform of a
component of the tangent cone of a rational surface singularity associated to A.
A subgraph A of Γ is a Tjurina component of the element T of E +(Γ) of Γ, if it is a maximal
(connected) subgraph of Γ made of vertices E such that (Z.E) = 0 (see [15] III Definition
3.1). Whenever A is a rational tree, a result of G. Tjurina ([17] or see [14]) implies that a
Tjurina component of the fundamental cycle Z of A is the tree of a desingularization of one of
the rational singularities which appear after the blowing-up of the singular point of a rational
surface singularity associated to A.
In the statement of the theorem ??, the glueing tree R of R1 and R2 at the vertices E1 and
E2 is rational if E1 or E2 is non-Tjurina for the fundamental cycle in R1 or R2 respectively.
Now, let us introduce the desingularization depth of a vertex in a rational tree.
Let E be a vertex in a rational tree A. Then, either E is non-Tjurina for the fundamental
cycle of A, or it is contained in a Tjurina component A1 for the fundamental cycle Z(A) of A.
In the first case, we say that the desingularization depth is zero, in the second case, either E is
non-Tjurina for the fundamental cycle of A1, or it is contained in a Tjurina component A2 for
the fundamental cycle Z(A1) of A1. By induction, we define the desingularization sequence of
the vertex E in A as the sequence A0 = A, A1,..., Ap of subgraphs of A, such that, for all i,
1 < i < p, E is a vertex of Ai, Ai is the Tjurina component of Ai-i for the fundamental cycle
of .4i_i and E is non-Tjurina in Ap for the fundamental cycle Z(Ap).
We also define the degree of the vertex E as —{Z{A).E)
Proof of Theorem ??. Let us suppose that the glueing tree R is rational. Let B0, B1,..., Bp be
the desingularization sequence of the vertex E1 in R1 and C0,C1, ... ,Cq be the desingularization
sequence of the vertex E2 in R2. As above, denote Z(A) the fundamental cycle of a singular
10
graph A. Then
0 0is a positive cycle of the tree R. The proposition ?? shows that the arithmetic genus p(U) of U
is < 0.
By theorem ??, the multiplicity of E1 (resp. E2) in the fundamental cycle Z(R1) (resp. Z(R2))
is one. The following lemma shows that the multiplicities of E1 (resp. E2) in the fundamental
cycles Z(Bi) (resp. Z(Cj)) are also 1, for any i, 0 < i < p (resp. any j , 0 < j < q).
L e m m a 4.12 Let Z = J27=iai^i and Z' = J2ieA' ai^i> A c {!>••• ,n}, be the fundamental
cycles of R and of a subtree TZ' of R respectively. Then we have ai < di for i e A'.
We shall give below a proof of this lemma.
Let us continue the proof of theorem ??. We have
p(U) =0 0 0<i<p,0<j<g
because, for all k / £, {Z{Bk).Z{BA) = 0 (resp. {Z{Ck).Z{CA) = 0), since for k < £, Be (resp.
Ci) is contained in a Tjurina component of Bk (resp. Ck). On the other hand, the remark made
above shows that the multiplicities of E1 (resp. E2) in the fundamental cycles Z(Bi) (resp.
Z(Cj)) are 1, so
(iJi)) = 1
for any i, j , 0 < i < p, 0 < j < q.
Now, any subtree of a rational tree being rational, we have p(Z(Bi)) = 0, for any i, 0 < i < p
(resp. p(Z(Cj)) = 0, for any j , 0<j<q). Therefore,
p(U) = (p + 1)(q + 1) - (p + 1 + q + 1) + 1.
Since p(U) < 0, we have pq < 0, which implies either p = 0 or q = 0. This proves theorem ??.
It remains to prove lemma ??.
Proof of lemma ?? : Let Z1 be the positive cycle defined as Z \A'= Z1 = J2ieAr aiEi. For i e A',
we have :
(Z • Ei) = ai(Ei • Ei) + ] T aj{ErEi).
