Abstract

For a quadratic Jordan algebra J , we define the concept of a ring of fractions for J with respect to an Ore monad S ⊆ J . We then find necessary and sufficient conditions on J for the existence of such a ring of fractions.

1 Introduction

Let R be an associative ring and let S ⊆ R be a subset of R which is closed under multiplication and which contains only regular elements. (By regular elements here we mean non-zero divisors). If R ⊆ Q for some overring Q, then Q is called a right ring of fractions for R with respect to the subset S if the following two conditions hold:

(i) each s ∈ S has s^{−1} ∈ Q

(ii) for each q ∈ Q there exists a ∈ R and s ∈ S such that q = as^{−1}.

Ore found a necessary and sufficient condition on R for the existence of a right ring of fractions [6]. This condition is referred to as the right Ore condition and can be stated as follows:

The Ore condition If R is an associative ring and S ⊆ R is a subset of R which is closed under multiplication and which contains only regular elements, then R is said to satisfy the right Ore condition with respect to S if for a ∈ R and s ∈ S, there exists a′ ∈ R and s′ ∈ S such that as′ = sa′.

Building on the work of Jacobson, McCrimmon, and Parvathi [2], and new concepts of Zel’manov, Martinez exhibited a necessary and sufficient Ore-type condition for an arbitrary linear Jordan algebra (with no 2 or 3 torsion) to have a ring of fractions [4]. In this paper, we extend these ideas and exhibit a necessary and sufficient Ore-type condition for a quadratic Jordan algebra to have a ring of fractions.

2 Background Material

Crucial to this paper will be concepts relating to quadratic Jordan algebras, Lie algebras, and Jordan pairs. We pause here to discuss the basic facts about these three structures.

2.1 Quadratic Jordan Algebras

A unital quadratic Jordan algebra J = (J, U, 1) over an arbitrary ring of scalars Φ, is a Φ-module J containing a distinguished element 1, and having a quadratic map U : J → End_{Φ}(J) such that

(QJ1) U_{1} = Id

(QJ2) U_{x}V_{y,x} = V_{x,y}U_{x} = U_{U_{x}y,x}

(QJ3) U_{U_{x}y} = U_{x}U_{y}U_{x} hold strictly in the sense that they continue to hold in all (free) scalar extensions. Equivalently, J is a quadratic Jordan algebra if (QJ1)-(QJ3) and all linearizations hold in J. To clarify the above notation, we have

V_{x,y}(z) = {x, y, z} = U_{x,z}(y)

where

U_{x,z} = U_{x+z} − U_{x} − U_{z}.

A quadratic Jordan algebra is just a subspace J = (J, U, 2) of some unital quadratic Jordan algebra, closed under the U-operator U_{x}y, and a squaring operation x^{2}, where U_{x}1 = x^{2}. We should mention that, given any quadratic Jordan algebra J, we can imbed J in its free unital hull Ĵ where

Ĵ = Φ1 ⊕ J

and

U_{α1⊕x}(β1 ⊕ y) = α^{2}β ⊕ {α^{2}y + 2αβx + αx ∘ y + βx^{2} + U_{x}y}.

Here the ∘ product denotes the linearization of the square

V_{x}y = x ∘ y = U_{x,y}1.

Unless otherwise stated, we will deal with quadratic Jordan algebras not necessarily containing an element 1. When needed, we will introduce the free unital hull Ĵ as described above.

There is a special type of subspace we will consider in the quadratic Jordan setting, called an inner ideal. An inner ideal B of a quadratic Jordan algebra J is a linear subspace of J closed under inner multiplication by Ĵ:

U_{B}Ĵ ⊆ B.

We will be especially interested in the principal inner ideal

K_{s} = Φs + Φs^{2} + U_{s}J = Φs + U_{s}Ĵ

where s ∈ J. In this case notice that

U_{K_{s}} Ĵ ⊆ U_{s}Ĵ

since

U_{αs+U_{s}â} = U_{s}B_{α,a,s} = B_{α,s,a}U_{s}

for

B_{α,a,b} = α^{2}I + αV_{a,b} + U_{a}U_{b}.

2.2 Lie Algebras

Let Φ be a unital, commutative associative ring and let L be a Φ-module, with an operation L × L → L, denoted (x, y) ↦ [x, y] and called the bracket of x and y. L is called a Lie algebra over Φ if

(L1) [ , ] is bilinear

(L2) [x, x] = 0 for all x ∈ L.

(L3) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 (x, y, z ∈ L).

A linear map Δ : L → L is called a derivation of Lie algebra L if, for all x, y ∈ L,

Δ[x, y] = [Δ(x), y] + [x, Δ(y)].

In the Jordan pair discussion that follows, we will detail the specific Tits-Kantor-Koecher construction of a Lie algebra from a given Jordan pair.

2.3 Jordan Pairs

Let Φ be a unital, commutative associative ring and let V = (V^{+}, V^{−}) be a pair of Φ-modules. Assume that there exist compositions

(x, y) ↦ Q_{x}(y) = Q(x)(y) ∈ V^{σ}

for (x, y) ∈ V^{σ} × V^{−σ}, σ = ±1, which are quadratic in x and linear in y. We set

Q(x, z) = Q_{x,z} = Q_{x+z} − Q_{x} − Q_{z}

which gives the trilinear product {xyz} = Q_{x,z}(y) with {V^{σ}V^{−σ}V^{σ}} ⊆ V^{σ}. We also set

D(x, y)(z) = D_{x,y}(z) = {xyz}.

If Ω is an extension ring of Φ, we can form

V_{Ω} = (V_{Ω}^{+}, V_{Ω}^{−}) = (Ω ⊗_{Φ} V^{+}, Ω ⊗_{Φ} V^{−}) ,

extending the compositions (x, y) ↦ Q_{x}(y) to V_{Ω}^{σ} × V_{Ω}^{−σ}. We say that V = (V^{+}, V^{−}) is a Jordan pair if

(JP1) D_{x,y}Q_{x} = Q_{x}D_{y,x}

(JP2) D(Q_{x}(y), y) = D(x, Q_{y}(x))

(JP3) Q(Q_{x}(y)) = Q_{x}Q_{y}Q_{x}

hold in V_{Ω} for all extensions Ω of Φ. Equivalently, V is a Jordan pair if (JP1)-(JP3) and all linearizations hold in V.

A pair of linear maps D = (D_{+}, D_{−}) with D_{σ} : V^{σ} → V^{σ} is a derivation of V if

D_{σ}Q_{x} = Q_{x}D_{−σ} + Q(D_{σ}(x), x).

The set Der(V) of all derivations of V forms a Lie algebra. Furthermore, if

δ_{x,y} = (D_{x,y}, −D_{y,x}),

then δ_{x,y} ∈ Der(V) and the subspace Inder(V) of inner derivations spanned by all δ_{x,y}, where x ∈ V^{+} and y ∈ V^{−}, is an ideal in Der(V).

Our definition of derivations as pairs (D_{+}, D_{−}) and inner derivations as

δ_{x,y} = (D_{x_{+},y_{−}}, −D_{y_{−},x_{+}})

is asymmetric in the indices ±. It is often convenient to treat the indices symmetrically (really this means working in the polarized triple system T (V) = V^{+} ⊕ V^{−} ) and to think of a derivation as

D = D_{+} ⊕ D_{−}

and

δ_{x_{+},y_{−}} = V_{x,y} ⊕ (−V_{y,x})

as a polarized derivation. In this context we define

δ_{y_{−},x_{+}} = \{ \begin{matrix}D_{y_{−},x_{+}} & on V^{−} \\ −D_{x_{+},y_{−}} & on V^{+} \end{matrix}

so we have a symmetric formula

δ_{x_{σ},y_{−σ}} = −δ_{y_{−σ},x_{σ}} = \{ \begin{matrix}D_{x_{σ},y_{−σ}} & on V^{σ} \\ −D_{y_{−σ},x_{σ}} & on V^{−σ}. \end{matrix}

i.e.

δ_{x_{σ},y_{−σ}}(a_{σ}) = {x_{σ}, y_{−σ}, a_{σ}}

and

δ_{x_{σ},y−−σ}(b_{−σ}) = −{y_{−σ}, x_{σ}, b_{−σ}}

for a_{σ} ∈ V^{σ} and b_{−σ} ∈ V^{−σ}.

If J is a quadratic Jordan algebra then we can form the Jordan pair

V(J) = (J_{+}, J_{−})

with

Q_{x} = U_{x}, Q_{x,y} = U_{x,y}, D_{x,y} = V_{x,y}.

We define operators on (Ĵ_{+} ⊕ Ĵ_{−} by

δ_{x_{σ},y_{−σ}} = (V_{x_{σ},y_{−σ}}^{σ}, −V_{y_{−σ},x_{σ}}^{−σ}),

where

V_{x_{σ},y_{−σ}}^{σ} : J_{σ} → J_{σ} and V_{y_{−σ},x_{σ}}^{−σ} : J_{−σ} → J_{−σ}

for σ = ±1. Each δ_{x,y} is a derivation for the Jordan pair and we can form D_{0}, the space of operators spanned by all δ_{x,y}. We can then construct the TKK-Lie algebra

£ = T KK(J, D_{0}) = J_{−} ⊕ D_{0} ⊕ J_{+}

with skew-symmetric bracket product defined by

(1) [x_{σ}, y_{σ}] = 0, σ = ±1

(2) [x_{σ}, y_{−σ}] = −δ_{x_{σ},y_{−σ}}, σ = ±1

(3) [D, x_{σ}] = D_{σ}(x_{σ}) for D = (D_{σ}, D_{−σ}) ∈ D_{0}, σ = ±1

(4) [D_{1}, D_{2}] = D_{1}D_{2} − D_{2}D_{1} ∈ D_{0} for D_{1}, D_{2} ∈ D_{0}

(5) [a, b] = −[b, a] for a, b ∈ £. For a quadratic Jordan algebra J (but not for Jordan pairs in general) we have a unital hull Ĵ, with J ◁ Ĵ. This leads to a Lie algebra

ˆ£ = T KK(Ĵ, Struc(Ĵ)) = (Ĵ)_{−1} ⊕ Struc(Ĵ) ⊕ (Ĵ)_{+1}

for

δ_{Ĵ,Ĵ} ⊆ Struc(Ĵ) ⊆ End(Ĵ_{+1} ⊕ Ĵ_{−1}).

The subalgebra

£ = (J)_{−1} ⊕ D̂_{0} ⊕ (J)_{+1} ◁ ˆ£

where

D̂_{0} = Φξ + δ_{J} + δ_{J,J}

for

ξ = 1_{(J)_{+1}} ⊕ (−1_{(J)_{−1}}) and δ_{x} = V_{x} ⊕ (−V_{x}),

is always an ideal of ˆ£. For quadratic Jordan algebras we will take this as our definition of T KK(J).

NOTE: This is not quite the same as

δ_{Ĵ,Ĵ} = δ_{Φ1_{+}+(J)_{+1},Φ1_{−}+(J)_{−1}} = Φδ_{1_{+},1_{−}} + δ_{J} + δ_{J,J}.

Observe,

δ_{1_{+},x_{−1}} = (V_{1,x} − V_{x,1}) = δ_{x}, δ_{x_{+1},1_{−}} = (V_{x,1}, −V_{1,x}) = δ_{x},

and

δ_{1,1} = 2ξ

so we never recover ξ itself. More important is our convention that

δ_{J,J} ⊆ End(Ĵ_{+1} ⊕ Ĵ_{−1}).

The usual T KK applied to the pair V(J) = (J, J) gives D_{00} spanned by all

δ_{x,y} = (V_{x,y}, V_{y,x}) ∈ End(J_{+1} ⊕ J_{−1}).

There is a natural epimorphism

D_{0} → D_{0}|_{J⊕J} = D_{00},

but this need not be an isomorphism since vanishing on J ⊕ J would not necessarily imply vanishing on 1⊕1. It will be convenient to use Struc(Ĵ) in place of Instruc(Ĵ) in defining D̂_{0}. This will allow maps δ_{x̃,y} where x̃ ∈ J̃ (for J ⊆ J̃), and y ∈ J have

{x̃, y, Ĵ} + {y, x̃, Ĵ} ⊆ J.

Since then

δ_{x̃,y} = (W, W^{∗})

lies in

Instruc(J̃) ∩ Hom(Ĵ, Ĵ) ⊆ Struc(Ĵ),

but not in Instruc(Ĵ).

3 The Ore Condition

In this section we discuss the terms and definitions which are fundamental to the work that follows. Along the way we prove some useful lemmas and one important proposition relating to the proposed Ore condition.

3.1 Definitions

We begin by describing the elements that will function as the denominators of our desired fractions. It is worth mentioning that these elements may, at times, exhibit abilities far beyond those of mortal denominators. In such cases, these elements will be called strong denominators.

Definition 1 (3) Let J and Q be two quadratic Jordan algebras and K an inner ideal of J with K ⊆ J ⊆ Q. An element x ∈ K is called a K-denominator for q ∈ Q if

(i) x ∘ q ∈ K

(ii) U_{x}q ∈ K

NOTE: if q ∈ J ⊆ Q, then (ii) is automatic by the innerness of K.

Definition 2 (3) An element x ∈ K is called a strong K-denominator for q ∈ Q if

(i) U_{x}q ∈ K

(ii) U_{q}x ∈ K

(iii) U_{x}U_{q}K̂ ⊆ K

(iv) U_{q}U_{x}K̂ ⊆ K

(v) {x, q, K̂} ⊆ K

(vi) {q, x, K̂} ⊆ K.

At times it may prove helpful to know the following: once one denominator for q ∈ Q is found, many others are immediately available. The following lemma explains why.

Lemma 1 Let J and Q be two quadratic Jordan algebras and K an inner ideal of J with K ⊆ J ⊆ Q. If x ∈ K is a K-denominator for q ∈ Q, then for y = x^{2} and any y ∈ U_{x^{2}} Ĵ, we have the following:

(i) y is a K-denominator for q ∈ Q

(ii) V_{y,q}, V_{q,y} ∈ V_{K,K̂} = V_{K,K} + V_{K}.

Proof Let q ∈ Q and let x ∈ K be a K-denominator for q. Then x ∘ q ∈ K and U_{x}q ∈ K. Therefore,

x^{2} ∘ q = x ∘ (x ∘ q) − 2U_{x}q ∈ K

and

U_{x^{2}}q = U_{x}U_{x}q ∈ U_{x}K ⊆ K,

by the innerness of K. Also, by ??

V_{x^{2},q} = V_{x,x∘q} − V_{U_{x}q} ∈ V_{K,K} − V_{K} ⊆ V_{K,K̂}

and dually

V_{q,x^{2}} ∈ V_{K,K̂}.

Hence, y = x^{2} is a K-denominator for q and has V_{y,q}, V_{q,y} ∈ V_{K,K̂}, as desired. Now, let y = U_{x^{2}} â ∈ U_{x^{2}} Ĵ. Then

y ∘ q = U_{x^{2}}(â) ∘ q
= V_{q}U_{x}(U_{x}â)
= U_{q∘x,x}U_{x}â − U_{x}V_{q}U_{x}â
∈ {K, J, x} − U_{U_{x}q,x^{2}} â
⊆ U_{K,K} Ĵ ⊆ K

and

U_{U_{x^{2}} â}q = U_{x^{2}}UâU_{x^{2}}q = U_{x}(U_{x}UâU_{x})U_{x}q ∈ U_{K}J ⊆ K.

Hence, y = U_{x^{2}} â is a K-denominator for q, for any â ∈ Ĵ. What about V_{y,q}, V_{q,y} ∈ V_{K,K̂} ? Well,

{U_{x}â, x, q} = U_{x}â ∘ (x ∘ q) − V_{x,a}(U_{x}q) ∈ K.

So,

V_{y,q} = V_{U_{x^{2}} â,q}
= V_{U_{x}(U_{x}â),q}
(J9) = −V_{U_{x}q,U_{x}â} + V_{x,{U_{x}â,x,q}}
∈ V_{K,K}
⊆ V_{K,K̂}.

Dually

V_{q,y} ∈ V_{K,K̂},
■

The lemma above concerns our concept of a “normal” denominator. As we continue, strong denominators play an increasingly important role. So, at this point it is convenient to introduce some notation relating to strong denominators. For a particular q ∈ Q, we will let D_{K}^{stg}(q) denote the collection of all strong K-denominators for q. One should note that membership in D_{K}^{stg}(q) is not as restricted as it may appear.

Remark 1 In order for an element x ∈ K to be in D_{K}^{stg}(q), it only need satisfy (i), (ii), (iii) ( or (iv) ) , and (v) ( or (vi) ) of definition 2.

Proof Let q ∈ Q and suppose that x ∈ K satisfies (v) in definition 2. We show that x also satisfies (vi). ( Dually, (vi) ⇒ (v). )

Choosing 1 ∈ K̂, we have

x ∘ q = {x, q, 1} ⊆ K.

Then, by identity (J22),

{q, x, K̂} = V_{x∘q}K̂ − {x, q, K̂} ⊆ K.

This shows that x satisfies (vi).

Now, let q ∈ Q and suppose that x ∈ K satisfies (iii) and (v) ( and hence (vi) ) in definition 2. We show that x also satisfies (iv). ( Dually, (iv) ⇒ (iii). ) Using identity ? we have

U_{q}U_{x}K̂ ⊆ U_{x∘q}K̂ − U_{x}U_{q}K̂ − V_{U_{x}U_{q}1}K̂ − V_{x,q}V_{q,x}K̂.

Looking at the terms on the right hand side of this equation, we can conclude the following:

(1) U_{x∘q}K̂ ⊆ K, since x ∘ q ∈ K.

(2) U_{x}U_{q}K̂ ⊆ K by (iii).

(3) V_{U_{x}U_{q}1}K̂ ⊆ K, since U_{x}U_{q}1 ∈ K by (iii).

(4) V_{x,q}V_{q,x}K̂ ⊆ K by (v) and by (vi).

Since all of these terms are in K, we conclude that U_{q}U_{x}K̂ ⊆ K, and x satisfies (iv) as desired.

■

Another useful fact about D_{K}^{stg}(q) is this: For a given q ∈ Q, D_{K}^{stg}(q) is an inner ideal of K. We prove this now.

Lemma 2 Let J and Q be two quadratic Jordan algebras and K an inner ideal of J with K ⊆ J ⊆ Q. If q ∈ Q, then D_{K}^{stg}(q) is an inner ideal of K.

Proof First we must show that D_{K}^{stg}(q) is a linear subspace of K. If x ∈ D_{K}^{stg}(q) then it is clear that αx ∈ D_{K}^{stg}(q) for α ∈ Φ, the coefficient ring.

Now, suppose that x, y ∈ D_{K}^{stg}(q). We must show that x + y ∈ D_{K}^{stg}(q). We will show this by showing that x + y satisfies (i), (ii), (iii), and (v) of definition 2.

(i): Since x, y ∈ D_{K}^{stg}(q) ⊆ K ⊆ K̂, we know that U_{x,y}(q) = {x, q, y} ∈ K, and U_{x}q, U_{y}q ∈ K. Hence,

U_{x+y}(q) = U_{x,y}q − U_{x}q − U_{y}q ∈ K − K − K ⊆ K.

(ii): U_{q}(x + y) = U_{q}x + U_{q}y ∈ K, since x, y ∈ D_{K}^{stg}(q).

(iii): Using identity (J10), we see that

U_{x+y}U_{q}K̂ = U_{x,y}U_{q}K̂ − U_{x}U_{q}K̂ − U_{y}U_{q}K̂
= V_{x,q}V_{y,q}K̂ − V_{x,U_{q}y}K̂ − U_{x}U_{q}K̂ − U_{y}U_{q}K̂.

We now examine the last four terms in this series of equations.

(1) V_{x,q}V_{y,q}K̂: Since x, y ∈ D_{K}^{stg}(q), we know that

V_{x,q}V_{y,q}K̂ ⊆ V_{x,q}K ⊆ V_{x,q}K̂ ⊆ K.

(2) V_{x,U_{q}y}K̂: Since y ∈ D_{K}^{stg}(q), we know that U_{q}y ∈ K. Hence,

V_{x,U_{q}y}K̂ ⊆ V_{x,K} K̂ ⊆ K.

(3) U_{x}U_{q}K̂: Since x ∈ D_{K}^{stg}(q), it is clear that U_{x}U_{q}K̂ ⊆ K.

(4) U_{y}U_{q}K̂: Since y ∈ D_{K}^{stg}(q), it is clear that U_{y}U_{q}K̂ ⊆ K.

Since each of these terms is in K, we conclude that U_{x+y}U_{q}K̂ ⊆ K as desired.

(v): Since x, y ∈ D_{K}^{stg}(q), we know that {x, q, K̂}, {y, q, K̂} ⊆ K. Therefore,

{x + y, q, K̂} = {x, q, K̂} + {y, q, K̂} ⊆ K + K ⊆ K.

So, x + y ∈ D_{K}^{stg}(q) as desired and we can conclude that D_{K}^{stg}(q) is indeed a linear subspace.

Now, we must verify the innerness of D_{K}^{stg}(q). Let x ∈ D_{K}^{stg}(q) and let a ∈ K̂. We will show that U_{x}a ∈ D_{K}^{stg}(q) by showing that it satisfies (i), (ii), (iii), and (v) of definition 2.

(i): U_{U_{x}a}(q) = U_{x}U_{a}U_{x}q ∈ K, since U_{x}q, x ∈ K, and a ∈ K̂.

(ii): Since x ∈ D_{K}^{stg}(q), we know that U_{q}U_{x}K̂ ⊆ K. Hence U_{q}U_{x}a ∈ K.

(iii): Since x ∈ D_{K}^{stg}(q), we know that U_{x}U_{q}K̂ ⊆ K. Therefore,

U_{U_{x}a}U_{q}K̂ = U_{x}U_{a}U_{x}U_{q}K̂ ⊆ U_{x}U_{a}K ⊆ K,

since x ∈ K and a ∈ K̂.

(v): Since x ∈ D_{K}^{stg}(q), we know that x, {q, x, a}, U_{x}q, a ∈ K̂. Coupling this with identity (J9), we have

{U_{x}a, q, K̂} = V_{U_{x}a,q}K̂ = V_{x,{q,x,a}}K̂ − V_{U_{x}q,a}K̂ ⊆ K.
■

We have made it a point to distinguish between denominators and strong denominators, but this distinction is not terribly crucial. Any plain denominator can become strong by taking higher powers of itself. This transformation is explained in the following lemma.

Lemma 3 Let J and Q be two quadratic Jordan algebras and K an inner ideal of J with K ⊆ J ⊆ Q. If x ∈ K is a K-denominator for q ∈ Q, then x^{4} ∈ K will be a strong K-denominator for this same q ( i.e. x^{4} ∈ D_{K}^{stg}(q) ) .

Proof We are given that x, x ∘ q ∈ K and U_{x}q ∈ K. We must show that x^{4} ∈ D_{K}^{stg}(q) by showing that it satisfies conditions (i), (ii), (iii), and (v) of definition 2.

(i): U_{x^{4}}q = U_{x^{3}}(U_{x}q) ∈ U_{x^{3}}(K) ⊆ K, since x ∈ K.

(ii): First, we will show that U_{x^{2}}q^{2} ∈ K. For simplicity we will adopt the following notation: If p − r ∈ K, then we write p ≡ r. Taking identity (J24) and multiplying by U_{x}, we obtain the following:

U_{x^{2}}q^{2} + U_{x}U_{q}x^{2} + U_{x}(x ∘ U_{q}x) = U_{x}(q ∘ x)^{2}.

Since q ∘ x ∈ K, and hence U_{x}(q ∘ x)^{2} ≡ 0, the above equation tells us that

U_{x^{2}}q^{2} ≡ −U_{x}U_{q}x^{2} − U_{x}(x ∘ U_{q}x)
(J25) = −U_{x}U_{q}x^{2} + 2(U_{x}q)^{2} − {x, x ∘ q, U_{x}q}
= −(U_{x}q)^{2} + 2(U_{x}q)^{2} − {x, x ∘ q, U_{x}q}
≡ 0,

since x, x ∘ q, U_{x}q ∈ K. So,

U_{x^{2}}q^{2} ∈ K. (1)

(Recall, we are working towards showing that U_{q}x^{4} ∈ K.) If z ∈ K̂, then identity (J29) gives us

U_{q∘x^{2}}z = U_{q}U_{x^{2}}z + U_{x^{2}}U_{q}z + V_{x,q∘x}V_{q∘x,x}z + V_{U_{x}q}^{2}z
−V_{x,x∘q}V_{U_{x}q}z − V_{U_{x}q}V_{x∘q,x}z − V_{U_{x^{2}}q^{2}}z

Hence, we have

U_{q∘x^{2}}z − (U_{q}U_{x^{2}}z + U_{x^{2}}U_{q}z) = V_{x,q∘x}V_{q∘x,x}z + V_{U_{x}q}^{2}z − V_{x,x∘q}V_{U_{x}q}z
−V_{U_{x}q}V_{x∘q,x}z − V_{U_{x^{2}}q^{2}}z.

Look at the subscripts on the right hand side of this equation: x, x ∘ q, U_{x}q, U_{x^{2}}q^{2}. Each of these is in K, including U_{x^{2}}q^{2} by (1) above. Therefore,

U_{q∘x^{2}}z ≡ U_{q}U_{x^{2}}z + U_{x^{2}}U_{q}z. (2)

Now

q ∘ x^{2} = (q ∘ x) ∘ x − 2U_{x}q ∈ K. (3)

Hence, U_{q∘x^{2}}z ∈ U_{K} Ĵ ⊆ K. That is,

U_{q∘x^{2}}z ∈ U_{K} Ĵ ⊆ K ≡ 0.

So,

U_{q}U_{x^{2}}z ≡ −U_{x^{2}}U_{q}z.

Letting z = 1 we obtain U_{q}x^{4} ≡ −U_{x^{2}}q^{2} ≡ 0 (by (1) above), which gives (ii).

(iii): For z ∈ K,

U_{x^{4}}U_{q}z = U_{x^{2}}(U_{x^{2}}U_{q}z)
(2) ≡ U_{x^{2}}(U_{q∘x^{2}}z − U_{q}U_{x^{2}}z)
= U_{x^{2}}U_{q∘x^{2}}z − U_{x^{2}}U_{q}U_{x^{2}}z
= U_{x^{2}}U_{q∘x^{2}}z − U_{x}(U_{U_{x}q})U_{x}z
≡ 0.

This gives (iii).

(v): By identity (J28),

{x^{4}, q, K̂} = {x^{2}, q ∘ x^{2}, K̂} − (U_{x^{2}}q) ∘ K̂ ⊆ K

since U_{x^{2}}q = U_{x}(U_{x}q) ∈ K, x^{2} ∈ K, and q ∘ x^{2} ∈ K by (3).

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At times it will be useful to have a denominator whose level of strength is beyond that of even a strong denominator. Next we define the notion of a super denominator and prove (similar to the last lemma) that any strong denominator can be boosted into a super one.

