K 3 SURFACES, CYCLOTOMIC POLYNOMIALS AND ORTHOGONAL GROUPS

. Let X be a complex projective K 3 surface, and let T X be its transcendental lattice; the characteristic polynomials of the isometries of T X induced by automorphisms of X are powers of cyclotomic polynomials. Which powers of cyclotomic polynomials occur ? The aim of this note is to answer this question, as well as related ones, and give an alternative approach to some results of Kondo, Machida, Oguiso, Vorontsov, Xiao and Zhang; this leads to questions and results concerning orthogonal groups of lattices.


Introduction
If X is a projective K3 surface over the complex numbers; we denote by S X its Picard lattice and by T X its transcendental lattice; if a : X → X is an automorphism, then a induces an isometry a * of the lattice H 2 (X, Z), and the characteristic polynomial of the restriction of a * to T X is a power of a cyclotomic polynomial (see Proposition 1.1).
Let m, r be integers with m 3 and r 1, and let C = Φ r m (where Φ m is the m-th cyclotomic polynomial).
Proposition 1. Assume that deg(C) 20.Then there exists an automorphism a : X → X of a projective K3 surface X such that the characteristic polynomial of the restriction of a * to T X is equal to C.
We denote by Aut(X) the group of automorphisms of the K3 surface X, and by Aut s (X) the subgroup of Aut(X) acting trivially on T X .We have the exact sequence 1 → Aut s (X) → Aut(X) → M X → 1, where M X is a finite cyclic group (see Nikulin,[N 83], Theorem 10.1.2);we denote by m X the order of M X .
Corollary 1.Let m ≥ 4 be an even integer such that ϕ(m) 20.Then there exists a projective K3 surface X with m X = m.
It is well-known that there exist K3 surfaces X with m X = 1, 2 (see for instance [H 16], Corollary 15.2.12), but as far as I know, the following question is open : Question 1.Let m > 1 be an odd integer.Does there exist a projective K3 surface X with m X = m ?Date: May 2, 2024. 1 In [MO 98], Machida and Oguiso obtain several results on related topics; see Remark 10.7, Proposition 11.1 and Remark 12.2 for details.
Following Vorontsov [V 83] and Kondo [K 92], we consider automorphisms that act trivially on the Picard lattice.Let N X be the kernel of Aut(X) → O(S X ); this is a finite cyclic group, that can be identified with a subgroup of M X ; we denote by n X the order of N X .The following proposition is proved in §4 : Proposition 2. There exists an automorphism a : X → X of a projective K3 surface X such that a * is the identity on S X and that the characteristic polynomial of the restriction of a * to T X is equal to C if and only if the following conditions hold (i) C(−1) is a square.
The possible values of n X can be deduced from Proposition 2, extending previous results of Vorontsov [V 83], Kondo [K 92] and Oguiso-Zhang [OZ 00]; see §12.Note that n X divides m X , since N X can be identified with a subgroup of M X .This suggests the following question Question 2. What are the possible values of the pairs (m X , n X ) ?
The proofs of the above propositions use some arithmetic results (see below), as well as the surjectivity of the period map, the strong Torelli theorem, and some results of McMullen [McM 16].
The arithmetic results are valid in a greater generality than the one needed for the applications to K3 surfaces.For instance, in order to prove Proposition 2, we introduce the following property, called property (P 2) : Let R, S 0 be integers such that such that R ≡ S (mod 8), and set N = R + S; suppose that deg(C) < N.

Let c
0 be an even integer with c deg(C) such that c R and deg(C) − c S.
Definition.We say that property (P 2) holds if there exists an even, unimodular lattice L of signature (R, S) and an isometry t : L → L such that • The characteristic polynomial of t is C(X)(X − 1) N −deg(C) .
In the application to K3 surfaces, we have R = 3, S = 19 and c = 2, and Proposition 2 is a consequence of the following result : Theorem 2. Property (P 2) holds if and only if the following conditions are satisfied : (i) C(−1) is a square.
In particular, Property 2 always holds if C(−1) is a square and C(1) > 1.
If m is an odd prime number, this can be deduced from a result of Brandhorst and Cattaneo,[BC 23], Theorem 1.1.Note that Theorem 2 gives a partial answer to a question of this paper (see [BC 23], Outlook).
With the same notation, we introduce property (P 1) : Definition.We say that property (P 1) holds if there exists an even, unimodular lattice L of signature (R, S) and an isometry t : L → L such that • The characteristic polynomial of t is divisible by C.
If C(1) = C(−1) = 1 and R = 0 or S = 0, then Property (P 1) does not always hold, but the indefinite case seems to be open : Question 3. Does Property (P 1) always hold when R > 0 and S > 0 ?
A modified version of this property is used to prove Proposition 1.A more tractable question is to ask for isometries of finite order; this leads to the following definition : Definition.We say that property (P 1') holds if there exists an even, unimodular lattice L of signature (R, S) and an isometry t : L → L of finite order such that • The characteristic polynomial of t is divisible by C.
Note that Properties (P 1) and (P 1') are equivalent if R = 0 or S = 0, since then all isometries are of finite order.If R > 0 and S > 0, it is possible that Property (P 1) always holds -however, this is not the case for Property (P 1'), as shown by the following example : Example.Let C = Φ 60 , and set R = 3, S = 19 and c = 2. Then • Property (P 1) holds (see Example 8.6), • Property (P 1') does not hold (see Proposition 11.2).
Note that the fact that property (P 1') does not hold implies the wellknown fact that there does not exist any projective K3 surface X having an automorphism a ∈ Aut(X) of finite order such that a * induces multiplication by a primitive 60-th root of unity on T X ; see Machida and Oguison [MO 98], Xiao [X 95], and Zhang [Z 07]; see also Proposition 11.1.
It can be useful to replace "of finite order" by "of order m".This point of view is taken by several authors; see the paper of Brandhorst [Br 19] and the references therein; see also §5, §6 and §7.
The paper is organized as follows.The first two sections are mainly preliminary, recalling notions and results on K3 surfaces and isometries of lattices; §2 and §8 also recall some results of

