Jordan Pairs and Hopf Algebras

John R. Faulkner Department of Mathematics University of Virginia Kerchof Hall Charlottesville VA 22903 USA

Abstract A (quadratic) Jordan pair is constructed from a ℤ-graded Hopf algebra having divided power sequences over all primitive elements and with three terms in the ℤ-grading of the primitive elements. The notion of a divided power representation of a Jordan pair is introduced and the universal object is shown to be a suitable Hopf algebra. This serves a replacement for the Tits-Kantor-Koecher construction.

1 Introduction

A major tool in transferring results between Jordan theory and Lie theory is the Tits-Kantor-Koecher (TKK) construction of a Lie algebra. This construction was first formulated for linear Jordan algebras and was modified as Jordan theory expanded to quadratic Jordan algebras, Jordan triple systems, and (quadratic) Jordan pairs. The reverse construction starts with a three-term ℤ-graded Lie algebra and recovers a Jordan structure. However, since the Lie algebra involves only linear operations, it is impossible to recover the quadratic operations unless the base ring contains \frac{1}{2}. The main purpose of this paper is to give constructions which substitute for the TKK construction and its reverse construction and which work for the quadratic operations for all base rings. The role of the Lie algebra is replaced by a certain kind of Hopf algebra. The primitive elements form a three-term ℤ-graded Lie algebra, but the Hopf algebra also contains divided power sequences which capture the quadratic properties of the Jordan pair.

In Sections 2 and 3, we review the basic properties of Hopf algebras and Jordan pairs, including the TKK construction. In Section 4, we show that starting with a ℤ-graded Hopf algebra with divided power sequences over all primitive elements and with a three-term ℤ-grading of the primitive elements, we can construct a (quadratic) Jordan pair. In Section 5, we introduce the notion of a divided power representation of a Jordan pair. This gives a new way of imbedding a Jordan pair within an associative algebra which generalizes the usual imbedding of a special Jordan pair, but which works for any Jordan pair via the TKK construction. The universal object for divided power representations is a ℤ-graded cocommutative Hopf algebra with the required divided power sequences. However, it takes considerably more work to show that the ℤ-grading of the primitive elements has only three terms.

In Section 6, we show that any divided power representation has the “exponential property,” which states that a certain product of exponentials is “diagonal.” The rather involved induction proof also requires a Poincaré-Birkhoff-Witt type of result for the universal divided power representation. As a corollary, one get that the ℤ-grading of the primitive elements has only three terms. In Section 7, we show, in characteristic zero, that the universal divided power representation of a Jordan pair whose grading derivation is inner coincides with the universal enveloping algebra of the universal cover of a Lie algebra given by a TKK construction. Finally, in Appendix A, we give an independent treatment of homogeneous maps and their linearizations, which we use throughout the paper.

2 Basics on Hopf algebras

We recall first some basic concepts and results about Hopf algebras. For details not given here, see Abe [1]. Associated with an algebra A with unit 1 over a unital, commutative associative ring Φ are the linear maps μ : A ⊗ A → A and η : k → A given by μ(a ⊗ b) = ab and η(λ) = λ1. Similarly, a coalgebra A has linear maps Δ : A → A ⊗ A (coproduct) and ε : A → Φ (counit) satisfying

(Id ⊗ ε) ∘ Δ = Id = (ε ⊗ Id) ∘ Δ

where we have identified A ⊗ Φ and Φ ⊗ A with A Unless otherwise stated, we will assume that an algebra is associative and that a coalgebra is coassociative; i.e.

(Id ⊗ Δ) ∘ Δ = (Δ ⊗ Id) ∘ Δ.

Also, a coalgebra is cocommutative if

τ ∘ Δ = Δ

where τ(a⊗b) = b⊗a. If A is both an (associative) algebra and a (coassociative) coalgebra, and both Δ and ε are Φ-algebra homomorphisms, then we say that A is a bialgebra. If A is a bialgebra, then a ∈ A is a group-like element if

ε(a) = 1 and Δ(a) = a ⊗ a

and a ∈ A is a primitive element if

Δ(a) = a ⊗ 1 + 1 ⊗ a.

Note if a is primitive, then a = (ε⊗Id)(Δ(a)) = ε(a)1+a implies (η ∘ε)(a) = 0. The set P(A) of all primitive elements forms a Lie algebra under [ab] = ab − ba.

A finite or infinite sequence (1 = a_{0}, a_{1}, a_{2}, a_{3}, ...) of elements from a bialgebra A is a divided power sequence (d.p.s.) if

Δ(a_{n}) = ∑_{i+j=n} a_{i} ⊗ a_{j}.

It is clear that an infinite sequence (a_{0}, a_{1}, a_{2}, a_{3}, ...) is a d.p.s. if and only if each finite sequence (a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n}) is a d.p.s. Note also that a is primitive if and only if (1, a) is a d.p.s. If a is primitive, we say that a d.p.s. (1, a, a_{2}, a_{3}...) is a d.p.s. over a. We shall assume from now on that a d.p.s is infinite unless otherwise stated.

An antipode S : A → A on a bialgebra A is a linear map satisfying

μ ∘ (S ⊗ Id) ∘ Δ = μ ∘ (Id ⊗ S) ∘ Δ = η ∘ ε.

We say a bialgebra with an antipode is a Hopf algebra. If a is primitive, we see that 0 = (η ∘ ε)(a) = S(a) + a, so S(a) = −a.

Lemma 1 If A is a Hopf algebra, then (i) S(ab) = S(b)S(a) for a, b ∈ A, (ii) ε ∘ S = ε, S ∘ η = η, (iii) Δ ∘ S = τ ∘ (S ⊗ S) ∘ Δ, (iv) the set G(A) of all group-like elements of A is a subgroup of the group

of units of A.

Proof. The first three statements are proven in [1] Theorem 2.1.4. For (iv), we note a, b ∈ G(A) implies that Δ(ab) = Δ(a)Δ(b) = ab ⊗ ab and ε(ab) = ε(a)ε(b) = 1, so ab ∈ G(A). Also, by (iii), we see that S(a) ∈ G(A), and by the defining condition for an antipode that S(a)a = aS(a) = η(ε(a)) = 1, so S(a) is the inverse of a.

Lemma 2 If A is a Hopf algebra and if A = (a_{0}, a_{1}, a_{2}, ...) and B = (b_{0}, b_{1}, b_{2}, ...) are divided power sequences, then so is C = (c_{0}, c_{1}, c_{2}, ...) where

c_{n} = ∑_{i+j=n}a_{i}b_{j}.

Also, A = (a_{0}, a_{1}, a_{2}, ...) is a d.p.s. if and only if D = (d_{0}, d_{1}, d_{2}, ...) is a d.p.s. where

d_{n} = \{ \begin{matrix}a_{r} & if n = kr \\ 0 & if n ≠ kr. \end{matrix}

Proof. We have

Δ(c_{n}) = ∑_{i+j=n}Δ(a_{i})Δ(b_{j})

= ∑_{p+q+g+h=n} a_{p}b_{g} ⊗ a_{q}b_{h}

= ∑_{k+l=n} c_{k} ⊗ c_{l}.

Also,

∑_{p+q=n} d_{p} ⊗ d_{q} = ∑_{ki+kj=n} a_{i} ⊗ a_{j}

= \{ \begin{matrix}∑_{i+j=r} a_{i} ⊗ a_{j} & if n = kr \\ 0 & if n ≠ kr \end{matrix} .

3 Basics on Jordan pairs

For more details about Jordan pairs see Loos [5]. Let Φ be a unital, commutative associative ring and let V = (V^{+}, V^{−}) be a pair of Φ-modules. Assume that there are compositions

(x, y) → Q_{x}(y) = Q(x)(y) ∈ V^{σ}

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

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

and get 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}.

For an extension ring Ω of Φ, we can form

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

and extend the compositions (x, y) → Q_{x}(y) to V_{Ω}^{σ} × V_{Ω}^{−σ} (see Appendix A). We say that V = (V^{+}, V^{−}) is a Jordan pair if

D_{x,y}Q_{x} = Q_{x}D_{y,x}  (JP1)
D(Q_{x}(y), y) = D(x, Q_{y}(x))  (JP2)
Q(Q_{x}(y)) = Q_{x}Q_{y}Q_{x}  (JP3)

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. It is clear that if V is a Jordan pair then so V_{Ω}. Moreover, using Theorem 35, we can extend the compositions (x, y) → Q_{x}(y) to V^{σ}[[s]] × V^{−σ}[[s]]. Using density arguments as in proof of Theorem 35, we can show that (JP1)-(JP3) and their linearizations hold for V[[s]]; i.e., V[[s]] is a Jordan pair.

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

D_{σ}Q_{x} = Q_{x}D_{−σ} + Q(D_{σ}(x), x).  (1)

It is easy to check that the set Der(V) of all derivations forms a Lie algebra. Moreover,

δ_{x,y} = (D_{x,y}, −D_{y,x}) ∈ Der(V)

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

If x ∈ V^{σ}, y ∈ V^{−σ}, we set

B_{x,y} = Id − D_{x,y} + Q_{x}Q_{y}.

We say that (x, y) is quasi-invertible if B_{x,y} is invertible and call

x^{y} = B_{x,y}^{−1}(x − Q_{x}y)

the quasi-inverse of (x, y). If (x, y) ∈ V^{+} × V^{−} is quasi-invertible, then so is (y, x) and

β_{x,y} = (B_{x,y}, B_{y,x}^{−1})

is an automorphism of V.

If D is any Lie subalgebra of Der(V) containing Inder(V), the Tits-Kantor-Koecher Lie algebra is

T KK(V, D) = V^{+} ⊕ D ⊕ V^{−}  (2)

with multiplication

[V^{σ}, V^{σ}] = 0, [D, z] = D_{σ}(z), [x, y] = −δ_{x,y}

for D = (D_{+}, D_{−}) ∈ D, x ∈ V^{+}, y ∈ V^{−}, and z ∈ V^{σ}. We shall generally use D = D_{0} = Φζ + Inder(V) where

ζ = (Id_{V^{+}}, −Id_{V^{−}}).

Using the decomposition (2) of T KK(V, D_{0}), we can define the automorphisms

exp_{+}(x) = ( \begin{matrix}1 & ad x & Q_{x} \\ 0 & 1 & ad x \\ 0 & 0 & 1 \end{matrix} ) , exp_{−}(y) = ( \begin{matrix}1 & 0 & 0 \\ ad y & 1 & 0 \\ Q_{y} & ad y & 1 \end{matrix} )

of T KK(V, D_{0}). Also, for h = (h_{+}, h_{−}) ∈ Aut(V), we have the automorphism

h̃ = ( \begin{matrix}h_{+} & 0 & 0 \\ 0 & h_{0} & 0 \\ 0 & 0 & h_{−} \end{matrix} )

of T KK(V, D_{0}), where h_{0} is conjugation by h acting on D_{0}. From Loos [6], we have Theorem 3 Let V be a Jordan pair. If x ∈ V^{+}, y ∈ V^{−}, then (x, y) is quasiinvertible if and only if there are z ∈ V^{+}, w ∈ V^{−}, and h ∈ Aut(V) with

exp_{+}(x) exp_{−}(y) = exp_{−}(w) h̃ exp_{+}(z).

In this case,

z = x^{y}, w = y^{x}, and h = β_{x,y}.

4 From Hopf algebras to Jordan pairs

We now assume that A is a ℤ-graded Hopf algebra over Φ. This means that

A = ⨁_{n∈ℤ} A_{n} with A_{i}A_{j} ⊂ A_{i+j}

and that Δ, ε, and S are graded homomorphisms using the gradings

(A ⊗ A)_{n} = ∑_{i+j=n} A_{i} ⊗ A_{j},
Φ_{0} = Φ,
(A^{op})_{n} = A_{n}.

We note that if x ∈ P(A) and x = ∑_{i} x_{i} with x_{i} ∈ A_{i}, then

∑_{i} Δ(x_{i}) = ∑_{i}(x_{i} ⊗ 1 + 1 ⊗ x_{i}).

Comparing terms in (A ⊗ A)_{n} shows that x_{n} ∈ P(A). Thus,

P(A) = ⨁_{n∈ℤ} P_{n} where P_{n} = P(A) ∩ A_{n}

and P(A) is a ℤ-graded Lie algebra.

We say that a d.p.s (a_{0}, a_{1}, a_{2}, a_{3}, ...) is homogeneous if for some k each a_{n} ∈ A_{kn}.

Lemma 4 Let A be a ℤ-graded Hopf algebra.

(i) If there is a d.p.s. over every x ∈ P(A), then there is a homogeneous d.p.s. over every x ∈ P_{k}.

(ii) If P(A) = P_{−1}⊕P_{0}⊕P_{1} and there is a homogeneous d.p.s. over x ∈ P_{±1}, then it is unique.

