N IELS O. N YGAARD
Higher de Rham-Witt complexes of supersingular K3 surfaces
Compositio Mathematica, tome 42, n
o2 (1980), p. 245-271
<http://www.numdam.org/item?id=CM_1980__42_2_245_0>
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HIGHER DE RHAM-WITT COMPLEXES OF SUPERSINGULAR K3 SURFACES
Niels O.
Nygaard
Printed in the Netherlands
Contents
Introduction
Let
XI k
be a proper smoothvariety
of dimension r over aperfect
field k
characteristic p
> 0. We define thehigher
de Rham-Wittcomplexes by
in
particular W03A9x(0)
is theordinary
de Rham-Wittcomplex.
Thesecomplexes
were used in[6]
and[9]
tostudy
the torsion incrystalline cohomology.
In this paper we
study
thehigher
de Rham-Wittcomplexes
in thecase where X is a
supersingular
K3 surface defined over an al-gebraically
closed field k of characteristic p > 2. We take the termsupersingular
to mean thatÊrx = Ga,
but do not assume thatrank
NS(X)
= 22 =b2.
It turns out that for
sufficiently small n,
thehypercohomology H2(X, W03A9X(n))
is related to thecupproduct pairing
onH2fl(X, Zp(l»,
in fact for n
sufficiently
smallH2(X, W03A9x(n)) injects
into the dual ofH2fl(X, Zp(1))~Zp W(k),
thus the spaces involved inOgus’
0010-437X/81/02245-27$00.20/0
of the
higher
de Rham-Wittcomplexes.
Ogus
shows thatH2fl(X, Zp(1)) splits
as anorthogonal
sum under thecupproduct pairing
where the
pairings ( , )
and( , )
areperfect.
Our main result is that there is a naturalisomorphism
where 03C30 is the Artin invariant of
X, i.e.,
is determinedby
rankTo = 2cro.
Thisisomorphism
endows ker F03C30d with a naturalpairing namely ( , )
mod p onTo0
k. Thanks to the results of[10]
we have anatural basis of ker
pUod,
we compute the matrices of thepairing ( , )
and of the frobenius
f
=idTo Q9 Frobk
onT0~k
in thisbasis,
andgive
a
proof along
the same lines asOgus
thatH2fl(X,ZP(1))
has Hasseinvariant ep = - 1.
In Section 2 we
give
acouple
ofapplications
of thistheory.
Weconsider the action of an
endomorphism g : X ~ X
on theglobal
2-forms
H°(X, 03A92X).
Such anendomorphism
induces an endomor-phism
on ker pUod which is either zero ororthogonal
with respect to thepairing ( , ),
this proves that the actionof g
onHO(X, 03A92X)
iseither 0 or
given by multiplication by
a p03C30+ l’st root ofunity. (More precise
results can be obtained in terms of the coordinates of thecorresponding point
in theperiod
space ofsupersingular
K3crystals.)
A
corollary
of this result is that anyautomorphism
of X has finiteorder on
H 2 DR (Xlk).
It is
conjectured
that 03C30~ 10(it
follows from linearalgebra
that03C30~
11),
this holds if rankNS(X)
=b2,
henceby
Artin[1]
if X iselliptic.
We consider the case where X has an involution 03B8 which preserves theglobal
2-form(e.g.
the Fermat surfaceX41
+X42
+3
+X44 = 0 p ~ 3
mod 4 or moregenerally
asupersingular
reduction of asingular
K3 surface over a numberfield[15] (p ~ 3
mod4)), using
thetheory
in Section 1 combined with the Woods Hole fixedpoint
formula we show that X has 03C30 ~ 10.
0. Review of the
slope spectral
sequence ofa
supersingular
K3 surfaceWe state without
proof
thefollowing result,
which isCorollary
3.6of
[10].
(0.1)
THEOREM: LetXlk
be asupersingular
K3surface
over analgebraically
closedfield of
characteristic p >0,
then(i) H0(X, f2’ )
= 0.(ii) H2crys(X/ W)
is torsionfree.
(iii)
Thedifferential
in theslope spectral
sequenceis
surjective.
