The Scalar Center for Quadratic Jordan Algebras

Bernard Fulgham Department of Mathematics University of Virginia Charlottesville, VA 22903 U.S.A.

Abstract We propose that an element of a Jordan algebra J should be considered “central” if it really is a scalar multiple of 1 in some tight unital extension of J. For Jordan algebras with no trivial ideals, this yields an acceptable center. Here it is important that an algebra with no extreme elements has a unique tight unital hull. Lopez has observed that a necessary condition for centrality of an element c is that the operators U_{c}, V_{c} must be in the centroid. When the algebra is nondegenerate, we show this condition is actually sufficient, but we give a counterexample for general Jordan algebras.

1 Background

Before we begin, a review of the background material is in order. A unital quadratic Jordan algebra J = (J, U, 1) over an arbitrary ring of scalars Φ consists of a Φ−module J, an element 1 ∈ J, and a quadratic map U : J → End_{Φ}(J) (its linearization U_{x, z} := U_{x+z} − U_{x} − U_{z} induces a Jordan triple product {x, y, z} = U_{x, z} y, and we denote “left triple multiplication by x, y” as V_{x, y}: V_{x, y} z = {x, y, z}) such that the following relations

U_{1} = Id
U_{x} V_{y, x} = V_{x, y} U_{x} = U_{U_{x}y, x}
U_{U_{x}y} = U_{x} U_{y} U_{x}

hold strictly; that is, they hold not only in J but also in all free scalar extensions J_{Ω} := Ω ⊗_{Φ} J, where the U operator is defined in the following manner:

U_{∑_{i} ω_{i}⊗x_{i}} := ∑_{i} ω_{i}^{2} ⊗ U_{x_{i}} + ∑_{i<j} ω_{i} ω_{j} ⊗ U_{x_{i}, x_{j}} .

An important extension is the polynomial extension J[T] := Φ[T] ⊗_{Φ} J, where Φ[T] is the polynomial ring with coefficients from Φ and indeterminates from T. We will use J[T] to construct a tight extension of J: an extension J̃ of J is tight if every nontrivial ideal of J̃ intersects nontrivially with J. In other words, if Ĩ ≠ 0 is an ideal of J̃, then Ĩ∩ J ≠ 0. For the matter at hand, tight extensions are useful because they inherit nondegeneracy, where a nondegenerate Jordan algebra is simply an algebra such that U_{x} = 0 implies x = 0. For a proof that a tight extension of a nondegenerate Jordan algebra is itself nondegenerate, see [3, (2.9)].

A nonunital Jordan algebra is just a subspace J = (J, U, 2) of some unital Jordan algebra closed under the products U_{x} y and the square x^{2} := U_{x} 1. In this case, J imbeds in the free unital hull Ĵ := Φ 1 ⊕ J, and the operators U_{x}, V_{x} extend naturally to operators on Ĵ (U_{x} 1 = x^{2} and V_{x} 1 = 2x). In Ĵ, the U operator is defined as

U_{α1⊕x}(β1 ⊕ y) := (α^{2} Id + α V_{x} + U_{x})(β1 ⊕ y)
:= α^{2}β 1 ⊕ (α^{2} y + 2αβ x + α x ∘ y + β x^{2} + U_{x} y) .

Note that the product x∘y is simply the linearization of the square, also denoted by

V_{x} y = V_{y} x = x ∘ y := (x + y)^{2} − x^{2} − y^{2} (= U_{x, y} 1 in Ĵ) .

For an arbitrary Jordan algebra J, the unital hull of J is simply defined as

Ĵ := \{ \begin{matrix}J & if J is unital \\ Φ 1 ⊕ J & if J is nonunital \end{matrix} .

The ring Φ is said to act faithfully on J if for ϕ ∈ Φ, ϕ J = 0 implies ϕ = 0. Every Jordan algebra over Φ is automatically an algebra over Φ′ = Φ / J^{⊥} (where J^{⊥} := {ϕ ∈ Φ : ϕ J = 0} is the set of scalars acting trivially on J), and clearly Φ′ acts faithfully on J. Thus there is no loss of generality in assuming that Φ is faithful. Note that if Φ acts faithfully on J it acts even more faithfully on any larger J̃ ⊃ J: J̃^{⊥} ⊂ J^{⊥} = 0.

If \frac{1}{2} ∈ Φ, a quadratic Jordan algebra is actually a linear Jordan algebra with product x · y := \frac{1}{2}(x ∘ y) satisfying

x · y = y · x and (x^{2} · y) · x = x^{2} · (y · x) .

In a linear Jordan algebra J, the center is easily understood as the elements of the algebra that behave like scalars. Hence, the center is simply the nucleus

C(J) = Nuc (J) := {c ∈ J : [c, x, y] = [x, c, y] = [x, y, c] = 0 ∀ x, y ∈ J}

where [x, y, z] := (x · y) · z − x · (y · z).

For a quadratic Jordan algebra, the center must be defined in terms of the centroid Γ(J), the set of all Φ−linear transformations T : J → J that behave like scalar multiplication:

\begin{matrix}(1) & T U_{x} = U_{x} T & (3) & U_{T (x)} = T^{2} U_{x} \\ (2) & T V_{x} = V_{x} T & (4) & T (x)^{2} = T^{2}(x^{2}) \end{matrix}

for all x ∈ J. Since U_{x, y} = U_{x+y} − U_{x} − U_{y}, it follows directly that T U_{x, y} = U_{x, y} T for all x, y ∈ J . Hence, in unital algebras, (2) and (4) are redundant because V_{x} = U_{x, 1} and y^{2} = U_{y} 1. Also note that T V_{x, y} = V_{x, y} T for all x, y ∈ J by the well known identity V_{x, y} = V_{x} V_{y} − U_{x, y}. Thus, we also have

T ({x, y, z}) = {T (x), y, z} = {x, T (y), z} = {x, y, T (z)} .

Note that the scalar multiples of the identity map lie in the centroid. The map Φ → Φ Id has kernel J^{⊥}, and so passage from Φ to the faithful action of Φ′ is equivalent to identifying the scalars Φ with the scalar multiplications Φ Id ∈ Γ(J).

We can’t really expect a center to behave very well if the centroid is not well behaved. For γ_{1}, γ_{2} ∈ Γ(J), it is immediately clear that [(γ_{1} + γ_{2}), U_{x}] = [(γ_{1} + γ_{2}), V_{x}] = 0 and [(γ_{1} γ_{2}), U_{x}] = [(γ_{1} γ_{2}), V_{x}] = 0 for all x ∈ J. Thus Γ(J) is a ring if and only if U_{(γ_{1}+γ_{2})(x)} = (γ_{1} + γ_{2})^{2} U_{x} and U_{(γ_{1} γ_{2})(x)} = (γ_{1} γ_{2})^{2} U_{x} for all γ_{1}, γ_{2} ∈ Γ(J) and all x ∈ J. Since

U_{(γ_{1} + γ_{2})(x)} = U_{γ_{1}(x)} + U_{γ_{1}(x), γ_{2}(x)} + U_{γ_{2}(x)} = (γ_{1}^{2} + 2 γ_{1} γ_{2} + γ_{2}^{2}) U_{x}

and (γ_{1} + γ_{2})^{2} U_{x} = (γ_{1}^{2} + γ_{1} γ_{2} + γ_{2} γ_{1} + γ_{2}^{2}) U_{x}, U_{(γ_{1}+γ_{2})(x)} = (γ_{1} + γ_{2})^{2} U_{x} if and only if [γ_{1}, γ_{2}] U_{x} = 0. Since U_{(γ_{1} γ_{2})(x)} = γ_{1}^{2} γ_{2}^{2} U_{x} and (γ_{1} γ_{2})^{2} U_{x} = (γ_{1} γ_{2} γ_{1} γ_{2}) U_{x}, U_{(γ_{1} γ_{2})(x)} = (γ_{1} γ_{2})^{2} U_{x} if and only if (γ_{1} [γ_{1}, γ_{2}] γ_{2}) U_{x} = 0. Now notice that (γ_{1} [γ_{1}, γ_{2}] γ_{2}) U_{x} = γ_{1} ( [γ_{1}, γ_{2}] U_{x}) γ_{2}, and so [γ_{1}, γ_{2}] U_{x} = 0 implies that (γ_{1} [γ_{1}, γ_{2}] γ_{2}) U_{x} = 0. Hence, Γ(J) is a ring if and only if

[γ_{1}, γ_{2}] U_{x} = 0 for all γ_{1}, γ_{2} ∈ Γ(J) and all x ∈ J .

We will only consider algebras whose centroids satisfy this property. Once the centroid is a ring, J is an algebra over Γ(J), and so we can replace Φ with Γ(J) and assume from the start that the ring of scalars Φ is actually Γ(J). In [4], it is proven that Γ(J) is commutative if J contains no extreme elements, i.e., the extreme radical

ExtRad(J) := {x ∈ J : U_{x} y = x ∘ y = {x, y, z} = 0 ∀ y, z ∈ J}

vanishes.

Now we are ready to propose a definition of the weak center of a Jordan algebra:

C_{w}(J) := {c ∈ J : U_{c}, V_{c} ∈ Γ(J)} .

For an arbitrary unital Jordan algebra J, it is clear that C_{w}(J) contains Γ(J) 1 (since V_{T (1)} = 2 T, U_{T (1)} = T^{2} ∈ Γ(J)), but rarely can we expect equality. Hence, we cannot assume that the central elements of J behave like scalar multiples of 1 in all cases, but this observation leads us to our primary guiding “meta-principle”: an element c in a quadratic Jordan algebra J over a ring Φ is considered to be a central element if there is an algebra J̃ containing J over a ring Φ̃ containing Φ such that c ∈ Φ̃ 1. We formalize this notion in the definition of the scalar center for J:

C_{s}(J) := {c ∈ J : c ∈ Φ̃ 1 for some J̃ ⊃ J over Φ̃ ⊃ Φ}
= ⋃ {J ∩ Φ̃ 1 : J ⊂ J̃ unital over Φ̃ ⊃ Φ} .

The key point is the following: if c is a scalar multiple of 1 in the larger algebra, then c will behave like a scalar multiple of 1 in the original algebra.

First, let us consider the center in the context of special algebras. Remember that a Jordan algebra is special if it is a subalgebra of A^{+}, the Jordan algebra induced by the associative algebra A. In terms of the associative product on A, the Jordan products on A^{+} are defined as follows: U_{x} y = xyx, x^{2} = xx, {x, y, z} = xyz + zyx and x ∘ y = xy + yx. Notice that C(A) ⊂ C_{w}(A^{+}), and if A is semiprime and contains a unit, the converse is also true. (An associative algebra A is semiprime if the intersection of all prime ideals is zero; semiprimeness is equivalent to the property that zero is the only nilpotent ideal.) We can, in fact, prove a slightly stronger result.

Lemma 1 Let A be a unital, semiprime, associative algebra over a ring Φ, and let c ∈ A^{+} with [V_{c}, V_{x}] y = 0 and [U_{c}, V_{x}] y = 0 for all x, y in a generating set G for A. Then c ∈ C(A).

Proof Note that V_{a} = l_{a} + r_{a} and [l_{A}, r_{A}] = 0 together imply

[V_{a}, V_{b}] = [l_{a}, l_{b}] + [r_{a}, r_{b}] = l_{[a, b]} + r_{[b, a]} = ad [a, b] .

Since [V_{c}, V_{x}] y = 0 for all x, y ∈ G, we have (ad [c, x]) y = 0 for all x, y ∈ G, and so [c, x] commutes with all elements of G. Thus [c, x] commutes with all of A, and [c, x] ∈ C(A) for all x ∈ G. Now look at [U_{c}, V_{x}] y = 0:

U_{c} V_{x} y − V_{x} U_{c} y = c x y c + c y x c − (x c y c + c y c x)
= (c x − x c) y c − c y (c x − x c)
= [c, x] y c − c y [c, x] .

From the above result, we know that [c, x] ∈ C(A) for all x ∈ G, and so for all y ∈ A, 0 = [c, x] y c − [c, x] c y = [c, x] (y c − c y) = [c, x] [y, c]. In particular, we have [c, x]^{2} = 0 for all x ∈ G. Write z = [c, x], and consider the principal ideal (z). Since z ∈ C(A), (z) = A z, and (z)^{2} = A z A z = A^{2} z^{2} = A^{2} 0 = 0. Thus (z) is a nilpotent ideal of A, thereby implying that (z) = 0 because A is semiprime. Since z = 1 z ∈ (z) = {0}, z = [c, x] = 0 for all x in the generating set G. Since c commutes with the generating set G, c commutes with all of A, and c ∈ C(A). ■

Thus C(A) = C_{w}(A^{+}) when A is unital and semiprime, and so in this case the central elements for the special Jordan algebra behave like scalar multiples of 1. We conclude that our notion of center works particularly well for a special Jordan algebra derived from a semiprime associative algebra with a unit. Since we can think of A as an associative algebra over its center C(A), the center of A^{+} is actually equal to all the scalar multiples of 1. Now the question that arises is why our notion of center works so well in this particular case. An associative algebra A is semiprime if and only if the quadratic Jordan algebra A^{+} is nondegenerate. This leads us to conjecture that nondegeneracy (in the Jordan sense) will insure a well behaved center. In the next section, we will show that this conjecture is correct.

2 The Nondegenerate Case

Theorem 1 Let J be a unital, nondegenerate Jordan algebra over a faithful ring of scalars Φ. Then there exists a Jordan algebra J̃ containing J over a ring Ω̃ containing Φ such that ˜1 = 1 and C_{w}(J) ⊂ Ω̃ 1. In particular, C_{w}(J) ⊂ C_{s}(J) is a Jordan subalgebra of J, closed under addition and Jordan multiplication, and there is a fixed extension J̃ ⊃ J such that C_{w}(J) = J ∩ Ω̃ 1.

Proof Since J is nondegenerate, it is an algebra over a faithful ring Γ(J), and since Φ is faithful, Φ ≅ Φ Id ⊂ Γ(J). Thus we can assume that the ring of scalars is Φ = Γ(J). For each c ∈ C_{w}(J), U_{c} = μ_{c} Id and V_{c} = ν_{c} Id for some μ_{c}, ν_{c} ∈ Φ. Define

Ω = Φ [{ω_{c} : c ∈ C_{w}(J)}] / ⟨{μ_{c} − ω_{c}^{2}, ν_{c} − 2 ω_{c}}⟩

where ⟨S⟩ denotes the ideal generated by S. Now choose an ideal M of Ω ⊗_{Φ} J which is maximal with respect to M ∩ J = 0. Then J̃ = (Ω ⊗_{Φ} J) / M is a tight extension of J over a faithful ring of scalars Ω̃ = Ω / J̃^{⊥}, and J ⊂ J̃ is imbedded faithfully. To see that Φ ⊂ Ω̃ faithfully, consider ω ∈ J̃^{⊥}. Then ωJ̃ = 0 implies that ωJ = 0, and so ω ∉ Φ because Φ acts faithfully on J. Hence, Φ ∩ J̃^{⊥} = 0, and Φ is imbedded faithfully in Ω̃.

