C OMPOSITIO M ATHEMATICA
A LEXANDRU D IMCA
On the homology and cohomology of complete intersections with isolated singularities
Compositio Mathematica, tome 58, n
o3 (1986), p. 321-339
<http://www.numdam.org/item?id=CM_1986__58_3_321_0>
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ON THE HOMOLOGY AND COHOMOLOGY OF COMPLETE INTERSECTIONS WITH ISOLATED SINGULARITIES
Alexandru Dimca
C Martinus Nijhoff Publishers, Dordrecht - Printed in the Netherlands
Let V be a
complex projective complete
intersectionhaving only
isolatedsingularities
at thepoints al
fori = 1,...,
k.In this paper we
investigate
to what extent theintegral homology
andthe
cohomology algebra
of V can be determined from the local informa- tion associated to thesingularities ( h, a,).
The relevant localproperties
of the
singularities (V, al)
turn out to be contained in their Milnor latticesL,.
In the first section we recall some facts on the
homology
of a smoothcomplete
intersection. Theonly
new result here ismaybe
the observation that the middlehomology
group of such acomplete
intersection(re- garded
as a latticewith the
intersectionform)
contains in a natural way a reduced Milnor latticeL
=L/Rad
L(see (1.4), (1.5)).
In the next section we prove the main result of the paper
(Theorem 2.1)
which says that thehomology
of V is determinedby
amorphism
oflattices
more
precisely, by
its kernel and cokernel.(We
denoteby
~orthogonal
direct
sums.)
In
fact,
the source and thetarget
latticesdepend only
on the localinformation at the
singular points,
the dimension and themultidegree
ofV. The
global
unknown information on which thehomology
of Vdepends (e.g.
theposition
of thesingularities
onV)
is contained in theapplication ~V
itself.The third section is arithmetic in nature.
Assuming
the latticesL, nondegenerate,
we show that the torsionpart T(V)
of thehomology
ofV
belongs
to a finite set of groups which iscomputable entirely
in termsof the lattices
Li’ using
the formalism of discriminant(bilinear)
formsassociated to lattices.
In
particular,
wecompute
the discriminant forms associated to thehypersurface simple singularities Am, Dm
andEm. Using
thesecomputa- tions,
we are able to prove thatT(V)
= 0 in a lot of concrete situations.The results in this section were worked out
jointly
with my student A.Nemethi.
In the forth section we
compute
thehomology
of two classes ofvarieties: the cubic surfaces
in p3
and some 2 m-dimensionalcomplete
intersections of
two quadrics.
In these favourable cases the arithmetic of the latticesL,
and L determinescompletely
theapplication
~V.In the next section we consider two
complete
intersectionsV0 and V,
with isolated
singularities
at thepoints ai
such that thesingularities (VO, al )
and(Vi, ai)
areisomorphic
fori = 1,..., k.
Wegive
in thissituation a sufficient condition such that
Tlo and VI
have the samehomology (Proposition 5.1).
Also we show how this result can be used to relate themorphisms Wv
tothe adjacency
ofsingularities.
This device is basic when the latticesLi
and Lgive
too little information on ~V, as for instance in thesimple
case of an odd-dimensionalA1-singularity.
In the final section we
point
out how themorphism T v
determines thecohomology algebra
of V with coefficients in anunitary ring.
Thisprovides
us with a finer tool forshowing
thenon-homotopy equivalence
of some varieties
(see (6.3)). Conversely,
in the case ofsurfaces,
thisresult combined with Whitehead classification of
simply
connected 4-di- mensionalCW-complexes,
can be used to show that certain surfaces arehomotopy equivalent (see (6.4), (6.5)).
1. The
homology
of smoothcomplete
intersectionsIn this section we recall some facts
concerning
the(integral) homology
groups
Hi(V)
of a smoothcomplete
intersection Vc cpn+p with dim V = n andmultidegree (d1,..., dp).
It is well known that
Hi(V)
=Hi(CPn) except
forHn(V)
which is afree group of rank
bn(V), computable
in terms ofd1,...,dp [13].
If
Pl
= ...= Pp
= 0 are thehomogeneous equations of V,
then thereis a natural
S1-bundle p : K ~ V,
whereK = {x ~ Cn+p+1; |x| =
1, P1(x)= ... = Pp(x) = 0}.
We
call p
the MilnorS’-bundle
of thevariety
V and note that itgives
a
Gysin
sequence inhomology
For n
odd,
Un+1 1 ismultiplication by d = d1 · .. · dp = deg
V. For neven, the same is true for Un 0 un+2.
Take now a
hyperplane
H in CPn+p such that W = V ~ H is smooth and let us denoteby
N a tubularneighborhood
of W in V. It is easy tosee that the associated
S1-bundle
8N - W isequivalent
to the MilnorS1-bundle
of thevariety
W(use
theequivalence
betweenSl-bundles
andcomplex
linebundles).