K3 surfaces
We recall some notation and basic facts on K3 surfaces and their automorphisms; see [H 16] and [K 20] for details.
A K3 surface X is a simply-connected compact complex surface with trivial canonical bundle.We have the Hodge decomposition The intersection form H 2 (X, Z) × H 2 (X, Z) → Z of X is an even unimodular lattice of signature (3,19).Such a form is unique up to isomorphism (see for instance [S 77], Chap.V, th. 5).The transcendental lattice T X is by definition the primitive sublattice of H 2 (X, Z) of minimal rank such that T X ⊗ Z C contains H 2,0 (X) ⊕ H 0,2 (X).Assume that X is projective; then lattices S X and T X are orthogonal to each other, and the orthogonal sum , where ρ X is the rank of S X .
If a : X → X is an automorphism, then a * : H 2 (X, C) → H 2 (X, C) respects the Hodge decomposition and is an isometry of the intersection form; hence a * is also an isometry of the lattices S X and T X .
The following is a result of Oguiso, [O 08], Theorem 2.4.
Proposition 1.1.Let a : X → X be an automorphism of a projective K3 surface.Then the minimal polynomial of the restriction of a * to T X is a cyclotomic polynomial.

Isometries of lattices
In this section we summarize some notions and results from [GM 02], [B 21] and [B 22] in the special cases needed in this paper.
A lattice is a pair (L, q), where L is a free Z-module of finite rank, and q : L × L → Z is a symmetric bilinear form; it is unimodular if det(q) = ±1, and even if q(x, x) is an even integer for all x ∈ L. The following lemma is well-known : Lemma 2.1.Let L be an even unimodular lattice of signature (r, s).Then r ≡ s (mod 8).
Let n 1 be an integer and let F ∈ Z[x] be a monic polynomial of degree 2n such that F (x) = x 2n F (x −1 ).We say that F satisfies condition Lemma 2.2.(Gross-McMullen) Let L be an even unimodular lattice and let t : L → L be an isometry with characteristic polynomial F .Then F satisfies condition (C 1).
Assume that F is a product of cyclotomic polynomials, and that Following [B 22], we associate to F a finite group G F ; we start by defining a set Π f,g for all f, g ∈ I, as follows.We say that a monic polynomial h is (±)symmetric if h(x) = ±x deg(h) h(x −1 ).We also use the terminology symmetric for (+)-symmetric.
If f, g ∈ I are such that deg(f ) 2, deg(g) 2, then Π f,g is the set of prime numbers p such that f (mod p) and g (mod p) have a common irreducible (±)symmetric factor in When F has no linear factors, then G F is already defined in [B 21], and several examples are given in [B 21], §25.Here is another example, that will be used in the proof of Proposition 9.5 : Example 2.4.Let F = Φ 60 Φ 12 .The resultant of Φ 60 and Φ 12 is 5 4 , and the polynomials Φ 60 and Φ 12 (mod 5) have the common irreducible factors x 2 + 2x+ 4 and x 2 + 3x+ 4 in F 5 [x].These polynomials are not (±)-symmetric, hence Π Φ 60 ,Φ 12 = ∅, and G F = Z/2Z.
The following is proved in [B 22], Corollary 12.4 : Theorem 2.5.Let r, s 0 be integers such that r ≡ s (mod 8) and that r + s = deg(F ).If G F = 0, then there exists an even unimodular lattice of signature (r, s) having an isometry with characteristic polynomial F .
We need a more precise result : it is not enough to fix the signature of the lattice, we also need information about the signature map of the isometry.We recall this notion from [B 22], §3 and §4.
Definition 2.6.Let V be a finite dimensional vector space over R, and let q : V × V → R be a non-degenerate quadratic form.Let t : V → V be an isometry of q.If f ∈ R[X], set V f = Ker(f (t)) and let q f be the restriction of q to V f .Let sign t : R[X] → N × N be the map sending f ∈ R[X] to the signature of (V f , q f ), where N is the set of all nonnegative integers; it is called the signature map of the isometry t.The signature of q is called the maximum of the signature map, and the characteristic polynomial of t the polynomial associated to the signature map.
Example 2.7.Let a : X → X be an automorphism of a projective K3 surface, and suppose that a * |S X is the identity and that the characteristic polynomial of the restriction of a * to T X is C. Let ρ X be the rank of S X , and let τ be the signature map of a * .Then we have τ (x − 1) = (1, ρ X − 1) and Theorem 2.8.Let r, s 0 be integers such that r ≡ s (mod 8) and that r + s = deg(F ).Let τ be a signature map of maximum (r, s) and associated polynomial F .If G F = 0, then there exists an even unimodular lattice having an isometry with signature map τ .
Proof.This is the statement of [B 22], Corollary 12.3.

