Embedding of lattices and K3-covers of an Enriques surface

Theorem 1.1

Let L and M be even integral lattices of rank ( L ) and rank ( M ) , and let rank 2 ( L ) and rank 2 ( M ) denote their ranks over Z ⁄ 2 . Let ϕ be an embedding of L into M. Then, one of the following conditions holds:

1. If rank 2 ( M ) = 0 , then there exists a lattice T such that L ≅ T ( 2 ) .

2. If rank 2 ( M ) > 0 and rank 2 ( L ) = 0 , then

   rank ( L ) ≤ rank ( M ) − 1 2 rank 2 ( M ) ,

   and there exists a lattice T such that L ≅ T ( 2 ) .

3. If rank 2 ( M ) > 0 and rank 2 ( L ) > 0 , then

   rank ( L ) ≤ rank ( M ) − 1 2 rank 2 ( M ) + 1 2 rank 2 ( L ) ,

   and there exists an even lattice T such that L ≅ T and its associated Gram matrix must have the form

   G T = 2 a 11 a 12 … a 1 λ a 12 2 a 22 ⋮ ⋮ ⋱ ⋮ a 1 λ … … 2 a λ λ ,

   where a 2 k − 1,2 k is odd for each 1 ≤ k ≤ 1 2 rank 2 ( L ) , and the remaining off-diagonal entries are even.

4. If rank(L) = rank(M), L ≅ M , provided that the embedding is primitive.

Theorem 1.2

(Keum’s criterion) [2, Theorem 1] A K3 surface X with transcendental lattice T X covers an Enriques surface if and only if there exists a primitive embedding of T X into Λ − ≔ U ⊕ U ( 2 ) ⊕ E 8 ( 2 ) such that there exists no vector v ∈ T X ⊥ with v 2 = − 2 .

Theorem 1.3

Let X be a K3 surface with 10 ≤ ρ ( X ) ≤ 20 whose transcendental lattice T X is of rank λ and signature ( 2 , λ − 2 ) . Then, X covers an Enriques surface if and only if one of the following conditions holds:

1. 11 ≤ ρ ( X ) ≤ 20 , and there exists an even lattice T such that T X ≅ T ( 2 ) .

2. 11 ≤ ρ ( X ) ≤ 20 , and there exists an odd lattice T such that T X ≅ T ( 2 ) except when T X is co-idoneal.

3. 11 ≤ ρ ( X ) ≤ 20 , and there exists an even lattice T such that T X ≅ T and its associated Gram matrix must be in the following form:

   G T = 2 a 11 a 12 … a 1 λ a 12 2 a 22 ⋮ ⋮ ⋱ ⋮ a 1 λ … … 2 a λ λ ,

   such that a i j is even for each 1 ≤ i , j ≤ λ except a 11 , a 12 .

4. ρ ( X ) = 10 , and T X ≅ Λ − .

Theorem 2.1

[3] A lattice embedding is primitive if and only if the greatest common divisor of the maximal minors of the embedding matrix with respect to any choice of basis is 1.

Theorem 2.2

[8, Theorem 1] Let L be an odd indefinite Z -lattice of rank l ≥ 3 with square free discriminant d. Then,

where B is a lattice of rank 2, and m , n ≥ 0 . Moreover, if d is even, then B can be chosen to be definite or indefinite.

Theorem 2.3

[9, Theorem 6] Let L be an odd indefinite Z -lattice of rank l ≥ 3 and cube-free discriminant d ≢ 0 ( mod 4 ) . Then, L has an orthogonal splitting

where B is an indefinite odd lattice of rank 3, and m , n ≥ 0 .

Lemma 2.4

The map G L n ( Z ) → G L n ( Z ⁄ 2 Z ) is surjective.