Since J is nondegenerate and J̃ is a tight extension, J̃ is also nondegenerate. Now consider the element

z_{c} = ω_{c} ⊗ 1 − 1 ⊗ c ∈ Ω ⊗_{Φ} J

and note that

U_{z_{c}} = U_{ω_{c}⊗1 − 1⊗c} = ω_{c}^{2} ⊗ U_{1} + 1 ⊗ U_{c} − ω_{c} ⊗ U_{1, c}
= ω_{c}^{2} ⊗ Id + 1 ⊗ μ_{c} Id − ω_{c} ⊗ V_{c}
= ω_{c}^{2} ⊗ Id + μ_{c} ⊗ Id − ω_{c} ⊗ ν_{c} Id = (ω_{c}^{2} + μ_{c} − ν_{c} ω_{c}) ⊗ Id .

In Ω, ω_{c}^{2} = μ_{c} and 2 ω_{c} = ν_{c}, which implies that

U_{z_{c}} = (ω_{c}^{2} + ω_{c}^{2} − 2 ω_{c}^{2}) ⊗ Id = 0 ⊗ Id = 0 .

Since U_{z_{c}} = 0 in Ω ⊗_{Φ} J, we must have \overline{U}_{\overline{z}_{c}} = ¯0 in J̃ = (Ω ⊗_{Φ} J) / M. Furthermore, nondegeneracy of J̃ implies that \overline{z}_{c} = ¯0 in J̃. Thus \overline{ω_{c} ⊗ 1} = \overline{1 ⊗ c}, and C_{w}(J) ⊂ Ω̃ 1. ■

Note that if c = ϕ 1 is already a scalar multiple of 1 in J, the new scalar ω_{c} collapses to the old scalar ϕ under the formation of Ω̃ = Ω / J̃^{⊥} (ϵ = ω_{c} − ϕ ∈ Ω lies in J̃^{⊥}: ϵJ̃ vanishes because it is a trivial ideal in the nondegenerate J̃ since ϵ^{2} = ω_{c}^{2} − 2ω_{c}ϕ + ϕ^{2} = 0 by μ_{c} = ϕ^{2}, ν_{c} = 2ϕ).

The theorem for nonunital, nondegenerate Jordan algebras will follow as a corollary, but first we need to prove some essential lemmas.

Lemma 2 Let J be a Jordan algebra with c ∈ J satisfying [V_{c}, V_{x}] = 0 for all x ∈ J, where [V_{c}, V_{x}] = V_{c} V_{x} − V_{x} V_{c}. Then

{c, x, y} = {x, c, y} = {x, y, c} ∀ x, y ∈ J .

In terms of operators, this result can be written as

V_{c, x} = V_{x, c} = U_{x, c} ∀ x ∈ J .

Proof Use the identity V_{x, y} = V_{x} V_{y} − U_{x, y} to get

V_{x, c} = V_{x} V_{c} − U_{x, c} = V_{c} V_{x} − U_{c, x} = V_{c, x} ,

which implies that {x, c, y} = {c, x, y} for all x, y ∈ J. Furthermore, {x, c, y} = {y, c, x} = {c, y, x} for all x, y ∈ J, and we are done. ■

Lemma 3 Let J be a Jordan algebra, and let c ∈ C_{w}(J). For n ∈ ℕ ∪ {0}, define c^{n} inductively by c^{0} := 1 ∈ Ĵ, c^{1} := c, c^{n+2} := U_{c} c^{n}. Then c^{n} ∈ C_{w}(J) for all n ∈ ℕ ∪ {0}.

Proof It is known [1] that U_{x^{n}}, V_{x^{n}} ∈ Φ[U_{x}, V_{x}] for all n ∈ ℕ ∪ {0}. Let c ∈ C_{w}(J), then clearly U_{c^{n}}, V_{c^{n}} ∈ Φ[U_{c}, V_{c}] ⊂ Γ(J), and so c^{n} ∈ C_{w}(J). ■

Lemma 4 Let J be a Jordan algebra, with c ∈ J satisfying [V_{c}, U_{x}] = 0 for all x ∈ J. Then

U_{U_{x_{1}} ··· U_{xn} c, U_{x_{1}} ··· U_{x_{n−1}} x_{n}^{2}} = V_{c} U_{U_{x_{1}} ··· U_{x_{n−1}} x_{n}^{2}} ∀ x_{i} ∈ J, n ∈ ℕ .

Proof In the unital hull Ĵ of J, x^{2} = U_{x} 1. Linearizing the fundamental formula gives U_{U_{x}y, U_{x}z} = U_{x} U_{y, z} U_{x}, which implies that

U_{U_{x_{1}} ··· U_{xn} c, U_{x_{1}} ··· U_{x_{n−1}} x_{n}^{2}} = U_{U_{x_{1}} (U_{x_{2}} ··· U_{xn} c), U_{x_{1}} (U_{x_{2}} ··· U_{xn} 1)}

= U_{x_{1}}U_{U_{x_{2}} ··· U_{xn} c, U_{x_{2}} ··· U_{xn} 1} U_{x_{1}} = U_{x_{1}} ⋯ U_{x_{n}}U_{c, 1} U_{x_{n}} ⋯ U_{x_{1}}
= U_{x_{1}} ⋯ U_{x_{n}}V_{c} U_{x_{n}} ⋯ U_{x_{1}} = V_{c} U_{x_{1}} ⋯ U_{x_{n}}U_{1} U_{x_{n}} ⋯ U_{x_{1}}
= V_{c} U_{U_{x_{1}} ··· U_{xn} 1} = V_{c} U_{U_{x_{1}} ··· U_{x_{n−1}} x_{n}^{2}} . ■

Lemma 5 Let J be a nondegenerate Jordan algebra, and let c ∈ C_{w}(J). Then

U_{c} x^{2} = U_{x} c^{2} ∀ x ∈ J .

Proof Note that this result is trivial for unital Jordan algebras. Let x ∈ J and define z = U_{c} x^{2} − U_{x} c^{2}, then we have the following:

U_{z} = U_{U_{c}x^{2} − U_{x}c^{2}} = U_{U_{c}x^{2}} + U_{U_{x}c^{2}} − U_{U_{c}x^{2}, U_{x}c^{2}}
= U_{c}^{2} U_{x^{2}} + U_{x} U_{c^{2}} U_{x} − U_{c} U_{x^{2}, U_{x}c^{2}} .

By Lemma 3, c^{2} ∈ C_{w}(J), and by Lemma 4, U_{x^{2}, U_{x}c^{2}} = V_{c^{2}} U_{x^{2}}, while V_{c^{2}} = V_{c, c} = U_{c, c} = 2 U_{c} by Lemma 2. These results imply that

U_{z} = U_{c^{2}} U_{x^{2}} + U_{c^{2}} U_{x}^{2} − U_{c} (2 U_{c}) U_{x^{2}}
= 2 U_{c^{2}} U_{x^{2}} − 2 U_{c^{2}} U_{x^{2}} = 0 .

Since J is nondegenerate, z = 0, and so U_{c} x^{2} = U_{x} c^{2}. ■

Proposition 1 Let J be a nonunital Jordan algebra with T ∈ Γ(J) and t ∈ J, and define an extension T̂ : Ĵ → Ĵ of T by T̂(ϕ1 ⊕ x) := ϕ t + T (x). Then T̂ ∈ Γ(Ĵ) if and only if the following are true for all x ∈ J:

(i) V_{t} = 2 T (ii) U_{t} = T^{2} (iii) T (x^{2}) = U_{x} t (iv) T (t) = t^{2}

Furthermore, (i) and (ii) are sufficient when J is nondegenerate, and this completely characterizes {T̂ ∈ Γ(Ĵ) : T̂(Ĵ) ⊂ J}.

Proof First suppose that T̂ ∈ Γ(Ĵ), then as transformations on J we have (i) V_{t} = V_{T̂(1)} = T V_{1} = 2 T, (ii) U_{t} = U_{T̂(1)} = T^{2} U_{1} = T^{2}, (iii) T (x^{2}) = T Û_{x} 1 = U_{x} T̂(1) = U_{x} t, and (iv) t^{2} = T̂(1)^{2} = T̂ ^{2}(1^{2}) = T̂(t) = T (t). Now assume that (i)-(iv) hold, then in order to show that T̂ ∈ Γ(Ĵ), it is sufficient to show that T̂ U_{ϕ1⊕x}(ẑ) = U_{ϕ1⊕x} T̂(ẑ) and U_{T̂(ϕ1⊕x)}(ẑ) = T̂^{2} U_{ϕ1⊕x}(ẑ) for all ϕ ∈ Φ, x ∈ J, ẑ ∈ Ĵ. By linearity, if suffices to check ẑ = 1 and ẑ = y ∈ J. Applying (i) and (iii) gives

T̂ U_{ϕ1⊕x} 1 = T̂(ϕ1 ⊕ x)^{2} = T̂(ϕ^{2}1 ⊕ (2 ϕ x + x^{2}))
= ϕ^{2} T̂(1) + ϕ (2 T)(x) + T (x^{2})
= ϕ^{2} t + ϕ (t ∘ x) + U_{x} t
= (ϕ^{2} Id + ϕ V_{x} + U_{x}) t = U_{ϕ1⊕x} T̂(1) ,

while T ∈ Γ(J) implies that

T̂ U_{ϕ1⊕x} y = T (ϕ^{2} Id + ϕ V_{x} + U_{x}) y
= (ϕ^{2} Id + ϕ V_{x} + U_{x}) T (y) = U_{ϕ1⊕x} T̂(y) .

Hence, we conclude that T̂ U_{ϕ1⊕x} = U_{ϕ1⊕x} T̂ on Ĵ.

Combining (i), (ii) and (iv) provides the following:

U_{T̂(ϕ1⊕x)} 1 = (ϕ t + T (x))^{2} = ϕ^{2} t^{2} + ϕ V_{t} T (x) + T (x)^{2}
= ϕ^{2} T (t) + ϕ (2 T) T (x) + T^{2}(x^{2})
= T̂^{2} (ϕ^{2} Id + ϕ V_{x} + U_{x}) 1 = T̂^{2} U_{ϕ1⊕x} 1 ,

while (ii) and T ∈ Γ(J) insure that

U_{T̂(ϕ1⊕x)} y = U_{ϕt + T (x)} y = (U_{ϕt} + U_{ϕt, T (x)} + U_{T (x)}) y
= ϕ^{2} U_{t} y + ϕ {t, y, T (x)} + U_{T (x)} y
= ϕ^{2} T^{2}(y) + ϕ T {t, y, x} + T^{2} U_{x} y .

By (i), [V_{t}, V_{x}] = [2 T, V_{x}] = 0 for all x ∈ J, and so Lemma 2 implies that {t, y, x} = {x, t, y} = U_{x, y} t = T (x ∘ y) (by linearizing (iii)). Thus {t, y, x} = T V_{x} y, which gives the following:

U_{T̂(ϕ1⊕x)} y = ϕ^{2} T^{2}(y) + ϕ T {t, y, x} + T^{2} U_{x} y
= T^{2}(ϕ^{2} Id + ϕ V_{x} + U_{x}) y = T̂^{2} U_{ϕ1⊕x} y .

Combining the above results gives U_{T̂(ϕ1⊕x)} = T̂^{2} U_{ϕ1⊕x} on Ĵ and T̂ ∈ Γ(Ĵ).

To see that (iii) and (iv) are redundant when J is nondegenerate, first consider

U_{T (x^{2}) − U_{x}t} = U_{T (x^{2})} + U_{U_{x}t} − U_{T (x^{2}), U_{x}t}
= T^{2} U_{x^{2}} + U_{x} U_{t} U_{x} − T U_{x^{2}, U_{x}t}
= T^{2} U_{x^{2}} + U_{x} T^{2} U_{x} − T U_{x^{2}, U_{x}t}
= 2 T^{2} U_{x^{2}} − T U_{x^{2}, U_{x}t} .

Since [V_{t}, U_{x}] = [2 T, U_{x}] = 0 for all x ∈ J, we can apply Lemma 4 to get U_{x^{2}, U_{x}t} = V_{t} U_{x^{2}} = 2 T U_{x^{2}}. Thus we have

U_{T (x^{2}) − U_{x}t} = 2 T^{2} U_{x^{2}} − T U_{x^{2}, U_{x}t} = 2 T^{2} U_{x^{2}} − 2 T^{2} U_{x^{2}} = 0 ,

and so nondegeneracy of J implies that T (x^{2}) = U_{x} t. For (iv), T (t) = t^{2} follows similarly from the computation

\begin{aligned}U_{T (t)−t^{2}} & = & U_{T (t)} − U_{T (t), t^{2}} + U_{t^{2}} = T^{2} U_{t} − T U_{t, t^{2}} + U_{t}^{2} \\ & = & T^{2} T^{2} − T U_{t} V_{t} + (T^{2})^{2} = 2 T^{4} − T (T^{2}) (2 T) = 0 . \end{aligned}

Once again, nondegeneracy of J implies that T (t) = t^{2}. ■

Corollary 1 Let J be a nondegenerate, nonunital Jordan algebra, and let Ĵ = Φ 1 ⊕ J, the free unital hull of J. Then C_{w}(J) ⊂ C_{w}(Ĵ).

Proof Let c ∈ C_{w}(J), then t = Û_{c} 1 = c^{2} satisfies (i) V_{t} = V_{c^{2}} = V_{c, c} = U_{c, c} = 2 U_{c}, and (ii) U_{t} = U_{c^{2}} = U_{c}^{2}. Also, s = V_{c} 1 = 2c satisfies (i) V_{s} = V_{2c} = 2 V_{c}, and (ii) U_{s} = U_{2c} = 4 U_{c} = 2 U_{c, c} = U_{c, c} + V_{c, c} = V_{c}^{2}. Hence, U_{c}, V_{c} ∈ Γ(J) by Lemma 1, and c ∈ C_{w}(Ĵ). ■

Remark: it is easy to see that centroidal T = U_{c}, V_{c} always satisfy (iv): (c^{2})^{2} = U_{c} c^{2} and (2c)^{2} = V_{c} (2c), V_{c} always satisfies (iii): V_{c} x^{2} = {c, x, x} = {x, c, x} = 2 U_{x} c = U_{x} (2c), but U_{c} satisfies (iii) only if U_{c} x^{2} = U_{x} c^{2}. Thus this is a natural additional condition to require of a central element, which is stronger than just U_{c}, V_{c} ∈ Γ(J) alone.

The stage is set; we are ready to prove the theorem for nonunital, nondegenerate Jordan algebras.