Proof. Suppose that (x_{0}, x_{1}, x_{2}, x_{3}, ...) is a d.p.s. over x ∈ P_{k}, and that x_{n} ∈ A_{kn}, for all n < m. Write x_{m} = ∑ a_{p} with a_{p} ∈ A_{p}. Comparing the km-components of

Δ(x_{m}) = ∑_{i+j=m} x_{i} ⊗ x_{j},

we get

Δ(a_{km}) = a_{km} ⊗ 1 +_{∑_{i+j=m} ; i,j≠0}x_{i} ⊗ x_{j} + 1 ⊗ a_{km}.

Letting z = a_{km} − x_{m}, we see that Δ(z) = z ⊕ 1 + 1 ⊗ z, i.e. z ∈ P(A). Let (z_{0}, z_{1}, z_{2}, z_{3}, ...) be a d.p.s. over z. By Lemma 2, we have (w_{0}, w_{1}, w_{2}, w_{3}, ...) is a d.p.s. where w_{n} = ∑_{i+mj=n} x_{i}z_{j}. Now w_{n} = x_{n} for n < m and w_{m} =

x_{m} + z_{1} = a_{km}. Induction on m shows (i). Suppose that (x_{0}, x_{1}, x_{2}, x_{3}, ...) and (x′_{0}, x′_{1}, x′_{2}, x′_{3}, ...) are homogeneous d.p.s. over x ∈ P_{±1}, and that x′_{n} = x_{n} for all n < m. We see that x′_{m} − x_{m} ∈ P_{±m} = 0, provided m > 1. Induction on m shows (ii).

Theorem 5 If A is a ℤ-graded Hopf algebra, such that

(i) P(A) = P_{−1} ⊕ P_{0} ⊕ P_{1},

(ii) there is a homogeneous divided power sequence (x^{(0)}, x^{(1)}, x^{(2)}, x^{(3)}, ...) over x ∈ P_{σ}, σ = ±1,

then (P_{1}, P_{−1}) is a Jordan pair with Q_{x}(y) = x^{(2)}y − xyx + yx^{(2)}.

Proof. Suppose that x, z ∈ P_{σ}, σ = ±1, and λ ∈ Φ. Since (..., λ^{n}x^{(n)}, ...) is the unique homogeneous d.p.s. over λx, we see that

(λx)^{(n)} = λ^{n}x^{(n)}.

Similarly, we get

(x + z)^{(n)} = ∑_{i+j=n} x^{(i)}z^{(j)},

using Lemma 2, and

S(x^{(n)}) = (−1)^{n}x^{(n)},

using the fact that S(x) = −x. Define

A = m ∘ (Id ⊗ S) ∘ Δ

where m : A ⊗ A^{op} → End(A) is the homomorphism given by

m(a ⊗ b) = ℓ_{a} r_{b}

where

ℓ_{a}(b) = ab = r_{b}(a).

We see that A : a → A_{a} is an algebra homomorphism A → End(A). We shall show if x ∈ P_{±1}, then

Δ ∘ A_{x^{(n)}} = ( ∑_{i+j=n} A_{x^{(i)}} ⊗ A_{x^{(j)}}) ∘ Δ.  (3)

We compute

Δ(A_{x^{(n)}}(c)) = Δ( ∑_{i+j=n} x^{(i)}c S(x^{(j)}))

= ∑_{i+j=n} Δ(x^{(i)})Δ(c)Δ(S(x^{(j)}))

= T (Δ(c))

where

T (a ⊗ b) = ∑_{i+j=n} Δ(x^{(i)})(a ⊗ b)Δ(S(x^{(j)}))

= ∑_{p+q+r+s=n}(x^{(p)} ⊗ x^{(q)})(a ⊗ b)(S(x^{(r)}) ⊗ S(x^{(s)}))

= ∑_{p+q+r+s=n} x^{(p)}aS(x^{(r)}) ⊗ x^{(q)}bS(x^{(s)})

= ∑_{l+k=n}(A_{x^{(l)}} ⊗ A_{x^{(k)}})(a ⊗ b)

showing (3). We next show for x ∈ P_{±1} that

A_{x^{(n)}}(P) ⊂ P  (4)

Indeed, since

A_{x^{(n)}}(1) = ∑_{i+j=n} x^{(i)}S(x^{(j)}) = (x + S(x))^{(n)}  (5)

= 0^{(n)} = \{ \begin{matrix}1 & if n = 0 \\ 0 & if n ≠ 0, \end{matrix}

we have by (3) for a ∈ P,

Δ(A_{x^{(n)}}(a)) = ( ∑_{i+j=n} A_{x^{(i)}} ⊗ A_{x^{(j)}})(a ⊗ 1 + 1 ⊗ a)
= A_{x^{(n)}}(a) ⊗ 1 + 1 ⊗ A_{x^{(n)}}(a).

Now let x ∈ P_{σ}, y ∈ P_{−σ} where σ = ±1. We shall verify that

A_{x^{(k)}}(y^{(l)}) = \{ \begin{matrix}0 & if k > 2l \\ (A_{x^{(2)}}(y))^{(l)} & if k = 2l. \end{matrix}  (6)

We prove the case k > 2l by induction on l. If l = 0, this follows from (5). In general, by (3),

Δ(A_{x^{(k)}}(y^{(l)})) = _{∑_{i+j=k} ; p+q=l}A_{x^{(i)}}(y^{(p)}) ⊗ A_{x^{(j)}}(y^{(q)}).  (7)

Since k > 2l, then either i > 2p or j > 2q. By the induction hypothesis, in the first case A_{x^{(i)}}(y^{(p)}) = 0 for p < l, and in the second case A_{x^{(j)}}(y^{(q)}) = 0 for q < l. Moreover, if p = l, then q = 0 and j > 2q unless j = 0. Similarly, if q = l, then i > 2p unless i = 0. Thus for k > 2l, (7) reduces to

Δ(A_{x^{(k)}}(y^{(l)})) = A_{x^{(k)}}(y^{(l)}) ⊗ 1 + 1 ⊗ A_{x^{(k)}}(y^{(l)}),

so A_{x^{(k)}}(y^{(l)}) ∈ P_{σ(k−l)} = 0, since k − l > l ≥ 1. We now consider (7) in the case k = 2l. Since the terms with either i > 2p or j > 2q are 0 by the first case, we are left with the terms with i = 2p and j = 2q, i.e.

Δ(A_{x^{(2l)}}(y^{(l)})) = ∑_{p+q=l} A_{x^{(2p)}}(y^{(p)}) ⊗ A_{x^{(2q)}}(y^{(q)}).

We see that (..., A_{x^{(2l)}}(y^{(l)}), ...) is the unique homogeneous d.p.s. over A_{x^{(2)}}(y), so A_{x^{(2l)}}(y^{(l)}) = (A_{x^{(2)}}(y)^{(l)}.

We now endow the pair (P_{−1}, P_{1}) with the composition Q_{x}(y) = A_{x^{(2)}}(y) for x ∈ P_{σ}, y ∈ P_{−σ}. Note that

Q_{x}(y) = x^{(2)}y − xyx + yx^{(2)},

since S(x^{(k)}) = (−1)^{k}x^{(k)}. Clearly, this is linear in y and quadratic in x. We have

(x + z)^{(2)} = x^{(2)} + xz + z^{(2)},

for x, z ∈ P_{δ}, so in particular, xz = zx, i.e. [xz] = 0. Since A is a homomorphism, we see that

Q_{x,z}(y) = A_{xz}(y) = A_{x}A_{z}(y).

Since A_{z}(y) = zy − yz = −A_{y}(z), we get

D_{x,y}(z) = Q_{x,z}(y) = −A_{x}A_{y}(z) = −[x[yz]] = −[[xy]z].

Applying A to

0 = A_{x^{(3)}}(y) = x^{(3)}y − x^{(2)}yx + xyx^{(2)} − yx^{(3)}

yields

A_{x^{(3)}}A_{y} − A_{x^{(2)}}A_{y}A_{x} + A_{x}A_{y}A_{x^{(2)}} − A_{y}A_{x^{(3)}} = 0,

while applying this to w ∈ P_{−δ}, and using A_{y}(w) = 0 = A_{x^{(3)}}(w) gives

Q_{x}D_{y,x} − D_{x,y}Q_{x} = 0

which is (JP1). Since yy = (y + y)^{(2)} − 2y^{(2)} = 2y^{(2)}, we can compute

[Q_{x}(y), y] = [x^{(2)}y − xyx + yx^{(2)}, y]
= 2x^{(2)}y^{(2)} − xyxy + yxyx − 2y^{(2)}x^{(2)}
= −[Q_{y}(x), x] = [x, Q_{y}(x)].

Thus, D(Q_{x}y, y) = D(x, Q_{y}(x)), which is (JP2). From (6), we have

(Q_{x}(y))^{(2)} = A_{x^{(4)}}(y^{(2)})
= x^{(4)}y^{(2)} − x^{(3)}y^{(2)}x + x^{(2)}y^{(2)}x^{(2)} − xy^{(2)}x^{(3)} + y^{(2)}x^{(4)}.

Applying A, we get Q_{Q_{x}(y)} = Q_{x}Q_{y}Q_{x}, which is (JP3).

It remains to show that (JP1)-(JP3) hold in any extension (P_{1}, P_{−1})_{Ω}. This is the same as saying that all linearizations of (JP1)-(JP3) are valid in (P_{1}, P_{−1}). Thus, it would suffice to consider just the scalar extension Ω = Φ[λ_{1}, λ_{2}, ...] for indeterminants λ_{i}. However, it will be convenient to allow any extension Ω which is a free Φ-module. Identifying A_{Ω} ⊗_{Ω} A_{Ω} with Ω ⊗_{Φ} A ⊗_{Φ} A = (A ⊗_{Φ} A)_{Ω}, we

can extend the ℤ-graded Hopf algebra structure to A_{Ω}. We can write P(A) = ker(Δ − ξ) where ξ(a) = a ⊗ 1 + 1 ⊗ a, so P(A_{Ω}) = ker(Id ⊗ (Δ − ξ)). It is easy to see that if T : V → W is a linear map of Φ-modules and Ω an extension of Φ which is a free Φ-module, then for Id ⊗ T : V_{Ω} → W_{Ω}, we have

ker(Id ⊗ T) = Ω ⊗ ker(T) = ker(T)_{Ω}.

Thus,

P(A_{Ω}) = P(A)_{Ω} = (P_{−1})_{Ω} ⊕ (P_{0})_{Ω} ⊕ (P_{1})_{Ω}
= (P(A_{Ω}))_{−1} ⊕ (P(A_{Ω}))_{0} ⊕ (P(A_{Ω}))_{1}.

Moreover, in any ℤ-graded Hopf algebra B, if

(x^{(0)}, x^{(1)}, x^{(2)}, x^{(3)}, ...) and (z^{(0)}, z^{(1)}, z^{(2)}, z^{(3)}, ...)

are homogeneous d.p.s. with x^{(n)}, z^{(n)} ∈ B_{kn}, then the same holds for (..., λ^{n}x^{(n)}, ...) and (..., ∑_{i+j=n}x^{(i)}z^{(j)}, ...). Since (ii) holds in A_{Ω} for x ∈ 1 ⊗ P_{σ} and since the

Ω-span of 1 ⊗ P_{σ} is (P(A_{Ω}))_{σ}, we see that (ii) holds for A_{Ω}. Thus, (JP1)-(JP3) hold for (P_{1}, P_{−1})_{Ω}, as desired.

5 Divided power representations of Jordan pairs

We shall now introduce another notion of divided powers, not involving a coproduct, although we shall eventually see a connection between the concepts.

Let V be a Φ-module and A a unital (associative) algebra over Φ. We say that a sequence ρ = (ρ_{0}, ρ_{1}, ρ_{2}, ...) of maps ρ_{n} : V → A is a sequence of binomial divided power (b.d.p.) maps provided ρ_{0}(v) = 1 and ρ_{n} is a homogeneous map of degree n whose (i, j)-linearization is (u, v) → ρ_{i}(u)ρ_{j}(v) (see Appendix A). We shall sometimes suppress the mention of ρ, and write simply v → v^{(n)} for a sequence of b.d.p. maps.

Lemma 6 If ρ is a sequence of maps ρ_{n} : v → v^{(n)} with v^{(0)} = 1, then ρ is a sequence of b.d.p. maps if and only if for u, v ∈ V

(i) (λv)^{(n)} = λ^{n}v^{(n)},

(ii) (v + u)^{(n)} = ∑_{k+l=n} v^{(k)}u^{(l)},

(iii) [v^{(k)}, u^{(l)}] = 0,

(iv) v^{(k)}v^{(l)} = \binom{k + l}{k}v^{(k+l)},

Proof. Conditions (A1), (A2), (A5), and (A4) of Appendix A reduce to (i)-(iv). (A6) is trivial, while (A3) follows by induction.

Corollary 7 If ρ is a sequence of maps ρ_{n} : v → v^{(n)} with v^{(0)} = 1, then ρ is a sequence of b.d.p. maps if and only if for each extension Ω of Φ, there is an extension ρ′_{n} : V_{Ω} → A_{Ω} satisfying

(i) (λv)^{(n)} = λ^{n}v^{(n)},

(ii) (v + u)^{(n)} = ∑_{k+l=n} v^{(k)}u^{(l)},

for all v ∈ V_{Ω}.