Next we prove the
following
basic result.LEMMA 1.2: The
affine variety
U =VB
W ishomeomorphic
to the Milnorfiber of the singularity (X, 0) = ( cone
overW, vertex).
PROOF: Assume that xo = 0 is an
equation
for thehyperplane
H. ThenU ~ Cn+p
isgiven by equations fi(x)=0
fori = 1, ... , p,
where X =(x1, ..., Xn+p)’ fi(x) = P1(1, x)
=Qi(x)
+Q’i(x)
withQ,
ahomogeneous polynomial
ofdegree d
l andQ’i
apolynomial
ofdegree d,.
Note that
QI
= ... =Qp
= 0 are theequations
of thesingularity (X, 0).
Next
take y
= je -r-1
for a real number r and denoteIt follows that
Ur ~ B~ ~ U ~ B~r,
whereB~ = {y ~ Cn+p; |y| ~}
and m is
multiplication by
r. We havealso y E Ur
if andonly
ifFor r
big enough,
theequation Q; = 0
isnothing
else but a smalldeformation of the
equation Qi
= 0.It follows that
Ur ~ B~
is the Milnor fiber of the isolatedsingularity
ofcomplete
intersection(X, 0) [16].
On the other
hand,
for anyalgebraic (possibly singular)
setZee N,
itis known that Z
r1 Br
ishomeomorphic
to Z for rbig enough.
0The
Mayer-Vietoris
sequencecorresponding
to the cover V = U U Ncontains the
morphism
and we want to
interpret
thecomponents jl
of thismorphism
after wereplace U ~ N by
aN = K’ = the total space of the MilnorS’-bundle
ofW,
Uby
X = the Milnor fiber of(X, 0) and
Nby
thevariety
W.We denote
by
L the Milnor lattice(Hn(X),( , ))
where the intersec- tion form( , ) is symmetric
for n even andskew-symmetric
for n odd.For any such lattice
L,
we denoteby
RadL the sublattice {x~L;
(x, y)
= 0 forany y ~ L}
and call thequotient
L =L/Rad
L with theinduced biliniar form the reduced Milnor lattice of the
singularity (X, 0).
With these
notations, j,
can be identified with the natural inclusionand j2
can be read off thecorresponding Gysin
exact sequence(1.1).
We
get
thus the exact sequenceCOROLLARY 1.4: For n
odd,
the lattice(Hn(V), ~ , ~),
where( , ~
denotes the
intersection form,
isisomorphic
to the reduced Milnor lattice L.In
particular,
L is unimodular.PROOF: In this case
Hn(W)=0=ker p*,
as can be seen from thecorresponding
exact sequence(1.1).
DNote that
by
recent work of Chmutov[6],
the fact that an odd-dimen- sionalhypersurface singularity
has unimodular reducedMilnor
latticehas
interesting
consequences on themonodromy
group.For n even, let h E
Hn(V)
be the class of thecycle corresponding
tothe intersection of V with a
generic ( p
+n/2)-plane. Alternatively,
h =
un+2(g)
where g is the canonicalgenerator
ofHn+2(V).
Then it is well-known
that ~h, h~ = d
and let us denote h~ theorthogonal complement
of h withrespect
to the intersectionforum ).
COROLLARY 1.5: With the notations
above,
there is a latticeisomorphism L ~ h~.
PROOF: When n is even, in the exact sequence
(1.3)
one hasHn ( W )
=Z,
ker p* =
Z/dZ.
The inclusion W c V
gives
an identifications(Hn(W))=Z.h.
Thegeometric description
of thecycle
h shows thats(L)~h~.
Asimple computation
proves that the lattice Z h +h~
has index d inHn(V)
andhence s(L) = h~.
0Note that the obvious consequences of
(1.5)
that h~ is an even latticeand that any
cycle
c E h~ can berepresented by
an embeddedsphere
S" ~ V have been
proved by completely
differentmeans in [15] (2.1).
On
the otherhand,
it follows from(1.5)
thatsign L = sign V - 1,
rkL =
bn(V) -
1 and these relations can be used toget
upper bounds for the number ofsingularities
which may occur on acomplete
intersection ofgiven
dimension andmultidegree [7].
2.