Furthermore, since (Z • Ei) < 0 for all i, we have —<ii(Ei • Ei) > J2j^iaj(Ej • Ei).
because aj > 0 and (Ej • Ei) > 0 for i / j . Then we obtain
> ]T]T aj{ErEi
11
So we have (Z1 • Ei) < 0 for all i. The fact that Z' < Z1 gives a- < ai for any i e A'. •
Theorem ?? gives the following corollary :
Corollary 4.13 Let R1 and R2 be rational trees and Z(R1) and Z(R2) be their fundamental
cycles respectively. Suppose that (Z(R1).E1) < 0 and the degree d1 = — (Z(R1).E1) of E1 in R1
is strictly greater than the desingularization depth of E2 in R2, then the glueing tree R of R1
and R2 at the vertices E1 and E2 is rational.
Proof. We shall prove that under the hypotheses of this corollary, the positive cycle
0
of the glueing tree R is in fact the fundamental cycle of the singular tree R and p(U) = 0.
First, let us show that R is a singular tree. It is enough to notice that, for any vertex E of
R, (U.E) < 0.
For this purpose, we need to prove the following lemma :
Lemma 4.14 Let A be a rational tree and let E be a vertex of A with multiplicity 1 in the
fundamental cycle of A. Let p be the desingularization depth of E and A0 = A, A1,... ,Ap be
its desingularization sequence. For any i, 1 <i <p, the cycle J2oZ(Ak) belongs to E+(A) and
it is the smallest cycle in E +(A) greater than Yl%o~l Z(Ak) + E.
We shall prove this lemma below.
Let us apply this lemma to the situation of corollary ??. For any vertex E of R2 , we have
0
Since (Z(R1).E) = 0 for any vertex E coming from R2 but E2, we have (U.E) < 0, for any
vertex E / E2 of R2 . For similar reasons, (U.E) < 0, for any vertex E / E\ of the subtree R1.
Therefore, it remains to estimate (U.E1) and (U.E2). We have
(U.E1) = (Z(R1).E1) + (£Z^j)-Ei)-0
However —(Z(7Zi).E\) is the degree E1 in R1 and, by corollary ?? and lemma ??, we have, for
any j , 0 < j < q, (Z(Cj).E1) = 1, so (J2l Z(Cj).E1) = q + 1. Therefore (U.E1) < 0. On the
other hand, we have
(U.E2) = (Z(R1).E2) + (J2Z(C3).E2).0
Since (Z(R1).E2) = 1 and (J%Z(Cj).E2) = (Z(Cq).E2) < - 1 , we have (U.E2) < 0. This gives
(U.U) < 0, because, either (Z(Cq).E2) = —1 and, since the weights of Cq are > 2, there is
12
necessarily another non-Tjurina vertex F / E2 of Cq, such that (Z(Cq).F) < 0 and one has
(U.F) < 0, or (Z(Cq).E2) < - 1 and (U.E2) < 0. Therefore, by a result of O. Zariski already
quoted above (see [20], theorem 7.1), R is a singular tree.
We prove that the cycle U is the fundamental cycle of the tree R. In fact, by lemma ?? Laufer
algorithm shows that the restriction of the fundamental cycle Z(R) to R2 belongs to E + (R2)
and is greater than Z(R2) + E = Z(C0) + E, since the desingularization depth of E is > 1. By
lemma ?? the restriction of the fundamental cycle Z(R) to R2 is greater than Z(C0) + Z(C1).
By induction, we show that the restriction of the fundamental cycle Z(R) to R2 is greater than
£o^(Cj). Therefore, Z(R) is greater than U = £0 Z(Cj) + Z(R1). Since we showed that U
belongs to E+(R), we have Z(R) = U as expected.
Now, we prove that the arithmetic genus p(U) = 0. In fact, we have
p(U) =0 0
Since p(Z(R1)) = 0 and p(Z(Cj)) = 0, for any j , 0 < j < q, because these are arithmetic genera
of fundamental cycles of rational trees, we have
0
B y c o r o l l a r y ? ? a n d l e m m a ? ? , t h e m u l t i p l i c i t i e s of E1 i n Z ( R 1 ) a n d E 2 i n Z(Cj), for a n y j ,
0<j<q, are equal to 1. So, (Z(R1).Z(Cj)) = 1 and
p(U) = 0.