Proof. Using Ω = Φ[t], and comparing coefficients of t^{l} in

∑_{k+l=n} t^{l}v^{(k)}u^{(l)} = (v + tu)^{(n)}

= (tu + v)^{(n)}

= ∑_{k+l=n} t^{l}u^{(l)}v^{(k)}

gives (iii). Similarly, (iv) follows from

(1 + t)^{n}v^{(n)} = (v + tv)^{(n)}

= ∑_{k+l=n} t^{l}v^{(k)}v^{(l)}.

The name “divided power” is explained by the following lemma.

Lemma 8 If ρ_{n} : v → v^{(n)} is a sequence of b.d.p. maps from V to A, then ρ_{1} is a linear map and

k! v^{(k)} = (v^{(1)})^{k}.

Conversely, if ℚ ⊂ Φ and φ : V → A is a linear map such that [φ(V), φ(V)] = 0, then

v^{(n)} = \frac{1}{n!}(φ(v))^{n}

defines a sequence of b.d.p. maps.

Proof. Since ρ_{1} is homogeneous of degree 1, it is linear. By Lemma 6(iv), v^{(k)}v^{(1)} = (k + 1)v^{(k+1)}, so k! v^{(k)} = (v^{(1)})^{k} follows by induction. For the converse, in any extension A_{Ω}, we can use the binomial theorem to write

(φ(v) + φ(u))^{n} = ∑_{k+l=n} \binom{n}{k}(φ(v))^{k}(φ(u))^{l},

so

(v + u)^{(n)} = ∑_{k+l=n} v^{(k)}u^{(l)},

as desired.

The kernel of a sequence of b.d.p. maps ρ_{k} : v → v^{(k)} is ker ρ = {v ∈ V : v^{(k)} = 0 for all k > 0} (see Appendix A). We say that a sequence of b.d.p. maps ρ_{k} : v → v^{(k)} is nilpotent of index n if for all k ≥ n, ρ_{k} and all of its linearizations are 0.

Lemma 9 Let ρ be a sequence of maps ρ_{k} : v → v^{(k)} and let m = 2 or 3. Assume that v^{(0)} = 1 and v^{(k)} = 0 for v ∈ V and k ≥ m. The sequence ρ is a sequence of b.d.p. maps nilpotent of index m if and only if for v, u, w ∈ V,

(i) for m = 2,

ρ_{1} is linear and v^{(1)}u^{(1)} = 0,  (8)

(ii) for m = 3,

ρ_{1} is linear and ρ_{2} is quadratic, (v + u)^{(2)} = v^{(2)} + v^{(1)}u^{(1)} + u^{(2)}, v^{(2)}u^{(1)} = u^{(1)}v^{(2)} = v^{(2)}u^{(2)} = v^{(1)}u^{(1)}w^{(1)} = 0.

(9)

Proof. We first note that if v^{(2)} = 0 for v ∈ V, then (ii) reduces to (i). Thus, it suffices to consider just the case m = 3. If ρ is a sequence of b.d.p. maps nilpotent of index 3, then linearizations of v^{(3)} = 0 and v^{(4)} = 0 show that (9) holds. Conversely, (9) shows that Lemma 6(i) holds. Both sides of Lemma 6(ii) are 0 if n > 4, for then either k > 2 or l > 2. Also, the cases n = 0, 1 are trivial. Thus, Lemma 6(ii) reduces to

(v + u)^{(2)} = v^{(2)} + v^{(1)}u^{(1)} + u^{(2)},
(v + u)^{(3)} = 0 = v^{(2)}u^{(1)} + v^{(1)}u^{(2)},
(v + u)^{(4)} = 0 = v^{(2)}u^{(2)},

which hold by (9). We have v^{(k)}u^{(l)} = 0 for k + l ≥ 3, so Lemma 6(iii) and (iv) reduce to

v^{(1)}u^{(1)} = u^{(1)}v^{(1)},
v^{(1)}v^{(1)} = 2v^{(2)},

which are easy consequences of (9). Thus, by Lemma 6, ρ is a sequence of b.d.p. maps. It is easy to see that all linearizations of ρ_{k} are 0 for k ≥ 3, so ρ is nilpotent of index 3.

Suppose that ρ_{k} : v → v^{(k)} is a nilpotent sequence of b.d.p. maps. We can form

exp_{ρ}(v) = ∑_{k=0}^{∞} v^{(k)}

as an element of A. More generally, for any sequence of b.d.p. maps ρ_{k} : v → v^{(k)}, we can form

exp_{ρ}(tv) = ∞ ; ∑_{k=0}v^{(k)}t^{k}

as an element of the formal power series algebra A[[t]]. Since

exp_{ρ}(tv + tu) = ∑_{k=0}^{∞}(v + u)^{(k)}t^{k}

= ∑_{k=0}^{∞} ∑_{i+j=k} v^{(i)}u^{(j)}t^{k}

= exp_{ρ}(tv) exp_{ρ}(tu),

we see that v → exp_{ρ}(tv) is a group homomorphism from V to the group of units of A[[t]] with kernel ker ρ. Similarly, if ρ is nilpotent, then v → exp_{ρ}(v) is a group homomorphism from V to the group of units of A.

Lemma 10 If ρ_{n} : v → v^{(n)} is a sequence of b.d.p. maps from V to A, then ρ̂_{n} : v → ad_{v}^{(n)} is a sequence of b.d.p. maps from V to End_{Φ}(A) where

ad_{v}^{(n)}(a) = ∑_{k+l=n} v^{(k)}a(−v)^{(l)}.

ad_{v}^{(n)}(ab) = ∑_{k+l=n} ad_{v}^{(k)}(a) ad_{v}^{(l)}(b).

Proof. It is easy to see that ad_{v}^{(0)} = Id and that Lemma 6(i),(iii) hold for

ρ̂_{n}. Now,

ad_{v+u}^{(n)}(a) = ∑_{k+l=n}(v + u)^{(k)}a(−u − v)^{(l)}

= ∑_{p+q+r+s=n} v^{(p)}u^{(q)}a(−u)^{(r)}(−v)^{(s)}

= ∑_{i+j=n} ad_{v}^{(i)}(ad_{u}^{(j)}(a))

so Lemma 6(ii) holds. Passing to A_{Ω} with Ω = Φ[t] and replacing u by tv, we also get Lemma 6(iv), so ρ̂ is a sequence of b.d.p. maps. We next compute

∑_{k+l=n} ad_{v}^{(k)}(a) ad_{v}^{(l)}(b) = ∑_{p+q+r+s=n} v^{(p)}a(−v)^{(q)}v^{(r)}bv^{(s)}

= ∑_{p+s=n} v^{(p)}abv^{(s)}

= ad_{v}^{(n)}(ab),

since

∑_{q+r=t}(−v)^{(q)}v^{(r)} = (−v + v)^{(t)}

= 0^{(t)} = \{ \begin{matrix}1 & if t = 0 \\ 0 & if t ≠ 0. \end{matrix}

Let V = (V^{+}, V^{−}) be a Jordan pair. A divided power representation (d.p. representation) of V is a pair ρ = (ρ^{+}, ρ^{−}) of sequences of b.d.p. maps ρ_{n}^{σ} : v → v^{(n)} from V^{σ} to A such that for all extensions Ω of Φ and for all x ∈ V_{Ω}^{σ}, y ∈ V_{Ω}^{−σ},

ad_{x}^{(k)}(y^{(l)}) = \{ \begin{matrix}0 & for k > 2l, \\ (Q_{x}(y))^{(l)} & for k = 2l. \end{matrix}  (10)

Clearly, this is equivalent to having (10) and all linearizations hold in V.

Let ρ be a d.p. representation of the Jordan pair V in A. We set ker ρ = (ker ρ^{+}, ker ρ^{−}) and observe that ker ρ is an ideal of V. Indeed, for l > 0,

(Q_{x}(y))^{(l)} = ∑_{i+j=2l} x^{(i)}y^{(l)}(−x)^{(j)}

= 0 if either x ∈ ker ρ^{σ} or y ∈ ker ρ^{−σ}. Also, the (l, l)-linearization of (10) gives

ad_{x}^{(l)}ad_{z}^{(l)}(y^{(l)}) = ∑_{2m+n=l}(Q_{x}(y))^{(m)}(Q_{z}(y))^{(m)}{xyz}^{(n)}

which reduces to 0 = {xyz}^{(l)} if x ∈ ker ρ^{σ}. We say that the d.p. representation ρ is faithful if ker ρ = (0, 0). Also, if both ρ^{+} and ρ^{−} are nilpotent of index n, we say that ρ has index n.

Example 1. The maps ψ_{n}^{σ} : v → \{ \begin{matrix}1 & if n = 0 \\ 0 & if n ≥ 1 \end{matrix} from V^{σ} to Φ form a d.p.

representation of index 1 called the trivial representation.

Example 2. If ρ_{n}^{σ} : V^{σ} → A are nilpotent sequences of b.d.p. maps of index 2, then ρ = (ρ^{+}, ρ^{−}) form a d.p. representation of index 2 if and only if

(Q_{x}(y))^{(1)} = −x^{(1)}y^{(1)}x^{(1)}.  (11)

Indeed,

ad_{x}^{(2)}(y^{(1)}) = x^{(2)}y^{(1)} − x^{(1)}y^{(1)}x^{(1)} + y^{(1)}x^{(1)}
= −x^{(1)}y^{(1)}x^{(1)}

while ad_{x}^{(k)}(y^{(l)}) = 0 if l ≥ 2 or k ≥ 3, and these hold for all extensions Ω of Φ.

Recall [6] that V is a special Jordan pair if it is isomorphic to a Jordan subpair of an associative pair S = (S^{+}, S^{−}) where S is viewed as a Jordan pair with Q_{x}(y) = xyx. The standard imbedding (see [6]) imbeds S in an associative algebra A with S^{σ}S^{σ} = 0. If θ = (θ^{+}, θ^{−}) is an isomorphism of V with a Jordan subpair of S, then ρ_{1}^{σ} = σθ^{σ} satisfy (8) and (11) and thereby give rise to a d.p. representation of index 2. On the other hand, if ρ is any faithful d.p. representation of index 2, we can view S = (A, A) as an associative pair and see that θ = (ρ^{+}, −ρ^{−}) is an isomorphism of V with a Jordan subpair of S; i.e. V is special.

Example 3. If ρ_{n}^{σ} : V^{σ} → A are nilpotent sequences of b.d.p. maps of index 3, then ρ = (ρ^{+}, ρ^{−}) form a d.p. representation of index 3 if and only if

(Q_{x}(y))^{(1)} = x^{(2)}y^{(1)} − x^{(1)}y^{(1)}x^{(1)} + y^{(1)}x^{(2)},  (12)
(Q_{x}(y))^{(2)} = x^{(2)}y^{(2)}x^{(2)},  (13)
0 = −x^{(2)}y^{(1)}x^{(1)} + x^{(1)}y^{(1)}x^{(2)},  (14)
0 = x^{(2)}y^{(1)}x^{(2)},  (15)

and all linearizations hold. Indeed, these are the conditions on ad_{x}^{(k)}(y^{(l)}) for (k, l) = (2, 1), (4, 2), (3, 1), and (4, 1), while ad_{x}^{(k)}(y^{(l)}) = 0 and its linearizations hold for l ≥ 3 or k ≥ 5.

Now consider the Tits-Kantor-Koecher Lie algebra T KK(V, D_{0}) with D_{0} = Φζ + Inder(V) and let A = End_{Φ}(T KK(V, D_{0})). We set x^{(1)} = ad x ∈ A for x ∈ V^{σ}, and using the decomposition

T KK(V, D_{0}) = V^{+} ⊕ D_{0} ⊕ V^{−}

define

x^{(2)} = ( \begin{matrix}0 & 0 & Q_{x} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ) , y^{(2)} = ( \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ Q_{y} & 0 & 0 \end{matrix} )

for x ∈ V^{+}, y ∈ V^{−}. Letting x^{(0)} = Id and x^{(n)} = 0 for n ≥ 3, x ∈ V^{σ}, we can easily check that (9) holds. Indeed, ad x ad z(y) = −[[zy]x] = {zyx} = Q_{x,z}(y), for x, z ∈ V^{σ}, y ∈ V^{−σ}. Thus, ρ_{n}^{σ} : x → x^{(n)} is a sequence of b.d.p. maps by Lemma 9. We next check (12)-(15). Both sides of (12) are trivial on V^{σ}. On D = (D_{+}, D_{−}), (12) becomes

[Q_{x}(y), D] = Q_{x}([y, D]) − [x[y[x, D]]]

i.e.