Cyclotomic polynomials and property (P 2)
The aim of this section is to prove Theorem 2 of the introduction.We start by recalling some basic properties of cyclotomic polynomials.Recall that Φ m denotes the m-th cyclotomic polynomial, and that deg(Φ m ) = ϕ(m).
(iv) If Φ m (1) = Φ m (−1) = 1, then by (i) and (ii) the integer m is not of the form p k or 2p k for some prime number p. Therefore m is divisible by 4 and by an odd prime number, or by two distinct odd prime numbers.This implies that ϕ(m) is divisible by 4, hence deg(Φ m ) ≡ 0 (mod 4).Proof.If C(1) = C(−1) = 1, then Lemma 3.1 (iv) implies that deg(C) is divisible by 4. Suppose that C(1)C(−1) = 1.Then by Lemma 3.1 we have m = p k or m = 2p k for some prime number p, and hence Φ m (1) = p or Φ m (−1) = p (see Lemma 3.1 (i) and (ii)).This implies that r is even, and since deg(Φ m ) is even, the degree of C is divisible by 4.
We now recall some notation from the introduction : m where m, r are integers with m 3 and r 1, • R, S 0 are integers such that R ≡ S (mod 8) and deg(C Theorem 2 is a consequence of lemmas 3.3 and 3.5 : Lemma 3.3.Let L be an even unimodular lattice of signature (R, S), and let t : L → L be an isometry with characteristic polynomial C(x)(x − 1) N −deg (C) .Let (c, deg(C) − c) be the signature of the sublattice Ker(C(t)).we have C(−1) 0 (cf. Lemma 3.1), hence C(−1) is  a square, and hence (i) holds.Set L 1 = Ker(C(t)), and let L 2 be the orthogonal complement of L 1 in L. If C(1) = 1, then the polynomials x − 1 and C are relatively prime over Z (i.e. the resultant of x − 1 and C is equal to 1); this implies that L = L 1 ⊕ L 2 , and hence the lattices L 1 and L 2 are both even and unimodular.Therefore mod 8), hence deg(C) ≡ 2c (mod 8), and therefore (ii) holds.   Lemma 3.4. (i) If C(1) and C(−1) are both squares, then deg(C) ≡ 0 (mod 4) Then there exists an even unimodular lattice L of signature (R, S) and an isometry t : L → L such that C) .By (i), the polynomial F satisfies condition (C 1).
Suppose that C(1) > 1, and note that by Lemma 3.1 this implies that m = p k for some prime number p.If C(1) > 1 and deg(C) < N − 2, then with the notation of §2 we have n + = 2, hence Π Φm(x),x−1 = {p}.Suppose now that C(1) > 1, that deg(C) = N − 2, and that C(1) is not a square.Then we have x−1 = {p} in this case as well.Therefore in both cases we have G F = 0.By Theorem 2.8, there exists an even, unimodular lattice L of signature (R, S) and an isometry t : L → L with characteristic polynomial F and signature map τ satisfying τ  Lemma 3.4 (i) and (iii).The polynomial C is a power of an irreducible polynomial, hence the group G C is trivial, and therefore in both cases we can apply Theorem 2.5, and conclude that there exists an even unimodular lattice L 1 of signature (c, deg(C)−c) and an isometry t 1 : L 1 → L 1 of characteristic polynomial C (note that this also follows from [BT 20], Theorem A).Let L 2 be an even unimodular lattice of signature The lattice L is even unimodular and of signature (R, S), and t is an isometry of L with the required properties.This completes the proof of the lemma.
The following is a reformulation of Theorem 2 of the introduction : Theorem 3.6.There exists an even unimodular lattice L of signature (R, S) and an isometry t : L → L such that if and only if the following conditions are satisfied : Proof.This is an immediate consequence of Lemmas 3.3 and 3.5.Notation 3.7.If L is a lattice, we denote by L ♯ its dual lattice, and set ∆(L) = L ♯ /L.Definition 3.8.Let p be a prime number.We say that a lattice L is pelementary if p∆(L) = 0. Proposition 3.9.Let L be a unimodular lattice of rank N and let t : L → L be an isometry with characteristic polynomial C(x)(x − 1) N −deg (C) .Set L C = Ker(C(t)), and let L 0 be the orthogonal complement of L C in L. Then we have (ii) The action of t endows L with a structure of Z[Γ]-module with Γ infinite cyclic; this action stabilizes L C and L 0 , hence also m with m = p k for some integer k.Therefore t acts on L C by t(x) = α.xwhere α is the image of x in Z[x]/(Φ p k ).On the other hand, t acts as the identity on L 0 .Since the Z[Γ]-modules ∆(L C ) and ∆(L 0 ) are isomorphic, this implies that p∆(L C ) = 0 and p∆(L 0 ) = 0.

Automorphisms acting trivially on Picard lattices
We now prove Proposition 2 of the introduction.Let m, r be integers with m 3 and r 1, and let C = Φ r m .Assume that deg(C) 20.Proposition 4.1.There exists an automorphism a : X → X of a projective K3 surface X such that a * is the identity on S X and that the characteristic polynomial of the restriction of a * to T X is equal to C if and only if the following conditions hold If C(1) = 1, then the lattice T X is unimodular.Moreover, the K3 surface is unique up to isomorphism if and only if C is a cyclotomic polynomial (i.e.r = 1).
Proof.Let a : X → X be an automorphism of a projective K3 surface such that a * |S X is the identity, and that the characteristic polynomial of a * |T X is equal to C. Applying Lemma 3.3 with L = H 2 (X, Z), t = a * , N = 22, R = 3, S = 19 and c = 2 shows that (i) and (ii) hold.Conversely, assume that (i) and (ii) hold, and set By Lemma 3.5 with N = 22, R = 3, S = 19 and c = 2 there exists an even unimodular lattice L of signature (3, 19) and an isometry t : L → L with characteristic polynomial F such that the signature of the sublattice , and let L 2 be the sublattice of L of fixed points by t.
subspace of signature (2, 0) and stable by t; for a generic choice of V , the intersection of L with the orthogonal of V is equal to L 2 .The restriction of t to V has determinant 1.Since t 2 is the identity, it is positive in the sense of McMullen, [McM 16], §2.By [McM 16], Theorem 6.1, there exists a projective K3 surface X and an automorphism a : X → X such that a * = t, that S X = L 2 and T X = L 1 .Therefore a * is the identity on S X and the restriction of a * to T X has characteristic polynomial C. If moreover C(1) = 1, then by Lemma 3.3 the lattice T X is unimodular.
If r = 1, then the uniqueness of the K3 surface up to isomorphism follows from a result of Brandhorst, [Br 19], Theorem 1.2.If r > 1, then varying the choice of the subspace V gives rise to an infinite family of K3 surfaces.This completes the proof of the proposition.Corollary 4.2.Let a : X → X be an automorphism of a projective K3 surface X such that a * is the identity on S X and that the characteristic polynomial of the restriction of a * to T X is equal to C.

Then one of the following holds :
(i) T X and S X are unimodular.
(ii) T X and S X are p-elementary, where p is a prime number such that Φ m (1) = p.
Proof.This is an immediate consequence of Proposition 3.9.
Note that this implies that p 19, hence we have the following : Corollary 4.3.If a projective K3 surface X has a non-trivial automorphism that induces the identity on S X , then the lattices T X and S X are either both unimodular, or both p-elementary with p 19.

