The Zassenhaus Decomposition

for the Orthogonal Group:

Properties and Applications

Alexander Hahn^{1}

Received: May 29, 2001

Communicated by Ulf Rehmann

Abstract. Zassenhaus [17] constructed a decomposition for any element in the orthogonal group of a non-degenerate quadratic space over a field of characteristic not 2 and used it to provide an alternative description of the spinor norm. This decomposition played a central role in the study of question of the length of an element in the commutator subgroup of the orthogonal group with respect to the generating set of all elementary commutators of hyperplane reflections. See Hahn [6]. The current article develops the fundamental properties of the Zassenhaus decomposition, e.g., those of uniqueness and conjugacy, and applies them to sharpen and expand the analysis of [6].

2000 Mathematics Subject Classification: 20G15, 20G25, 20F05.

1. Introduction. We begin with a discussion of the length question just mentioned. For the moment, consider any group G along with a set of generators A (not containing the identity element of G) that satisfies A^{−1} = A. Of all the factorizations of an element σ ∈ G as a product of elements from A choose one that involves the smallest number of factors. This smallest number is defined to be the length ℓ(σ) of σ. One very basic question - necessarily in the context of specific examples - is this: are there parameters attached to σ from which ℓ(σ) can be read off?

A number of theorems have responded to this question. For G a Weyl group - or more generally a Coxeter group - and A an appropriate set of hyperplane reflections, see Humphreys [7]. Refer to Dyer [3] for a recent result in this context. For G a classical group and A a set of canonical elements coming from the underlying geometry, see Hahn-O’Meara [5] for a comprehensive treatment of the theorems of Dieudonné, Wall, and others. For G a classical group and A a set of generators coming from a single conjugacy class of elements, see Ellers-Malzan [2] and Knüppel [10]. In a related direction, interesting codes have been constructed starting with G = SL_{2}(ℤ) and carefully selected A. See Margulis [12, 13] and Rosenthal-Vontobel [16] for details. The connection with the length problem is provided by the associated Cayley graph and its diameter.

1The author wishes to thank the algebraists of Louisiana State University for their splendid organization of Quadratic Forms 2001 and their warm hospitality throughout the conference.

Documenta Mathematica · Quadratic Forms LSU 2001 · 165–181

Example 1. Let G be the symmetric group on {1, …, n} and let A be the set of transpositions. Let k(σ) be the the number of cycles of σ including the trivial cycles. Then ℓ(σ) = n − k(σ).

The fact that ℓ(σ) ≤ n − k(σ) follows from the decomposition of σ into its disjoint cycles. The other inequality is a consequence of the fact that k(στ) = k(σ) ± 1 for any transposition τ. A similar (but more complicated) argument provides

Example 2. Let G be the alternating group on {1, …, n} and let A be the set of three cycles, or equivalently, the set of elementary commutators of transpositions. This time, let k(σ) be the number of cycles of odd cardinality again including the trivial cycles. Then n − k(σ) is even and ℓ(σ) = \frac{1}{2}(n − k(σ)).

We now turn to the orthogonal group and begin by recalling some of the basics. For the details, see [5], especially Sections 5.2A, 5.2B, Chapter 6 (all specialized to the orthogonal case Λ = 0) and Section 8.2A.

Let V be a non-zero, non-degenerate, n−dimensional quadratic space with symmetric bilinear form B over a field F with char(F) ≠ 2. Denote B(x, x) by Q(x) and \frac{1}{2}Q(x) by q(x). Check that B(x, y) = q(x + y) − q(x) − q(y). Two vectors x and y are orthogonal if B(x, y) = 0. A non-zero vector x in V that is orthogonal to itself is isotropic and it is anisotropic otherwise. A non-degenerate plane that contains isotropic vectors is a hyperbolic plane and an orthogonal sum of hyperbolic planes is a hyperbolic space. If U and W are orthogonal subspaces that intersect trivially, then U ⊕ W is denoted U ⊥ W. The orthogonal complement of a subspace U of V is denoted by U^{⊥}, and the radical of U is defined by Rad U = U ∩ U^{⊥}. If W is a complement of Rad U in U, then W is non-degenerate and U = Rad U ⊥ W is a radical splitting of U. Any two such complements of Rad U are isometric.

Zassenhaus Decomposition for the Orthogonal Group 167

Let O_{n}(V) be the orthogonal group of V. For σ ∈ O_{n}(V), let S be the subspace S = (σ − 1_{V})V of V. This S is the space of σ. Intuitively, this is where the ”action” of σ is. In particular, there is no action on the orthogonal complement S^{⊥} of S; the fact is that S^{⊥} = {x ∈ V | σ(x) = x}. Clearly, σ = 1_{V} if and only if S = 0. It turns out that dim S is even if and only if σ ∈ O_{n}^{+}(V), the subgroup of O_{n}(V) consisting of the elements of determinant 1. If η ∈ O_{n}(V) commutes with σ, then ηS = S. We will ”transfer” properties of S to σ. For example, σ is non-degenerate, degenerate, or totally degenerate, if S is nondegenerate, degenerate, or totally degenerate, that is, if the radical Rad S of S is, respectively, zero, non-zero, or S. In the same way, σ is anisotropic if S is anisotropic.

An element σ ∈ O_{n}(V) is an involution if σ^{2} = 1. It is easy to see that σ is an involution if and only if σ_{S} = −1_{S}. In particular, involutions have the form σ = −1_{S} ⊥ 1_{S^{⊥}} and are non-degenerate. Let v be an anisotropic vector and define τ_{v} in O_{n}(V) by

τ_{v}(x) = x − B(x, v)q(v)^{−1}v for all x ∈ V .

Check that the space of τ_{v} is F v and that τ_{v} _{F v} = −1_{F v}. So τ_{v} is an involution. These involutions are the hyperplane reflections or symmetries.