−D_{σ}(Q_{x}(y)) = −Q_{x}(D_{−σ}(y)) − {D_{σ}(x), y, x}

which is the definition of a derivation (1). On w ∈ V^{−σ}, (12) becomes

[Q_{x}(y), w] = −[x[y[x, w]]] + [y, Q_{x}(w)]

which is the linearization of

[Q_{x}(y), y] = [x, Q_{y}(x)]

i.e.

−δ(Q_{x}(y), y) = −δ(x, Q_{y}(x))

which is (JP2). On D_{0} ⊕ V^{σ}, (13) is 0, while on V^{−σ}, it becomes (JP3). On D_{0} ⊕ V^{σ}, (14) is 0, while on w ∈ V^{−σ}, it becomes

0 = −Q_{x}[y, [x, w]] + [x[y, Q_{x}(w)]

i.e.

0 = Q_{x}{yxw} − {x, y, Q_{x}(w)}

which is (JP1). Finally, (15) is 0 on all of T KK(V, D_{0}). Since all linearizations of (12)-(15) also hold, we see that ρ = (ρ^{+}, ρ^{−}) is a d.p. representation of index 3. We call this the TKK representation of V. Note that since ad(x)(ζ) = −σx, ρ_{1}^{σ} is injective and the TKK representation is faithful.

The following lemma is easily verified, and we omit its proof.

Lemma 11 If ρ is a d.p. representation of the Jordan pair V in A, if ω : V′→ V is a homomorphism of Jordan pairs, and if ξ : A → A′ is a homomorphism of associative algebras, then ρ′ defined by (ρ′)_{n}^{σ} = ξ∘ρ_{n}^{σ}∘ω_{σ} is a d.p. representation of V′ in A′.

Lemma 12 If ρ is a d.p. representation of the Jordan pair V in A and ξ is a d.p. representation of V in B, then

(ρ ⊗ ξ)_{n}^{σ} : x → ∑_{k+l=n} ρ_{k}^{σ}(x) ⊗ ξ_{l}^{σ}(x)

defines a d.p. representation in A ⊗ B.

Proof. We write ρ_{n}^{σ}(x) = x^{(n)}, ξ_{n}^{σ}(x) = x^{(n)}, and (ρ ⊗ ξ)_{n}^{σ}(x) = x^{<n>}, so

x^{<n>} = ∑_{k+l=n} x^{(k)} ⊗ x^{(l)}.

We shall use Corollary 7 to show x → x^{<n>} is a sequence of b.d.p. maps. Clearly, x^{<0>} = 1 ⊗ 1. Also, for any extension Ω, (λx)^{<n>} = λ^{n}x^{<n>} and

(x + z)^{<n>} = ∑_{k+l=n}(x + z)^{(k)} ⊗ (x + z)^{(l)}

= ∑_{p+q+r+s=n} x^{(p)}z^{(q)} ⊗ x^{(r)}z^{(s)}

= ∑_{i+j=n}( ∑_{p+r=i} x^{(p)} ⊗ x^{(r)})( ∑_{q+s=j} z^{(q)} ⊗ z^{(s)})

= ∑_{i+j=n} x^{<i>}z^{<j>}.

Next we compute

ad_{x}^{<n>}(a ⊗ b) = ∑_{i+j=n} x^{<i>}(a ⊗ b)(−x)^{<j>}

= ∑_{p+q+r+s=n} x^{(p)}a(−x)^{(r)} ⊗ x^{(q)}b(−x)^{(s)}

= ∑_{k+l=n} ad_{x}^{(k)}(a) ⊗ ad_{x}^{(l)}(b).

Thus,

ad_{x}^{<n>}y^{<m>} = _{∑_{k+l=n} ; p+q=m}ad_{x}^{(k)}y^{(p)} ⊗ ad_{x}^{(l)}y^{(q)}.

If n > 2m, then either k > 2p or l > 2q so ad_{x}^{<n>}y^{<m>} = 0. If n = 2m, then k ≥ 2p or l ≥ 2q, so

ad_{x}^{<2m>}y^{<m>} = ∑_{p+q=m} ad_{x}^{(2p)}y^{(p)} ⊗ ad_{x}^{(2q)}y^{(q)}

= ∑_{p+q=m}(Q_{x}(y))^{(p)} ⊗ (Q_{x}(y))^{(q)}

= (Q_{x}(y))^{<m>}.

Since these hold for any extension Ω, ρ ⊗ ξ is a d.p. representation.

Lemma 13 If ρ is a d.p. representation of the Jordan pair V in A, then χ_{n}^{σ} = ρ_{n}^{σ} ∘ (−Id) defines a d.p. representation in the opposite algebra A^{op}.

Proof. By Lemma 33(iii), χ^{σ} is a sequence of homogeneous polynomial maps. Moreover, the (i, j)-linearization of χ_{n}^{σ} is

(x, z) → ρ_{i}^{σ}(−x)ρ_{j}^{σ}(−z)
= ρ_{j}^{σ}(−z)ρ_{i}^{σ}(−x)
= χ_{i}^{σ}(x) · χ_{j}^{σ}(z)

where · denotes the product in A^{op}. Thus, χ^{σ} is a sequence of b.d.p. maps. Moreover,

χ_{n}^{σ}(x)(a) = ∑_{k+l=n}(−x)^{(k)} · a · (x)^{(l)}
= ad_{x}^{(n)}(a),

so χ is a d.p. representation.

Lemma 14 If ρ is a d.p. representation of a Jordan pair V in A, then ρ̂ with ρ̂_{n}^{σ}(x) = ad_{x}^{(n)} is a d.p. representation of V in End(A).

Proof. Let m be the homomorphism m : A ⊗ A^{op} → End(A) given by m(a ⊗ b) = l_{a}r_{b} where l_{a}(b) = ab = r_{b}(a), and let χ be as in Lemma 13. Since

((m ∘ (ρ ⊗ χ)_{n}^{σ})(x))(a) = ∑_{k+l=n} x^{(k)}a(−x)^{(l)}
= ad_{x}^{(n)}(a),

the result follows from Lemmas 13, 12, and 11.

For a Jordan pair V, let B be the associative algebra with 1 generated by symbols x^{(n)} for x ∈ V^{σ}, σ = ± subject to the relations

x^{(0)} = 1

(λx)^{(n)} = λ^{n}x^{(n)},

(x + z)^{(n)} = ∑_{k+l=n} x^{(k)}z^{(l)},

[x^{(k)}, z^{(l)}] = 0,

x^{(k)}x^{(l)} = \binom{k + l}{k}x^{(k+l)},

for x, z ∈ V^{σ}. Hence, x → x^{(n)} is a sequence of b.d.p. maps. We now let U(V) be B with the further relations

ad_{x}^{(k)}(y^{(l)}) = \{ \begin{matrix}0 & for k > 2l, \\ (Q_{x}(y))^{(l)} & for k = 2l. \end{matrix}

for x ∈ V^{σ}, y ∈ V^{−σ}, and their linearizations. Thus, γ defined by γ_{n}^{σ}(x) = x^{(n)} is a d.p. representation. Moreover, since any d.p. representation ρ of V in A satisfies the defining relations of U(A), there is a unique associative algebra homomorphism φ : U(V) → A such that ρ_{n}^{σ} = φ ∘ γ_{n}^{σ}. Therefore, we call γ the universal representation of V.

Theorem 15 U(V) is a ℤ-graded cocommutative Hopf algebra and γ_{1}^{σ} is injective.

Proof. For simplicity of notation, we set U = U(V). Let ρ be the TKK representation and let

φ : U → End_{Φ}(T KK(V, D))

be the homomorphism satisfying ρ_{n}^{σ} = φ ∘ γ_{n}^{σ}. Since ρ_{1}^{σ} is injective, so is γ_{1}^{σ}. We have the maps μ : U ⊗ U → U and η : Φ → U given by the product and unit in U. By Lemma 12, γ ⊗ γ is a d.p. representation of V in U ⊗ U, so there is a homomorphism Δ : U → U ⊗ U with (γ ⊗ γ)_{n}^{σ} = Δ ∘ γ_{n}^{σ}; i.e.,

Δ(x^{(n)}) = ∑_{k+l=n} x^{(k)} ⊗ x^{(l)}.

Also, by Lemma 13, γ_{n}^{σ} ∘ (−Id) gives a d.p. representation of V in U^{op}, so there is a homomorphism S : U → U^{op} with γ_{n}^{σ} ∘ (−Id) = S ∘ γ_{n}^{σ}; i.e.,

S(x^{(n)}) = (−x)^{(n)}.

Finally, as in Example 1, we have the trivial representation ψ of V in Φ, so there is a homomorphism ε : U → Φ with ψ_{n}^{σ} = ε ∘ γ_{n}^{σ}; i.e.,

ε(x^{(n)}) = \{ \begin{matrix}1 & if n = 0 \\ 0 & if n ≥ 1 \end{matrix} .

We now show that with these maps, U is a cocommutative Hopf algebra. Since

((Δ ⊗ Id) ∘ Δ)(x^{(n)}) = ∑_{i+j+l=n} x^{(i)} ⊗ x^{(j)} ⊗ x^{(l)}
= ((Id ⊗ Δ) ∘ Δ)(x^{(n)}),

since U is generated by the x^{(n)}, and since Δ is a homomorphism, we see Δ is coassociative. Since

((ε ⊗ Id) ∘ Δ)(x^{(n)}) = ∑_{k+l=n} ε(x^{(k)}) ⊗ x^{(l)}
= Id(x^{(n)})
= ((Id ⊗ ε) ∘ Δ)(x^{(n)}),

since U is generated by the x^{(n)}, and since Δ and ε are homomorphisms, we see ε is a counit. Again let m be the homomorphism m : A ⊗ A^{op} → End(A) given by m(a ⊗ b) = l_{a}r_{b}, so A = m ∘ (Id ⊗ S) ∘ Δ) is a homomorphism A → End(A) with

A(x^{(n)})(a) = ∑_{k+l=n} x^{(k)}a(−x)^{(l)}
= ad_{x}^{(n)}(a).

Note A(x^{(n)})(1) = ad_{x}^{(n)}(1) = ε(x^{(n)})1. If A(a)(1) = ε(a)1 and A(b)(1) = ε(b)1, then

A(ab)(1) = A(a)A(b)(1) = ε(ab)1.

Since U is generated by the x^{(n)}, we see that

A(a)(1) = ε(a)1 = (η ∘ ε)(a)

for all a ∈ U. Since μ(u) = m(u)(1), we have

(μ ∘ (Id ⊗ S) ∘ Δ) = η ∘ ε,

and similarly

(μ ∘ (S ⊗ Id) ∘ Δ) = η ∘ ε.

Thus, S is an antipode. Since

Δ(x^{(n)}) = ∑_{k+l=n} x^{(k)} ⊗ x^{(l)}

= ∑_{k+l=n} x^{(l)} ⊗ x^{(k)}

= (τ ∘ Δ)(x^{(n)}), Δ is cocommutative. Thus, U is a cocommutative Hopf algebra.

If A is the free associative algebra with 1 generated by the x^{(n)}, x ∈ V^{σ}, σ = ±, then A has a ℤ-grading A = ∑ A_{k} with x^{(n)} ∈ A_{σn}. We see that if a ∈ A_{k}, then ad_{x}^{(n)}(a) ∈ A_{σn+k}. In particular, ad_{x}^{(2l)}(y^{(l)}) ∈ A_{σl} for y ∈ V^{−σ}. We see that the defining relations for U are homogeneous, so U is ℤ-graded as an associative algebra. It is easy to check that Δ, ε, and S are graded homomorphisms, so U is a ℤ-graded Hopf algebra.

6 The exponential property

Let V be a Jordan pair and let ρ be a divided power representation of V in A. For x ∈ V^{+}, y ∈ V^{−}, we recursively define

x_{1} = x, y_{1} = y,  (16)
x_{n} = Q_{x}(y_{n−1}), y_{n} = Q_{y}(x_{n−1}).

In the Jordan pair of formal power series V[[s, t]] , we can form

u = u(x, y) = ∑_{i=1}^{∞} s^{i}t^{i−1}x_{i},

v = v(x, y) = ∑_{i=1}^{∞} s^{i−1}t^{i}y_{i}.

Using Theorem 35, we can extend the map ρ_{n}^{σ} to a homogeneous map ρ̃_{n}^{σ} : V^{σ}[[s, t]] → A[[s, t]] of degree n. Also, density arguments as in the proof of Theorem 35 show that this is a divided power representation. Recall that in A[[s, t]], we can also form exp(sx) = exp_{ρ^{+}}(sx) and exp(ty) = exp_{ρ^{−}}(ty). Moreover, we can form

exp(u) = ∑_{n=0}^{∞} u^{(n)},

exp(v) = ∑_{n=0}^{∞} v^{(n)},

because s^{i}t^{j} occurs only finitely often in the expansions of u^{(n)} and v^{(n)}. We now set

h(x, y) = exp(v) exp(−sx) exp(−ty) exp(u)  (17)

= ∑_{p,q=0}^{∞} h_{pq}(x, y)s^{p}t^{q}.