Cyclotomic polynomials and isometries of finite order
We keep the notation of §3; in particular, C = Φ r m where m, r are integers with m 3 and r 1.The following result implies Theorem 1 of the introduction; it is actually a strengthening of Theorem 1, since it implies the existence of an isometry is of order m.
Theorem 5.1.Suppose that C(1)C(−1) > 1.Then there exists an even, unimodular lattice L of signature (R, S) and an isometry t : L → L of order m such that Proof.If C(1) > 1 and C(−1) is a square, then this follows from Theorem 3.6.Suppose that C(−1) > 1, and that C(1) is a square.In this case, set and that C(−1) is not a square.Note that by Lemma 3.1 this implies that m = 2p k for some odd prime number p; with the notation of §2, we have This implies that G F = 0 in this case as well.By Theorem 2.8, there exists an even, unimodular lattice L of signature (R, S) and an isometry t : L → L with characteristic polynomial F and signature map τ satisfying τ (C) = (c, deg(C) − c).
Suppose now that C(−1) is a square and deg(C) = N − 2; then Lemma 3.4 implies that condition (C 1) holds for C and that deg(C) − c ≡ c (mod 8).The polynomial C is a power of an irreducible polynomial, hence the group G C is trivial, and therefore we can apply Theorem 2.5, and conclude that there exists an even unimodular lattice L 1 of signature (c, deg(C) − c) and an isometry t 1 : L 1 → L 1 of characteristic polynomial C (note that this also follows from [BT 20], Theorem A).Let L 2 be an even unimodular lattice of signature (R − c, S − deg(C) − c), and let t 2 : L 2 → L 2 be the identity.Set L = L 1 ⊕ L 2 , and t = (t 1 , t 2 ); then t : L → L has the required properties.
Finally, suppose that C(1) and C(−1) are both non-squares.In this case, C(1) = C(−1) = 2, and C = Φ 2 k for some integer k (see Lemma 3.1).Suppose We treat these cases in the next sections; the following results will be useful : m is contained in G p .Let f : G → (Z/mZ) × be an isomorphism; then f (G p ) is the subgroup of (Z/mZ) × generated by p.This implies the equivalence of (b) and (c).
Corollary 5.5.Let m, p be distinct prime numbers.If p is not a square modulo m then the polynomial Φ m has a symmetric irreducible factor mod p.
Proof.We have This implies that the subgroup of (Z/mZ) × generated by p contains −1, and hence by Proposition 5.4 the polynomial Φ m has a symmetric irreducible factor mod p.
We start by noting that if N is sufficiently large, then Property (P 1') holds.
The polynomial F satisfies condition (C 1).Since Π Φmp,Φm = {p}, we have G F = 0. Therefore by Theorem 2.8 there exists a lattice L and an isometry t with the required properties.
Note that Proposition 5.4 implies that there exist infinitely many prime numbers p such that Π Φmp,Φm = {p}.In the following sections, we give conditions on N for the existence of an isometry of order m.
Corollary 5.2 implies that if deg(C) ≡ 0 (mod 8), then there exists an even, unimodular lattice L of signature (R, S) having an isometry t : L → L of order m such that the signature of Ker(C(t)) is (c, deg(C) − c).
Suppose that deg(C) ≡ 4 (mod 8); then r is odd, and Lemma 3.1 implies that m is of one of the following forms • m = 4p k where p is a prime number with p ≡ 3 (mod 4) and k 1 is an integer; • m = p k q s where p and q are distinct prime numbers with ≡ 3 (mod 4) and k, s 1 are integers; • m = 2p k q s where p and q are distinct prime numbers with ≡ 3 (mod 4) and k, s 1 are integers.Proposition 6.2.Suppose that m = 4p k where p is a prime number with p ≡ 3 (mod 4) and k 1 is an integer.Then there exists an even unimodular lattice L of signature (R, S) and an isometry t : L → L of order m such that Proof.Set F = CΦ 2 4 (x − 1) N −deg(C)−4 ; note that Lemma 6.1 implies that N −deg(C)−4 0, and that F satisfies condition (C 1).Since p ≡ 3 (mod 4), Φ 4 is irreducible mod p, and therefore Π Φm,Φ 4 = {p}.This implies that G F = 0, and hence by Theorem 2.8 there exists a lattice L and an isometry t with the required properties.
If p and q are distinct prime numbers ≡ 3 (mod 4), then by quadratic reciprocity either p is a square modulo q, or q is a square modulo p, and these cases are mutually exclusive.Therefore we may assume that p is a square modulo q.Proposition 6.3.Suppose that m = p k q s or 2p k q s where p and q are distinct prime numbers with p, q ≡ 3 (mod 4) and k, s 1 are integers, and assume that p is a square modulo q.Suppose that N deg(C) + (p − 1)p k−1 + 2.
Then there exists an even unimodular lattice L of signature (R, S) and an isometry t : L → L of order m such that The polynomial F satisfies condition (C 1), and Lemma 3.1 implies that G F = 0. Theorem 2.8 implies that there exists a lattice L and an isometry t with the required properties.Proof.Indeed, we are assuming that c ≡ 2 (mod 4), hence the hypothesis deg(C) ≡ 4 (mod 8) implies that deg(C) ≡ 2c (mod 8).Therefore by Corollary 5.2 there exists an even, unimodular lattice L and an isometry t : L → L of order m such that the signature of Ker(C(t)) is (c, deg(C) − c).By Lemma 5.3, such a lattice is indefinite, hence L is isomorphic to Λ R,S .This concludes the proof of the lemma.
Suppose that deg(C) ≡ 0 (mod 8).Recall that N = R + S. Using the results of [B 22] (in particular, Theorem 2.8) and Proposition 5.4 it is possible to determine the values of N for which Λ R,S has an isometry t of order m such that the signature of Ker(C(t)) is (c, deg(C) − c); since this would be quite long, we only give some partial results that will be useful for the for the proof of Proposition 1.
Proposition 7.2.Let m = 2 n p with n 2 and p a prime number such that p ≡ 3, 5 (mod 8), or m = pq with p and q distinct prime numbers such that p is not a square modulo q.Suppose that N deg Then Λ R,S has an isometry t with characteristic polynomial F such that the signature of Ker(C(t)) is (c, deg(C)−c) and that the signature of Ker((t−1) M ) is (1, M − 1).
Proposition 7.3.Let m = 2pq with p and q distinct prime numbers such that p is not a square modulo q.Suppose that N deg Then Λ R,S has an isometry t with characteristic polynomial F such that the signature of Ker(C(t)) is (c, deg(C)−c) and that the signature of Ker((t−1) M ) is (1, M − 1).

