Isometries of quadratic spaces over global fields

Eva Bayer–Fluckiger

Introduction

Let k be a field of characteristic not 2. An quadratic space is a non–degenerate symmetric bilinear form q : V × V → k defined on a finite dimensional k–vector space V, and an isometry of (V, q) is an element of SO(q), in other words an isomorphism t : V → V such that q(tx, ty) = q(x, y) for all x, y ∈ V and that det(t) = 1. In [7], Milnor investigated isometries of quadratic spaces, and raised the following question :

Question. Let q be a quadratic space over k, and let f ∈ k[X] be an irreducible polynomial. How can we tell whether q has an isometry with minimal polynomial f ?

He gave an answer in the case of local fields. The main purpose of the present paper is to study this question for global fields.

Suppose that k is a global field, let q be an quadratic space, let f ∈ k[X] be an irreducible polynomial, and let F = f^{m} for some m ∈ N. The following Hasse principle is proved in section 5 :

Theorem. The quadratic space q has an isometry with characteristic polynomial F and minimal polynomial f if and only if such an isometry exists over all the completions of k.

For m = 1, this is a consequence of a result of Prasad and Rapinchuk, cf. [9], th. 7.3. In order to obtain a necessary and sufficient criterion, we need to consider the case of separable minimal polynomials over local fields and the field of real numbers. This is done in §3 and §4, and leads for two necessary conditions over global fields, the hyperbolicity condition and the signature condition (cf. §5).

Let us suppose that F is symmetric, that F (1)F (−1) ≠ 0 and that dim(V) = deg(F). We have (see 5.2) :

Theorem. The quadratic space q has an isometry with minimal polynomial f if and only if the signature condition and the hyperbolicity conditions are satisfied, and

det(q) = F (1)F (−1) ∈ k^{∗}/k^{∗2}.

The first two sections are devoted to some definitions and useful lemmas, most of them based on Milnor’s paper [7]. Section 6 contains an application of the results of the paper to isometries of quadratic spaces extended from unimodular, even lattices, in relationship with a question of Gross and McMullen [5].

§1. Definitions, notation and basic facts.

Let k be a field of characteristic not 2. An quadratic space is a pair (V, q), where V is a finite dimensional k–vector space, and q : V × V → k is a symmetric bilinear form of non–zero determinant. The determinant of (V, q) is denoted by det(q); it is an element of k^{∗}/k^{∗2}. For basic facts concerning quadratic spaces, see [8] and [9].

An isometry of the quadratic space (V, q) is an isomorphism t : V → V such that q(tx, ty) = q(x, y) for all x, y ∈ V and that det(t) = 1.

The answer to Milnor’s question is obvious if f(X) = X + 1 or X − 1, hence we can exclude these cases. This means that we can assume that f(1)f(−1) ≠ 0. Therefore, throughout the paper the isometries are supposed to have neither 1 nor -1 as an eigenvalue.

The following results are well–known, see for instance Levine [6] and Milnor [7], as well as the appendix of [5].

A polynomial f ∈ k[X] is said to be symmetric if f(X) = X^{deg(f)}f(X^{−1}).

Proposition 1.1 The minimal polynomials of isometries of quadratic spaces are symmetric of even degree.

Proof. Let (V, q) be an quadratic space, and let t : V → V be an isometry of q. By definition, we have q(tx, y) = q(x, t^{−1}y) for all x, y ∈ V. This implies that for any polynomial p ∈ k[X], we have q(p(t)x, y) = q(x, p(t^{−1})y) for all x, y ∈ V. Let f ∈ k[X] be the minimal polynomial of t. Applying the above equality to p = f, we see that the endomorphism t^{degf}f(t^{−1}) annihilates V. As f is the minimal polynomial of t, this implies that f divides X^{degf}f(X^{−1}), therefore we have f(X) = ϵX^{degf}f(X^{−1}) for some ϵ = ±1. Note that if ϵ = −1, then f(1) = 0, hence f is divisible by X − 1. This is impossible, as we have supposed that 1 is not an eigenvalue of t. Therefore ϵ = 1, hence f is symmetric. If deg(f) is odd, then this implies that f(−1) = 0, which contradicts the assumption that -1 is not an eigenvalue of t. Therefore deg(f) is even.

