An Enriques involution of a supersingular K3 surface over odd characteristic
1 Introduction
Let be an algebraically closed field. A projective minimal smooth surface over is an Enriques surface if and . We assume the characteristic of the base field is not 2 and is an Enriques surface over . It is known that and is nontrivial divisor. Since is a line bundle of order 2, has a double étale cover such that . Here is a K3 surface. Since the (local or global) moduli space of algebraic K3 surfaces is of 19 dimension and the moduli space of Enriques surfaces is of 10 dimension, not every K3 surface is the double cover of an Enriques surface. Over the field of complex numbers, the criterion for a K3 surface to be the double étale cover of an Enriques surface comes from the Torelli theorem for K3 surfaces. We put is the unique even unimodular lattice of rank 10 of signature (1,9). has a decomposition , where is the hyperbolic lattice of rank 2 and is the negative definite lattice induced by the root system . The torsion free NeronSeveri lattice of an Enriques surface is isomorphic to . The criterion is following.
Theorem ([10]) A complex K3 surface is the double cover of an Enriques surface or equivalently, has a fixed point free involution if and only if there exists a primitive embedding such that the orthogonal complement of in has no vector of self intersection .
Note that here the orthogonal complement of in is even negative definite. But over a field of positive characteristic, of course, the Torelli theorem is meaningless. However, if the characteristic of is and is a supersingular surface, we have the crystalline Torelli theorem. ([19]) In this paper, using the crystalline Torelli theorem, we prove that for a supersingular K3 surface over an algebraically closed filed of characteristic , the same criterion to be the double cover of an Enriques surface as the complex case holds.
Theorem 4.1. Assume is a superingular K3 surface defined over an algebraically closed field of characteristic . is an Enriques K3 surface if and only if there exists a primitive embedding of into such that the orthogonal complement of in does not contain a vector of self intersection .
This theorem depend only on the lattice structure of the NeronSeveri group. The NeronSeveri lattice of a supersingular K3 surface is determined only by the characteristic of the base field and the Artin invariant. (see section 2) Therefore the criterion to have an Enriques involution is applied to the family of supersingular K3 surfaces of the same Artin invariant.
Over an odd characteristic, a supersingular K3 surface of Artin invariant 1 is the Kummer sufrace associated to the product of two supersingular elliptic curves. Hence it has an Enriques involution. Using the above theorem, we prove over almost all characteristic, a supersingular K3 surface has an Enriques involution if and only if the Artin invariant is less than 6.
Corollary 4.7.
Let be an algebraically closed field of characteristic or and be a supersingular K3 surface over . is an Enriques K3 surface
if and only if the Artin invariant of is less than 6.
Acknowledgment
The author appreciates to J.Keum and M.Schütt for helpful comments and correcting mistakes.
This research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(20110011428).
2 FCrystals of K3 surfaces
In this section, we review some properties of crystalline cohomologies of K3 surfaces and the crystalline Torelli theorem of supersingular K3 surfaces.
Let be an algebraically closed field of characteristic , be the ring of Witt vectors of and be the fraction field of . Assume is a K3 surface defined over . does not have a nonvanishing 1form ([21],[15]), so the Hodge diagram of is
Hence the HodgeDerham spectral sequence of degenerates at and
Considering the crystalline universal coefficient theorem ([4], 7.26)
and is a free module of rank 22. is equipped with a canonical Frobenius linear endomorphism induced by the absolute Frobenius morphism of . We denote this morphism by
The rational crystalline cohomology with the induced Frobenius linear morphism is an Fisocrystal over . An Fisocrystal is determined by its Newton slopes up to isomorphism as follow. Let be the Frobenius morphism on . Put is a Frobenius commutative polynomial ring over such that for any . Then we have a noncanonical isomorphism
of Fisocrystal. Here is a positive integer and is a nonnegative integer. In the given isomorphism, corresponds to the multiplication by . The Newton slopes of are the rational number with multiplicity . A refined information of Newton slopes of the crystalline cohomology can be obtained using the DerhamWitt complex
We recall some properties of the DerhamWitt complex from [8]. The crystalline cohomology is realized as the hypercohomology of the DerhamWitt complex. The naive filtration of the DerhamWitt complex give raise to the slope spectral sequence
Here the sheaf cohomology is not a finitely generated module in general, but the rank of its free part is finite and the spectral sequence
degenerates at level. The sheaf has two operators and which are Frobenius linear and inverseFrobenius linear respectively satisfying . Hence with the induced operator is an Fisocrystal. Here and is topologically nilpotent, so the Newton slopes of are contained in .