Salem polynomials and isometries of lattices
A Salem polynomial is a monic irreducible polynomial S ∈ Z[X] such that S(X) = X deg(S) S(X −1 ) and that S has exactly two roots outside the unit circle, both positive real numbers.
Example 8.1.Let n be an integer 0, and set We recall some notions and results from [B 22], §7 and §12.Notation 8.2.Let S be a Salem polynomial such that S(1) = −1 and S(−1) = 1, and let C be a cyclotomic polynomial.Let Π S,C be the set of prime numbers p such that S (mod p) and C (mod p) have a common irreducible symmetric factor in Proposition 9.1.There exists an automorphism a : X → X of a projective K3 surface X such that the characteristic polynomial of the restriction of a * to T X is equal to C.
The proof of the proposition is divided into several parts, according to the value of m.Note first that if m = p k for some prime number p = 2, then Proposition 9.1 follows from Proposition 4.1 : Proposition 9.2.Suppose that m = p k where p is a prime number, p = 2, and k 1 is an integer.Then there exists an automorphism a : X → X of a projective K3 surface X such that the characteristic polynomial of the restriction of a * to T X is equal to C.
Proof.We have C(1) = p r and C(−1) = 1, hence Proposition 4.1 implies the existence of an automorphism a : X → X of a projective K3 surface X with the required properties.
Proposition 9.3.Suppose that m = 2p k where p is a prime number, and k 1 is an integer.Suppose that r is even, and that deg(C) ≡ 4 (mod 8).Then there exists an automorphism a : X → X of a projective K3 surface X such that the characteristic polynomial of the restriction of a * to T X is equal to C.
Proof.We have C(1) = 1 and C(−1) = p r ; since r is even, C(−1) is a square, hence the conditions of Proposition 4.1 are satisfied; therefore this implies the existence of an automorphism a : X → X of a projective K3 surface X with the required properties.
In the remaining cases, the proofs use modified versions of the results of the previous sections.The following lemma is based on results of McMullen in [McM 16], and will be used in the proof of Proposition 9.1.Recall from [McM 16], §2 that an isometry of a hyperbolic lattice is said to be positive if it stabilizes a chamber.
Lemma 9.4.Let (L, q) be an even unimodular lattice of signature (3,19), and let t : L → L be an isometry of L. Let L 1 and L 2 be mutually orthogonal The lattice L has an isometry t ′ : L → L such that the restriction of t ′ to L 2 is positive, and that t ′ and t coincide on L 1 .
(ii) Let V ⊂ L 1 ⊗ Z R be a 2-dimensional subspace of signature (2, 0) and stable by t such that the intersection of L with the orthogonal of V is equal to L 2 and that the restriction of t to V is in SO(V ).Then there exists a projective K3 surface X and an automorphism a : X → X such that T X ≃ L 1 , S X ≃ L 2 , and a * |T X = t.
Proof.(i) Set t 1 = t|L 1 and t 2 = t|L 2 .For i = 1, 2, set L i = L ♯ i /L i , and let q i and t i be the induced symmetric bilinear forms and isometries; since L is unimodular, we have (L 1 , q 1 , t 1 ) ≃ (L 2 , −q 2 , t 2 ).If L 2 has no roots, then t 2 is a positive isometry in the sense of McMullen [McM 16], §2; otherwise, let ρ be an element of the Weyl group of L 2 such that ρ • t 2 is positive.Set t ′ 2 = t 2 in the first case, and t ′ 2 = ρ•t 2 in the second one.The elements of the Weyl group of L 2 inducent the identity on L 2 , hence we have This implies that there exists an isometry t ′ : L → L such that t ′ |L 1 = t 1 and t ′ |L 2 = t ′ 2 ; the isometry t ′ 2 is positive, and this implies (i).(ii) Applying [McM 16], Theorem 6.1 to the isometry t ′ : L → L constructed in part (i), we conclude that there exists a projective K3 surface X with T X ≃ L 1 , S X ≃ L 2 , and an automorphism a : X → X such that a * = t ′ .By construction, we have t ′ |L 1 = t|L 1 , hence the restriction of a * to T X is equal to t|L 1 .This concludes the proof of the lemma.
Proposition 9.5.Suppose that C(1) = C(−1) = 1.Then there exists an automorphism a : X → X of a projective K3 surface X such that the characteristic polynomial of the restriction of a * to T X is equal to C.
Proof.If deg(C) ≡ 4 (mod 8), then this follows from Proposition 4.1.Suppose that deg(C) ≡ 0 (mod 8), and that m = 30, 60.Then we have m = 15, 20, 24 and r = 1 or 2, or m = 40, 48 and r = 1.By Proposition 7.2 the lattice Λ 3,19 has an isometry t such that the characteristic polynomial of t is divisible by C and by (x − 1) 4 , and that the signature of Ker(C(t)) is (2, deg(C) −2).The same property holds for m = 30 and r = 1 by Proposition 8.7.If m = 60, then by Example 8.6, the lattice Λ 3,19 has an isometry t such that the characteristic polynomial of t is CS 2 ; if m = 30 and r = 2, then this holds for CS 1 by Example 8.7.
Set L 1 = Ker(C(t)), and let L 2 be the orthogonal complement of L 1 in L. The hypotheses of Lemma 9.4 are fulfilled; hence by Lemma 9.4 there exists a projective K3 surface X with T X = L 1 , and an automorphism a : X → X such that the characteristic polynomial of the restriction of a * to T X is equal to C. Proposition 9.6.Let p be a prime number, let r, k 0 be integers, and let 16, or r is even and deg(C) 20.Then there exists an automorphism a : X → X of a projective K3 surface X such that the characteristic polynomial of the restriction of a * to T X is equal to C.

Proof. The hypothesis implies that
5 , set C = S 0 (cf.Example 8.1).Then CC ′ satisfies condition (C 1) and G CC ′ = 0, hence by Theorem 2.8 the lattice Λ 3,19 has an isometry with characteristic polynomial CC ′ such that the signature of Ker(C(t)) is (2, deg(C) − 2).If r is even and p = 2 or deg(C) ≡ 4 (mod 8), then the existence of such an isometry (with C ′ a power of x − 1) follows from Proposition 4.1.We conclude as in the proof of Proposition 9.5.