It remains to prove lemma ??.
Proof of lemma ??. We shall give a proof by induction on i. For i = 0, it is the definition of the
fundamental cycle of A0 := A For i = 1 it is the result of the proposition of §14 in [14] (p.165).
Now, let i > 2. We assume the result true for £, 0 < £ < i — 1.
Denote Ui = Eo Z(Ak). Let F be a vertex of A Suppose that (Ui.F) > 0 :
0 0
Since, by induction, we have (Eo"1 Z(Ak).F) < 0, necessarily (Z(Ai).F) > 0, so that F is
adjacent to Z(Ai) and is linked to Z(Ai) by a vertex F' of A- Assume that F belongs to A- i -
Then, (Z(A-i) + Z(Ai).F) < 0 by applying the result of the proposition of §14 in [14] (p.165).
Since (Eo~2 Z(Ak).F) < 0 by the induction hypothesis, we obtain (J^Z{Ak).F) < 0, which
contradicts the hypothesis (Ui.F) > 0. So, the vertex F does not belong to A- i - Since A iis a
Tjurina component of A - i , F is adjacent to A - i at F' also.
The restriction Z»_i := Z(A)|A-i of the fundamental cycle of Z(A) to A - i belongs to
£+(A-i) (see the proof of ??). On the other hand F' belongs to a Tjurina component of A- i -
13
Since it is linked to F in A, Z*_i > Z{Ai-i) + F'. Proposition of §14 in [14] (p.165) implies that
Zi-\ > Z(Ai-i) + Z(Ai), which implies that the multiplicity of E in the fundamental cycle Z(A)
is > 2. This, again, contradicts the hypothesis on the multiplicity of E in Z(A). Therefore, for
any vertex F of A, we must have (Ui.F) < 0.
It remains to prove that, for any i, 1 < i < p, the cycle J2o Z(Ak) is the smallest cycle in
E+(A) greater than Eo" 1 Z(Ak) + E. Let Y = Y^~l Z(Ak) + E + T be a positive cycle in E+(A).
For any F in Ai we have (Y.F) < 0. So, by lemma ?? the restriction T" of the cycle E + T to
the subtree Ai is in E+(Ai), then T' > Z(Ai), which implies Y > J2oZ(Ak)- Since we proved
that Eo Z{Ak) is in E+(A), it is the smallest in E+(A) which is greater than J2o~l Z(Ak) + E.
This ends the proof of lemma ??. •
Now, we prove an assertion of M. Spivakovsky which allows to obtain many rational trees once
one of them is known.
Theorem 4.15 Let R be a rational tree. Let TZ' be a tree obtained from R by increasing the
weights. Then, TV is a rational tree.
Proof : Let R and TV be trees defined as in the theorem. First we will show that TV is a
singular tree : Let us denote by (.) and (.)' the bilinear forms defined on the free abelian group
generated by the vertices of R and TV respectively. Let Y = J27=i hEi be a positive cycle such
that (Y -Ei)< 0 for any i. Since (Ei • Erf < (Ei • Ei) for any i, we have (Y • Erf < (Y • Ei) < 0
for any i. By the same way, we obtain (Y • Y)'2 < (Y • Y)2 < 0. By (II) of Definition ??, TV is
singular.
Now let us denote p(Y) and p'(Y) the arithmetic genus of a positive cycle Y defined by taking
the bilinear forms in R and TZ' respectively. Let Z' = Y17=i ai^i be the fundamental cycle of
TZ'. We will show that p'(Z') = 0 : We have p'(Z') > 0 (see [1], theorem 3), and since Z' can be
considered as a positive cycle with support on R, Theorem ?? implies p(Z') < 0. So it will be
sufficient to show that p'(Z') < p(Z').