Theorem 2 Let J be a nonunital, nondegenerate Jordan algebra over a faithful ring Φ. Then there exists a unital Jordan algebra J̃ containing J over a ring Ω̃ containing Φ such that C_{w}(J) ⊂ Ω̃ 1.

Proof By Corollary 1, C_{w}(J) ⊂ C_{w}(Ĵ), where Ĵ = Φ 1 ⊕ J, the free unital hull of J. Choose an ideal M of Ĵ maximal with respect to M ∩ J = 0. Then J̄ = Ĵ/ M is a tight unital extension of J, and therefore nondegenerate. Since C_{w}(J) ⊂ C_{w}(Ĵ), we must also have C_{w}(J) ⊂ C_{w}(J̄). We observed in Section 1 that since Φ acts faithfully on J, it also acts faithfully on J̄. Now apply Theorem 1 to J̄: there is a unital Jordan algebra J̃ containing J̄ (with ˜1 = 1) over a ring Ω̃ containing Φ such that C_{w}(J) ⊂ C_{w}(J̄) ⊂ Ω̃ 1. ■

Now we truly know how central elements behave in a nondegenerate Jordan algebra. First notice that this result gives rise to an obvious but very useful corollary.

Corollary 2 Let J be a nondegenerate Jordan algebra with χ(Φ) = 2 (i.e., 2J = 0), and let c ∈ C_{w}(J); then V_{c} = 0.

Furthermore, identities which were difficult to prove are now trivial. For example,

U_{{x, c, y}} = U_{c} U_{x∘y}
U_{U_{x}c, U_{y}c} = U_{c} U_{x^{2}, y^{2}}
U_{c, x} U_{y_{1}} ⋯ U_{y_{n}} U_{d, z} = U_{d, x} U_{y_{1}} ⋯ U_{y_{n}} U_{c, z}

for all x, y, z ∈ J and c, d ∈ C_{w}(J). In short, the center of a nondegenerate Jordan algebra behaves exactly as we hoped it would. The center of a degenerate Jordan algebra is still not well understood, however. Clearly, the most significant results derived above depend strongly on nondegeneracy. Even though a few of the preliminary results are true for all Jordan algebras, they are very specific. In the following section, we will see that there is little hope for an analogous theorem for degenerate algebras.

3 A Degenerate Counterexample

We will introduce a degenerate Jordan algebra for which the central elements (as defined above) don’t behave like scalar multiples of 1 for any extension of the original algebra. Begin with a set of four generators G = {x, y, c, d }. (In order to avoid confusion with triple products, brackets for sets will be boldfaced.) Then construct ⟨G⟩, the free nonunital quadratic algebra over a unital commutative ring Φ with χ(Φ) = 2. ⟨G⟩ is simply the set of all linear combinations over Φ of all possible squares, circle products, triple products, and quadratic U products built out of x, y, c and d. For example, ⟨G⟩ contains x^{2}, x∘y, {x, y, c}, U_{x} y, etc. First we will consider the free algebra ⟨G⟩ modulo the following relations:

(i) r^{2} = 0 for all r ∈ ⟨G⟩,

(ii) r ∘ s = 0 for all r, s ∈ ⟨G⟩,

(iii) {r, s, t} = {r, t, s} = {s, r, t} = {t, s, r} for all r, s, t ∈ ⟨G⟩, (iv) U_{c} = U_{d} = 0,

(v) a monomial r = 0 if deg_{g}(r) > 2 for any g ∈ {x, y, c} or deg_{d}(r) > 1.

Note that (ii) and (iv) together imply that c, d ∈ Γ(J) because U_{c}, V_{c} = U_{d}, V_{d} = 0. Another result of (i) and (ii) is that there are no degree 2 monomials, and induction will reveal that there are also no monomials of degree 2n for all n ∈ ℕ. We already see that the statement is true for n = 1. Now suppose it is true for 1 ≤ k < n. If deg(U_{r} s) = 2n for monomials r and s, then deg(s) = 2k for some 1 ≤ k < n, implying that s = 0 and U_{r} s = 0. If deg({r, s, t}) = 2n for monomials r, s and t, then at least one term must have even degree 2k with 1 ≤ k < n, and so {r, s, t} = 0. Hence, (i), (ii) and (v) together imply that we are left only with elements of degree 1, 3, 5 and 7.

Within this framework, we will construct a Jordan algebra. Below is a list of 25 monomials and their multiplication table.

b_{1 1} = x b_{3 1} = U_{x} y b_{3 4} = {x, c, d} b_{5 1} = U_{x} U_{y} c b_{5 3} = {U_{x} y, c, d}
b_{1 1}^{∗} = y b_{3 1}^{∗} = U_{y} x b_{3 4}^{∗} = {y, c, d} b_{5 1}^{∗} = U_{y} U_{x} c b_{5 3}^{∗} = {U_{y} x, c, d}
h_{1 c} = c b_{3 2} = U_{x} c h_{3 c} = {x, y, c} b_{5 2} = U_{x} U_{y} d b_{5 4} = {U_{x} c, y, d}
h_{1 d} = d b_{3 2}^{∗} = U_{y} c h_{3 d} = {x, y, d} b_{5 2}^{∗} = U_{y} U_{x} d b_{5 4}^{∗} = {U_{y} c, x, d}
b_{3 3} = U_{x} d  b_{5 5} = {U_{x} d, y, c}
b_{3 3}^{∗} = U_{y} d  b_{5 5}^{∗} = {U_{y} d, x, c}
h_{7} = U_{{x, y, c}} d

We have the following general rules for all a, b, z ∈ B = { b_{i j}, b_{i j}^{∗}, h_{k l}, h_{7}}.

b^{2} = 0  a ∘ b = 0  {a, b, b} = 0
{a, b, z} = {a, z, b} = {b, a, z} = {z, b, a}

All nonzero products of specific monomials are listed below.

\begin{matrix}U_{b_{1 1}} b_{1 1}^{∗} = b_{3 1} & U_{b_{1 1}^{∗}} b_{1 1} = b_{3 1}^{∗} \\ U_{b_{1 1}} h_{1 c} = b_{3 2} & U_{b_{1 1}^{∗}} h_{1 c} = b_{3 2}^{∗} \\ U_{b_{1 1}} h_{1 d} = b_{3 3} & U_{b_{1 1}^{∗}} h_{1 d} = b_{3 3}^{∗} \\ U_{b_{1 1}} b_{3 2}^{∗} = b_{5 1} & U_{b_{1 1}^{∗}} b_{3 2} = b_{5 1}^{∗} \\ U_{b_{1 1}} b_{3 3}^{∗} = b_{5 2} & U_{b_{1 1}^{∗}} b_{3 3} = b_{5 2}^{∗} \\ U_{b_{1 1}} b_{3 4}^{∗} = b_{5 3} + b_{5 4} + b_{5 5} & U_{b_{1 1}^{∗}} b_{3 4} = b_{5 3}^{∗} + b_{5 4}^{∗} + b_{5 5}^{∗} \end{matrix}

U_{h_{3 c}} h_{1 d} = h_{7}

\begin{matrix}\{b_{1 1}, b_{1 1}^{∗}, h_{1 c}\} = h_{3 c} & \{b_{1 1}, b_{1 1}^{∗}, h_{1 d}\} = h_{3 d} \\ \{b_{1 1}, b_{1 1}^{∗}, h_{3 c}\} = b_{5 1} + b_{5 1}^{∗} & \{b_{1 1}, b_{1 1}^{∗}, h_{3 d}\} = b_{5 2} + b_{5 2}^{∗} \\ \{b_{1 1}, b_{1 1}^{∗}, b_{3 4}\} = b_{5 4} + b_{5 5} & \{b_{1 1}, b_{1 1}^{∗}, b_{3 4}^{∗}\} = b_{5 4}^{∗} + b_{5 5}^{∗} \\ \{b_{1 1}, h_{1 c}, h_{1 d}\} = b_{3 4} & \{b_{1 1}^{∗}, h_{1 c}, h_{1 d}\} = b_{3 4}^{∗} \\ \{b_{1 1}, h_{1 c}, b_{3 1}^{∗}\} = b_{5 1} + b_{5 1}^{∗} & \{b_{1 1}^{∗}, h_{1 c}, b_{3 1}\} = b_{5 1} + b_{5 1}^{∗} \\ \{b_{1 1}, h_{1 c}, b_{3 3}^{∗}\} = b_{5 5}^{∗} & \{b_{1 1}^{∗}, h_{1 c}, b_{3 3}\} = b_{5 5} \end{matrix}

\begin{matrix}\{b_{1 1}, h_{1 c}, h_{3 d}\} = b_{5 3} + b_{5 5} & \{b_{1 1}^{∗}, h_{1 c}, h_{3 d}\} = b_{5 3}^{∗} + b_{5 5}^{∗} \\ \{b_{1 1}, h_{1 c}, b_{5 4}^{∗}\} = h_{7} & \{b_{1 1}^{∗}, h_{1 c}, b_{5 4}\} = h_{7} \\ \{b_{1 1}, h_{1 d}, b_{3 1}^{∗}\} = b_{5 2} + b_{5 2}^{∗} & \{b_{1 1}^{∗}, h_{1 d}, b_{3 1}\} = b_{5 2} + b_{5 2}^{∗} \\ \{b_{1 1}, h_{1 d}, b_{3 2}^{∗}\} = b_{5 4}^{∗} & \{b_{1 1}^{∗}, h_{1 d}, b_{3 2}\} = b_{5 4} \\ \{b_{1 1}, h_{1 d}, h_{3 c}\} = b_{5 3} + b_{5 4} & \{b_{1 1}^{∗}, h_{1 d}, h_{3 c}\} = b_{5 3}^{∗} + b_{5 4}^{∗} \\ \{b_{1 1}, b_{3 2}^{∗}, b_{3 4}\} = h_{7} & \{b_{1 1}^{∗}, b_{3 2}, b_{3 4}^{∗}\} = h_{7} \\ \{b_{1 1}, b_{3 4}^{∗}, h_{3 c}\} = h_{7} & \{b_{1 1}^{∗}, b_{3 4}, h_{3 c}\} = h_{7} \\ \{h_{1 c}, h_{1 d}, b_{3 1}\} = b_{5 3} & \{h_{1 c}, h_{1 d}, b_{3 1}^{∗}\} = b_{5 3}^{∗} \\ \{h_{1 c}, h_{1 d}, b_{5 1}\} = h_{7} & \{h_{1 c}, h_{1 d}, b_{5 1}^{∗}\} = h_{7} \end{matrix}

{h_{1 d}, b_{3 2}, b_{3 2}^{∗}} = h_{7}

Our degenerate Jordan algebra will be J = span (B). Notice that the map on the generators defined by x^{∗} = y, y^{∗} = x, c^{∗} = c and d^{∗} = d extends naturally to an involution on J. For the spanning elements we have (b_{ij})^{∗} = b_{ij}^{∗}, (b_{ij}^{∗})^{∗} = b_{ij}, (h_{kl})^{∗} = h_{kl} and (h_{7})^{∗} = h_{7}. From the multiplication table, it is evident that (U_{a} b)^{∗} = U_{a^{∗}} b^{∗} and {a, b, z}^{∗} = {a^{∗}, b^{∗}, z^{∗}} for all a, b, z ∈ B. This involution will be useful because once we know that some expression M(x, y, c, d) = 0, we can immediately conclude that M(y, x, c, d) = M(x, y, c, d)^{∗} = 0.

To see that B spans a Jordan algebra, the axioms for a quadratic Jordan algebra must be verified. In [5], it is shown that a quadratic algebra is a Jordan algebra if and only if the following identities hold for all scalar extensions:

V_{a, a} = V_{a^{2}} ,  (1)
U_{a} V_{a} = V_{a} U_{a} ,  (2)
U_{a} (a^{2}) = (a^{2})^{2} ,  (3)
(a^{3})^{2} = (a^{2})^{3} ,  (4)
U_{a^{2}} = U_{a}^{2} ,  (5)
U_{a^{3}} = U_{a}^{3} .  (6)

Since r^{2} = 0 and r ∘ s = 0 for all r, s ∈ J, it follows immediately that (2), (3) and (4) are true. Also, V_{b, b} = 0 for all b ∈ B implies that (1) is true. Hence, we only need to prove that U_{a}^{2} = 0 and U_{a^{3}} = 0 for all scalar extensions of J. Note that it will suffice to show that all linearizations of U_{a}^{2} = 0 and U_{a^{3}} = 0 hold for elements of B.

Lemma 6 U_{a}^{2} = 0 and its linearizations hold for all elements of B.

Proof First we will compute all of the linearizations of U_{a}^{2} = 0, then each will be verified in order. Let {a_{1}, …, a_{25}} be an ordering of the elements of the spanning set B, and consider a = ∑ λ_{i} a_{i} ∈ J_{Φ[λ_{1}, ... , λ_{25}]}. Then

U_{a}^{2} = (U_{∑ λ_{i} a_{i}})^{2} = (∑_{i} λ_{i}^{2} U_{a_{i}} + ∑_{i<j} λ_{i} λ_{j} U_{a_{i}, a_{j}} )^{2} .

We get 5 linearizations by gathering the terms associated with λ_{i}^{4}, (λ_{i}^{3} λ_{j}), (λ_{i}^{2} λ_{j}^{2}), (λ_{i}^{2} λ_{j} λ_{k}) and (λ_{i} λ_{j} λ_{k} λ_{l}) respectively, where i, j, k and l are distinct:

\begin{matrix}U_{a_{i}}^{2} = 0 , & (5) \\ [U_{a_{i}}, U_{a_{i}, a_{j}}] = 0 , & (7) \\ [U_{a_{i}}, U_{a_{j}}] + U_{a_{i}, a_{j}}^{2} = 0 , & (8) \\ [U_{a_{i}}, U_{a_{j}, a_{k}}] + [U_{a_{i}, a_{j}}, U_{a_{i}, a_{k}}] = 0 , & (9) \\ [U_{a_{i}, a_{j}}, U_{a_{k}, a_{l}}] + [U_{a_{i}, a_{k}}, U_{a_{j}, a_{l}}] + [U_{a_{i}, a_{l}}, U_{a_{j}, a_{k}}] = 0 . & (10) \end{matrix}

Proof of (5) and (7): For any generator g ∈ G = {x, y, c, d } occurring as part of the monomial a_{i}, deg_{g}(r) ≥ deg_{a_{i}}(r) for any monomial r. Hence, by relation (v) deg_{a_{i}}(r) > 2 implies that r = 0. Since deg_{a_{i}} (U_{a_{i}}^{2}) = 4 and deg_{a_{i}} ([U_{a_{i}}, U_{a_{i}, a_{j}}]) = 3, both are zero.