Complète
intersections with isolatedsingularities
We assume from now on that is a
complete
intersection incpn+p
with dimV = n, multidegree (d1,..., dp)
andhaving only
isolatedsingu-
larities at the
points ai
for i =1,...,
k. If H is ahyperplane
such thatW = V ~ H is
smooth,
we denoteagain by ( X, 0)
thesingularity
definedby
the cone over W.Then, exactly
as in theproof
of(1.1),
one can show that U =VBW
ishomeomorphic
to asingular _fiber
in a deformation of thesingularity
(X, 0), arbitrary
close to thespecial
fiber X.By
a result in the new book ofLooijenga [16] (7.13),
we deduce that U has thehomotopy type
of abouquet
ofn-spheres
and there is a natural exact sequencewhere
L, (resp. L)
is the Milnor lattice of thesingularity (V, a 1 )
fori =
1,..., k (resp. (X, 0))
and theinclusion iV respects
the intersection forms i.e. it is amorphism
of lattices.Let N be a tubular
neighbourhood
of W inVB{a1,...,ak}.
TheMayer-Vietoris
sequencecorresponding
to the cover V = U U N contains themorphism
As in the smooth case treated in section
1,
we canidentify
U ~ N = aNwith the total space K’ of the Milnor
S1-bundle
of W and then thecomponent
can be read off from the
corresponding
sequence(1.1).
The other
component jl corresponds
to thecomposition
In
particular, j1
has the same kernel and cokernel as themorphism
~V defined
by the composition
of theinclusion iV
with the naturalprojection
L ~ L =L/Rad
L. Theadvantage
ofreplacing
themorphism il by
T v is that the latter is amorphism
of lattices.We can state now our main result.
THEOREM 2.1:
denotes direct sum
of
groups.
PROOF: The
point (i)
is clear from the fact that U is abouquet
ofn-spheres
and from theGysin
sequence(1.1) applied
to W.The second
part (ü)
must beproved separately
for n odd and for neven. We
give
the detailsonly
for the case n even, which isslightly
moredelicate than the other case.
Assuming n
even, one hasHn+1(W)=0
and theMayer-Vietoris
sequence above
gives Hn+1(V)
=ker j
and the exact sequenceNow, i2
= 0 and henceker j
= keril
= ker ~V, which proves the secondpart
of(ü).
Moreover, coker i
=coker ~V
+Hn(W)
and ker p* =Z/dZ.
There is a natural
S1-bundle q :
K ~ V constructed as in section 1 andby
a result of Hamm[11] ]
K is( n - 1)-connected.
If a =i*b,
wherei : V ~ CPn+p and
b ~ H2(CPn+p)
is the standardgenerator,
then themorphism
um :Hm(V) ~ Hm-2(V)
which occurs in theGysin
sequenceof q
isprecisely
thecap-product
with a. Inparticular,
fromit follows that there is an element
h o
EHn(V)
such thatu(h0)
= 1. Wededuce that
Hn(V)
= ker u +Z ho.
Let
g~Hn(W)
be agenerator
andh=i0*(g)
wherei0:W~V.
Then, by comparing
theGysin
sequences for Wand V,
it follows thatu ( h )
=± d
and hence h =± dh o +
h’ for some h’ ~ ker u.On the other
hand,
it is clear thats(coker (pv)
is contained in ker uand hence
But these two inclusions must be
equalities
since the index of ker u + Z h inHn(V)
isequal
tod,
which is the index of im s inHn(V) by (2.2).
nAs trivial consequences of the Theorem we have the
following
COROLLARY 2.3:
where X denotes the Euler-Poincaré
characteristic, ho
is a smoothcomplete
intersection with the same dimension and
multidegree
as V and03BC(V, a l )
=rkLi
= the Milnor numberof
thesingularity (V, a,).
The
difficulty
inapplying
Theorem(2.1)
in concrete cases consists in the fact that one does not known how toidentify
theembedding iV
starting,
let’s say, from theequations
of V. The rest of the paper is devoted to various devices to circumvent thisdifficulty.
REMARK 2.4: The reduced Milnor lattice L is the same for all the
complete
intersections ofgiven
dimension andmultidegree (and
for allchoices of a smooth
hyperplane
sectionWl by p-constant
deformations arguments.Sometimes,
information about L can be obtainedusing (1.4)
and
(1.5).
3.
Singularities
withnondegenerate
Milnor latticesFirst we recall some basic definitions
concerning
lattices.By
a lattice(M, ( , ))
we mean afinitely generated
free abelian group Mtogether
with a bilinear form
( , ) on
M which is either(i) symmetric
and even i.e.(x, x )
= 0(mod 2)
for any x EM,
or(ii)
skew -symmetric.
We call the lattice M
nondegenerate
if Rad M = 0. In this case the naturalhomomorphism
of groupsis an
embedding
and the discriminant groupD(M)
of M isby
definitionthe finite group coker
iM.