Proof

Indeed, GL n ( Z ⁄ 2 Z ) ≅ SL n ( Z ⁄ 2 Z ) is generated by transvections (elementary), and these obviously lift to GL n ( Z ) .□

Theorem 3.1

Let F be a finite field of characteristic 2. Then, every symmetric matrix M n over F with zero diagonal is congruent to ⊕ i = 1 q H ⊕ ⊕ i = 1 n − 2 q [ 0 ] or ⊕ i = 1 n [ 0 ] , where q ∈ N + and

Furthermore, two symmetric matrices with zero diagonal over the field F are congruent if and only if they have the same rank.

Proof

Suppose that M = 0 , the result is trivial.

Suppose that M ≠ 0 , so that there exists some non-zero element a i j of M . Since GL n ( Z ⁄ 2 Z ) ≅ SL n ( Z ⁄ 2 Z ) is generated by the elementary matrices E i j ( α ) , where α ∈ Z ⁄ 2 , performing the elementary congruent transformations, we may write

Performing successive elementary congruent transformations of E i j ( α ) , i varying from 3 to n , we obtain a symmetric matrix such that

A 1 is an ( n − 2 ) -rowed symmetric matrix with zero diagonal. Therefore, we may proceed recursively and obtain a symmetric matrix of the form ⊕ i = 1 q H ⊕ ⊕ i = 1 n − 2 q [ 0 ] congruent to the given matrix M .

Because of the fact that the congruence of matrices is an equivalence relation possessing symmetry and transitivity, and by the above reasoning, two symmetric matrices of M n with zero diagonal over the field F are congruent if and only if they have the same rank. This completes the proof.□

Corollary 3.2

The number o ( n ) of orbits of the even symmetric matrix M n of size n × n over Z ⁄ 2 Z with a zero diagonal under the action of G L n ( Z ⁄ 2 Z ) by the transposition is given by

Definition 3.3

Let L be an integral even lattice of rank λ and G L be its associated Gram matrix. Let L ′ be an induced Z ⁄ 2 -module of L by restriction of ring of integers Z to Z ⁄ 2 and their associated Gram matrices G L = ( a i j ) of L and G L ′ = ( a i j ( mod 2 ) ) of L ′ . The rank of G L of L over the field of characteristic 2, will be the rank of G L ′ of L ′ , denoted by rank 2 ( L ) .

Lemma 3.4

Let L be an integral even lattice of rank λ and G L be its associated Gram matrix, with rank 2 ( L ) = 2 q . Then,

such that a 2 k − 1,2 k is odd for each 1 ≤ k ≤ q , and the remaining off-diagonal entries are even.

Proof

Let L be an even lattice. The Gram matrix G L of L is symmetric, and its diagonal entries are even.

Reduction of G L modulo 2 yields a symmetric matrix over Z ⁄ 2 Z . The diagonal entries, being even, reduce to zero, so the reduced matrix has zero diagonal entries, and the off-diagonal entries are either 0 or 1.

By Lemma 2.4, the map G L n ( Z ) → G L n ( Z ⁄ 2 Z ) is surjective. Therefore, we can use the action of G L n ( Z ⁄ 2 Z ) to bring G L mod 2 into a canonical form.

Over Z ⁄ 2 Z , by Theorem 3.1, every symmetric matrix with zero diagonal can be reduced, via congruence, to a block diagonal form consisting of 2x2 blocks of the type:

and possibly zero blocks.

The number of such blocks corresponds to the rank of the matrix modulo 2. Let rank 2 ( L ) = 2 q , which means G L mod 2 can be transformed into a block diagonal matrix with q blocks of the form H , while the remaining part of the matrix consists of zero blocks.

Once we have the canonical form of G L mod 2 , we lift this back to an integral matrix over Z . For each block H in the reduced matrix, the corresponding entries in the lifted matrix G L will be odd. That is, the off-diagonal terms a 2 k − 1,2 k are odd for each k = 1,2 , … , q . The remaining off-diagonal terms, which were zero modulo 2, will be even in G L . This completes the proof.□

Proof of Theorem 1.1

Let L have a Gram matrix G L and let M have a Gram matrix G M . The induced Z ⁄ 2 -modules are given by

where { x i } i and { u i } i are basis for L and M of ranks l and m , respectively.