(iv) H2(X, W03A91X)
=k03C3[[x]]
where the actionof
F isgiven by F(xi) = Xi-1 for
i > 1 andF(1)
=0, V(xi)
= 0for
all i.(v)
We have thefollowing description of
theEl
termof
theslope spectral
sequence :where d is
given by
with 03C30 determined
by 2uo
=ordp (discr. H2fl(X, Zp))
withcupproduct).
(vi)
F is ap-linear automorphism of H’(X, W03A91X).
1. The
cohomology
ofW03A9·X(n)
In this section we compute the
hypercohomology
groupsH2(X, W03A9X(n))
f or X asupersingular
K3 surface. It turns out thatthey
are torsion free for n :5 03C30, and that there are natural inclusionsMoreover
H2(W03A9X(03C30))
isisomorphic
to the dual ofH1(W03A91x)
whichgives
theisomorphism
alluded to in the introduction.
In order to
compute H2(W03A9x(n))
we need aduality
theorem in-volving
thehigher cycles
and boundaries in the de Rhamcomplex.
later studies of the
multiplicative
structure of the de Rham-Wittcomplex.
(1.1)
DEFINITION: LetXlk
be a scheme over aperfect
field k ofcharacteristic p
> 0. Defineinductively
X(p") as thepull-back
ofX(pn-1) along
the frobenius ofSpec k.
It follows that for all n, m with n ~ m we have a commutativediagram:
where the interior square is cartesian and Fn-m
(the
relative fro-benius)
is definedby
thecommutativity
of the exteriordiagram.
Assume from now on that X is smooth. Recall that the Cartier operator is a
p-’-linear surjective
mapwhere
Z1.
The kernel of C isB103A9iX =
Im d :
03A9i-1X ~ 03A9iX,
so C induces anisomorphism
We also have the inverse of the Cartier operator, which is a
p-linear
map
(1.2)
DEFINITION: We defineinductively
abelian sheavesZn03A9iX
andBn03A9iX by
and
Bnfl 9 by requiring
thatis an
isomorphism.
It is immédiate that thèse sheaves are the same as those defined
by
Illusie in[6] 0(2.2.2)
and[5]
2.3 sothey
arelocally free OX(pn)
submodules of
Fn*03A9iX.
AsOx
modules their structure isgiven by f·03C9=fpn03C9,
(1.3)
LEMMA: There are exact sequenceso f Ox(pn)
modules :If
n - 1 ~ m then there are exact sequencesof Ox(pn)
modulesPROOF:
(i)
and(ii)
followimmediately
from the definitionsplus
thef act that F is finite and flat
(X being smooth), (iii)
and(iv)
followf rom
(i)
and(ii)
in an obvious manner.(1.4)
THEOREM: LetXlk
be a proper and smoothvariety of
dimen-sion r. Assume n a m then
for
alli, j ~
r there is aperfect pairing of
k vectorspacesPROOF: Recall first the
f ollowing
lemma due to Milne[8].
LetM,
N belocally
free(jx
modules with aperfect pairing
then the
pairing
ofOx(p)
modulesis
perfect.
We use induction to construct
perfect pairings
ofOx(Pn)
modulesAssume first m = 0.
n = 0: The
pairing
given by
thewedge product
isperfect.
n = 1:
(See
also Milne[8]).
Consider thefollowing
commutativediagram
the
diagram
commutesby
thefollowing computation:
Let EF,f2’x
and T E
F*03A9
thenBy
Milne’s lemma the two lowerpairings
areperfect
hence we have induced aperfect pairing
between the kernel and the cokernel(since
dis
OX(P)-linear).
n = 2:
By (i)
and(ii)
of(1.3)
we have adiagram
with exact rowsThe
diagram
commutesby
thefollowing computation:
B y
thepreceding
case andby
Milne’s lemma the two lowerpairings
areperfect
and since dC andC-’ d
areOX(p2)
linear we have induced aperfect pairing
where
( , ) = C2().
Assume now that the
pairing
given by Cn-1()
isperfect,
then the commutativediagram
shows that the induced
pairing
is
perfect.
For m ~ 0 we use the exact sequences
(iii)
and(iv)
of(1.3)
and get a commutativediagram
ofOx(Pn)-modules:
it follows
using
induction that we have aperfect pairing
ofOx(Pn)
modulesNow use Grothendieck
duality
on the smooth propervariety
X(P") todeduce the statement of the theorem.
