C OMPOSITIO M ATHEMATICA
H. P INKHAM
Appendix to : “smoothings of cusp singularities via triangle singularities”
Compositio Mathematica, tome 53, n
o3 (1984), p. 317-324
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317
APPENDIX TO: "SMOOTHINGS OF CUSP SINGULARITIES VIA TRIANGLE SINGULARITIES"
H. Pinkham
© 1984 Martinus Nijhoff Publishers, Dordrecht. Printed in The Netherlands.
The
goal
of thisappendix
is to prove thefollowing slightly improved
version of Theorems 1 and 4 of the
preceding
paper(which
will bereferred to as
[FP],
all other referencesbeing
those of[FP]).
THEOREM: Let D be a cusp
singularity
with r 3 components and m = r + 9.Assume the dual cusp
b
liesanticanonically
on a rationalsurface.
Thenfor
every
good embedding
T: R ~Tpqr
039Bexcept possibly
theprimitive
em-bedding of T3,6,12
there is afamily
W ~ 0394 as in[FP],
Theorem4,
where thesingularity
D’of Wo
is infact
D. Furthermore(- 2, - 5, -11)
has agood,
non
primitive embedding.
By
the discussionfollowing
Theorem 4 of[FP]
the result isalready proved
in all casesexcept D = ( - 4, - 5, - 9)
and( - 2, - 5, -11).
Be-fore
dealing
with these cases we make somegeneral remarks,
valid for allTp,q,r.
DEFINITION: Let T be an over lattice of
Tpqr.
Then thegeometric
sectionof the map T ~
T/Tpqr
is the section a:T/Tqqr ~
T which to t ETITPI,
associates the
unique 03C3(t) ~
T all of whose coefficients in the natural basis ofTpqr
are > 0 and 1. The support ofa(t)
consists of the sublattice ofTPqr spanned by
the vertices where the coefficients ofa(t)
are non zero.
We will construct branched covers
using 03C3(t),
whence theterminology.
DEFINITION: If T: R - L is an
embedding
oflattices,
then theprimitive
sublattice of L
generated by R
is called the saturation of R for cp.We make the
following hypothesis,
which holds for all the overlattices considered in this note:(H)
for every non-trivial t ~Tpqr/T, , ,
thesupport
of thegeometric
sectiona(t)
has connected components withnegative
definite intersection matrix oftype An-
It seems that( H )
holdsfor all overlattices of
TPqr,
where p+ q
+ r =21,
which is the caseconsidered in this note, but J. Wahl has
given
anexample
of an318
overlattice T of
T4,7,9
for which it fails. Wahl shows that T is the saturation of agood embedding
ofT4,7,9,
so that Lemma 1 below fails without(H).
LEMMA 1: Let (p:
Tpqr
A be agood embedding
in the senseof Looijenga,
and let T be the saturation
of Tpqr for
cp. Assumehypothesis (H).
Thenfor
any non-trivial t E
TIT
thegeometric
section0’( t) of
t has square03C3(t)2 -4.
PROOF: Since cp is
good
there is a K - 3 surface X with Picard group T and aT pqr configuration ([FP],
Lemma2). Apply
Riemann-Roch to03C3(t)
considered as an element of Pic X. The
hypotheses imply
thatH0(X, O(03C3(t)))
andH2(X, O(03C3(t)))
aretrivial,
so that the desiredinequality
isequivalent
to the assertion that dimH1(X, O(03C3(t)))
0.LEMMA 2:
Let V by a
rationalsurface containing
the minimal resolutionof
either the cusp D or the associated
Dpqr singularity,
and assume theorthogonal complement Tpqr of
the componentsof
theexceptional
divisorcontains a lattice
Tp’,q’,r’ of finite
index such that(i)
the verticesof
theTp’,q’,r’
arerepresented by P
1’ s, so V has aTp’,q’,r’
configuration,
in thelanguage of [FP].
(ii) hypothesis (H)
holds.Then
03C3(t)2 = -2 for
every non trivial t, and thesingularity
is infact
thecusp
if TpqlTp’,q’,r’
is non-trivial.PROOF: Contract the connected
components
of thesupport
of03C3(t),
considered as a
fractional
divisoron V,
to rational doublepoints.
Callthe
resulting
surface V. As inNikulin,
"Finiteautomorphism
groups ofKâhler K 3 surfaces"
(Trans.
Moscow Math.Soc. 38(2), (1980)), §8.2,
wecan construct
using 03C3(t)
acyclic
cover Z of V ramifiedonly
above theserational double
points
and with Z smooth above the doublepoints.
Thedualizing
sheaf on V isanti-effective,
so that on Z is too. If the resolutionon V is that of
Dpqrl then ’171
of theexceptional locus
is trivial so thatDp qr
splits
into severalisomorphic singularities
in the cover, if the cover is nontrivial.
By
the classification ofsingular algebraic
surfaces with trivial canonical divisor due to J.Y. Mérindol(C.R.A.S.