We prove this last assertion by induction on the number of vertices where the two weighted
trees R and TZ' differ. Therefore it is sufficient to prove the assertion when they differ only at
one vertex E1. The condition (iv) of Definition ?? gives :
p'(Z') - p(Z') = a'i(Ei-Eiy + a>i _ a?(El-El)+a'lwl
Since (E1 • E\)' = —w[ and, by hypothesis, w[ = w1 + k, k G N*, we obtain
Since a\ > 1, we have p'(Z') —p(Z') < 0. This gives the theorem and furthermore a\ = 1. •
Remark . A "geometrical" proof of the preceding result is obtained as follows : blowing-up a
general point on a component of the exceptional divisor of a resolution of a rational singularity
14
gives a new rational tree in which a subtree is the one obtained from R by increasing one of the
weights (see the proof of ?? below). By induction, Proposition ?? shows that TZ' is rational.
5 Complexity of rational trees
We say that the rational tree R has multiplicity m if its fundamental cycle satisfies Z2 = —m.
According to [1] (theorem 4), the number — Z2 is the multiplicity of a rational singularity having
TZ as dual graph of a desingularization.
In this paragraph we want to prove that the complexity of a rational tree of given multiplicity
is bounded. In the cases of multiplicity 2, 3 and 4, it has been observed (in [4], [1] and [16]) that
there are a finite number of types of rational trees with weights > 2. It is therefore natural to
understand this result. The first problem is to give a proper definition of the complexity. Since
rupture vertices of the dual graph of the minimal good resolution measure the local topological
complexity of the link of a complex normal surface singularity (see [13]), it seems natural to
define the complexity of a rational tree whose all vertices have weights > 2 to be the number of
rupture vertices.
This idea is enhanced by the following result :
Theorem 5.1 Let R be a rational tree of multiplicity m whose all vertices have weights > 2.
Then the number of rupture vertices of R is bounded by m — 2 if m > 3.
Remark 5.2 The multiplicity of a rational tree is equal to 2 if and only if all its vertices have the
weight 2 (see [4]). As we said in a preceding Remark ??, these rational trees are A n , D n , E6, E7
and E8 , and the number of rupture vertices in these trees is equal to 0 or 1 (see [1]). Of course,
in the Theorem, we could replace the bound by m — 1 to include the cases of multiplicity 2, but
in the following example, we give an infinite class of rational trees with arbitrary multiplicity
m > 3 for which the bound m — 2 is reached.
Example 5.3 Let us consider the following tree R with n vertices :
where " o " and "x" denote the vertices with weights 2 and 3 respectively. The fundamental
cycle of this tree is Z = YA=\ Ei. If the number of vertices of weight 3 is k, then Z2 = —{k + 2).
We have p(Z) = 0. So R is indeed rational with multiplicity (k + 2) and k rupture vertices. •
Lemma 5.4 Assume that R is a rational tree of multiplicity m whose all vertices have weights
> 2 and which contains a unique vertex with weight > 3. Then the number of rupture vertices
ofTZis<m — 2.
15
Proof : Denote by E the vertex which has the weight > 3 and by E1, • • •, En the vertices with
weight 2, in R. Let Z = AEE + J27=i aiEi be the fundamental cycle of R. Since Z2 = —m, we
will show that :
(1) m - 2 = aE(wE - 2) > s
where s is the number of rupture vertices of R. In what follows, the case s = 1, for which
inequality is trivial, is excluded.
Let us denote R1, • • • ,Rp the maximal subtrees of R — {E}. Obviously, all the vertices of
Rj, (j = 1, • • • ,p), has weight 2, and so these are among the rational trees of type A n , D n ,
E6, E7, since E8 has been excluded by Corollary ??. Let Ej be the vertex of Rj adjacent to E
(j = 1,...,p). We have υR(E) = p. Then, the rupture vertices of R are the possible rupture
vertices of the subtrees Rj, (1 < j <p), and the rupture vertices obtained by glueing E and all
the Rj’s. We know, by Remark ??, that the glueing of E and one Rj can give a rupture vertex
only in the case where E is attached to an interior vertex of Rj which is of type A n .
Let us denote α the number of subtrees among R1, • • • ,Rp which has a rupture vertex or
which gives a rupture vertex at Ej when they are attached to E and β the number of subtrees
among R1, • • ,Rp which are of type A n and which are attached to E by a extremity vertex.