Theorem 1. (Cartan-Scherk-Dieudonné) Let G be the group O_{n}(V) and let A be the set of hyperplane reflections. If σ is not totally degenerate, then ℓ(σ) = dim S. If σ is totally degenerate, then ℓ(σ) = dim S + 2.

Theorem 1 in combination with Examples 1 and 2 calls attention to the length problem in the situation where G is the commutator subgroup Ω_{n}(V) of O_{n}(V) and A the set of elementary commutators of symmetries. It seems surprising that this question did not receive scrutiny until recently. John Hsia first called attention to it in the case of a non-dyadic local field and it was solved in this context in Hahn [6]. The answer is not simply a modification of the conclusion of Theorem 1, as a comparison of Examples 1 and 2 might suggest. We will see that, unlike Theorem 1, it depends critically on the arithmetic of the field F.

2. The Zassenhaus Decomposition. An element σ in O_{n}(V) is unipotent if its minimal polynomial has the form (X − 1)^{m} for some positive integer m. A non-trivial unipotent element is degenerate and can, therefore, exist only if V is isotropic. The elements with minimal polynomial (X − 1)^{2} are precisely the non-trivial totally degenerate elements. A degenerate element σ with dim S = 2 is an Eichler transformation. Let S be a degenerate plane and put S = F u ⊥ F v with u ∈ Rad S and v ∈ S. Define Σ_{u,v} ∈ O_{n}(V) by

Σ_{u,v}(x) = x + B(x, v)u − B(x, u)v − q(v)B(x, u)u for all x ∈ V .

Then Σ_{u,v} is an Eichler transformation and all Eichler transformations have this form. A totally degenerate Eichler transformation has minimal polynomial (X−1)^{2} and one that is not totally degenerate has minimal polynomial (X−1)^{3}. In particular, all Eichler transformations are unipotent.

Let σ be any element in O_{n}(V). Consider the subspace

X = {x ∈ V | (σ − 1_{V})^{j}x = 0 some j}

of V. This unique largest space on which σ acts as a unipotent transformation turns out to be non-degenerate. Let R = X^{⊥}. Then R is non-degenerate and X = R^{⊥}. Notice that σR^{⊥} = R^{⊥}. So σR = R and hence σ = σ_{R^{⊥}} ⊥ σ_{R} . Put μ = σ_{R^{⊥}} ⊥ 1_{R} and ρ = 1_{R^{⊥}} ⊥ σ_{R}. Then

σ = μ · ρ

with μ unipotent and ρ non-degenerate with space R. This is the Zassenhaus decomposition or splitting of σ. Note that μ and ρ commute.

To develop the essential properties of the Zassenhaus splitting, we need the Wall form. Let σ ∈ O_{n}(V). Define

( , )_{σ} : S × S ⟶ F

by the equation (σx − x, σy − y)_{σ} = B(σx − x, y) for all σx − x and σy − y in S. This is the Wall form on S. It is non-degenerate and bilinear, but it is almost never symmetric. In fact, ( , )_{σ} is symmetric if and only if σ is an involution, and in this case, (s, s′)_{σ} = −\frac{1}{2}B(s, s′) for all s, s′ in S. Also, ( , )_{σ} is alternating if and only if σ is totally degenerate.

The space S is now equipped with both the Wall form ( , )_{σ} and the restriction of B. When the focus is on ( , )_{σ}, then S is denoted by S_{σ}. Similarly, the space S_{1} of σ_{1} in O_{n}(V) is written S_{σ_{1}} when ( , )_{σ_{1}} is under consideration, and analogously for σ_{2}. The spaces of orthogonal transformations μ, ρ, μ′, ρ′ and so on, will be denoted by U, R, U′, R′ and so on, with appropriate subscripts when the focus is on the Wall form.

The key facts are these. Let S_{1} be a non-degenerate subspace of S_{σ}. Then there is a unique σ_{1} ∈ O_{n}(V) - the transformation belonging to S_{1} - such that S_{σ_{1}} = S_{1}. Let S_{2} be the right complement of S_{1} in S_{σ}. Then S_{2} is non-degenerate. If σ_{2} is the transformation belonging to S_{2}, then σ = σ_{1}σ_{2}. Conversely, if σ = σ_{1}σ_{2} with S_{1} ∩ S_{2} = 0, then S_{σ} = S_{σ_{1}} ⊥ S_{σ_{2}}. This means that the Wall forms of both S_{σ_{1}} and S_{σ_{2}} are obtained by restricting the Wall form ( , )_{σ} and that (s_{1}, s_{2})_{σ} = 0 for all s_{1} ∈ S_{1} and s_{2} ∈ S_{2} (but it is not required that (s_{2}, s_{1})_{σ} = 0). For example, if σ = μρ is the Zassenhaus splitting of σ, then because μ and ρ commute,

S_{σ} = U_{μ} ⊥ R_{ρ} = R_{ρ} ⊥ U_{μ} .

Another important fact asserts that elements σ and σ_{1} in O_{n}(V) are conjugate in O_{n}(V) if and only if the spaces S_{σ} and S_{σ_{1}} are isometric.

To conclude this discussion of the Wall form, we note that the map

Θ : O_{n}^{+}(V) ⟶ F^{∗} /F^{2}^{∗}

defined by Θ(σ) = (disc S_{σ})F^{2}^{∗}, where disc S_{σ} is the discriminant of the space S_{σ}, provides one of the (equivalent) definitions of the spinor norm. Its kernel is denoted by O′_{n}(V). All unipotent elements are in O′_{n}(V). It is clear that O′_{n}(V) ⊇ Ω_{n}(V) and it is a standard fact that if V is isotropic, then O′_{n}(V) = Ω_{n}(V). A formula useful for computations is

Θ(σ) = Θ(ρ) = det (ρ − 1_{V})_{R} disc R ,

where ρ is the non-degenerate component of the Zassenhaus decomposition of

σ and disc R ∈ F^{∗} /F^{2}^{∗} is the discriminant of the space R relative to the form B.