Proof of (8): If deg(a_{i}) ≥ 3, then deg([U_{a_{i}}, U_{a_{j}}] + U_{a_{i}, a_{j}}^{2}) ≥ 8, and

[U_{a_{i}}, U_{a_{j}}] + U_{a_{i}, a_{j}}^{2} = 0. Otherwise, deg(a_{i}) = 1 (it must be odd), and so both a_{i}, a_{j} ∈ G by the symmetry of the expression. If a_{i} = d, then the d-degree of the expression is 2, and everything is zero. Hence, by distinctness of a_{i}, a_{j} there are exactly three possibilities: {a_{i}, a_{j}} = {x, c}, {y, c} or {x, y}. Notice that if a_{j} = c, then [U_{a_{i}}, U_{c}] z + U_{a_{i}, c}^{2} z = U_{a_{i}, c}^{2} z. Once we check {a_{i}, a_{j}} = {x, c}, the case {a_{i}, a_{j}} = {y, c} will follow immediately by interchanging the roles of x and y with the involution. For {a_{i}, a_{j}} = {x, c}, the argument z must have total degree no greater than 3 with deg_{x}(z) = deg_{c}(z) = 0, deg_{y}(z) ≤ 2 and deg_{d}(z) ≤ 1. Thus we only have to check z ∈ {y, d, U_{y} d }.

{x, c, {x, c, y}} = {b_{1 1}, h_{1 c}, h_{3 c}} = 0 ,
{x, c, {x, c, d}} = {b_{1 1}, h_{1 c}, b_{3 4}} = 0 ,
{x, c, {x, c, U_{y} d}} = {b_{1 1}, h_{1 c}, {b_{1 1}, h_{1 c}, b_{3 3}^{∗}}}
= {b_{1 1}, h_{1 c}, b_{5 5}^{∗}} = 0 .

Notice that we have just proved the following:

U_{a_{i}, c}^{2} = 0 for all a_{i} ∈ B. (∗)

For {a_{i}, a_{j}} = {x, y}, deg(z) ≤ 3 with deg_{x}(z) = deg_{y}(z) = 0, deg_{c}(z) ≤ 2 and deg_{d}(z) ≤ 1. Notice that deg(z) = 3 is ruled out because U_{c} d = {c, c, d} = 0. Thus deg(z) = 1 and we only have to check z ∈{c, d }.

U_{x} U_{y} c + U_{y} U_{x} c + {x, y, {x, y, c}} = b_{5 1} + b_{5 1}^{∗} + {b_{1 1}, b_{1 1}^{∗}, h_{3 c}}
= b_{5 1} + b_{5 1}^{∗} + (b_{5 1} + b_{5 1}^{∗}) = 0 ,
U_{x} U_{y} d + U_{y} U_{x} d + {x, y, {x, y, d}} = b_{5 2} + b_{5 2}^{∗} + {b_{1 1}, b_{1 1}^{∗}, h_{3 d}}
= b_{5 2} + b_{5 2}^{∗} + (b_{5 2} + b_{5 2}^{∗}) = 0 .

Proof of (9): If deg(a_{i}) ≥ 3, then deg([U_{a_{i}}, U_{a_{j}, a_{k}}] + [U_{a_{i}, a_{j}}, U_{a_{i}, a_{k}}]) ≥ 8, and so the expression must be zero. Otherwise, a_{i} ∈ G, and since the a_{i}-degree of (9) is 2, a_{i} ≠ d. Hence, a_{i} ∈ {x, y, c}, and we only have to check a_{i} = c and a_{i} = x because y = x^{∗}.

CASE 1: (a_{i} = c) For a_{i} = c, [U_{c}, U_{a_{j}, a_{k}}]+[U_{c, a_{j}}, U_{c, a_{k}}] = [U_{c, a_{j}}, U_{c, a_{k}}], and so we only have to show that [U_{c, a_{j}}, U_{c, a_{k}}] = 0. We can, in fact, prove a slightly stronger result.

Lemma 7 {c, a, {c, b, z}} = 0 for all a, b, z ∈ B.

Proof Notice that a, b, z must be distinct because {a, a, b} = 0 and we saw that U_{c, b}^{2} = 0 for all a, b ∈ G by (∗). Also, either: (1) a, b, z are all generators or (2) exactly one has degree 3 and the other two are generators. If all three are generators, the only possibility is {a, b, z} = {x, y, d }. Thus we have three cases: a = x and {b, z} = {y, d }, a = y and {b, z} = {x, d }, a = d and {b, z} = {x, y}. Since the case with a = y is analogous to a = x, we only have to check a = x and a = d.

\begin{matrix}\{c, x, \{c, y, d\}\} = \{b_{1 1}, h_{1 c}, b_{3 4}^{∗}\} = 0 , \\ \{c, d, \{c, x, y\}\} = \{h_{1 c}, h_{1 d}, h_{3 c}\} = 0 . \end{matrix}

Otherwise, {a, b, z} consists of two generators and an element of degree 3. Degree considerations leave us with only the following cases: {a, b, z} = {x, y, {x, y, d}}, {x, d, U_{y} x} or {y, d, U_{x} y}. We only have to check the first two because {y, d, U_{x} y} = {x^{∗}, d^{∗}, U_{y^{∗}} x^{∗}}. First we will look at the three subcases for {a, b, z} = {x, y, {x, y, d}}.

{c, x, {c, y, {x, y, d}}} = {b_{1 1}, h_{1 c}, {b_{1 1}^{∗}, h_{1 c}, h_{3 d}}}
= {b_{1 1}, h_{1 c}, b_{5 3}^{∗} + b_{5 5}^{∗}} = 0 + 0 = 0 ,
{c, y, {c, x, {x, y, d}}} = {c, x, {c, y, {x, y, d}}}^{∗} = 0 ,
{c, {x, y, d}, {c, x, y}} = {h_{1 c}, h_{3 d}, h_{3 c}} = 0 .

Now consider the three subcases for {a, b, z} = {x, d, U_{y} x}.

{c, x, {c, d, U_{y} x}} = {b_{1 1}, h_{1 c}, {h_{1 c}, h_{1 d}, b_{3 1}^{∗}}}
= {b_{1 1}, h_{1 c}, b_{5 3}^{∗}} = 0 ,
{c, d, {c, x, U_{y} x}} = {h_{1 c}, h_{1 d}, {b_{1 1}, h_{1 c}, b_{3 1}^{∗}}}
= {h_{1 c}, h_{1 d}, b_{5 1} + b_{5 1}^{∗}}
= h_{7} + h_{7} = 0 ,
{c, U_{y} x, {c, x, d}} = {h_{1 c}, b_{3 1}^{∗}, b_{3 4}} = 0 . ■

Thus (9) is proven for a_{i} = c. Before we look at a_{i} = x, it will be helpful if we prove a simple lemma first.

Lemma 8 {c, U_{a} c, b} = 0 for all a, b ∈ B.

Proof Notice that we must have a ∈ {x, y} and deg(b) ≤ 3 with deg_{c}(b) = 0. If a = x, then b ∈ {y, d, U_{y} d }, and the case a = y follows directly from the symmetry between x and y.

{c, U_{x} c, y} = {b_{1 1}^{∗}, h_{1 c}, b_{3 2}} = 0 ,
{c, U_{x} c, d } = {h_{1 c}, h_{1 d}, b_{3 2}} = 0 ,
{c, U_{x} c, U_{y} d } = {h_{1 c}, b_{3 2}, b_{3 3}^{∗}} = 0 . ■

Now we can resume our proof of (9).

CASE 2: (a_{i} = x) If a_{i} = x, then a_{j}, a_{k}, z ∈ {y, c, d, U_{y} c, U_{y} d, {y, c, d }} because the a_{i}-degree of (9) is 2. Notice that (9) is symmetric in a_{j} and a_{k} with a_{j} and a_{k} distinct, and so we only have to check cases where j < k. Our choices for a_{j}, a_{k} are:

\begin{matrix}(1) & (2) & (3) & (4) & (5) & (6) & (7) & (8) \\ y, c & y, d & y, \{y, c, d \} & c, d & c, U_{y} c & c, U_{y} d & c, \{y, c, d \} & d, U_{y} c \end{matrix}

Taking into consideration degree restrictions, we are left with 14 nontrivial cases.

(1i)  a_{j} = y  a_{k} = c  z = y
(1ii)  y  c  c
(1iii)  y  c  d
(1iv)  y  c  {y, c, d }
(2i)  y  d  y
(2ii)  y  d  c
(3)  y  {y, c, d }  c
(4i)  c  d  y
(4ii)  c  d  c
(4iii)  c  d  U_{y} c
(5)  c  U_{y} c  d
(6)  c  U_{y} d  c
(7)  c  {y, c, d }  y
(8)  d  U_{y} c  c

Notice that Lemma 7 and Lemma 8 will be used often in the following computations (denoted by L 7 and L 8, respectively), along with the fact (from the general rules of multiplication on B) that

U_{b, b} = 0 for all b ∈ B. (U 2)

(1i) U_{x} {y, c, y} + {y, c, U_{x} y} + {x, y, {x, c, y}} + {x, c, {x, y, y}}

= 0 + {b_{1 1}^{∗}, h_{1 c}, b_{3 1}} + {b_{1 1}, b_{1 1}^{∗}, h_{3 c}} + 0 (by U 2 twice) = (b_{5 1} + b_{5 1}^{∗}) + (b_{5 1} + b_{5 1}^{∗}) = 0 ,

(1ii) U_{x} {y, c, c} + {y, c, U_{x} c} + {x, y, {x, c, c}} + {x, c, {x, y, c}}

= 0 + 0 + 0 + 0 = 0 (by U 2 twice, L 7 and L 8 once each) ,

(1iii) U_{x} {y, c, d} + {y, c, U_{x} d} + {x, y, {x, c, d}} + {x, c, {x, y, d}}

= U_{b_{1 1}} b_{3 4}^{∗} + {b_{1 1}^{∗}, h_{1 c}, b_{3 3}} + {b_{1 1}, b_{1 1}^{∗}, b_{3 4}} + {b_{1 1}, h_{1 c}, h_{3 c}} = (b_{5 3} + b_{5 4} + b_{5 5}) + b_{5 5} + (b_{5 4} + b_{5 5}) + (b_{5 3} + b_{5 5}) = 0 ,

(1iv) U_{x} {y, c, {y, c, d}} + {y, c, U_{x} {y, c, d}}
+ {x, y, {x, c, {y, c, d}}} + {x, c, {x, y, {y, c, d}}}

= 0 + {b_{1 1}^{∗}, h_{1 c}, U_{b_{1 1}} b_{3 4}^{∗}}

+ 0 + {b_{1 1}, h_{1 c}, {b_{1 1}, b_{1 1}^{∗}, b_{3 4}^{∗}} (by L 7 twice) = {b_{1 1}^{∗}, h_{1 c}, b_{5 3} + b_{5 4} + b_{5 5}} + {b_{1 1}, h_{1 c}, b_{5 4}^{∗} + b_{5 5}^{∗}} = 0 + h_{7} + 0 + h_{7} + 0 = 0 ,

(2i) U_{x} {y, d, y} + {y, d, U_{x} y} + {x, y, {x, d, y}} + {x, d, {x, y, y}}

= 0 + {b_{1 1}^{∗}, h_{1 d}, b_{3 1}} + {b_{1 1}, b_{1 1}^{∗}, h_{3 d}} + 0 (by U 2 twice) = (b_{5 2} + b_{5 2}^{∗}) + (b_{5 2} + b_{5 2}^{∗}) = 0 ,

(2ii) U_{x} {y, d, c} + {y, d, U_{x} c} + {x, y, {x, d, c}} + {x, d, {x, y, c}}

= U_{b_{1 1}} b_{3 4}^{∗} + {b_{1 1}^{∗}, h_{1 d}, b_{3 2}} + {b_{1 1}, b_{1 1}^{∗}, b_{3 4}} + {b_{1 1}, h_{1 d}, h_{3 c}} = (b_{5 3} + b_{5 4} + b_{5 5}) + b_{5 4} + (b_{5 4} + b_{5 5}) + (b_{5 3} + b_{5 4}) = 0 ,

(3)  U_{x} {y, {y, c, d}, c} + {y, {y, c, d}, U_{x} c}}
+ {x, y, {x, {y, c, d}, c}} + {x, {y, c, d}, {x, y, c}}

= 0 + {b_{1 1}^{∗}, b_{3 4}^{∗}, b_{3 2}} + 0 + {b_{1 1}, b_{3 4}^{∗}, h_{3 c}} (by L 7 twice) = h_{7} + h_{7} = 0 ,

(4i) U_{x} {c, d, y} + {c, d, U_{x} y} + {x, c, {x, d, y}} + {x, d, {x, c, y}}

= U_{b_{1 1}} b_{3 4}^{∗} + {h_{1 c}, h_{1 d}, b_{3 1}} + {b_{1 1}, h_{1 c}, h_{3 d}} + {b_{1 1}, h_{1 d}, h_{3 c}} = (b_{5 3} + b_{5 4} + b_{5 5}) + b_{5 3} + (b_{5 3} + b_{5 5}) + (b_{5 3} + b_{5 4}) = 0 ,

(4ii) U_{x} {c, d, c} + {c, d, U_{x} c} + {x, c, {x, d, c}} + {x, d, {x, c, c}}

= 0 + 0 + 0 + 0 = 0 (by U 2 twice, L 7 and L 8 once each) ,

(4iii) U_{x} {c, d, U_{y} c} + {c, d, U_{x} U_{y} c}
+ {x, c, {x, d, U_{y} c}} + {x, d, {x, c, U_{y} c}}

= 0 + {h_{1 c}, h_{1 d}, b_{5 1}} + {b_{1 1}, h_{1 c}, {b_{1 1}, h_{1 d}, b_{3 2}^{∗}}} + 0 (by L 8 twice)

= h_{7} + {b_{1 1}, h_{1 c}, b_{5 4}^{∗}} = h_{7} + h_{7} = 0 ,

(5)  U_{x} {c, U_{y} c, d} + {c, U_{y} c, U_{x} d}
+ {x, c, {x, U_{y} c, d}} + {x, U_{y} c, {x, c, d}}

= 0 + 0 + {b_{1 1}, h_{1 c}, b_{5 4}^{∗}} + h_{7} (by L 8 twice)

= h_{7} + h_{7} = 0 ,

(6)  U_{x} {c, U_{y} d, c} + {c, U_{y} d, U_{x} c}
+ {x, c, {x, U_{y} d, c}} + {x, U_{y} d, {x, c, c}}

= 0 + 0 + 0 + 0 = 0 (by U 2 twice, L 7 and L 8 once each) .