By Theorem 3.1 and Corollary 3.2, we know that:

where rank 2 ( L ) = 2 p . Similarly,

where rank 2 ( M ) = 2 q . They are uniquely determined by their ranks over Z and Z ⁄ 2 by Theorem 3.1 and Corollary 3.2.

Consider the embedding ϕ : L → M and the induced embedding ϕ ′ : L ′ → M ′ . Since ϕ and ϕ ′ are embeddings, the necessary conditions for the embedding of L ′ into M ′ are that

Therefore, each embedding ϕ of L into M must satisfy

Now we can analyze each case.

Case I: rank 2 ( M ) = 0 .

Since rank 2 ( M ) = 0 , G M ′ ≅ ⊕ i = 1 m [ 0 ] . It implies that

where l ≤ m .

By Lemma 3.4, there exists a lattice T such that L ≅ T ( 2 ) . Thus, we have:

for some lattice T . This completes the proof for Case I.

Case II: rank 2 ( M ) > 0 and rank 2 ( L ) = 0 .

Since rank 2 ( L ) = 0 , we conclude that G L ′ ≅ ⊕ i = 1 l [ 0 ] . Suppose that

By Theorem 3.1, Corollary 3.2, we have

for some t ∈ N + . This contradicts the condition that rank 2 ( L ) = 0 . Therefore,

The existence of a lattice T such that L ≅ T ( 2 ) follows from Lemma 3.4. This completes the proof for Case II.

Case III: rank 2 ( M ) > 0 and rank 2 ( L ) > 0 .

Suppose that

By Theorem 3.1 and Corollary 3.2, we have

for some t ∈ N + such that t > rank 2 ( L ) . This leads to a contradiction with the uniqueness of its canonical form. Hence, we obtain the following rank condition:

By Lemma 3.4, we can express the Gram matrix G L of L in the specified form:

where a 2 k − 1,2 k is odd for each 1 ≤ k ≤ 1 2 rank 2 ( L ) , and the remaining off-diagonal entries are even.

This completes the proof for Case III.

Case IV: rank ( L ) = rank ( M ) .

If rank(L) = rank(M), then L ≅ M . The claim trivially follows by the definition of a primitive embedding.

By addressing each case, we have proved that for any embedding ϕ of L into M , one of the conditions stated in the theorem must hold. Therefore, the proof is complete.□

Theorem 4.1

Let T X be transcendental lattice of signature ( 2 , λ − 2 ) . If there exists an embedding of T X into Λ − , then rank 2 ( T X ) = 0 or 2. Particularly, the associated Gram matrix of each embedding of T X into Λ − must be one of the following types:

1. T X ≅ T ( 2 ) , where T is an even lattice,

2. T X ≅ T ( 2 ) , where T is an odd lattice,

3. rank 2 ( T X ) = 2 ,

   G T X ≅ 2 a 11 a 12 … a 1 λ a 12 2 a 22 ⋮ ⋮ ⋱ ⋮ a 1 λ … … 2 a λ λ ,

   such that a i j is even for each 1 ≤ i , j ≤ λ except a 11 , a 12 .

Proof

Let { x i } i be a basis of the transcendental lattice T X and let { u 1 , u 2 } and { v 1 , v 2 } be the standard basis of U and U ( 2 ) , respectively.

If ϕ : T X ↪ Λ − is an embedding defined generically by

where a i j ′ are integers and w i ∈ E 8 ( 2 ) for 1 ≤ i ≤ λ and 1 ≤ j ≤ 4 then, we have that,

for 1 ≤ i ≤ λ and

for 1 ≤ i < k ≤ λ .

By Theorem 1.1, equations (4.2) and (4.3) are solvable over Z ⁄ 2 if and only if rank 2 ( T X ) ≤ rank 2 ( Λ − ) . Since rank 2 ( Λ − ) = 2 , rank 2 ( T X ) is either 0 or 2.

Case I: rank 2 ( T X ) = 0 .