Proposition 1. Let σ = μρ be the Zassenhaus splitting of an element σ ∈ O_{n}(V). Then S = U ⊥ R, and

i) σ is in Ω_{n}(V) if and only if both μ and ρ are in Ω_{n}(V).

ii) An element in O_{n}(V) commutes with σ if and only if it commutes with both μ and ρ.

Proof: Recall that O_{n}(V) has non-trivial unipotent elements only if V is isotropic. So any non-trivial unipotent element of O_{n}(V) is in O′_{n}(V) = Ω_{n}(V). This implies (i). As to (ii), observe that if η ∈ O_{n}(V) commutes with σ, then η stabilizes X = R^{⊥}. So η = η_{R^{⊥}} ⊥ η_{R} and it follows that η commutes with both μ and ρ. QED.

We next consider the question of the uniqueness of the Zassenhaus splitting. It is not difficult to construct situations of the following sort: a non-degenerate element σ and a non-trivial unipotent element μ_{0} with U_{0} ⊆ S such that μ_{0} commutes with σ and the space of ρ_{0} = μ_{0}^{−1}σ is S. In such a situation, σ = 1_{V} σ = μ_{0}ρ_{0} are two different ways of writing σ as a commuting product of a unipotent element and a non-degenerate element. We will see that such situations are in essence the only obstruction to the uniqueness of the Zassenhaus splitting.

Let σ = μρ be the Zassenhaus splitting of σ ∈ O_{n}(V). Suppose that σ = μ′ρ′ is any factorization of σ with μ′ unipotent, ρ′ non-degenerate, and such that μ′ and ρ′ commute.

Denote by W the orthogonal complement W = R^{′^{⊥}} of the space R′ of ρ′. By an application of Proposition 1 (ii), the elements μ, μ′, ρ, and ρ′ all commute with each other. In particular, σ commutes with ρ′. So σR′ = R′ and hence σW = W. Therefore, σ = σ_{W} ⊥ σ_{R′}. The fact that ρ′ _{W} = 1_{W}, tells us that σ_{W} = μ′ _{W}. So σ is unipotent on W and hence W ⊆ R^{⊥}. Therefore, R′ = W^{⊥} ⊇ R. Let T be the orthogonal complement of R in R′. Because R′ and R are both non-degenerate, R′ = T ⊥ R with T non-degenerate. Because R^{⊥} is the largest space on which σ is unipotent, T is the largest space on which σ_{R′} is unipotent. So

σ_{R′} = (μ_{T} ⊥ 1_{R})(1_{T} ⊥ ρ_{R})

is the Zassenhaus splitting of σ_{R′}. Notice that 1_{T} ⊥ ρ_{R} = ρ_{R′} and hence that μ_{T} ⊥ 1_{R} = μ_{R′}. Since μ′ and ρ′ commute with with both ρ′ and ρ, it follows that μ′ and ρ′ stabilize the spaces R′, R, and therefore T. So

μ_{T} = σ_{T} = (μ′ _{T})(ρ′ _{T}) .

Therefore, ρ′ _{T} is a product of two commuting unipotent transformations. So ρ′ _{T} is unipotent. If T where to be non-zero, then ρ′ would fix a non-zero vector in T. But this is impossible, because ρ′ is non-degenerate with space R′. So T = 0. Hence R′ = R and W = R^{⊥}. This means that μ′ _{R^{⊥}} = σ_{R^{⊥}} = μ_{R^{⊥}} and hence that μ′ = μ_{R^{⊥}} ⊥ μ′ _{R}. Because μ′ _{R} · ρ′ _{R} = σ_{R} = ρ_{R}, it follows that ρ′ = 1_{R^{⊥}} ⊥ (μ′ _{R})^{−1}(ρ_{R}). Therefore the obstruction to the uniqueness of the Zassenhaus splitting is as described earlier.

Notice that U′ ∩ R′ = U′ ∩ R = 0 if and only if μ′ _{R} = 1_{R}. In this case, μ′ = μ and ρ′ = ρ. If R is anisotropic, then O_{r}(R) has no non-trivial unipotent elements, and this condition is met. The following uniqueness criterion is a special case of our discussion.

Proposition 2. (Uniqueness) Let σ = μρ be the Zassenhaus splitting of σ ∈ O_{n}(V). Suppose that σ = μ′ρ′ where μ′ is unipotent, ρ′ non-degenerate, and S = U′ ⊥ R′. Then

μ′ = μ and ρ′ = ρ .

Proposition 3. (Conjugacy) Let σ and σ_{1} be elements in O_{n}(V) and let σ = μρ and σ_{1} = μ_{1}ρ_{1} be their Zassenhaus splittings. Then σ_{1} is conjugate to σ if and only if μ_{1} is conjugate to μ and ρ_{1} is conjugate to ρ.

Proof: If σ_{1} is conjugate to σ then by an application of Proposition 2, μ_{1} is conjugate to μ and ρ_{1} is conjugate to ρ. As to the converse, observe first that S_{σ} = U_{μ} ⊥ R_{ρ} = R_{ρ} ⊥ U_{μ} and similarly for S_{σ_{1}}. If μ_{1} is conjugate to μ and ρ_{1} is conjugate to ρ, then U_{μ_{1}} is isometric to U_{μ} and R_{ρ_{1}} is isometric to R_{ρ}. Therefore S_{σ_{1}} is isometric to S_{σ}, and hence σ_{1} is conjugate to σ. QED.

3. Application to the Length Problem. Our study of the length problem for the group Ω_{n}(V) and its set of generators

A = {τ_{v}τ_{w}τ_{v}τ_{w}|τ_{v} and τ_{w} non-commuting hyperplane reflections in O_{n}(V)} will expand on the results of Hahn [6].