A basic observation is that there is a relationship between the determinant of an quadratic space, and the values of the characteristic polynomials of its isometries, as follows :

Proposition 1.2 Let (V, q) be an quadratic space, and let F ∈ k[X] be the characteristic polynomial of an isometry of q.Then

det(q) = F (1)F (−1) ∈ k^{∗}/k^{∗2}.

Proof. Let us define q′ : V × V → k by q′(x, y) = q(x, (t − t^{−1})(y)). Then q′ is skew– symmetric, hence det(q′) ∈ k^{2}. On the other hand, we have det(q′) = det(q)F (1)F (−1), so the proposition is proved.

Proposition 1.3 The characteristic polynomials of isometries of quadratic spaces are symmetric of even degree.

Proof. This follows from a straightforward computation, see for instance [6], I. 7, Lemma (a).

§2. Primary decomposition and transfer

The aim of this section is to recall some results of Milnor [7], as well as to introduce some terminology that will be used later on. Let f ∈ k[X] be a monic, irreducible polynomial. Let us define the dual of f as the monic, irreducible polynomial f^{∗} ∈ k[X] defined by f^{∗}(X) = \frac{1}{f(0)}X^{m}f(X^{−1}). Note that f is symmetric if and only if it is equal to its dual polynomial.

Let (V, q) be an quadratic space of dimension 2n, let t be an isometry of q and let F be the characteristic polynomial of t. For each monic, irreducible factor f of F, set

V_{f} = {v ∈ V | f^{i}(t)(v) = 0 for some i ∈ N}.

Let U and W be two subspaces of V. We say that U and W are orthogonal to each other if q(u, w) = 0 for all u ∈ U and w ∈ W. We say that (V, q) is hyperbolic if V has a self–orthogonal subspace of dimension n.

Proposition 2.1 Let f and g be two monic, irreducible factors of F. If f ≠ g^{∗}, then V_{f} and V_{g} are orthogonal to each other.

Proof. See Milnor [7], Lemma 3.1.

Corollary 2.2 If f is not symmetric, then (V_{f} ⊕ V_{f^{∗}}, q) is hyperbolic.

Proof. See [7], §3, Case 3.

Proposition 2.3 We have the following orthogonal decomposition

(V, q) ≃ ⨁(V_{f}, q) ⊕ H

where the sum is taken over all distinct monic, symmetric and irreducible factors of F, and where H is a hyperbolic space.

Proof. This follows from prop. 2.1 and cor. 2.2.

Definition. A symmetric polynomial is said to be hyperbolic if none of its irreducible factors is symmetric.

Corollary 2.4 A quadratic space having an isometry with hyperbolic characteristic polynomial is hyperbolic

Proof. This is an immediate consequence of 2.3.

Proposition 2.5 Let f ∈ k[X] be a monic, symmetric, separable and irreducible polynomial, and set K = k[X]/(f). Then sending X to X^{−1} induces a k–linear involution : K → K. Moreover, for every quadratic space (V, q) over k and every isometry having minimal polynomial f, there exists a non–degenerate hermitian form (V, h) over K such that for all x, y ∈ V

q(x, y) = Tr_{K/k}(h(x, y)).

Conversely, if V is a finite dimensional vector space over K and if h : V → V is a non–degenerate hermitian form, then setting

q(x, y) = Tr_{K/k}(h(x, y))

for all x, y ∈ V we obtain an quadratic space (V, q) over k together with an isometry with minimal polynomial f.

Proof. See [7], Lemma 1.1 and Lemma 1.2.