Definition 3 Let J and Q be two quadratic Jordan algebras and K an inner ideal of J with K ⊆ J ⊆ Q. An element x ∈ K is called a super K-denominator for q ∈ Q if

(i) U_{x}q ∈ U_{K} K̂

(ii) U_{q}x ∈ U_{K} K̂

(iii) U_{x}U_{q}K̂ ⊆ U_{K} K̂

(iv) U_{q}U_{x}K̂ ⊆ U_{K} K̂

(v) {x, q, K̂} ⊆ U_{K} K̂

(vi) {q, x, K̂} ⊆ U_{K} K̂.

Lemma 4 Let J and Q be two quadratic Jordan algebras and K an inner ideal of J with K ⊆ J ⊆ Q. If x ∈ K is a strong K-denominator for q ∈ Q, then x^{3} ∈ K will be a super K-denominator for this same q.

Proof We are given that x ∈ K satisfies (i)-(vi) of definition 2 and we must show that x^{3} satisfies (i)-(vi) of definition 3.

(i): U_{x^{3}}q = U_{x}U_{x}(U_{x}q) ∈ U_{K}U_{K}K ⊆ U_{k}K̂.

(ii): U_{q}x^{3}

= U_{q}U_{x}x
(J20) = (U_{q∘x} + U_{U_{q}x,x} − U_{x}U_{q} − V_{q}U_{x}V_{q})x
∈ U_{K}K + U_{K,K}K − U_{K}K − q ∘ U_{x}(q ∘ x)
= U_{K}K + U_{K,K}K − U_{K}K − q ∘ U_{x^{2},x}q
(J21) = U_{K}K + U_{K,K}K − U_{K}K − (x^{2} ∘ U_{q}x + x ∘ U_{q}x^{2})
∈ U_{K}K + U_{K,K}K − U_{K}K − K ∘ (K + U_{q}U_{x}1)
⊆ U_{K}K + U_{K,K}K − U_{K}K − K ∘ K
⊆ U_{K} K̂.

(iii): U_{x^{3}}U_{q}K̂ = U_{x}U_{x}(U_{x}U_{q}K̂) ⊆ U_{K}U_{K}K ⊆ U_{K} K̂.

(iv): U_{q}U_{x^{3}} K̂

= U_{q}U_{x}U_{x^{2}} K̂
(J23) = (U_{q∘x} + U_{U_{q}x,x} − U_{x}U_{q} − V_{q}U_{x}V_{q})U_{x^{2}} K̂
∈ (U_{K} + U_{K,K})K − U_{K}U_{q}U_{x}K − V_{q}U_{x}V_{q}U_{x}(U_{x}K̂)
= (U_{K} + U_{K,K})K − U_{K}K
−(U_{U_{q}x,x}U_{x} + U_{{q,x,x},U_{x}q}U_{x} − U_{x,U_{x}q} + U_{x,U_{x}q} + U_{x^{2},(U_{x}q)^{2}})K̂
⊆ (U_{K} + U_{K,K})K − U_{K}K
−(U_{K,K}K − U_{K,K}U_{K,K} + U_{K,K})K̂
⊆ U_{K} K̂.

(v): {x^{3}, q, K̂}

= {x, {x, x, q}, K̂} − {U_{x}q, x, K̂}
⊆ {K, K, K̂} − {K, K, K̂}
⊆ U_{K}K + K ∘ K
⊆ U_{K} K̂.

(vi): {q, x^{3}, K̂}

= (q ∘ x^{3}) ∘ K̂ − {x^{3}, q, K̂}
⊆ ({q ∘ x, x, x} − (U_{x}q) ∘ x) ∘ K̂ − U_{K} K̂
⊆ (U_{K}K − K ∘ K) ∘ K̂ − U_{K} K̂
⊆ U_{K} K̂.
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3.2 Ore Conditions

As we develop the results that follow, denominators will be the key players. We have furthered our understanding of them with the statement and proof of some relevant lemmas. An important part will also be played by a special type of subset of J. Next we describe exactly what type of subset we mean and then introduce the all important Ore condition.

Definition 4 A Jordan monad S of a quadratic Jordan algebra J is a nonempty subset of J consisting only of injective elements such that for s, t ∈ S we have

s^{2}, U_{s}(t) ∈ S.

Definition 5 An Ore monad S of a quadratic Jordan algebra J is a Jordan monad of J with the additional property that for s, t ∈ S we have

U_{s}(S) ∩ U_{t}(S) ≠ ϕ.

NOTE: Throughout the rest of the paper this will be referred to as the common inner multiple property, or CIMP.

Definition 6 A quadratic Jordan algebra J is said to satisfy the Ore condition with respect to an Ore monad S ⊆ J if for all s ∈ S, a ∈ J there exists s′ ∈ U_{s}(S) such that

s′ ∘ a ∈ K_{s}

for K_{s} = Φs + U_{s}(Ĵ) where Ĵ is the unital hull of J.

NOTE: U_{s′}a ∈ K_{s} automatically since

U_{s′}a = U_{U_{s}(s′′)}a = U_{s}U_{s′′}U_{s}a ∈ U_{s}(J) ⊆ K_{s}.

Also, based on lemma ?, if s′ ∈ U_{s}(S) satisfies

s′ ∘ a ∈ K_{s},

then we can actually find s′′ ∈ U_{s}(S) such that

s′′ ∘ a ∈ U_{s}Ĵ ⊆ K_{s}.

In the above definition, s′ is a K_{s}-denominator in the sense of definition ? but has the additional restriction that s′ ∈ U_{s}(S) (in the definition of a K-denominator, s′ is only restricted to being an element of K_{s}). In particular these denominators are always injective since S consists of only injective elements. This more restricted type of denominator is given its own name in the next definition.

Definition 7 Let J and Q be two quadratic Jordan algebras with J ⊆ Q. If S ⊆ J is an Ore monad in J and if s ∈ S, then an element x ∈ U_{s}(S) satisfying the conditions in definitions 1, 2, or 3 (with K = K_{s}) is called respectively an s-denominator, a strong s-denominator or a super s-denominator for q ∈ Q.

As we continue, we will need to examine the interactions between the multiplication operators built out of arbitrary elements of J and those built out of elements of an Ore monad S ⊆ J. Below, we introduce some terminology describing a special form these interactions may take.

Notation: For a quadratic Jordan algebra J, we let M(J) denote the unital multiplication algebra generated over Φ by I and all operators of the form U_{a}, V_{a} where a ∈ J.

Definition 8 Let W ∈ M(J) and let s ∈ S ⊆ J. We say that W is s-movable if there exists s′ ∈ S and W′ ∈ M(J) such that W U_{s′} = U_{s}W′. When specification is needed, we will say that W is movable past s using s′.

Remark 2 If we can move W past s using s′ then we can also use any s′′ ∈ U_{s′} Ĵ∩ S since W U_{s′′} = W U_{s′}Ux̂U_{s′} = U_{s}W′Ux̂U_{s′} = U_{s}W′′.

Definition 9 We say that W ∈ M(J) is S − movable if it is s-movable for each s ∈ S.

This is an appropriate place to state and prove a lemma concerning the idea of “movability” and the structure group of Ĵ, Struc(Ĵ).

Lemma 5 Let s ∈ J generate U_{s}Ĵ. If W ∈ Struc(Ĵ) then we have the following:

(i) If W (s) = U_{s}r ∈ U_{s}Ĵ then W U_{s} = U_{s}W̃ on Ĵ for W̃ = V_{r,s} − W^{∗}

(ii) If W (s) ∈ U_{s}Ĵ then W (K_{s}) ⊆ U_{s}Ĵ.

If W^{∗}(Ĵ) ⊆ J (eg. if W = V_{x,y} for at least one of x, y ∈ J) then we have the following:

(iii) If W (s) ∈ U_{s}Ĵ then W (U_{s}Ĵ) ⊆ U_{s}J

(iv) If W (s) ∈ U_{s}J then W (K_{s}) ⊆ U_{s}J.

Proof (i): U_{W (s),s} = U_{U_{s}r,s} = U_{s}V_{r,s}. Hence, by Lie-structurality,

W U_{s} = U_{W (s),s} − U_{s}W^{∗} = U_{s}(V_{r,s} − W^{∗}) = U_{s}W̃ .

(ii): W (K_{s}) = W (Φs + U_{s}Ĵ) ⊆ U_{s}Ĵ by (i) and by hypothesis on W (s).

(iii): W (U_{s}Ĵ) = U_{s}W̃ (Ĵ) by (i). Furthermore, since

V_{r,s}Ĵ ⊆ J for s ∈ J,

we have

W̃ (Ĵ) = (V_{r,s} − W^{∗})(Ĵ) ⊆ J

by hypothesis. Hence, W (U_{s}Ĵ) ⊆ U_{s}J.

(iv): W (K_{s}) = W (Φs + U_{s}Ĵ) = ΦW (s) + W (U_{s}Ĵ) ⊆ J by (iii) and by hypothesis on W (s).

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As it turns out, the Ore condition on quadratic Jordan algebras can be stated in a variety of different ways. Depending on the setting, there are times when some ways prove more useful than others. In the following lemma, we state these various forms the Ore condition can take and prove their equivalency.

Proposition 1 Let J be a quadratic Jordan algebra and let S ⊆ J be an Ore monad in J. Then the following are equivalent:

(1) Each W ∈ M(J) is S-movable.

(2) For each a ∈ J , U_{a} and V_{a} are S-movable.

(3) Each a ∈ J has an s-denominator for each s ∈ S.

(this is the way the Ore condition on quadratic Jordan algebras was originally stated)

(4) Each a ∈ J has an s-denominator s′ for each s ∈ S with V_{s′,a}, V_{a,s′} ∈ V_{K_{s},K_{s}}

(5) Each a ∈ J has a strong s-denominator for each s ∈ S.

(6) Each a ∈ J has a super s-denominator for each s ∈ S.

Proof

(1) ⇒ (2) This implication is clear since U_{a}, V_{a} ∈ M(J) for all a ∈ J.

(2) ⇒ (1) Let W ⊆ M(J) denote the set of all operators in M(J) which are Smovable. We will show that W is a subalgebra of M(J).

Let s ∈ S and suppose that W_{1}, W_{2} are movable past s using s′_{1}, s′_{2} ∈ S. Since S is an Ore monad, by Remark 1, we know that we can move both W_{1} and W_{2} past s using some s′_{3} ∈ U_{s′_{1}}S ∩ U_{s′_{2}}S ≠ ϕ. As a result, W_{1} + W_{2} is movable past s using s′_{3}:

(W_{1} + W_{2})U_{s′_{3}} = W_{1}U_{s′_{3}} + W_{2}U_{s′_{3}} = U_{s}W′′_{1} + U_{s}W′′_{2} ∈ U_{s}(M(J)).

Hence, W_{1} + W_{2} ∈ W. Furthermore, if α ∈ Φ then αW_{1} is movable past s using s′_{1}:

αW_{1}U_{s′_{1}} = αU_{s}W′_{1} = U_{s}αW′_{1} ∈ U_{s}(M(J)).

And so αW_{1} ∈ W as well.

Clearly the identity operator I is S-movable since IU_{s} = U_{s}I for all s ∈ S. What about W_{1}W_{2}?

Since s′_{1} ∈ S, we know we can move W_{2} past s′_{1} using some s′_{4} ∈ S:

W_{2}U_{s′_{4}} = U_{s′_{1}}M′_{1},

for some M′_{1} ∈ M(J). Thus, we can move W_{1}W_{2} past s using this same s′_{4}:

W_{1}W_{2}U_{s′_{4}} = W_{1}U_{s′_{1}}M′_{1} = U_{s}W′_{1}M′_{1} ∈ U_{s}(M(J)).

Hence W_{1}W_{2} ∈ W.

Together these facts give us that W is a subalgebra of M(J). Now, condition (2) tells us that U_{a}, V_{a} ∈ W for each a ∈ J. So, W is a subalgebra of M(J) containing all of the generators of M(J). We conclude that we must have W = M(J) which gives condition (1) as desired.

(2) ⇒ (3) Let a ∈ J and s ∈ S. By condition (2) and Remark 1, we know that we can move V_{a} past s using some s′_{1} ∈ U_{s′}S ∩ U_{s}S ≠ ϕ:

V_{a}U_{s′_{1}} = U_{s}W′_{1}.

Then, s′ = U_{s′_{1}}s ∈ U_{s}S has

a ∘ s′ = V_{a}s′ = V_{a}U_{s′_{1}}s = U_{s}W′_{1}s ∈ K_{s}.

This gives condition (3).

(3) ⇒ (2) Let a ∈ J and s ∈ S. By condition (3) there exists s′ ∈ U_{s}S such that s′ ∘ a ∈ K_{s}. Thus s′, s′ ∘ a ∈ K_{s}. Keeping this in mind while considering the identity

V_{a}U_{s′} + U_{s′}V_{a} = U_{s′∘a,s′},

we obtain the following string of equalities and containments:

V_{a}U_{s′} = U_{s′∘a,s′} − U_{s′}V_{a}
∈ U_{K_{s},K_{s}} − U_{K_{s}}V_{a}
⊆ U_{K_{s}}M(J)
⊆ U_{s}B_{Φ,Ĵ,S}M(J)
⊆ U_{s}M(J),

where

B_{Φ,Ĵ,S} = Φ^{2}I + V_{Ĵ,S} + UĴU_{S}.

This tells us that V_{a} is S-movable. What about U_{a} ?

Consider the following identity:

U_{a}U_{s^{′^{2}}} = −U_{s^{′^{2}}}U_{a} + V_{a}U_{s^{′^{2}}}V_{a} + U_{s^{′^{2}}∘a} + U_{U_{a}s^{′^{2}},s^{′^{2}}}.

We have just seen that V_{a}U_{s′} ∈ U_{s}M(J). Hence, we know that V_{a}U_{s^{′^{2}}} ∈ U_{s}M(J). Clearly, we also have U_{s^{′^{2}}}U_{a} ∈ U_{s}M(J). We would like to show further that:

(i) s^{′^{2}} ∈ K_{s} (ii) s^{′^{2}} ∘ a ∈ K_{s} (iii) U_{a}s^{′^{2}} ∈ K_{s}.

(i) is clearly true since s′ ∈ K_{s}, and (ii) is true by Lemma 1 . For (iii) we notice that

U_{a}s^{′2} = U_{s′}a^{2} + (s′ ∘ a)^{2} − U_{s′∘a,s′}(a)
∈ U_{s′}J + K_{s}^{2} − U_{K_{s},K_{s}}(J)
⊆ K_{s}.

Looking back at our original identity, we have:

U_{a}U_{s^{′^{2}}} = −U_{s^{′^{2}}}U_{a} + V_{a}U_{s^{′^{2}}}V_{a} + U_{s^{′^{2}}∘a} + U_{U_{a}s^{′^{2}},s^{′^{2}}}
∈ U_{s}M(J) − U_{s}M(J) + U_{K_{s}} + U_{K_{s},K_{s}}
⊆ U_{s}M(J),

which tells us that U_{a} is S-movable.

So given condition (3) we have shown that both U_{a} and V_{a} are S-movable for each a ∈ J. This gives condition (2).

(3) ⇒ (4) Let a ∈ J and s ∈ S. By condition (3), there exists s′ ∈ U_{s}S such that s′ ∘ a ∈ K_{s}. By Lemma 1, we can immediately conclude that the element (s^{′)^{2}} will satisfy the conditions of (4).

(4) ⇒ (5) ⇒ (6) Let a ∈ J and s ∈ S. By condition (4) we know there exists an s-denominator s′ ∈ U_{s}S for a ∈ J. By Lemma 3, we know that (s′)^{4} will be a strong s-denominator for a, which gives condition (5); and by Lemma 4 we know that (s′)^{12} will be a super s-denominator for a, which gives (6).

(6) ⇒ (5) ⇒ (4) ⇒ (3) These implications follow immediately.

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4 Necessity of the Ore Condition

We will show that the quadratic Ore condition is a necessary condition for the existence of ring of fractions for an arbitrary quadratic Jordan algebra. Before doing this, we best define what we mean by a ring of fractions in the quadratic Jordan setting.

Definition 10 Let J be a quadratic Jordan algebra and let S ⊆ J be an Ore monad in J. We say that Q, with J ⊆ Q, is a ring of S-fractions for J if

(i) Every s ∈ S is invertible in Q

(ii) Every q ∈ Q has a J-denominator s ∈ S.

NOTE: By J-denominator we mean a K-denominator where we are considering the trivial inner ideal K = J.

4.1 Proving Necessity

Now that we have a working definition for a quadratic Jordan ring of fractions, we are almost ready to show that our Ore condition is necessary for the existence of such a ring. As a last step of preparation, we need to state and prove three helpful lemmas, the third being very crucial.

Lemma 6 Let J and Q be two quadratic Jordan algebras and S an Ore monad in J with S ⊆ J ⊆ Q. For an element q ∈ Q, the following two conditions are equivalent:

(i) q has a J-denominator s ∈ S

(ii) There exists t ∈ S with

(A) q ∘ K_{t} ⊆ J

(B) U_{K_{t}}q ⊆ J.

Proof

(i) ⇒ (ii) Condition (i) tells us that q has a J-denominator s ∈ S. By Lemma 1, the element t = s^{2} will satisfy both (A) and (B) of condition (ii).

(ii) ⇒ (i) Condition (ii) implies that every element of K_{t} ∩ S is a J-denominator for q in S. To be specific, we could select the element t^{2}.

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Lemma 7 Let J and Q be two quadratic Jordan algebras and S an Ore monad in J with S ⊆ J ⊆ Q. Suppose that q ∈ Q has a J-denominator x ∈ S. Then q ∈ Q has a J-denominator in U_{s}S, for any s ∈ S.

Proof Let s ∈ S, q ∈ Q, and suppose that x ∈ S is a J-denominator for q. By Lemma 1, with K = J, we know that any element of U_{x^{2}}S will also function as a J-denominator for q. Since S is an Ore monad, we know that there exists x′ ∈ U_{x^{2}}S ∩ U_{s}S ≠ ϕ. Hence, this x′ is of the form U_{s}t for some t ∈ S and is a Jdenominator for q. Lemma is proved.

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Lemma 8 Let J and Q be two quadratic Jordan algebras and S an Ore monad in J with S ⊆ J ⊆ Q. Also, suppose that J satisfies the Ore condition with respect to S. If q ∈ Q has a J-denominator x ∈ S then q has an s-denominator for any s ∈ S.

Proof Let q ∈ Q, s ∈ S, and let x ∈ S be a J-denominator for q. Since S is an Ore monad we know that there exists r ∈ U_{x^{2}}S ∩ U_{s}S ⊆ K_{s}. By Lemma 1, this r is still a J-denominator for q. Hence, a_{1} = r ∘ q and a_{2} = U_{r}q both lie in J.

Since J satisfies the Ore condition with respect to S, the elements r ∘ a_{1} and a_{2} (both in J ) have strong s-denominators t_{1}, t_{2} ∈ U_{s}S. Again, since S is an Ore monad, we can select t ∈ U_{t_{1}}S ∩ U_{t_{2}}S ≠ ϕ and t will be a strong s-denominator for both r ∘ a_{1} and a_{2} by Lemma 2.

If we then take x_{s} = U_{r^{2}}t ∈ U_{r}S ⊆ U_{s}S, x_{s} will be the desired s-denominator for q:

U_{x_{s}}q ∈ U_{r}U_{J}U_{r}q = U_{r}U_{J}a_{2} ⊆ U_{r}J ⊆ K_{s}.

and

x_{s} ∘ q = U_{r^{2}}t ∘ q
= {r, U_{r}t, r ∘ q} − {r^{2}, t, U_{r}q}
= {r, U_{r}t, a_{1}} − {r^{2}, t, a_{2}}
= {r^{2}, t, r ∘ a_{1}} − U_{r}{r, t, a_{1}} − {r^{2}, t, a_{2}}
∈ {K_{s}, t, r ∘ a_{1}} − U_{K_{s}}J − {K_{s}, t, a_{2}}
⊆ K_{s}.

Lemma is proved.

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Now we have all the tools to show that the Ore condition is necessary for the existence of a quadratic Jordan ring of fractions for an arbitrary quadratic Jordan algebra J.

Proposition 2 If quadratic Jordan algebra J has a ring of S-fractions with respect to an Ore monad S ∈ J, then J must satisfy the Ore condition with respect to S.

Proof Let Q be a ring of S-fractions for J with respect to Ore monad S. We will show that J satisfies the Ore condition with respect to S by showing that J meets condition (3) of Proposition 1: each a ∈ J has an s-denominator for each s ∈ S.

Let a ∈ J and s ∈ S. Then {s^{−1}, a, s} ∈ Q has a J-denominator r ∈ S and by Lemma 1 we can assume that r = U_{s}t for t ∈ S. Then,

U_{s^{−1}}(a ∘ U_{s}U_{s}t) = {s^{−1}, a, s} ∘ U_{s}t − U_{s}(a ∘ t).

Since U_{s}t is a J-denominator for {s^{−1}, a, s}, we conclude that U_{s^{−1}}(a ∘ U_{s}U_{s}t) ∈ J. This implies that a∘U_{s}U_{s}t ∈ U_{s}J ⊆ K_{s}. Therefore, s′ = U_{s}U_{s}t ∈ U_{s}S and a∘s′ ∈ K_{s}. Thus J satisfies the Ore condition with respect to S via condition (3) of Proposition 1.

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4.2 Sufficient?

While it was fairly easy to show that the Ore condition is necessary for the exsistence of a ring of fractions for an arbitrary quadratic Jordan algebra, it appears that proving the sufficiency of the condition is another story. In fact, in order to show that the Ore condition is sufficient for the existence of a quadratic Jordan ring of fractions, we will need to restrict down to the collection of quadratic Jordan algebras satisying some additional axioms. In order to state these axioms, we will need to develop some special operators. We do this now.

Let J be a quadratic Jordan algebra satisfying the Ore condition with respect to an Ore monad S ⊆ J. Let y ∈ J, s ∈ S, and let t be a super s-denominator for y. Then

we have the following:

(i) t = U_{s}t′, for t′ ∈ S

(ii) t ∘ y = U_{s}ŷ_{0}, for ŷ_{0} ∈ Ĵ

(iii) U_{y}t = U_{s}ŷ_{2}, for ŷ_{2} ∈ Ĵ.

Now, given the above, we define two operators on J in the following way:

M_{y} := U_{s}Uŷ_{0} + U_{s}U_{ŷ_{2},t′} + U_{y}U_{s}U_{t′} + V_{y^{2}}U_{s}U_{t′} − V_{y}U_{s}U_{ŷ_{0},t′}
M_{y}^{∗} := Uŷ_{0}U_{s} + U_{ŷ_{2},t′}U_{s} + U_{t′}U_{s}U_{y} + U_{t′}U_{s}V_{y^{2}} − U_{ŷ_{0},t′}U_{s}V_{y}.

These operators will play a crucial role in reaching our results because of some useful properties they possess. We state one of these properties as the next lemma.

Lemma 9 Let M_{y} and M_{y}^{∗} be as defined above. Then,

(i) U_{t}U_{y} = M_{y}U_{s}

(ii) U_{y}U_{t} = U_{s}M_{y}^{∗}.

Proof

M_{y}U_{s} = (U_{s}Uŷ_{0} + U_{s}U_{ŷ_{2},t′} + U_{y}U_{s}U_{t′} + V_{y^{2}}U_{s}U_{t′} − V_{y}U_{s}U_{ŷ_{0},t′})U_{s}
= U_{U_{s}ŷ_{0}} + U_{U_{s}ŷ_{2},U_{s}t′} + U_{y}U_{U_{s}t′} + V_{y^{2}}U_{U_{s}t′} − V_{y}U_{U_{s}ŷ_{0},U_{s}t′}
= (U_{t∘y} + U_{U_{y}t,t}) + U_{y}U_{t} + V_{y^{2}}U_{t} − V_{y}U_{y∘t,t}
= (U_{t}U_{y} + U_{y}U_{t} + V_{y}U_{t}V_{y}) + U_{y}U_{t} + (V_{y}^{2} − 2U_{y})U_{t} − V_{y}(V_{y}U_{t} + U_{t}V_{y})
= U_{t}U_{y}.

The argument follows dually for M_{y}^{∗}.

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Now, given quadratic Jordan algebra J, we can construct

J̃ = J[λ_{1}, …, λ_{n}]

and

J̄ = Ĵ[λ_{1}, …, λ_{n}],

where

{λ_{1}, …, λ_{n}}

is a finite set of indeterminants. If S ⊆ J is an Ore monad and J satisfies the Ore condition with respect to S, then the same will continue to be true for S in relation to J̃ and ˜ Ĵ. Furthermore, if

{y_{i}}_{i=1}^{n}

is a finite set of elements of J and s ∈ S, then for the element

ỹ = ∑_{i} λ_{i}y_{i} ∈ J̃,

we can choose t ∈ S to be a super s-denominator so that the following are true:

(i) t = U_{s}t′ for some t′ ∈ S

(ii) t ∘ ỹ = U_{s}ȳ_{0} for some ȳ_{0} ∈ ˜ Ĵ

(iii) Uỹt = U_{s}ȳ_{2} for some ȳ_{2} ∈ ˜ Ĵ

(iv) {ỹ, t, s} = U_{s}ȳ_{1} for some ȳ_{1} ∈ ˜ Ĵ.

Then, for Mỹ and M ∗ ; ỹ defined in the same manner as M_{y} and M_{y}^{∗} above, we have

U_{t}Uỹ = MỹU_{s} and UỹU_{t} = U_{s}M ∗ ; ỹ. (1)

Now U_{s} is injective. Hence the elements ȳ_{k}, k = 0, 2 , are uniquely determined and have zero constant term. Since ỹ is homogeneous of degree 1 in each of the λ_{i}’s, so are ȳ_{0} and ȳ_{1} while ȳ_{2} is quadratic:

ȳ_{0} = ∑_{i} λ_{i}ŷ_{i,0}, ȳ_{1} = ∑_{i} λ_{i}ŷ_{i,1} ȳ_{2} = ∑_{i} λ_{i}^{2}ŷ_{i,2} + ∑_{i≤j} λ_{i}λ_{j}ŷ_{ij,2}.