Let σ ∈ Ω_{n}(V) be arbitrary. Note that the typical element τ_{v}τ_{w}τ_{v}τ_{w} in A is equal to τ_{v}τ_{τ_{w}(v)} = τ_{v}τ_{v′} where F v ≠ F v′ and Q(v) = Q(v′). Conversely, any such product is an element in A. It is a direct consequence of this fact and Theorem 1, that

ℓ(σ) ≥ \frac{1}{2} dim S.

We will therefore define σ ∈ Ω_{n}(V) to be short if ℓ(σ) = \frac{1}{2} dim S and long if ℓ(σ) > \frac{1}{2} dim S.

Our goal is the same as that of Theorem 1, namely the complete description of the long elements of Ω_{n}(V) and the determination of their lengths.

Let σ in Ω_{n}(V) be an involution. By an application of the Wall form, σ is short if and only if S = W_{1} ⊥ ⋯ ⊥ W_{k} with dim W_{i} = 2 and disc W_{i} = 1. Totally degenerate elements are in Ω_{n}(V). It follows from Theorem 1 that they are long. We now focus on the elements in Ω_{n}(V) that are neither involutions nor totally degenerate.

Theorem 2. Let σ ∈ Ω_{n}(V) be long with σ neither totally degenerate nor an involution. Let σ = μρ be the Zassenhaus splitting of σ. Then

i) The space of μ satisfies U = Rad U ⊥ T with T anisotropic. The element μ is a product of \frac{1}{2}(dim U) commuting Eichler transformations, exactly dim T of which are not totally degenerate. In particular, (μ − 1_{V})^{3} = 0.

ii) The element ρ is long and its space R is anisotropic.

iii) (Splicing Condition) The space T ⊥ R is anisotropic.

Finally, if V is isotropic, then ℓ(σ) = \frac{1}{2} dim S + 1.

Proof: In view of Hahn [6] and in particular Proposition 15, only the existence of the factorization in (i) requires proof. By the same proposition, we know that (μ − 1_{V})U ⊆ Rad U. If T = 0, then μ is totally degenerate. By Hahn-O’Meara [5], μ is a product of \frac{1}{2}(dim U) totally degenerate commuting Eichler transformations. So we may assume that T ≠ 0. Let w_{1} ∈ T be non-zero. If μw_{1} = w_{1}, then w_{1} is in the fixed space U^{⊥} of μ. But this implies that w_{1} ∈ U ∩ U^{⊥} = Rad U, a contradiction. So μw_{1} − w_{1} is a non-zero vector in Rad U. Put μw_{1} = u_{1} + w_{1} with u_{1} ∈ Rad U. Note that μ(F u_{1} ⊥ F w_{1}) = F u_{1} ⊥ F w_{1} and (because μ is unipotent) that the restriction of μ to this plane has determinant 1. Let α_{1} = B(w_{1}, w_{1})^{−1} and consider the Eichler transformation Σ_{u_{1},α_{1}w_{1}}. Check that Σ_{u_{1},α_{1}w_{1}}(u_{1}) = u_{1} and Σ_{u_{1},α_{1}w_{1}}(w_{1}) = w_{1} + u_{1}. Observe that μΣ_{u_{1},α_{1}w_{1}}^{−1} _{(F u_{1}⊥F w_{1})} = 1_{(F u_{1}⊥F w_{1})}. By 8.2.16 of [5],

μ commutes with Σ_{u_{1},α_{1}w_{1}}^{−1}. Put μ_{1} = Σ_{u_{1},α_{1}w_{1}}^{−1}μ. By general facts, U_{1} ⊆ (F u_{1} ⊥ F w_{1}) + U ⊆ U. Because μ_{1} fixes w_{1} while μ does not, the fixed space U_{1}^{⊥} of μ_{1} strictly contains the fixed space U^{⊥} of μ. It follows that dim U_{1} = dim U − 2. Because μ = Σ_{u_{1},α_{1}w_{1}} · μ_{1}, U ⊆ (F u_{1} ⊥ F w_{1}) + U_{1}. By dimensions, U = (F u_{1} ⊥ F w_{1}) ⊕ U_{1}. Since Σ_{u_{1},α_{1}w_{1}} commutes with μ, it commmutes with μ_{1}. Therefore, U = (F u_{1} ⊥ F w_{1}) ⊥ U_{1}. Note that Rad U = F u_{1} ⊥ Rad U_{1}. Put U_{1} = Rad U_{1} ⊥ T_{1}. Because

U = (F u_{1} ⊥ Rad U_{1}) ⊥ (F w_{1} ⊥ T_{1})

is a radical splitting of U we know that F w_{1} ⊥ T_{1} is isometric to T and hence that T_{1} is anisotropic. The element μ_{1} is unipotent because it is a product of two commuting unipotent elements. An induction completes the proof. QED.

Remark: By dimension considerations, the spaces of the Eichler transformations in Theorem 2 (i) are planes with trivial intersection. Because these Eichler transformations commute, these planes are orthogonal. Observe also that \frac{1}{2}dim U ≥ dim T, and hence that dim Rad U ≥ dim T.

The next two results will show that the limitations that Theorem 2 imposes on the components μ and ρ of the Zassenhaus splitting of a long element σ are considerable.

Let p(X) = a_{k}X^{k} + ⋯ + a_{1}X + a_{0} be a polynomial in F [X]. We call p(X) symmetric if the two sequences of coefficients a_{k}, …, a_{0} and a_{0}, …, a_{k} are identical.

Theorem 3. Let σ ∈ Ω_{n}(V) be long. Then the prime decomposition of the minimal polynomial of σ has the form

(X − 1)^{m} p_{1}(X) ⋯ p_{j}(X)

where 0 ≤ m ≤ 3 and the p_{i}(X) are distinct, monic, symmetric, and irreducible.