Lemma 2.6 Let (V, q) be a hyperbolic quadratic space of dimension 2n. Let f ∈ k[X] be monic, separable and irreducible polynomial of degree d, and suppose that f ≠ f^{∗}. Set m = \frac{n}{d}. Then (V, q) has an isometry with minimal polynomial ff^{∗} and characteristic polynomial (ff^{∗})^{m}.

Proof. Let A = k[X]/(ff^{∗}), and let : A → A be the involution induced by X ↦ X^{−1}. Let us define a quadratic space q_{0} : A × A → k by setting q_{0}(x, y) = Tr_{A/k}(x\overline{y}). This space is hyperbolic of dimension 2d, and has an isometry of minimal (and characteristic) polynomial ff^{∗} by construction. Note that the quadratic space q is isomorphic to the orthogonal sum of m copies of q_{0}, hence q has an isometry with minimal polynomial ff^{∗} and characteristic polynomial (ff^{∗})^{m}.

Finally, we give a construction and a terminology that will be used repeatedly in the paper.

Let F ∈ k[X] be a monic, symmetric polynomial such that F (1) ≠ 0 and F (−1) ≠ 0. Let f_{1}, …, f_{p} ∈ k[X] be the distinct monic, irreductible and symmetric factors of F, and let g_{1}, …, g_{q} be the monic, irreducible factors of F such that g_{i}∗ ≠ g_{i} for all i = 1, …, q. Set f = f_{1} … f_{p}g_{1}g_{1}^{∗} … g_{q}g_{q}^{∗}, and note that F = f_{1}^{n_{1}} … f_{r}^{n_{p}}(g_{1}g_{1}^{∗})^{m_{1}} … (g_{q}g_{q}^{∗})^{m_{q}} for some integers n_{i}, m_{j}.

Set A = k[X]/(f), and let : K → K be the involution induced by X ↦ X^{−1}. We have A = K_{1} × … × K_{r} × A_{1} × … × A_{s}, where K_{i} = k[X]/(f_{i}) and A_{i} = k[X]/(g_{i}g_{i}^{∗}).

Then the K_{i}’s and the A_{i}’s are stable by the involution. The fields K_{i} will be called the fields with involution associated to F.

Set V_{i} = K_{i}^{n_{i}} for i = 1, …, p. Let (V_{0}, q_{0}) be a hyperbolic quadratic space with minimal polynomial g_{1}g_{1}^{∗} … g_{s}g_{s}^{∗} and characteristic polynomial (g_{1}g_{1}^{∗})^{m_{1}} … (g_{s}g_{s}^{∗})^{m_{s}}. This is possible by lemma 2.6. The vector spaces V_{i} will be called the vector spaces associated to F, and the (V_{0}, q_{0}) the hyperbolic space associated to F.

§3. The field of real numbers

In this section the ground field k is the field of real numbers R. Let (V, q) be a quadratic space over R. It is well–known that q is isomorphic to

X_{1}^{2} + … + X_{r}^{2} − X_{r+1}^{2} − … − X_{r+s}^{2}

for some natural numbers r and s. These are uniquely determined by q, and we have r + s = dim(V). The couple (r, s) is called the signature of q.

Let F ∈ R[X] be a symmetric polynomial, and suppose that F (1) ≠ 0 and F (−1) ≠ 0. Let deg(F) = 2n, and let 2σ be the number roots of F off the unit circle. The following result is proved by Gross and McMullen in [5], cor. 2.3, in the case where F is separable.

Proposition 3.1 The quadratic space (V, q) has an isometry with characteristic polynomial F and separable minimal polynomial if and only if r + s = 2n, (r, s) ≥ (σ, σ), and (r, s) ≡ (σ, σ) (mod 2).

Proof. Suppose that q has an isometry with characteristic polynomial F and separable minimal polynomial. Let f_{1}, …, f_{p} ∈ R[X] be the distinct monic, irreducible and symmetric factors of F. We have F = F_{1}F_{2}, where F_{1} is a product of powers of the f_{i}’s and F_{2} is a product of non–symmetric polynomials. By definition, we have deg(F_{2}) = 2σ.