(1.5)
DEFINITION: LetW03A9X(n)
denote thecomplex
where the
W(k)-module
structure onWÙX
is definedthrough a--",
the n’th power inverse frobenius map, such that all the differentials in the
complex
becomesW(k)-linear.
Let now
X/k
denote asupersingular
K3 surface over analgebraic- ally
closedfield,
letordp (H2fl(X, Zp(1)))
=2ao.
(1.6)
THEOREM: TheW(k)-module H2(X, W03A9X(n))
istorsion free for n s
0"0.PROOF: Consider the exact sequence of
(pro-) complexes
we get an exact squence of
hypercohomology
groupsso it is
enough
to show thatH~(W03A9x(n)/p)
= 0 for n ~ 0"0.We show first that
W03A9x(n)/p
isquasi-isomorphic
to thecomplex
We follow the same method as Illusie
[6]
13.15. Let m = n and denoteby Wm03A9X(n)
thecomplex
we want to show that the natural
projection
is a
quasi-isomorphism
for all m~ n.Illusie shows
[6]
I 3.15 that there is an exact sequencewhere gr is the
graded object
associated to the filtration definedby
and p
is defined in[6] I.
3.4.We have
grm W03A9iX = Vm03A9ix + dVm03A9i-1x ([6]
13.2)
so we get a fourterm exact sequence of
complexes
so it is
enough
to show that the map between the two firstcomplexes
is a
quasi-isomorphism
orequivalently
that thecomplex
is
acyclic.
Now look at the
following
commutativediagram
the
commutativity
of the top square followsby
the relation FdV = d.It is clear that V" is
bijective
and since the left hand sidecomplex
is
acyclic ([6] I 3.13)
theright
hand side is as well and soWm03A9x(n)/P
-Wm-103A9x(n)/p
is aquasi-isomorphism.
This reduces the theorem to
proving
that Hl of thecomplex
vanishes.
Consider
again
an exact sequence ofcomplexes
Since
H1(Ox)
=0,
H’ of the left handcomplex
maps onto H’ of the middlecomplex
and so it isenough
to show thatone computes this
hypercohomology
to beH0(Z103A91x/Bn03A91X). By
1.4H0(Z103A91X/Bn03A91X)
is dual toH2(Zn03A91X/B1603A91X).
In[10] (3.6)
we have shown(see 0.1)
thatdVn : H2(WOX) ~ H2(W03A91X)
issurjective
forn :5 03C30;
using
the commutativediagram
it follows that
is
surjective
for n ~ 03C30.By [6]
3.11.3 F" :Wn+103A91X~03A91X
induces anisomorphism
Now look at
we conclude that
H2(B103A91X)~H2(Zn03A91X)
issurjective
henceH2(Zn03A9X1/B103A91X)
= 0 for n ~ cro and the theorem isproved.
(1.7)
DEFINITION: Define maps ofcomplexes
and
It is clear from the definition of the
W(k)-module
structure onWtx
that
V
is linear andfi
isp-linear.
(1.8)
PROPOSITION: The mapsare
injective for
n + 1 ~ oo.PROOF:
By
the definition ofWf2* x (n)
we have an exact sequence(we
useH0(W03A92X) = 0)
withker Fnd ~H2 (WOx).
This fits into acommutative
diagram
V.
Theinjectivity
offi
is shownsimilarly (1.9)
COROLLARY: For n + 1 :5 ao we havePROOF: This follows from the commutative
diagram
From now on we will suppress
V
andidentify H2(W03A9x(n))
with itsimage
inH2( Wilx(n
+1)),
so we have anascending
chainand
by (1.9)
we have for n :5 03C30Now we define a map
as follows:
Since
H1(W03A91X)=H2fl(X, Zp
we have anondegenerate pairing
onH1(W03A91x)
inducedby
thecupproduct pairing
onH2fl(X,Zp(1)).
Let
x ~ H2(W03A9X(n))
thenpx ~ H1(W03A91X)
and hence we get an element inH1(W03A91X)v by
theassignment z~(px, z). By (1.9)
we can find yo, y,, ..., yn EH2crys(X/W)
such that x=03A3ni=0 FiYi.