Proof: If σ is an involution or totally degenerate, this is clear. So assume that Theorem 2 applies to σ. Consider ρ_{R}. Because R is anisotropic, any non-zero subspace W of R is non-degenerate. It follows that R = W_{1} ⊥ ⋯ ⊥ W_{j} where each W_{i} is invariant under ρ, but ρ_{W_{i}} has no non-trivial invariant subspaces. By applying the results of Huppert, e.g., Satz 2.4 of [8] and Satz 4.1 of [9], (also see the references to Cikunov in Milnor [14]), we see that the minimal polynomial of ρ_{W_{i}} is symmetric and irreducible. QED.

Proposition 4. Let i be the Witt index of V. Let σ ∈ Ω_{n}(V) be a unipotent element with minimal polynomial (X − 1)^{m}. i) If V is hyperbolic, then m ≤ 2i − 1.

ii) If V is not hyperbolic, then m ≤ 2i + 1.

In either case, there exist unipotent elements such that equality holds.

Proof: The inequalities follow by induction on i. The case i = 0 is the anisotropic case, where we already know that 1_{V} is the only unipotent element. So assume that i ≥ 1. Now refer to case (2) of the proof of Theorem 2.4 of [4] and in particular to the unipotent element τ = σ_{Y} ∈ Ω_{n−2}(X). Let k be the degree of the minimal polynomial of τ. Notice that V = Z ⊥ X with Z a hyperbolic plane. So the Witt index of X is i − 1. Applying the induction hypothesis to τ, provides the inequality k ≤ 2(i − 1) − 1 = 2i − 3 if X is hyperbolic and k ≤ 2(i − 1) + 1 = 2i − 1 if not. It follows from the way σ and τ are related that m ≤ k + 2. This completes the proof of the inequalities. The construction of the required elements also follows inductively. If V is a hyperbolic plane, then 1_{V} is the only unipotent element and it satisfies the equality trivially. Note next that a hyperbolic space of Witt index i contains a non-degenerate space of Witt index i − 1 that is not hyperbolic. This implies that it suffices to carry out the construction of the required unipotent element in case (ii). So suppose that V is not hyperbolic with Witt index i. To get the induction off the ground, take i = 1. Let σ = Σ_{u,v} be a non-degenerate Eichler transformation. Then m = 3 = 2i + 1 as required. It is also easy to check that (σ − 1_{V})^{2}V = F u. Because V is spanned by isotropic vectors, there is an isotropic vector w in V such that (σ − 1_{V})^{2}w = u. Because i = 1, it follows that B(σ(σ − 1_{V})^{2}w, w) = B(u, w) ≠ 0. Suppose that i ≥ 2 and let V = H ⊥ W with H a hyperbolic plane. Note that W is not hyperbolic and that it has Witt index i − 1. For the induction hypothesis, assume that τ is a unipotent element in Ω_{n−2}(W) and that the minimal polynomial of τ is (X − 1)^{k} with k = 2(i − 1) + 1 = 2i − 1. Assume further that (τ −1_{W})^{k−1}W is a line spanned by (τ −1_{W})^{k−1}w with w isotropic and B(τ(τ − 1)^{k−1}w, w) ≠ 0. Put H = F u ⊕ F v with u and v isotropic and B(u, v) = 1. To complete the proof, we will show that σ = Σ_{u,w} · (1_{H} ⊥ τ) is a unipotent element in Ω_{n}(V) that satisfies all the properties of τ with k + 2 in place of k. ¿From the defining equation of Σ_{u,w} we see that σu−u = 0, σv−v = w, and that

σx − x = τx − x + B(τx, w)u for all x ∈ W .

This formula and an induction shows that

(σ − 1_{V})^{j}x = (τ − 1_{W})^{j}x + B(τ(τ − 1_{W})^{j−1}x, w)u for all x ∈ W and j ≥ 1 .

We claim that σ has minimal polynomial (X − 1)^{k+2}, that (σ − 1_{V})^{k+1}V is spanned by (σ−1_{V})^{k+1}v, and that B(σ(σ−1_{V})^{k+1}v, v) ≠ 0. To see this, observe first that (σ − 1_{V})^{k+1}x = 0 for all x ∈ W. Because (σ − 1_{V})u = 0, it follows that (σ − 1_{V})^{k+1}V is spanned by (σ − 1_{V})^{k+1}v. Recall that (σ − 1_{V})v = w. Therefore,

(σ − 1_{V})^{k+1}v = (σ − 1_{V})^{k}w
= (τ − 1_{W})^{k}w + B(τ(τ − 1_{W})^{k−1}w, w)u
= B(τ(τ − 1_{W})^{k−1}w, w)u ≠ 0 .

Because σu = u, we see that (σ − 1_{V})^{k+2}v = 0 and hence that σ has minimal polynomial (X − 1)^{k+2}. Finally, B(σ(σ − 1_{V})^{k+1}v, v) = B(B(τ(τ − 1_{W})^{k−1}w, w)u, v) = B(τ(τ − 1_{W})^{k−1}w, w)B(u, v) ≠ 0. The proof is complete. QED.

The elements of Theorem 2 can be constructed as follows: Start with a long anisotropic ρ in Ω_{n}(V). Choose a subspace U = Rad U ⊥ T in R^{⊥} such that T ⊥ R is anisotropic and dim Rad U ≥ dim T. Split U into an orthogonal sum of degenerate planes. For each plane choose an Eichler transformation that has the plane as its space. Let μ be the product of these Eichler transformations and set σ = μρ. This - by its uniqueness property - is the Zassenhaus decomposition of σ. Therefore, the description of the long elements of Ω_{n}(V) has been reduced to the following two problems:

A. Classify all long anisotropic elements ρ in Ω_{n}(V) and compute their lengths in the case of an anisotropic V.

B. Determine which of the elements in Theorem 2 are actually long.

Notice that the conjugates of a long element in Ω_{n}(V) are long elements of Ω_{n}(V). If the long element is anisotropic, then the conjugates are also anisotropic. Thus, the problem of classifying long elements calls for the classification of their conjugacy classes.