Proof of Proposition 1 -continued
The aim of this section is to prove Proposition 1 (that is, Proposition 9.1) in the remaining cases; the results are stated in a more general setting than needed.
Notation 10.1.Let q be a prime number.
Lemma 10.2.Let V be a quadratic form over Q 2 .Then V contains an even, unimodular Z 2 -lattice if and only if Proof.Let H = 1, −1 and N = 2, 6 .By [BG 05], Proposition 5.2, we see that V contains an even, unimodular Z 2 -lattice if and only if V is an orthogonal sum of copies of H and N; the lemma follows by computing the invariants of these orthogonal sums.
Notation 10.3.Let K be a field, and let E be an étale K-algebra with a K-linear involution σ : We denote by b λ the quadratic form b λ : E × E → K given by b λ (x, q) = Tr E/K (λxσy).
Proof.Set 2n = 22 − deg(C), and let U be the Q-vector space with basis e 1 , . . ., e n , f 1 , . . ., f n ; let Q : U ×U → Q be the orthogonal sum of the quadratic form equal to 2, −2p r on the subspace generated by e 1 and f 1 , and of the diagonal form −2, . . ., −2 on the subspace generated by e i , f j for i, j = 1.Let T : U → U be the isometry of Q given by T (f 1 ) = −f 1 and by T (e i = e i for all i, T Suppose first that r is even.Since C(1) = 1, C(−1) = p r and deg(C) ≡ 0 (mod 4), the polynomial satisfies condition (C 1).Moreover, G F = 0, since C is a power of an irreducible polynomial.We are assuming that deg(C) ≡ 4 (mod 8), hence deg(C) − 2 ≡ 2 (mod 8).Therefore Theorem 2.8 implies that Λ 2,deg(C)−2 has an isometry T ′ with characteristic polynomial C. Note that (U, Q) contains a lattice isomorphic to Λ 1,2n−1 stable by T , hence Λ 3,19 has an isometry with the required properties.
Assume now that r is odd.Set F = Q[x]/(Φ 2p k ), and let α ∈ F be the image of x.Let σ F : F → F be the involution induced by α → α −1 , and let F 0 be the fixed field of this involution.Let E 0 be an extension of F 0 of degree r that is linearly independent of F over F 0 , and set E = E 0 ⊗ F 0 F .Then E is a field, and the characteristic polynomial of α ∈ E is equal to C = Φ r 2p k .Let σ be the extension of σ F to E.
If q is a prime number, set 0 be such that ∂((E q , b λq , α) = 0 and that (E q , b λq ) contains an even, unimodular Z q -lattice stabilized by α; this is possible by [BT 20], Proposition 7.1, Proposition 9.1 and the fact that det(E q , b λq ) = p and deg(C) ≡ p − 1 (mod 4).Let λ ∞ ∈ R × be such that the signature of For all prime numbers q, set U q = (U, Q) ⊗ Q Q q and W q = (E q , b λq ); we have d(U q ) = p and d(W q ) = −p.Note that this implies that w(W q ⊕ U q ) = w(W q ) + W (U q ).If q = 2, p, we have w(W q ) = w(U q ) = 0.
Together with Lemma 10.2, this allows us to compute w(W q ) for all q, as follows.
Since w(W q ) = 0 if q = 2, p, the sum of the invariants w(W q ) (for q a prime number) and w(W ∞ ) is 0.
Again, since w(W q ) = 0 if q = 2, p, the sum of the invariants w(W q ) (for q a prime number) and w(W ∞ ) is 0.
We have w 2 (E q , b λq ) = w(b 1 ) + cor Eq/Qq (λ q , d) for all prime numbers q (see [B 22], Proposition 5.4).Let V be the set of all places of Q; since b 1 is a global form, the above argument shows that v∈V cor Ev/Qv (λ p , d) = 0.
The quadratic form V has determinant −1, signature (3, 19), w(V 2 ) = 1 and all the other Hasse-Witt invariants of V are trivial.This implies that V is isomorphic to Λ 3,19 ⊗ Z Q.The characteristic polynomial of t is CC ′ .The quadratic form V contains an even unimodular lattice stabilized by t everywhere locally; this is clear by construction at all prime numbers q = 2, and for q = 2 it follows from Takada, [T 22], Theorem 4.2.The intersection of these lattices is an even unimodular lattice stabilized by the isometry t; this lattice is isomorphic to Λ 3,19 .By construction, t has the required properties.This concludes the proof of the Proposition.4.2.Let L be the intersection of these lattices; L is stabilized by t, and we have L ≃ Λ 3,19 .This concludes the proof of the proposition.
Proof of Proposition 9.1.If m = p k where p is a prime number with p = 2, then the proposition follows from Proposition 9.2; if C(1) = C(−1), then from Proposition 9.5.Suppose that C = Φ r 2p k .If r is even of if deg(C) 16, it is a consequence of Proposition 9.6.Suppose that r is odd and that deg(C) = 18 or 20; apply Proposition 10.4 f p = 2, and Proposition 10.5 if p = 2. Proposition 9.1 implies the following result : Corollary 10.6.Let m be an integer such that m 1 and that ϕ(m) 20.Then there exists an automorphism a : X → X of a projective K3 surface X inducing multiplication by a primitive m-th root of unity on T X .
Proof.For m = 1, we can take the identity, and there are many examples of automorphisms of projective K3 surfaces X inducing −id on T X (see for instance [H 16], Corollary 15.2.12).Suppose that m 3, and let C = Φ m ; Proposition 9.1 implies that there exists an automorphism a : X → X of a projective K3 surface such that the characteristic polynomial of a * |T X is equal to C; hence a * |T X acts by multiplication by a primitive m-th root of unity.
Remark 10.7.Corollary 10.6 follows from results of Machida-Oguiso [MO 98], Xiao [X 95] and Zhang [Z 07] when m = 60; more precisely, they prove the existence of an automorphism a : X → X of finite order inducing multiplication by a primitive m-th root of unity on T X .They also show that this is not the case for m = 60; in the next section we give another proof of this result.