The natural number det M
equal
to the order of the groupD( M )
isan
important
numerical invariant of the lattice M.In
particular,
the lattice M is called unimodular if det M = 1. If M isa sublattice of another lattice N such that rk M = rk N
(equivalently,
theindex
[N : M]
isfinite)
then one has theequality
Now we come back to our main
problem (and
to the notations from theprevious section).
LEMMA 3.2:
(i) If
the Milnor latticesof
all thesingular points
al E V arenondegenerate,
thenHn+1(V)
=Hn+1(CPn)
and the rankof Hn(V)
isbn(V) = bn(V0) - 03BC(V, al).
(ii) If
the Milnor latticesof
all thesingular points
a, E V are unimodu-lar,
then in additionHn(V)
is afree
group.PROOF: This result can be
easily
derived either from(2.1)
orusing
thesimple
observation that V is aQ-homology (resp.
aZ-homology)
mani-fold in case
(i) (resp.
in case(ii)).
Then the Poincaréduality
over 0(resp.
over
Z)
and the formula for~(V)
in(2.3) give
the result.EXAMPLES 3.3: If n = dim V is even, then the Milnor lattices of the
simple hypersurface singularities An, D,,, E6, E7
andE8
arenondegen-
erate. More
precisely:
detAm
= m +1,
detDm
=4,
detEm
= 9 - m andhence,
inparticular, E8
is unimodular.(Note
that we use the samesymbol
for asingularity
and its Milnorlattice!).
These and many otherexamples
of even-dimensionalhypersurface singularities
withnondegen-
erate Milnor lattices can be found in
Ebeling
paper[9].
In odd
dimensions,
we can use the stabilization ofsingularities (i.e.
addition of a square to the
defining equation
of ahypersurface singular- ity)
and the relation between thecorresponding
intersection matrices[10]
to show that the lattices
A2m, E6
andE8
areunimodular,
while thelattices
corresponding
to the othersimple hypersurface singularities
aredegenerate.
The first
examples
in this range of dimensions ofnondegenerate
lattices which are not unimodular are those
corresponding
to thesingu-
larities
Tp,q,r(p,
q, rodd)
andQk, (i even)
which have detTp,q,r
= detQk,i=4 [6].
Further
examples
can be found in the papers of Brieskorn[3]
andHamm
[12].
We assume from now on in this section that all the
singularities (V, a, )
havenondegenerate
Milnor latticesL,.
Then themorphism ~V
isan
embedding
and theonly
unknownpart
of thehomology
of V is thefinite torsion group
We show now that the lattices
L, put strong
arithmetic restrictions onthe group
T(V).
The arithmetic
problem
is thefollowing: given
anondegenerate
latticeM,
to describe the set of finite groupsN is a
supralattice
of jIt is a
simple
observation that in this definition we can takeonly supralattices
N with rkN = rkM.Using
this fact and the formula(3.1)
we
get
thefollowing simple,
but useful result.COROLLARY 3.4:
If F ~ T(M), then 1 F 2
divides detM, where 1 F |
denotes the order
of
F.Moreover, if
we are in thesymmetric
case andsign (M)~0 (mod 8), then 1 F 12 =A det
M.PROOF:
Since |F| = [N : M],
the first assertion is clear. The secondpart
follows from the fact that an evensymmetric
lattice N withsign(N) ~
0(mod 8)
cannot be unimodular[17].
0To describe more
accurately
the setT(M)
we can use the bilinear discriminant form ofM,
which we define now. The bilinear form on Mcan be extended in a natural way to a bilinear form M* X M* ~ Q and this induces a bilinear form b :
D(M) D(M)~Q/Z
which is called the bilinear discriminantform
ofM[8].
If N is a
supralattice
of M such thatF( N )
=N/M
is a finite group, then there is a natural inclusion ofF(N)
inD(M)
as anisotropic subgroup
i.e.b|F(N) F(N)=0. Moreover,
thiscorrespondence
de-fines a
bijection
betweensupralattices
N of M withF(N)
finite andisotropic subgroups
inD( M ) [8] (1.4.4).
HenceT(M)
is the same as theset of
isotropic subgroups
inD(M)
and can becomputed
if the bilinear discriminant form of M is known. And this discriminant form can becomputed
in most of the casesusing
the obvious factD( Ml ~ M2) = D(M1) ~ D(M2).
In the
skew-symmetric
case, anynondegenerate
lattice M is anorthogonal
direct sum ofelementary
latticesMd
=Z~e1, e2~
with(e,, el )
= 0 and
( e 1, e2 ) =
d. Directcomputations
show thatD(Md) = (Z/dZ)2
and
b((âl, bl), (â2, 2)) = (a,b2 - a2b1)d-1 E Q/Z.
Inparticular,
itfollows that
T(M)
= 0 if andonly
if M is unimodular.In the
symmetric
case the results are much moreinteresting.