Let
z ~H1(W03A91x)
thensince the restriction of
F
toH1(W03A91X)
is theautomorphism
F. Nowwrite z =
F’zi,
zi EH1(W03A91X),
we getthe last
equality
holds since F isorthogonal
with respect to thepairing
onH’(Wf2’).
We haveand so
(px, z) = p 03A3ni=1 (yi, zi)H2crys, i.e., (Px, -) ~ PH1(W03A91)v;
nowdefine
(1.10)
PROPOSITION:~n defines
aninjection
PROOF: This follows from
(1.5)
and from the fact that thepairing
on
H1(W03A91x)
isnon-degenerate.
(1.11)
LEMMA: ker F"d CH2(WOX)
=k03C3 [[V]]
has rank n + oo with basis{1
V,...,Vo+n-il.
PROOF: Since Im F nd D Im d for all n ? 0 and since d is
surjective
it follows that Fnd is
surjective,
so Im Fnd = Im Fn+1 d =H1(W03A92X).
Consider the commutative
diagram
with
11, V,..., V’O-’l
as basis the lemma follows.In the
following
we shall use thequadratic
residuesymbol,
hencewe assume
p ~
2.(1.12)
THEOREM: Let( To, p ~, ~) (9 (TI, (, ))
be anorthogonal splitting of H2(X, Zp)) with (, )
and( , ) perfect
onTo
andTl
resp.(Such
asplitting
existsby Ogus [11] 3.13.3)
then there is a canonicalisomorphism
PROOF: We have
H2fl(X, Zp (1))v=
soH1(W03A91x)v=
(1/p)T0~ W(k) ~ T1 ~
and soH1(W03A91X)v/H1(W03A91x) = (1/p)T0 ~ W(k)/T0 ~ W(k) ~ To ~ k,
hence there is a commutativediagram
Since
cPuo
isinjective
the induced mapis
injective
and since both spaces have dimension2ao
it is anisomorphism.
To Q9 k
cornesequipped
with aperfect pairing namely ~ , ~
mod p, and a frobeniusf
=idTo ~ Frobk,
the aboveisomorphism
transports these structures to kerFeod.
On this space we have the natural basis{1,
V,...,V203C30-1},
we aregoing
to compute the matricesof ( , )
andf
in this basis.
(1.13)
PROPOSITION: There is ap-linear automorphism :H2(W03A9x(03C30))~H2(W03A9X(03C30)) extending
theautomorphism
F onH1(W03A91x).
PROOF: Assume first that
H2(Wilx(uo+ 1))
is torsion free. Then the argument of(1.10)
shows thatis
injective,
and it follows from the commutativediagram
below and(1.12)
that
~03C30+1
is anisomorphism,
henceV
issurjective
which is acontradiction. Let T denote the torsion submodule of
H2(W03A9X(03C30+1))
so T ~
0;
consider the commutativediagram
Since
V
is additive it is clear that T~ H2(W03A9X(03C30))
= 0 so Tinjects
into
H2(OX)
= k hence a is anisomorphism,
it follows that03B2
is alsoan
isomorphism.
Now we havepH2(W03A9·X(03C30
+1))
=pM
= pVH2(Wilx(uo» = pH2(W03A9·x(03C30)).
Let x E
H2(Wax(uo»,
thenÊx
EH2(Wilx(uo + 1))
hence there is ay E
H2(W03A9·x(03C30))
such thatpÉx
= py, this y isuniquely
determinedsince
H2(W03A9·X(03C30))
is torsionfree,
now putFx
= y. Let z be anelement of
H1(X03A91X),
thenpFz
=pF,z
=Fpz
=Fpz
=pFz
sop (Fz -
Fz)
=0, i.e.,
Fz = Fz so F extendsF,
and thep-linearity
followseasily.
below is commutative
The map
H2(W03A9·X(03C30))~ To 0
k can be described as follows: Takex E
H2(W03A9·X(03C30))
then px EH1(W03A91X)
and(1/P)(Px, -)
EH1(W03A91x)v= ((1/P)T0 ~ T1) ~ W(k),
take(1/p)(px, -)
modH1(W03A91X)
in
H1(W03A91)v/H1(W03A91) =(1/p)T0T0~k,
thenmultiply by
p to getpx E
ToQ9 k.