Since rank 2 ( T X ) = 0 , T X ′ ≅ ⊕ i = 1 λ [ 0 ] , λ ≤ 11 using Theorem 1.1. By Lemma 3.4, there are two types of associated Gram matrices of T X rising after lifting up to Z :

• T X ≅ T ( 2 ) , where T is an even lattice,

• T X ≅ T ( 2 ) , where T is an odd lattice.

Case II: rank 2 ( T X ) = 2

If rank 2 ( T X ) = 2 , by Theorem 1.1 and Lemma 3.4,

such that a 12 is odd and the remaining off-diagonal entries are even.

To determine the parities of diagonal entries of T X , we need to consider the equations (4.2) and (4.3). If both a 11 and a 22 are odd, it contradicts with equations (4.2) and (4.3). Suppose both a 11 and a 22 are even. Then, under the action of element g ∈ GL ( λ , Z ) where all diagonal entries are 1, the ( 2,1 ) -entry is 1, and all other off-diagonal entries are 0, both a 11 ′ and a 12 ′ will be odd, the parities of remaining entries of T X will be invariant. Hence, without loss of generality, assume that a 11 is odd and a 22 is even. Since a 11 is odd, it enforces that a 11 ′ and a 12 ′ are odd by equation (4.2). a 12 is odd, it also enforces that a 21 ′ and a 22 ′ have a different parity by equation (4.3). Thus, by equations (4.2) and (4.3), both a i 1 ′ and a i 2 ′ are even for 3 ≤ i ≤ λ , it implies that a i i ∈ 2 Z for 3 ≤ i ≤ λ . This completes the proof.□

Lemma 4.2

Let T X be an even lattice of signature ( 2 , λ − 2 ) , rank 2 ( T X ) = 2 . Then, for each embedding of T X into Λ − , there exists no v ∈ T X ⊥ with v 2 = − 2 .

Proof

Let { x i } i be a basis of the transcendental lattice T X , and let { u 1 , u 2 } and { v 1 , v 2 } be the standard basis of U and U ( 2 ) , respectively. Consider the embedding ϕ : T X ↪ Λ − defined generically by

where a i j ′ are integers and w i ∈ E 8 ( 2 ) for 1 ≤ i ≤ λ and 1 ≤ j ≤ 4 .

By Theorem 4.1, a 11 is odd, it enforces that a 11 ′ and a 12 ′ are odd. Similarly, a 12 is odd, it also enforces that a 21 ′ and a 22 ′ have a different parity.

To prove the orthogonal complement of the image of ϕ in Λ − contains no self-intersection − 2 vector, let f = X u 1 + x ′ u 2 + Y v 1 + y ′ v 2 + e ∈ Λ − , where e ∈ E 8 ( 2 ) with e ⋅ e = − 4 k , k ≥ 0 . From the equation,

we obtain that

Since a 21 ′ and a 22 ′ have a different parity, x ′ or X must be even. We obtain

Therefore, the orthogonal complement of the image of ϕ in Λ − contains no self-intersection − 2 vector.□

Lemma 4.3

Let T X be an even lattice of signature ( 2 , λ − 2 ) and T X ≅ T ( 2 ) , where T is an even lattice. Then, for each embedding of T X into Λ − , there exists an induced embedding such that there exists no v ∈ T X ⊥ with v 2 = − 2 .

Proof

Let { x i } i be a basis of the transcendental lattice T X , and let { u 1 , u 2 } and { v 1 , v 2 } be the standard basis of U and U ( 2 ) , respectively. Consider the embedding ϕ : T X ↪ Λ − defined generically by

where a i j ′ are integers and w i ∈ E 8 ( 2 ) for 1 ≤ i ≤ λ and 1 ≤ j ≤ 4 .

Suppose T X ≅ T ( 2 ) , where T is an even lattice, a i j is even for 1 ≤ i , j ≤ λ ≤ 11 by Theorem 4.1.

Since T X is an even lattice of signature ( 2 , λ − 2 ) , there exists a i j ′ such that a i j ′ ≠ 0 for 1 ≤ i ≤ λ and 1 ≤ j ≤ 2 .