Paris293,
417-420(1981),
théorème1.4)
this cannot occur. So the resolution on V is that of D(where
theexceptional locus has ’171 === Z). By
Mérindol’s classificationagain,
we see that Z isrational,
so77BZ, OZ)
= 0. Use standard ramifica- tiontheory
and Riemann-Roch on V tocompute H1(Z, OZ):
it is zero ifand
only
if03C3(t)2
= - 2 for all non-trivial t.We now check the theorem in the two
remaining
cases.Case I:
( - 4, - 5, - 9)
T’s,6,lo
has one overlatticeTl
in which it has index 4. Werepresent
itby listing
the non-zero coefficients of thegeometric
section of agenerator
ofT1/T5,6,10:
T5,6,10
also has one overlatticeT2
in which it has index2, corresponding
tothe
subgroup
of order 2 ofT1/T5,6,10.
These are theonly
overlattices ofTS,6,IO
as can be checkedusing [P2],
Lemmas 1 and 2. In both cases, forall non-trivial
t, 03C3(t)2 = -4 so
thatby
Lemma 2(and [FP]
Lemma9, (ii)) applied
to( p’, q’, r’) = (5, 6, 10)
theonly
cusp orDolgachev singularity, corresponding
as in[FP]
Lemma 2 toTpqr,
with rankT5,6,10
=19,
and inthe
negative part
of the versal deformation ofD5,6,10
is(
-4, - 5, - 9)
orDS,6,IO
itself. The theorem now followsby examining
the statement of[FP],
theorem 4: all otherpossibilities
for D" have been ruled out.Case II:
(-2, -5, -11)
T3,6,12
has twooverlattices,
and has index 3 in both. Thefirst,
which wecall
Ml corresponds
to thegeometric
section320
The
other, M2,
toNote that for
Ml (resp. M2 )
every non-trivialgeometric
section hassquare - 4 (resp. - 2),
so thatM2
is not the saturation of agood embedding
ofT3.16,12 (Lemma 1)
andMI
cannotcorrespond
to aTp,q,r
asin Lemma
2,
where( p’, q’, r’) = (3, 6, 12).
On thepositive
side we have:LEMMA 3:
(i)
There exists agood embedding
99:T3,6,12
- A with saturationMl.
(ii)
There exists a smooth rationalsurface
Vcontaining
a nodal rationalcurve C
of self-intersection -10
and aT3,6,12 configuration of
rationalcurves
orthogonal
to C. The saturationof T3,6,12
inH2(V, Z)
is thennecessarily M2.
PROOF: The existence of 99 in
(i)
isstandard, using [N];
thecomputation
that T is
good
isgiven
in Lemma 4.For
(ii)
we exhibit Vexplicitly:
take two inflectionpoints
on a nodalcubic C
in P2,
and blow up 15 times at one, and 4 times at the other. The proper transforms of the two inflection lines and of theexceptional
curves with self-intersection - 2 form a
T3,6,12 diagram.
We now conclude the
proof
of the theorem.By
Lemma3, i)
weconstruct
using
T afamily
W - à as in Theorem 4 of[FP].
We want toshow that the
special
fiberWo
has a( - 2, - 5, -11) singularity.
If notWo
has a(-10)
cuspsingularity:
indeed theonly
otherpossibility
allowed
by
theorem 4 is aD2,3,16 singularity,
but this is ruled outby
Lemma 2.
So assume
Wo
has a( -10)
cuspsingularity.
Call thesingular point p.
As in Lemma 11 of
[FP]
let B be a Milnor ballaround p,
and letYt = Wt - B.
All the(Y, ~Yt)
arediffeomorphic,
andaY
is a certainS1
XS’1
bundleover S’ (described
forexample
in[P3],
but we will notneed more detailed
information).
Let L =
H2 (Y,).
Note that L has rank20,
and has a kernel of dim. 1.We have a natural inclusion of lattices for t ~ 0: L =
H2(Yt) H2(Wt) =
A. Let L’ be the saturation of L in A. Since
T3,6,12
L and the saturation ofTp,q,r by
theembedding
into A isMl,
we haveM1 L’;
and sinceMl
is non
degenerate M, -
L’ where indicates thequotient by
thekemel.
_
But we also have L =
H2(Y0),
andby
construction we haveM2 ~
L cL’.
Sofinally
we havebut this is absurd since
L’
has aninteger
valuedquadratic
form andMl
and
M2
are distinct maximal overlattices ofT3,6,12.
To conclude we must show that the
embedding
of Lemma3(i)
isgood.
Using
the definition ofgood given by Looijenga
in[L2]
it ispossible
tocheck,
via astraightforward
but involvedcomputation,
whether anygiven embedding
of aTpqr
isgood
or not.(I
thankLooijenga
forexplaining
tome how to do
this,
and forshowing
me someunpublished manuscripts
on this
subject).
There is onesituation, covering
the case we are inter- estedin,
which can be treatedusing
ageneralization
ofLooijenga’s proof
that
primitive embeddings
aregood:
LEMMA 4: Let T:
Tpqr ~ 039B
and let T be the saturation.If for
every non-trivial t ET/Tpqr
thegeometric
sectiona(t) satisfies :
(i) 03C3(t)2 = -4;
(ii)
the supportof 0’( t)
is contained in 2 branchesof Tpqr;
then the
embedding
isgood.