Considering (i)-(iv) and identifying coefficients of λ_{i}, λ_{i}^{2}, and λ_{i}λ_{j} we obtain the following:

\begin{matrix}t ∘ y_{i} = U_{s}y_{i,0} & \{y_{i}, t, s\} = U_{s}y_{i,1} \\ U_{y_{i}}t = U_{s}y_{i,2} & U_{y_{i}.y_{j}}t = U_{s}y_{ij,2}. \end{matrix}

Identifying coefficients of λ_{i}^{2} and λ_{i}λ_{j} in (1), gives

\begin{matrix}U_{t}U_{y} = M_{y_{i}}U_{s} & U_{t}U_{y_{i},y_{j}} = M_{y_{i},y_{j}}U_{s} \\ U_{y}U_{t} = U_{s}M_{y_{i}}^{∗} & U_{y_{i},y_{j}}U_{t} = U_{s}M_{y_{i},y_{j}}^{∗} \end{matrix}

where M_{y_{i}} and M_{y_{i}}^{∗} are defined as before and

M_{y_{i},y_{j}} := U_{s}U_{y_{i,0},y_{j,0}} + U_{s}U_{y_{ij,2},t′} + U_{y_{i},y_{j}}U_{s}U_{t′}
+V_{y_{i}∘y_{j}}U_{s}U_{t′} − (V_{y_{i}}U_{s}U_{y_{j,0},t′} + V_{y_{j}}U_{s}U_{y_{i,0},t′})

M_{y_{i},y_{j}}^{∗} := U_{y_{i,0},y_{j,0}}U_{s} + U_{y_{ij,2},t′}U_{s} + U_{t′}U_{s}U_{y_{i},y_{j}}
+U_{t′}U_{s}V_{y_{i}∘y_{j}} − (U_{y_{j,0},t′}U_{s}V_{y_{i}} + U_{y_{i,0},t′}U_{s}V_{y_{j}}).

In addition to these M operators that we have introduced, we want to define a V̄

operator. If s ∈ S, and y ∈ J, and a, b ∈ K_{s} with a = U_{s}a′, and t is an s-denominator for y so that {y, t, a} = U_{s}ã, then define V̄_{a} by

V̄_{a} := V_{U_{s}U_{t}y,a′} − V_{U_{s}t,ã}.

5 Proving Sufficiency

For the remainder of this paper, let us assume that J is a quadratic Jordan algebra satisfying the following: If s ∈ S, and y, d ∈ J, and a, b ∈ K_{s}, and t is an s-

denominator for y, then

(AX1) M_{y}U_{s}M_{d,y}^{∗} − M_{y}U_{d,s}M_{s,y}^{∗} − M_{d,y}U_{s}M_{y}^{∗} + M_{s,y}U_{d,s}M_{y}^{∗} = 0 on Ĵ.

(AX2) V̄_{a}V̄_{b} = V_{U_{s}U_{t}b,M_{y}^{∗}a} + U_{s}U_{t}U_{b,a}M_{y}.

(AX3) V_{t,a}M_{y} + M_{y}V_{U_{s}t,a′} = M_{y,s}V_{s,ã} − U_{t}V_{y,ã}.

NOTE: These additional Ore conditions were suggested by Kevin McCrimmon after realizing that proving sufficiency was not as easy as hoped.

For such a J, the Ore condition does imply the existence of a quadratic Jordan ring of fractions. We will not prove this directly. We will instead argue in the following way. After introducing two more lemmas, we will show that in order for a ring of S-fractions for J to exist, we only need find a Jordan overring for J where the elements of S are invertible. Once we have established this fact, we argue that the Ore condition always allows us to construct such an overring.

5.1 S^{−1}

Lemma 10 Let J ⊆ Q̃ where Q̃ is a Jordan overring of J in which all elements of S are invertible. Then for each t ∈ S, t^{−1} has a J-denominator in S.

Proof Let t ∈ S. Then t^{2} ∈ S. Furthermore,

t^{2} ∘ t^{−1} = 2t ∈ J

and

U_{t^{2}}(t^{−1}) = U_{t}U_{t}(t^{−1}) = U_{t}t ∈ J.

Hence, t^{−1} has J-denominator t^{2} ∈ S, as desired.

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Lemma 11 Let J ⊆ Q̃ where Q̃ is a Jordan overring of J in which all elements of S are invertible. If q_{1}, q_{2} ∈ Q̃ each have a J-denominator in S, then U_{q_{1}}q_{2} also has a J-denominator in S.

Proof Let q_{1}, q_{2} ∈ Q̃ and suppose that each has a J-denominator in S. By Lemma 3, we know that each then has a strong J-denominator in S. By the CIMP of S, and by the innerness of D_{J}^{stg}(q_{i}), i = 1, 2, we can choose an element s such that s ∈ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) and K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}). Now, by Proposition 1, both q_{1} and q_{2} have a strong s-denominator in S. By using the same reasoning seen above, we can choose an element t such that t ∈ D_{s}^{stg}(q_{1})∩D_{s}^{stg}(q_{2}) and K_{t} ⊆ D_{s}^{stg}(q_{1}) ∩ D_{s}^{stg}(q_{2}). Continuing in the same fashion, we know that each of q_{1} and q_{2} has a strong tdenominator in S. We again argue that we can choose an element x such that x ∈ D_{t}^{stg}(q_{1}) ∩ D_{t}^{stg}(q_{2}). Based on how we have selected the element x, we have the following for i, j = 1, 2:

(i) x ∘ q_{i} ∈ K_{t} ⊆ D_{s}^{stg}(q_{j}) (ii) U_{x}q_{i} ∈ K_{t} ⊆ D_{s}^{stg}(q_{j}) (iii) U_{q_{i}}x ∈ K_{t} ⊆ D_{s}^{stg}(q_{j}) (iv) U_{q_{i}}x^{2} ∈ K_{t} ⊆ D_{s}^{stg}(q_{j})

(v) U_{x}q_{i}^{2} ∈ K_{t} ⊆ D_{s}^{stg}(q_{j}) (vi) {x, q_{i}, K_{t}} ⊆ K_{t} ⊆ D_{s}^{stg}(q_{j}) (vii) {q_{i}, x, K_{t}} ⊆ K_{t} ⊆ D_{s}^{stg}(q_{j}). These containments give rise to : (i)′ U_{x}q_{i} ∘ q_{j} ∈ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J (ii)′ U_{q_{i}}x^{2} ∘ q_{j} ∈ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J (iii)′ U_{x∘q_{i}}q_{j} ∈ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J (iv)′ U_{U_{x}q_{i}}q_{j} ∈ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J (v)′ {q_{j}, x ∘ q_{i}, K_{s}} ⊆ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J (vi)′ {x ∘ q_{i}, q_{j}, K_{s}} ⊆ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) (vii)′ {U_{q_{i}}x, q_{j}, K_{s}} ⊆ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J (vii)′ {U_{x}q_{i}^{2}, q_{j}, K_{s}} ⊆ K_{s} ⊆ D_{J}^{stg}(q_{1}) ∩ D_{J}^{stg}(q_{2}) ⊆ J. Lastly, we have: (i)′′ (U_{x}q_{i} ∘ q_{j}) ∘ q_{i} ⊆ J (ii)′′ {x ∘ q_{i}, q_{j}, K_{s}} ∘ q_{i} ⊆ J

(iii)′′ {q_{j}, x ∘ q_{i}, K_{s}} ∘ q_{i} ⊆ J.

We are now prepared to show that the element x^{2} functions as the desired J-denominator for U_{q_{1}}q_{2}: since x ∈ K_{s} we have

x^{2} ∘ U_{q_{1}}q_{2} = −q_{2} ∘ U_{q_{1}}x^{2} + {x^{2}, q_{1}, q_{2}} ∘ q_{1}
= −q_{2} ∘ U_{q_{1}}x^{2} + {x, x ∘ q_{1}, q_{2}} ∘ q_{1} − (U_{x}q_{1} ∘ q_{2}) ∘ q_{1}
∈ J

by (ii)′, (iii)′′, (i)′′, and since x, x^{2} ∈ K_{t} ⊆ K_{s} we have

U_{x^{2}}U_{q_{1}}q_{2} = U_{x}U_{x}U_{q_{1}}q_{2}
= U_{x} (U_{x∘q_{1}} + U_{x,U_{q_{1}}x} + U_{q_{1}}U_{x} − V_{q_{1}}U_{x∘q_{1},x} + V_{q_{1}^{2}}U_{x}) q_{2}
= (U_{x}U_{x∘q_{1}} + U_{x}U_{x,U_{q_{1}}x} + U_{U_{x}q_{1}} − (U_{x}V_{q_{1}})U_{x,x∘q_{1}} + U_{U_{x}q_{1}^{2},x^{2}}) q_{2}
∈ U_{x}(U_{x∘q_{1}}q_{2}) + U_{x}({K_{s}, q_{2}, U_{q_{1}}x}) + U_{U_{x}q_{1}}q_{2} − (U_{x}V_{q_{1}}){K_{s}, q_{2}, x ∘ q_{1}} + {U_{x}q_{1}^{2}, q_{2}, K_{s}}
⊆ U_{x}J + U_{x}J + J − U_{x}(q_{1} ∘ {K_{s}, q_{2}, x ∘ q_{1}}) + J
⊆ J − U_{x}J
⊆ J.

by (iii)′, (vi)′, (iv)′, (vii)′, (ii)′′. Lemma is proved.

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Proposition 3 Let J ⊆ Q̃ where Q̃ is a Jordan overring of J in which all elements of S are invertible. Then

Q = U_{S^{−1}}J = {U_{s^{−1}}x : s ∈ S, x ∈ J}

is a ring of S-fractions for J.

Proof If q ∈ Q̃ has a J-denominator s ∈ S, then U_{s}q ∈ J which implies q = U_{s^{−1}}x ∈ Q. So

Q′ = {q ∈ Q̃ : q has a J-denominator in S } ⊆ Q.

Lemma 10 tells us S^{−1} ⊆ Q′. We further claim that Q′ is actually a subalgebra of Q̃ so U_{S^{−1}}J ⊆ Q′ and hence

Q′ = U_{S^{−1}}J = Q.

If q_{1}, q_{2} ∈ Q′ then q_{1} and q_{2} have J-denominators s_{1}, s_{2} ∈ S. Hence, we can choose s′ ∈ U_{s_{1}}(S)∩U_{s_{2}}(S) and s′ will serve as a J- denominator for q_{1}+q_{2} so that q_{1}+q_{2} ∈ Q′. For α ∈ Φ, it is clear that s_{1} will be a denominator for αq and hence αq ∈ Q′. By Lemma 11, we conclude that U_{q_{1}}q_{2} ∈ Q′, and clearly 1 ∈ Q′. In fact, choose any s ∈ S and we have s ∘ 1 = s ∈ J and U_{s}1 = s^{2} ∈ J.

Thus Q′ is a unital subalgebra of Q̃. Hence, Q = Q′ and Q is a ring of S-fractions for J.

■

Now what remains is to show that if J satisfies the Ore condition with respect to the Ore monad S ⊆ J then we can construct an overring for J in which all of the elements of S will be invertible. The construction itself is a complicated process to which we devote the entire next section.

6 Constructing the Overring

6.1 Gathering the Tools

Constructing the required overring will require a lot of machinery. One of the more important tools we will use is the following proposition suggested by John Faulkner [1].

Proposition 4 Let A = A_{−2} ⊕ A_{−1} ⊕ A_{0} ⊕ A_{+1} ⊕ A_{+2} be a ℤ-graded associative algebra over Φ and suppose we have the following:

(1) V_{−1} and V_{+1} are subspaces with V_{σ} ⊆ A_{σ}, σ = ±1

(2) x ↦ x^{2} is a quadratic map from V_{σ} to A_{2σ} such that the corresponding bilinear map has (x_{1}, x_{2}) ↦ x_{1}x_{2}

(3) For x ∈ V_{σ} and y ∈ V_{−σ}, σ = ±1,

(a) Q_{x}y := x^{2}y + yx^{2} − xyx ∈ V_{σ}

(b) (Q_{x}y)^{2} = x^{2}y^{2}x^{2}

(c) x^{2}yx = xyx^{2}.

(4) The conditions 3(b) and 3(c) hold in all (free) scalar extensions of Φ.

Then (V_{+}, V_{−}) is a Jordan pair. Proof First notice that the above hypotheses give rise to a few useful facts:

(a) xx = 2x^{2} (hence, x^{2} = xx − x^{2})

(b) {x, y, z} = xzy − xyz − zyx + yzx = [x, [z, y]]

(c) x^{2}yz + xzyx = zyx^{2} + xyxz (by linearization of x^{2}yx = xyx^{2}).

(d) xz = zx for x, z ∈ V_{σ}, σ = ±1.

(e) x^{2}z = zx^{2} = xxz = xzx = 0 for x, z ∈ V_{σ}, σ = ±1.

We must verify that V = (V_{+}, V_{−}) with Q_{x}y = x^{2}y + yx^{2} − xyx satisfies (JP1)-(JP3). (JP1) Let x ∈ V^{σ} and y, w ∈ V^{−σ}, for σ = ±1. Then

D_{x,y}Q_{x}w = {x, y, Q_{x}w}
= x(Q_{x}w)y − xy(Q_{x}w − (Q_{x}w)yx + y(Q_{x}w)x
= x(x^{2}w + wx^{2} − xwx)y − xy(x^{2}w + wx^{2} − xwx)

−(x^{2}w + wx^{2} − xwx)yx + y(x^{2}w + wx^{2} − xwx)x

(3c) = (xx^{2}wy + x^{2}wxy − xxwxy) − (x^{2}yxw + xywx^{2} − xyxwx)

−(x^{2}wyx + wxyx^{2} − xwxyx) + (yx^{2}wx + ywx^{2}x − yxwxx)

(a) = (xx^{2}wy + x^{2}wxy − 2x^{2}wxy) − (x^{2}yxw + xywx^{2} − xyxwx)

−(x^{2}wyx + wxyx^{2} − xwxyx) + (yx^{2}wx + ywx^{2}x − 2yxwx^{2})

(a), (3c) = (xx^{2}wy − x^{2}wxy) − (x^{2}yxw + xywx^{2} − xyxwx)

−(x^{2}wyx + wxyx^{2} − xwxyx) + (−yxwx^{2} + ywxx^{2})

(a) = (xx^{2}wy − x^{2}wxy) − (x^{2}yxw + xywxx − xywx^{2} − xyxwx)

−(xxwyx − x^{2}wyx + wxyx^{2} − xwxyx) + (−yxwx^{2} + ywxx^{2})

(a) = x^{2}(xwy − wxy − yxw + wyx) + (xyw − wxy − yxw + ywx)x^{2}

+x(yxw + wxy − wyx − xwy)x

= x^{2}({yxw}) + ({yxw})x^{2}

−x(wyx + xwy − yxw − wxy)

(d) = x^{2}({yxw}) + ({yxw})x^{2}

−x(ywx + xwy − yxw − wxy)

= Q_{x}({yxw}).

(JP2) Let x, z ∈ V_{σ} and y ∈ V_{−σ}, σ = ±1. Then

D_{Q_{x}y,y}(z) = {Q_{x}y, y, z}
= (x^{2}y + yx^{2} − xyx)zy + yz(x^{2}y + yx^{2} − xyx)

−(x^{2}y + yx^{2} − xyx)yz − zy(x^{2}y + yx^{2} − xyx)

= (x^{2}yzy + yx^{2}zy − xyxzy) + (yzx^{2}y + yzyx^{2} − yzxyx)

−(x^{2}yyz + yx^{2}yz − xyxyz) − (zyx^{2}y + zyyx^{2} − zyxyx)

= (x^{2}yz − xyxz)y + (yx^{2}z + yzx^{2})y + y(zyx^{2} − xzyx)

−(x^{2}yyz + yx^{2}yz − xyxyz) − (zyx^{2}y + zyyx^{2} − zyxyx)

(e), (a) = (x^{2}yz − xyxz)y + 0 + y(zyx^{2} − xzyx)

−(2x^{2}y^{2}z + yx^{2}yz − xyxyz) − (zyx^{2}y + 2zy^{2}x^{2} − zyxyx)

(a) = (x^{2}yz − xyxz)y + y(zyx^{2} − xzyx)

−(xxy^{2}z + yx^{2}yz − xyxyz) − (zyx^{2}y + zy^{2}xx − zyxyx)

= (x^{2}yz − xyxz)y + y(zyx^{2} − xzyx)

−(xxy^{2})z − y(x^{2}yz) + x(yxy)z −(zyx^{2})y − z(y^{2}xx) + z(yxy)x

(c) = y(−xyxz) + (−xzyx)y − (xxy^{2})z − y(x^{2}yz) + x(yxy)z

−(zyx^{2})y − z(y^{2}xx) + z(yxy)x

= xz(−yxy) + (−yxy)xz

+x(−xy^{2} + yxy)z + z(−y^{2}x + yxy)

= xz(−yxy) + (−yxy)xz

+x(−xy^{2} + yxy)z + z(−y^{2}x + yxy) +0 + 0 + xzy^{2}x − xzy^{2}x + xy^{2}xz − xy^{2}xz

(d), (e) = xz(−yxy) + (−yxy)xz

+x(−xy^{2} + yxy)z + z(−y^{2}x + yxy) +xzxy^{2} + y^{2}xxz + xz(y^{2}x) − z(xy^{2})x + (xy^{2})xz − x(y^{2}x)z

= xz(y^{2}x + xy^{2} − yxy) + (y^{2}x + xy^{2} − yxy)xz

−x(y^{2}x + xy^{2} − yxy)z − z(y^{2}x + xy^{2} − yxy)x

= {x, Q_{y}x, z}.

(JP3) Let x ∈ V_{σ} and y, z ∈ V_{−σ}, σ = ±1. We want to show

Q_{x}Q_{y}Q_{x}(z) = Q_{Q_{x}y}(z).

Well,

Q_{x}Q_{y}Q_{x}(z) = Q_{x}Q_{y}(x^{2}z + zx^{2} − xzx)
= Q_{x}(y^{2}x^{2}z − yx^{2}zy + x^{2}zy^{2} − y^{2}xzx + yxzxy

−xzxy^{2} + y^{2}zx^{2} − yzx^{2}y + zx^{2}y^{2})

(e) = x^{2}y^{2}x^{2}z − xy^{2}x^{2}zx + y^{2}x^{2}zx^{2} − x^{2}yx^{2}zy

+xyx^{2}zyx − yx^{2}zyx^{2} − x^{2}y^{2}xzx + xy^{2}xzxx −y^{2}xzx^{2} + x^{2}yxzxy − xyxzxyx + yxzxyx^{2} −x^{2}xzxy^{2} + xxzxy^{2}x − xzxy^{2}x^{2} − x^{2}yzx^{2}y +xyzx^{2}yx − yzx^{2}yx^{2} + x^{2}zx^{2}y^{2} − xzx^{2}y^{2}x +zx^{2}y^{2}x^{2}

(a), (deg) = x^{2}y^{2}x^{2}z + zx^{2}y^{2}x^{2} − yx^{2}zyx^{2} − x^{2}yzx^{2}y

−xyxzxyx − xy^{2}x^{2}zx − xzx^{2}y^{2}x + xyzx^{2}yx +xyx^{2}zyx − xzxy^{2}x^{2} − x^{2}y^{2}xzx + yxzxyx^{2} +x^{2}yxzxy + 2x^{2}zxy^{2}x + 2xy^{2}xzx^{2}

= x^{2}y^{2}x^{2}z + zx^{2}y^{2}x^{2} − yx^{2}zyx^{2} − x^{2}yzx^{2}y

−xyxzxyx − xy^{2}x^{2}zx − xzx^{2}y^{2}x + xyzx^{2}yx +xyx^{2}zyx − xzxy^{2}x^{2} − x^{2}y^{2}xzx + yxzxyx^{2}

(3)  +x^{2}yxzxy + 2xzx^{2}y^{2}x + 2xy^{2}x^{2}zx
= x^{2}y^{2}x^{2}z + zx^{2}y^{2}x^{2} − yx^{2}zyx^{2} − x^{2}yzx^{2}y

−xyxzxyx + xy^{2}x^{2}zx + xzx^{2}y^{2}x + xyzx^{2}yx +xyx^{2}zyx − xzxy^{2}x^{2} − x^{2}y^{2}xzx + yxzxyx^{2} +x^{2}yxzxy.

While,

Q_{Q_{x}y}(z) = (Q_{x}y)^{2}z + z(Q_{x}y)^{2} − (Q_{x}y)z(Q_{x}y)
= x^{2}y^{2}x^{2}z + zx^{2}y^{2}x^{2} − x^{2}yzx^{2}y + x^{2}yzxyx

−x^{2}yzyx^{2} + xyxzx^{2} − xyxzxyx + xyxzyx^{2} −yx^{2}zx^{2}y + yx^{2}zxyx − yx^{2}zyx^{2}

(deg) = x^{2}y^{2}x^{2}z + zx^{2}y^{2}x^{2} − x^{2}yzx^{2}y + x^{2}yzxyx

−0 + xyxzx^{2} − xyxzxyx + xyxzyx^{2} −0 + yx^{2}zxyx − yx^{2}zyx^{2}.

Based on the conclusions of these two long strings of equalities, we have

Q_{x}Q_{y}Q_{x}(z) − Q_{Q_{x}y}(z) = xy^{2}x^{2}zx + xzx^{2}y^{2}x
+xyzx^{2}yx + xyx^{2}zyx
−xzxy^{2}x^{2} − x^{2}y^{2}xzx
−x^{2}yzxyx − xyxzyx^{2}
(3) = xy^{2}xzx^{2} + x^{2}zxy^{2}x
+xyzxyx^{2} + x^{2}yxzyx
−xzxy^{2}x^{2} − x^{2}y^{2}xzx
−x^{2}yzxyx − xyxzyx^{2}
= x^{2}(zxy^{2} + yxzy − y^{2}xz − yzxy)x
+x(y^{2}xz + yzxy − zxy^{2} − yxzy)x^{2}
(c) = x^{2}(0)x + x(0)x^{2}
= 0.

So (JP1)-(JP3) all hold in V. By hypothesis (4) in the statement of the proposition, we conclude that (JP1)-(JP3) will continue to hold in all free scalar extensions of Φ. Therefore, V is a Jordan pair and the proposition is proved.

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We will use this proposition to create the Jordan overring in which all of the elements of our Ore monad will be invertible. In order to use the proposition though, we must first find an appropriate 5-graded associative algebra and the appropriate subspaces. Before beginning our search we should define some useful concepts and take note of another useful proposition.

Definition 11 Let B be a vector space and let

𝔅 = {B_{α}}_{α∈Ω}

be a collection of subspaces of B. We say that 𝔅 is a downwardly directed set of subspaces if the following condition holds:

For any α, ω ∈ Ω there exists γ ∈ Ω such that B_{γ} ⊆ B_{α} ∩ B_{ω}.

Definition 12 Let B be a vector space and let

𝔅 = {B_{α}}_{α∈Ω}

be a downwardly directed set of subspaces of B. Also, let f be a linear map on B

f : B_{α} → B

for some B_{α}. We say that the map f is shrinkable if, given any γ ∈ Ω, there exists γ′ ∈ Ω such that f(B_{γ′}) ⊆ B_{γ}.

Proposition 5 Let B = B_{−1} ⊕ B_{0} ⊕ B_{+1} be a vector space and let

𝔅 = {B_{α}}_{α∈Ω}

be a set of downwardly directed graded subspaces of B where

B_{α} = (B_{α})_{−1} ⊕ (B_{α})_{0} ⊕ (B_{α})_{+1}.

Also, let

A = ⋃_{α∈Ω} Hom(B_{α}, B)) = ⋃_{α∈Ω} ( ∑_{k=−2}^{2} ∑_{k=j−i} Hom((B_{α})_{i}, B_{j}))

for i, j ∈ {±1, 0}, and set

Ā = {f ∈ A : f is shrinkable}.

On Ā define an equivalence relation by

f ∽ g if there exists γ ∈ Ω such that f_{|B_{γ}} = g_{|B_{γ}}.

Then Ã = Ā/(∽) forms a five-graded associative algebra under the usual addition and composition of maps.

(The proof is a lengthy but straightforward verification.)

6.2 Defining Ã_{J}

First, we define the 5-graded associative algebra we will need. If J is a quadratic

Jordan algebra then we can form the Jordan pair (J_{+}, J_{−}) with

Q_{x} = U_{x}, Q_{x,y} = U_{x,y}, D_{x,y} = V_{x,y}.

We define operators on (J_{+}, J_{−}) by

δ_{x_{σ},y_{−σ}} = (V_{x_{σ},y_{−σ}}^{σ}, −V_{y_{−σ},x_{σ}}^{−σ}),

where

V_{x_{σ},y_{−σ}}^{σ} : J_{σ} → J_{σ} and V_{y_{−σ},x_{σ}}^{−σ} : J_{−σ} → J_{−σ}

for σ = ±1. As mentioned in the background section of this paper, we can construct the T KK-Lie algebra

ˆ£ = Ĵ_{−} ⊕ StrucĴ⊕ Ĵ_{+}.

If S ⊆ J is an Ore monad in J then for each s ∈ S we can also form the Lie algebra

£_{s} := (K_{s})_{−1} ⊕ D̂_{0}^{s} ⊕ (K_{s})_{+1} ⊆ £(J) = £ ⊆ ˆ£

where

D̂_{0}^{s} = δ_{K_{s},K_{s}} ⊆ δ_{J,J} ⊆ D̂_{0}

is a Lie algebra of linear transformations on J.

Ignoring the Lie structure for the moment, we can think of £ as a Φ-module. Furthermore, since S ⊆ J is an Ore monad in J,

{£_{s}}_{s∈S}

forms a downwardly directed set of graded subspaces of £. Then, as described in Proposition 5, we can form

A_{J} = ⋃_{s∈S} Hom(£_{s}, £)

and set

Ā_{J} = {f ∈ A_{J} : f is shrinkable }.

We define an equivalence relation on Ā_{J} by

f ∽ g if there exists t ∈ S such that f_{|£_{t}} = g_{|£_{t}}

and obtain the 5-graded associative algebra

Ã_{J} = Ã_{−2} ⊕ Ã_{−1} ⊕ Ã_{0} ⊕ Ã_{+1} ⊕ Ã_{+2} = Ā_{J}/(∽),

where elements of Ã_{J} are equivalence classes of shrinkable linear maps.

Ã_{J} will function as the 5-graded algebra we need in order to apply Faulkner’s proposition. Within the ±1 spaces of Ã_{J} , we still need to define the appropriate subspaces.

6.3 Defining Ṽ_{σ}

Before defining the necessary subspaces, we introduce some terminology.

Definition 13 If L_{1} is a Lie algebra over Φ and L_{2}, L_{3} are Lie sub-algebras with

L_{3} ⊆ L_{2} ⊆ L_{1},

then a linear map Δ : L_{2} → L_{1} is called a derivation if

Δ[a, b] = [Δ(a), b] + [a, Δ(b)]

for a, b ∈ L_{2}. A pair of linear maps (Λ, Λ^{2}), with

Λ : L_{2} → L_{1}

and

Λ^{2} : L_{3} → L_{1},

is called a squared-derivation of L_{2} into L_{1} if Λ is a derivation and Λ^{2} is a divided square for Λ, meaning

Λ^{2}[a, b] = [Λ^{2}(a), b] + [a, Λ^{2}(b)] + [Λ(a), Λ(b)]

for a, b ∈ L_{3}, and

ΛΛ = 2Λ^{2}.

If L_{1} is ℤ-graded so that

L_{1} = ⨁_{k∈ℤ}L_{k},

then a linear map Θ : L_{2} → L_{1} is said to be of degree i if

Θ(a_{k}) ∈ L_{k+i}

for all k ∈ ℤ and a_{k} ∈ L_{k}. A squared derivation (Θ, Θ^{2}) is said to be of degree i if Θ is of degree i and Θ^{2} is of degree 2i.