Then the number of rupture vertices of R is equal to α if p < 2, and it is equal to α + 1 if p > 3
(i.e. according to that E becomes a rupture vertex in R by glueing it with Rj).
Let AE et a1, • • •, ap the multiplicities of E and E1, • • •, Ep in Z respectively. The fact that
(Z • E) < 0 gives :
(2)
Lemma 5.5 If a subtree Rj has a rupture vertex or if it contributes to a rupture vertex at Ej
when it is glued to E, then we have aj > 2.
Proof : It is obvious that the multiplicity of a bad vertex in the fundamental cycle of a rational
tree is > 2. Since the rupture vertex of Rj is a bad vertex and there are only the vertices with
weight 2 on the geodesic from the rupture vertex of Rj to E, we can see easily that the vertex
of Rj adjacent to E have the multiplicity > 2 in the fundamental cycle of R. •
We deduce that AEWE >2a + β, so 12AEWE > Α. Lemma ?? will be proved if we show that:
AE(WE - 2) > 12AEWE
because this implies AE(WE — 2) > α + 1. Moreover, by the preceding inequality, we have
(WE — 2) > 21WE or WE > 4. So the Lemma ?? is true for WE > 5.
R e m a r k 5.6 If β > 0 then the lemma ?? is true for WE > 4.
By using essentially Theorem ?? and Lemma ??, we finish the proof of Lemma ?? :
For WE = 3 : Since p < 4 by ??, Theorem ?? gives that the number of vertices s < 2. The only
case to be proved is s = 2, which happens when p = 2 or p = 3. If p = 2, R is constructed by
16
two subtrees Rj (j = 1,2) attached to E and each one of these subtrees has at most one rupture
vertex in R, so s = α ; then if s = α = 2, since AEWE = 3aE > 2α + β > 2α = 4, we have
AE > 2, so AE(WE — 2) = aE > 2 = S.
If p = 3, R is constructed by three subtrees Rj attached to E, since E becomes a good vertex
without being very good, only one at most of these subtrees Rj has a rupture vertex in R and
s = α + 1; if s = 2, α = 1 and β = 2, so
AEWE = 3aE > 2α + β = 4
and aE > 2. This also implies AE(WE — 2) = aE > 2 = S.
For WE = 4 : We have p < 5. The inequality (1) is obvious when p <2 and 4 < p < 5 for which
we have s < 2. When p = 3, we obtain s < 4. If s = 3 (resp. s = 4), E being a rupture vertex,
there exist two (resp. three) subtrees Rj which have a rupture vertex in R. Then,
> 2α + β = 5
(resp. AEWE = 4aE > 2α + β = 6).
In both cases, we have aE > 2. So we have AE(WE — 2) = 2aE > 4 > s. •
Proof of the theorem ?? : Assume that R contains several vertices with weights > 3 and with
weights 2. By using lemma ??, we will describe the subtrees of R and the glueing between them
to obtain R.
Let Z = Y2=i aiEi be the fundamental cycle of R and E1, • • •, Ek be the vertices with weights
> 3 in R. Observe that m > 3 implies k > 1. So we have to show :fc
(3) m - 2 = ^ a i ( w i - 2 ) > s
where s is the number of rupture vertices in R.
For each vertex Ei, (1 < i < k), with weight > 3, we consider the maximal subtree Bi of R
which contains Ei and all subtrees Rj, j G Ji, whose vertices have all their weights equal to 2
and which are adjacent to Ei. The subtree Bi is rational and contains a unique vertex of weight
> 3. By lemma ??, Bi satisfies a'i(wi — 2) > si where Zi = a^Ei + Y.Ei&Z' ai^i is the fundamental
cycle of Bi and si is the number of rupture vertices of Bi. By lemma ??, we havek k k
m-2 = Ydai(wi-2)>Yda'i(wi-2)>Ydsii=1 i=1 i=1
Since R is obtained by the glueing of the subtrees Bi's, the valency in R of the vertex of Bi by
which the glueing is made, might not be the same as in Bi. So we cannot assert that YA=I Si> S
since it is possible to create some new rupture vertices by glueing the Bi's between them to
obtain R.