Thebilinear discriminant form b of an even lattice M is determined
by
thecorresponding quadratic form q : D(M) ~ Q/2Z, q(x + M) = (x, x) +
2Z and the
isotropic subgroups F ~ D(M)
are thesubgroups
F forwhich 1 F=
0.In the next results we determine the
quadratic
formscorresponding
tothe
singularities Am, Dn
andEm
andgive
some direct consequences.The
proofs
consist ofsimple
but tediouscomputations involving
thecorresponding
intersection forms and are notgiven
here.PROPOSITION 3.5:
(i) D(Am)=Z/(m+1)Z and q() = -m(m + 1)-1.
(ii) T(Am) = {Z/eZ; e m
+ 1 andm(m
+1)e-2
is an eveninteger).
As an
example,
for1 m
20 theonly
nontrivial setsT( Am )
are thefollowing
PROPOSITION 3.6:
(i)
For m even,D(Dm) = (Z/2Z)2
and the generators u1, u2of D( Dm )
can be chosen such thatq(u1)
= -1 andq(u2)
=q(U1
+u2)
=0, 6/4, 1, 1/2 according
to the cases m~ 0, 2, 4, 6 (mod 8).
(ii)
For modd, D(Dm) = Z/4Z
andq() = 7/4, 5/4, 3/4, 1/4 according
to the cases m~ 1, 3, 5,
7(mod 8).
In
particular, T(Dm) = {0, Z/2 7-1 for
m ~ 0(mod 8)
andT(Dm) =
{0}
otherwise.PROPOSITION 3.7:
(i) D(E6)
=Z13Z
andq(Î)
=2/3 (ii) D(E7) = Z/2Z
andq(Î)
=1/2 (iii) D(E8) = 0
In
particular, T(Em) = {0}.
These results
give obviously
information onT(V)
when the even-di- mensionalvariety
hasprecisely
onesingular point
oftype An (resp.
Dm
orEm )
and some othersingularities
with unimodular Milnor lattices.These unimodular lattices and the
corresponding singularities
will not bementioned at all in what
follows,
since their presence do not affect the groupT(V)
as follows from(2.1).
But we can also handle with thesetechniques
the case of severalsingularities,
as is shown in the next twoexamples.
EXAMPLE 3.8: Assume that has p
singularities
oftype Al, q singulari-
ties of
type Dmi (for
various evenintegers ml)
and rsingularities
oftype E7.
ThenT(V) = (Z/2Z)S,
where 2S p +2q
+ r.Moreover, if p
+Lm,
+ 7r ~ 0
(mod 8),
theinequality
above is strict.PROOF: Let M be the lattice
pA1
~rE7
~Dm1
~ ...~ Dmq.
ThenD(M)
=(Z/2Z)t,
wheret = p + 2q + r.
Since anysubgroup F ~ D(M)
is aZ/2Z-vector
space, it follows that F =(Zl2Z)s.
Then use(3.4).
0EXAMPLE 3.9: If the
singularities
on V are describedby
asymbol
in thefollowing list,
thenT(V)
= 0:PROOF: Let us treat for
example
the caseA1A3.
First note that we cannotuse
only (3.4)
toget
the result!Using (3.5)
we find out thatD(A1A3) = Z/2Z ~ Z/4Z
andq(â, b ) = - 1 2 a2 - 3 4b2.
Asimple
verification shows thatq( â, b )
= 0 in0/27-
if andonly
if à= b
= 0. Hence theonly isotropic subgroup
inD(A1A3)
is the trivial group.4. Two
geometric examples
In this section we show that Theorem
(2.1)
can be used to determineexactly
the(nontrivial)
torsion groupT(V)
for the cubic surfacesin CP3
and for some
types
ofcomplete
intersections of twoquadrics.
The
singularities
which may occur on a cubicsurface V ~ CP3 (with
isolated
singularities)
are thefollowing [4]
All the groups of
singularities
in thislist, except
the four underlined ones, cangive
no torsionby
the results in theprevious
section. The nextproposition
shows that these four casesgive
indeed nontrivial torsion inhomology.
PROPOSITION 4.1: The
homology of
a cubicsurface V ~ CP3
with isolatedsingularities
has no torsion, except thefollowing
cases.Singularities
on VT(V)
= TorsH2 (V)
PROOF: We
give
the detailsonly
for the case4A,,
the other casesbeing
similar.
_
The reduced Milnor lattice L is in this situation
E6.
We prove that anembedding (p: 4A1 ~ E6
isunique
up to anautomorphism
u of thelattice
E6.