Note that replacing x by −x and y by −y, replaces x_{n} by −x_{n} and y_{n} by −y_{n}. Hence,

h(−x, −y) = exp(−v) exp(sx) exp(ty) exp(−u)

so

exp(sx) exp(ty) = exp(v)h(−x, −y) exp(u)  (18)

From (17) we get

h_{pq}(x, y) = ∑ (∏_{j} y_{j}^{(l_{j})}) (−x)^{(k)}(−y)^{(l)} (∏_{i} x_{i}^{(k_{i})})  (19)

where the sum is over all solutions to

p = ∑_{j}(j − 1)l_{j} + k + ∑_{i} ik_{i},  (20)
q = ∑_{j} jl_{j} + l + ∑_{i}(i − 1)k_{i}.

In particular,

h_{pq}(−x, −y) = (−1)^{p+q}h_{pq}(x, y).

From (18) we get

x^{(p)}y^{(q)} = ∑ (∏_{j} y_{j}^{(l_{j})}) (−1)^{k+l}h_{kl}(x, y) (∏_{i} x_{i}^{(k_{i})})  (21)

where the sum is again over all solutions to (??). We shall be investigating the exponential property for the representation ρ:

h_{pq}(x, y) = 0 for all p ≠ q and all x ∈ V^{+}, y ∈ V^{−}.  (E)

We shall also be interested in the restricted exponential property:

h_{pq}(x, y) = 0 for all p ≠ q with min(p, q) ≤ N,  (E_{N})
and all x ∈ V^{+}, y ∈ V^{−}.

Lemma 16 The TKK representation satisfies the exponential property.

Proof. Since B_{sx,ty} has constant term Id, it is invertible, so (sx, ty) is quasi-invertible. Also,

D_{x,y}(x_{n}) = D_{x,y}Q_{x}y_{n−1} = Q_{x}D_{y,x}y_{n−1}

so by induction

D_{x,y}(x_{n}) = 2x_{n+1}, D_{y,x}(y_{n}) = 2y_{n+1}.

Thus,

B_{sx,ty}(u) = u − stD_{x,y}(u) + s^{2}t^{2}Q_{x}Q_{y}(u)

= ∑_{i=1}^{∞} s^{i}t^{i−1}x_{i} − 2 ∑_{i=1}^{∞} s^{i+1}t^{i}x_{i+1} + ∑_{i=1}^{∞} s^{i+2}t^{i+1}x_{i+2}

= sx_{1} − s^{2}tx_{2} = sx − Q_{sx}(ty).

Thus, u = (sx)^{ty} and similarly v = (ty)^{sx}. By Theorem 3 and (18), we have h(−x, −y) = β̃_{sx,ty}. Since β̃_{sx,ty} preserves the grading of T KK(V[[s, t]], D_{0}[[s, t]]), we see h_{pq}(−x, −y) = 0 for p ≠ q.

Lemma 17 (E_{0}) holds for any d.p. representation ρ of V in A. Moreover, h_{00}(x, y) = 1 for all x ∈ V^{+}, y ∈ V^{−}.

Proof. If p = 0 in (??), then k = k_{i} = l_{j} = 0 for all i > 0 and all j > 1. Thus, (19) becomes

h_{0q}(x, y) = ∑_{l_{1}+l=q} y^{(l_{1})}(−y)^{(l)} = 0^{(q)}.

Thus, h_{00}(x, y) = 1 and h_{0q}(x, y) = 0 for q > 0. Similarly, h_{p0}(x, y) = 0 for p > 0.

Let X denote the subalgebra of A generated by all x^{(n)} with x ∈ V^{+}. Similarly, let Y be generated by all y^{(n)} with y ∈ V^{−} and H be generated by all h_{nn}(x, y) with x ∈ V^{+}, y ∈ V^{−}. Also, let ⟨X , Y⟩ be the subalgebra generated by X and Y.

Lemma 18 If (E_{N}) holds for the universal representation γ of V in U(V), then

h_{k} = h_{kk}(x, y), for 0 ≤ k ≤ N + 1, x ∈ V^{+}, y ∈ V^{−}

is a finite divided power sequence. Also, h_{ij}(x, y) is primitive for i ≠ j and min(i, j) ≤ N + 1. If (E) holds for γ, then H is a Hopf subalgebra of U(V).

Proof. Since ε(x^{(n)}) = 0 for n > 0, we see that ε(exp(sx)) = 1. Also, since x^{(n)} is a d.p. sequence,

Δ(exp(sx)) = ∑_{n=0}^{∞} ∑_{i+j=n} s^{i}x^{(i)} ⊗ s^{j}x^{(j)}
= exp(sx) ⊗ exp(sx),

and we see that exp(sx) is group-like. Similarly, the other factors in (17) are group-like, so h(x, y) is group-like. Comparing the s^{i}t^{j} coefficients of

Δ(h(x, y)) = h(x, y) ⊗ h(x, y)

gives

Δ(h_{ij}(x, y)) = _{∑_{p+q=i} ; r+s=j}h_{pr}(x, y) ⊗ h_{qs}(x, y).  (22)

For i = j = k ≤ N + 1, then p ≠ r implies that either p ≤ N or r ≤ N, so h_{pr}(x, y) = 0. Thus,

Δ(h_{kk}(x, y)) = ∑_{p+q=k} h_{pp}(x, y) ⊗ h_{qq}(x, y).

If i ≠ j and min(i, j) ≤ N + 1, then by symmetry, we can assume i ≤ N + 1. In (22), if p ≠ 0, i, then p, q ≤ N. Moreover, p + q < r + s implies either p ≠ r or q ≠ s. Thus, h_{pr}(x, y) ⊗ h_{qs}(x, y) = 0 for p ≠ 0, i. Also, h_{0l}(x, y) = 0 for l > 0 and h_{00}(x, y) = 1, by Lemma 17. Thus, (22) becomes

Δ(h_{ij}(x, y)) = 1 ⊗ h_{ij}(x, y) + h_{ij}(x, y) ⊗ 1.

If (E) holds, then h_{k}, k ≥ 0, is a divided power sequence, so Δ(h_{k}) ∈ H ⊗ H. Since the h_{k}’s generate H, we have Δ(H) ⊂ H ⊗ H.

For a monomial m = ∏_{i} u_{i}^{(n_{i})} with u_{i} ∈ V^{σ_{i}}, σ_{i} = ±, we define the σ-degree

of m by

deg_{σ}(m) = ∑_{{i:σ_{i}=σ}} n_{i}.

We also define the level of m by

λ(m) = ∑_{(i,j)∈L} n_{i}n_{j}

where

L = {(i, j) : i < j, σ_{i} = +, σ_{j} = −}.

We observe that

λ(m_{1}m_{2}) = λ(m_{1}) + λ(m_{2}) + deg_{+}(m_{1}) deg_{−}(m_{2}).

Thus, if deg_{σ}(m′_{2}) ≤ deg_{σ}(m_{2}), σ = ±, and λ(m′_{2}) < λ(m_{2}), then

λ(m_{1}m′_{2}m_{3}) < λ(m_{1}m_{2}m_{3}).  (23)

We let M_{rs}(n) be the span of all monomials m with deg_{+}(m) ≤ r, deg_{−}(m) ≤ s, and λ(m) ≤ n.

Lemma 19 If ρ is a d.p. representation of V in A, then for x ∈ V^{+}, y ∈ V^{−},

h_{pq}(x, y) ≡ (−1)^{p+q}x^{(p)}y^{(q)} modM_{pq}(pq − 1).

Proof. For the monomial

m = (∏_{j} y_{j}^{(l_{j})}) (−x)^{(k)}(−y)^{(l)} (∏_{i} x_{i}^{(k_{i})})

appearing in (19), we have deg_{+}(m) = k + ∑_{i} k_{i} ≤ p, deg_{−}(m) = l + ∑_{j} l_{j} ≤ q,

and λ(m) = kl ≤ pq. Moreover, λ(m) < pq unless k_{i} = l_{j} = 0 for all i, j.

Lemma 20 If ρ is a d.p. representation of V in A, then for x, z ∈ V^{+}, y, w ∈ V^{−}, the following elements are in M_{pq}(pq − 1):

(i) x^{(p)}y^{(q)}  if p ≥ 2q or q ≥ 2p,
(ii) x^{(p_{1})}z^{(p_{2})}y^{(q)} if p_{1} + p_{2} = p ≥ 2q,
(iii) x^{(p)}y^{(q_{1})}w^{(q_{2})} if q_{1} + q_{2} = q ≥ 2p.

Proof. Note that (i) is just the special case of (ii) and (iii) with p_{2} = q_{2} = 0. Any linearization of (Q_{x}(y))^{(q)} in the variable x lies in X and hence in M_{2q,q}(0). Thus, the (p_{1}, p_{2})-linearization of (10) gives

ad_{x}^{(p_{1})}ad_{z}^{(p_{2})}y^{(q)} = ∑_{\begin{matrix}i+j=p_{1} \\ k+l=p_{2} \end{matrix}} x^{(i)}z^{(k)}y^{(q)}(−z)^{(l)}x^{(j)} ∈ M_{pq}(pq − 1).

Moreover,

λ(x^{(i)}z^{(k)}y^{(q)}(−z)^{(l)}x^{(j)}) = (i + k)q < pq

unless i = p_{1}, k = p_{2}. This shows (ii) and the proof of (iii) is similar.

We shall need the following result on binomial coefficients which we include for completeness.

Lemma 21 If d = gcd\{\binom{n}{i} : 1 < i < n\}, then d = \{ \begin{matrix}p & if n = p^{e}, p a prime \\ 1 & otherwise. \end{matrix}

Proof. For any prime p, write n = p^{e}m with p ∤ m. In ℤ_{p}[t], we have (1 + t)^{n} = (1 + t^{p^{e}})^{m}, so

\binom{n}{p^{e}} ≡ m mod p.

If n ≠ p^{e}, this shows that p ∤ d. Thus, d = 1 if n is not a power of a prime. If n = p^{e}, let o_{p}(k) denote the highest power of p dividing k. Note that for 0 ≤ k ≤ n, we have o_{p}(n − k) = o_{p}(k). Since

\binom{n}{k} = \binom{n}{k − 1}\frac{n − k + 1}{k},

we see that

o_{p} (\binom{n}{k}) = o_{p} (\binom{n}{k − 1}) + o_{p}(k − 1) − o_{p}(k)
= e − o_{p}(k)

by induction on k ≥ 1. Thus, o_{p}(d) = min{e − o_{p}(k) : 1 < k < n} = 1. Since d divides p^{e} = \binom{n}{1}, the result follows.

Lemma 22 If (E_{N}) holds for the a d.p. representation ρ of V in A, and if p ≠ q, min(p, q) ≤ N + 1, x ∈ V^{+}, y ∈ V^{−}, then

x^{(p)}y^{(q)} ∈ M_{pq}(pq − 1).

Proof. By Lemma 20 we can assume that max(p, q) < 2 min(p, q). If min(p, q) ≤ N, we apply (E_{N}) to (21) to get

x^{(p)}y^{(q)} = ∑ (∏_{j} y_{j}^{(l_{j})}) h_{kk}(x, y) (∏_{i} x_{i}^{(k_{i})})  (24)

with

p = ∑_{j}(j − 1)l_{j} + k + ∑_{i} ik_{i},
q = ∑_{j} jl_{j} + k + ∑_{i}(i − 1)k_{i}.

Since

h_{kk}(x, y) ∈ M_{kk}(k^{2})

by Lemma 19, and since k^{2} ≤ (min(p, q))^{2} < pq, we have finished this case. We can now assume that min(p, q) = N + 1, say N + 1 = q < p < 2q. If 1 < l < q, then min(p, l) ≤ N, so (23) shows that

\binom{q}{l}x^{(p)}y^{(q)} = x^{(p)}y^{(l)}y^{(q−l)} ∈ M_{pq}(pq − 1).

Similarly, if 1 < k < p, then min( k, p − k) ≤ N so

\binom{p}{k}x^{(p)}y^{(q)} = x^{(p−k)}x^{(k)}y^{(q)} ∈ M_{pq}(pq − 1).

We note that q < p < 2q implies that p and q cannot be powers of the same prime. Thus, using Lemma 21, we get

x^{(p)}y^{(q)} ∈ M_{pq}(pq − 1).

The case, N + 1 = p < q < 2p is similar.

Lemma 23 If the d.p. representation ρ of V in A satisfies the exponential property (E), then

⟨X , Y⟩ = YHX .

If ρ satisfies the restricted exponential property (E_{N}), then

h_{rs}(x, y) ∈ YHX

for min(r, s) ≤ N + 1, x ∈ V^{+}, y ∈ V^{−}.