4. Local Fields. The study of the arithmetic theory of quadratic forms flows classically via the progression

ℂ, ℝ, finite fields, local fields, and global fields

from the easy situations to the hard ones. The theory over local fields makes use of that over finite fields (via the residue class field) and the theory over global fields - in characteristic zero these are the finite extensions of ℚ - is based via local/global principles on the theory over local fields and ℂ and ℝ.

The benefit of hindsight, namely that the length problem that is being considered depends on the arithmetic of the field, suggests that its analysis should proceed along the same path. Theorem 4 below is a routine application of Theorem 1. See Hahn [6] for the details. Observe that it applies at once to ℂ, ℝ, and finite fields.

Theorem 4. Suppose that card F^{∗} /F^{2}^{∗} ≤ 2. Then the totally degenerate elements σ are the only long elements in Ω_{n}(V), and for these ℓ(σ) = \frac{1}{2} dim S + 1.

Let’s turn next to the case of a local field. ℝ^{+} be the set of positive real numbers. A local field is a field F that has a valuation

| | : F ⟶ ℝ^{+} ∪ {0}

which satisfies the strong triangle inequality and with respect to which |F^{∗} | is discrete and F is complete. Let

𝔬 = {α ∈ F | |α| ≤ 1}

be the valuation ring of F and 𝔭 = {α ∈ F | |α| < 1} its unique maximal ideal. As part of the definition of local field, the residue class field 𝔬/𝔭 is assumed to be finite. We continue the assumption that char F ≠ 2. Denote by 𝔲 = {ε ∈ 𝔬 | |ε| = 1} the group of invertible elements of 𝔬. Because the maximal ideal 𝔭 is principal, 𝔭 = 𝔬π for some π ∈ 𝔬. Any such π is a prime element in 𝔬. Note that |π| is the largest value such that |π| < 1.

We refer to O’Meara [15] for the notation and the basic properties of local fields, their quadratic forms and orthogonal groups. Two important facts about quadratic forms over local fields are these: any non-degenerate quadratic space of dimension five or more is isotropic, and there is, up to isometry, a unique anisotropic four dimensional quadratic space.

It will be necessary to distinguish non-dyadic local fields from dyadic local fields. The local field F is non-dyadic if 2 is invertible in 𝔬 and dyadic if not. So F is non-dyadic if |2| = 1 and dyadic if |2| < 1. If V is the unique 4dimensional anisotropic quadratic space, then Ω_{4}(V) has index two in O′_{4}(V) if F is non-dyadic, and Ω_{4}(V) = O′_{4}(V) if F is dyadic.

Consider a long element σ ∈ Ω_{n}(V) that is neither totally degenerate nor an involution and return to the properties of the Zassenhaus decomposition σ = μρ provided by Theorem 2. The fact that ρ is long implies that dim R ≥ 4. Therefore,

4 ≤ dim R ≤ dim (T ⊥ R) ≤ 4 .

So T = 0, and μ is totally degenerate. In reference to Theorem 3, it follows that the bound on m is 0 ≤ m ≤ 2. Also, dim R = 4 and R is the unique 4-dimensional anisotropic space over F. What else can be said about ρ?

Proposition 5. Let ρ in Ω_{n}(V) be anisotropic with dim R = 4. Then ρ_{R} ∈ O′_{4}(R), and

i) If F is non-dyadic, then ρ is long if and only if n ≥ 5 and ρ_{R} ∈ O′_{4}(R) − Ω_{4}(R) .

ii) If F is dyadic, then ρ is long if and only if ρ_{R} ∈ O′_{4}(R) = Ω_{4}(R) is long.

This initial answer to Question A is proved in [6]. Suppose that F is non-dyadic. Then Proposition 5 together with Theorem 2 tell us that ℓ(σ) = \frac{1}{2}dim S +1 for any long element σ. Proposition 5 also provides a complete answer to Question B. See Theorem 3 of [6]. It asserts that the long elements in Ω_{n}(V) that are not involutions and not totally degenerate are those of Theorem 2, namely they are precisely the elements with Zassenhaus decomposition σ = μρ, where μ = 1_{V} or μ is totally degenerate and ρ satisfies (i) above. If F is dyadic, then Question B is as yet not resolved. However, it is known that ℓ(σ) = \frac{1}{2}dim S + 1 for all long elements σ.

Let V be the anisotropic 4-dimensional space over F. In view of Proposition 5 we will analyze the elements σ in O′_{4}(V) that satisfy

(a) σ in O′_{4}(V) − Ω_{4}(V) if F is non-dyadic, and

(b) σ a long element in O′_{4}(V) = Ω_{4}(V) if F is dyadic.

Is there a criterion that pinpoints when an element in O′_{4}(V) satisfies (a) or (b)? A theorem of Milnor [14] tells us where to look.

Theorem 5. Let V be an n-dimensional, non-degenerate quadratic space over a local field F. Let m(X) be a monic, irreducible polynomial in F [X] and let deg m(X) = k. Assume that m(X) is neither X − 1 nor X + 1.

i) m(X) is the minimal polynomial of an element of O_{n}(V) if and only if k is

even and divides n, m(X) is symmetric, and disc V = (m(1)m(−1))^{\frac{n}{k}} F^{2}^{∗}.

ii) Given such a polynomial m(X) there is precisely one conjugacy class of elements in O_{n}(V) with minimal polynomial m(X).