(7)  U_{x} {c, {y, c, d}, y} + {c, {y, c, d}, U_{x} y}}
+ {x, c, {x, {y, c, d}, y}} + {x, {y, c, d}, {x, c, y}}

= 0 + 0 + {b_{1 1}, h_{1 c}, {b_{1 1}, b_{1 1}^{∗}, b_{3 4}^{∗}}} + {b_{1 1}, h_{3 c}, b_{3 4}^{∗}} (by L 7 twice)

= {b_{1 1}, h_{1 c}, b_{5 4}^{∗} + b_{5 5}^{∗}} + h_{7} = h_{7} + 0 + h_{7} = 0 ,

(8)  U_{x} {d, U_{y} c, c} + {d, U_{y} c, U_{x} c}
+ {x, d, {x, U_{y} c, c}} + {x, U_{y} c, {x, d, c}}

= 0 + {h_{1 c}, b_{3 2}, b_{3 2}^{∗}} + 0 + {b_{1 1}, b_{3 2}^{∗}, b_{3 4}} (by L 8 twice)

= h_{7} + h_{7} = 0 .

Proof of (10):

[U_{a_{i}, a_{j}}, U_{a_{k}, a_{l}}] + [U_{a_{i}, a_{k}}, U_{a_{j}, a_{l}}] + [U_{a_{i}, a_{l}}, U_{a_{j}, a_{k}}] = 0

Either: (1) {a_{i}, a_{j}, a_{k}, a_{l}} = G or (2) exactly 3 are generators and one has degree 3.

CASE 1: If {a_{i}, a_{j}, a_{k}, a_{l}} = G, then without loss of generality, we can assume a_{i} = x, a_{j} = y, a_{k} = c and a_{l} = d. If deg(z) = 3, then z = U_{r} s or z = {r, s, t} for some r, s, t ∈ G. If z = U_{r} s, then the r-degree of (10) is 3, and so everything is zero. If z = {r, s, d }, then the d-degree of the expression is 2, and again everything is zero. Otherwise, z = {x, y, c}. Thus we see that degree restrictions eliminate all cases but z ∈ {x, y, c, {x, y, c}}, and the case z = y will follow from z = x:

[U_{x, y}, U_{c, d}] x + [U_{x, c}, U_{y, d}] x + [U_{x, d}, U_{y, c}] x
= {x, y, {c, d, x}} + {c, d, {x, y, x}} + {x, c, {y, d, x}}
+ {y, d, {x, c, x}} + {x, d, {y, c, x}} + {y, c, {x, d, x}}
= {b_{1 1}, b_{1 1}^{∗}, b_{3 4}} + 0 + {b_{1 1}, h_{1 c}, h_{3 d}}
+ 0 + {b_{1 1}, h_{1 d}, h_{3 c}} + 0 (by U 2 three times)
= (b_{5 4} + b_{5 5}) + (b_{5 3} + b_{5 5}) + (b_{5 3} + b_{5 4}) = 0 .

Now check z = c:

{x, y, {c, d, c}} + {c, d, {x, y, c}} + {x, c, {y, d, c}} + {y, d, {x, c, c}} + {x, d, {y, c, c}} + {y, c, {x, d, c}}

= 0 + 0 + 0 + 0 + 0 + 0 = 0 (by U 2 and L 7 three times each) .

Finally check z = {x, y, c}:

{x, y, {c, d, {x, y, c}}} + {c, d, {x, y, {x, y, c}}}
+ {x, c, {y, d, {x, y, c}}} + {y, d, {x, c, {x, y, c}}}
+ {x, d, {y, c, {x, y, c}}} + {y, c, {x, d, {x, y, c}}}
= 0 + {h_{1 c}, h_{1 d}, {b_{1 1}, b_{1 1}^{∗}, h_{3 c}}} + {b_{1 1}, h_{1 c}, {b_{1 1}^{∗}, h_{1 d}, h_{3 c}}}
+ 0 + 0 + {b_{1 1}^{∗}, h_{1 c}, {b_{1 1}, h_{1 d}, h_{3 c}}} (by L 7 three times)
= {h_{1 c}, h_{1 d}, b_{5 1} + b_{5 1}^{∗}} + {b_{1 1}, h_{1 c}, b_{5 3}^{∗} + b_{5 4}^{∗}}
+ {b_{1 1}^{∗}, h_{1 c}, b_{5 3} + b_{5 4}}
= h_{7} + h_{7} + 0 + h_{7} + 0 + h_{7} = 0 .

Thus for all cases where {a_{i}, a_{j}, a_{k}, a_{l}} = G, (10) has been verified.

CASE 2: (a_{i}, a_{j}, a_{k}, z ∈ G and deg(a_{l}) = 3) Since a_{i}, a_{j} and a_{k} are distinct, it is sufficient to consider i < j < k. Thus our choices for a_{i}, a_{j}, a_{k} are:

(1) x, y, c  (2) x, y, d  (3) x, c, d  (3^{∗}) y, c, d

Notice that once a_{i}, a_{j}, a_{k} and z have been chosen, a_{l} is uniquely determined by the degree restrictions on the generators. We are left with the remaining cases to check (in the order a_{i}, a_{j}, a_{k}, z, a_{l}):

(1i) x y c x {y, c, d}  (1i^{∗}) x y c y {x, c, d}
(1ii) x y c c {x, y, d}
(1iii) x y c d {x, y, c}
(2)  x y d c {x, y, c}
(3i) x c d x  U_{y} c  (3i^{∗}) y c d y  U_{x} c
(3ii) x c d y {x, y, c}  (3ii^{∗}) y c d x {x, y, c}
(3iii) x c d c  U_{y} x  (3iii^{∗}) y c d c  U_{x} y

Due to the symmetries in x and y, we only have to check the cases in the left column. For the remaining cases, the expression has degree 7, and so the only possibilities are 0 and h_{7}. Thus if deg(r) = 7 for some monomial r, we can conclude that r + r^{∗} = r + r = 2r = 0. (This fact will simplify a couple of the calculations.)

(1i) [U_{x, y}, U_{c, {y, c, d}}] x + [U_{x, c}, U_{y, {y, c, d}}] x + [U_{x, {y, c, d}}, U_{y, c}] x

= {x, y, {c, {y, c, d}, x}} + {c, {y, c, d}, {x, y, x}}} + {x, c, {y, {y, c, d}, x}} + {y, {y, c, d}, {x, c, x}} + {x, {y, c, d}, {y, c, x}} + {y, c, {x, {y, c, d}, x}}

= 0 + 0 + {b_{1 1}, h_{1 c}, {b_{1 1}, b_{1 1}^{∗}, b_{3 4}^{∗}}} + 0 + {b_{1 1}, b_{3 4}^{∗}, h_{3 c}} + 0 (by L 7 and U 2 twice each)

= {b_{1 1}, h_{1 c}, b_{5 4}^{∗} + b_{5 5}^{∗}} + h_{7} = h_{7} + 0 + h_{7} = 0 ,

(1ii) [U_{x, y}, U_{c, {x, y, c}}] c + [U_{x, c}, U_{y, {x, y, d}}] c + [U_{x, {x, y, d}}, U_{y, c}] c

= {x, y, {c, {x, y, d}, c}} + {c, {x, y, d}, {x, y, c}}} + {x, c, {y, {x, y, d}, c}} + {y, {x, y, d}, {x, c, c}} + {x, {x, y, d}, {y, c, c}} + {y, c, {x, {x, y, d}, c}}

= 0 + 0 + 0 + 0 + 0 + 0 (by L 7 and U 2 three times each) = 0 ,

(1iii) [U_{x, y}, U_{c, {x, y, c}}] d + [U_{x, c}, U_{y, {x, y, c}}] d + [U_{x, {x, y, c}}, U_{y, c}] d

= {x, y, {c, {x, y, c}, d}} + {c, {x, y, c}, {x, y, d}}} + {x, c, {y, {x, y, c}, d}} + {y, {x, y, c}, {x, c, d}} + {x, {x, y, c}, {y, c, d}} + {y, c, {x, {x, y, c}, d}}

= 0 + 0 + {x, c, {y, {x, y, c}, d}} + {y, {x, y, c}, {x, c, d}} + {y, {x, y, c}, {x, c, d}}^{∗}

+ {x, c, {y, {x, y, c}, d}}^{∗} (by L 7 twice)

= 0 ,

(2)  [U_{x, y}, U_{d, {x, y, c}}] c + [U_{x, d}, U_{y, {x, y, c}}] c + [U_{x, {x, y, c}}, U_{y, d}] c

= {x, y, {d, {x, y, c}, c}} + {d, {x, y, c}, {x, y, c}}} + {x, d, {y, {x, y, c}, c}} + {y, {x, y, c}, {x, d, c}} + {x, {x, y, c}, {y, d, c}} + {y, d, {x, {x, y, c}, c}}

= 0 + 0 + 0 + {y, {x, y, c}, {x, d, c}} + {y, {x, y, c}, {x, d, c}}^{∗} + 0 (by L 7 three times and U 2 once)

= 0 ,

(3i) [U_{x, c}, U_{d, U_{y} c}] x + [U_{x, d}, U_{c, U_{y} c}] x + [U_{x, U_{y} c}, U_{c, d}] x

= {x, c, {d, U_{y} c, x}} + {d, U_{y} c, {x, c, x}}} + {x, d, {c, U_{y} c, x}} + {c, U_{y} c, {x, d, x}}

+ {x, U_{y} c, {c, d, x}} + {c, d, {x, U_{y} c, x}}

= {b_{1 1}, h_{1 c}, {b_{1 1}, h_{1 d}, b_{3 2}^{∗}}} + 0 + 0

+ 0 + {b_{1 1}, b_{3 2}^{∗}, b_{3 4}} + 0 (by U 2 and L 8 twice each) = {b_{1 1}, h_{1 c}, b_{5 4}^{∗}} + h_{7} = h_{7} + h_{7} = 0 ,

(3ii) [U_{x, c}, U_{d, {x, y, c}}] y + [U_{x, d}, U_{c, {x, y, c}}] y + [U_{x, {x, y, c}}, U_{c, d}] y

= {x, c, {d, {x, y, c}, y}} + {d, {x, y, c}, {x, c, y}}} + {x, d, {c, {x, y, c}, y}} + {c, {x, y, c}, {x, d, y}} + {x, {x, y, c}, {c, d, y}} + {c, d, {x, {x, y, c}, y}}

= {b_{1 1}, h_{1 c}, {b_{1 1}^{∗}, h_{1 d}, h_{3 c}}} + 0 + 0 + 0 + {b_{1 1}, h_{3 c}, b_{3 4}^{∗}} + {h_{1 c}, h_{1 d}, {b_{1 1}, b_{1 1}^{∗}, h_{3 c}}} (by L 7 twice and U 2)

= {b_{1 1}, h_{1 c}, b_{5 3}^{∗} + b_{5 4}^{∗}} + h_{7} + {h_{1 c}, h_{1 d}, b_{5 1} + b_{5 1}^{∗}} = (0 + h_{7}) + h_{7} + (h_{7} + h_{7}) = 0 ,

(3iii) [U_{x, c}, U_{d, U_{y} x}] c + [U_{x, d}, U_{c, U_{y} x}] c + [U_{x, U_{y} x}, U_{c, d}] c

= {x, c, {d, U_{y} x, c}} + {d, U_{y} x, {x, c, c}}} + {x, d, {c, U_{y} x, c}} + {c, U_{y} x, {x, d, c}}

+ {x, U_{y} x, {c, d, c}} + {c, d, {x, U_{y} x, c}}

= 0 + 0 + 0 + 0 + 0 + 0 (by L 7 and U 2 three times each ) = 0 .

Thus we conclude that (10) is true, and therefore all linearizations of U_{a}^{2} = 0 have been verified for elements of B. ■ Lemma 9 U_{a^{3}} = 0 and its linearizations hold for all elements of B. Proof Once again, first we will compute all of the linearizations of U_{a^{3}} = 0, then each will be verified in turn. As before, consider a = ∑ λ_{i} a_{i} ∈ J_{Φ[λ_{1}, ... , λ_{25}]}, and first notice that since deg (U_{a^{3}}) = 6 deg(a), we can assume that deg (a_{i}) = 1 for each a_{i} occurring as part of a. Hence, we only have to consider a_{1} = x, a_{2} = y, a_{3} = c and a_{4} = d. Then we can write

a^{3} = U_{a} a = (∑_{i=1}^{4} λ_{i}^{2} U_{a_{i}} + ∑_{i<k} λ_{i} λ_{k} U_{a_{i}, a_{k}}) (∑_{j=1}^{4} λ_{j} a_{j})

= ∑_{i≠j} λ_{i}^{2} λ_{j} U_{a_{i}}a_{j} + ∑_{i<k, j≠i, k} λ_{i} λ_{j} λ_{k} U_{a_{i}, a_{k}} a_{j}

because V_{a_{i}, a_{i}} = 0 and deg_{a_{i}}(r) > 2 implies that r = 0 for any monomial r. Since χ(Φ) = 2, U_{a_{i}, a_{j}} = V_{a_{i}, a_{j}} and {a_{i}, a_{j}, a_{k}} is symmetric in i, j, k, we can rewrite the second sum as

∑_{i<k, j≠i, k} λ_{i} λ_{j} λ_{k} U_{a_{i}, a_{k}} a_{j}

= ∑_{i<j<k} λ_{i} λ_{j} λ_{k} (U_{a_{i}, a_{k}} a_{j} + U_{a_{i}, a_{j}} a_{k} + U_{a_{j}, a_{k}} a_{i})

= ∑_{i<j<k} 3 λ_{i} λ_{j} λ_{k} {a_{i}, a_{j}, a_{k}} = ∑_{i<j<k} λ_{i} λ_{j} λ_{k} {a_{i}, a_{j}, a_{k}} .

Thus we have

U_{a^{3}} = U_{(∑_{i≠j} λ_{i}^{2} λ_{j} U_{a_{i}} a_{j} + ∑_{i<j<k} λ_{i} λ_{j} λ_{k} {a_{i}, a_{j}, a_{k}})} ,

and so U_{a^{3}} is a linear combination of the following types of monomials (with i, j, k distinct and l, m, n distinct):

\begin{aligned}U_{U_{a_{i}} a_{j}} & U_{\{a_{i}, a_{j}, a_{k}\}} & U_{U_{a_{i}} a_{j}, U_{a_{l}} a_{m}} \\ & U_{U_{a_{i}} a_{j}, \{a_{l}, a_{m}, a_{n}\}} & U_{\{a_{i}, a_{j}, a_{k}\}, \{a_{l}, a_{m}, a_{n}\}} . \end{aligned}

Since deg_{a_{i}}(U_{U_{a_{i}} a_{j}}) = 4, U_{U_{a_{i}} a_{j}} = 0. We know that i, j and k are distinct for U_{{a_{i}, a_{j}, a_{k}}}. For U_{U_{a_{i}} a_{j}, U_{a_{l}} a_{m}}, i = l or m implies that the a_{i}-degree of the monomial is at least 3, which forces U_{U_{a_{i}} a_{j}, U_{a_{l}} a_{m}} = 0. Hence, we can assume that i ≠ l, m, and dually l ≠ i, j. This leaves only the following cases: U_{U_{a_{i}} a_{j}, U_{a_{l}} a_{j}} and U_{U_{a_{i}} a_{j}, U_{a_{l}} a_{m}} for i, j, l, m distinct.