Lemma 5.7 A rupture vertex in R which is not a rupture vertex in any Bi, is one of the
following types :
17
(a) it is the vertex Ei of weight > 3 with valency < 2 in Bi.
(b) it is a vertex of weight 2 which is the extremity of one of the subtrees Rj of Bi which is of
type A n .
Proof: This follows from the construction of Bi and from theorem ??. In fact, if it is a vertex
of weight 2, it is necessarily a bad vertex in R. By Theorem ??, the subtree Rj of the maximal
subtree Bi, which contains it, must be of type A n . Moreover, it is an extremity of this subtree
of type A n , because if it was not, it would already be a rupture vertex in one of the Bi's which
contain it. •
So, we shall consider the following three cases of vertices of a subtree Bi which are rupture
vertices in R without being rupture in the subtree Bi :
(A) the vertex Ei of Bi which has weight > 3,
(B) vertices of weight 2 which are extremities of subtrees Rj of type A n ,
(C) vertices of the preceding two types.
5.8 To find the rupture vertices of R which are not rupture vertices in Bi, it is sufficient to
consider the attachement to Bi of the vertices in TZ—Bi which are adjacent to Bi. By construction
of Bi, these vertices, which are adjacent to Bi in R, have weights > 3. Let us denote by B[ a
minimal subtree of R which contains Bi and in which the vertices, which become rupture vertices
in R without being of rupture vertices in Bi, are rupture vertices. Let s\ be the number of rupture
vertices in B[. If Bi does not contain rupture vertices of R which are not rupture vertices in Bi,
we set B\ = Bi and s\ = si.
Let a!'E. be the multiplicity of Ei in the fundamental cycle of B\. By lemma ?? we have
aEi > O>'E-- Since J2i=i si > s , Theorem ?? will be proved if we show, in these three cases above,
that :
(4) a%{wEz-2)>s't
(A) Let us consider a subtree Bi of R which contains a rupture vertex of weight > 3 ofR which
is not a rupture vertex in that Bi.
Denote by B such a subtree Bi and by E its unique vertex with weight > 3. Since E is not
a rupture vertex in B, B — {E} has at most 2-connected components. By hypothesis, we have
s' = s + 1 where s' is the number of rupture vertices of a minimal subtree B' of R which contains
B and in which E is a rupture vertex. So there are three cases to be proved depending on the
valency of E in B :
(1) If ΥB(E) = 0, we have B = {E}. A minimal subtree B' is obtained by glueing three vertices
with weight > 3 of R to E. The inequality (4) is obvious for WE > 3.
(2) If ΥB(E) = 1, the rational tree B consists of the vertex E attached to a tree of type A n ,
D n , E6 or E7 attached to the vertex E. A minimal subtree B' of R is obtained by glueing two
18
vertices of R with weight > 3 to the vertex E of B. We have s' < 2. Since the inequality (4) is
obvious for WE > 4, the only case to consider is when WE = 3 and s' = 2. In this case, by lemma
?? and inequality (2), we obtain 3aE > 4. This gives aE > 2, and therefore we have inequality
(4).
(3) If ΥB(E) = 2, B is obtained by glueing two trees of type A n , D n , E 6 or E 7 to the vertex
E. A minimal subtree B' contains B and a vertex with weight > 3 of R attached to E. Since
s' < 3, our result is obvious when WE > 5. Now, we have s' < 2 (resp. s' < 3) if WE = 3 (resp.
WE = 4). When s' = 2 (resp. s' = 3), by lemma ?? and the inequality (2), we obtain 3a'E > 4
(resp. 4aE > 5). This gives aE > 2, and so we also have inequality (4) for WE = 3 and 4.
(see lemme 7.1 and lemme 7.4 in [18] for all possibles rational trees of the type given in (2) and
(3) respectively)
(B) Let us consider a subtree Bi of R which contains rupture vertices of R which are not the
rupture vertices in Bi and assume that these vertices have weight 2 :
Denote by B such a subtree Bi. As in the proof of lemma ??, among the maximal subtrees of
vertices of weight 2 of B, we have :
(i) the α maximal subtrees with vertices of weight 2 which contain a rupture vertex of B,
(ii) the γ maximal subtrees with vertices of weight 2 which contain a rupture vertex of R which
is not a rupture vertex in B. We saw that these subtrees are of type A n and they are attached
to the vertex with weight > 3 by one extremity.