Proposition 2.1 ([8], p.616).
The Newton slopes of are the part of the Newton slopes of .
By the Poincare duality, the Newton slopes of are symmetric. In other words, for , the multiplicity of the slope is equal to the multiplicity of the slope . Since is fixed, the Newton slopes of are determined by the Newton slopes of . On the other hand, is the Dieudonné module of the formal Brauer group of , . ([2]) Since and , is a 1dimensional smooth formal group and the Dieudonné module of is
= , or
= , .
In the first case, we say the height of is and in the second case, we say the height of is . If the height is , is a free module of rank and the all the Newton slopes are . In this case is the only slope of less than 1 and the multiplicity of is . Also the only Newton slope greater than 1 is with multiplicity . If , we say is ordinary. If is ordinary, the Newton polygon and the Hodge polygon of coincide. If the height is , and all the slopes of are 1.
Now consider the Kummer sequence on the flat topology of ,
From the associated long exact sequence, we obtain the short exact sequence
Since all the terms of the above sequence are of finite length, they satisfy the MittagLeffler condition and after the inverse limit of the sequence, we obtain an exact sequence
Since , the Picard number of is equal to or less than the rank of the free part of . By the way, there exists an exact sequence
Since is the slope part of Fisocrystal and on , , from the above we obtain
The kernel of the right map is space whose dimension is the multiplicity of slope 1 of . Therefore the Picard number of is equal to or less than the multiplicity of slope 1 of . Since the Picard number of is positive, the multiplicity of slope 1 of should be positive. If the height of is , the multiplicity of slope 1 is , so . The Newton polygon and the Hodge polygon of a K3 surface are following.
Note that here the Newton polygon lies above the Hodge Polygon.([4], 8.39)
When the height of is , we say is Artinsupersingular. If the Picard number of is maximal, 22, we say is supersingular. By the above argument, if is supersingular, is Artinsupersingular. It is conjectured that every Artinsupersingular K3 surface is supersingular. If the base field is a finite field this conjecture is equivalent to the Tate conjecture for Artinsupersingular K3 surfaces. The Tate conjecture holds for a K3 surface if the characteristic of the base field is . ([16],[17], [5]) It follows that if the base field is of characteristic , an Artinsupersingular K3 surface is supersingular by a deformation argument. (cf. [1]) It is also known that an Artin supersingular K3 surface equipped with an elliptic fibration is supersingular. ([3]) If the Picard number of a K3 surface is grater than 4, the NeronSeveri lattice has an isotropic vector, so it has an elliptic fibration. Hence an Artinsupersingular K3 surface with a Picard number greater than 4 is supersingular.
When is a supersingular K3 surface, is a lattice of rank 22 and the inclusion
gives an isomorphism
This isomorphism preserves the lattice structures. The signature of is and the discriminant of is with . We call this value the Artininvariant of . It is known that the isomorphism class of the lattice is determined by . ([22]) We denote this lattice by . It is also known that supersingular K3 surfaces of Artin invariant forms dimensional family and a supersingular K3 surface of Artin invariant 1 is unique up to isomorphism. For an explicit construction of , see [23].
Proposition 2.2.
When is a supersingular K3 surface over , there exists a surjection .
Proof.
For a projective smooth surface, the slope spectral sequence degenerates at level. Moreover the only possibly nontrivial map on level of the slope spectral sequence is . ([8], p.619) When is a supersingular K3 surface does not have slope 2, so . It follows that , here means the filtration given by the slope spectral sequence. Therefore
Considering the discriminant, the length of is . Since , is surjective and the length of is . Therefore
By the way, from the exact sequence
we have . Therefore there exists a canonical morphism . ∎
If the height of a K3 surface is , is a finite free module, so the slope spectral sequence of degenerates at level. ([8]) And since the Newton polygon of the Fcrystal touches the Hodge polygon at the two edge points, the Fcrystal has a decomposition
. ([9], 1.6.1)
Here Fcrystal and . The Fcrystal has constant Newton slopes 1, and the Hodge polygon of coincides with its Newton polygon. Hence is times of a unit crystal. The cup product induces a perfect paring on which is compatible with . For this paring, and are isotropic and dual to each other. with the restricted pairing is a unimodular lattice. Put . is a free module of rank . with the induced paring is a unimodular lattice. A unimodular lattice is determined up to isomorphism by the rank and the discriminant. The discriminant of = . ([19], p.363) Here is the Legendre symbol. We conclude that if the height of is , the Fcrystal with the lattice structure is determined by and it has no more information. However, if is supersingular, this information actually determines the isomorphism class of .