Milnor’s result no longer holds when m(X) is reducible. Any Eichler transformation that is not totally degenerate has minimal polynomial (X − 1)^{3} and provides an example showing that (i) no longer holds. The nontrivial totally degenerate elements - all of which have minimal polynomial (X − 1)^{2} - show that (ii) fails. Let μ and μ_{1} be totally degenerate elements. Then μ is conjugate to μ_{1} if and only if their respective spaces U and U_{1} have the same dimension. This follows from the conjugacy criterion given by the Wall form and the fact U_{μ} and U_{μ_{1}} are both alternating.

Let σ ∈ O′_{4}(V) and let m(X) be its minimal polynomial. Milnor’s theorem suggests that it should be possible to look at m(X) and decide whether σ satisfies condition (a) or (b) or not. In reference to Theorem 3, we are interested in precise information about the product p_{1}(X) ⋯ p_{k}(X). In the discussion that follows, only the arithmetic aspects of the proofs will be provided.

Proposition 6. Let V be anisotropic with dim V = 4 and let σ ∈ O′_{4}(V). Then σ satisfies (a) or (b) if and only if σ satisfies the long criterion:

Q(σx − x) = −β_{x}^{2}Q(x) for all x ∈ V and some β_{x} ∈ F^{∗} .

Suppose that m(X) has a factor of the form X − a. So σx = ax for some non-zero x in V. Because x is anisotropic, a = ±1. Since σx = x violates the long criterion, we must have σx = −x. So a = −1. Therefore, X + 1 is the only possible monic linear factor of m(X). If (X + 1)^{2} is a factor of m(X), then −σ is a non-trivial unipotent element on some subspace of V. But this is impossible, because V is anisotropic.

1. Suppose deg m(X) = 1. This implies that m(X) = X + 1 and hence that σ = −1_{V}. Because disc V = 1, −1_{V} ∈ O′_{4}(V). Check that σ = −1_{V}

satisfies the long criterion precisely when −1 ∈ F^{2}^{∗}.

2. Suppose deg m(X) = 2. Observe that m(X) must be irreducible. By Theorem 5, m(X) = X^{2} − cX + 1 for some c ∈ F. Notice that c ≠ ±2. Every line of V contains a plane that is invariant under σ. Let

W be any such plane. By Theorem 5, disc W = −(c − 2)(c + 2)F^{2}^{∗}. Because V is anisotropic, W is not a hyperbolic plane, and therefore,

(c−2)(c+2) ∉ F^{2}^{∗}. Again by Theorem 5, there is precisely one conjugacy class of such elements σ for a given c. A spinor norm computation shows

that Θ(σ) = (c − 2)^{2}F^{2}^{∗} = F^{2}^{∗}. So any σ with minimal polynomial of this form is in O′_{4}(V). It turns out that σ satisfies the long criterion if and

only if c − 2 ∈ F^{2}^{∗}.

3. Suppose deg m(X) = 3. By Theorem 5, m(X) is reducible. It follows that m(X) = (X + 1)(X^{2} − cX + 1) with X^{2} − cX + 1 irreducible. Again, c ≠ ±2. Let p_{1}(X) = X + 1 and p_{2}(X) = X^{2} − cX + 1. Put U = p_{2}(σ)V and W = p_{1}(σ)V. Observe that U and W are planes that are invariant under σ, that V = U ⊥ W, that σ_{U} = −1_{U}, and that σ_{W} has minimal polynomial X^{2} − cX + 1. As in the previous case,

disc W = −(c − 2)(c + 2)F^{2}^{∗} and (c − 2)(c + 2) ∉ F^{2}^{∗}. By Theorem 63:20 of [15], there are two isometry classes of anisotropic planes of a given discriminant. An application of Theorem 5 implies that there are two conjugacy classes of σ for a given c. By a spinor norm computation,

Θ(σ) = −(c − 2)F^{2}^{∗}. So σ ∈ O′_{4}(V) if and only −(c − 2) ∈ F^{2}^{∗} and

−(c + 2) ∉ F^{2}^{∗}. The analysis of the long criterion for σ ∈ O′_{4}(V) will

follow shortly. We will see that if it holds, then −1 ∈ F^{2}^{∗} and c − 2 ∈ 4𝔲^{2}. This implies in turn that c ∈ 2𝔲. If F is non-dyadic, the converse is true.

Namely, −1 ∈ F^{2}^{∗} and c − 2 ∈ 𝔲^{2} together imply the long criterion.

4. Suppose deg m(X) = 4. In this case, either

m(X) = (X^{2} − cX + 1)(X^{2} − dX + 1) with distinct irreducible factors, or

m(X) = X^{4} − cX^{3} − dX^{2} − cX + 1 is irreducible.

The first case is very similar to case (3). The second seems complicated and is as yet not completely understood.

We now return to case (3) and to the analysis of the long criterion. Let U̇ and Ẇ denote the non-zero elements of U and W and let

C = Q(Ẇ)/Q(U̇) .

The set C is closed under multiplication by squares and hence under taking inverses.

Assume that the long criterion holds. Applying it to U and W we get that

(i) − 1 and c − 2 are both in F^{2}^{∗}.

Put −1 = i^{2} and c − 2 = s^{2} and let t = −2is^{−1}. Applying the long criterion to the vectors x = u + w with u ∈ U and w ∈ W, tells us that

(ii) \frac{1 + γt^{2}}{1 + γ} ∈ F^{2}^{∗} for all γ ∈ C .

Conversely, the long criterion is equivalent to the combination of (i) and (ii).

We assume that (i) and (ii) hold and consider the consequences for the constant c. We show first that t ∈ 𝔲. It follows from the discussion in paragraph 63.C of O’Meara [15] that C contains a prime element π. Therefore C contains π^{i} for any odd i either positive or negative. Put t = δπ^{k} with δ ∈ 𝔲. Taking γ = π we get,

\frac{1 + γt^{2}}{1 + γ} = \frac{1 + δ^{2}π^{2k+1}}{1 + π} .