The primitive 60-th roots of unity
The following result was proved by Machida-Oguiso [MO 98], Xiao [X 95] and Zhang [Z 07] : Proposition 11.1.There does not exist any automorphism of finite order of a projective K3 surface inducing multiplication by a primitive 60-th root of unity on its transcendental lattice.
The aim of this section is to give another proof of Proposition 11.1.Set C = Φ 60 .
Proposition 11.2.Let L be an even unimodular lattice of signature (3,19).The lattice L does not have any isometry t : L → L having the following properties : (i) The characteristic polynomial of t is CC ′ , where C ′ is a product of cyclotomic polynomials.
(ii) The signature of the sublattice The main ingredient of the proof of Proposition 11.2 is the following lemma : Lemma 11.3.Set C ′ = Φ 12 , and let M be an even unimodular lattice of signature (2, 18).The lattice M does not have any isometry t : M → M having the following properties : We give two proofs of this lemma; the first one is based on some results of [B 22], the second one is a direct proof.Let τ be a signature map with characteristic polynomial F and maximum (2, 18) such that τ (C) = (2, 14) and τ (C ′ ) = (0, 4), and let ǫ ∞ τ : C(I) → Z/2Z be the associated homomorphism (see [B 22], §9 and §12).We obtain a homomorphism ǫ τ : ], Theorem 12.1, there exists an even unimodular lattice M having an isometry t : M → M with properties (i) and (ii) if and only if ǫ τ = 0.

With the notation of [
Similarly, let τ ′ be a signature map with characteristic polynomial F and maximum (2, 18) such that τ (C) = (0, 16) and τ (C ′ ) = (2, 2), and let ǫ ∞ τ : C(I) → Z/2Z be the associated homomorphism.We have a Both C and C ′ satisfy condition (C 1), and since they are irreducible, we have G C = G C ′ = 0. Therefore by Theorem 2.5 there exists an even unimodular lattice N of signature (0, 16) having an isometry with characteristic polynomial C, and an even unimodular lattice N ′ of signature (2, 2) having an isometry with characteristic polynomial C ′ .The lattice N ⊕ N ′ is even unimodular of signature (2, 18), and has an isometry of characteristic polynomial F and of signature map τ ′ .Applying [B 22], Theorem 12.1, this implies that On the other hand, we have ǫ τ = ǫ finite F + ǫ ∞ τ , and this implies that ǫ τ = 0; therefore there does not exist any even unimodular lattice M having an isometry t : M → M with properties (i) and (ii).
The second proof of Lemma 11.3 uses the notion of Hasse-Witt invariant of a quadratic form.
Notation 11.4.Let K be a field of characteristic = 2, let V be a finite dimensional K-vector space, and let q : V × V → K be a non-degenerate quadratic form.The Hasse-Witt invariant of q is denoted by w 2 (q); it is an element of Br 2 (K).
If K is a p-adic field or R, then Br 2 (K) is a group of order two, that we identify with {0, 1}; see [S 77], Chapitre IV for the properties of Hasse-Witt invariants of quadratic forms that are needed here.
Second proof of Lemma 11.3.Set F = CC ′ .Suppose that the even unimodular lattice (M, q) of signature (2, 18) has an isometry t : M → M with characteristic polynomial F , and let us show that (ii) does not hold.
Set M 1 = Ker(C(t)) and let q 1 be the restriction of q to M 1 × M 1 ; similarly, set M 2 = Ker(C ′ (t)) and let q 2 be the restriction of q to M 2 × M 2 .Set Since C and C ′ are distinct irreducible polynomials, the quadratic space (V, q) is the orthogonal sum of (V 1 , q 1 ) and (V 2 , q 2 ).
The resultant of C and C ′ is 5 4 .This implies that if p is a prime number with p = 5, then (M, q) ⊗ Z Z p is the orthogonal sum of (M 1 , q 1 ) ⊗ Z Z p and (M 2 , q 2 ) ⊗ Z Z p ; therefore the Z p -lattices (M 1 , q 1 ) ⊗ Z Z p and (M 2 , q 2 ) ⊗ Z Z p are unimodular.
Let K be the cyclotomic field of the 12-th roots of unity, and let K 0 be its maximal real subfield The prime 5 is inert in the extension K 0 /Q, and splits in K/K 0 .This implies that there exists a degree 2 polynomial f ∈ Z 5 [x] such that C ′ = f f * in Z 5 [x], where f * (x) = x 2 f (x −1 ), and such that f = f * .Let us denote by t 2 the restriction of t to V 2 , and note that Ker(f (t 2 )) is an isotropic subspace of dimension 2 of V 2 ⊗ Q Q 5 ; this implies that (V 2 , q 2 ) ⊗ Q Q 5 ≃ U 2 ⊗ Q Q 5 , and therefore w 2 (q 2 ) = 0 in Br 2 (Q 5 ).
Proof of Proposition 11.2.Let C ′ be a product of cyclotomic polynomials and let t : L → L be an isometry with characteristic polynomial CC ′ , and set F = CC ′ .Suppose first that C ′ is not divisible by Φ 12 .Then C and C ′ are relatively prime over Z. Indeed, deg(C ′ ) = 6, and if Φ is a cyclotomic polynomial of degree 6 such that Φ is not relatively prime to C over Z, then Φ = Φ 12 ; this follows from the values of the resultants of cyclotomic polynomials, see for instance [Ap 70].Since C and C ′ are relatively prime over Z, the lattice L is the orthogonal sum of the even unimodular lattices L C and Ker(C ′ (t)).If (ii) holds, then the signature of L C is (2, 14); this contradicts the fact that L C is an even unimodular lattice.
The possible values of n X can be deduced from Proposition 4.1.As we will see, it is enough to consider K3 surfaces X such that rank(T X ) = ϕ(n X ) or rank(T X ) = 2ϕ(n X ).
Proposition 12.3.Let m be an integer with m 1 and ϕ(m) 20.The following are equivalent (i) There exists a projective K3 surface X such that n X = m.
(ii) There exists a projective K3 surface X such that n X = m and rank(T X ) = ϕ(m) or rank(T X ) = 2ϕ(m).
This will be proved at the end of the section.
We start with the case where C is a cyclotomic polynomial, i.e. when the rank of T X is ϕ(n X ).
Corollary 12.4.Let m be an integer such that m 3 and that ϕ(m) 20; set C = Φ m .There exists a projective K3 surface X such that n X = m and the rank of T X is ϕ(n X ) if and only if the following conditions hold : The K3 surface is unique up to isomorphism.Moreover, the lattice T X is unimodular if and only if C(1) = 1, and p-elementary with p = C(1) if C(1) > 1.
Proof.Suppose that conditions (i) and (ii) hold.Then by Proposition 4.1, there exists an automorphism a : X → X of a projective K3 surface such that the restriction of a * is the identity and that the characteristic polynomial of a * |T X is equal to C; this implies that m divides n X .Since C = Φ m , the rank of T X is equal to ϕ(m); hence ϕ(m) = ϕ(n X ).Suppose that C(1) > 1; then m = p r , where p is an odd prime number (cf.Lemma 3.1).Then either n X = m or n X = 2m; but Φ 2p r does not satisfy (i), hence n X = 2p r , and this implies that n X = m.Assume now that C(1) = 1.By (ii), we have m ≡ 0 (mod 2); since m divides n X and ϕ(m) = ϕ(n X ), this implies that n X = m.
Conversely, suppose that there exists a projective K3 surface X such that n X = m and rank(T X ) = ϕ(m), and let a be a generator of the cyclic group N X .Then the restriction of a * to S X is the identity and the characteristic polynomial of a * |T X is equal to C. Therefore Proposition 4.1 (i) implies that C(−1) is a square; since C is a cyclotomic polynomial, we have C(−1) = 1, hence (i) holds.Proposition 4.1 (ii) implies that if C(1) = 1, then deg(C) ≡ 4 (mod 8).Moreover, the hypothesis C(1) = 1 implies that T X is unimodular; this implies that n X ≡ 0 (mod 2), hence m ≡ 0 (mod 2), and therefore (ii) holds.
The uniqueness of the K3-surface follows from Proposition 4.1.If C(1) = 1 then T X is unimodular (cf.Proposition 4.1).Conversely, suppose that T X is unimodular; then C satisfies condition (C 1).Since C is a cyclotomic polynomial, this implies that C(1) = 1.If C(1) > 1, then by Proposition Lemma 3.2.Let C = Φ r m where m, r are integers with m 3 and r 1.If C(1) and C(−1) are both squares, then deg(C) is divisible by 4.
. (ii) If C(1) and C(−1) are both squares, then Condition (C 1) holds for C. Let c 0 be an even integer such that c deg(C), c R and deg(C) − c S. We have (iii) If N = deg(C) + 2 and deg(C) ≡ 0 (mod 4), then deg(C) ≡ 2c (mod 8).Proof.If C(1) and C(−1) are both squares, then Lemma 3.2 implies that deg(C) is divisible by 4, hence (i) holds.Set deg(C) = 2n; then n is even, hence C(1), C(−1) and (−1) n C(1) and C(−1) are all squares, therefore condition (C 1) holds for C, and this implies (ii).Let us prove (iii).Since deg(C) is divisible by 4 and deg(C) = N − 2, we have N ≡ 2 (mod 4), and therefore R and S are both odd integers.Since deg(C) = N − 2, the inequalities c R and deg(C) − c S imply that R − 2 c R; since R and S are odd, this implies that c = R − 1 and deg(C) − c = S − 1. W have R ≡ S (mod 8) by hypothesis, hence deg(C) − c ≡ c (mod 8), as claimed.Lemma 3.5.Let c 0 be an even integer such that c deg(C), c R and deg(C) − c S. Suppose that the following conditions hold