The frobenius
f
onTo (D
k is the reduction mod p of the restriction of F onH1(W03A91x)
toTo 0 W(k),
sof(px)
=F(px)
=F(px)
=pF(x)
which shows the
commutativity
of thediagram.
(1.14)
PROPOSITION: Consider theimage of
thecomposite
mapThis is a maximal
isotropic subspace
with basis1 Vo,
...,V203C30-1}.
PROOF: Let Je denote this
image,
then thediagram
below shows that Je isgenerated by 1 Vo,
...,V203C30-1}.
The
f ollowing computation
shows that Je isisotropic :
Let x, y EH2crys(X/W),
then we have~px,py~ = (1/p)(px,py)
mod p =(1/p)(px, py)H2crys mod p
=p(x, y)H2crys mod p
= 0. Since dim ker Fuod =2uo,
dim Je = Uo and thepairing (, )
isnon-degenerate
Je must be maximalisotropic.
(1.15)
PROPOSITION: Leti : ker F03C30-1d ~ ker F03C30d
denote the in- clusion then thefollowing diagram is
commutativePROOF: It suffices to show that
commutes. Let x E
H2(W03A9·X(03C30-1))
thensince
H2(Wilx(up»
is torsion free it follows thatx
=Éx. (Here
wehave used that FV =
VF,
which is evident from thedefinition.) (1.16)
COROLLARY:dim K ~ f (K) = 03C30-1
and03A303C30-1i=0 fi(K)
=ker F03C30d.
PROOF: This is clear from
(1.14)
and(1.15).
(1.17)
COROLLARY: The matrixof f
in thebasis {V203C30-1, V203C30-2, ..., V, 1} is given by
PROOF: It follows from
(1.15)
thatf(Vi)
=Vi-1
for i ~ 1.Since Je is maximal
isotropic
we musthave (y2uo-l, V03C30-1~~ 0,
indeed
(V203C30-n, V03C30-1~ = ~fn-1V203C30-1, fn-1V03C30+n-2~
=~V203C30-1, vuo+n-2) =
0n > 1 so if
(V2°°-’, V03C30-1)
= 0 K +kV"0-’would
be anisotropic subspace
assume that we have a basis
{x, f(x), f2(x),..., f203C30-1(x)}
with(x, f03C30(x)~
= 1. Putbi = (x, f03C30+1(x)~
i =1,
..., 03C30- 1.(1.18)
PROPOSITION: The matrixof
thepairing (, )
in the basis{x, f(x),fi(x),...,f203C30-1(x)} is given by
PROOF: The i’th row is
given by
Assume i ~ cro then
fi-1(x)
E 3if and(1.19) LEMMA:
The matrixof f
in the basis{x, f(x), f2(x),...,f203C30-1(x)}
isgiven by
PROOF: It suffices to show that
Write
we see from
(1.18)
that(f ’O(x), f "(x»
= 0 for all n :52oo -
1 exceptn =
0,
so we getsince
f
isorthogonal
we have(1.20)
THEOREM: Let d denote the discriminantof
thepairing ( , )
on
To,
then theLegendre symbol (d) satisfies
The Hasse invariant of
H2fl(X, Zp(1))
isequal
to -1.PROOF: Let ei, ..., e203C30 be an
orthogonal
basis ofTo/pTo
then d ~~e1, el)
...(e2uo, e203C30~
mod p.In
To ~ k
we haveand
fn(x)=03A3203C30i=1 Af"ei,
so the coordinate transformation between the bases{e1,..., e203C30}
and{x, f(x),..., f203C30-1(x)}
isgiven by
the matrixand so we have
Let F denote the matrix obtained
by raising
all the entries in ll to thep’th
power then039BF
is the matrix of the coordinate transformation{e1,..., e203C30}~{f(x), f 2(x), ..., f203C30(x)},
so~F=f·A and hence
det F =
(det )P
=det f ·
det ,by (1.19)
we get detf
= -1. This shows that det~ Fp,
hence (det A)-2
isnot a square in
Fp
Let
dl
=discr.( Ti, ( , ))
then we have thefollowing
formula for the Hasse invariant ep:where
(p2Uod, d1)
is the Hilbertsymbol.