Suppose that a i 1 ′ and a i 2 ′ have different parity for 1 ≤ i ≤ λ , then by the same reasoning as in Lemma 4.2, the orthogonal complement of the image of ϕ in Λ − contains no self-intersection − 2 vector.

Suppose that a i 1 ′ and a i 2 ′ are of the form a i j ′ = 2 k i j . m i j , where k i j ∈ Z + , m i j ∉ 2 Z for all a i j ′ ≠ 0 , 1 ≤ i ≤ λ , 1 ≤ j ≤ 2 . Let k be a minimum among the all υ 2 ( a i 1 ′ ) the exponent of the largest power of 2 of a i 1 ′ ≠ 0 . Without loss of generality, let k = k 11 . If we insert a i 1 ″ = 2 k i 1 − k . m i 1 and a i 2 ″ = 2 k i 2 + k . m i 2 in the place of a i 1 ′ ≠ 0 , a i 2 ′ ≠ 0 , respectively; equations (4.2) and (4.3) are satisfied for the new embedding ϕ ′ induced by the embedding ϕ for i , j , 1 ≤ i ≤ λ , 1 ≤ j ≤ 2 . Particularly, if ϕ is primitive, then the induced embedding ϕ ′ is also primitive by Theorem 2.1.

Since a 11 ″ and a 12 ″ have different parity, then, again by the same reasoning as in Theorem 4.2, the orthogonal complement of the image of ϕ in Λ − contains no self-intersection − 2 vector.□

Proof of Theorem 1.3

Let X be a K3 surface of 11 ≤ ρ ( X ) ≤ 20 with its transcendental lattice T X of rank λ and signature ( 2 , λ − 2 ) . Each primitive embedding of T X into Λ − is also an embedding, by Theorem 4.1, the associated Gram matrix of T X of each embedding of T X into Λ − must be one of the following types:

• T X ≅ T ( 2 ) , where T is an even lattice,

• T X ≅ T ( 2 ) , where T is an odd lattice,

• rank 2 ( T X ) = 2 ,

  G T X ≅ 2 a 11 a 12 … a 1 λ a 12 2 a 22 ⋮ ⋮ ⋱ ⋮ a 1 λ … … 2 a λ λ ,

  such that a i j is even for each 1 ≤ i , j ≤ λ except a 11 , a 12 .

If T X ≅ T ( 2 ) , where T is an even lattice, by Lemma 4.3, for each embedding of T X into Λ − , there exists an induced embedding such that there exists no v ∈ T X ⊥ with v 2 = − 2 . By Theorem 1.2, the claim follows.

If T X ≅ T ( 2 ) , where T is an odd lattice, and T X is not a co-idoneal lattice, by the definition of the co-idoneal lattice and Theorem 1.2, the claim follows.

If rank 2 ( T X ) = 2 , by Lemma 4.2, for each embedding of T X into Λ − , there exists no v ∈ T X ⊥ with v 2 = − 2 . Hence, by Theorem 1.2, the claim follows.

Finally, if ρ ( X ) = 10 , then rank ( T X ) = rank ( Λ − ) , T X ≅ Λ − , so the claim is trivial.□

Theorem 5.1

If T X is a co-idoneal lattice, T X ≅ T ( 2 ) , where T is an odd lattice.

Proof

The proof of this theorem is a direct consequence of Theorem 4.1 and Lemmas 4.2 and 4.3.□

Theorem 5.2

If T X is a co-idoneal lattice such that T X ≅ T ( 2 ) , where T has a square free discriminant d, then T X must be the following forms: m [ 2 ] ⊕ n [ − 2 ] ⊕ B ( 2 ) for some m , n ∈ N , where B is a lattice of rank 2, 0 ≤ m ≤ 2 . Moreover, if d is even, T X must be the one of the following forms: 2 [ 2 ] ⊕ n [ − 2 ] ⊕ B ( 2 ) for some n, where B is a negative definite lattice of rank 2, or n [ − 2 ] ⊕ B ( 2 ) , where B is a positive definite lattice of rank 2 .