By prop. 2.3, we have

(V, q) ≃ ⨁_{i=1,...,m}(V_{f_{i}}, q_{f_{i}}) ⊕ H

with H hyperbolic of dimension 2σ. The signature of H is (σ, σ), so this implies that (r, s) ≥ (σ, σ).

By prop. 2.5, there exists a hermitian form (V_{f_{i}}, h_{f_{i}}) such that

q_{f_{i}}(x, y) = Tr_{K/k}(h_{f_{i}}(x, y))

for all x, y ∈ V_{f_{i}}. Let (u_{i}, v_{i}) be the signature of h_{f_{i}}. Then the signature of q_{f_{i}} is (2u_{i}, 2v_{i}). This implies that (r, s) ≡ (σ, σ) (mod 2).

Conversely, suppose that r + s = 2n, (r, s) ≥ (σ, σ), and (r, s) ≡ (σ, σ) (mod 2). Let K_{i} be the fields with involution, V_{i} the K_{i}–vector spaces, and (V_{0}, q_{0}) the hyperbolic quadratic space associated to F (cf. end of §2). Then dim_{k}(V_{0}) = 2σ. Let 2u = r − σ, and 2v = s − σ. Note that u + v = n − σ. Let u_{i}, v_{i} ∈ N such that 0 ≤ u_{i}, v_{i} ≤ dim_{K_{i}}(V_{i}) and that u_{1} + … + u_{p} = u and v_{1} + … + v_{q} = v.

Let h_{i} : V_{i} × V_{i} → K_{i} be the hermitian form of signature (u_{i}, v_{i}) over V_{i}, and let q_{h_{i}} : V_{i} × V_{i} → k the quadratic space defined by q_{h_{i}}(x, y) = Tr_{K_{i}/k}(h_{i}(x, y)) for all x, y ∈ V_{i}. Then the signature of q_{h_{i}} is (2u_{i}, 2v_{i}). Let W = V_{0} ⊕ V_{1} ⊕ … ⊕ V_{p}, and let q′ : W × W → k be the orthogonal sum of q_{0}, q_{1}, …, q_{p}. Then q′ has an isometry with separable minimal polynomial and characteristic polynomial F by construction. The signature of q′ is (r, s), hence q′ ≃ q. This completes the proof of the proposition.

4. Local fields

Suppose that k is a local field, and let F ∈ k[X] be a monic, symmetric polynomial. If F is hyperbolic, then we know from cor. 2.4 that only hyperbolic quadratic spaces admit isometries with characteristic polynomial F. Hence, from now on we suppose that F is not hyperbolic. Let q be a quadratic space.

Theorem 4.1 The quadratic space q has an isometry with characteristic polynomial F and separable minimal polynomial if and only if deg(F) = dim(V), and

det(q) = F (1)F (−1) ∈ k^{∗}/k^{∗2}.

The proof of th. 4.1 relies on the following result of Milnor. Let K be a separable extension of k of finite degree endowed with a non–trivial k–linear involution : K → K. For any non–degenerate hermitian form h : U × U → K, let us denote by q_{h} : U × U → k the quadratic space defined by q_{h}(x, y) = Tr_{K/k}(h(x, y)) for all x, y ∈ U. We have

Theorem 4.2 If the hermitian spaces h and h′ have the same dimension but different determinants, then the quadratic spaces q_{h} and q_{h′} have the same dimension and determinant but different Hasse invariants.

Proof. See Milnor, [7], th. 2.7.