Assume that4A1
=Z~f1,..., f4~
withf2i = - 2, (fi, fj) = 0
fori ~ j
and thatE6 = Z~e1, ... , e6~
with theproducts given by
the usualDynkin diagram
in which el, ... , esstay
in a lineand e6
isbelow e3 [2].
Since the
Weyl
groupAut( E6 )
is transitive on the set of vectors v withv2 = - 2,
we cantake ~(f1)=e6. It
follows thatl(f2’ f3, f4~
is em-bedded in
e6 ,
theorthogonal complement
of e6.A basis for
e~6
isgiven by
the vectors e2, e 1,è 3,
e., e4 wheree3
=2e3
+ e2 + e4 + e6 and thecorresponding Dynkin diagram
is pre-cisely A5. Using
theexplicit description
of the lattice(or
rootsystem) A 5
and of its
symmetries [2],
it follows that theembedding ~:3A1 ~ A5
induced
by
(p isequivalent
up to anautomorphism
û EAut(A5)
with theobvious
embedding f2
~ e2,f3
~é3, f4
- e4.Moreover,
M extends to an u EAut( E6 )
since il is acomposition
ofreflexions and
u(e6)
= e6.This
gives
us Tors(coker T)
=Z/2Z
asrequired.
0REMARK 4.2: The
homology
of somesingular
cubic surfaces have been determinedby
Barthel andKaup (see
pp.136,
275-276 in[1]) using completely
different methods(e.g. desingularisations
and localhomology sheaves).
Now we consider two classes of
complete
intersections V of twoquadrics in CP2m+2.
PROPOSITION 4.3:
(i) If
theSegre symbol corresponding
to V is[(1, 1), ... , (1, 1), 2( m -
k) + 3] k = 0, 1, 2,..., m + 1,
then on thevariety
V there are 2ksingular
points of type AI
and asingular point of
typeA2(m-k)+2 for
k =1= m + 1.Moreover, T(V) = (Z/2Z)k-1 for k
2 and is trivialfor
k =0,
1.(ii) If
theSegre symbol corresponding
to V is[(1,1),...,(1,1), (2(m-k)+2,1)] k = 0,..., m
then on the
variety
V there are 2 ksingular points of
typeA,
and asingular point of
typeD2(m-k)+3 (with
the conventionD3
=A3).
MoreoverT(V) =
(7-12Z)k.
PROOF: The connection between the
Segre symbol
of acomplete
inter-section of 2
quadrics
V and thetype
of thesingularities
on V isexplained
in
[14].
The reduced Milnor lattice L is in this situation
D2m+3 [14], [15]
andthe
computation
of the torsion groupT(V)
goesessentially
as theproof
of
(4.1), using
thefollowing simple
observation.Assume that the lattice
D2m+3
is the latticeZ~e1,...,e2m+3~
with thecorresponding Dynkin diagram having
e2l e3, ..., e2m+3 in a line and elsitting
under e3. Then note thatei
is theorthogonal
direct sum ofZ e2
and
Z~e3,
e4, ... ,e2m+3~
wheree3
=2e3
+ el + e2 + e4.Moreover,
thislast lattice is
obviously D2m+ 1 (recall
thatD3 = A3).
This fact allows one to prove(4.3) by
induction on k. DThe interested reader can check that similar arguments
give
thehomology
of some othertypes
ofcomplete
intersections of twoquadrics.
The
simplest
case which cannot be decided with these methods is thecase of a 4-fold
(m
=2)
with 4singular points
oftype Al.
5. Déformations of
complète
intersectionsFirst we prove a technical result which
gives
a sufficient condition for the inclusion of Milnor latticesiV : L,
~ ...~ Lk ~
L to beindependent
of V in the
following
sense.Consider two
complete
intersectionsVo, V,
Ccpn+p
of the samedimension n and
multidegree (d1, ... , dp).
Moreover,
assumethat V0 and V,
haveisomorphic
isolatedsingulari-
ties at the
points
al, ... , am and(possibly)
some other isolatedsingulari-
ties.
Let us denote
by i0 (resp. il)
the restriction of theinclusion iV0 (resp.
iV1)
defined in section 2 to the direct sum of the Milnor latticescorresponding
to thesingularities
al, ... , am.It is natural to ask when
io = il (under
some identification of thecorresponding
Milnorlattices).
Consider the vector spaces of
polynomials
homogeneous
and the natural
isomorphism
We can choose the coordinates on
CPn+p
such thatH ~ Vt
are smoothfor t =
0,
1 where H is thehyperplane
xo = 0. LetFo
c F be the Zariski open setcorresponding
to thecomplete
intersections W such that H ~ W is smooth. Then asystem
ofequations for V, gives
apoint ft ~ Fo (t = 0, 1).