If k < 0, then 2k + 1 < 0, and \frac{|1+δ^{2}π^{2k+1}|}{|1+π|} = |π|^{2k+1} by the Principle of Domination. Because 2k + 1 is odd, the element above cannot be a square. This contradicts (ii). If k > 0, a similar contradiction is obtained by taking γ = π^{−1}. Therefore, k = 0 and t ∈ 𝔲 as required. Let ε = −it^{−1} ∈ 𝔲. Because s = 2ε, we get

c − 2 = 4ε^{2}.

Therefore, c − 2 ∈ 4𝔲^{2} as asserted earlier. Note that c = 2(2ε^{2} + 1). If F is dyadic, then by domination, c ∈ 2𝔲. This is also true in the non-dyadic case. If c ∉ 𝔲, then c ∈ 𝔭. But this would imply by Hensel’s Lemma that X^{2} − cX + 1 is reducible.

We now explore the converse. Assume both −1 ∈ F^{2}^{∗} and c − 2 ∈ 4𝔲^{2}. Does σ ∈ O′_{4}(V) with such a c satisfy the long criterion, or equivalently, conditions (i) and (ii)? Condition (i) holds trivially, so the focus is on (ii). Put c−2 = 4ε^{2} with ε ∈ 𝔲. Set s = 2ε and t = −2is^{−1} = −iε^{−1}. Notice that t ∈ 𝔲. Because C

is closed under taking inverses, (ii) is equivalent to 1 + \frac{t^{2}−1}{1+γ} ∈ F^{2}^{∗} for all γ in

C. Check that t^{2} − 1 = −\frac{4+s^{2}}{s^{2}} = −\frac{c+2}{c−2} = −\frac{c+2}{4ε^{2}} . So the question is this: Is it the case that

(iii) 1 − \frac{c + 2}{4ε^{2}(1 + γ)} ∈ F^{2}^{∗}

for all γ ∈ C?

The first step toward the answer is the observation that {|1 + γ| | γ ∈ C} is bounded below by |4|. For suppose that |1 + γ| ≤ |4π| for some γ ∈ C. Then

1 + γ = 4απ for some α ∈ 𝔬. But this means that −γ = 1 − 4απ ∈ F^{2}^{∗} by the Local Square Theorem. Because C is closed under multiplication by squares, −γ ∈ C. But this implies that the intersection Q(U̇) ∩ Q(Ẇ) is not empty.

This would mean that V contains a plane of discriminant F^{2}^{∗} = −F^{2}^{∗}, i.e., a hyperbolic plane. This is not possible because V is anisotropic. Now assume that |c + 2| < |4|^{3}. Given the bound just established, |\frac{c+2}{4ε^{2}(1+γ)}| < |4| for all γ ∈ C. Therefore by another application of the Local Square Theorem,

1 − \frac{c + 2}{4ε^{2}(1 + γ)} ∈ F^{2}^{∗}

for all γ in C.

We conclude the discussion of the converse by assuming that F is non-dyadic. In this case (iii) is satisfied for any c (such that c−2 ∈ 4𝔲^{2}). Because |4| = 1, we already know that (iii) holds when c+2 ∈ 𝔭. Since c+2 ∈ 𝔬, only the case c+2 ∈ 𝔲 remains. Instead of (iii), we will verify the equivalent condition (ii). Recall from the beginning of the analysis of case (3) that disc W = −(c − 2)(c + 2)F^{2}^{∗}

and −(c+2) ∉ F^{2}^{∗}. So disc W = −(c+2)F^{2}^{∗} and, by an application of Example

63:15 of [15], C = π𝔲F^{2}^{∗}. Let γ = πδα^{2} ∈ C with δ ∈ 𝔲 and α ∈ F^{∗} be arbitrary.

If |α| ≤ 1, then |γ| < 1. So 1 + γ and 1 + t^{2}γ are both in F^{2}^{∗} by the Local Square Theorem. Therefore (ii) holds. Suppose |α| > 1. Now |γ| > 1 and the Local Square Theorem tells us that 1 + γ^{−1} and 1 + t^{−2}γ^{−1} are both squares. So \frac{1+t^{−2}γ^{−1}}{1+γ^{−1}} is a square. Therefore \frac{1+γt^{2}}{1+γ} is a square as well and (ii) holds in this case also. The proof of the converse in the non-dyadic case is complete. The dyadic situation is much more delicate and is not completely settled.

5. Global Fields. Let F be a global field, let V be a non-degenerate quadratic space over F, and consider the group Ω_{n}(V). Not much is known about the length question in this situation, but it is clear that local-global considerations are relevant. Let 𝔭 be a prime - Archimedean or not - and consider the completion V_{𝔭}. The first indication is the theorem that tells us that σ ∈ Ω_{n}(V) if and only if σ_{𝔭} ∈ Ω_{n}(V_{𝔭}) for all 𝔭. Another is the fact (analogous to what was observed in the local case) that the analysis of the anisotropic long elements in Ω_{n}(V) reduces to the 4-dimensional anisotropic long elements. This is true not only in the situation where F is a function field or a totally complex number field (in these situations there are no anisotropic spaces of dimension 5 or more) but in general. More precisely, if σ is an anisotropic long element, then

σ = ω_{1} ⋯ ω_{k}σ_{1},

where all ω_{i} are elementary commutators of hyperplane reflections, the space S = W_{1} ⊕ ⋯ ⊕ W_{k} ⊕ S_{1}, and σ_{1} is long with dim S_{1} = 4.

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Alexander Hahn

Department of Mathematics University of Notre Dame Notre Dame, IN 46556 USA hahn.1@nd.edu 182

Documenta Mathematica · Quadratic Forms LSU 2001