Proof of 4.1 The necessity of the condition follows from prop. 1.2. Conversely, suppose that dim(V) = deg(F), and that det(q) = F (1)F (−1) ∈ k^{∗}/k^{∗2}. Let K_{i} be the fields with involution and V_{i} the K_{i}–vector spaces associated to F (cf. §2). Let h_{i} : V_{i} × V_{i} → K_{i} be the unit hermitian form over V_{i}, and let q_{h_{i}} : V_{i} × V_{i} → k the quadratic space defined by q_{h_{i}}(x, y) = Tr_{K_{i}/k}(h_{i}(x, y)) for all x, y ∈ V_{i}. Let (V_{0}, q_{0}) be the hyperbolic quadratic space associated to F. Let W = V_{0} ⊕ V_{1} ⊕ … ⊕ V_{p}, and let q′ : W × W → k be the orthogonal sum of q_{0}, q_{1}, …, q_{p}. Then q′ has an isometry with separable minimal polynomial and characteristic polynomial F by construction, and we have det(q′) = F (1)F (−1). If the Hasse invariants of q and q′ are equal, then we are finished. If not, let E_{1} be the fixed field of the involution in K_{1}, and let α ∈ E_{1}^{∗} be such that α ∉ N_{K_{1}/E_{1}}(K_{1}^{∗}). Let h′_{1} :

V_{1} × V_{1} → K_{1} be a hermitian form over K_{1} of determinant α, and let q_{h′_{1}} : V_{1} × V_{1} → k the quadratic space defined by q_{h′_{1}}(x, y) = Tr_{K_{1}/k}(h′_{1}(x, y)) for all x, y ∈ V_{1}.Then by th. 4.2 the quadratic spaces q_{h′_{1}} and q_{h_{1}} have equal determinants but different Hasse invariants. Let q′′ : W × W → k be the orthogonal sum of q_{0}, q′_{1}, q_{2}, …, q_{r}. Then q′′ has an isometry with separable minimal polynomial and characteristic polynomial F. Moreover, q and q′′ have equal dimension, determinant and Hasse invariant, hence they are isomorphic. This completes the proof .

§5. Global fields

Suppose that k is a global field. The aim of this section is to give an answer to the question of Milnor quoted in the introduction in the case of global fields.

Let (V, q) be an quadratic space over k, and let f ∈ k[X] be a monic, symmetric, irreducible polynomial. We have the following local-global principle :

Theorem 5.1 The quadratic space q has an isometry with minimal polynomial f if and only if such an isometry exists over every completion of k.

Note that a result of Prasad and Rapinchuk (cf. [9], 7.3) implies 5.1 in the case of an irreducible characteristic polynomial. Before proving th. 5.1, let us use the results of the previous two sections to obtain necessary and sufficient conditions for an isometry to exist. Let F be a power of f such that deg(F) = dim(V) = 2n.

For every real place v of k, let (r_{v}, s_{v}) denote the signature of q over k_{v}, and let σ_{v} be the number of roots of F that are not on the unit circle.

We say that the signature condition is satisfied for q and F if for every real place v of k, we have (r_{v}, s_{v}) ≥ (σ_{v}, σ_{v}), and (r_{v}, s_{v}) ≡ (σ_{v}, σ_{v}) (mod 2).

We say that the hyperbolicity condition is satisfied for q and F if for all places v of k such that F ∈ k_{v}[X] is a hyperbolic polynomial, the quadratic form q_{v} over k_{v} is hyperbolic.

Corollary 5.2 The quadratic space q has an isometry with minimal polynomial f if and only if the signature condition and the hyperbolicity conditions are satisfied, and

det(q) = F (1)F (−1) ∈ k^{∗}/k^{∗2}.

Proof. The necessity of the conditions follows from prop. 3.1 and prop. 1.2. Conversely, suppose that the signature condition is satisfied and that det(q) = F (1)F (−1) ∈ k^{∗}/k^{∗2}. Then by prop. 3.1 and th. 4.1, the quadratic space q has an isometry with minimal polynomial f over k_{v} for every place v of k. By th. 5.1, this implies that q has an isometry with minimal polynomial f.

The following reformulation of cor. 5.2 shows that it suffices to check a finite number of conditions. Let q and F be as above, with dim(q) = deg(F) = 2n. Let S be the set of places of k at which the Hasse–Witt invariant of q is not equal to the Hasse–Witt invariant of the 2n dimensional hyperbolic form. Note that S is a finite set.