Theorem 2.3 ([19], Crystalline Torelli theorem).
and are supersingular K3 surface defined over . Let be an isomorphism compatible with the Frobenius morphism and the cup product. Then and are isomorphic.
For a supersingular K3 surface , the lattice has a decomposition
Here is a unimodular lattice of rank and is the times of a unimodular lattice of rank . Recall that . Composing with , we have the induced morphism . We call the period space of a supersingular K3 surface . is a dimensional isotropic subspace of . Now we state another form of crystalline Torelli theorem which we will use later in this paper.
Theorem 2.4 ([19], p.371).
and are supersingular K3 surfaces defined over . is an isometry. If takes the ample cone of into the ample cone of and onto , then is induced by an isomorphism
3 Lattices
In this section, we recall some facts on lattices.
We denote an even unimodular lattice of rank 2 with signature
by . We denote the Cartan matrix of the root system by again. is a negative definite even unimodular lattice of rank 8. An even unimodular lattice of signature exists if and only if (mod 8) and it is a direct sum of finite number of and . In particular, it is unique up to isomorphism. When is an Enriques surface over an algebraically closed field , the numerical lattice of is a unimodular even lattice of signature (1,9). ([6], p.117) It is isomorphic to . We denote this lattice by and call the Enriques lattice. If is a K3 surface defined over , the singular cohomology is a free abelian group of rank 22 equipped with an even unimodular lattice structure of signature . Hence the lattice is isomorphic to . We denote this lattice by and call the K3 lattice.
Let be an even lattice and be the dual of . There is a natural embedding and the cokernel is a finite abelian group. We call this finite group the discriminant group of and denote by . The order of is the absolute value of , the discriminant of . The discriminant group is equipped with a nondegenerated bilinear form with values in and with the associated quadratic form with values in .
Let be an inclusion of lattices of same rank. We say is an over lattice of . Then there is a chain of embeddings of lattices of same rank . We denote . Then is an isotropic subgroup of and by definition.
Proposition 3.1 ([14], p.110).
The correspondence gives a bijection between the over lattices of and the isotropic subgroups of .
We assume is a primitive embedding of even lattices and , . is the orthogonal complement of in . is also a primitive sublattice of and . is an over lattice of .
Proposition 3.2 ([14], p.111).
For , the projection and are injective. Let and be the image of by the above two projections and be the isomorphism compatible with the projections. Let and be isometries. extends to an isometry of if and only if and preserves and respectively and .
Proposition 3.3 ([25], p.915).
Let be a primitive embedding of even lattices. If and are relatively prime, then
Moreover and the projection is bijective.
Now we recall Shimada’s result on a comparison of a primitive embedding of an even lattice into and . Let us consider a condition for lattices and ,
Suppose is an integer, is a prime number which does not divide and is an positive integer less than 11. We consider the following condition
Here means the Legendre symbol.
Proposition 3.4 ([24], p.396).
Assume is an even lattice of rank and of signature such that , . is the discriminant of and is a prime number which does not divide .
(1) If , does not hold.
(2) If , is equivalent to .
(3) If , is equivalent to and .
Let be an algebraically closed field of characteristic, be an Enriques surface over and be the K3 covering of . The pull back gives a primitive embedding . ([13], p.206) In particular the Picard number of , .
Proposition 3.5.
Assume the characteristic of is . If is of finite height, the height . If the height of is infinite, is supersingular and the Artin invariant of , .
Proof.
Recall that the Picard number of a smooth algebraic surface is equal to or less than the dimension of the slope 1 part of . When is of finite height , the dimension of the slope 1 part of , so . Assume the height of is infinite. Since the Picard number of is greater than 4, has an elliptic fibraiton, so is supersingular. Let be the Artin invariant of . Since the discriminant of is , by proposition 3.4 (1), . When , is a perfect square, so fails. By proposition 3.4 (3), there is no primitive embedding of into . ∎
Remark 3.6.
The above proposition states that for an Enriques K3 surface, there are 11 possibilities of the Frobenius invariants. Let us consider the moduli space of Enriques surfaces (local or global) which is 10 dimensional and the K3 cover of the universal family. Since generally in a family of K3 surfaces, the one step degeneration of Frobenius invariants is of codimension 1, it is reasonable to expect that all the possible Frobenius invariants, in the above proposition, actually occur for an Enriques K3 surface.