(iii) the β maximal subtrees with vertices of weight 2 of B which contain no rupture vertex of
R.
By lemma ?? and inequality (2), we have AEWE > 2α + 2γ + β where E is the vertex with
weight > 3 in B. As in the case (A), let B' be a minimal subtree of R which contains B and
in which the vertices of weight 2 in B, which are rupture vertices in B without being rupture
vertices in B, are rupture vertices. Since s = α or α + 1, we have s' = α + γ or s' = α + γ + 1. If
WE > 5, theorem ?? is proved in the case (B), because we have AE(WE — 2) > 12AEWE > α + γ.
It only remains to prove it for the cases WE = 3 et WE = 4.
WE = 3 : case (B) concerns γ > 1. Lemma ?? shows that necessarily the valency of E is < 3.
If the valency is 3, by ?? again, the only possibility is s' = 1.
If s' = 1, aE(wE - 2) = aE > 1 = s'.
If the valency of E is < 2, s' < 2. Case s' = 1 has just been considered. If s' = 2, α + γ = 2,
and AEWE = 3aE > 2α + 2γ + β > 4, which implies aE > 2, so AE(WE — 2) = 2 > 2 = s'.
WE = 4 : as above γ > 1. Then, by ??, the valency of E is < 4 and, if it is 4, s' = 1, in which
case, the inequality AE(WE — 2) > s' is true.
We may assume that the valency of E is < 3. Then, s' < 4. The case s' = 1 has already
been treated. In the same way, for s' = 2, our inequality holds. Thus, we may suppose that the
19
valency of E is 3. Then, for α + γ = 2 and β = 1 (resp. α + γ = 3 and β = 0), we have s' = 3
(resp. s' = 4). We have
aEwE = 4a E > 2 (α + γ) + β > 5
(resp. AEWE = 4aE > 2(α + γ) + β > 6),
which implies in both cases aE = 2 and AE(WE — 2) = 2aE > 4 > s'.
(C) Let us consider a subtree Bi ofR containing rupture vertices with weights 2 and > 3 ofR
which are not the rupture vertices in that Bi.
Denote by B such a subtree Bi of R. By definition, B contains only one vertex with weight > 3.
Let E denote this vertex. Since E is not a rupture vertex in B and it becomes a rupture vertex
in R, we have necessarily the valency ΥB(E) < 2 (see the case (A) above). This implies that B
contains at most two rupture vertices with weights 2 of R which are not rupture vertices in B.
Therefore we will consider a minimal subtree C of R which contains B and in which E and the
vertices of B which are rupture vertices in R and not rupture vertices in B are rupture vertices.
Then, we only have the cases s = 0 or s = 1, since s = 2 would imply γ = 0, contradicting our
hypothesis C. So s' = 2 or s' = 3. This means that we have O,E(WE — 2) > s' for WE > 5. Again,
it remains to prove this inequality for WE = 3 and WE = 4.
w = 3 Let s' = 2, then γ = 1, α = 0 and β = 1. Since there are two vertices of weight > 3
adjacent to E, we have so a'EWE = 3aE > 2(α + γ) + β + 2 = 5. So aE > 2.
If s' = 3, then, α + γ = 2. Then, O,EWE = 3aE > 2(α + γ) + β + 2 = 6, which gives aE > 2.
In both of these cases, O,E(WE — 2) > s' as expected.
w = 4 The inequality is obvious for s' = 2. Suppose that s' = 3. Then, α + γ = 2 and
flWB = 4aE > 2(α + γ) + β + 2 = 6, which gives aE > 2 and the searched inequality.
This ends the proof of theorem ??. •
6 Some classes of Rational Trees
There exist some interesting classes of rational trees having some nice properties. In this section
we define these classes and give few of their properties.
Definition 6.1 ([8],4.4.1) Let O be a normal local ring of dimension 2 and let S := Spec(O).