Corollary 5.3 The quadratic space q has an isometry with minimal polynomial f if and only if the following conditions are satisfied :

(i) F (1)F (−1) = det(q) ∈ k^{∗}/k^{∗2}; (ii) The signature condition holds; (iii) If v ∈ S, then F ∈ k_{v}[X] is not hyperbolic.

Proof. It suffices to prove that the conditions (i) and (iii) imply the hyperbolicity condition. Let v be a place of k such that F ∈ k_{v}[X] is hyperbolic. Then there exists a polynomial G ∈ k_{v}[X] such that F = GG^{∗}. Note that deg(G) = n. We have F (1) = G(1)^{2}, and F (−1) = (−1)^{n}G(−1). By condition (i), we have F (1)F (−1) = det(q), hence (−1)^{n}det(q) = disc(q) ∈ k_{v}^{2}. On the other hand, condition (iii) implies that v ∉ S. Therefore over k_{v}, the quadratic space q has the same dimension, discriminant and Hasse– Witt invariant as the 2n–dimensional hyperbolic form. Hence q is hyperbolic over k_{v}, in other words the hyperbolicity condition is satisfied.

The following will be used in the proof of th. 5.1. Let f ∈ k[X] be an irreducible, symmetric polynomial, and let K = k[X]/(f). Let : K → K be the k–linear involution induced by X ↦ X^{−1}, and let E be the fixed field of this involution. Let (W, Q) be a quadratic space over k.

Lemma 5.4 Let v be a finite place of k such that every place of E above v splits in K. Suppose that over k_{v} the quadratic space Q has an isometry with minimal polynomial f. Then Q is hyperbolic over k_{v}.

Proof. Let w_{1}, …, w_{r} be the places of E above v. Then k_{v}[X]/(f) ≃ K_{1} × … × K_{r}, where K_{i} is a field if w_{i} is inert or ramified in K and a product of two fields if w_{i} is split in K. Here we are supposing that every w_{i} splits in K, hence all the K_{i}’s are products of two fields. This implies that f = f_{1}f_{1}^{∗} … f_{r}f_{r}^{∗} with f_{i} ∈ k_{v}[X] monic and irreducible and f_{i} ≠ f_{i}^{∗} for all i = 1, …, r. By prop. 2.4, the quadratic space Q is hyperbolic over k_{v}.

Proof of th. 5.1 Let F = f^{m} and deg(f) = 2d. Let K = k[X]/(f), and let : K → K be the involution induced by X ↦ X^{−1}. Let E be the fixed field of the involution. Let θ ∈ E^{∗} such that K = E(\sqrt{θ}), and for any place w of E, let ( , )_{w} denote the Hilbert symbol at E_{w}.

Let v be a real place of k. Then the signature (r_{v}, s_{v}) of q at k_{v} satisfies (r_{v}, s_{v}) ≥ (σ_{v}, σ_{v}) and (r_{v}, s_{v}) ≡ (σ_{v}, σ_{v}) (mod 2). In particular, s_{v} − σ_{v} is even. Set s_{v} − σ_{v} = 2u_{v}, and note that 0 ≤ u_{v} ≤ n − σ_{v}. Let us denote by 2τ_{v} the number of roots of f that are not on the unit circle. Then we have σ_{v} = mτ_{v}. Let us write u_{v} = u_{v}^{1} + … + u_{v}^{m} for some integers u_{v}^{i} such that 0 ≤ u_{v}^{i} ≤ d − τ_{v}.

Let w_{1}, …, w_{d−τ_{v}} be the real places of E above v that extend to a complex place of K. Let α_{i} ∈ E^{∗} such that (α_{i}, θ)_{w_{j}} = −1 if j = 1, … u_{i}, and that (α_{i}, θ)_{w_{j}} = 1 if j = u_{i} + 1, …, d − τ_{i}.

Let h′ : V × V → K be the hermitian form defined by h′ =< α_{1}, …, α_{m} > and let let q_{h′} : V × V → k be the quadratic space defined by q_{h′}(x, y) = Tr_{K/k}(h′(x, y)) for all x, y ∈ V. By construction, the signature at v of q_{h′} is (r_{v}, s_{v}).