Earlier we talked about some different types of denominators that can exist for elements of the quadratic Jordan algebra J. We now define a new notion of denominator for a squared-derivation in A_{J}.

Definition 14 Let (Δ, Δ^{2}) be a squared derivation of £_{s} into ˆ£ of degree σ = ±1, where Δ, Δ^{2} ∈ A_{J} for A_{J} as described above. We say that s ∈ S is a denominator for (Δ, Δ^{2}) if there exsists

d_{−σ} ∈ J_{−σ}

such that

(D1) for Δ(s) = (−D_{σ}, D_{−σ}) ∈ D̂_{0} we have Δ(U_{s}â) = δ_{â,d_{−σ}} − δ_{D_{σ}(â),s}

(D2) U_{s}D_{σ}(â) = {d_{−σ}, â, s} ∈ J_{−σ}

(D3) U_{s}Δ^{2}U_{s} = U_{d_{−σ}}

where we think of â ∈ Ĵ_{σ} and s ∈ S ⊆ J_{−σ}.

NOTE: As we deal with derivations, it will be helpful to keep the following fact in mind. Any derivation of a Lie algebra is completely determined by its action on a set of generators. More particularly, if Δ : £_{s} → ˆ£ is a derivation of degree σ = ±1, for some s ∈ S, then Δ is completely determined by its action on (K_{s})_{−σ} since

Δ(δ_{a_{σ},b_{−σ}}) = −Δ[a_{σ}, b_{−σ}] = −([Δ(a_{σ}), b_{−σ}] + [a_{σ}, Δ(b_{−σ})]) = 0 + Δ(b_{−σ})[a_{σ}]

for a_{σ} ∈ (K_{s})_{σ} and b_{−σ} ∈ (K_{s})_{−σ}.

We pause now to state and prove a lemma and a corollary concerning denominators of squared-derivations.

Lemma 12 Let (Δ, Δ^{2}) be a squared derivation of degree σ with denominator s ∈ S, where Δ, Δ^{2} ∈ A_{J}. If y = U_{s}z ∈ U_{s}J then the following hold:

(1)′ for Δ(y) = (−D_{σ}^{y}, D_{−σ}^{y}) ∈ D̂_{0} we have Δ(U_{y}â) = δ_{â,d_{−σ}^{y}} − δ_{D_{σ}^{y}(â),y}

(2)′ U_{y}D_{σ}^{y}(â) = {d_{−σ}^{y}, â, y} ∈ J_{−σ}

(3)′ U_{y}Δ^{2}U_{y} = U_{d_{−σ}^{y}}

where

d_{−σ}^{y} = U_{s}U_{z}d_{−σ},

and

D_{σ}^{y}(â) = {D_{σ}(z), s, â} − {z, d_{−σ}, â},

and we think of â ∈ Ĵ_{σ} and s ∈ S ⊆ J_{−σ}.

Proof We claim that if we set

d_{−σ}^{y} = U_{s}U_{z}d_{−σ},

and

D_{σ}^{y}(â) = {D_{σ}(z), s, â} − {z, d_{−σ}, â}
D_{σ}^{y}(â) = {s, D_{σ}(z), â} − {d_{−σ}, x, â}

then the following hold: (1)′ Δ(y) = (−D_{σ}^{y}, D_{−σ}^{y}) ∈ D̂_{0} and Δ(U_{y}â) = δ_{â,d_{−σ}^{y}} − δ_{D_{σ}^{y}(â),s} (2)′ U_{s}D_{σ}^{y}(â) = {d_{−σ}^{y}, â, y} ∈ J_{−σ} (3)′ U_{s}Δ^{2}U_{s} = U_{d_{−σ}} where we are thinking of â ∈ Ĵ_{σ}, z ∈ J_{σ}, and s ∈ J_{−σ}. Well,

Δ(y)[b̂_{σ}] = (δ_{z,d_{−σ}} − δ_{D_{σ}(z),s})[b̂_{−σ}] = −{d_{−σ}, z, b̂_{−σ}} + {s, D_{σ}(z), b̂_{−σ}} = D_{−σ}^{y}(b̂)

Δ(y)[â_{σ}] = (δ_{z,d_{−σ}} − δ_{D_{σ}(z),s})[â_{σ}] = {z, d_{−σ}, â_{σ}} − {D_{σ}(z), s, â_{σ}} = −D_{σ}^{y}(â) and

Δ(U_{y}â) = Δ(U_{s}U_{z}U_{s}â) = δ_{U_{z}U_{s}â,d_{−σ}} − δ_{D_{σ}(U_{z}U_{s}â),s}

by virtue of the fact that s ∈ S is a denominator for Δ. When acting on an element b_{σ} ∈ J_{σ} we have

Δ(U_{y}â)[b_{σ}] = Δ(U_{s}U_{z}U_{s}â)[b_{σ}]
= δ_{U_{z}U_{s}â,d_{−σ}}(b_{σ}) − δ_{D_{σ}(U_{z}U_{s}â),s}(b_{σ})
= {U_{z}U_{s}a, d_{−σ}, b_{σ}} − {D_{σ}(U_{z}U_{s}â), s, b_{σ}}

Applying U_{s} we obtain

U_{s}Δ(U_{y}â)[b_{σ}] = U_{s}{U_{z}U_{s}â, d_{−σ}, b_{σ}} − U_{s}{D_{σ}(U_{z}U_{s}â), s, b_{σ}}
(J2) = U_{s}{U_{z}U_{s}â, d_{−σ}, b_{σ}} − {U_{s}D_{σ}(U_{z}U_{s}â), b_{σ}, s}
(D2) = U_{s}{U_{z}U_{s}â, d_{−σ}, b_{σ}} − {{d_{−σ}, U_{z}U_{s}â, s}, b_{σ}, s}
(J6) = −{d_{−σ}, U_{z}U_{s}â, U_{s}b_{σ}}.

We hope that the above is equal to

U_{s} ((δ_{â,d_{−σ}^{y}} − δ_{D_{σ}^{y}(â),y})[b_{σ}]) .

First, notice the following:

U_{s}D_{σ}^{y}(â) = U_{s}({D_{σ}(z), s, â} − {z, d_{−σ}, â})
= {U_{s}D_{σ}(z), â, s} − U_{s}{z, d_{−σ}, â}
(D2) = {{d_{−σ}, z, s}, â, s} − U_{s}{z, d_{−σ}, â}
(J6) = {d_{−σ}, z, U_{s}â}.

We label this equality:

U_{s}D_{σ}^{y}(â) = {d_{−σ}, z, U_{s}â}. (A)

Then,

U_{s} ((δ_{â,d_{−σ}^{y}} − δ_{D_{σ}^{y}(â),y})[b_{σ}])
= U_{s}{â, U_{s}U_{z}d_{−σ}, b_{σ}}
−U_{s}{D_{σ}^{y}(â), U_{s}z, b_{σ}}
= U_{s}U_{â,b_{σ}}U_{s}(U_{z}d_{−σ}) − U_{s}U_{D_{σ}^{y}(â),b_{σ}}U_{s}z
(J3) = U_{U_{s}â,U_{s}b_{σ}}U_{z}d_{−σ} − U_{U_{s}D_{σ}^{y}(â),U_{s}b_{σ}}z
(A) = {U_{s}â, U_{z}d_{−σ}, U_{s}b_{σ}} − {{U_{s}â, z, d_{−σ}}, z, U_{s}b_{σ}}
= {U_{s}â, U_{z}d_{−σ}, U_{s}b_{σ}} − {{d_{−σ}, z, U_{s}â}, z, U_{s}b_{σ}}
= −{d_{−σ}, U_{z}U_{s}â, U_{s}b_{σ}}

Therefore, we do have

U_{s}Δ(U_{y}â)[b_{σ}] = U_{s}(δ_{â,d_{−σ}^{y}} − δ_{D_{σ}^{y}(â),U_{s}z})[b_{σ}].

Since U_{s} is injective we conclude that

Δ(U_{y}â)[b_{σ}] = δ_{a,d_{−σ}^{y}} − δ_{D_{σ}^{y}(â),U_{s}z})[b_{σ}]

which gives (1)′. What about (2)′? For â ∈ Ĵ_{σ}

U_{y}D_{σ}^{y}(â) = U_{s}U_{z}U_{s}({D_{σ}(z), s, â} − {t, d_{−σ}, a})
= U_{s}U_{z}(U_{s}{D_{σ}(t), s, â} − U_{s}{z, d_{−σ}, â})
(J2) = U_{s}U_{z}({U_{s}D_{σ}(z), â, s} − U_{s}V_{z,d_{−σ}} â)
= U_{s}U_{z}(U_{V_{d_{−σ},z}s,s}â − U_{s}V_{z,d_{−σ}} â)
(J13) = U_{s}U_{z}(V_{d_{−σ},z}U_{s}â)
(J2) = U_{s}U_{U_{z}d_{−σ},z}U_{s}â
(J3) = U_{U_{s}U_{z}d_{−σ},U_{s}z}â
= {U_{s}U_{z}d_{−σ}, â, U_{s}z}
= {d_{−σ}^{y}, â, y}.

This gives (2)′. As for (3)′,

U_{y}Δ^{2}U_{y} = U_{s}U_{z}(U_{s}Δ^{2}U_{s})U_{z}U_{s}
(3) = U_{s}U_{z}(U_{d_{−σ}})U_{z}U_{s}
= U_{s}(U_{z}d_{−σ})U_{s}
= U_{U_{s}U_{z}d_{−σ}}
= U_{d_{−σ}^{y}}

This proves the lemma.

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Corollary 1 Let (Δ, Δ^{2}) be a squared derivation of degree σ = ±1 with denominator s ∈ S, where Δ, Δ^{2} ∈ A_{J}. Then U_{s}t is a denominator for (Δ, Δ^{2}), for any t ∈ S.

Proof This follows immediately from lemma?.

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Now our 5-graded associative algebra is Ã_{J} which is a collection of equivalence classes of shrinkable maps from A_{J}. We stop now to check that both Δ and Δ^{2} are shrinkable for any squared-derivation (Δ, Δ^{2}) having denominator.

Lemma 13 Let (Δ, Δ^{2}) be a squared-derivation of degree σ = ±1 where

Δ, Δ^{2} : £_{s} → ˆ£

for s ∈ S a denominator for (Δ, Δ^{2}). Then both Δ and Δ^{2} are shrinkable. That is, given any t ∈ S there exists s′ ∈ U_{s}S such that

Δ(£_{s′}) ⊆ £_{t} and Δ^{2}(£_{s′}) ⊆ £_{t}.

Proof Fix an arbitrary t ∈ S. If s ∈ S is a denominator for (Δ, Δ^{2}) we have seen

that any s̃ ∈ U_{s}S will also be a denominator, and by the CIMP we can choose s̃ so that

s̃ ∈ U_{s}S ∩ U_{t}S.

So, we will assume from the start that s ∈ S is a denominator for (Δ, Δ^{2}) and that we actually have

s ∈ U_{t}S ⊆ U_{t}J ⊆ K_{t}.

Since s is a denominator for (Δ, Δ^{2}), we know there exists

d_{−σ} ∈ J_{−σ}

satisfying (D1)-(D3).

By the Ore condition we can choose an element r ∈ U_{t^{3}}S ⊆ U_{t}S to be a super t-denominator for the element

d̃= s ∘ D_{σ}(t)

so that

r ∘ (s ∘ D_{σ}(t)) ∈ U_{t}J.

Also, for â ∈ Ĵ, we have

U_{r}â ∘ d̃ = (U_{d̃∘r,r} − U_{r}Vd̃)â
∈ (U_{U_{t}J,U_{t}J} − U_{U_{t}J})Ĵ
⊆ U_{t}J.

Hence,

K_{r} ∘ d̃⊆ U_{t}J. (A)

Set y = U_{s}t and s′_{1} = U_{y}r. We claim that

Δ((K_{s′_{1}})_{−σ}) ⊆ δ_{U_{t}J,U_{t}J} ⊆ δ_{K_{t},K_{t}}.

Notice that

K_{s′_{1}} = ΦU_{y}r + U_{U_{y}r}Ĵ ⊆ U_{y}(Φr + U_{r}J) ⊆ U_{y}K_{r}.

So, let us consider

Δ(U_{y}k) for k ∈ K_{r},

where we think of U_{y}k as a degree −σ element. From Corollary 1 we know that y = U_{s}t is a denominator for Δ and specifically that

Δ(U_{y}k) = δ_{k,U_{s}U_{t}d_{−σ}} − δ_{{D_{σ}(t),s,k}−{t,d_{−σ},k},y}.

Look at the subscripts on the δ terms in this equation. Since r, s ∈ U_{t}J, we know that

k ∈ K_{r} ⊆ U_{t}J, y = U_{s}t ∈ U_{s}J ⊆ U_{t}J U_{s}U_{t}d_{−σ} ∈ U_{t}J.

What about

{D_{σ}(t), s, k} − {t, d_{−σ}, k}?

Well

{t, d_{−σ}, k} ∈ V_{t,J}U_{t}J ⊆ U_{t}J.

As for {D_{σ}(t), s, k}, we have

{D_{σ}(t), s, k} = (D_{σ}(t) ∘ s) ∘ k − {s, D_{σ}(t), k} ∈ U_{t}J − {U_{t}J, J, U_{t}J} ⊆ U_{t}J

by (A).

So all of the subscripts of the δ terms are in U_{t}J. We conclude that

Δ((K_{s′_{1}})_{−σ}) ⊆ δ_{U_{t}J,U_{t}J} ⊆ δ_{K_{t},K_{t}}

as desired.

Now, since Δ is of degree σ, we know that

Δ((K_{s′_{1}})_{σ}) = 0.

Furthermore, if

δ_{a_{σ},b_{−σ}} ∈ D̂_{0}^{s′_{1}}

for a_{σ} ∈ (K_{s′_{1}})_{σ} and b_{−σ} ∈ (K_{s′_{1}})_{−σ}, then

Δ(δ_{a_{σ},b_{−σ}}) = −Δ([a_{σ}, b_{−σ}]
= −[Δ(a_{σ}), b_{−σ}] − [a_{σ}, Δ(b_{−σ})]
= −[0, b_{−σ}] − [a_{σ}, Δ(b_{−σ})]
= 0 + Δ(b_{−σ})[a_{σ}]
∈ δ_{K_{t},K_{t}}(K_{t})
⊆ K_{t}.

Therefore

Δ(£_{s′}) ⊆ £_{t}

as desired.

Now we consider Δ^{2}. Since J satisfies the Ore condition with respect to S, Proposition 1 allows us to select a super U_{s}t-denominator for d_{−σ}. Set

s′_{2} = U_{U_{s}t_{1}}t.

We claim that

Δ^{2}(£_{s′_{2}}) ⊆ £_{t}.

Let x̂ ∈ Ĵ_{σ} and think of s′_{2} ∈ S ⊆ J_{−σ}, so that U_{s′_{2}} x̂ ∈ J_{−σ}. Then, since t_{1} is a super denominator we have

U_{s}Δ^{2}(U_{s′_{2}} x̂) = U_{s}Δ^{2}U_{s}U_{t_{1}}U_{s}U_{t}U_{s}U_{t_{1}}U_{s}[x̂]
= (U_{s}Δ^{2}U_{s})U_{t_{1}}U_{s}U_{t}U_{s}U_{t_{1}}U_{s}[x̂]
(3) = U_{d_{−σ}}U_{t_{1}}U_{s}U_{t}U_{s}U_{t_{1}}U_{s}[x̂]
= U_{d_{−σ}}U_{t_{1}}U_{U_{s}t}U_{t_{1}}U_{s}[x̂]
∈ U_{U_{s}t}Ĵ
⊆ U_{s}U_{t}J.

Since s ∈ S is injective, this gives

Δ^{2}(U_{s′_{2}} x̂) ∈ U_{t}J. (B)

Similarly, looking at Δ^{2}(s′_{2}), we have

U_{s}Δ^{2}(s′_{2}) = U_{s}Δ^{2}U_{s}U_{t_{1}}U_{s}t
= (U_{s}Δ^{2}U_{s})U_{t_{1}}U_{s}t
(3) = U_{d_{−σ}}U_{t_{1}}U_{s}t
∈ U_{U_{s}t}Ĵ
⊆ U_{s}U_{t}J

Again, since s ∈ S is injective we conclude

Δ^{2}(s′_{2}) ∈ (U_{t}J)_{σ}. (C)

Together (B) and (C) give

Δ^{2}((K_{s′_{2}})_{−σ}) ⊆ (U_{t}J)_{σ}.

Now Δ^{2} is of degree 2σ so

Δ^{2}((K_{s′_{2}})_{σ}) = 0 and Δ^{2}(D̂_{0}^{s′_{2}}) = 0.

Therefore we conclude

Δ^{2}(£_{s′_{2}})) ⊆ £_{t}

as desired.

So we have shown that Δ is shrinkable using s′_{1} while Δ^{2} is shrinkable using s′_{2}. By the CIMP we can select an element

s′ ∈ U_{s′}S ∩ U_{s′′}S

and conclude that both Δ and Δ^{2} are shrinkable using this same element.

■

We are now ready to explicitly define the subspaces of Ã_{J} we will use. We define

V_{σ} ⊆ A_{J}

for σ = ±1, to be the set of all Δ ∈ A_{J} such that the following holds:

Δ has a divided square Δ^{2} such that (Δ, Δ^{2}) is a squared-derivation of degree σ having denominator in S.

We have just shown above that the elements of V_{σ} are shrinkable for σ = ±1. Therefore, using earlier notation, we do have

V_{σ} ⊆ Ā_{J}

for σ = ±1. Then, if we set

Ṽ_{σ} = V_{σ}/(∽)

for σ = ±1, where ∽ is as defined before, we have Ṽ_{σ} ⊆ Ā_{J}/(∽) = Ã_{J}.

It turns out that elements of Ṽ_{σ}, σ = ±1, commute with each other. This fact will be helpful later on so we stop here to prove it after introducing a necessary lemma.

Lemma 14 Let (Δ, Δ^{2}) be a squared-derivation of degree σ = ±1, with denominator s ∈ S. Then, for b̂, ẑ ∈ Ĵ,

(U_{s}Δ(U_{s}ẑ))[b̂] = −{d_{−σ}, ẑ, U_{s}b̂}

where d_{−σ} is as described in the definition of denominator of a squared-derivation and we think of s ∈ S ⊆ J_{−σ}, and b̂, ẑ ∈ Ĵ_{σ}.

Proof Since s ∈ S is a denominator for (Δ, Δ^{2}), we know there exists d_{−σ} satisfying (D1)-(D3). So,

U_{s} (Δ(U_{s}ẑ)[b̂]) = U_{s}(δ_{ẑ,d_{−σ}}[b̂] − δ_{D_{σ}(ẑ),s}[b̂])
= U_{s}V_{ẑ,d_{−σ}}[b̂] − U_{s}V_{D_{σ}(ẑ),s}[b̂]
(J13), (J2) = U_{{d_{−σ},ẑ,s},s}[b̂] − V_{d_{−σ},ẑ}U_{s}[b̂] − U_{U_{s}D_{σ}(ẑ),s}[b̂]
(2) = U_{{d_{−σ},ẑ,s},s}[b̂] − V_{d_{−σ},ẑ}U_{s}[b̂] − U_{{d_{−σ},ẑ,s},s}[b̂]
= −{d_{−σ}, ẑ, U_{s}b̂}.
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Lemma 15 Let Δ, Λ ∈ Ṽ_{σ}, σ = ±1. Then ΔΛ = ΛΔ.

Proof Recall, since we are looking at Δ and Λ as elements of Ã_{J}, we only need show that

ΔΛ = ΛΔ

on £_{s′} for some s′ ∈ S.

By Corollary 1, and since S is an Ore monad, we can find an element s ∈ S such that s is a denominator for both (Δ, Δ^{2}) and (Λ, Λ^{2}). That is, there exist d_{−σ} such that (D1)-(D3) hold for (Δ, Δ^{2}) and e_{−σ} such that (D1)-(D3) hold for (Λ, Λ^{2}). Let ẑ ∈ Ĵ_{σ},

t ∈ S, and set

s′_{1} = U_{s}t.

Then by Lemma 12

U_{s} ((Δ(Λ(U_{s′_{1}} ẑ))) = U_{s} (Δ(δ_{ẑ,U_{s}U_{t}e_{−σ}} − δ_{{E_{σ}(t),s,ẑ}−{t,e_{−σ},ẑ},U_{s}t})) .

Set

A = U_{s} (Δ(δ_{ẑ,U_{s}U_{t}e_{−σ}}))
B = U_{s} (Δ(δ_{{E_{σ}(t),s,ẑ}−{t,e_{−σ},ẑ},U_{s}t})) .

Then

A = U_{s} (Δ(U_{s}U_{t}e_{−σ})[ẑ])
(lem14) = −(V_{d_{−σ},U_{t}e_{−σ}}U_{s})[ẑ]
= −V_{U_{s}ẑ,U_{t}e_{−σ}}[d]
(J8) = (V_{e_{−σ},U_{t}U_{s}ẑ} − V_{{U_{s}ẑ,t,e_{−σ}},t})[d_{−σ}]
= V_{d_{−σ},U_{t}U_{s}ẑ}(e_{−σ}) − V_{d_{−σ},t}U_{e_{−σ},U_{s}ẑ}(t)
= {d_{−σ}, U_{t}U_{s}ẑ, e_{−σ}} − {d_{−σ}, t, {e_{−σ}, t, U_{s}ẑ}}

and

B = U_{s}Δ(U_{s}t)[{E_{σ}(t), s, ẑ}] − U_{s}Δ(U_{s}t)[{t, e_{−σ}, ẑ}]
(lem14) = −{d_{−σ}, t, U_{s}({E_{σ}(t), s, ẑ})} + {d_{−σ}, t, U_{s}({t, e_{−σ}, ẑ})}
(J2) = −{d_{−σ}, t, {U_{s}E_{σ}(t), ẑ, s} − U_{s}({t, e_{−σ}, ẑ})}
(D2) = −{d_{−σ}, t, {{e_{−σ}, t, s}, ẑ, s} − U_{s}({t, e_{−σ}, ẑ})}
(J6) = −{d_{−σ}, t, {e_{−σ}, t, U_{s}ẑ}}.

So,

U_{s}(Δ(Λ(U_{s′} ẑ))) = A − B
= {d_{−σ}, U_{t}U_{s}ẑ, e_{−σ}}
= U_{s}(Λ(Δ(U_{s′} ẑ)))

by symmetry. Since s ∈ S is injective, we conclude that

ΔΛ(U_{s′_{1}} ẑ) = ΛΔ(U_{s′_{1}} ẑ)

and hence

ΔΛ(U_{s′_{1}} Ĵ) = ΛΔ(U_{s′_{1}} Ĵ).

Now, let

s′ = U_{s′_{1}}s.

Then,

(K_{s′})_{−σ} ⊆ U_{s′_{1}} Ĵ

and therefore

(ΔΛ − ΛΔ)[(K_{s′})_{−σ}] = 0.

Since both Δ and Λ are of degree σ, we know that

ΔΛ(D̂_{0}^{s′}) = ΛΔ(D̂_{0}^{s′}) = 0

and

ΔΛ((K_{s′})_{σ}) = ΛΔ((K_{s′})_{σ}) = 0.

Therefore, we have

ΔΛ = ΛΔ on £_{s′}

as desired.

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6.4 Meeting the Requirements

We should pause for a moment and review exactly where we are. Given a quadratic Jordan algebra J that satisfies (AX1)-(AX3) and that satisfies the Ore condition with respect to an Ore monad S ⊆ J, we hope to imbed J in a Jordan overring in which all of the elements of S are invertible. The Jordan overring we will use ultimately is one we will construct by employing Proposition 4 as proposed by John Faulkner. In order to use this proposition we need two things:

(1) a 5-graded associative algebra

(2) two subspace lying within the ±1 subspaces of our algebra, that satisfy the axioms mentioned in John’s proposition.

Once we have found these needed ingredients we must then verify that J can be imbedded in the overring and that all of the elements of S are invertible in the overring.

We have found the required 5-graded associative algebra by constructing Ã_{J}. We have also constructed Ṽ_{σ}, σ = ±1, but have not yet verified that they indeed have the properties we need. We will now carry out that verification in the following manner:

(1) we will show that Ṽ_{σ}, σ = ±1, are linear subspaces.

(2) we will show that Ṽ_{σ}, σ = ±1, contain the elements we will need to establish an imbedding of J and to “invert” the elements of S.

(2) we will show that Ṽ_{σ}, σ = ±1, satisfy the axioms stipulated in Faulkner’s proposition.

6.4.1 Linear Subspaces

We need one fact before proving that Ṽ_{σ}, σ = ±1, are linear subspaces.

Lemma 16 Let (Δ, Δ^{2}) be a squared derivation for Δ, Δ^{2} ∈ Ã_{J}, and let s ∈ S be a denominator for (Δ, Δ^{2}) so that there exists d_{−σ} satisfying (D1)-(D3). If t is a super s-denominator for d_{−σ}, and b = U_{s}ĉ for some ĉ ∈ Ĵ, and c̃ ∈ Ĵ is such that

{d_{−σ}, t, b} = {d_{−σ}, t, U_{s}ĉ} = U_{s}c̃,

then,

D_{σ}(U_{t}U_{s}ĉ) = {t, s, c̃} − {ĉ, s, U_{t}d_{−σ}}.

Proof

U_{s}D_{σ}(U_{t}U_{s}ĉ) = {d_{−σ}, U_{t}U_{s}ĉ, s}
= {s, U_{t}U_{s}ĉ, d_{−σ}}
(J8) = {s, t, {d, t, U_{s}ĉ}} − {s, U_{t}d_{−σ}, U_{s}ĉ}
= {s, t, U_{s}c̃} − {U_{s}ĉ, U_{t}d_{−σ}, s}
= U_{s}({t, s, c̃} − {ĉ, s, U_{t}d_{−σ}}).

By injectivity, we cancel the U_{s} and obtain

D_{σ}(U_{t}U_{s}ĉ) = {t, s, c̃} − {ĉ, s, U_{t}d_{−σ}},

as desired.

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Lemma 17 Ṽ_{σ} is a linear subspace of Ã_{J} for σ = ±1.

Proof If Δ ∈ Ṽ_{σ} and α ∈ Φ, then it is clear that αΔ ∈ Ṽ_{σ} where αΔ has the associated divided square α^{2}Δ^{2}.

It is not so clear that Δ + Λ ∈ Ṽ_{σ} if Δ, Λ ∈ Ṽ_{σ}. So, let Δ, Λ ∈ Ṽ_{σ}. By Corollary 1 and the CIMP we can choose s ∈ S so that s is a denominator for both (Δ, Δ^{2}) and (Λ, Λ^{2}). That is, there exist d_{−σ} such that (D1)-(D3) hold for (Δ, Δ^{2}), and e_{−σ} such that (D1)-(D3) hold for (Λ, Λ^{2}). Now set

(Δ + Λ)^{2} := Δ^{2} + Λ^{2} + ΔΛ.