Hypothesis (ii)
isprobably
unnecessary. It is ratherrestrictive,
sincefor instance it
forces |T/Tpqr|
3.PROOF: We label the vertices of
Tpqr
as in[L2]:
322
Let n be the fundamental
isotropic
element([L2], §3):
e.g., if3 p q
r then n = Ap-2 + 2Ap-1 + 3E + 2Bq-1 + Bq-2 + 2Cr-1 + Cr-2.
According
to[L2],
we must show that for any t E T such that t. n = 0and t2
= -2,
then either t isorthogonal
to allcomponents
in thesupport
of n, or issupported
on thesupport
of n.Write t = s - u, where s E
Tp qr
and u is thegeometric
section of anelement of
T/Tpqr. By Looijenga’s
result in theprimitive
case we mayassume u ~ 0.
By hypothesis
u issupported
on 2 branches ofTpqr,
say forconcreteness on
( E, Bj
1j q - 1, Ck 1 k
r -1).
It is easy to seethat |T/Tpqr|
3 and u is of the form:if the order is 2.
Thus q
and r are even. The conditionu 2
= - 4implies
q + r = 16. This case was
already
treated in[P2].
Or if the order is 3:(or interchange 1/3 ~ 2/3).
So q = r ~ 0 mod3,
and q + r = 18.Note that t is
obviously
notsupported
on n.We will need the
following
two facts onthe -Ak
rootsystem.
Let«1, ... , ak be the standard
basis,
so thata, 2 = -2,
al’aj
= 1if i - j|
= 1and 0 if
|i - j| > 1.
SUBLEMMA 1 :
(Looijenga) If z
=03A3kJ=1 zjaj
is in theintegral
spanof
theaJ’
then z - z +
z21
0 withequality iff
z=
0 and z - z +zi 2
= -1iff
± z = al +...
+a, for
some 1 k.SUBLEMMA 2:
If
k = mp - 1( m
and ppositive integers)
andthen
for
any w in theintegral
spanof
theaJ’
PROOF: We
only
prove Sublemma 2. Use the standardrepresentation
of- Ak
inR"’
1 with basis e1,...,ek+1 and bilinear formel · eJ = -03B4ij.
Then
aJ
=eJ - eJ+1,
W =03A3k+1J=1tJeJ
with03A3k+1J=1tJ
=0,
andtJ
~ Z.So that
Now make the
change
of variable tocomplete
the square. Theequation
becomes 2u · w -
w2
=03A3k+1J=1s2j - m
with constraint03A3k+1J=1sJ
= m.Clearly
this is
always > 0,
asrequired.
We go back to the
original
situation: t = s - u. Note thatreplacing t by
t + an for anyinteger a
affects neither thehypothesis
nor theconclusion of what we are
trying
toshow,
so we may assume the coefficient of s inA p -1 1 (the
vertex inTpqr,
not the rootsystem.... )
is0
andstrictly
less than thecorresponding
coefficient of n.Write s = z + w, where z is
supported
on(Al,
1i p - 1)
and w on( E, BJ,
1j q - 1, Ck, 1 k r - 1). Let z1
be the coefficient ofAp-
1in z
and we
that of E in w. NowIf z, =
0,
sincez2 0, u2 -4 by hypothesis
andw 2 - 2a · w
0by
sublemma
2,
there is no solution to thisequation.
The rest of theproof
consists in
checking
caseby
case theremaining
valuesof zl :
0 z, coefficient ofAp- 1 in
n.For concreteness we
only
do the case( p,
q,r)
=(3, 6, 12)
and T theoverlattice
M,
defined above. Then there isonly
one case to check:zl = 1. Note that z · n = u · n = 0 so the
hypothesis
on timplies w - n
= 0.Using
sublemma1, equation (* *)
becomes:Identify
the chainB1,..., Bq-1, E, CI-1, .... CI
with theAI,
rootsystem
in the obvious way and use the standardrepresentation
as in theproof
ofSublemma 2. Then
324
and the condition w - n = 0 is
Thus we are
trying
to solve inintegers (use equation ( * )
in( *
** )) :
with constraints
03A318J=1tJ
= 0 and(****).
Make the obvious
change
of variable tocomplete
the square. Theequation
becomes :with constraints
and
It is obvious there are no solutions in
integers.
Therefore there are nocandidates
for t,
and the Lemma isproved.
Acknowledgements
This
appendix
was written at the Centre deMathématiques
of the EcolePolytechnique.
1 thank its members and staff for theirhospitality
andfinancial
support.
1 also thank EduardLooijenga
forhelp concerning
"good" embeddings.
Reference
[FP] R. FRIEDMAN and H. PINKHAM: Smoothings of cusp singularities via triangle singular-
ities. Comp. Math. 53 (1984) 303-316.
(Oblatum 26-IV-1983)
Columbia University
New York, NY 10027 USA