4 Enriques involution of a supersingular K3 surface
Let be an algebraically closed field of characteristic .
Theorem 4.1.
Assume is a superingular K3 surface defined over . is an Enriques K3 surface if and only if there exists a primitive embedding of into such that the orthogonal complement of in does not contain a vector of self intersection .
Proof.
Assume is the K3 cover of an Enriques surface and is the covering map. Let be the associated involution. is a primitive sublattice of . Let be the orthogonal complement of in . is negative definite. and are the eigenspace of in for eigenvalues 1 and 1 respectively. Assume there is a vector such that . Then by the RiemannRoch theorem, either of or is effective. Suppose is effective and is not. But and it contradicts. Therefore does not contain a vector of self intersection 2.
Conversely we assume there exists a primitive embedding and the orthogonal complement of in , does not contain a vector of self intersection 2. Let .
Lemma 4.2.
extends to .
Proof.
We denote the extension of to by again. and
is the positive cone of . , the set of roots of . For any , we denote the reflection with respect to by . Let us denote the subgroup of the orthogonal group of generated by all the refections and by . We set . It is known that the acts simply transitively on the set of connected components of . Moreover we can prove that the connected component which contains an ample divisor is the ample cone of . ([19], [13]) In other words, represents an ample divisor if and only if and for all . Equivalently, if and for any , there exists a unique element in such that represents an ample divisor.
Lemma 4.3.
There exists such that is induced by an automorphism of
Proof.
Since the signature of is and , there exists such that . By the above argument there exists such that is an ample class. We may assume itself contains an ample class. Because fixes an ample class, it preserves the ample cone. By the crystalline Torelli theorem, it is enough to show that preserves the period space of ,
Since , is unimodular in , so has a decomposition
where is a unimodular overlattice of . We have
Therefore induces a multiplication by on and is preserved by . ∎
We assume is identity and contains an ample class. Let be the automorphism of which induces on . is an involution since so is . Now we show has no fixed point. We will follow the argument in [13]. By the above decomposition , we obtain
Hence induces on and 1 on by proposition 2.2. The Serre duality implies induces on and also on . Assume is a fixed point of . Because acts on by 1, there exists an étale neighborhood of , such that , . Therefore the fixed locus of is a divisor of and is a smooth surface. Let be the projection. Since does not contain a vector with self intersection , is a minimal surface. Let be the fixed locus of and . is the ramification locus of . It follows that and is effective, so is a rational surface. However is of 10 dimensional. It contradicts since is 2 or 3. Therefore does not have a fixed point. is an étale double cover of , so is an Enriques surface and has an Enriques involution. ∎
Remark 4.4.
Theorem 4.1 is only concerned with the lattice structure of . The NeronSeveri lattice of a supersingular K3 surface is determined by the characteristic of the base field and the Artin invariant. Assume and be algebraically closed fields of same characteristic , if one supersingular K3surface of Artin invariant over has an Enriques involution, every supersingular K3surface of Artin invariant over has an Enriques involution.
is an algebraically closed filed of characteristic . Since the unique supersingular K3 surface of Artin invariant 1 over is the Kummer surface of the product of two supersingular elliptic curves, it has an Enriques involution. (cf. [12], p.383) In the following we prove a supersingular K3 surface has an Enriques involution if and only if the Artin invariant is less than 6 over almost all characteristics.
Theorem 4.5.
Let be an algebraically closed field of characteristic and be a supersingular K3 surface of Artin invariant .
 (1)

If or and = 3 or 5, has an Enriques involution
 (2)

If , and =2 or 4, has an Enriques involution.
Proof.
For any primitive embedding , the orthogonal complement is isomorphic to . ([7], p.81) Let be a standard basis of such that the matrix corresponding to the basis is
We also fix a basis of , for which the corresponding matrix of the lattice is
For a positive integer , we put
, ,
and
We denote a basis of and for the given matrix presentation by . We also denote a basis of and for the given matrix presentation by . We give a primitive embedding of each lattice into as follow.
,
[0.1cm]
,
,
,
We denote the orthogonal complements of and , and in by , , and respectively. For each and , is negative definite of rank and has no vector of self intersection 2. We can check that in each case, the discriminant group of is
is an even lattice of signature . Hence an embedding is unique up to isometry and there exist a K3 surface over such that the transcendental lattice of