We claim that (Δ + Λ, (Δ + Λ)^{2}) is a squared derivation with denominator s.

It is easy to verify that, as defined, (Δ + Λ)^{2} is a pseudo-derivation. What remains to be seen is that s is in fact a denominator for (Δ + Λ, (Δ + Λ)^{2}). Set

h_{−σ} = d_{−σ} + e_{−σ}

and note that

(Δ + Λ)[s] = Δ(s) + Λ(s) = (−(D_{σ} + E_{σ}), D_{−σ} + E_{−σ}) =: (−H_{σ}, H_{−σ}).

Then:

(A)

(Δ + Λ)[U_{s}â] = Δ[U_{s}â] + Λ[U_{s}â]
= δ_{â,d_{−σ}} − δ_{D_{σ}(â),s} + δ_{â,e_{−σ}} − δ_{E_{σ}(â),s}
= δ_{â,d_{−σ}+e_{−σ}} − δ_{(D_{σ}+E_{σ})[â],s}
= δ_{â,h_{−σ}} − δ_{H_{σ}(â),s}.

(B)

U_{s}H_{σ}(â) = U_{s}D_{σ}(â) + U_{s}E_{σ}(â)
= {d_{−σ}, â, s} + {e_{−σ}, â, s}
= {d_{−σ} + e_{−σ}, â, s}
= {h_{−σ}, â, s}.

(C) If t is a super s-denominator for e_{−σ}, b = U_{s}U_{t}ĉ for ĉ ∈ Ĵ, and c̃ ∈ Ĵ is such

that

{e_{−σ}, t, b} = {e_{−σ}, t, U_{s}ĉ} = U_{s}c̃,

then

U_{s}(ΔΛ)U_{s}[U_{t}U_{s}ĉ] = U_{s}Δ(Λ[U_{s}U_{t}U_{s}ĉ])
(lem12) = U_{s}Δ [δ_{U_{t}U_{s}ĉ,e_{−σ}} − δ_{E_{σ}(U_{t}U_{s}ĉ),s}]
= U_{s}Δ [(−δ_{U_{t}e_{−σ},U_{s}ĉ} + δ_{t,{U_{s}ĉ,t,e_{−σ}}})
−δ_{{t,s,c̃}−{ĉ,s,U_{t}e_{−σ}},s}]
(J8), (J9) = U_{s}Δ [−δ_{U_{t}e_{−σ},U_{s}ĉ} + δ_{t,U_{s}c̃} − δ_{c̃,U_{s}t} − δ_{t,U_{s}c̃}
+δ_{ĉ,U_{s}U_{t}e_{−σ}} + δ_{U_{t}e_{−σ},U_{s}ĉ}]
= U_{s}Δ[−δ_{c̃,U_{s}t} + δ_{ĉ,U_{s}U_{t}e_{−σ}}]
= U_{s} [−Δ(U_{s}t)[c̃] + Δ(U_{s}U_{t}e_{−σ})[ĉ]]
= {d_{−σ}, t, U_{s}c̃} − {d_{−σ}, U_{t}e_{−σ}, U_{s}ĉ}
= {d_{−σ}, t, {e_{−σ}, t, U_{s}ĉ}} − {d_{−σ}, U_{t}e_{−σ}, U_{s}ĉ}
= {d_{−σ}, U_{t}U_{s}ĉ, e_{−σ}}
= U_{d_{−σ},e_{−σ}}[U_{t}U_{s}ĉ].

So

U_{s}(ΔΛ)U_{s}[U_{t}U_{s}ĉ] = U_{d_{−σ},e_{−σ}}[U_{t}U_{s}ĉ]. (∗)

Therefore,

U_{s}(Δ + Λ)^{2}U_{s}[U_{t}U_{s}ĉ] = (U_{s}Δ^{2}U_{s} + U_{s}Λ^{2}U_{s} + U_{s}(ΔΛ)U_{s})[U_{t}U_{s}ĉ]
(D3) = (U_{d_{−σ}} + U_{e_{−σ}})[U_{t}U_{s}ĉ] + U_{s}(ΔΛ)U_{s}[U_{t}U_{s}ĉ]
(∗) = (U_{d_{−σ}} + U_{e_{−σ}})[U_{t}U_{s}ĉ] + U_{d_{−σ},e_{−σ}}[U_{t}U_{s}ĉ]
= U_{d_{−σ}+e_{−σ}}[U_{t}U_{s}ĉ]
= U_{h_{−σ}}[U_{t}U_{s}ĉ].

We conclude that

U_{s}(Δ + Λ)^{2}U_{s} = U_{h_{−σ}}

on U_{s′} Ĵ for

s′ = U_{t}s.

This, coupled with the fact that U_{s}(Δ+Λ)^{2}U_{s} and U_{h_{−σ}} are of degree −2σ, tells us that

U_{s}(Δ + Λ)^{2}U_{s}[(K_{s′})_{−σ}] = U_{h_{−σ}}[(K_{s′})_{−σ}] = 0
U_{s}(Δ + Λ)^{2}U_{s}[D̂_{0}^{s′}] = U_{h_{−σ}}[D̂_{0}^{s′}] = 0

and

U_{s}(Δ + Λ)^{2}U_{s}[(K_{s′})_{σ}] = U_{h_{−σ}}[(K_{s′})_{σ}].

Hence

U_{s}(Δ + Λ)^{2}U_{s}[£_{s′}] = U_{h_{−σ}}[£_{s′}]

and thus

U_{s}(Δ + Λ)^{2}U_{s} = U_{h_{−σ}} in Ṽ_{σ}.

Together (A),(B), and (C) tell us that s is indeed a denominator for

(Δ + Λ, (Δ + Λ)^{2}).

Therefore we do have (Δ + Λ) ∈ Ṽ_{σ}, and Ṽ_{σ} is a subspace of Ã_{J} as desired.

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6.4.2 Λ_{s}, ad(x) ∈ Ṽ_{σ}

For each s ∈ S define

Λ_{s} : £_{s} → ˆ£

to be a degree σ (σ = − ± 1) linear map by

Λ_{s}[(U_{s}â)_{−σ}] := δ_{s,â} = −δ_{â,s} for â ∈ Ĵ_{σ}, s ∈ J_{−σ}
Λ_{s}[(K_{s})_{σ}] := 0
Λ_{s}[(s)_{−σ}] := 2σξ
Λ_{s}[δ_{x_{σ},y_{−σ}}] := Λ_{s}(y_{−σ})[x_{σ}] for x_{σ} ∈ (K_{s})_{σ}, y_{−σ} ∈ (K_{s})_{−σ}.

and define

Λ_{s}^{2} : £_{s^{2}} → £

to be a degree 2σ (σ = ±1) linear map, by

Λ_{s^{2}}[s_{−σ}^{2}] := 2ξ
Λ_{s}^{2}[U_{s}â] := â for â ∈ Ĵ_{σ}, s ∈ J_{−σ}
Λ_{s}^{2}[(K_{s})_{σ}] := 0
Λ_{s}^{2}[D̂_{0}^{s}] := 0.

We must check that the Λ_{s} maps, as defined above, are actually in our subspaces Ṽ_{σ}, σ = ±1. We introduce several lemmas to help us establish this fact.

Lemma 18 For each s ∈ S, Λ_{s} is a derivation from £_{s} into ˆ£.

Proof We need to show that, for a, b ∈ £_{s} we have

(∗) Λ_{s}[a, b] = [Λ_{s}(a), b] + [a, Λ_{s}(b)].

Taking a, b ∈ £_{s}, gives six different cases depending on whether a and b individually lie in (K_{s})_{+1}, (K_{s})_{−1}, or D_{0}^{s}. Let

[+, +]

denote the case where a ∈ (K_{s})_{+1} and b ∈ (K_{s})_{+1}, and

[0, +]

denote the case where a ∈ D_{0}^{s} and b ∈ (K_{s})_{+1}, and so on. Then (∗) holds trivially in

the following cases by degree arguments:

[+, +], [+, 0].

This leaves four cases that need to be checked.

[+, −]:

Λ_{s}[a, b] = Λ_{s}[−δ_{a,b}]
= −Λ_{s}(b)[a]
= 0 − [Λ_{s}(b), a]
= [Λ_{s}(a), b] + [a, Λ_{s}(b)].

[−, −]: Assume a = U_{s}x and b = U_{s}y.

Λ_{s}[a, b] = 0
= {U_{s}x, y, s} − {U_{s}x, y, s}
(J2) = {s, x, U_{s}y} − {s, y, U_{s}x}
= δ_{s,x}(U_{s}y) − δ_{s,y}(U_{s}x)
= Λ_{s}(U_{s}x)[U_{s}y] − Λ_{s}(U_{s}y)[U_{s}x]
= [Λ_{s}(U_{s}x), U_{s}y] + [U_{s}x, Λ_{s}(U_{s}y)]
= [Λ_{s}(a), b] + [a, Λ_{s}(b)].

[0, 0]: Assume a = δ_{y,U_{s}x} and b = δ_{y′,U_{s}x′}. Then, [Λ_{s}(a), b] + [a, Λ_{s}(b)]

= [Λ_{s}(δ_{y,U_{s}x}), δ_{y′,U_{s}x′}] + [δ_{y,U_{s}x}, Λ_{s}(δ_{y′,U_{s}x′}]
= [{y, s, x}, δ_{y′,U_{s}x′}] + [δ_{y,U_{s}x}, {x′, s, y′}]
= −{y′, U_{s}x′, {x, s, y}} + {y, U_{s}a, {x′, s, y′}}.

While

Λ_{s}[a, b]
= Λ_{s}(δ_{{y,U_{s}a,y′},U_{s}x′} − δ_{y′,{U_{s}x,y,U_{s}x′}})
= Λ_{s}(δ_{U_{y,y′}U_{s}x,U_{s}x′} − δ_{y′,U_{s}U_{x,x′}U_{s}y}
= −{x′, s, U_{y,y′}U_{s}x} − {U_{x,x′}U_{s}y, s, y′}.

For sake of simpler notation, we move to the s-homotope and obtain: Λ_{s}[a, b]

= −{x′, s, U_{y,y′}U_{s}x} − {U_{x,x′}U_{s}y, s, y′}
= −x′ ∘ {y, x, y′} + {x, y, x′} ∘ y′
(J26) = −{y′, x′, x ∘ y} + {y, x, x′ ∘ y′}
= −{y′, U_{s}x′, {x, s, y}} + {y, u_{s}x, {x′, s, y′}}
= [Λ_{s}(a), b] + [a, Λ_{s}(b)].

[0, −]: Assume a = δ_{y,U_{s}x} and b = U_{s}c. Then

Λ_{s}[a, b]

= Λ_{s}(δ_{y,U_{s}x}[U_{s}c])
= Λ_{s}(−{U_{s}x, y, U_{s}c})
= −Λ_{s}(U_{s}{x, U_{s}y, c})
= δ_{{x,U_{s}y,c},s}.

While

[Λ_{s}(a), b] + [a, Λ_{s}(b)]
= [Λ_{s}(δ_{y,U_{s}x}), U_{s}c] + [δ_{y,U_{s}x}, Λ_{s}(U_{s}c)]
= [−{x, s, y}, U_{s}c] + [δ_{y,U_{s}x}, −δ_{c,s}]
= δ_{{x,s,y},U_{s}c} − δ_{{y,U_{s}x,c},s} + δ_{c,{U_{s}x,y,s}}
= δ_{{x,U_{s}y,c},s} − δ_{c,U_{s}{x,s,y}} + δ_{{y,U−sx,c},s}
−δ_{{y,U_{s}x,c},s} + δ_{c,U_{s}V_{y,s}a}
= δ_{{x,U_{s}y,c},s}.

Therefore, (∗) holds in all cases and the lemma is proved.

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Lemma 19 For each s ∈ S, Λ_{s}^{2} is a divided square for Λ_{s} on £_{s^{2}}. Proof We must show

(1) Λ_{s}Λ_{s} = 2Λ_{s}^{2} on £_{s^{2}} (2) Λ_{s}^{2}[a, b] = [Λ_{s}^{2}(a), b] + [a, Λ_{s}^{2}(b)] + [Λ_{s}(a), Λ_{s}(b)] for a, b ∈ £_{s^{2}}.

(1):

Λ_{s}Λ_{s}[U_{s^{2}} â] = Λ_{s}(Λ_{s}[U_{s}U_{s}â])
= Λ_{s}[δ_{s,U_{s}â}]
= −Λ_{s}(s)[U_{s}â]
= −(−2U_{s}â)
= 2U_{s}â
= 2Λ_{s}^{2}[U_{s}(U_{s}â)]
= 2Λ_{s}^{2}[U_{s^{2}} â].

Λ_{s}Λ_{s}(s^{2}) = Λ_{s}(Λ_{s}[U_{s}ˆ1]
= Λ_{s}(δ_{1_{σ},s_{−σ}}
= Λ_{s}(s_{−σ})[1_{σ}]
= 2σξ(1_{σ})
= 2σ^{2}(1_{σ})
= 2Λ_{s}^{2}(U_{s}ˆ1)
= 2Λ_{s}^{2}(s^{2}).

Hence, Λ_{s}Λ_{s} = 2Λ_{s}^{2} on U_{s^{2}} Ĵ and on Φs^{2}, and hence on all of K_{s^{2}}.

Now, since Λ_{s} is of degree σ and Λ_{s}^{2} is of degree 2σ, we have

Λ_{s}Λ_{s}[D̂_{0}^{t}] = Λ_{s}Λ_{s}[(K_{t})_{σ}] = 0 = 2Λ_{s}^{2}[D̂_{0}^{t}] = 2Λ_{s}^{2}[(K_{t})_{σ}].

Therefore we can conclude that

Λ_{s}Λ_{s} = 2Λ_{s}^{2}

on £_{s^{2}}, which gives (1).

(2): Let a, b ∈ £_{s^{2}}. Using the notation introduced in lemma?, (2) holds immediately in the following cases because all of the terms reduce to zero by degree arguments:

[+, +], [+, −], [0, 0], [0, +].

This leaves only two non-trivial cases that need to be checked:

[−, −] and [0, −].

[−, −]: Let

a = U_{s}x̂ and b = U_{s}ŷ ∈ U_{s}Ĵ

for

s ∈ S ⊆ J_{−σ} and x̂, ŷ ∈ Ĵ_{σ}.

Then,

Λ_{s}^{2}[a, b] = Λ_{s}^{2}[0] = 0,

since a, b ∈ (K_{s^{2}})_{−σ}. Looking at the other side of the equation for (2), we have:

[Λ_{s}^{2}(a), b] + [a, Λ_{s}^{2}(b)] + [Λ_{s}(a), Λ_{s}(b)]

= [Λ_{s}^{2}(U_{s}x̂), U_{s}ŷ] + [U_{s}x̂, Λ_{s}^{2}(U_{s}ŷ)] + [Λ_{s}(U_{s}x̂), Λ_{s}(U_{s}ŷ)]
= [x̂, U_{s}ŷ] + [U_{s}x̂, ŷ] + [δ_{s,x̂}, δ_{s,ŷ}]
(J15) = −δ_{x̂,U_{s}ŷ} − δ_{U_{s}x̂,ŷ} − δ_{ŷ,{s,x̂,s}} + δ_{{x̂,s,ŷ},s}
(J8) = −δ_{x̂,U_{s}ŷ} − δ_{U_{s}x̂,ŷ} − δ_{U_{s}x̂,ŷ} + δ_{x̂,U_{s}ŷ} + 2δ_{U_{s}x̂,ŷ}
= 0.

Therefore, (2) holds in the case [−, −]. [0, −]: Since K_{s^{2}} ⊆ U_{s}Ĵ let

a = δ_{U_{s}x̂,U_{s}ẑ}

and

b = U_{s}ŷ

for U_{s}ŷ, U_{s}x̂ ∈ J_{−σ} and U_{s}ẑ ∈ J_{σ}. Then

Λ_{s}^{2}[a, b] = Λ_{s}^{2}[δ_{U_{s}x̂,U_{s}ẑ}, U_{s}ŷ]
= Λ_{s}^{2} [{U_{s}x̂, U_{s}ẑ, U_{s}ŷ}]
= Λ_{s}^{2}[U_{s}U_{x̂,ŷ}U_{s}(U_{s}ẑ)]
= U_{x̂,ŷ}U_{s}U_{s}ẑ
= {x̂, U_{s}U_{s}ẑ, ŷ}.

Looking at the other side of the equation for (2), we have:

[Λ_{s}^{2}(a), b] + [a, Λ_{s}^{2}(b)] + [Λ_{s}(a), Λ_{s}(b)]

= [0, U_{s}ŷ] + [δ_{U_{s}x̂,U_{s}ẑ}, Λ_{s}^{2}(U_{s}ŷ] + [Λ_{s}(δ_{U_{s}x̂,U_{s}ẑ}), Λ_{s}(U_{s}ŷ)]
= 0 + [δ_{U_{s}x̂,U_{s}ẑ}, ŷ] − [Λ_{s}(U_{s}x̂)[U_{s}ẑ], δ_{s,ŷ}]
= −{U_{s}ẑ, U_{s}x̂, ŷ} − [δ_{s,x̂}(U_{s}ŷ), δ_{s,ŷ}]
= −{U_{s}ẑ, U_{s}x̂, ŷ} + [δ_{x̂,s}(U_{s}ŷ), δ_{s,ŷ}]
= −{U_{s}ẑ, U_{s}x̂, ŷ} + [{x̂, s, U_{s}ẑ}, δ_{s,ŷ}]
= −{U_{s}ẑ, U_{s}x̂, ŷ} + [δ_{ŷ,s}, {x̂, s, U_{s}ẑ}
= −{U_{s}ẑ, U_{s}x̂, ŷ} + {ŷ, s, {x̂, s, U_{s}ẑ}}
(J8) = −{U_{s}ẑ, U_{s}x̂, ŷ} + {U_{s}ẑ, U_{s}x̂, ŷ} + {x̂, U_{s}U_{s}ẑ, ŷ}
= {x̂, U_{s}U_{s}ẑ, ŷ}

Therefore, (2) holds in the case [0, −]. Thus we conclude that Λ_{s}^{2} is a divided square for Λ_{s}, as desired.

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Lemma 20 For each s ∈ S, (Λ_{s}, Λ_{s}^{2}) is a squared derivation of degree σ, σ = ±1, with denominator s^{2}.

Proof Lemmas 18 and 19 show that (Λ_{s}, Λ_{s}^{2}) is a squared derivation of degree σ from £_{s^{2}} → ˆ£. So all we have to check is that s^{2} ∈ S is indeed a denominator for (Λ_{s}, Λ_{s}^{2}). Let

d_{−σ} = s^{3} ∈ S ⊆ J_{−σ}.

Then, for â ∈ Ĵ_{σ}, note the following: (1)

Λ_{s}[U_{s^{2}} â] = Λ_{s}(U_{s}(U_{s}â))
= −δ_{U_{s}â,s^{2}}
= δ_{â,s^{3}} − δ_{s∘â,s^{2}}
= δ_{â,d_{−σ}} − δ_{D_{σ}(â),s^{2}}

for D_{σ}(â) = V_{s}â since

(−D_{σ}, D_{−σ}) = Λ_{s}(s^{2}) = −δ_{s_{σ},1_{−σ}} = −(V_{s}, −V_{s})

implies D_{σ} = V_{s} = D_{−σ}.

(2) U_{s^{2}}(s ∘ â) = U_{s^{3},s^{2}} â = {s^{3}, â, s^{2}} = {d_{−σ}, â, s^{2}}.

(3) U_{s^{2}}Λ_{s}^{2}U_{s^{2}}[â] = U_{s^{3}} â = U_{d_{−σ}} â.

(1),(2), and (3) tell us exactly that s is a denominator for (Λ_{s}, Λ_{s}^{2}) and the lemma is proved.

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Together, Lemmas 18,19, and 20 tell us that the Λ_{s} maps are elements of Ṽ_{σ}, σ = ±1, as was desired.

As mentioned above, our overring must not only contain all of the inverses of the elements of S, but must be a suitable place for an imbedding of J. For each x ∈ J we will ultimately use the map ad(x) to play the role of x once the imbedding has been carried out. As with the Λ_{s} maps, we need to verify that the ad(x) maps actually do lie in our subspaces and hence will be available to us in our constructed overring. We do this now by mirroring the lemmas used above for the Λ_{s} maps.

Lemma 21 For each x ∈ J, ad(x) is a derivation from £ into £.

Proof We present no proof here as this is a well established fact in Lie algebras.

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Lemma 22 For each x ∈ J, U_{x} is a divided square for ad(x).

Proof We mirror the proof of Lemma 19.

(1): Think of x ∈ J_{σ} so that ad(x) is degree σ, σ = ±1. Let â ∈ Ĵ_{−σ}. Then

ad(x)ad(x)[â] = ad(x)[x, â]
= ad(x)[δ_{â,x}]
= [x, δ_{â,x}]
= [δ_{x,â}, x]
= {x, â, x} = 2U_{x}(â).

(2): We need to show that

U_{x}[a, b] = [U_{x}a, b] + [a, U_{x}b] + [ad(x)[a], ad(x)[b]]

for a, b ∈ ˆ£. Using the same type of notation as in the proof of Lemma 18, this equality holds trivially in the following cases:

[+, +], [+, −], [0, 0], [0, +].

So we only need check the two cases:

[−, −] and [0, −].

[−, −]: Let a = ẑ ∈ Ĵ_{−σ} and let b = ŷ ∈ Ĵ_{−σ}. Then,

U_{x}[a, b] = U_{x}[0] = 0.

On the other hand:

[U_{x}a, b] + [a, U_{x}b] + [ad(x)[a], ad(x)[b]]
= [U_{x}ẑ, ŷ] + [ẑ, U_{x}ŷ] + [ad(x)[ẑ], ad(x)[ŷ]]
= −δ_{U_{x}ẑ,ŷ} − δ_{ẑ,U_{x}ŷ} + [δ_{x,ẑ}, δ_{x,ŷ}]
(J15) = −δ_{U_{x}ẑ,ŷ} − δ_{ẑ,U_{x}ŷ} + δ_{{x,ẑ,x},ŷ} − δ_{x,{ẑ,x,ŷ}}
(J8) = −δ_{U_{x}ẑ,ŷ} − δ_{ẑ,U_{x}ŷ} − δ_{U_{x}ẑ,ŷ} + δ_{ẑ,U_{x}ŷ} + 2δ_{U_{x}ẑ,ŷ}
= 0.

So the necessary identity holds in the case [−, −]. [0, −]: Let a = δ_{ŵ,ẑ} for ŵ ∈ Ĵ_{−σ} and ẑ ∈ Ĵ_{σ}, and let b = ŷ ∈ Ĵ_{−σ}. Then

U_{x}[a, b] = U_{x}[δ_{ŵ,ẑ}, ŷ]
= U_{x}{ŵ, ẑ, ŷ}.

On the other hand:

[U_{x}(a), b] + [a, U_{x}(b)] + [ad(x)[a], ad(x)[b]]
= [U_{x}(δ_{ŵ,ẑ}), ŷ] + [δ_{ŵ,ẑ}, U_{x}ŷ] + [ad(x)[δ_{ŵ,ẑ}], ad(x)[ŷ]]
= 0 + δ_{ŵ,ẑ}[U_{x}ŷ] + [δ_{ẑ,ŵ}(x), δ_{ŷ,x}]
= −δ_{ẑ,ŵ}[U_{x}ŷ] + [{ẑ, ŵ, x}, δ_{ŷ,x}]
= −{ẑ, ŵ, U_{x}ŷ} + [δ_{x,ŷ}, {ẑ, ŵ, x}]
= −{ẑ, ŵ, U_{x}ŷ} + {x, ŷ, {ẑ, ŵ, x}}
(J13) = −{ẑ, ŵ, U_{x}ŷ} + {ẑ, ŵ, U_{x}ŷ} + U_{x}{ŵ, ẑ, ŷ}
= U_{x}{ŵ, ẑ, ŷ}

So the necessary identity holds in the case [0, −].

Thus we conclude that U_{x} is a divided square for ad(x), as desired.

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Lemma 23 Let s ∈ S. For each x ∈ J, (ad(x), U_{x}) is a squared derivation of degree σ, σ = ±1, with denominator s for any s ∈ S.

Proof Lemmas 21 and 22 show that (ad(x), U_{x}) is a squared derivation of degree σ. So all we have to check is that s ∈ S is indeed a denominator for (ad(x), U_{x}).

Let

d_{−σ} = U_{s}x ∈ J_{−σ}

and

D_{σ}(â) = V_{x,s}â

where

x ∈ J_{σ} and s ∈ S ⊆ J_{−σ}.

Then, for â ∈ Ĵ_{σ}, note the following:

(1) ad(x)[U_{s}â] = [x, U_{s}â] = −δ_{x,U_{s}â} = δ_{â,U_{s}x} − δ{x, s, â}, s = δ_{â,d_{−σ}} − δ_{D_{σ}(â,s}.

(2) U_{s} [{x, s, â}] = U_{s}V_{x,s}[â] = U_{U_{s}x,s}[â] = {U_{s}x, â, s} = {d_{−σ}, â, s}.

(3) U_{s}U_{x}U_{s}[â] = U_{U_{s}x}â = U_{d_{−σ}} â.

(1),(2), and (3) tell us exactly that s is a denominator for (ad(x), U_{x}).

■

Together, Lemmas 21,22, and 23 tell us that the ad(x) maps are elements of Ṽ_{σ}, σ = ±1, as we desired.

We have verified that Ṽ_{σ}, σ = ±1, contain all of the elements we will need to carry out an embedding of J and to invert the elements of S. Recall, the next step we need to carry out is verifiying that Ṽ_{σ}, σ = ±1, satisfy all of the axioms required of them by Faulkner’s proposition.

For

Δ ∈ Ṽ_{σ} and Θ ∈ Ṽ_{−σ}

define

Q_{Δ}Θ := Δ^{2}Θ + ΘΔ^{2} − ΔΘΔ

and

(Q_{Δ}Θ)^{2} := Δ^{2}Θ^{2}Δ^{2}.

We must prove that the following are true:

(1) Q_{Δ}Θ ∈ Ṽ_{σ}

(2) (Q_{Δ}Θ)^{2} = Δ^{2}Θ^{2}Δ^{2}

(3) Δ^{2}ΘΔ = ΔΘΔ^{2}

(4) (1)-(3) continue to hold in all (free) scalar extensions of Φ.

NOTE: The equality in (2) is true by definition but what we really need to show is that Δ^{2}Θ^{2}Δ^{2} is a divided square for Q_{Δ}Θ.

We will address these 4 properties in the following order: (3), (2), (1), (4). Before doing this we introduce two more general facts about the elements of our subspaces.

Lemma 24 For Δ ∈ Ṽ_{σ} and Θ ∈ Ṽ_{−σ}, Q_{Δ}Θ is a derivation of degree σ.