For U_{U_{a_{i}} a_{j}, {a_{l}, a_{m}, a_{n}}}, degree considerations force i ≠ j, l, m, n. Once a_{i} is chosen, the distinct a_{l}, a_{m}, a_{n} use up the three remaining generators. Hence, we must have j = l, m or n. Without loss of generality, we can assume that j = l, leaving only U_{U_{a_{i}} a_{j}, {a_{j}, a_{m}, a_{n}}} for distinct i, j, m, n.

Finally, for U_{{a_{i}, a_{j}, a_{k}}, {a_{l}, a_{m}, a_{n}}}, there are only four choices with which to fill six positions. Hence, degree considerations and symmetry of triple products allow us to assume that a_{i} = a_{l} and a_{j} = a_{m}, again leaving only a single type of monomial U_{{a_{i}, a_{j}, a_{k}}, {a_{i}, a_{j}, a_{n}}} for distinct i, j, k, n.

Thus U_{a^{3}} is a linear combination of the following types of monomials (changing some indices in the above cases) with i, j, k and l distinct (each monomial type is listed under its associated scalar):

\underline{λ_{i}^{2} λ_{j}^{2} λ_{k}^{2}} :  \underline{λ_{i}^{2} λ_{j}^{2} λ_{k} λ_{l}} :
U_{{a_{i}, a_{j}, a_{k}}}  U_{U_{a_{i}} a_{k}, U_{a_{j}} a_{l}}
U_{U_{a_{i}} a_{k}, U_{a_{j}} a_{k}}  U_{U_{a_{i}} a_{j}, {a_{j}, a_{k}, a_{l}}}
U_{{a_{i}, a_{j}, a_{k}}, {a_{i}, a_{j}, a_{l}}} .

Hence, we have two linearizations of U_{a^{3}} to check:

U_{{a_{i}, a_{j}, a_{k}}} + U_{U_{a_{i}} a_{k}, U_{a_{j}} a_{k}} + U_{U_{a_{i}} a_{j}, U_{a_{k}} a_{j}} + U_{U_{a_{j}} a_{i}, U_{a_{k}} a_{i}} = 0 ,  (11)

U_{U_{a_{i}} a_{k}, U_{a_{j}} a_{l}} + U_{U_{a_{i}} a_{l}, U_{a_{j}} a_{k}} + U_{U_{a_{i}} a_{j}, {a_{j}, a_{k}, a_{l}}}

+ U_{U_{a_{j}} a_{i}, {a_{i}, a_{k}, a_{l}}} + U_{{a_{i}, a_{j}, a_{k}}, {a_{i}, a_{j}, a_{l}}} = 0 .  (12)

Proof of (11): The only product that isn’t zero by degree restrictions is the case when {a_{i}, a_{j}, a_{k}} = {x, y, c} acting on z = d:

U_{{x, y, c}} d + U_{U_{x}y, U_{c}y} d + U_{U_{x}c, U_{y}c} d + U_{U_{y}x, U_{c}x} d
= U_{{x, y, c}} d + 0 + {U_{x}c, U_{y}c, d} + 0 (U_{c} = 0)
= U_{h_{3 c}} h_{1 d} + {h_{1 d}, b_{3 2}, b_{3 2}^{∗}}
= h_{7} + h_{7} = 0 .

Proof of (12): Due to degree restrictions, we can assume that a_{i}, a_{j} ∈ {x, y, c} and a_{k}, a_{l} ∈ G\ {a_{i}, a_{j}} (otherwise, everything is zero and (12) is true). Hence, we only have three remaining cases:

{a_{i}, a_{j}} = {x, y} , {a_{k}, a_{l}} = {c, d } , z = c ,
= {x, c} ,  = {y, d } ,  = y ,
= {y, c} ,  = {x, d } ,  = x .

Notice that it is only necessary to check the first two cases, due to the symmetry in x and y.

{U_{x} c, U_{y} d, c} + {U_{x} d, U_{y} c, c} + {U_{x} y, {c, y, d}, c}

+ {U_{y} x, {x, c, d}, c} + {{x, c, y}, {x, y, d}, c}

= 0 + 0 + 0 + 0 + 0 = 0 (by L 7 three times and L 8 twice) ,

{U_{x} y, U_{c} d, y} + {U_{x} d, U_{c} y, y} + {U_{x} c, {y, c, d}, y}

+ {U_{c} x, {x, y, d}, y} + {{x, y, c}, {x, c, d}, y}

= 0 + 0 + {b_{1 1}^{∗}, b_{3 2}, b_{3 4}^{∗}} + 0 + {b_{1 1}^{∗}, h_{3 c}, b_{3 4}} (U_{c} = 0)

= h_{7} + h_{7} = 0 . ■

We can finally conclude that J = span (B) is indeed a quadratic Jordan algebra, but we can see that our definition of weak center which works so well for nondegenerate algebras fails miserably when considering this particular algebra. Since U_{c} = 0 (and x ∘ y = 0, for that matter), U_{c} U_{x∘y} = 0 ≠ U_{{x, y, c}}, and the weak central element c does not behave like a scalar multiple of 1. (Also notice that {U_{x} c, y, d} ≠ {U_{x} d, y, c}, another relation that must hold for scalar central elements c and d.) Hence, we must conclude that our definition of the weak center is inadequate for dealing with degenerate algebras. Note, however, that this counterexample does not rule out the possibility of an acceptable axiomatic definition for the center of a general algebra. Maybe only a few more axioms are necessary in order to construct a tenable idea of center for all algebras. That is, perhaps there is a finite list of relations that will insure an element c lies in the center. At this point, however, the hope for such a finite list that works for all algebras is very dim. It seems very likely that one can construct an algebra J_{n} for each n ∈ ℕ containing elements x, y_{1}, …, y_{n+1}, c such that U_{c}, V_{c} ∈ Γ(J_{n}) and

U_{x, c} U_{y_{1}} ⋯ U_{y_{i}} U_{x, c} = U_{c} V_{x} U_{y_{1}} ⋯ U_{y_{i}} V_{x}  (C_{i})

for all 1 ≤ i ≤ n, while at the same time

U_{x, c} U_{y_{1}} ⋯ U_{y_{n+1}} U_{x, c} ≠ U_{c} V_{x} U_{y_{1}} ⋯ U_{y_{n+1}} V_{x} .  (C_{n+1})

Such a collection of algebras {J_{n}} would not eliminate the possibility of a workable definition of center for degenerate algebras (there are possibly other types of axioms which would insure that C_{n+1} is true for all n ∈ ℕ), but it would lead us to believe that a concise list of axioms for the center for degenerate algebras is unlikely. This is in no way a proof, of course, but we conjecture that degenerate algebras are simply too pathological to have an easily understood center.

4 The Scalar Center

At the heart of the matter is a very important problem with our notion of weak center when looking at general Jordan algebras: given c, d ∈ C_{w}(J), it is not necessarily true that c + d ∈ C_{w}(J). For nondegenerate algebras this worry turned out to be unfounded, but it clearly is a problem when looking at arbitrary algebras. This is exactly the problem that arises in the previous counterexample: U_{c+d} = U_{c} + U_{c, d} + U_{d} = U_{c, d}, and U_{x} U_{c, d} y = 0 but U_{c, d} U_{x} y ≠ 0, implying that c + d is not a central element even though both c and d are. There appears to be a similar problem with our “meta-principle” mentioned earlier: if there exist J̃_{1} ⊃ J over Φ̃_{1} ⊃ Φ and J̃_{2} ⊃ J over Φ̃_{2} ⊃ Φ such that c_{1} ∈ Φ̃_{1} 1 and c_{2} ∈ Φ̃_{2} 1 (assuming both J̃_{1} and J̃_{2} share the same unit), can we find J̃_{3} ⊃ J over Φ̃_{3} ⊃ Φ such that c_{1} + c_{2} ∈ Φ̃_{3} 1? In other words, if c_{1} and c_{2} behave like scalar multiples of 1, does this necessarily imply that c_{1} + c_{2} behaves like a scalar multiple of 1? Stated more generally, the question we are asking becomes: when is C_{s}(J) closed under addition? Recall that this was the same problem for the centroid Γ(J) in general, and the answer was positive only when J had zero extreme radical. It seems reasonable to conjecture that the scalar center as defined above should be well behaved whenever the centroid is well behaved. When J has no trivial ideals, this turns out to be true. (An ideal I of J is trivial if U_{I} Ĵ = 0, and it is easy to show that the extreme radical is always a trivial ideal.) The next result is useful because it reduces analysis of the scalar center of an algebra with no trivial ideals and no unit to that of the scalar center of a tight unital hull. The result is also very interesting in its own right because it shows that any algebra with no extreme elements has a unique tight unital hull.

Theorem 3 Let J be a nonunital, quadratic Jordan algebra over Φ with ExtRad(J) = 0, and let Ĵ be a tight unital hull of J. Then

(i) ExtRad(Ĵ) = 0;

(ii) if Φ acts faithfully on J then Φ also acts faithfully on Ĵ;

(iii) C_{s}(J) ⊂ C_{s}(Ĵ) = Φ ˆ1 + C_{s}(J);

(iv) Ĵ is unique.

Proof (i) For J^{(ˆ1)} = Φ ˆ1 ⊕ J, let Ĵ = J^{(ˆ1)} / I be any tight unital extension of J over Φ (i.e., any I ◁ J^{(ˆ1)} is maximal with respect to I ∩ J = 0). Since ExtRad(Ĵ)∩J ⊂ ExtRad(J) = 0, ExtRad(Ĵ) = 0 because Ĵ is a tight extension of J.

(ii) If Φ acts faithfully on J, it also acts faithfully on any larger Ĵ.

(iii) In order to show that C_{s}(J) ⊂ C_{s}(Ĵ), it will simplify matters if we use an equivalent definition of C_{s}(J): c ∈ C_{s}(J) if there exists J̃ such that c ∈ Φ̃ 1 ⊂ J̃ where

(1) J̃ ⊃ J over Φ̃ ⊃ Φ, where Φ̃ acts faithfully,

(2) J̃ = Φ̃ [1, J ] (J̃ is a unital Φ̃−envelope of J),

(3) J̃ is a tight Φ̃−cover of J: if 0 ≠ Ĩ◁ J̃, then Ĩ∩ J ≠ 0.

Clearly, if such a J̃ exists, then c ∈ C_{s}(J). Conversely, for each c ∈ C_{s}(J), we can construct J̃_{c} satisfying conditions (1)−(3): for any unital J̃ ⊃ J over Φ̃ ⊃ Φ with c ∈ Φ̃ 1, J ⊂ Φ̃ [1, J ]. Furthermore, for any unital Φ̃−envelope Φ̃ [1, J ] and any maximal M̃ ◁ Φ̃ [1, J ] missing J (M̃ ∩ J = 0), J̃_{c} = Φ̃ [1, J ] / M̃ is a tight unital Φ̃−envelope of J with c ∈ Φ̃ 1. As usual, we can assume Φ̃ is faithful by passing to Φ̃ / J̃^{⊥}.

We claim that any J̃ satisfying conditions (1) − (3) already “contains” Ĵ.

Lemma 10 If ExtRad(J) = 0 and J̃ is a tight unital Φ̃−envelope of J, then the natural homomorphism ψ̃ : J^{(ˆ1)} = Φ ˆ1 ⊕ J → J̃ has ker (ψ̃) ⊃ K for all ideals K ◁ J^{(ˆ1)} with K ∩ J = 0. Thus there is a unique maximal ideal I ◁ J^{(ˆ1)} with I ∩ J = 0, and I = ker (ψ̃) for all tight envelopes J̃, so ψ̃ induces a natural unital imbedding of Ĵ = J^{(ˆ1)} / I into J̃.

Proof It is easy to verify that the map ψ̃ : J^{(ˆ1)} → J̃ defined by ϕ ˆ1 ⊕ x ↦ ϕ 1 + x is a homomorphism of Jordan algebras. Also note that ψ̃(x) = x for all x ∈ J. If K ◁ J^{(ˆ1)} with K ∩ J = 0, define K̃ = Φ̃ ψ̃(K). Then K̃ is an ideal of J̃ such that

UK̃ J = Φ̃ ψ̃(U_{K} J) ⊂ Φ̃ ψ̃(K ∩ J) = 0 ,
VK̃ J = Φ̃ ψ̃(V_{K} J) ⊂ Φ̃ ψ̃(K ∩ J) = 0 ,
U_{K̃, J} J = Φ̃ ψ̃(U_{K, J} J) ⊂ Φ̃ ψ̃(K ∩ J) = 0 .

By the above calculations, K̃ ∩ J ⊂ ExtRad(J) = 0, and so K̃ = 0 because J̃ is a tight extension of J. Hence, ψ̃(K) = 0 because ψ̃(K) ⊂ K̃, implying that K ⊂ ker (ψ̃) while ker (ψ̃) ∩ J = 0. Thus ker (ψ̃) is the unique maximal ideal I of J^{(ˆ1)} missing J. Therefore, J ⊂ Ĵ = J^{(ˆ1)} / I = J^{(ˆ1)} / ker (ψ̃) ⊂ J̃ where Ĵ is the unique tight unital hull of J. ■

Thus for each c ∈ C_{s}(J), there exists J̃ ⊃ Ĵ over Φ̃ ⊃ Φ such that c ∈ Φ̃ 1, and so C_{s}(J) ⊂ C_{s}(Ĵ). Next we will show that C_{s}(Ĵ) ⊂ Φ ˆ1 + C_{s}(J). Let ϕ ∈ Φ and x ∈ J, then ϕ ˆ1 + x ∈ C_{s}(Ĵ) implies that there exists J̃ ⊃ Ĵ over Φ̃ ⊃ Φ such that ϕ ˆ1 + x = ϕ̃ˆ1 for some ϕ̃ ∈ Φ̃. Thus x = ϕ̃ˆ1 − ϕ ˆ1 = (ϕ̃ − ϕ) ˆ1 ∈ Φ̃ ˆ1, which implies that x ∈ C_{s}(J). Hence, C_{s}(Ĵ) ⊂ Φ ˆ1 + C_{s}(J). To see that Φ ˆ1 + C_{s}(J) ⊂ C_{s}(Ĵ), notice that c ∈ C_{s}(J) ⊂ C_{s}(Ĵ) implies that there exists J̃ ⊃ Ĵ over Φ̃ ⊃ Φ such that c = ϕ̃ˆ1 for some ϕ̃ ∈ Φ̃. Thus ϕ ˆ1 + c = ϕ ˆ1 + ϕ̃ˆ1 = (ϕ + ϕ̃) ˆ1 ∈ Φ̃ ˆ1, which implies that ϕ ˆ1 + c ∈ C_{s}(Ĵ). Finally, we conclude that C_{s}(J) ⊂ C_{s}(Ĵ) = Φ ˆ1 + C_{s}(J).