Proof. We first shall show by induction on n that if (E_{N}) holds and min(r, s) ≤ N + 1, then

M_{rs}(n) ⊂ YHX.  (25)

Clearly, M_{rs}(0) ⊂ YX. Let m be a monomial with deg_{+}(m) ≤ r, deg_{−}(m) ≤ s, and λ(m) = n, but m ∉ M_{rs}(n − 1). Let x, z ∈ V^{+} and y, w ∈ V^{−}. If m factors as m_{1}x^{(p)}y^{(q)}m_{3}, with p, q ≠ 0, then Lemma 22 and (23) imply p = q. Moreover, m cannot factor as m_{1}x^{(p_{1})}z^{(p_{2})}y^{(q)}m_{3}, with p_{1}, p_{2}, q ≠ 0, for then p_{1} = q = p_{2} contradicting Lemma 20. Similarly, m cannot factor as m_{1}x^{(p)}y^{(q_{1})}w^{(q_{2})}m_{3}, with p, q_{1}, q_{2} ≠ 0. Thus, we have

m = m_{1} (∏_{i} v_{i}^{(p_{i})}u_{i}^{(p_{i})}) m_{3}

with v_{i} ∈ V^{+}, u_{i} ∈ V^{−}, m_{1} ∈ Y, and m_{3} ∈ X. By Lemma 19 and (23), we see

m ≡ m_{1} (∏_{i} h_{p_{i}p_{i}}(v_{i}, u_{i})) m_{3} mod M_{rs}(rs − 1),

so m ∈ YHX by the induction hypothesis. This shows (25). If (E) holds, then (25) implies ⟨X , Y⟩ = YHX, since we always have YHX ⊂ ⟨X , Y⟩. Also, if (E_{N}) holds then Lemma 19 and (25) give

h_{rs}(x, y) ∈ M_{rs}(rs) ⊂ YHX.

We now assume that V^{±} are free modules over Φ with ordered bases {v_{i} : i ∈ I} and {u_{j} : j ∈ J}, respectively. Let k be a map k : I → ℕ = {n ∈ ℤ : n ≥ 0} with k : i → k_{i} and k_{i} = 0 for all but finitely many i. Define

m_{k}^{+} = ∏_{i∈I} v_{i}^{(k_{i})},

and make a similar definition of m_{l}^{−} for l : J → ℕ. For b ∈ H, define

m_{kl}(b) = m_{l}^{−}bm_{k}^{+} ∈ YHX.

We also let {v_{i}^{∗} : i ∈ I} be the dual basis which we extend to T KK(V, D_{0}) by setting v_{i}^{∗}(V^{−} + D_{0}) = 0, and similarly for {u_{j}^{∗} : j ∈ J}.

Lemma 24 Let V be a Jordan pair with V^{±} free as modules over Φ. Let γ be the universal representation of V in U(V), let ρ be the TKK representation, and let φ : U(V) → End_{Φ}(T KK(V, D_{0})) be the homomorphism with ρ_{k} = φ ∘ γ_{k}. The maps λ_{i}^{+} : U(V) → Φ and λ_{j}^{−} : U(V) → Φ defined by

λ_{i}^{+}(a) = v_{i}^{∗}(φ(S(a))(ζ)),
λ_{j}^{−}(a) = u_{j}^{∗}(φ(a)(ζ)),

satisfy

λ_{i}^{+}(bx^{(1)}) = ε(b)v_{i}^{∗}(x),
λ_{i}^{+}(YH) = 0,
λ_{j}^{−}(y^{(1)}b) = ε(b)u_{j}^{∗}(y),
λ_{j}^{−}(HX) = 0,

for b ∈ H, x ∈ V^{+}, y ∈ V^{−}.

Proof. Since h(x, y) is group-like, ε(h(x, y)) = 1 and S(h(x, y)) = h(x, y)^{−1}.

Also,

φ(h(x, y)) = β̃_{−sx,−ty} = β̃_{sx,ty}

and

β̃_{sx,ty}(ζ) = ζ = ε(h(x, y))ζ.

Hence,

φ(b)(ζ) = ε(b)ζ = φ(S(b))(ζ)

holds for b = h_{kk}(x, y) and therefore for all b ∈ H. We have

λ_{i}^{+}(bx^{(1)}) = v_{i}^{∗}(φ(−x^{(1)})φ(S(b))(ζ))
= v_{i}^{∗}(−ad x(ε(b)ζ))
= ε(b)v_{i}^{∗}(x),

and

λ_{i}^{+}(YH) ⊂ v_{i}^{∗}(φ(S(H))φ(Y)(ζ))
⊂ v_{i}^{∗}(D_{0} + V^{−}) = 0.

The calculations for λ_{j}^{−} are similar.

For k, k′ : I → ℕ, define k + k′ by (k + k′)_{i} = k_{i} + k′_{i} and set |k| = ∑_{i} k_{i}. If

k ≠ 0, let s = s(k) be minimal in I with k_{s} ≠ 0, and define k̄ by

k̄_{i} = \{ \begin{matrix}k_{i} & if i ≠ s \\ k_{s} − 1 & if i = s \end{matrix} .

Similarly, we define l + l′, |l|, and l̄. Since

Δ(v_{i}^{(k_{i})}) = ∑_{k′_{i}+k′′_{i}=k_{i}} v_{i}^{(k′_{i})} ⊗ v_{i}^{(k′′_{i})},

we also have

Δ(m_{k}) = ∑_{k′+k′′=k} m_{k′} ⊗ m_{k′′}

and

Δ(m_{kl}(b)) = _{∑_{k′+k′′=k} ; l′+l′′=l ; t}m_{k′l′}(b′_{t}) ⊗ m_{k′′l′′}(b′′_{t})  (26)

if

Δ(b) = ∑_{t} b′_{t} ⊗ b′′_{t}.

We recursively define β_{pq} : U(V) → U(V) by

β_{pq} = \{ \begin{matrix}Id & if p = q = 0 \\ (λ_{s(q)}^{−} ⊗ β_{0q̄}) ∘ Δ & if p = 0, q ≠ 0 \\ (λ_{s(p)}^{+} ⊗ β_{p̄q}) ∘ Δ & if p ≠ 0 \end{matrix}  (27)

Lemma 25 Let V be a Jordan pair with V^{±} free as modules over Φ and let γ be the universal representation of V in U(V). If |k| ≤ |p| and |l| ≤ |q|, then

β_{pq} ∘ m_{kl} = \{ \begin{matrix}0 & if (p, q) ≠ (k, l) \\ Id & if (p, q) = (k, l) \end{matrix} .

Proof. We consider the case p ≠ 0. The case p = 0 is similar. By (26), we

have

β_{pq}(m_{kl}(b)) = _{∑_{k′+k′′=k} ; l′+l′′=l ; t}λ_{s(p)}^{+}(m_{k′l′}(b′_{t}))β_{p̄q}(m_{k′′l′′}(b′′_{t})).  (28)

Clearly, k′ + k′′ = k implies |k′′| ≤ |k| ≤ |p|. If |k′′| = |p|, then k′ = 0, and

λ_{s(p)}^{+}(m_{k′l′}(b′_{t})) ∈ λ_{s(p)}^{+}(YH) = 0,

by Lemma 24. Thus, we can assume that |k′′| ≤ |p| − 1 = \left|{ \atop p̄}\right|, and we can apply induction to get

β_{p̄q}(m_{k′′l′′}(b′′_{t})) = \{ \begin{matrix}0 & if (p̄, q) ≠ (k′′, l′′) \\ b′′_{t} & if (p̄, q) = (k′′, l′′) \end{matrix} .

Moreover, if (p̄, q) = (k′′, l′′), then |l| ≤ |q| implies l = q and l′ = 0. Also,

|k′| + \left|{ \atop p̄}\right| = |k| ≤ |p| and \left|{ \atop p̄}\right| = |p| − 1, so |k′| ≤ 1. Thus, k′ = 0 or k′ : i → δ_{ei}

for some e. In the first case, by Lemma 24,

λ_{s(p)}^{+}(m_{k′l′}(b′_{t})) = λ_{s(p)}^{+}(b′_{t}) = 0,

while in the second

λ_{s(p)}^{+}(m_{k′l′}(b′_{t})) = λ_{s(p)}^{+}(b′_{t}v_{e}^{(1)}) = ε(b′_{t})v_{s(p)}^{∗}(v_{e}).

We see that terms in (28) are 0, unless p̄ = k′′, q = l = l′′, and e = s(p). Moreover, if this holds, then k = p̄ + k′ = p. Thus, β_{pq}(m_{kl}(b)) = 0, if (p, q) ≠ (k, l), while

β_{pq}(m_{pq}(b)) = ∑_{t} ε(b′_{t})b′′_{t}
= (ε ⊗ Id)(Δ(b))
= b.

Lemma 26 Let V be a Jordan pair with V^{±} free as modules over Φ and let γ be the universal representation of V in U(V). Every a ∈ YHX can be written uniquely as

a = ∑_{k,l} m_{kl}(c_{kl}), c_{kl} ∈ H.

Also, if a ∈ YHX is primitive, then a ∈ (V^{+})^{(1)} + H + (V^{−})^{(1)}.

Proof. Clearly,

a = ∑_{k,l} m_{kl}(c_{kl}),

for some c_{kl} ∈ H. To show uniqueness, it suffices to show that a = 0 implies all c_{kl} = 0. To prove both statements, we now let a be primitive (including a = 0). We also can assume a and each of the terms m_{kl}(c_{kl}) belong to U_{d}, in the ℤ-grading on U(V). If some c_{kl} ≠ 0, choose p, q with c_{pq} ≠ 0 and |p| maximal. Thus, c_{kl} ≠ 0 implies that |k| ≤ |p|. However, since H ⊂ U_{0} by (19), we have

d = |k| − |l| = |p| − |q| ,

so |l| ≤ |q|, as well. Thus, by Lemma 25, c_{pq} = β_{pq}(a). In particular, a = 0 implies all c_{kl} = 0.

We now consider the three cases in the definition of β_{pq} (see (27)). If p = q = 0, then a = β_{00}(a) = c_{00} ∈ H. For the remaining two cases, we shall show that m_{pq}(c_{pq}) ∈ (V^{±})^{(1)}, and use induction on the number of terms to handle a − m_{pq}(c_{pq}). If p = 0, q ≠ 0, then by Lemma 24

c_{0q} = β_{0q}(a)

= (λ_{s(q)}^{−} ⊗ β_{0q̄})(a ⊗ 1 + 1 ⊗ a)

= λ_{s(q)}^{−}(a)β_{0q̄}(1)

= \{ \begin{matrix}0 & if q̄ ≠ 0 \\ λ_{s(q)}^{−}(a) & if q̄ = 0 \end{matrix} .

Since c_{pq} ≠ 0, we see that p = 0, q ≠ 0, implies q̄ = 0 and m_{0q}(c_{0q}) = λ_{s(q)}^{−}(a)u_{s(q)}^{(1)} ∈ (V^{−})^{(1)}. Similarly, if p ≠ 0, then

c_{pq} = β_{pq}(a)

= (λ_{s(p)}^{+} ⊗ β_{p̄q})(a ⊗ 1 + 1 ⊗ a)

= λ_{s(p)}^{+}(a)β_{p̄q}(1)

= \{ \begin{matrix}0 & if (p̄, q) ≠ (0, 0) \\ λ_{s(p)}^{+}(a) & if (p̄, q) = (0, 0) \end{matrix}

so p̄ = 0 and m_{pq}(c_{pq}) = λ_{s(p)}^{+}(a)v_{s(p)}^{(1)} ∈ (V^{+})^{(1)}.

Theorem 27 If ρ is a divided power representation of a Jordan pair V in A, then ρ satisfies the exponential property and ⟨X , Y⟩ = YHX.

Proof. By Lemma 23, it suffices to prove (E). We first assume that V^{±} are free modules over Φ and that ρ = γ, the universal representation in U(V). We shall show (E_{N}) by induction on N. If (E_{N}) holds, if p ≠ q with min(p, q) ≤ N + 1, and if x ∈ V^{+}, y ∈ V^{−}, then by Lemmas 23 and 18, h_{pq}(x, y) is a primitive element of YHX. Thus, by Lemma 26, h_{pq}(x, y) ∈ γ_{1}(V^{±}) (since h_{pq}(x, y) ∈ U_{p−q} ≠ U_{0} ⊃ H). If ξ is the TKK representation and φ : U(V) → End_{Φ}(T KK(V, D)) with φ ∘ γ_{k} = ξ_{k}, then φ(h_{pq}(x, y)) = 0 since the TKK representation satisfies the exponential condition (Lemma 16). However, ξ_{1} is injective, so h_{pq}(x, y) = 0, i.e. (E_{N+1}) holds. To prove (E) for ρ ≠ γ, we again let φ : U(V) → A with φ ∘ γ_{k} = ρ_{k} and note that φ maps h_{pq}(x, y) for γ to h_{pq}(x, y) for ρ. To drop the assumption that V^{±} are free modules over Φ, it suffices to find a Jordan pair homomorphism ω : V′ → V with ω_{+}(x′) = x, ω_{−}(y′) = y where V^{′±} are free modules over Φ. Indeed, by Lemma 11, (ρ′)_{n}^{σ} = ρ_{n}^{σ} ∘ ω_{σ} defines a d.p. representation of V′ in A and h_{pq}(x′, y′) for ρ′ coincides with h_{pq}(x, y) for ρ. Using induction and the definition (16), it is easy to verify that

Q_{x_{k}}(y_{l}) = x_{2k+l−1}, Q_{y_{k}}(x_{l}) = y_{2k+l−1},
{x_{k}y_{l}x_{n}} = 2x_{k+l+n−1}, {y_{k}x_{l}y_{n}} = 2y_{k+l+n−1}.