Let si
be the order of5Kdeterminacy
of thesingularity ( ho, al ) [18]
for i =
1,..., m
and consider theproduct
ofjet
spacesand let S c J denote the
corresponding product
of .%orbits.There is a linear
map ~ :
F ~ Jgiven by
thecomposition
of h withthe map G - J which associates to a
polynomial
map g itssi-jet
at thepoint â,
fori = 1, ... , m.
Here we assume thatai = (1, ai)
for i =1,..., m.
With these
notations,
we have thefollowing.
PROPOSITION 5.1:
If
the constructibleset ~(F)
~ S isirreducible,
thenio = il.
PROOF:
Since ~
is a linear map, it follows thatF1=~-1(~(F)~S)
isirreducible. Hence the open Zariski set
Fol
=Fo
~FI
is connected. Takea
path ft E Fol
for t E[0, 1], joining
the twopoints f0
andf1.
The
complete intersection V, corresponding
to theequation ft =
0 hasonly
isolatedsingularities (since Vt
~ H issmooth!)
and itssingularity
atthe
point
ai isisomorphic
to(Vo, a, )
for i =1,...,
m.To this
variety V, corresponds
anembedding
of latticesil’
definedsimilarly to i0
andi1.
In this way we
get
a continuousfamily
ofembeddings
and since theset of all
homomorphism
between two lattices iscountable,
we must haveit
=io
for any t E[0, 1].
0REMARK 5.2: Here are two
simple
cases when the condition in(5.1)
isfulfilled.
(i)
If themultidegree (dl, ... , d p )
isbig enough compared
to thedeterminacy
orders(sl, ..., sm),
then themap 0
issurjective.
Since S is asmooth connected submanifold in
J,
it follows that~(F) ~ S = S
isirreducible. For
example,
in thehypersurface
case( p
=1)
it isenough
tohave
(ii)
Assume that we are in thehypersurface
case and that thesingular-
ities
(V0, a, )
are all oftype Al.
Then S is a Zariski open subset of a vector space in
J,
the same istrue
for ~(F)
~ S and hence this last set is irreducible.Now we
present
twoapplications
of(5.1).
The first one is a directconsequence of Theorem
(2.1).
With the notationabove,
assume that allthe
singularities on %
andVI
different from thepoints
al’...’ am have unimodular Milnor lattices. If the condition in(5.1)
isfulfilled,
then itfollows
easily
thatho and V,
have the same Betti numberbn+1
and thesame torsion
part T(V0)
=T(V1). Moreover,
note that instead ofstarting
with the set of
points
al’...’ am we can start with two sets ofpoints
al’’’.’ am
(singular points
onV0)
andbl, ... , bm (singular points
onVi )
which are
projectively equivalent
i.e. there is anautormorphism
u ofCPn+p such that
u(al) = bl
fori = 1, ... , m.
The second
application
is more subtle and combines(5.1)
and theadjacency
ofsingularities [16]
to obtain information on thehomology
inthe presence of
singularities
withdegenerate
Milnor lattices. This method will become clear from thefollowing
twoexamples.
EXAMPLE 5.3: Assume that the odd dimensional
complete
intersection V has asingular point al
oftype A1
and that all the othersingular points
of V
(if any)
have unimodular Milnor lattices.Then:
(i) ker ~V = 0, except
the case when V is ahypersurface
ofdegree 2,
when ker ~V = Z.
(ii)
coker ggv is torsion free.PROOF: When V is a
hypersurface
ofdegree 2,
one has L = 0 and henceeverything
is clear.Assume now that V is a
hypersurface
ofdegree
d > 3.Then, by (5.1) essentially, ker ~V
and Tors coker cp v are the same as for thehyper-
surface W with affine
equation
Indeed,
for a suitable choice of the constants03B1i, 03B2
thehypersurface
W has
only
onesingular point (1 : 0: ... :0)
which is oftype Al.
On the other
hand,
if wetake al
= 0 we can arrange that Wacquires
asingle A 2 singularity.
This shows that the inclusioniW : A1 ~
L factorizesas a
composition A1 ~ A2 ~
L. SinceA2 is unimodular,
it follows that itsimage
in L is a direct summand and this ends theproof
in this case.The case of
complete
intersections can be treatedsimilarly (see [16], (7.18)).
0EXAMPLE 5.4: Assume that on the n-dimensional
(n
3odd) hyper-
surface V of
degree
dd(S),
thesingularities correspond
to one of thefollowing symbols
Then ker ~V = Tors coker ~V = 0.
PROOF: The value of
d(S)
iscomputed by
the formula in(5.2.i)
so thatwe may
apply Proposition (5.1).