Let ξ denote a singular point of S. We say that ξ is a minimal singularity if
multξS = emb.dimξS + 1
and the tangent cone of S at ξ is reduced.
As in ([15], p. 425) :
20
Definition 6.2 A rational tree is called minimal rational tree if all its weights are > 2 and its
fundamental cycle is reduced (i.e. the multiplicities of all vertices are 1)
J. Kollar in ([8],4.4.10) proved that :
T h e o r e m 6.3 A normal surface singularity is minimal if and only if the dual graph associated
with the exceptional divisor of the minimal resolution of the singularity is rational minimal.
We have a simple characterization of rational minimal trees given by M. Spivakovsky :
Proposition 6.4 ([15], remark 2.3) A tree R is minimal rational if and only if, for any vertex
Ei, (1 < i <n), of R, we have wi > υR(i).
Proof: The first implication follows from the fact that we have (Z-Ei) < 0 for any i, (1 < i < n).
If R is a tree such that we have wi > υR(i) for any vertex Ei in R, by Laufer algorithm, we
obtain Z = J27=i Ei, and so Z2 < 0. Since we have n — 1 = 2 J2 υR(Ei), we obtain
p ( Z ) = T^i(-Wi
This ends the proof. •
We have the immediate corollary
Corollary 6.5 Any subtree of a rational minimal tree is a rational minimal tree.
Notice that there is no bad vertices in a rational minimal tree. Now :
Proposition 6.6 Let R1 and R2 be two rational minimal trees and let E1 (resp. F1) be a vertex
of R1 (resp. R2) such that w1 > υR1(E1) (resp. w2 > υR2(E2). Then the glueing tree of R1
and R2 at E1 and F1 is a rational minimal tree.
Proof: Let Z1 = J2t=i Ei and Z 2 = Y?j=i Fj be the fundamental cycles of R1 and R2 respec-
tively. Let E1 and F1 be defined as in the proposition. Let R denote the glueing tree obtained
by attaching E1 to F1. It is easy to show, by Laufer algorithm, that the fundamental cycle of
TZ exists and it is exactly Z = Z1 + Z2. So we have Z2 < 0 and
Et=1 υR2(Fj) + 2-2s-2t2 '
Then this implies p(Z) = 0. Therefore the proposition. •
Another class of rational tree is the class of non-singular trees :
Definition 6.7 A non-singular tree is the dual tree of an embedded resolution of a complex
plane curve germ.
21
An interesting characterization of non-singular tree is given by M. Artin (see [1] theorem 4):
Theorem 6.8 A weighted tree is non-singular if and only if it is rational and the self-intersection
of its fundamental cycle equals — 1.
However a subtree of a non-singular tree is not non-singular in general. In fact, following M.
Spivakovsky (see [15] definition 1.9), we define
Definition 6.9 A weighted tree is called Sandwich if it is the subtree of a non-singular tree.
Since a non-singular tree is rational, any sandwich tree is a rational tree. There is an interesting
characterization of Sandwich trees by M. Spivakovsky (see [15] p. 420) :
Theorem 6.10 A weighted tree is sandwich if and only if by attaching a finite number of vertices
of weight 1, it gives a non-singular tree.
This result leads immediately to :
Theorem 6.11 Let R be a sandwich tree. Let Tt1 be a tree obtained from R by increasing the
weights. Then TV is a sandwich tree.
Proof. Let A be a non-singular tree which contains R as subtree. Let E be a vertex of A of
weight w which belongs to R. Consider the tree A1 obtained from A by attaching a vertex E1 of
weight 1 to E. It is easy to see that A1 is the dual tree of the exceptional divisor of the embedded
resolution of a plane complex curve. In fact, let E be the component of the exceptional divisor
of an embedded resolution of a plane complex curve associated to A. Then, by blowing-up a
general point of E, we obtain another exceptional divisor of an embedded resolution of a plane
complex curve whose dual graph is precisely A1. By proceeding by induction on the difference
between the sums of weights of TV and R, we prove our theorem. •
References
[1] M. Artin, On isolated rational singularities of surfaces, Amer. J. Math. 88 (1966), pp.129-
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