Now (Φ[T], Φ[T]) is a Jordan pair under Q_{a}(b) = aba and (Φ[T]′, Φ[T]′) is a subpair where Φ[T]′ denotes the free submodule over Φ with basis {T^{2k−1} : k ≥ 1}. It is easy to check that ω_{+}(T^{2k−1}) = x_{k}, ω_{−}(T^{2k−1}) = y_{k}, define a homomorphism of Jordan pairs.

Corollary 28 If V is a Jordan pair with V^{±} free as modules over Φ and γ is the universal representation of V in U(V), then

P(U(V)) = (V^{+})^{(1)} + P(H) + (V^{−})^{(1)}.

If H is also a free module over Φ with basis {b_{k} : k ∈ K}, if {v_{i} : i ∈ I} is a basis for V^{+}, and if {u_{j} : j ∈ J} is a basis for V^{−}, then

m_{pq}(b_{k}) = (∏_{i∈I} v_{i}^{(p_{i})}) b_{k} (∏_{j∈J} u_{j}^{(q_{j})})

is a basis for U(V).

Proof. This follows immediately from Lemma 26 and Theorem 27.

7 Universal properties

The main purpose of this section is to show, in characteristic zero, that the universal divided power representation of a Jordan pair V is isomorphic with the universal enveloping algebra of a certain Lie algebra L(V). If ζ is an inner derivation, L(V) is just the universal covering of T KK(V, Inder(V)). For Jordan algebras, the constructions of {V^{+}, V^{−}} and L(V) and the results of Lemmas 30 and 31 are due to Allison and Gao [2] within the context of structurable algebras and Steinberg unitary Lie algebras. For Jordan pairs, they are due to Benkart and Smirnov [3] within the context of Jordan-Kantor pairs and Lie algebras of type BC_{1}. Because the formulation for Jordan pairs is much simpler, we include the proofs here for the convenience of the reader.

We shall need the following construction of a Lie algebra.

Lemma 29 If L is a Lie algebra and φ : M → L is a homomorphism of L-modules, then

(i) M_{0} = span{φ(x) · x : x ∈ M} is an L-submodule of M contained in ker φ,

(ii) if M_{0} = 0, then M is a Lie algebra under [xy] = φ(x) · y and φ is a Lie algebra homomorphism with ker φ contained in the center of M.

Proof. For x, y ∈ M, set x ∗ y = φ(x) · y. Since

φ(x ∗ y) = φ(φ(x) · y)
= [φ(x), φ(y)],

φ is a homomorphism. Note that φ(x ∗ x) = 0, so M_{0} ⊂ ker φ. If l ∈ L, then

l · (x ∗ y) = l · (φ(x) · y)
= [l, φ(x)] · y + φ(x) · (l · y)
= φ(l · x) · y + φ(x) · (l · y)
= (l · x) ∗ y + x ∗ (l · y)

so l acts as a derivation on M. Since x ∗ y + y ∗ x ∈ M_{0} by linearization, we see that l · (x ∗ x) ∈ M_{0}, showing (i). We also have that ad(x) given by ad(x)(y) = x ∗ y is a derivation of M. Thus, if M_{0} = 0, then [xy] = x ∗ y is a Lie product. Clearly, [xy] = φ(x) · y = 0 for x ∈ ker φ.

Let V be a Jordan pair over a commutative associative ring Φ containing \frac{1}{2}. The linear map

δ : V^{+} ⊗ V^{−} → Inder(V)

with

δ(x ⊗ y) = δ_{x,y} = (D_{x,y}, −D_{y,x})

is a homomorphism of Inder(V)-modules. Since

δ_{x,y}(x ⊗ y) = 2(Q_{x}(y) ⊗ y − x ⊗ Q_{y}(x)),

we see that M_{0} of Lemma 29 is the Φ-span of all

Q_{x}(y) ⊗ y − x ⊗ Q_{y}(x)  (29)

for x ∈ V^{+}, y ∈ V^{−}. We let {x, y} denote the image of x ⊗ y in the Inder(V)- module

{V^{+}, V^{−}} := (V^{+} ⊗ V^{−})/M_{0},

and continue to use δ to denote the induced map {V^{+}, V^{−}} → Inder(V). By Lemma 29, {V^{+}, V^{−}} is a Lie algebra with [u, v] = δ(u) · v and δ is a Lie homomorphism with ker δ contained in the center of {V^{+}, V^{−}}. We now form

L(V) = V^{+} ⊕ {V^{+}, V^{−}} ⊕ V^{−}

with multiplication

[V^{σ}, V^{σ}] = 0,
[u, z] = −[z, u] = δ(u)_{σ}(z),
[x, y] = −[y, x] = −{x, y},

for x ∈ V^{+}, y ∈ V^{−}, z ∈ V^{σ}, u ∈ {V^{+}, V^{−}}. (Compare with the definition (2) of T KK(V, D)). Clearly, L_{1} = V^{+}, L_{−1} = V^{−}, L_{0} = {V^{+}, V^{−}} gives a grading on L(V). Also, κ : L(V) → T KK(V, Inder(V)) with

κ(x + u + y) = x + δ(u) + y

is a homomorphism. To check the Jacobi identity for L(V), it suffices to take the elements in the graded components. Moreover, if [a[bc]] ∈ L_{±1}, the Jacobi identity follows from its image under κ. Also, if a, b, c ∈ L_{0}, we know the Jacobi identity holds. We are left with the case a ∈ L_{0}, b ∈ L_{1}, c ∈ L_{−1}. We have

[a[bc]] = −[a, {b, c}]
= −δ(a) · {b, c}
= −{δ(a) · b, c} − {b, δ(a) · c}
= [[ab]c] + [b[ac]].

We see that L(V) is a Lie algebra and that ker(κ) = ker(δ) is contained in the center of L(V).

If

L = L_{1} ⊕ L_{0} ⊕ L_{−1}

is a ℤ-grading of a Lie algebra L over Φ containing \frac{1}{2}, it is well-known (and easy to see) that (L_{1}, L_{−1}) is a Jordan pair with

Q_{x}(y) = \frac{1}{2}[x[xy]]

We now show that L(V) is universal in the following sense.

Lemma 30 If V is a Jordan pair over Φ containing \frac{1}{2}, if L = L_{1}⊕L_{0}⊕L_{−1} is a ℤ-graded Lie algebra, and if θ : V → (L_{1}, L_{−1}) is a Jordan pair homomorphism, then there is a unique Lie algebra homomorphism θ̂: L(V) → L which extends θ.

Proof. If u ∈ L_{1}, v ∈ L_{−1}, then

[Q_{u}(v), v] = \frac{1}{2}[[u[uv]]v]

= \frac{1}{2}[[uv][uv]] − \frac{1}{2}[u[v[uv]]]

= [u, Q_{v}(u)]

so the map x ⊗ y → [θ_{+}(x), θ_{−}(y)] of V^{+} ⊗ V^{−} to L_{0} induces a map θ_{0} : {V^{+}, V^{−}} → L_{0}. It is easy to see that

θ̂: x + u + y → θ_{+}(x) + θ_{0}(u) + θ_{−}(y)

is the desired map.

Lemma 31 If V is a Jordan pair over Φ containing \frac{1}{2} and if ζ ∈ Inder(V), then L(V) is the universal covering of T KK(V, Inder(V)).

Proof. It is clear that L(V) is a central extension of T KK(V, Inder(V)). Moreover, if κ(ζ̃) = ζ, then [ζ̃, x] = σx for x ∈ L_{σ}, so L(V) is perfect and a covering of T KK(V, Inder(V)). Now let π : L̃ → T KK(V, Inder(V)) be any central extension. Again choose ζ̃ with π(ζ̃) = ζ. For x ∈ V^{σ}, let π(u) = x and set x̃ = σ[ζ̃, u]. We see π(x̃) = σ[ζ, x] = x. If also π(v) = x, then

[ζ̃, v] ∈ [ζ̃, u + Z(L̃)] = [ζ̃, u] = σx̃.

Thus, x̃ is the unique element of L̃ with π(x̃) = x and [ζ̃, x̃] =σx̃. If z ∈ V^{σ}, then [ζ̃, [x̃, z̃]] = 2[x̃, z̃], so π([x̃, z̃]) = 0, [x̃, z̃] ∈ Z(L̃), 2[x̃, z̃] = 0 and [x̃, z̃] = 0. Setting L̃_{σ} = {x̃ : x ∈ L_{σ}}, we see that

L̃_{+} + [L̃_{+}, L̃_{−}] + L̃_{−}

is a ℤ-graded subalgebra of L̃. Now θ_{σ}(x) = x̃ gives a Jordan pair isomorphism θ : V → (L̃_{+}, L̃_{−}) which extends to a Lie algebra homomorphism θ̂ : L(V) → L̃_{+} + [L̃_{+}, L̃_{−}] + L̃_{−}. Clearly, π ∘ θ̂ = κ on V^{σ} and hence on L(V).

Theorem 32 If V is a Jordan pair over a field of characteristic zero, then the universal divided power representation U(V) is isomorphic with the universal enveloping algebra of L(V).

Proof. We identify L(V) with its image in its universal enveloping algebra U(L(V)). We define

ρ_{n}^{σ}(v) = v^{(n)} = \frac{1}{n!}v^{n}

for v ∈ V^{σ}. Since [V^{σ}, V^{σ}] = 0, Lemma 8 shows that ρ_{n}^{σ} is a sequence of binomial divided power maps, and

ad_{v}^{(n)} = \frac{1}{n!}(ad v)^{n}.

If x ∈ V^{σ}, y ∈ V^{−σ}, then

ad_{x}^{(2)}y = \frac{1}{2}[x[xy]] = Q_{x}(y),
ad_{x}^{(k)}y = 0 for k ≥ 3,

since [x, Q_{x}(y)] = 0. Thus, by Lemma 10,

ad_{x}^{(k)}y^{(l)} = \frac{1}{l!} ∑ ∏_{i=1}^{l} ad_{x}^{(k_{i})}y

where k = ∑_{i=1}^{l} k_{i}.

If k > 2l, then some k_{i} > 2 and ad_{x}^{(k)}y^{(l)} = 0. If k = 2l and all k_{i} ≤ 2, then all k_{i} = 2 and

ad_{x}^{(2l)}y^{(l)} = \frac{1}{l!}(Q_{x}(y))^{l} = (Q_{x}(y))^{(l)}.

Thus, ρ is a d.p. representation of V, and there is an associative homomorphism φ : U(V) → U(L(V)) with φ ∘ γ_{n}^{σ} = ρ_{n}^{σ}.

On the other hand, by Corollary 28,

P(U(V)) = (V^{+})^{(1)} + P(H) + (V^{−})^{(1)}

is a ℤ-graded Lie algebra, so by Lemma 30 there is a Lie algebra homomorphism

γ̂_{1} : L(V) → P(U(V))

extending the Jordan pair homomorphism γ_{1} = (γ_{1}^{+}, γ_{1}^{−}). By the properties of the universal enveloping algebra, there is an associative homomorphism θ : U(L(V)) → U(V) extending γ̂_{1}. On V^{σ},

φ ∘ θ = φ ∘ γ_{1}^{σ} = ρ_{1}^{σ} = Id,
θ ∘ φ ∘ γ_{1}^{σ} = θ = γ_{1}^{σ}.

Since V^{±} generate U(L(V)) and (V^{±})^{(1)} generate U(V), φ and θ are inverse isomorphisms.

A Appendix: Homogeneous maps

We develop here properties of homogeneous maps and their linearizations modeled on the treatment by Jacobson [4] of quadratic maps. For a different approach see Loos [5], Appendix.

Let V and W be modules over a unital, commutative associative ring Φ. If f : V → W is constant, we say f is homogeneous of degree 0. For each n ≥ 1, we shall recursively define f : V → W to be homogeneous of degree n with (i, j)-linearization f_{ij} : V × V → W, where i + j = n, i, j ≥ 1, provided for all λ ∈ Φ, u, v, w ∈ V,

f(λv) = λ^{n}f(v),  (A1)
f(u + v) = f(u) + f(v) + ∑_{\begin{matrix}i+j=n \\ i,j≥1 \end{matrix}} f_{ij}(u, v),  (A2)

u → f_{lk}(u, w) is homogeneous of degree l with  (A3)
(i, j)-linearization (u, v) → f_{ijk}(u, v, w),
f_{ij}(v, v) = \binom{n}{i}f(v) for i + j = n, i, j ≥ 1,  (A4)
f_{ij}(u, v) = f_{ji}(v, u) for i + j = n, i, j ≥ 1,  (A5)
f_{ijk}(u, v, w) = f_{ikj}(u, w, v) for i + j + k = n, i, j, k ≥ 1.  (A6)

We also recursively define f_{i_{1}...i_{k}} where i_{1} + ... + i_{k} = n, i_{l} ≥ 1, to be the

(i_{1}, ..., i_{k})-linearization of f provided (v_{1}, v_{2}) → f_{i_{1}...i_{k}}(v_{1}, v_{2}, v_{3,}..., v_{k}) is the (i_{1}, i_{2})-linearization of v → f_{i_{1}+i_{2},i_{3}...i_{k}}(v, v_{3}, ..., v_{k}).