(iv) The lemma above proved that the tight unital hull Ĵ is unique. ■

Theorem 4 Let J be a quadratic Jordan algebra (over a faithful ring Φ) with no trivial ideals. Then there exists J̃ ⊃ J over Ω̃ ⊃ Φ such that C_{s}(J) ⊂ Ω̃ 1. In particular, C_{s}(J) is a Jordan subalgebra of J, and there is a fixed extension J̃ of J such that C_{s}(J) = J ∩ Ω̃ 1.

Proof By the above lemma, there is a unique tight unital hull Ĵ of J with Φ acting faithfully on Ĵ and C_{s}(J) ⊂ C_{s}(Ĵ). Furthermore, Ĵ has no trivial ideals because it is a tight extension of J. Hence, we only need to prove the theorem when J contains a unit. Since J has no trivial ideals, ExtRad(J) = 0, implying that Γ(J) is commutative. Furthermore, Φ is faithful, and so we can replace Φ with Γ(J) and assume from the start that Φ = Γ(J). Thus for all c ∈ C_{s}(J), U_{c} = μ_{c} Id and V_{c} = ν_{c} Id for some μ_{c}, ν_{c} ∈ Φ. We will follow the same approach as in Theorem 1. Now define

\begin{matrix}Ω = Φ [ω_{c} : c ∈ C_{s}(J)] / ⟨\{μ_{c} − ω_{c}^{2}, ν_{c} − 2 ω_{c}\}⟩ , \\ K̃ = J_{Ω} = Ω ⊗_{Φ} J . \end{matrix}

Also, for each c ∈ C_{s}(J) define

Ω_{c} = Φ [ω_{c}] / ⟨{μ_{c} − ω_{c}^{2}, ν_{c} − 2 ω_{c}}⟩ and J̃_{c} = J_{Ω_{c}} = Ω_{c} ⊗_{Φ} J ,

and consider ⟨z_{c}⟩ in J̃_{c} where z_{c} = ω_{c} ⊗ 1 − 1 ⊗ c. Since

U_{z_{c}} = ω_{c}^{2} ⊗ Id − ω_{c} ⊗ V_{c} + 1 ⊗ U_{c} = (ω_{c}^{2} − ω_{c} ν_{c} + μ_{c}) ⊗ Id = 0 ,

⟨z_{c}⟩ = {M(z_{c}) : M ∈ 𝔐(J̃_{c})}, where 𝔐(J̃_{c}) is the multiplication algebra for J̃_{c}. Since ω_{c}^{2} = μ_{c}, each M ∈ 𝔐(J̃_{c}) can be written in the form M = 1 ⊗ M_{1} + ω_{c} ⊗ M_{c} where M_{1}, M_{c} ∈ 𝔐(J). Thus M(z_{c}) can be written as

M(z_{c}) = (1 ⊗ M_{1} + ω_{c} ⊗ M_{c}) (ω_{c} ⊗ 1 − 1 ⊗ c)
= ω_{c} ⊗ M_{1}(1) − 1 ⊗ M_{1}(c) + ω_{c}^{2} ⊗ M_{c}(1) − ω_{c} ⊗ M_{c}(c)
= 1 ⊗ [ μ_{c} M_{c}(1) − M_{1}(c)] + ω_{c} ⊗ [M_{1}(1) − M_{c}(c)] .

Since c = ϕ̃1 ∈ J̃ ⊃ J with ϕ̃^{2} = μ_{c}, 2 ϕ̃ = ν_{c} by faithfulness of Ω̃, we have the following relations:

U_{M(c)} = μ_{c} , U_{M(1)} ,  U_{M(c), N(c)} = μ_{c} U_{M(1), N(1)} ,
U_{M(1), M(c)} = ν_{c} U_{M(1)} ,  U_{M(1), N(c)} = U_{M(c), N(1)} ,
2 U_{x, M(c)} = ν_{c} U_{x, M(1)}

for all M ∈ 𝔐(J), x ∈ J. Now we can compute U_{M(z_{c})}:

U_{M(z_{c})} = 1 ⊗ U_{μ_{c}M_{c}(1) − M_{1}(c)} + ω_{c} ⊗ U_{μ_{c}M_{c}(1)−M_{1}(c), M_{1}(1) − M_{c}(c)}
+ ω_{c}^{2} ⊗ U_{M_{1}(1)−M_{c}(c)}
= 1 ⊗ [μ_{c}^{2} U_{M_{c}(1)} − μ_{c} U_{M_{c}(1), M_{1}(c)} + U_{M_{1}(c)}]
+ ω_{c} ⊗ [μ_{c} U_{M_{c}(1), M_{1}(1)} − μ_{c} U_{M_{c}(1), M_{c}(c)} − U_{M_{1}(c), M_{1}(1)}
+ U_{M_{1}(c), M_{c}(c)}]
+ 1 ⊗ μ_{c} [U_{M_{1}(1)} − U_{M_{1}(1), M_{c}(c)} + U_{M_{c}(c)}]
= 1 ⊗ [μ_{c}^{2} U_{M_{c}(1)} − μ_{c} U_{M_{1}(1), M_{c}(c)} + μ_{c} U_{M_{1}(1)}]
+ ω_{c} ⊗ [μ_{c} U_{M_{1}(1), M_{c}(1)} − μ_{c} ν_{c} U_{M_{c}(1)} − ν_{c} U_{M_{1}(1)}
+ μ_{c} U_{M_{1}(1), M_{c}(1)}]
+ 1 ⊗ μ_{c} [U_{M_{1}(1)} − U_{M_{1}(1), M_{c}(c)} + μ_{c} U_{M_{c}(1)}]
= 1 ⊗ [2 μ_{c}^{2} U_{M_{c}(1)} − 2 μ_{c} U_{M_{1}(1), M_{c}(c)} + 2 μ_{c} U_{M_{1}(1)}]
+ ω_{c} ⊗ [2 μ_{c} U_{M_{1}(1), M_{c}(1)} − μ_{c} ν_{c} U_{M_{c}(1)} − ν_{c} U_{M_{1}(1)}]
= 1 ⊗ [2 μ_{c}^{2} U_{M_{c}(1)} − μ_{c} ν_{c} U_{M_{1}(1), M_{c}(1)} + 2 μ_{c} U_{M_{1}(1)}]
+ ω_{c} ⊗ [2 μ_{c} U_{M_{1}(1), M_{c}(1)} − μ_{c} ν_{c} U_{M_{c}(1)} − ν_{c} U_{M_{1}(1)}]
= (2 μ_{c}^{2} − μ_{c} ν_{c} ω_{c}) ⊗ U_{M_{c}(1)} + (2 μ_{c} ω_{c} − μ_{c} ν_{c}) ⊗ U_{M_{1}(1), M_{c}(1)}
+ (2 μ_{c} − ν_{c} ω_{c}) ⊗ U_{M_{1}(1)} = 0

because 2 μ_{c} = 2 ω_{c} ω_{c} = ν_{c} ω_{c} so 2 μ_{c}^{2} = μ_{c} ν_{c} ω_{c}, and 2 μ_{c} ω_{c} = ν_{c} ω_{c} ω_{c} = ν_{c} μ_{c}. Thus ⟨z_{c}⟩ is a trivial ideal of J̃_{c}. Now consider Ω ⟨z_{c}⟩. This is a trivial ideal of

Ω ⊗_{Ω_{c}} J̃_{c} = Ω ⊗_{Ω_{c}} (Ω_{c} ⊗_{Φ} J) ≅ Ω ⊗_{Φ} J = J_{Ω} = K̃

because any scalar extension of a trivial ideal is a trivial ideal: U_{J_{Ω}} I_{Ω} = Ω (U_{J} I) ⊂ I_{Ω} and U_{I_{Ω}} J_{Ω} = Ω (U_{I} J) = 0.

Choose a maximal ideal M̃ of K̃ that misses J (i.e., M̃∩J = 0), then K̃ / M̃ is a tight extension of J, thereby implying that K̃ / M̃ has no trivial ideals. Hence, Ω ⟨z_{c}⟩ = ¯0 ∈ K̃ / M̃ implies that ⟨\overline{z}_{c}⟩ = ¯0 in K̃ / M̃ and \overline{ω_{c} ⊗ 1} = \overline{1 ⊗ c} for all c ∈ C_{s}(J). As usual, this is faithful over Ω̃ = Ω / J̃^{⊥}. ■

The theorem above proves that for any quadratic Jordan algebra J with no trivial ideals, C_{s}(J) is closed under addition and Jordan products. What happens when we weaken the condition to simply ExtRad(J) = 0? This is an open question; it looks as if the scalar center will not be a subalgebra in this case, but constructing a counterexample has proven to be problematic. The difficulty lies in insuring that the algebra in question does in fact have zero extreme radical. Simply modding out by an algebra’s extreme radical does not insure that the resulting quotient algebra is free of extreme elements. Nevertheless, it appears unlikely that the scalar center for an algebra with vanishing extreme radical is well behaved.

We are beginning to develop a general understanding of when the center of a quadratic Jordan algebra can be properly defined. Our original notion of center C_{w}(J) works well for nondegenerate algebras but is clearly insufficient for general algebras. The scalar center C_{s}(J) is more restrictive than C_{w}(J): for general algebras, C_{s}(J) ⊂ C_{w}(J), but Theorem 1 shows that for nondegenerate algebras they are equivalent. The scalar center works well for algebras with no trivial ideals. In this case, Theorem 4 tells us that C_{s}(J) is a subalgebra of J. However, it is unknown if the scalar center is satisfactory when ExtRad(J) = 0. A positive answer seems doubtful, even though ExtRad(J) = 0 insures that the centroid is a commutative ring.

References

[1] R. Lewand and K. McCrimmon, Macdonald’s Theorem for quadratic Jordan algebras, Pacific J. Math. 35 (1970), 681-706.

[2] O. Loos, “Jordan Pairs”, Lecture Notes in Mathematics, Vol. 460, Springer-Verlag, New York, 1975.

[3] K. McCrimmon, A characterization of the nondegenerate radical in quadratic Jordan triple systems, Algebras Groups Geom. 4 (1987), 145-164.

[4] K. McCrimmon, Jordan centroids, Comm. Algebra 27 (1999), 933-954.

[5] K. McCrimmon, Quadratic Jordan algebras and cubing operations, Trans. AMS 153, (Jan 1971), 265-278.

[6] K. McCrimmon and E. I. Zel’manov, The structure of strongly prime quadratic Jordan algebras, Adv. Math. 69 (1988), 133-222.

BASIS FOR DEGENERATE COUNTEREXAMPLE: b_{1 1} = x b_{1 1}^{∗} = y h_{1 c} = c h_{1 d} = d b_{3 1} = U_{x} y b_{3 1}^{∗} = U_{y} x b_{3 2} = U_{x} c b_{3 2}^{∗} = U_{y} c b_{3 3} = U_{x} d b_{3 3}^{∗} = U_{y} d b_{3 4} = {x, c, d} b_{3 4}^{∗} = {y, c, d} h_{3 c} = {x, y, c} h_{3 d} = {x, y, d} b_{5 1} = U_{x} U_{y} c b_{5 1}^{∗} = U_{y} U_{x} c b_{5 2} = U_{x} U_{y} d b_{5 2}^{∗} = U_{y} U_{x} d b_{5 3} = {U_{x} y, c, d} b_{5 3}^{∗} = {U_{y} x, c, d} b_{5 4} = {U_{x} c, y, d} b_{5 4}^{∗} = {U_{y} c, x, d} b_{5 5} = {U_{x} d, y, c} b_{5 5}^{∗} = {U_{y} d, x, c} h_{7} = U_{{x, y, c}} d

B = {b_{i j}, b_{i j}^{∗}, h_{k l}, h_{7}}

b^{2} = 0  a ∘ b = 0  {a, b, b} = 0

\begin{matrix}\{a, b, z\} = \{a, z, b\} = \{b, a, z\} = \{z, b, a\} \\ ∀ a, b, z ∈ B \end{matrix}

\begin{matrix}U_{b_{1 1}} b_{1 1}^{∗} = b_{3 1} & U_{b_{1 1}^{∗}} b_{1 1} = b_{3 1}^{∗} \\ U_{b_{1 1}} h_{1 c} = b_{3 2} & U_{b_{1 1}^{∗}} h_{1 c} = b_{3 2}^{∗} \\ U_{b_{1 1}} h_{1 d} = b_{3 3} & U_{b_{1 1}^{∗}} h_{1 d} = b_{3 3}^{∗} \\ U_{b_{1 1}} b_{3 2}^{∗} = b_{5 1} & U_{b_{1 1}^{∗}} b_{3 2} = b_{5 1}^{∗} \\ U_{b_{1 1}} b_{3 3}^{∗} = b_{5 2} & U_{b_{1 1}^{∗}} b_{3 3} = b_{5 2}^{∗} \\ U_{b_{1 1}} b_{3 4}^{∗} = b_{5 3} + b_{5 4} + b_{5 5} & U_{b_{1 1}^{∗}} b_{3 4} = b_{5 3}^{∗} + b_{5 4}^{∗} + b_{5 5}^{∗} \end{matrix}