Proof Let Δ be a derivation of £_{s_{1}} into ˆ£ of degree σ for some s_{1} ∈ S, and let Θ be a derivation of £_{s_{2}} into ˆ£ of degree −σ for some s_{2} ∈ S, where σ = ±1. We must show that Q = Q_{Δ}(Θ) is a derivation of £_{s} into ˆ£ for some s ∈ S. That is, we must show that

Q([a, b]) = [Q(a), b] + [a, Q(b)]

for Q = Q_{Δ}(Θ), where a, b ∈ £_{s} for some s ∈ S.

(NOTE: In the end we will not explicitly display a choice for what the element s should be. However, we can still be certain that such an s exists since Δ, Δ^{2}, Θ, and Θ^{2} are all shrinkable, as is any finite string constructed out of them.)

Well,

Q[a, b] = Δ^{2}Θ[a, b] + ΘΔ^{2}[a, b] − ΔΘΔ[a, b].

Set

A = Δ^{2}Θ[a, b]
B = ΘΔ^{2}[a, b]
C = ΔΘΔ[a, b]
D = [Q(a), b] + [a, Q(b)]

Then,

A = Δ^{2}([Θ(a), b] + [a, Θ(b)])
= Δ^{2}[Θ(a), b] + Δ^{2}[a, Θ(b)]
= [Δ^{2}Θ(a), b] + [Θ(a), Δ^{2}(b)] + [ΔΘ(a), Δ(b)]

+[Δ^{2}(a), Θ(b)] + [a, Δ^{2}Θ(b)] + [Δ(a), ΔΘ(b)]

B = Θ ([Δ^{2}(a), b] + [a, Δ^{2}(b)] + [Δ(a), Δ(b)])
= Θ[Δ^{2}(a), b] + Θ[a, Δ^{2}(b)] + Θ[Δ(a), Δ(b)]
= [ΘΔ^{2}(a), b] + [Δ^{2}(a), Θ(b)] + [Θ(a), Δ^{2}(b)]

+[a, ΘΔ^{2}(b)] + [ΘΔ(a), Δ(b)] + [Δ(a), ΘΔ(b)]

C = ΔΘ[Δ(a), b] + ΔΘ[a, Δ(b)]
= Δ ([ΘΔ(a), b] + [Δ(a), Θ(b)])

+Δ ([Θ(a), Δ(b)] + [a, ΘΔ(b)])

= Δ[ΘΔ(a), b] + Δ[Δ(a), Θ(b)]

+Δ[Θ(a), Δ(b)] + Δ[a, ΘΔ(b)]

= [ΔΘΔ(a), b] + [ΘΔ(a), Δ(b)]

+[ΔΔ(a), Θ(b)] + [Δ(a), ΔΘ(b)] +[ΔΘ(a), Δ(b)] + [Θ(a), ΔΔ(b)] +[Δ(a), ΘΔ(b)] + [a, ΔΘΔ(b)]

= [ΔΘΔ(a), b] + [ΘΔ(a), Δ(b)]

+2[Δ^{2}(a), Θ(b)] + [Δ(a), ΔΘ(b)] +[ΔΘ(a), Δ(b)] + 2[Θ(a), Δ^{2}(b)] +[Δ(a), ΘΔ(b)] + [a, ΔΘΔ(b)]

D = [Δ^{2}Θ(a), b] + [ΘΔ^{2}(a), b] − [ΔΘΔ(a), b]

+[a, Δ^{2}Θ(b)] + [a, ΘΔ^{2}(b)] − [a, ΔΘΔ(b)].

Carefully putting all of this together gives

Q([a, b]) − ([Q(a), b] + [a, Q(b)]) = (A + B − C) − D
= 0.
■

Lemma 25 Let (Δ, Δ^{2}) be a squared derivation of degree σ = ±1 with denominator s ∈ S so that

Δ, Δ^{2} : £_{s} → ˆ£.

Then

Δ = Q_{Λ_{s}}(ad(d_{−σ}))

in Ã_{J}, where d_{−σ} ∈ J_{−σ} is as mentioned in the definition for a denominator of a squared derivation.

Proof

Q_{Λ_{s}}(ad(d_{−σ}))[(U_{s}â)_{−σ}] = (Λ_{s}^{2}ad(d_{−σ}) + ad(d_{−σ})Λ_{s}^{2} − Λ_{s}(ad(d_{−σ}))Λ_{s})[U_{s}â]
= 0 + ad(d_{−σ})[â] − Λ_{s}ad(d_{−σ})[δ_{s,â}]
= δ_{â,d_{−σ}} + Λ_{s}{s, â, d_{−σ}}
(D2) = δ_{â,d_{−σ}} + Λ_{s}U_{s}D_{σ}(â)
= δ_{â,d_{−σ}} − δ_{D_{σ}(â),s}
(D1) = Δ[U_{s}â].

If we let

s′ ∈ U_{s}(S)

then the above allows us to conclude that

Q_{Λ_{s}}(ad(d_{−σ})) = Δ on (K_{s′})_{−σ}

while we already know

Q_{Λ_{s}}(ad(d_{−σ}))[(K_{s′})_{σ}] = 0 = Δ[(K_{s′})_{σ}].

Since both Q_{Λ_{s}}(ad(d_{−σ})) and Δ are derivations, this implies

Q_{Λ_{s}}(ad(d_{−σ})) = Δ on D̂_{0}^{s′} = δ_{K_{s},K_{s}}

and hence

Q_{Λ_{s}}(ad(d_{−σ})) = Δ on £_{s′}.

Therefore

Q_{Λ_{s}}(ad(d_{−σ})) = Δ

in Ã_{J}, and the lemma is proved.

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6.4.3 Δ^{2}ΘΔ = ΔΘΔ^{2}

Definition 15 Let (Δ, Δ^{2}) be a squared derivation of degree σ, σ = ±1, and (Θ, Θ^{2}) a squared derivation of degree −σ. We say that Δ Jordan commutes with Θ, denoted Δ ↬_{J} Θ, if

Δ^{2}ΘΔ = ΔΘΔ^{2}

on £_{z} for some z ∈ S.

Lemma 26 Let (Δ, Δ^{2}) be a squared derivation of degree σ, σ = ±1, with denominator s ∈ S. Then, for y ∈ J_{−σ}, we have

ad(y) ↬_{J} Δ.

Proof In the end we will want to employ several of the facts in Appendix A. Using these facts requires that the elements being acted on live in certain places. That is, there are several domain restrictions to consider. Ultimately, regardless of how complicated the expressions we deal with may seem, we will always be dealing with a finite number of strings of a finite number of shrinkable maps. Using the fact that these finite strings are shrinkable, and the CIMP, we can be sure of two things:

(1) there is some restricted domain available to us on which all of the facts in Appendix A hold true

(2) we can assume this domain has the form U_{r}Ĵ, for some r ∈ S.

Furthermore, to show that ad(y) ↬_{J} Δ we only need show that

U_{y}Δad(y) = ad(y)ΔU_{y}

on £_{t′} for some t′ ∈ S. So, without explicitly saying what the element r looks like, let us assume that a ∈ U_{r}Ĵ where we think of a ∈ J_{−σ}. Also, choose t to be a strong s-denominator for both y and d_{−σ}.

Then, recall the following (based on what was presented earlier regarding the M_{y}, M_{y}^{∗}, M_{x,y}, M_{x,y}^{∗} maps):

(∗)  \begin{matrix}U_{t}U_{y} = M_{y}U_{s} & U_{t}U_{d,y} = M_{d,y}U_{s} & U_{t}U_{s,y} = M_{s,y}U_{s} \\ U_{y}U_{t} = U_{s}M_{y}^{∗} & U_{d,y}U_{t} = U_{s}M_{d,y}^{∗} & U_{s,y}U_{t} = U_{s}M_{s,y}^{∗} \end{matrix}

Axiom ??? tells us that

0 = (M_{y}U_{s}M_{d,y}^{∗} − M_{y}U_{d,s}M_{s,y}^{∗} − M_{d,y}U_{s}M_{y}^{∗} + M_{s,y}U_{d,s}M_{y}^{∗})[a].

Now, set

A = (M_{y}U_{s}M_{d,y}^{∗} − M_{y}U_{d,s}M_{s,y}^{∗})[a]
B = (−M_{d,y}U_{s}M_{y}^{∗} + M_{s,y}U_{d,s}M_{y}^{∗})[a].

Then:

A = (M_{y}U_{d,y}U_{t} − M_{y}U_{d,s}Λ_{s}^{2}U_{s}M_{s,y}^{∗})[a]
(∗) = (M_{y}U_{d,y}U_{t} − M_{y}U_{d,s}Λ_{s}^{2}U_{s,y}U_{t})[a]
(A3) = (M_{y}U_{s}Λ_{s}^{2}U_{d,y}U_{t} − M_{y}U_{d,s}Λ_{s}^{2}U_{s,y}U_{t})[a]
(∗) = (U_{t}U_{y}Λ_{s}^{2}U_{d,y}U_{t} − U_{t}U_{y}Λ_{s}^{2}U_{d,s}Λ_{s}^{2}U_{s,y}U_{t})[a]
(A4) = U_{t}U_{y} (Λ_{s}^{2}ad(d)ad(y) − Λ_{s}^{2}ad(d)ad(s)Λ_{s}^{2}ad(s)ad(y)) [U_{t}a]
= U_{t}U_{y} (Λ_{s}^{2}ad(d) − Λ_{s}^{2}ad(d)ad(s)Λ_{s}^{2}ad(s)) ad(y)U_{t}[a]
(A5) = U_{t}U_{y} (Λ_{s}^{2}ad(d) − Λ_{s}^{2}ad(s)ad(d)Λ_{s}^{2}ad(s)) ad(y)U_{t}[a]
(A7) = U_{t}U_{y} (Λ_{s}^{2}ad(d) − Λ_{s}ad(d)Λ_{s}) ad(y)U_{t}[a]
= U_{t}U_{y} (Λ_{s}^{2}ad(d) + 0 − Λ_{s}ad(d)Λ_{s}) ad(y)U_{t}[a]
= U_{t}U_{y} (Λ_{s}^{2}ad(d) + ad(d)Λ_{s}^{2} − Λ_{s}ad(d)Λ_{s}) ad(y)U_{t}[a]
= U_{t}U_{y} (Q_{Λ_{s}}(ad(d))) ad(y)U_{t}[a]
(lem25) = U_{t}U_{y} (Δ) ad(y)U_{t}[a]

B = (−M_{d,y}U_{y}U_{t} + M_{s,y}U_{d,s}Λ_{s}^{2}U_{s}M_{y}^{∗})[a]
(A1) = (−M_{d,y}U_{s}Λ_{s}^{2}U_{y}U_{t} + M_{s,y}U_{d,s}Λ_{s}^{2}U_{s}M_{y}^{∗})[a]
(∗) = (−U_{t}U_{d,y}Λ_{s}^{2}U_{y}U_{t} + M_{s,y}U_{d,s}Λ_{s}^{2}U_{s}M_{y}^{∗})[a]
(∗) = (−U_{t}U_{d,y}Λ_{s}^{2}U_{y}U_{t} + M_{s,y}U_{d,s}Λ_{s}^{2}U_{y}U_{t})[a]
(A2) = (−U_{t}U_{d,y}Λ_{s}^{2}U_{y}U_{t} + M_{s,y}U_{s}Λ_{s}^{2}U_{d,s}Λ_{s}^{2}U_{y}U_{t})[a]
(∗) = (−U_{t}U_{d,y}Λ_{s}^{2}U_{y}U_{t} + U_{t}U_{s,y}Λ_{s}^{2}U_{d,s}Λ_{s}^{2}U_{y}U_{t})[a]
= −U_{t} (U_{d,y}Λ_{s}^{2} − U_{s,y}Λ_{s}^{2}U_{d,s}Λ_{s}^{2}) U_{y}U_{t}[a]
(A4) = −U_{t} (ad(y)ad(d)Λ_{s}^{2} − ad(y)ad(s)Λ_{s}^{2}ad(d)ad(s)Λ_{s}^{2}) U_{y}U_{t}[a]
= −U_{t}ad(y) (ad(d)Λ_{s}^{2} − ad(s)Λ_{s}^{2}ad(d)ad(s)Λ_{s}^{2}) U_{y}U_{t}[a]
(A6) = −U_{t}ad(y) (ad(d)Λ_{s}^{2} − Λ_{s}ad(d)Λ_{s}) U_{y}U_{t}[a]
= −U_{t}ad(y) (0 + ad(d)Λ_{s}^{2} − Λ_{s}ad(d)Λ_{s}) U_{y}U_{t}[a]
= −U_{t}ad(y) (Λ_{s}^{2}ad(d) + ad(d)Λ_{s}^{2} − Λ_{s}ad(d)Λ_{s}) U_{y}U_{t}[a]
= −U_{t}ad(y) (Q_{Λ_{s}}(ad(d))) U_{y}U_{t}[a]
(lem25) = −U_{t}U_{y} (Δ) ad(y)U_{t}[a]

Therefore, we have

0 = (M_{y}U_{s}M_{d,y}^{∗} − M_{y}U_{d,s}M_{s,y}^{∗} − M_{d,y}U_{s}M_{y}^{∗} + M_{s,y}U_{d,s}M_{y}^{∗})[a]
= A + B
= U_{t} (U_{y}Δad(y) − ad(y)ΔU_{y}) U_{t}[a].

Since t ∈ S is injective, this implies

(U_{y}Δad(y) − ad(y)ΔU_{y})[U_{t}a] = 0.

Then, if we let

t′ = U_{t}r,

so that

(£_{t′})_{σ} ⊆ (U_{t}Ĵ)_{σ},

then we conclude

U_{y}Δad(y) = ad(y)ΔU_{y}

on £_{t′}, and the lemma is proved.

■

Proposition 6 (Shifting Proposition)

(1) Let Θ ∈ Ṽ_{−σ}, σ = ±1, and Δ ∈ Ṽ_{σ}. Suppose that s ∈ S is a denominator for (Δ, Δ^{2}) and d_{−σ} is as mentioned in the definition for denominator of a squared derivation. Then Δ ↬_{J} Θ if ad(d_{−σ}) ↬_{J} Q_{Λ_{s}}(Θ).

(2) For y ∈ J_{−σ}, Q_{Λ_{s}}(ad(y)) ↬_{J} Θ for all Θ ∈ Ṽ_{−σ}, if ad(y) ↬_{J} Δ for all Δ ∈ Ṽ_{σ}.

Proof Note that it suffices to prove (1).

We want to show that

Δ^{2}ΘΔ = ΔΘΔ^{2}

on £_{t′} for some t′ ∈ S.

By Lemma 24 we know that

Δ = Q_{Λ_{s}}(ad(d_{−σ}))

on £_{s′} for s′ ∈ U_{s}(S).

In addition to this last equality, we will want to make use of the facts in Appendix B. Now, similar to what was said at the beginning of the proof of Lemma 26, we can assume that we are able to restrict down to a domain in which all of the requirements for these facts are met. Furthermore, we can assume that this domain has the form U_{q}Ĵ for some q ∈ S. So, without explicitly saying what the element q looks like, let us assume that b ∈ U_{q}Ĵ and choose t to be simultaneously a super U_{s}z-denominator and a super U_{s}r-denominator for d_{−σ} where t ∈ U_{U_{r}s}Ĵ, and z and r are as mentioned in (B1) and (B2) respectively. (We know that we can make this choice by (B3).)

Under the above conditions, for b′ = U_{s}b, we have

U_{s} (Δ^{2}ΘΔ − ΔΘΔ^{2}) [U_{U_{s}t}b]

= U_{s} (Δ^{2}ΘΔ − ΔΘΔ^{2}) [U_{s}U_{t}b′]
(B10) = U_{s} ((Λ_{s}^{2}U_{d}Λ_{s}^{2})Θ(ad(d)Λ_{s}^{2} + Λ_{s}^{2}ad(d) − Λ_{s}ad(d)Λ_{s})) [U_{s}U_{t}b′]
−U_{s} ((ad(d)Λ_{s}^{2} + Λ_{s}^{2}ad(d) − Λ_{s}ad(d)Λ_{s})Θ(Λ_{s}^{2}ΘΛ_{s}^{2})) [U_{s}U_{t}b′]
= U_{s} ((Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2})Θ(ad(d)Λ_{s}^{2} + 0 − Λ_{s}ad(d)Λ_{s})) [U_{s}U_{t}b′]
−U_{s} ((0 + Λ_{s}^{2}ad(d) − Λ_{s}ad(d)Λ_{s})Θ(Λ_{s}^{2}U_{d}Λ_{s}^{2})) [U_{s}U_{t}b′]
= U_{d}Λ_{s}^{2}Θad(d)[U_{t}b′] − U_{d}Λ_{s}^{2}ΘΛ_{s}ad(d)Λ_{s}U_{s}[U_{t}b′]
−ad(d)ΘΛ_{s}^{2}U_{d}[U_{t}b′] + U_{s}Λ_{s}ad(d)Λ_{s}ΘΛ_{s}^{2}U_{d}[U_{t}b′].

Now set

A = U_{d}Λ_{s}^{2}Θad(d)[U_{t}b′]
B = U_{d}Λ_{s}^{2}ΘΛ_{s}ad(d)Λ_{s}U_{s}[U_{t}b′]
C = ad(d)ΘΛ_{s}^{2}U_{d}[U_{t}b′]
D = U_{s}Λ_{s}ad(d)Λ_{s}ΘΛ_{s}^{2}U_{d}[U_{t}b′].

Then

B = U_{d}(Λ_{s}^{2}ΘΛ_{s})ad(d)(−δ_{U_{t}b′,s})
= U_{d}Λ_{s}^{2}ΘΛ_{s}(δ_{U_{t}b′,s}d)
= −U_{d}(Λ_{s}^{2}ΘΛ_{s}){s, U_{t}b′, d}
(B11) = −U_{d}(Λ_{s}ΘΛ_{s}^{2}){s, U_{t}b′, d}
= −U_{d}Λ_{s}ΘΛ_{s}^{2}δ_{d,U_{t}b′}(s)
(B12) = −U_{d}Λ_{s}ΘΛ_{s}(δ_{d,U_{t}b′})

D = U_{s}Λ_{s}ad(d)Λ_{s}ΘΛ_{s}^{2}U_{d}[U_{t}b′]
(B11) = U_{s}Λ_{s}ad(d)(Λ_{s}^{2}ΘΛ_{s})U_{d}U_{t}b′
(B11) = U_{s}(Λ_{s}^{2}ad(d)Λ_{s})ΘΛ_{s}U_{d}U_{t}b′
= ad(d)(Λ_{s}ΘΛ_{s})U_{d}U_{t}b′
A = U_{d} [Q_{Λ_{s}}(Θ) + Λ_{s}ΘΛ_{s}] ad(d)[U_{t}b′]
(hyp) = [ad(d)Q_{Λ_{s}}(Θ)U_{d}] (U_{t}b′) + U_{d}Λ_{s}ΘΛ_{s}δ_{U_{t}b′,d}
= ad(d) [ΘΛ_{s}^{2} − Λ_{s}ΘΛ_{s}] U_{d}U_{t}b′ − U_{d}Λ_{s}ΘΛ_{s}δ_{d,U_{t},b′}
= ad(d)ΘΛ_{s}^{2}U_{d}[U_{t}b′] − ad(d)Λ_{s}ΘΛ_{s}U_{d}U_{t}b′ − U_{d}Λ_{s}ΘΛ_{s}δ_{d,U_{t}b′}
= C − D + B.

So, we have

U_{s} (Δ^{2}ΘΔ − ΔΘΔ^{2}) [U_{U_{s}t}b]
= A − B − C + D
= (C − D + B) − B − C + D
= 0.

Since s is injective, this implies

(Δ^{2}ΘΔ − ΔΘΔ^{2}) [U_{U_{s}t}b]

which gives

Δ^{2}ΘΔ = ΔΘΔ^{2}

on £_{t′} for t′ = U_{s}t. This proves the lemma.

■

So, Lemma 26 implies Proposition 6 and hence, for Δ ∈ Ṽ_{σ} and Θ ∈ Ṽ_{−σ}, σ = ±1, we have

Δ^{2}ΘΔ = ΔΘΔ^{2}.

6.4.4 (Q_{Δ}Θ)^{2} = Δ^{2}Θ^{2}Δ^{2} As mentioned earlier, in order to show that

(Q_{Δ}Θ)^{2} = Δ^{2}Θ^{2}Δ^{2}

we really need to show that Δ^{2}Θ^{2}Δ^{2} is a divided square for Q = Q_{Δ}(Θ). That is, we need to show:

(i) QQ = 2Δ^{2}Θ^{2}Δ^{2} (ii) Q^{2}[a, b] = [Q^{2}(a), b] + [a, Q^{2}(b)] + [(Q)(a), (Q)(b)]. We prove (i), then introduce a lemma and corollary before proving (ii).

Lemma 27 Let Δ ∈ Ṽ_{σ} and Θ ∈ Ṽ_{−σ} for σ = ±1. Then Q_{Δ}(Θ)Q_{Δ}(Θ) = 2Δ^{2}Θ^{2}Δ^{2}.

Proof Take a moment to recall that we already know the following: (A) ΔΔ = 2Δ^{2} and ΘΘ = 2Θ^{2} (B) Δ^{2}ΘΔ = ΔΘΔ^{2}.

Then

(Q_{Δ}Θ)(Q_{Δ}Θ) = (Δ^{2}Θ + ΘΔ^{2} − ΔΘΔ)(Δ^{2}Θ + ΘΔ^{2} − ΔΘΔ)
= Δ_{2}ΘΔ^{2}Θ + Δ^{2}ΘΘΔ^{2} − Δ^{2}ΘΔΘΔ
+ΘΔ^{2}Δ^{2}Θ + ΘΔ^{2}ΘΔ^{2} − ΘΔ^{2}ΔΘΔ
−ΔΘΔΔ^{2}Θ − ΔΘΔΘΔ^{2} + ΔΘΔΔΘΔ
(deg) = 0 + Δ^{2}ΘΘΔ^{2} − Δ^{2}ΘΔΘΔ + 0 + 0 − 0 − 0 − ΔΘΔΘΔ^{2}
= Δ^{2}ΘΘΔ^{2} − Δ^{2}ΘΔΘΔ − ΔΘΔΘΔ^{2} + ΔΘΔΔΘΔ
(A) = 2Δ^{2}Θ^{2}Δ^{2} − Δ^{2}ΘΔΘΔ − ΔΘΔΘΔ^{2} + 2ΔΘΔΘΔ^{2}
(B) = 2Δ^{2}Θ^{2}Δ^{2} − ΔΘΔΘΔ^{2} − ΔΘΔΘΔ^{2} + 2ΔΘΔΘΔ^{2}
= 2Δ^{2}Θ^{2}Δ^{2}.
■

This lemma gives (i).

Lemma 28 Let Δ ∈ Ṽ_{σ}, σ = ±1, and suppose that s ∈ S is a denominator for (Δ, Δ^{2}). If s′ = U_{s}t for t ∈ S an s-denominator for d_{−σ}, then for all y, z ∈ U_{U_{s′}s}Ĵ we have

Δ^{2}ad(y)Δ[z_{−σ}] = Δad(y)Δ^{2}[z_{−σ}].

Proof Let y = U_{s′}a, and z = U_{s′}b for a, b ∈ U_{s}Ĵ ⊆ (K_{s})_{σ}. By Corollary 1 s′ = U_{s}t is a denominator for (Δ, Δ^{2}) with d′ = U_{s}U_{t}d. Also, note that we have the following for c ∈ J_{σ}:

(i) Δ(y_{−σ})[z_{−σ}] = U_{s′}{a, d′, b}_{σ} (ii) U_{s′} [Δ(y_{−σ})[c_{σ}]] = −{d′, a, U_{s′}c}_{−σ} (iii) Δ(ad(y_{−σ})[a_{σ}]) = Δ(δ_{a,y}) = Δ(y_{−σ})[a_{σ}] (iv) ad(y_{−σ})Δ[z_{−σ}] = −Δ(z_{−σ})[y_{−σ}] = −Δ(y_{−σ})[z_{−σ}].

Now,

U_{s′}Δ^{2}ad(y)Δ[z] = U_{s}U_{t}U_{s}Δ^{2}ad(y)Δ[z]
(B10) = U_{s}U_{t}U_{s}(Λ_{s}^{2}U_{d}Λ_{s}^{2})ad(y)Δ[z]
(iv) = −U_{s}U_{t}U_{s}(Λ_{s}^{2}U_{d}Λ_{s}^{2})(Δ(y)[z])
(i) = −U_{s}U_{t}U_{s}(Λ_{s}^{2}U_{d}Λ_{s}^{2})(U_{s′}{a, d′, b})
= −U_{s}U_{t}U_{s}(Λ_{s}^{2}U_{d}Λ_{s}^{2})(U_{s}U_{t}U_{s}){a, d′, b})
= −U_{s}U_{t}U_{s}(Λ_{s}^{2}U_{d}U_{t}U_{s}{a, d′, b})
= −U_{s}U_{t}U_{d}U_{t}U_{s}{a, d′, b};

and

U_{s′}Δad(y)Δ^{2}[z] = U_{s′}Δad(y)(Λ_{s}^{2}U_{d}Λ_{s}^{2})[z]
(iii) = U_{s′}Δ(U_{s′}a)(Λ_{s}^{2}U_{d}Λ_{s}^{2})[z]
(ii) = −{d′, a, U_{s′}Λ_{s}^{2}U_{d}Λ_{s}^{2}z}
= −{d′, a, U_{s}U_{t}U_{d}Λ_{s}^{2}U_{s′}b}
= −{d′, a, U_{s}U_{t}U_{d}U_{t}U_{s}b}
= −{U_{s}U_{t}d, a, U_{s}U_{t}U_{d}U_{t}U_{s}b}
= −U_{s}U_{t}{d, U_{t}U_{s}a, U_{d}U_{t}U_{s}b}
= −U_{s}U_{t}U_{d}{U_{t}U_{s}a, d, U_{t}U_{s}b}
= −U_{s}U_{t}U_{d}U_{t}U_{s}{a, U_{s}U_{t}d, b}
= −U_{s}U_{t}U_{d}U_{t}U_{s}{a, d′, b}).

Therefore

U_{s′}Δ^{2}ad(y)Δ[z] = U_{s′}Δad(y)Δ^{2}[z]

Since s′ is injective we conclude

Δ^{2}ad(y)Δ[z] = Δad(y)Δ^{2}[z]

and the lemma is proved.

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Corollary 2 Let Δ ∈ Ṽ_{σ}, σ = ±1, and let a, b ∈ J_{−σ}. Then [Δ^{2}(a), Δ(b)] = [Δ^{2}(b), Δ(a)].