Hence it is
enough
tocomputer
ker ç w and Tors coker ~W, where W is anyhypersurface
ofdegree
d with thesingularities prescribed by
oneof the
symbols
S.Note that the
(affine) equation
where C is a
generic
cubic form and theconstants 03BBi
1 are chosenconveniently
defines ahypersurface Wo
with asingle singularity
oftype
E6.
A
study
of the versal deformation ofE6 [5]
shows that thesingularity
E6
deforms to thesingularity E6
which in turns deforms to any of thesymbols S
in the list above(i.e.
for any suchsymbol S
there is a fiber inthe versal deformation of
E6
whosesingularities
areexactly
thoseprescribed by S ).
Moreover,
all these deformations can beperformed using only
mono-mials of
degree
3.It follows that the
corresponding morphisms cp w
factorizethrough
theunimodular lattice
E6
and thisgives
the result. DREMARK 5.5: We have said
nothing
about the case of curves(i.e.
whendim V =
1).
If V isirreducible,
itshomotopy type
can beeasily
obtainedfrom its
normalization h (the singularities
of Vcorresponding essentially
to the identification of some
points
inV).
We cansafely
leave the details for the reader.6. The
cohomology algebra
andapplications
Let R be a commutative
unitary ring. Assuming
theintegral homology H*(V)
of ourcomplete
intersectionknown,’ the
additive structure ofH*(V, R)
follows at onceby
the Universal Coefficient Theorem.As to the
cup-product,
theonly
difficultpoint
is toidentify
thepairing
inducedby
it on the middlecohomology
groupLet U be a
regular neighborhood
of Vin CPn+p (such
that V is adeformation retract of
U).
Let V denote a smoothcomplete
intersection of the same dimension andmultidegree as V,
which is contained in U.The natural
morphism
gives
ahomomorphism
athomology
level.Using (1.4), (1.5)
and(2.11
we see that via thishomomorphism
we canidentify Hn(V, R )
withHn(V, R)/I,
whereNote that the Poincaré
isomorphism
is
given by D(u)=~u,·~, where (, )
is the intersection form onHn(V, R ). Moreover,
D iscompatible
with thepairings
on the sourceand the
target (intersection
form andcup-product).
The above
description
ofHn(V, R )
and theequality Hn(V, R) = Hom(Hn(V, R), R)
shows that we canidentify Hn(V, R )
withThus we
get
thefollowing.
PROPOSITION 6.1: The
pairing
a isisomorphic
with thepairing induced by the intersection form ~ , ~
on theorthogonal complement of
I inHn(V, R).
In
particular, Hn(V)
=Hn(V, Z)
is torsion free and hence is a latticeas defined in section 3.
Some information on this lattice can be
easily
derived from(6.1),
using
somegeneral
facts on lattices:COROLLARY 6.2:
(i) If
the lattice(I, ~ , ~ 1 I)
isnondegenerate,
thenHn(V)
is alsonondegenerate,
and(ii) If
moreover n is even, thenWhen the Milnor lattices
Li
arenondegenerate,
then I =L1 ~
...~ Lk
and hence det I
= n
detL,
andsign
I= E sign L,.
In some
simple
cases(e.g.
those of section4,
or whenT(V)
= 0 andone has
uniqueness
results as inExample (6.4) below)
theProposition
6.1determines
effectively
the latticeHn(V).
As an
application,
we consider theproblem
ofdeciding
if certain varieties are or nothomotopy equivalent.
EXAMPLE 6.3: Consider the
complete
intersectionsV1, V2
andV3 (of
theeven dimension n and same
multidegree) having
thefollowing type
ofsingularities: 3A 1, A1 A2
andrespectively A 3 .
Then the varietiesVi
havethe same
homology
and even the same realcohomology algebras.
Butthese varieties have distinct
integral cohomology algebras,
as followsfrom the table
(use (6.2.i))
Hence V,
is nothomotopy equivalent to Jj
fori ~ j.
EXAMPLE 6.4: Let
Xl and X2
becomplete
intersections of even dimen- sion n and samemultidegree, having
both the sametype
ofsingularities, namely 3A
1(or A1A2,
orA3).
ThenXl
andX2
have the sameintegral cohomology algebra.
If moreover n =2, Xl
ishomotopy equivalent
toX2 .
PROOF : First note that
T(X1)
=T(X2)
= 0by
the results in section 4 and hence the inclusionsI(Xi)
cHn(Xi)
areprimitive.
By Proposition 6.1,
the discriminant form of the latticeH;
=Hn(Xi)
isequal
to the discriminant form(D, -q)
of the latticeI(Xi)
withreversed
sign ([17], 1.6.2).
The latticesHI
are even iff d =deg X,
is even.By
Nikulinuniqueness
results([17],
1.13.3 and1.16.10),
the latticesH,
are