We remark that the conditions for f to be homogeneous of degree 1 reduce to f(λv) = λf(v) and f(u + v) = f(u) + f(v); i.e. f is linear. Also, f is homogeneous of degree 2 if and only if f(λv) = λ^{2}f(v) and

f_{11}(u, v) = f(u + v) − f(u) − f(v)

is bilinear. This is the usual definition of a quadratic map. For n ≥ 3, f does not always uniquely determine its linearizations. Indeed, f : x → x^{3} is homogeneous of degree 3 as a map from ℤ_{2} to itself with either f_{12}(u, v) = f_{21}(u, v) = uv or with f_{12}(u, v) = f_{21}(v, u) = 0.

If f is homogeneous of degree n with linearizations f_{i_{1}...i_{k}}, we define the kernel of f and its linearizations to be all v ∈ V such that

f(v) = f_{i_{1}...i_{k}}(v, v_{2}, ..., v_{k}) = 0

for all linearizations f_{i_{1}...i_{k}} and all v_{i} ∈ V. We implicitly assume that the linearizations are given and write this as ker(f). The proof of the following lemma is direct, and we omit it.

Lemma 33 If f : V → W is homogeneous of degree n with linearizations f_{i_{1}...i_{k}}, then

(i) ker(f) is a submodule of V,

(ii) if U ⊂ ker(f) is a submodule of V , then

f̄(v + U) = f(v),
f̄_{i_{1}...i_{k}}(v_{1} + U, ..., v_{k} + U) = f_{i_{1}...i_{k}}(v_{1}, ..., v_{k})

define a homogeneous map f̄: V/U → W with linearizations f̄_{i_{1}...i_{k}}.

(iii) if g : M → V and h : W → N are linear maps, then h ∘ f ∘ g is homogeneous with linearizations (x_{1}, ..., x_{k}) → h(f_{i_{1}...i_{k}}(g(x_{1}), ..., g(x_{k}))).

We shall now relate the above definition to the notion of polynomial maps. Let Φ[σ] = Φ[σ_{α} : α ∈ I], the polynomial ring over Φ with indeterminants σ_{α}, and let φ(σ) ∈ Φ[σ] be a homogeneous polynomial of degree n; i.e., φ is a linear combination of monomials of total degree n. Clearly, replacing σ_{α} by λσ_{α} in φ(σ) gives

φ(λσ) = λ^{n}φ(σ).  (P1)

We also can replace σ_{α} by sσ_{α}+tτ_{α} in φ(σ) to obtain φ(sσ+tτ) ∈ Φ[s, t, σ_{α}, τ_{α} : α ∈ I]. We have

φ(sσ + tτ) = φ(σ)s^{n} + φ(τ)t^{n} + ∑_{\begin{matrix}i+j=n \\ i,j≥1 \end{matrix}} φ_{ij}(σ, τ)s^{i}t^{j}  (P2)

where

φ_{ij}(σ, τ) ∈ Φ[σ, τ] = Φ[σ][τ] is a homogeneous  (P3)

polynomial of degree i in σ and of degree j in τ.

Replacing τ by σ in (P2) gives

(s + t)^{n}φ(σ) = φ(sσ + tσ)

= φ(σ)s^{n} + φ(σ)t^{n} + ∑_{\begin{matrix}i+j=n \\ i,j≥1 \end{matrix}} φ_{ij}(σ, σ)s^{i}t^{j}

so

φ_{ij}(σ, σ) = \binom{n}{i}φ(σ) for i + j = n, i, j ≥ 1.  (P4)

Replacing s, σ, t, τ by t, τ, s, σ in (P2) gives

φ_{ij}(σ, τ) = φ_{ji}(τ, σ) for i + j = n, i, j ≥ 1,  (P5)

and similarly we get

φ_{ijk}(σ, τ, ξ) = φ_{ikj}(σ, ξ, τ) for i + j + k = n, i, j, k ≥ 1.  (P6)

For any Φ-algebra A and any map x : I → A, we have the algebra homomorphism φ(σ) → φ(x) of Φ[σ] to A given by σ_{α} → x_{α}. In particular, if V is a free module with basis {v_{α} : α ∈ I}, φ(σ) ∈ Φ[σ], and u = ∑_{α∈I} a_{α}v_{α}, we can define

f(u) = φ(a). From (P1)-(P6), we see that f is homogeneous of degree n with (i, j)-linearization f_{ij}(u, v) = φ_{ij}(a, b) for v = ∑_{α∈I} b_{α}v_{α}. More generally, if M

is the set of all monomials m(σ) ∈ Φ[σ] which are homogeneous of degree n, if W is any Φ-module, and w : M → W is any map, then m(a) = m_{ij}(a, b) = 0 for all but finitely many m, and we can form

f(u) = ∑_{m∈M} m(a)w_{m},
f_{ij}(u, v) = ∑_{m∈M} m_{ij}(a, b)w_{m}.

and see that f : V → W is homogeneous of degree n with (i, j)-linearization f_{ij}. We shall say f is a homogeneous polynomial map of degree n.

Theorem 34 Let f : V → W be homogeneous of degree n with linearizations f_{i_{1}...i_{k}}, let M be a free module with g : M → V a linear epimorphism, and identify M/ ker(g) with V. The map F = f ∘ g is a homogeneous polynomial map of degree n with linearizations F_{i_{1}...i_{k}}(x_{1}, ..., x_{k}) = f_{i_{1}...i_{k}}(g(x_{1}), ..., g(x_{k})). Moreover, ker(g) ⊂ ker(F), so F̄ = f and F̄_{i_{1}...i_{k}} = f_{i_{1}...i_{k}}; i.e., every homogeneous map on V of degree n and its linearizations are induced by a homogeneous polynomial map on M of degree n and its linearizations.

Proof. By Lemma 33, we know that F is homogeneous of degree n with linearizations F_{i_{1}...i_{k}}. It is also clear that ker(g) ⊂ ker(F), so F̄ = f and F̄_{i_{1}...i_{k}} = f_{i_{1}...i_{k}}. Thus, it suffices to show that F is a homogeneous polynomial map of degree n with linearizations F_{i_{1}...i_{k}}. For convenience of notation, we set F_{n} = F. Let {x_{α} : α ∈ I} be a basis of M. If m = m(σ) = σ_{α_{1}}^{i_{1}}...σ_{α_{k}}^{i_{k}} is a monomial of degree n (with distinct α_{l}), we set

w_{m} = F_{i_{1}...i_{k}}(x_{α_{1}}, ..., x_{α_{k}}).

By (A5) and (A6), we see that the definition of w_{m} does not depend on the order of the factors σ_{α_{i}}^{i_{i}} in m. Using the map w : m → w_{m}, we can define the homogeneous polynomial map F′ and its linearizations F′_{i_{1}...i_{k}}as above. To show F_{i_{1}...i_{k}} = F′_{i_{1}...i_{k}}(including F = F′), it suffices by (A1) and (A2) to show that these agree on all subsets of basis elements. If α_{1}, ..., α_{k} are distinct, then

F′_{i_{1}...i_{k}}(x_{α_{1}}, ..., x_{α_{k}}) = w_{m} = F_{i_{1}...i_{k}}(x_{α_{1}}, ..., x_{α_{k}})

for m = σ_{α_{1}}^{i_{1}}...σ_{α_{k}}^{i_{k}}. On the other hand, if α_{1} = α_{2}, then by (A3) and (A4), we have

F_{i_{1}...i_{k}}(x_{α_{1}}, ..., x_{α_{k}}) = \binom{i_{1} + i_{2}}{i_{1}}F_{i_{1}+i_{2},i_{3}...i_{k}}(x_{α_{1}}, x_{α_{3}}..., x_{α_{k}})

and similarly for F′, so we can reduce to the case with distinct α_{1}, ..., α_{k}.

Shortly, we shall prove an extension result for homogeneous maps for two types of extensions of the modules. First, let Ω be an extension ring of Φ and form the Ω-module V_{Ω} = Ω ⊗_{Φ} V. We shall identify V with 1 ⊗ V to view V_{Ω}

as an extension of V. Secondly, we can form the formal power series ring Φ[[s]] as well as V[[s]], the set of formal power series with coefficients from V. We shall view V[[s]] as a Φ[[s]]-module, and note that V[[s]] contains the Φ-module V and Φ[s]-module V[s] consisting of polynomials with coefficients from V. We can also identify V[s] with V_{Φ[s]}.

Theorem 35 Let f : V → W be homogeneous of degree n with linearizations f_{i_{1}...i_{k}} and let Ω be an extension ring of Φ. If (Ṽ, W̃) is either (V_{Ω}, W_{Ω}) or (V[[s]], W[[s]]), then there is a unique homogeneous map of degree n, f̃: Ṽ → W̃ with linearizations f̃_{i_{1}...i_{k}}, which extend f and f_{i_{1}...i_{k}}.

Proof. Let M be free with g : M → V a linear epimorphism. Let F = f ∘g, so F is a polynomial map with linearizations F_{i_{1}...i_{k}} such that F̄ = f and F̄_{i_{1}...i_{k}} = f_{i_{1}...i_{k}}, as in Theorem 34. It is easily seen that M_{Ω} is free, that Id ⊗ g : M_{Ω} → V_{Ω} is a linear epimorphism, and that ker(Id ⊗ g) = Ω ⊗ ker(g). We can identify the monomials in Φ[σ] with the monomials in Ω[σ] and use the map w : m → w_{m} = 1 ⊗ w_{m} ∈ W_{Ω} to get a polynomial map F̃ : M_{Ω} → W_{Ω} with linearizations F̃_{i_{1}...i_{k}} extending F and F_{i_{1}...i_{k}}. Since

ker(Id ⊗ g) = Ω ⊗ ker(g)

⊂ Ω ⊗ ker(F)

⊂ ker(F̃), we can form f̃ = \overline{F̃} : V_{Ω} → W_{Ω}, homogeneous of degree n with

f̃(1 ⊗ v) = F̃(1 ⊗ x) = 1 ⊗ F (x) = 1 ⊗ f(v)

if g(x) = v. Thus, f̃ extends f and similarly f̃_{i_{1}...i_{k}} extends f_{i_{1}...i_{k}}. Since V_{Ω} is the Ω-span of V, the extensions are unique.

In Zariski and Samuel [7], Chapter VII, it is shown that Φ[[s]] is a complete metric space relative the metric d(a, b) = 2^{−o(a−b)} for a ≠ b where o(a) = k for a = ∑ a_{i}s^{i} if k is minimal with a_{k} ≠ 0. Moreover, addition and multiplication in Φ[[s]] are continuous operations, and Φ[s] is dense in Φ[[s]]. Similarly, V[[s]] is a complete metric space, the module operations are continuous, and V[s] is dense in V[[s]]. By the first part of this theorem, we have a homogeneous map f′ : V[s] → W[s] ⊂ W[[s]] of degree n with linearizations f′_{i_{1}...i_{k}} which extend f and f_{i_{1}...i_{k}}. If v ∈ V[s] with o(v) = k, we can write v = s^{k}v_{0}, so f′(v) = s^{kn}f′(v_{0}) and o(f′(v)) ≥ n o(v). Thus, f′and similarly f′_{i_{1}...i_{k}} are uniformly continuous. Since V[s] is dense and W[[s]] is complete, f′ and f′_{i_{1}...i_{k}} extend to uniformly continuous maps f̃: V[[s]] → W[[s]] and f̃_{i_{1}...i_{k}}. We have that

(λ, v) → f̃(λv) − λ^{n}f̃(v)

is a continuous map Φ[[s]] × V[[s]] → W[[s]] which vanishes on the dense subset Φ[s] × V[s]. Thus, (A1) holds for f̃. Similar density arguments can be used to show (A2) and (A4)-(A6) for f̃, f̃_{ij}, and f̃_{ijk}, while (A3) follows by induction on degree. Thus, f̃ is homogeneous of degree n with linearizations f̃_{i_{1}...i_{k}}. If g : V[[s]] → W[[s]] is any homogeneous map of degree n extending f, then the restriction g′ of g to V[s] is a homogeneous map of degree n extending f and mapping V[s] into W[s]. Thus, g′ = f′. Since g is continuous and agrees with f̃ on a dense subset, g = f̃. Similarly, g_{i_{1}...i_{k}} = f̃_{i_{1}...i_{k}}.

References

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