U_{h_{3 c}} h_{1 d} = h_{7}

\begin{matrix}\{b_{1 1}, b_{1 1}^{∗}, h_{1 c}\} = h_{3 c} & \{b_{1 1}, b_{1 1}^{∗}, h_{1 d}\} = h_{3 d} \\ \{b_{1 1}, b_{1 1}^{∗}, h_{3 c}\} = b_{5 1} + b_{5 1}^{∗} & \{b_{1 1}, b_{1 1}^{∗}, h_{3 d}\} = b_{5 2} + b_{5 2}^{∗} \\ \{b_{1 1}, b_{1 1}^{∗}, b_{3 4}\} = b_{5 4} + b_{5 5} & \{b_{1 1}, b_{1 1}^{∗}, b_{3 4}^{∗}\} = b_{5 4}^{∗} + b_{5 5}^{∗} \\ \{b_{1 1}, h_{1 c}, h_{1 d}\} = b_{3 4} & \{b_{1 1}^{∗}, h_{1 c}, h_{1 d}\} = b_{3 4}^{∗} \\ \{b_{1 1}, h_{1 c}, b_{3 1}^{∗}\} = b_{5 1} + b_{5 1}^{∗} & \{b_{1 1}^{∗}, h_{1 c}, b_{3 1}\} = b_{5 1} + b_{5 1}^{∗} \\ \{b_{1 1}, h_{1 c}, b_{3 3}^{∗}\} = b_{5 5}^{∗} & \{b_{1 1}^{∗}, h_{1 c}, b_{3 3}\} = b_{5 5} \\ \{b_{1 1}, h_{1 c}, h_{3 d}\} = b_{5 3} + b_{5 5} & \{b_{1 1}^{∗}, h_{1 c}, h_{3 d}\} = b_{5 3}^{∗} + b_{5 5}^{∗} \\ \{b_{1 1}, h_{1 c}, b_{5 4}^{∗}\} = h_{7} & \{b_{1 1}^{∗}, h_{1 c}, b_{5 4}\} = h_{7} \\ \{b_{1 1}, h_{1 d}, b_{3 1}^{∗}\} = b_{5 2} + b_{5 2}^{∗} & \{b_{1 1}^{∗}, h_{1 d}, b_{3 1}\} = b_{5 2} + b_{5 2}^{∗} \\ \{b_{1 1}, h_{1 d}, b_{3 2}^{∗}\} = b_{5 4}^{∗} & \{b_{1 1}^{∗}, h_{1 d}, b_{3 2}\} = b_{5 4} \\ \{b_{1 1}, h_{1 d}, h_{3 c}\} = b_{5 3} + b_{5 4} & \{b_{1 1}^{∗}, h_{1 d}, h_{3 c}\} = b_{5 3}^{∗} + b_{5 4}^{∗} \\ \{b_{1 1}, b_{3 2}^{∗}, b_{3 4}\} = h_{7} & \{b_{1 1}^{∗}, b_{3 2}, b_{3 4}^{∗}\} = h_{7} \\ \{b_{1 1}, b_{3 4}^{∗}, h_{3 c}\} = h_{7} & \{b_{1 1}^{∗}, b_{3 4}, h_{3 c}\} = h_{7} \\ \{h_{1 c}, h_{1 d}, b_{3 1}\} = b_{5 3} & \{h_{1 c}, h_{1 d}, b_{3 1}^{∗}\} = b_{5 3}^{∗} \\ \{h_{1 c}, h_{1 d}, b_{5 1}\} = h_{7} & \{h_{1 c}, h_{1 d}, b_{5 1}^{∗}\} = h_{7} \end{matrix}

{h_{1 d}, b_{3 2}, b_{3 2}^{∗}} = h_{7}

\underline{ORDER 1}  \underline{ORDER 5}  \underline{ORDER 7} (all equal)
x  U_{x}U_{y}c  U_{{x, y, c}}d
y  U_{x}U_{y}d  {x, c, U_{y}{x, c, d}}
c  U_{y}U_{x}c  {x, c, {x, y, {y, c, d}}}
d  U_{y}U_{x}d  {x, c, {x, d, U_{y}c}}

U_{x}{y, c, d} {x, c, {y, d, {x, y, c}}}

\underline{ORDER 3}  U_{y}\{x, c, d\}  \{x, U_{y}c, \{x, c, d\}\}
U_{x}y  {x, y, {x, y, c}} = {x, c, U_{y}x}  {x, {x, y, c}, {y, c, d}}
U_{x}c  = {y, c, U_{x}y}  {y, c, U_{x}{y, c, d}}
U_{x}d  {x, y, {x, y, d}} = {x, d, U_{y}x}  {y, c, {x, y, {x, c, d}}}
U_{y}x  = {y, d, U_{x}y}  {y, c, {x, d, {x, y, c}}}
U_{y}c  {x, y, {x, c, d}}  {y, c, {y, d, U_{x}c}}
U_{y}d  {x, y, {y, c, d}}  {y, U_{x}c, {y, c, d}}
{x, y, c}  {x, c, U_{y}d}  {y, {x, y, c}, {x, c, d}}
{x, y, d}  {x, c, {x, y, d}}  {c, d, U_{x}U_{y}c}
{x, c, d}  {x, d, U_{y}c}  {c, d, U_{y}U_{x}c}
{y, c, d}  {x, d, {x, y, c}}  {d, U_{x}c, U_{y}c}
{y, c, U_{x}d}
{y, c, {x, y, d}}
{y, d, U_{x}c}
{y, d, {x, y, c}}
{c, d, U_{x}y}
{c, d, U_{y}x}

ORDER 5 RELATIONS: U_{x}U_{y}c + U_{y}U_{x}c = {x, y, {x, y, c}} U_{x}U_{y}d + U_{y}U_{x}d = {x, y, {x, y, d}}

U_{x}{y, c, d} + {y, c, U_{x}d} = {x, d, {x, y, c}} {y, d, U_{x}c} + {c, d, U_{x}y} = {x, d, {x, y, c}}

U_{x}{y, c, d} + {y, d, U_{x}c} = {x, c, {x, y, d}} {y, c, U_{x}d} + {c, d, U_{x}y} = {x, c, {x, y, d}}

U_{x}{y, c, d} + {c, d, U_{x}y} = {x, y, {x, c, d}} {y, d, U_{x}c} + {y, c, U_{x}d} = {x, y, {x, c, d}}

U_{y}{x, c, d} + {x, c, U_{y}d} = {y, d, {x, y, c}} {x, d, U_{y}c} + {c, d, U_{y}x} = {y, d, {x, y, c}}

U_{y}{x, c, d} + {x, d, U_{y}c} = {y, c, {x, y, d}} {x, c, U_{y}d} + {c, d, U_{y}x} = {y, c, {x, y, d}}

U_{y}{x, c, d} + {c, d, U_{y}x} = {x, y, {y, c, d}} {x, d, U_{y}c} + {x, c, U_{y}d} = {x, y, {y, c, d}}

Before we prove the above claim, we first need to prove the following lemma.

Lemma 11 Let M̃ = M_{0} + ϕ̃_{1} M_{1}, Ñ = N_{0} + ϕ̃_{1} N_{1} with M_{0}, M_{1}, N_{0}, N_{1} ∈ 𝔐(J). Then

(i) U_{M̃(1), Ñ(c_{2})} = U_{M̃(c_{2}), Ñ(1)},

(ii) U_{M̃(c_{2}), Ñ(c_{2})} = μ_{2} U_{M̃(1), Ñ(1)},

(iii) U_{M̃(1), M̃(c_{2})} = ν_{2} U_{M̃(1)},

(iv) U_{M̃(c_{2})} = μ_{2} U_{M̃(1)} .

Proof Notice that we have the following relations for M, N ∈ 𝔐(J) because c_{2} = ϕ̃_{2} 1 ∈ J̃_{2} ⊃ J over a ring Φ̃_{2} ⊃ Φ :

U_{M(1), N(c_{2})} = U_{M(1), N(ϕ̃_{2} 1)} = U_{M(ϕ̃_{2} 1), N(1)} = U_{M(c_{2}), N(1)} ,

U_{ϕ̃_{1} M(1), N(c_{2})} = ϕ̃_{1} (U_{M(1), N(c_{2})}) = ϕ̃_{1} (U_{M(c_{2}), N(1)}) = U_{ϕ̃_{1} M(c_{2}), N(1)} ,

U_{M(1), ϕ̃_{1} N(c_{2})} = ϕ̃_{1} (U_{M(1), N(c_{2})}) = ϕ̃_{1} (U_{M(c_{2}), N(1)}) = U_{M(c_{2}), ϕ̃_{1} N(1)} ,

U_{ϕ̃_{1} M(1), ϕ̃_{1} N(c_{2})} = ϕ̃_{1}^{2} (U_{M(1), N(c_{2})}) = ϕ̃_{1}^{2} (U_{M(c_{2}), N(1)}) = U_{ϕ̃_{1} M(c_{2}), ϕ̃_{1} N(1)} ,

The above relations prove (i), while (ii) and (iii) are proved in a similar manner. For (iv), we compute the following:

U_{M(c_{2}) + ϕ̃_{1} N(c_{2})} = U_{M(c_{2})} + U_{M(c_{2}), ϕ̃_{1} N(c_{2})} + U_{ϕ̃_{1} N(c_{2})}
= U_{M(ϕ̃_{2} 1)} + ϕ̃_{1} (U_{M(c_{2}), N(c_{2})}) + ϕ̃_{1}^{2} U_{N(ϕ̃_{2} 1)}
= ϕ̃_{2}^{2} U_{M(1)} + ϕ̃_{1} (U_{M(ϕ̃_{2} 1), N(ϕ̃_{2} 1)}) + ϕ̃_{2}^{2} μ_{1} U_{N(1)}
= μ_{2} U_{M(1)} + ϕ̃_{1} (ϕ̃_{2}^{2} U_{M(1), N(1)}) + μ_{2} μ_{1} U_{N(1)}
= μ_{2} U_{M(1)} + ϕ̃_{1} (μ_{2} U_{M(1), N(1)}) + μ_{2} μ_{1} U_{N(1)}
= μ_{2} (U_{M(1)} + U_{M(1), ϕ̃_{1} N(1)} + U_{ϕ̃_{1} N(1)})
= μ_{2} U_{M(1) + ϕ̃_{1} N(1)} . ■

In this case, we can exhibit a counterexample.

Lemma 12 There exists a Jordan algebra J with ExtRad(J) = 0 for which C_{s}(J) is not closed under addition.

Proof Begin with J = F J({c_{1}, c_{2}, x, y, z}), the free unital quadratic Jordan algebra over Φ = ℤ_{2}(μ_{1}) with generators {c_{1}, c_{2}, x, y, z}, along with the following relations: if deg_{c_{1}} P ≥ 3, deg_{c_{2}} P ≥ 2 or (deg_{c_{1}} P ≥ 1 and deg_{c_{2}} P ≥ 1), then P = 0; U_{c_{1}} = μ_{1}, μ_{1}^{2} = 0, and V_{c_{1}} = U_{c_{2}} = V_{c_{2}} = 0. For the extensions, Φ̃_{1} = ℤ_{2}(ϕ̃_{1}) where ϕ̃_{1} 1 = c_{1}, ϕ̃_{1}^{2} = μ_{1} and ϕ̃_{1}^{3} = 0, while Φ̃_{2} = ℤ_{2}(μ_{1}, ϕ̃_{2}) where ϕ̃_{2} 1 = c_{2}, ϕ̃_{2}^{2} = 0 and μ_{1} ϕ̃_{2} = 0. By the above relations, any expression with degree 2 in c_{2} or containing both c_{1} and c_{2} vanishes. Also, an expression with degree 2 in c_{1} is either of the form M(x, y, z) U_{c_{1}} P (x, y, z) = μ_{2} M(x, y, z) P (x, y, z) or N(c_{1}, x, y, z) (c_{1}) = μ_{2} N′(x, y, z) (1) where P (x, y, z) ∈ F J({x, y, z}), M(x, y, z), N′(x, y, z) ∈ 𝔐 (F J({x, y, z})), and N(c_{1}, x, y, z) ∈ 𝔐 (F J({c_{1}, x, y, z})). Thus anything in J can be written in the form

P (x, y, z) + M(x, y, z)(c_{1}) + N(x, y, z)(c_{2})

where P (x, y, z) ∈ F J({x, y, z}) and M(x, y, z), N(x, y, z) are elements of 𝔐 (F J({x, y, z})). Now consider the ideal ⟨U_{x} (c_{1} + c_{2})⟩ in J. Since U_{U_{x} (c_{1} + c_{2})} = μ_{1} U_{x^{2}} and any operator in 𝔐(J) can be written in the form

M_{0}(x, y, z) + M_{1}(c_{1}, x, y, z) + M_{2}(c_{2}, x, y, z) ,

any element in ⟨U_{x} (c_{1} + c_{2})⟩ can be written in the form

[M_{0}(x, y, z) + M_{1}(c_{1}, x, y, z) + M_{2}(c_{2}, x, y, z)] U_{x} (c_{1} + c_{2})
+ N(x, y, z) U_{U_{x} (c_{1} + c_{2})} P (x, y, z)
= M_{0}(x, y, z) U_{x} (c_{1} + c_{2}) + μ_{1} M_{1}(1, x, y, z) x^{2}
+ μ_{1} N(x, y, z) U_{x^{2}} P (x, y, z)

where M_{1}(1, x, y, z) is linear in the factor 1 and cannot be U_{1}.

In J̄ = J / ⟨U_{x} (c_{1} + c_{2})⟩, \overline{U_{x} (c_{1} + c_{2})} = ¯0 but

\overline{U_{x} (c_{1} + c_{2})^{2}} = \overline{U_{x} (c_{1}^{2} + c_{1} ∘ c_{2} + c_{2}^{2})} = \overline{U_{x} (c_{1}^{2})} = \overline{μ_{1} x^{2}} ≠ ¯0 .

Clearly, if c_{1} + c_{2} = ϕ 1, then (c_{1} + c_{2})^{2} = ϕ^{2} 1 and

\overline{U_{x} (c_{1} + c_{2})^{2}} = \overline{U_{x} (ϕ^{2} 1)} = \overline{ϕ U_{x} (ϕ 1)} = \overline{ϕ U_{x} (c_{1} + c_{2})} = \overline{ϕ 0} = ¯0 .

In order to exhibit a counterexample, we need to show that U_{x} c_{1} and μ_{1} x^{2} are not in ExtRad(J̄). Then the algebra J̄/ ExtRad(J̄) would have no extreme elements but still possess this undesirable property. It is easy to see that (U_{x} c_{1}) ∘ y = {U_{x} c_{1}, 1, y} = {x^{2}, c_{1}, y} and U_{μ_{1} x^{2}, z} y = μ_{1} {x^{2}, y, z} are not of the form

\begin{matrix}M_{0}(1, x, y, z) U_{x} (c_{1} + c_{2}) + μ_{1} M_{1}(1, x, y, z) x^{2} \\ + μ_{1} N(1, x, y, z) U_{x^{2}} P (1, x, y, z) . \end{matrix}

Since U_{x} c_{1} ∘ y, U_{μ_{1}x^{2}, z} y ∉ ⟨U_{x} (c_{1} + c_{2})⟩, \overline{U_{x} c_{1}} and \overline{μ_{1} x^{2}} are not extreme elements of J̄, and we are done. ■

It seems possible that an algebra with zero extreme radical could have elements x ∈ J, c_{1}, c_{2} ∈ C_{s}(J) with c_{1}^{2} = μ_{1} 1 ≠ 0, c_{2}^{2} = c_{1} ∘ c_{2} = 0 and U_{x} (c_{1} + c_{2}) = 0 even though

U_{x} (c_{1} + c_{2})^{2} = U_{x} (c_{1}^{2} + c_{1} ∘ c_{2} + c_{2}^{2}) = U_{x} (c_{1}^{2}) = μ_{1} x^{2} ≠ 0 . Clearly, if c_{1} + c_{2} = ϕ 1, then (c_{1} + c_{2})^{2} = ϕ^{2} 1 and

U_{x} (c_{1} + c_{2})^{2} = U_{x} (ϕ^{2} 1) = ϕ U_{x} (ϕ 1) = ϕ U_{x} (c_{1} + c_{2}) = ϕ 0 = 0 . It is relatively easy to construct an algebra with these properties, but the extreme radical is not zero.