Since
TI, ( , )
isperfect ep(T1)
=1,
and also(p203C30d, dl)
= 1 soNow
2.
Applications
toendomorphisms
of XIn this section we consider an
endomorphism g : X - X.
Such anendomorphism
induces anendomorphism
of ker F03C30d which is eitherzero or
orthogonal
with respect to thepairing ( , ),
this puts rather strict restrictions on the way g can act onH2(OX)
andH0(03A92x).
(2.1)
THEOREM: Letg : X ~ X be
anendomorphism
then theendomorphism g*:H2(Ox)~(~x) is
either zeroor multiplication by
a À E 03BCp03C30+1
(the p03C30+
l’stroots of unity in k).
PROOF: The
endomorphism g
induces anendomorphism
ofBrx
=Ga,
hence an elementg
ofEnd(a)=k03C3[[F]].
Letg = 03BB0 + 03BB1F + 03BB2F2 + ···
then the action on the covariant Dieudonné module ofGa
which isk03C3[[V]]
isgiven by right multiplication by
thepower series
= 03BB0 + 03BB1V + 03BB2V2 + ···.
We havek03C3[[V]] = H2(WOX)
and since the de Rham-Witt
complex
isfunctorial g
must stabilizeker d. But ker d is finite dimensional with basis
(1, V, ..., V03C30-1)
so inorder to stabilize this
subspace
we musthave 03BB1
= À2 = ... = 0 and so the actionof g
onH2(WOX)
isgiven by
Assume now that 03BB0 ~
0,
theisomorphism To 0
k = ker F"od isfunctorial,
and theendomorphism
inducedby g
onT0 ~ k
is neces-sarily orthogonal
with respect to thepairing, ( , ~.
We have~V203C30-1, V--’) 7é
0, and so we getand so
03BB1/p203C30-10 03BB1/p03C30-10 =
1 which is the same as03BBp003C30+1 =
1. Now theexact sequence
shows that the action on
H2(OX)
isgiven by multiplication by
Ào,by
Serre
duality
the action onH0(03A92X)
ismultiplication by Aj’
hence thetheorem.
(2.2)
REMARK: We get even stricter restrictions on Àoby
consider-ing
the coordinates of thepoint
in theperiod
space of supersingular
K3
crystals, A03C30-1/03BCp03C30+1 [11].
Let thepoint corresponding
toH2crys(X/W)
have coordinates(b,, ..., b,-,),
i.e. in the notation of(1.18) bi = ~x, f03C30+i(x)~.
Assume for instance thatb1 ~ 0
then we have03BBp03C300
=Ào. hence (À6uoy
=03BBp03C300
soÀguo
EFp
and so alsoko
EFp,
it followsthat
03BB20
= 1.EXAMPLE: Let Y be a
singular
K3 surface over C, i.e. the Neron- Severi group of Y has maximum rank = 20. Shioda and Inose[15]
have shown that Y is defined over a number field K. Moreover Y can
be constructed as a certain double
covering
of a Kummer surfaceX =
Km(C
xC)
where C is theelliptic
curveH2( Y, OY)/pr*H2(Y, Z).
Let e be a
prime
of k with Ne 3 mod 4 and assume that C hasgood supersingular
reduction at P. Let thesubscript
o denote reduc- tion modP;
it follows from the results of Shioda and Inose onthe C
function ofY,
thatYo and Xo
aresupersingular
K3 surfaces. SinceXo
is the Kummer surface associated to aproduct
ofsupersingular elliptic
curves it hasuo(Xo)
= 1 and henceby Ogus’
Torelli theorem isuniquely determined;
in fact it isisomorphic
to the Fermathypersur-
face
X40+X41+ X42+X43 = 1,
but this surface has anautomorphism
oforder 4 which
multiplies
theglobal
2-formby i, (i2 = -1 ).