Lemma 5. 3 .
Let t : L → L be an isometry of a lattice L such that • The characteristic polynomial of t is divisible by C, • The signature of the sublattice Ker(C(t)) is (c, deg(C) − c).If deg(C) ≡ 0 (mod 4) and c ≡ 2 (mod 4), then L is indefinite.Proof.Indeed, the sublattice Ker(C(t)) is indefinite : since deg(C) ≡ 0 (mod 4) and c ≡ 2 (mod 4), we have c = deg(C) and c = 0. Proposition 5.4.Let m 3 be an integer and let p be a prime number that does not divide m.The following are equivalent (a) The polynomial Φ m has a symmetric irreducible factor mod p.(b) The prime ideals of Q(ζ m + ζ −1 m ) above p are inert in Q(ζ m ).(c) The subgroup of (Z/mZ) × generated by p contains −1.Proof The equivalence of (a) and (b) follows from [W 97], Proposition 2.14.Let us prove that (b) and (c) are equivalent.Let G be the Galois group of Q(ζ m )/Q, and let P be a prime ideal of Q(ζ m ) above p.The decomposition group G P is by definition {g ∈ G | g(P ) = P }.Since G is abelian, this group only depends on the prime number p; set G P = G p .Condition (b) holds if and only if the element of G induced by ζ m → ζ −1 Proposition 5.6.Suppose that C(1) = C(−1) = 1.Let p be a prime number such that Π Φmp,Φm = {p}.If N > deg(C) + ϕ(mp), then there exists an even unimodular lattice L of signature (R, S) and an isometry t : L → L of order mp such that • The characteristic polynomial of t is divisible by C, • The signature of the sublattice Ker(C(t)) is (c, deg(C) − c).
The polynomials S n are Salem polynomials (see[McM 02], §4, or [GM 02], §7, Example 1); we have S n (1) = −1 and S n (−1) = 1.If a : X → X is an automorphism of a projective K3 surface, then the characteristic polynomial of a * : H 2 (X, C) → H 2 (X, C) is either a product of cyclotomic polynomials, or it is of the form SC, where S is a Salem polynomial and C a product of cyclotomic polynomials (see [McM 02], Theorem 3.2 and Corollary 3.3).
Proposition 2 in §4 (seeProposition 4.1).A stronger form of Theorem 1 is in §5, see Theorem 5.1.Proposition 1 is proved in §9 and §10.The last section concerns the possible values of m X and n X , a discussion of some results of Kondo and Vorontsov, as well as a generalization of these results, and some open questions.
[B 21] and [B 22], and give some examples that are used in the paper.Theorem 2 is proved in §3 (see Theorem 3.6), and