Proof

[Δ^{2}(a), Δ(b)] = Δ^{2}[a, Δ(b)] − [a, Δ^{2}Δ(b)] − [Δ(a), ΔΔ(b)]
(deg) = Δ^{2}[a, Δ(b)] − 0 − [Δ(a), ΔΔ(b)]
= Δ^{2}[a, Δ(b)] − 2[Δ(a), Δ^{2}(b)]
= Δ^{2}ad(a)Δ(b) − 2[Δ(a), Δ^{2}(b)]
= Δad(a)Δ^{2}(b) − 2[Δ(a), Δ^{2}(b)]
= Δ[a, Δ^{2}(b)] − 2[Δ(a), Δ^{2}(b)]
= ([Δ(a), Δ^{2}(b)] + [a, ΔΔ^{2}(b)]) − 2[Δ(a), Δ^{2}(b)]
= −[Δ(a), Δ^{2}(b)] + 0
= [Δ^{2}(b), Δ(a)].
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Lemma 29 Let Δ ∈ Ṽ_{σ} and Θ ∈ Ṽ_{−σ} for σ = ±1. If

Q := Q_{Δ}(Θ) and Q^{2} := Δ^{2}Θ^{2}Δ^{2}

then

(∗) Q^{2}[a, b] = [Q^{2}(a), b] + [a, Q^{2}(b)] + [(Q)(a), (Q)(b)]

for a and b in the required domains. Proof Using earlier notation, (∗) holds in the following cases because degree arguments reduce everything to zero:

[+, +], [+, −], [0, 0], [0, +].

This leaves only two non-trivial cases that need to be checked:

[−, −] and [0, −].

[−, −]: Suppose a, b ∈ J_{−σ} Set

A = [Q^{2}(a), b]
B = [Θ^{2}Δ^{2}(a), Δ^{2}(b)]
C = [ΔΘ^{2}Δ^{2}(a), Δ(b)]
D = [ΘΔ^{2}(a), ΔΘΔ(b)]
E = [Δ^{2}ΘΔ(a), ΘΔ(b)]
H = ΔΘ^{2}[Δ^{2}(a), Δ(b)]

and for each X ∈ {A, B, C, D, H} let

X′ = X with a, b switched.

Take a moment to notice that H = H′ by Corollary 2. Now, since a and b are both degree −σ, we have

Q[a, b] = Q(0) = 0.

What about the right hand side of (∗)? [Q^{2}(a), b] + [a, Q^{2}(b)] + [(Q)(a), (Q)(b)]

= A − A′ + [(ΘΔ^{2} − ΔΘΔ)[a], (ΘΔ^{2} − ΔΘΔ)[b]]
= A − A′ − [ΘΔ^{2}(a), ΔΘΔ(b)] − [ΔΘΔ(a), ΘΔ^{2}(b)] + [ΘΔ^{2}(a), ΘΔ^{2}(b)]

+[ΔΘΔ(a), ΔΘΔ(b)]

= A − A′ − D + D′ + (Θ^{2}[Δ^{2}(a), Δ^{2}(b)] − [Θ^{2}Δ^{2}(a), Δ^{2}(b)]

−[Δ^{2}(a), Θ^{2}Δ^{2}(b)])

+(Δ^{2}[ΘΔ(a), ΘΔ(b)] − [Δ^{2}ΘΔ(a), ΘΔ(b)] −[ΘΔ(a), Δ^{2}ΘΔ(b)])

= A − A′ − D + D′ + (0 − [Θ^{2}Δ^{2}(a), Δ^{2}(b)] − [Δ^{2}(a), Θ^{2}Δ^{2}(b)])
= +(0 − [Δ^{2}ΘΔ(a), ΘΔ(b)] − [ΘΔ(a), Δ^{2}ΘΔ(b)])
= A − A′ − D + D′ − B + B′ − E + E′
= (A + B + C) − (2B + C + D + E) − (A′ + B′ + C′) + (2B′ + C′ + D′ + E′).

We stop to note three things:

(i) (A + B + C) = Δ^{2}[Θ^{2}Δ^{2}(a), b] = 0 since both Θ^{2}Δ^{2}(a) and b are of the same degree

(ii) (A′ + B′ + C′) = Δ^{2}[Θ^{2}Δ^{2}(b), a] = 0 since both Θ^{2}Δ^{2}(b) and a are of the same degree

(iii) 2B + C + D + E = H = H′ = 2B′ + C′ + D′ + E′ since

(E + D)) + (C + 2B) = Δ[ΘΔ^{2}(a), ΘΔ(b)] + Δ[Θ^{2}Δ^{2}(a), Δ(b)]
= Δ ([ΘΔ^{2}(a), ΘΔ(b)] + [Θ^{2}Δ^{2}(a), Δ(b)])
= Δ (Θ^{2}[Δ^{2}(a), Δ(b)] − [Δ^{2}(a), Θ^{2}Δ(b)])
(deg) = ΔΘ^{2}[Δ^{2}(a), Δ(b)] − 0
= H
= H′
= E′ + D′ + C′ + 2B′.

Thus

[Q^{2}(a), b] + [a, Q^{2}(b)] + [(Q)(a), (Q)(b)]
= (A + B + C) − (2B + C + D + E) − (A′ + B′ + C′) + (2B′ + C′ + D′ + E′)
= 0 − H − 0 + H
= 0.

So (∗) holds in the case [−, −].

[−, 0]: Suppose b ∈ D̂_{0} and a ∈ J_{−σ}. Set the following:

A = [Δ^{2}Θ^{2}Δ^{2}(a), b]
B = [ΔΘ^{2}Δ^{2}(a), Δ(b)]
C = [ΘΔ^{2}(a), Δ^{2}Θ(b)]
D = [Δ^{2}ΘΔ(a), ΘΔ(b)]
E = [ΔΘΔ(a), ΔΘΔ(b)]
F = [Δ^{2}ΘΔ(a), ΔΘ(b)]
G = [Δ^{2}(a), ΔΘ^{2}Δ(b)].

Also, by degree arguments, note that [Θ^{2}Δ^{2}(a), Δ^{2}(b)] = 0

[Δ^{2}ΘΔ^{2}(a), Θ(b)] = 0

[Δ^{2}Δ(a), Θ^{2}Δ(b)] = 0

[ΘΔ(a), Δ^{2}ΘΔ(b)] = 0.

Then, looking at the left hand side of (∗), we have Δ^{2}Θ^{2}Δ^{2}[a, b]

= Δ^{2}Θ^{2} ([Δ^{2}(a), b] + [a, Δ^{2}(b)] + [Δ(a)Δ(b)])
= Δ^{2}Θ^{2} ([Δ^{2}(a), b] + 0 + [Δ(a)Δ(b)])
= Δ^{2} ([Θ^{2}Δ^{2}(a), b] + [Δ^{2}(a), Θ^{2}(b)] + [ΘΔ^{2}(a), Θ(b)]
+[Θ^{2}Δ(a), Δ(b)] + [Δ(a), Θ^{2}Δ(b)] + [ΘΔ(a), ΘΔ(b)])
= Δ^{2} ([Θ^{2}Δ^{2}(a), b] + 0 + [ΘΔ^{2}(a), Θ(b)]
+0 + [Δ(a), Θ^{2}Δ(b)] + [ΘΔ(a), ΘΔ(b)])
= (A + 0 + B) + (0 + C + [ΔΘΔ^{2}(a), ΔΘ(b)])
+(0 + [Δ(a), Δ^{2}Θ^{2}Δ(b)] + [ΔΔ(a), ΔΘ^{2}Δ(b)]) + D + 0 + E
= A + B + C + [Δ^{2}ΘΔ(a), ΔΘ(b)]
+[Δ(a), Δ^{2}Θ^{2}Δ(b)] + [ΔΔ(a), ΔΘ^{2}Δ(b)] + D + E
= A + B + C + [Δ^{2}ΘΔ(a), ΔΘ(b)]
+[Δ(a), Δ^{2}Θ^{2}Δ(b)] + 2[Δ^{2}(a), ΔΘ^{2}Δ(b)] + D + E
= A + B + C + F
+[Δ(a), Δ^{2}Θ^{2}Δ(b)] + 2[Δ^{2}(a), ΔΘ^{2}Δ(b)] + D + E
(cor2) = A + B + C + F
+[ΔΘ^{2}Δ(b), Δ^{2}(a)] + 2[Δ^{2}(a), ΔΘ^{2}Δ(b)] + D + E
= A + B + C + F
−[Δ^{2}(a), ΔΘ^{2}Δ(b)] + 2[Δ^{2}(a), ΔΘ^{2}Δ(b)] + D + E
= A + B + C + F − G + 2G + D + E.

Looking at the right hand side of (∗) we have

[Q^{2}(a), b] + [a, Q^{2}(b)] + [(Q)(a), (Q)(b)]

(deg) = [Q^{2}(a), b] + [a, Q^{2}(b)] + [(ΘΔ^{2} − ΔΘΔ)(a), (Δ^{2}Θ − ΔΘΔ)(b)]
= A + 0 + [ΘΔ^{2}(a), Δ^{2}Θ(b)] + [ΔΘΔ(a), ΔΘΔ(b)]

−[ΘΔ^{2}(a), ΔΘΔ(b)] − [ΔΘΔ(a), Δ^{2}Θ(b)]

= A + C + E − [ΘΔ^{2}(a), ΔΘΔ(b)] − [Δ(ΘΔ(a)), Δ^{2}(Θ(b))]
(cor2) = A + C + E − [ΘΔ^{2}(a), ΔΘΔ(b)] − [Δ(Θ(b)), Δ^{2}(ΘΔ(a))]
= A + C + E − [ΘΔ^{2}(a), ΔΘΔ(b)] + [Δ^{2}ΘΔ(a), ΔΘ(b)]
= A + C + E − [ΘΔ^{2}(a), ΔΘΔ(b)] + [ΔΘΔ^{2}(a), ΔΘ(b)]
= A + C + E − [ΘΔ^{2}(a), ΔΘΔ(b)] + F
= A + C + E − Δ ([ΘΔ^{2}(a), ΘΔ(b)]) + [ΔΘΔ^{2}(a), ΘΔ(b)] + F
= A + C + E − Δ ([ΘΔ^{2}(a), ΘΔ(b)]) + D + F
= A + C + E

−Δ(Θ^{2}[Δ^{2}(a), Δ(b)] − [Θ^{2}Δ^{2}(a), Δ(b)] − [Δ^{2}(a), Θ^{2}Δ(b)])

+D + F

= A + C + E

−Δ(Θ^{2}[Δ^{2}(a), Δ(b)]) + Δ[Θ^{2}Δ^{2}(a), Δ(b)] + Δ[Δ^{2}(a), Θ^{2}Δ(b)] +D + F

= A + C + E

−Δ(0) + [ΔΘ^{2}Δ^{2}(a), Δ(b)] + [Θ^{2}Δ^{2}(a), ΔΔ(b)] + Δ[Δ^{2}(a), Θ^{2}Δ(b)] +D + F

= A + C + E − 0 + [ΔΘ^{2}Δ^{2}(a), Δ(b)] + 0 + Δ[Δ^{2}(a), Θ^{2}Δ(b)]

+D + F

= A + C + E + B + Δ[Δ^{2}(a), Θ^{2}Δ(b)]

+D + F

= A + C + E + B

+[Δ^{2}(a), ΔΘ^{2}Δ(b)] + [ΔΔ^{2}(a), Θ^{2}Δ(b)]

+D + F

(deg) = A + C + E + B

+[Δ^{2}(a), ΔΘ^{2}Δ(b)] + 0

+D + F

= A + C + E + B + G + D + F.

So, the left hand side and right hand side are equal and hence (∗) holds in the case [−, 0].

This proves the lemma.

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7 Q_{Δ}(Θ) ∈ Ṽ_{σ}

To show that Q_{Δ}(Θ) ∈ Ṽ_{σ} we must show that Q_{Δ}(Θ) is a derivation of degree σ and that it has a denominator (meaning that the squared derivation (Q_{Δ}(Θ), Δ^{2}Θ^{2}Δ^{2}) has a denominator).

In order to show that Q_{Δ}(Θ) has a denominator we will need several lemmas. We introduce these after defining some terminology regarding denominators of Δ and Θ elements.

Definition 16 We call any element of Ṽ_{σ}, σ = ±1, a fractional derivation. Any fractional derivation Δ with

Δ = Q_{Λ_{s}}Θ

for s a denominator for (Δ, Δ^{2}), we call a proper fraction.

Lemma 30

(i) If {e, a, t} = U_{t}ã ∈ U_{t}Ĵ for some ã ∈ Ĵ and t ∈ S ⊆ J_{−σ} then

(▲) ((Q_{Λ_{t}}(ad(e)))(U_{t}a)) [U_{t}b] = U_{t}{a, e, b}

for all b ∈ Ĵ.

If U_{e,t} = U_{t}T̃ for T̃ defined on Ĵ, then (▲) holds for all a, b ∈ Ĵ.

(ii) If e = U_{t_{0}}U_{r_{0}}e_{0}, and t = U_{t_{0}}r_{0} (where {e_{0}, r_{0}, t_{0}} = U_{t_{0}} r̃_{0}) then U_{e,t} = U_{t}T̃ on Ĵ for

T̃(â) = (V_{r̃_{0},t_{0}} − V_{r_{0},e_{0}})(â),

and (▲) holds for Q_{Λ_{t}}(ad(e)) ≈ Q_{Λ_{t_{0}}}(ad(e_{0})).

(iii) If the Ore condition is satisfied then any fraction Q_{Λ_{t_{0}}}(ad(e_{0})) is equivalent to a proper fraction Q_{Λ_{t}}(ad(e)).

Proof (i):

Q_{Λ_{t}}(ad(e))[U_{t}a] = (Λ_{t}^{2}ad(e) + ad(e)Λ_{t}^{2} − Λ_{t}ad(e)Λ_{t})[U_{t}a]
= 0 + ad(e)[a] − Λ_{t}ad(e)δ_{t,a}
= −δ_{e,a} + Λ_{t}({t, a, e})
(hyp) = −δ_{e,a} + Λ_{t}(U_{t}ã)
= −δ_{e,a} + δ_{t,ã}.

Hence

Q_{Λ_{t}}(ad(e))[U_{t}a][U_{t}b] = −V_{e,a}U_{t}b + {t, ã, U_{t}b}
= (−U_{{e,a,t},t}b + U_{t}V_{a,e}b) + {U_{t}ã, b, t}
(hyp) = −{{e, a, t}, b, t} + U_{t}{a, e, b} + {{e, a, t}, b, t}
= U_{t}{a, e, b}.

This gives the formula (▲). The second statement in (i) is clear. (ii): If we apply lemma? (follow the leader) with W = V_{e_{0},r_{0}} and

ã = Ṽ_{e_{0},r_{0}}(a) = (V_{r̃_{0},t_{0}} − V_{r_{0},e_{0}})[a] = T̃(a),

then we have

{e, a, t} = {U_{t_{0}}U_{r_{0}}e_{0}, a, U_{t_{0}}r_{0}}
= U_{t_{0}}{U_{r_{0}}e_{0}, U_{t_{0}}a, r_{0}}
= U_{t_{0}}U_{r_{0}}{e_{0}, r_{0}, U_{t_{0}}a}
(hyp) = U_{t_{0}}U_{r_{0}}U_{t_{0}} ã
= U_{t}ã.

As far as

Q_{Λ_{t}}(ad(e)) ≈ Q_{Λ_{t_{0}}}(ad(e_{0})),

we have

Q_{Λ_{t}}(ad(e)) = Q_{Λ_{U_{t_{0}} r_{0}}}(ad(U_{t_{0}}U_{r_{0}}e_{0})
(C4)(C2) ≈ Q_{Λ_{t_{0}}}Q_{Λ_{r_{0}}}Q_{Λ_{t_{0}}}Q_{ad(t_{0})}Q_{ad(r_{0})}(ad(e_{0}))
(C1) ≈ Q_{Λ_{t_{0}}}(ad(e_{0})).

(iii): This follows immediately.

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Lemma 31 Let s ∈ S and let a ∈ U_{s}Ĵ with

a = U_{s}a′ and {y, t, a} = U_{s}ã

(eg. ã = Ṽ_{y,t}a when t is a super s-denominator for y). Then the element

ā := {t, s, ã} − {U_{t}y, s, a′} = M_{y,s}^{∗}(a)

satisfies

(i) U_{s}ā = {y, U_{t}a, s} and the operator

V̄_{a} := V_{U_{s}U_{t}y,a′} − V_{U_{s}t,ã}

satisfies (ii) V̄_{a} = V_{y,U_{t}a} − V_{s,ā} = V_{y,U_{t}a} − V_{s,M_{y,s}^{∗}(a)}.

Proof (i):

U_{s}ā = U_{s}({t, s, ã} − {U_{t}y, s, a′})
= {s, t, U_{s}ã} − {s, U_{t}y, U_{s}a′}
= {s, t, {y, t, a}} − {s, U_{t}y, a}
(J8) = {s, U_{t}a, y}.

Let us also note that we do have ā = M_{y,s}^{∗}(a). By above, we have

U_{s}ā = {s, U_{t}a, y}
= U_{y,s}U_{t}a
= U_{s}M_{y,s}^{∗}a

by the definition of M_{y,s}^{∗}.

(ii):

V_{y,U_{t}a} − V_{s,{t,s,ã}−{U_{t}y,s,a′}}

= V_{y,U_{t}a} − (V_{U_{s}t,ã} + V_{U_{s}ã,t}) + (V_{U_{s}U_{t}y,a′} + V_{U_{s}a′,U_{t}y})
= V_{y,U_{t}a} − V_{U_{s}t,ã} − V_{{y,t,a},t} + V_{U_{s}U_{t}y,a′} + V_{a,U_{t}y}
= V_{y,U_{t}a} − V_{U_{s}t,ã} − (V_{y,U_{t}a} + V_{a,U_{t}y})
+V_{U_{s}U_{t}y,a′} + V_{a,U_{t}y}
= V_{U_{s}U_{t}y,a′} − V_{U_{s}t,ã}
:= V̄_{a}.
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Lemma 32 Let Δ ∈ Ṽ_{σ}, σ = ±1, and let s ∈ S be a denominator for (Δ, Δ^{2}). If y, x, z ∈ J, and a ∈ K_{s}, and t ∈ S is a super s-denominator for y, then

U_{s}({Δ(U_{s}x)[z], y, U_{t}a}) = V̄_{a}{d, x, U_{s}z}.

Proof

U_{s}V_{U_{t}a,y}Δ(U_{s}x)[z]

= (−V_{y,U_{t}a}U_{s} + U_{{y,U_{t}a,s},s})Δ(U_{s}x)[z]
= (−V_{y,U_{t}a}U_{s} + U_{U_{s}ā,s})Δ(U_{s}x)[z]
= (−V_{y,U_{t}a} + V_{s,ā})U_{s}(Δ(U_{s}x)[z])
= (−V̄_{a})(−{d, x, U_{s}z})
= V̄_{a}{d, x, U_{s}z}.
■

From this point our logic will go as follows. We first will show that anything of the form

Q_{Λ_{t}}(ad(e))

for e ∈ J and t ∈ S, is a fractional derivation (ie. has denominator). Next we will show in Theorem 1 that anything of the form

Q_{ad(y)}Δ

for y ∈ J, is equivalent to something of the form

Q_{Λ_{t}}(ad(e)).

Finally, in Theorem 2 we will show that an arbitrary Q_{Δ}(Θ) has a denominator by showing that it is equivalent to something of the form

Q_{ad(y)}(Δ).

Lemma 33 Let x ∈ J and s, t ∈ S and suppose {x, t, s} = U_{s}t̃ for t̃ ∈ J (eg. if t is a super s-denominator for x). Then for any a ∈ Ĵ we have

{x, t, U_{s}a} = U_{s}ã

for ã = (V_{t̃,s} − V_{t,x})[a]. Thus

{x, t, s} ∈ U_{s}Ĵ ⇒ {x, t, K_{s}} ⊆ U_{s}Ĵ.

Proof

V_{x,t}U_{s}a = −U_{s}V_{t,x}a + U_{{x,t,s},s}a
= −U_{s}V_{t,x}a + U_{U_{s}t̃,s}a
= U_{s}[−V_{t,x} + V_{t̃,s}[a]
= U_{s}ã.

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Theorem 1 Let x ∈ J and s ∈ S and suppose that Θ = Q_{Λ_{s}}(ad(x)). Then Θ is a fractional derivation and s′ = U_{s}t is a denominator for (Θ, Θ^{2}) (for Θ^{2} = Λ_{s}^{2}U_{x}Λ_{s}^{2}) where t is a super s-denominator for x.

Proof Let a ∈ U_{s}Ĵ. Note that since t is a super s-denominator for x, we have

V_{s,t}x = U_{s}t̃.

Also, if

t_{a} = −U_{t}{s, a, x} + {t, U_{s}a, t̃}

then

{s, U_{t}U_{s}a, x} = U_{s}t_{a}.

So,

Θ[U_{s′}a] = (Λ_{s}^{2}ad(x) + ad(x)Λ_{s}^{2} − Λ_{s}ad(x)Λ_{s})[U_{s′}a]
= 0 + ad(x)Λ_{s}^{2}[U_{s}U_{t}U_{s}a] − Λ_{s}ad(x)(δ_{s,U_{t}U_{s}a})
= ad(x)(U_{t}U_{s}a) + Λ_{s}({s, U_{t}U_{s}a, x})
= −δ_{x,U_{t}U_{s}a} + Λ_{s}U_{s}t_{a}
= −δ_{x,U_{t}U_{s}a} + δ_{s,t_{a}}
= −δ_{x,U_{t}U_{s}a} − δ_{s,U_{t}{s,a,x}} + δ_{s,{t,U_{s}a,t̃}}.

Set a′ = {t, s, a} and set

A = −δ_{x,U_{t}U_{s}a}
B = −δ_{s,U_{t}{s,a,x}}
C = δ_{s,{t,U_{s}a,t̃}}.

Then

A = δ_{U_{s}a,U_{t}x} − δ_{{x,t,U_{s}a},t}
= δ_{U_{s}a,U_{t}x} − [δ_{x,t}, δ_{U_{s}a,t}] − δ_{U_{s}a,{t,x,t}}
= δ_{U_{s}a,U_{t}x} − [δ_{x,t}, δ_{U_{s}a,t}] − 2δ_{U_{s}a,U_{t}x}
= −δ_{U_{s}a,U_{t}x} − [δ_{x,t}, −δ_{U_{s}t,a} + δ_{s,{a,s,t}}]
= −δ_{U_{s}a,U_{t}x} + [δ_{x,t}, δ_{s′,a}] − [δ_{x,t}, δ_{s,a′}].
B = −δ_{s,[−V_{a,s}U_{t}+U_{{a,s,t},t}]x}
= δ_{s,{a,s,U_{t}x}} − δ_{s,{a′,x,t}}
= δ_{U_{s}a,U_{t}x} + δ_{U_{s}U_{t}x,a} − δ_{s,{t,x,a′}}.
C = δ_{s,−{t,U_{s}t̃,a}+{t,s,{a,s,t̃}}}
= −δ_{s,{t,{s,t,x},a}} + δ_{s,{t,s,{a,s,t̃}}}
= (δ_{{s,t,x},{t,s,a}} − δ_{{{s,t,x},a,s},t} − δ_{{{s,t,x},t,s},a})

+(δ_{U_{s}t,{a,s,t̃}} + δ_{U_{s}{a,s,t̃},t})

= δ_{{s,t,x},a′} − δ_{{s,a{s,t,x}},t} − (2δ_{U_{s}U_{t}x,a} + δ_{{x,t,U_{s}t},a})

+δ_{s′,{a,s,t̃}} + δ_{{s,a,U_{s}t̃},t}

= δ_{{x,t,s},a′} − 2δ_{U_{s}U_{t}x,a} − δ_{{x,t,s′},a} + δ_{s′,{a,s,t̃}}
= −2δ_{U_{s}U_{t}x,a} + δ_{{x,t,s},a′} − ([δ_{x,t}, δ_{s′,a}] + δ_{s′,{t,x,a}})

+δ_{s′,{a,s,T (x)}}

= −2δ_{U_{s}U_{t}x,a} − [δ_{x,t}, δ_{s′,a}] + δ_{{x,t,s},a′}

+δ_{s′,[{t̃,s,a}−{t,x,a}]}.

Therefore

Θ[U_{s′}a] = A + B + C
= −δ_{U_{s}U_{t}x,a} + δ_{s′,[V_{t̃,s}−V_{t,x}]a}
= δ_{a,e′} − δ_{E′(a),s′}

for

e′ = U_{s}U_{t}x and E′ = V_{t̃,s} − V_{t,x}.

Hence, s′ satisfies the first condition for being a denominator of (Θ, Θ^{2}).

Also, if ã = E′(a), then

{e′, a, s′} = {U_{s}U_{t}x, a, U_{s}t}
= U_{s}{U_{t}x, U_{s}a, t}
= U_{s}U_{t}{x, t, U_{s}a}
(lem33) = U_{s}U_{t}U_{s}ã
= U_{s′} ã
= U_{s′}E′(a).

Hence, s′ satisfies the second condition for being a denominator of (Θ, Θ^{2}).

The third condition requires that

Λ_{s′}Θ^{2}Λ_{s′} = U_{e′}.

Well,

U_{s′}Θ^{2}U_{s′}
= U_{s}U_{t}U_{s}Λ_{s}^{2}U_{x}Λ_{s}^{2}U_{s}U_{t}U_{s}
= U_{s}U_{t}U_{x}U_{t}U_{s}
= U_{U_{s}U_{t}x}
= U_{e′}.

So s′ satisfies this third condition as well and the theorem is proved.

■

Theorem 2 Let Δ be a fractional derivation with denominator s ∈ S, and let y ∈ J. Then the derivation Q_{ad(y)}(Δ) is again a fractional derivation. Indeed, if t is a super s-denominator for both y and d (where Δ = Q_{Λ_{s}}(ad(d)) is a proper fraction), then

Q_{ad(y)}(Δ) = Q_{Λ_{t}}(ad(e))

for e = M_{y}(d).

Proof Recall that Δ can be written as a proper fraction by Lemma 30 (iii). By degrees it suffices to show that these maps agree on £_{U_{t}s^{2}} ⊆ U_{t}U_{s}Ĵ. We hope to show specifically that for elements U_{t}a, U_{t}b (where a = U_{s}a′ ∈ U_{s}Ĵ) we have

[Q_{ad(y)}(Δ)(U_{t}a)][U_{t}b] = U_{t}{a, e, b} = [Q_{Λ_{t}}(ad(e))(U_{t}a)][U_{t}b]. (∗)

The left hand side of (∗) becomes:

Q_{ad(y)}(Δ)(U_{t}a)[U_{t}b] = (U_{y}Δ + ΔU_{y} − ad(y)Δad(y))(U_{t}a)[U_{t}b]
= (0 + Δ(U_{y}U_{t}a) − ad(y)Δ(δ_{U_{t}a,y}))[U_{t}b]
= (Δ(U_{y}U_{t}a) − δ_{Δ(δ_{U_{t}a,y}),y})[U_{t}b].