Since Y is constructed as a doublecovering
of X it follows thatYo
has anautomorphism
with theproperties above,
inparticùlar
we must havewhich
implies
thatao(Yo)
is odd. Ifoo(Yo)
= 1 thenYo
itself is aKummer surface hence
isomorphic
toXo,
if03C30(Y0) = 3
we can con- sider thecorresponding point
in theperiod
space(bh b2)
EA2/03BCp3+1,
but
by
the remarkf ollowing
theproof
of2.1,
we seethat b
1=0,
andso there is at most a 1-dimensional
family
ofsupersingular
K3’s with03C30 =
3’
which are réduction of asingular
one.Does this
family
exist? or do allsingular
K3 surfaces reduce to Kummer surfaces?It can be noted that for a
given singular
K3 surface Y there isonly
a finite number of
supersingular primes
in K where the réduction canbe
non-Kummer,
indeed if one considers the lattice of transcendentalcycles Ty,
and if P is such thatp = N9J,( det T y,
thenNP ~ det NS( Y)
henceNS(Y) ~ Zp
is unimodular. We have aninjective specialization
map1 Assuming the conjectural Torelli theorem of Ogus [11].
which
splits
sinceNS(Y) ~Zp
isunimodular,
it follows that if rankNS( Yo)
=22,
then thep-adic
valuation of the discriminant is 2 hence03C30(Y0)
= 1 soYo
is Kummer.(2.3)
COROLLARY: Letg : X ~ X be
anautomorphism
then theinduced
automorphism
g* :H2DR(X/k)~ H2DR(X/k)
hasfinite
order.PROOF: Consider the
diagram
where
E0,22
is theE2
term in theslope spectral
séquence, it follows that we have an exact sequenceWe have
H2fl(X,Zp(1))~k = H2fl(X, Zp)(1))/p~k
and sinceH2fl(X, ZP(1))/p
is a finite group it follows that theautomorphism
induced on
H2fl(X, Zp(1))~k
has finite order.By (2.1)
g* has finite order onE0,22
C ker F03C30d so the exact sequence shows that there is a power n of g* such that(gn*- id)2
= 0 onH2DR(X/k),
it f ollows then that(gn* - id)P
= 0 orgnp*
= id.As a final
application
of ourtheory
we show how we in aspecial
case can conclude uo:5 10.
(2.4)
THEOREM: Assume X has an involution 03B8 such that8*
= id onH0(03A92X) then 03C30~10.
REMARK: It is a conséquence of Artin’s
conjecture
that uo:5 10 for ailsupersingular
K3’s.PROOF OF
(2.4) :
C onsider the exact sequenceIf 03C30 =
11,
thenT0~=H2fl(X, Zp(1)) ~k,
and the condition(J*=id
implies
that 0 actstrivially
onH2( WOx)
hence onTo Q9
k and onE0,22.
The exact sequence above then shows that
and hence that the de Rham Lef shetz number
Since 0 has order
2,
the fixedpoint
scheme is smooth. Let x EX e,
and consider the action d8 of 0 on
TX,x.
The condition0*
= id onH0(03A92X) implies
that det d8 = 1. Assume now that Xe has a 1-dimen-sional component
passing through
x ; this wouldgive
a vector v ETX,x
such that
d03B8(03BD)
= v, and since d03B8 can bediagonalized
this wouldimply
d8 = id onTX,x
which is a contradiction to the smoothness of X e. It follows that X03B8 is discrete and hence finite.The Lefshetz fixed
point
formula in de Rhamcohomology
thengives
Let us compute the local factor
det(1- d03B8).
We candiagonalize d03B8,
let(a0 0b) be
the matrix indiagonal form;
then ab = det d03B8 =1,
and a2 =b2
= 1 it follows that a = b= -1,
sodet(1- d03B8)
= 4. Since03B8*
is trivial onH0(03A92X)
it is also trivial onH2(OX),
so the Wood’s Holefixed
point
formulagives
hence #X03B8 ~ 8 mod p, so we get 8 ~ 24
mod p ~
p = 2 contradiction.REFERENCES
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Scient. l’Ec. Norm. Sup. 4e serie, 10 (1977) 87-132.
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[11] A. OGUS: Supersingular K3 crystals. Journées de geometrie algébrique de Rennes
1978. Astérisque.
[12] I.I. PJATECKII-SAPIRO and I.R. SAFAREVICH: A Torelli theorem for algebraic
surfaces of type K3. Jzv. Akad. Nauk. SSSR, 35 (1971) 530-572.
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Ann.
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(Oblatum 4-X-1979 & 13-11-1980) Department of Mathematics Princeton University
Fine Hall
Princeton, NJ. 08544 U.S.A.