Let T be the set of finite places of k such that every place of E above v splits in K. The local conditions imply that q has an isometry of minimal polynomial f over every completion of k. By lemma 5.4 this implies that if v ∈ T, then q is hyperbolic over k_{v}.

Let S be the set of finite places of k at which the Hasse invariants of q and q_{h′} are not equal. This is a finite set of even cardinality. Note that the quadratic space q_{h′} has an isometry of minimal polynomial f by construction, therefore lemma 5.4 implies that q_{h′} is hyperbolic over k_{v} if v ∈ T. This implies T and S are disjoint.

For each v ∈ S, let us choose a place w of E which does not split in K; this is possible because S and T are disjoint. Let us denote by S_{E} the set of these places. Then S_{E} is a finite set of even cardinality.

For all w ∈ S_{E}, let β_{w} ∈ E_{w}^{∗} such that (β_{w}, θ)_{w} = −1; note that such an β_{w} exists as w is does not split in K. By Hilbert reciprocity, there exists β ∈ E^{∗} such that (β, θ)_{w} = (β_{w}, θ)_{w} = −1 if w ∈ S_{E}, and that (β_{w}, θ)_{w} = 1 for all the other places w of E. Let h : V ×V → K be the hermitian form defined by h =< βα_{1}, …, α_{d} > and let q_{h} : V ×V → k be the quadratic space defined by q_{h}(x, y) = Tr_{K/k}(h(x, y)) for all x, y ∈ V. Then by th. 4.2, the Hasse invariants of q_{h} and q are equal. This implies that q and of q_{h} have equal dimension, determinant, signatures and Hasse invariants, therefore these quadratic spaces are isomorphic. Note that q_{h} has an isometry with minimal polynomial f by construction, hence q also has such an isometry, and this concludes the proof of the theorem.

§6. Isometries of even, unimodular lattices

An integral lattice is a pair (L, q_{0}), where L is a free Z–module of finite rank, and q : L × L → Z is a symmetric bilinear form. We say that the lattice q_{0} is unimodular if det(q_{0}) = ±1, and even if q_{0}(x, x) ≡ 0 (mod 2) for all x ∈ L. Which polynomials F ∈ Z[X] can be realized as isometries of even, unimodular lattices ? This question was raised in [1], [2], [3], as well as in the paper of Gross and McMullen [5]. More precisely, Gross and McMullen [5] concentrate on the case of irreducible characteristic polynomials and indefinite lattices (see also [4]), and the case where F is a power of an irreducible polynomial and the lattice is definite is handled in [1].

Let (L, q_{0}) be an even, unimodular lattice, let V = L ⊗_{Z} Q, and let q : V × V → Q be the extension of q_{0} to Q. Let f ∈ Z[X] be an irreducible polynomial, let F = f^{m} and suppose that deg(F) = dim(V) = 2n. The results of §5 imply the following :

Theorem 6.1 The quadratic space q has an isometry with minimal polynomial f if and only if the signature condition is satisfied, and

F (1)F (−1) = (−1)^{n} ∈ Q^{∗}/Q^{∗2}.

Proof. This is a consequence of cor. 5.3. Indeed, it is well–known that the signature of an even, unimodular lattice is divisible by 8. This implies that q_{0} is an orthogonal sum of a definite form of rank divisible by 8 with a certain number of hyperbolic planes. The number of these planes is congruent to n modulo 2, hence det(q_{0}) = (−1)^{n}. Therefore the conditions of the theorem are necessary. On the other hand, as q_{0} is an orthogonal sum of a definite form of rank divisible by 8 and of a hyperbolic form, the Hasse–Witt invariant of q coincides with the Hasse–Witt invariant of the 2n–dimensional hyperbolic form. Therefore the set S of cor. 5.3 is empty. This shows that the conditions are also sufficient, hence the theorem is proved.

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