We can hit this with U_{s} to obtain:

U_{s}(Q_{ad(y)}(Δ)(U_{t}a)[U_{t}b])

= U_{s} (Δ(U_{y}U_{t}a) − δ_{Δ(δ_{U_{t}a,y}),y})[U_{t}b])
= U_{s} (Δ(U_{y}U_{t}a)[U_{t}b] − {Δ(δ_{U_{t}a,y}), y, U_{t}b})
= U_{s} (Δ(U_{y}U_{t}a)[U_{t}b] − {Δ(−δ_{U_{t}y,a} + δ_{t,{a,t,y}}), y, U_{t}b})
= U_{s} (Δ(U_{y}U_{t}a)[U_{t}b] − {Δ({a, t, y})(t), y, U_{t}b} + {Δ(a)(U_{t}y), y, U_{t}b})
= U_{s} (Δ(U_{s}M_{y}^{∗}a)[U_{t}b] − {Δ({a, t, y})(t), y, U_{t}b}

+{Δ(a)(U_{t}y), y, U_{t}b})

= U_{s}Δ(U_{s}M_{y}^{∗}a)[U_{t}b] − U_{s}{Δ({a, t, y})(t), y, U_{t}b}

+U_{s}{Δ(a)(U_{t}y), y, U_{t}b}

= −{d, M_{y}^{∗}(a), U_{s}U_{t}b} − U_{s}{Δ(U_{s}ã)(t), y, U_{t}b}

+U_{s}{Δ(U_{s}a′)(U_{t}y), y, U_{t}b}

(lem31) = −{d, M_{y}^{∗}(a), U_{s}U_{t}b} − {d, ã, U_{s}t}) + V̄_{b}({d, a′, U_{s}U_{t}y}
= −{U_{s}U_{t}b, M_{y}^{∗}(a), d} + V̄_{b}((V_{U_{s}U_{t}y,a′} − V_{U_{s}t,ã})[d])
(lem31) = (−V_{U_{s}U_{t}b,M_{y}^{∗}(a)} + V̄_{b}V̄_{a})[d]
(AX2) = U_{s}U_{t}{a, M_{y}d, b}
= U_{s}U_{t}{a, e, b}.

Cancelling the U_{s} gives us the first equality in (∗). Now, Lemma 30 allows us to immediately conclude that

[(Q_{Λ_{t}}ad(e))(U_{t}a)][U_{t}b] = U_{t}{a, e, b},

which will yield the second equality in (∗). However, we can only appeal to this lemma if we know that all

a ∈ U_{s}U_{t}Ĵ

satisfy

{e, a, t} = {M_{y}d, a, t} ∈ U_{t}J.

We will show that this is indeed the case. We assumed that Δ = Q_{Λ_{s}}(ad(d)) was proper. Hence,

(1) {d, a, s} = U_{s}D_{σ}(a)

for all a ∈ J. Now, writing a = U_{s}a′ for a′ ∈ U_{t}Ĵ, we have

{y, t, a} = U_{s}ã,

since t is a super s-denominator for y. By Lemma 5 we have

V_{t,a}M_{y}(d) = V_{t,U_{s}a′}M_{y}(d)
(AX3) = (−M_{y}V_{U_{s}t,a′} + M_{y,s}V_{s,ã} − U_{t}V_{y,ã})[d].

Look at the three terms in the last expression. Clearly the third term is in U_{t}J. By (1) we can look at the second term and conclude (by the definition of M_{y,s}):

M_{y,s}{s, a, d} ∈ M_{y,s}U_{s}Ĵ = U_{t}U_{y,s}Ĵ ⊆ U_{t}J.

As for the first term:

−M_{y}{U_{s}t, a′, d} = −M_{y}(V_{s,t}V_{s,a′} − U_{s}U_{t,a′})[d]
(1) ∈ −M_{y}(V_{s,t}U_{s}Ĵ+ U_{s}J)
⊆ M_{y}(U_{s}V_{t,s}Ĵ+ U_{s}J)
⊆ M_{y}U_{s}J
= U_{t}U_{y}J
⊆ U_{t}J.

Since all three terms are in U_{t}J, we conclude that

{e, a, t} = V_{t,a}M_{y}(d) ∈ U_{t}J

and hence we can appeal to Lemma 30. Thus both equalities in (∗) hold.

■

Theorem 3 Let Δ ∈ Ṽ_{σ} and Θ ∈ Ṽ_{−σ} for σ = ±1. Then Q_{Δ}(Θ) has a denominator.

Proof Let s ∈ S be a denominator for Δ so that

Δ = Q_{Λ_{s}}(ad(d))

and hence

Q_{ad(s)}(Δ) = ad(d).

Set Δ′ = Q_{Λ_{s}}(Θ). Then, appealing to the facts in Appendix C, we have

Q_{Δ}(Θ) = (Q_{Λ_{s}}Q_{ad(s)})Q_{Δ}(Q_{ad(s)}Q_{Λ_{s}})Θ
= Q_{Λ_{s}}(Q_{Q_{ad(s)}Δ})Δ′
= Q_{Λ_{s}}(Q_{ad(d)}Δ′)
(th2) = Q_{Λ_{s}}(Q_{Λ_{t}}(ad(e)))
= Q_{Λ_{s}}(Q_{Λ_{t}}(Q_{Λ_{s}}Q_{ad(s)}ad(e)))
= Q_{Λ_{Ust}}(ad(U_{s}e))
= Q_{Λ_{t′}}(ad(x′))

for t′ = U_{s}t and x′ = U_{s}e. By thm??, we know that Q_{Λ_{t′}}(ad(x′)) does have a denominator. Since

Q_{Δ}(Θ) = Q_{Λ_{t′}}(ad(x′))

we conclude that Q_{Δ}(Θ) does as well and we have our desired result.

■

8 Conclusion

We should note one more subtle point. Faulkner’s proposition also requires that the conditions we have just verified continue to hold in all (free) scalar extensions. If we have S ⊆ J ⊆ J̃ where J̃ is a scalar extension of J, then S will continue to be an Ore monad in J̃ and J̃ will continue to satisfy the Ore condition with respect to S. These two facts were all that were needed to verify that the other axioms in Faulkner’s proposition held. Therefore, those axioms will continue to hold for J̃ as well.

So, after all of this work we know the following: we can apply Faulkner’s proposition to Ã_{J} and the subspaces Ṽ_{σ} to construct the Jordan pair

(Ṽ_{σ}, Ṽ_{−σ}).

From here we can define an involution to form a Jordan triple and then use the unit element 1 to obtain a unital Jordan algebra (if 1 is not available then we can move to the unital hull). We then imbed J in this unital Jordan algebra by sending each x ∈ J to ad(x). Once imbedded in their new setting, the elements s ∈ S have inverses existing as the Λ_{s} maps. These details and the verification that the suggested map is an injective Jordan homomorphism, are reasonably straight forward and left to the reader. We should remark at this point that, in characteristic 2, the map

x → ad(x)

may not always faithfully imbed J. For all x in the extreme radical, (ad(x)U_{x}) = (0, 0). Hence, these extreme radical elements do not lead to the same nice theorems we obtained above.

A Facts I

Let s ∈ S be a denominator for (Λ_{s}, Λ_{s}^{2}). Also, let x, y ∈ J and choose t ∈ U_{s}S to be a super s-denominator for y. In this setting we note the following facts, some without proof.

(A1) U_{y}U_{t}â ∈ U_{s}Ĵ if â ∈ K_{s}.

(A2) U_{d_{−σ},s}Ĵ ∈ U_{s}D_{σ}(Ĵ).

(A3) U_{d_{−σ},y}U_{t} = U_{s}M_{d_{−σ},y}^{∗} on J.

(A4) U_{x,y} = ad(x_{σ})ad(y_{σ}) = ad(y_{σ})ad(x_{σ}).

Proof Let b̂ ∈ Ĵ. Then

ad(x_{σ})ad(y_{σ})[b̂_{−σ}] = ad(x_{σ})(−δ_{y_{σ},b̂_{−σ}})
= −[x_{σ}, δ_{y_{σ},b̂_{−σ}}]
= [δ_{y_{σ},b̂_{−σ}}, x_{σ}]
= {y, b̂, x}_{σ}
= {x, b̂, y}_{σ}
= [δ_{x,b̂}, y]
= −[y_{σ}, δ_{x_{σ},b̂_{−σ}}]
= ad(y_{σ})(−δ_{x_{σ},b̂_{−σ}})
= ad(y_{σ})ad(x_{σ})[b̂_{−σ}].
■

(A5) If Δ, Λ ∈ Ṽ_{σ}, then ΔΛ = ΛΔ on £_{t′} for some t′ ∈ S.

(A6) ad(s_{−σ})Λ_{s}^{2} = −Λ_{s} on (K_{s^{3}})_{−σ}. Proof If x̂ ∈ Ĵ and x = U_{s}U_{s}x̂, then

ad(s_{−σ})Λ_{s}^{2}[U_{s^{3}} x̂] = ad(s_{−σ})Λ_{s}^{2}[U_{s}x] = ad(s)[x] = −δ_{s,x} = −Λ_{s}[U_{s}x] = −Λ_{s}[U_{s^{3}} x̂].

ad(s_{−σ})Λ_{s^{2}}(s^{3}) = ad(s_{−σ})Λ_{s}^{2}[U_{s}s] = ad(s_{−σ})(s_{σ}) = −δ_{s_{−σ},s_{σ}} = −Λ_{s}[U_{s}s] = −Λ_{s}(s^{3}).

■

(A7) Let W = δ_{x,U_{s}y} for x ∈ Ĵ_{σ} and U_{s}y ∈ J_{−σ}. Then

Λ_{s}^{2}ad(s_{−σ})[W] = −Λ_{s}[W].

Proof

Λ_{s}^{2}ad(s_{−σ})[W] = Λ_{s}^{2}ad(s_{−σ})[δ_{x,U_{s}y}]
= Λ_{s}^{2}[δ_{U_{s}y,x}, s_{−σ}]
= Λ_{s}^{2}{U_{s}y, x, s}
= Λ_{s}^{2}U_{s}{y, s, x}
= {y, s, x}
= δ_{y,s}x
= −δ_{s,y}x
= −Λ_{s}(U_{s}y)[x]
= −Λ_{s}[δ_{x,U_{s}y}]
= −Λ_{s}[W].
■

B Facts II

Let Δ ∈ Ṽ_{σ}, σ = ±1, and Θ ∈ Ṽ_{−σ}. Also, let s ∈ S be a denominator for the squared derivation (Δ, Δ^{2}). In this setting we note the following facts, some without proof.

(B1) Since Θ is shrinkable, we can choose r ∈ U_{s}S such that

Θ[(K_{r})_{σ}] ∈ δ_{U_{s}Ĵ,U_{s}Ĵ}.

(B2) There exists w ∈ U_{U_{r}s}(S) ∩ U_{U_{s}r}(S). Since Θ is shrinkable, and by the CIMP, we can choose z ∈ U_{w}(S) such that

Θ[(K_{z})_{σ}] ⊆ δ_{U_{w}Ĵ,U_{w}Ĵ}.

(B3) We can choose t to be simultaneously a super U_{s}z-denominator and a super U_{s}r-denominator for d_{−σ}. Furthermore, by the CIMP and innerness of strong denominators, we can require that t ∈ U_{U_{r}s}Ĵ.

(B4) U_{d_{−σ}}U_{t}K̂_{U_{s}z} ⊆ U_{U_{s}z}Ĵ.

Proof This follows immediately from B3.

■

(B5) {d_{−σ}, t, K̂_{U_{s}z}} ⊆ U_{U_{s}r}Ĵ.

Proof This follows immediately from B3.

■

(B6) Since Θ, Λ_{s}, U_{d_{−σ}}, and U_{t} are all shrinkable, there exists an element b such that

ΘΛ_{s}U_{d_{−σ}}U_{t}[b] ∈ U_{U_{s}r}Ĵ.

Furthermore, by the CIMP, we can require that b ∈ U_{U_{s}r}Ĵ.

(B7) {d_{−σ}, t, b} ∈ {d_{−σ}, t, K̂_{U_{s}r}} ⊆ {U_{U_{s}r}Ĵ}. Proof This follows immediately from

B3.

■

(B8) δ_{d_{−σ},U_{t}b} ∈ δ_{U_{Usr}Ĵ,U_{Urs}Ĵ}.

Proof

δ_{d_{−σ},U_{t}b} = −δ_{b,U_{t}d_{−σ}} + δ_{{d_{−σ},t,b},t} ∈ δ_{U_{Usr}Ĵ,U_{Usr}Ĵ}

since

b ∈ U_{U_{s}r}Ĵ
U_{t}d_{−σ} ∈ U_{U_{r}s}Ĵ
{d_{−σ}, t, b} ∈ U_{U_{s}r}Ĵ (by B7)
t ∈ U_{U_{r}s}Ĵ.
■

(B9) {d_{−σ}, U_{t}b, s} ∈ U_{U_{s}r}Ĵ.

Proof

{d_{−σ}, U_{t}b, s} = δ_{d_{−σ},U_{t}b}(s)
∈ {U_{U_{s}r}Ĵ, U_{U_{r}s}Ĵ, s}
= {s, U_{U_{r}s}Ĵ, U_{U_{s}r}Ĵ}
= δ_{s,U_{Urs}Ĵ}(U_{U_{s}r}Ĵ)
= U_{s}δ_{U_{Urs}Ĵ,s}(U_{r}U_{s}Ĵ)
= U_{s}{U_{r}U_{s}Ĵ, s, U_{U_{r}s}Ĵ}
⊆ U_{s}U_{r}U_{s}Ĵ
= U_{U_{s}r}Ĵ.
■

(B10) Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2} = Δ^{2} on £_{s}.

Proof Let â ∈ Ĵ. Then

U_{s}Δ^{2}U_{s}[â] = U_{d_{−σ}}[â] = U_{d_{−σ}}Λ_{s}^{2}[U_{s}â].

If we hit this equality with Λ_{s}^{2} on the left, we obtain

Λ_{s}^{2}U_{s}Δ^{2}[U_{s}â] = Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2}[U_{s}â],

which implies

Δ^{2}[U_{s}â] = Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2}[U_{s}â].

So,

Δ^{2}[(K_{s})_{−σ}] = Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2}[(K_{s})_{−σ}],

and by degree we have

Δ^{2}[(K_{s})_{σ}] = 0 = Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2}[(K_{s})_{σ}].

and

Δ^{2}[D_{0}^{s}] = 0 = Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2}[D_{0}^{s}].

Therefore,

Δ^{2} = Λ_{s}^{2}U_{d_{−σ}}Λ_{s}^{2}

on all of £_{s} as desired.

■

(B11) Let Θ ∈ Ṽ_{−σ}, s ∈ S, and let c ∈ S be such that

Θ[(K_{c})_{σ}] ⊆ δ_{(U_{s}Ĵ)_{−σ},(U_{s}Ĵ)_{σ}}.

Then

Λ_{s} ↬_{J} Θ

on U_{U_{s}c}Ĵ. Proof Let â ∈ Ĵ_{σ} and suppose

Θ(U_{c}U_{s}â) = ∑_{i} δ_{U_{s}x̂_{i},U_{s}ŷ_{i}} ∈ δ_{(U_{s}Ĵ)_{−σ},(U_{s}Ĵ)_{σ}}.

Then

Λ_{s}^{2}ΘΛ_{s}[U_{s}U_{c}U_{s}â] = Λ_{s}^{2}Θ[δ_{s,U_{c}U_{s}â}]
= Λ_{s}^{2}Θ(U_{c}U_{s}â)[s]
= Λ_{s}^{2} ∑_{i}(δ_{U_{s}x̂_{i},U_{s}ŷ_{i}}[s])
= Λ_{s}^{2} ∑_{i}[{U_{s}x̂_{i}, U_{s}ŷ_{i}, s}]
= ∑_{i} Λ_{s}^{2}[U_{s}{x̂_{i}, s, U_{s}ŷ_{i}}
= ∑_{i}{x̂_{i}, s, U_{s}ŷ_{i}}
= ∑_{i} −δ_{s,x̂_{i}}[U_{s}ŷ_{i}]
= − ∑_{i} Λ_{s}(U_{s}x̂_{i}[U_{s}ŷ_{i}]
= − ∑_{i} Λ_{s}[δ_{U_{s}ŷ_{i},U_{s}x̂_{i}}]
= ∑_{i} Λ_{s}[δ_{U_{s}x̂_{i},U_{s}ŷ_{i}}]
= Λ_{s}Θ[U_{c}U_{s}â]
= Λ_{s}ΘΛ_{s}^{2}[U_{s}U_{c}U_{s}â]
■

(B12) Let W = δ_{U_{s}x̂,U_{s}ŷ} ∈ δ_{(U_{s}Ĵ)_{σ},(U_{s}Ĵ)_{−σ}}, for s ∈ S. Then Λ_{s}[W] = Λ_{s}^{2}W [s].

Proof

Λ_{s}[W] = Λ_{s}[δ_{U_{s}x̂,U_{s}ŷ}]
= Λ_{s}(U_{s}ŷ)[U_{s}x̂]
= δ_{s,ŷ}(U_{s}x̂)
= −δ_{ŷ,s}(U_{s}x̂)
= −{ŷ, s, U_{s}x̂}
= −Λ_{s}^{2}U_{s}[{ŷ, s, U_{s}x̂}]
= −Λ_{s}^{2}[{U_{s}ŷ, U_{s}x̂, s}]
= −Λ_{s}^{2}[δ_{U_{s}ŷ,U_{s}x̂}(s)]
= Λ_{s}^{2}[δ_{U_{s}x̂,U_{s}ŷ}(s)].
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C Facts III

(C1) Let Θ ∈ Ṽ_{−σ}, σ = ±1, and let s ∈ S_{σ}. Then

Q_{Λ_{s}}(Q_{ad(s)}(Θ)) = Θ

in Ã_{J}.

Proof Because all of the expressions we will deal with are shrinkable, we assume from the start that b is in a restricted domain £_{t} where all of the relevant strings of maps are defined.

Set

A = ad(s)Λ_{s}^{2}U_{s}ad(s)[b]
B = ΘΛ_{s}^{2}U_{s}[b]
C = Λ_{s}ΘΛ_{s}U_{s}[b]
D = ad(s)Λ_{s}ΘΛ_{s}ad(s)[b].

Then

A = ad(s)Λ_{s}^{2}Θ[s, b]
= −ad(s)Λ_{s}^{2}Θ(b)[s]
= ad(s)Λ_{s}^{2}[U_{s}ĉ](some ĉ ∈ Ĵ)
= −ad(s)[ĉ]
= δ_{s,ĉ}
B = ΘΛ_{s}^{2}[U_{s}b]
= Θ[b]
C = Λ_{s}ΘΛ_{s}[U_{s}b]
= Λ_{s}Θ[δ_{s,b}]
= Λ_{s}Θ(b)[s]
= Λ_{s}[U_{s}ĉ]
= δ_{s,ĉ}

D = ad(s)Λ_{s}ΘΛ_{s}[δ_{b,s}]
= ad(s)Λ_{s}Θ(Λ_{s}[s])[b]
= ad(s)Λ_{s}Θ[−2b]
= −2ad(s)[ĉ]
= 2δ_{s,ĉ}.

By degree arguments, this gives

Q_{ad(s)}(Q_{Λ_{s}}Θ)[b] = −A + B − C + D
= −δ_{s,ĉ} + Θ[b] − δ_{s,ĉ} + 2δ_{s,ĉ}
= Θ[b].

(Q_{Λ_{s}}(Q_{ad(s)}(Θ))^{2} = Λ_{s}^{2}(Q_{ad(s)}(Θ))^{2}Λ_{s}^{2}
= Λ_{s}^{2}U_{s}Θ^{2}U_{s}Λ_{s}^{2}
= Θ^{2}.
■

(C2) Let y ∈ J_{−σ}, σ = ±1, and e ∈ J_{σ}. Then Q_{ad(y)}(ad(e)) = ad(U_{y}e).

Proof Let b ∈ J_{σ}. Then,

Q_{ad(y)}(ad(e)) = [U_{y}ad(e) + ad(e)U_{y} − ad(y)ad(e)ad(y)][b]
= 0 + ad(e)U_{y}b − ad(y)ad(e)[y, b]
= [e, U_{y}b] − ad(y)(−δ_{e,y}[b])
= −δ_{e,U_{y}b} + ad(y)[{e, y, b}]
= −δ_{e,U_{y}b} − δ_{y,{e,y,b}}
(J9) = −δ_{U_{y}e,b}
= ad(U_{y}e)[b].

(Q_{ad(y)}(ad(e)))^{2} = U_{y}U_{e}U_{y}
= U_{U_{y}e}
= (ad(U_{y}e))^{2}.

■

(C3) Let t, r ∈ S. Then Q_{ad(U_{t}r)} = Q_{ad(t)}Q_{ad(r)}Q_{ad(t)}.

Proof By (C2), we have

Q_{ad(U_{t}r)} = Q_{Q_{ad(t)}(ad(r))};

and by Proposition 4 we know that

Q_{Q_{ad(t)}(ad(r))} = Q_{ad(t)}Q_{ad(r)}Q_{ad(t)}.

Also, given Δ,

(Q_{ad(U_{t}r)}(Δ))^{2} = U_{U_{t}r}Δ^{2}U_{U_{t}r}
= U_{t}U_{r}U_{t}Δ^{2}U_{t}U_{r}U_{t}
= (ad(t))^{2}(ad(r))^{2}(Q_{ad(t)}(Δ))^{2}(ad(r))^{2}(ad(t))^{2}
= (ad(t))^{2}(Q_{ad(r)}Q_{ad(t)}(Δ))^{2}(ad(t))^{2}
= (Q_{ad(t)}Q_{ad(r)}Q_{ad(t)}Δ)^{2}.
■

(C4) Let t, r ∈ S. Then Q_{Λ_{U_{t}r}} = Q_{Λ_{t}}Q_{Λ_{r}}Q_{Λ_{t}}.

Proof

Q_{Λ_{U_{t}r}} = Q_{Λ_{t}}Q_{Λ_{r}}Q_{Λ_{t}}Q_{ad(t)}Q_{ad(r)}Q_{ad(t)}Q_{Λ_{U_{t}r}}
= Q_{Λ_{t}}Q_{Λ_{r}}Q_{Λ_{t}}(Q_{ad(U_{t}r)}Q_{Λ_{U_{t}r}})
(C1) = Q_{Λ_{t}}Q_{Λ_{r}}Q_{Λ_{t}}.

Note,

Λ_{U_{t}r}^{2}(U_{U_{t}r}a) = a = Λ_{t}^{2}Λ_{r}^{2}Λ_{t}^{2}(U_{U_{t}r}a).

So,

(Q_{Λ_{U_{t}r}}(Δ))^{2} = Λ_{U_{t}r}^{2}Δ^{2}Λ_{U_{t}r}^{2}
= Λ_{t}^{2}Λ_{r}^{2}Λ_{t}^{2}Δ^{2}Λ_{t}^{2}Λ_{r}^{2}Λ_{t}^{2}
= Q_{Λ_{t}}Q_{Λ_{r}}Q_{Λ_{t}}(Δ)
■

D d E e F f G g H h I i J Jordan Identities

(J1) U_{1} = 1 (J2) V_{x,y}U_{x} = U_{x}V_{y,x} = U_{U_{x}y,x} (J3) U_{U_{x}y} = U_{x}U_{y}U_{x} (J4) V_{x,x} = V_{x^{2}} (J5) V_{U_{x}y,y} = V_{x,U_{y}x} (J6) U_{x,z}V_{y,x} + U_{x}V_{y,z} = U_{x,{x,y,z}} + U_{z,U_{x}y} = V_{x,y}U_{x,z} + V_{z,y}U_{x} (J7) V_{x,{y,x,z}} + U_{x}U_{y,z} = V_{x,z}V_{x,y} + V_{U_{x}y,z} = V_{x,y}V_{x,z} + V_{U_{x}z,y} (J8) V_{{x,y,z},y} = V_{z,U_{y}x} + V_{x,U_{y}z} (J9) V_{x,{y,x,z}} = V_{U_{x}y,z} + V_{U_{x}z,y} (J10) V_{x,y}V_{z,y} = U_{x,z}U_{y} + V_{x,U_{y}z} (J11) U_{x,z}V_{y,x} = U_{U_{x}y,z} + V_{z,y}U_{x} (J12) V_{x,y}U_{x,z} = U_{U_{x}y,z} + U_{x}V_{y,z} (J13) V_{x,y}U_{z} + U_{z}V_{y,x} = U_{z,{x,y,z}} (J14) V_{x,y}V_{x,z} = V_{U_{x}y,z} + U_{x}U_{y,z} (J15) [V_{x,y}, V_{u,v}] = V_{{x,y,u},v} − V_{u,{y,x,v}} (J16) V_{U_{x}y,z}U_{x} = U_{x}V_{y,U_{x}z} (J17) V_{U_{x}y,z}V_{x,y} = U_{x}U_{y}V_{x,z} + V_{x,U_{y}U_{x}z} (J18) U_{U_{x}y,{x,y,z}} = U_{x}U_{y}U_{x,z} + U_{x,z}U_{y}U_{x} (J19) U_{{x,y,z}} + U_{U_{x}y,U_{z}y} = U_{x}U_{y}U_{z} + U_{z}U_{y}U_{x} + U_{x,z}U_{y}U_{x,z} (J20) U_{y}U_{x} = U_{y∘x} + U_{U_{y}x,x} − U_{x}U_{y} − V_{y}U_{x}V_{y} (J21) y ∘ U_{x^{2},x}y = x^{2} ∘ U_{y}x + x ∘ U_{y}(x^{2}) (J22) V_{x∘y} = V_{x,y} + V_{y,x} (J23) U_{x}U_{y} + U_{y}U_{x} = U_{x∘y} + V_{U_{x}(y^{2})} − V_{x,y}V_{y,x} (J24) U_{x}(a^{2}) + U_{a}(x^{2}) + x ∘ (U_{a}x) = (a ∘ x)^{2} (J25) {x, x ∘ a, U_{x}a} = 2(U_{x}a)^{2} + U_{x}(x ∘ U_{a}x) (J26) In s-homotope:

−V_{x′}U_{y,y′}(x) = U_{y,y′}(x ∘ x′) − {x′ ∘ y, x, y′} − {x′ ∘ y′, x, y}

V_{y′}U_{x,x′}(y) = {y′, x, y ∘ x′} + {y′, x′, y ∘ x} − {y′, x ∘ x′, y}

(J27) x ∘ (U_{y}x) = y ∘ (U_{x}y) (J28) V_{x^{2},y} = V_{x,x∘y} − V_{U_{x}y} (J29)

U_{x∘y^{2}} = U_{x}U_{y^{2}} + U_{y^{2}}U_{x} + V_{y,x∘y}V_{x∘y,y}
+V_{U_{y}x}^{2} − V_{y,x∘y}V_{U_{y}x} − V_{U_{y}x}V_{x∘y,y}
−V_{U_{y^{2}}x^{2}}