The Symmetric and Antisymmetric Parts of an Algebra Product

Introduction

Every bilinear product carries two operations made from it, and over a field in which $2$ is invertible the two are the halves of the product:

$$ xy = \tfrac{1}{2}(xy + yx) + \tfrac{1}{2}(xy - yx). $$

The first half is commutative in every algebra, the second anticommutative in every algebra, and together they determine the product. The split is therefore not a choice of notation but a canonical decomposition, the one induced by the exchange $x \otimes y \mapsto y \otimes x$ of the tensor square. What the halves do not carry automatically is the identity each is named after: the antisymmetric half satisfies the Jacobi identity only when the algebra is Lie-admissible, the symmetric half satisfies the Jordan identity only when it is Jordan-admissible, and each of the two conditions is an identity in one half alone, so that either can fail while the other holds. Associativity implies both, the two conditions are otherwise independent of one another, and the derivation identity that links the two halves in the associative case is a third condition, not implied by the two together.

This article develops the split and its uniqueness, the two admissibility conditions, the criterion that reads Lie-admissibility off the associator, the two examples that separate the conditions, the passage to the opposite algebra and to the products with the same symmetric half, and the collapse of the construction when $2$ is not invertible. It is a general article: Symmetric and Antisymmetric Decomposition of the Four General Products carries out the same construction for the four general products of the biquaternion algebra, and the present article is the form of it that belongs to the general theory.

Throughout, $k$ is a field of characteristic not $2$ and $A$ is a $k$-algebra in the broad sense of Algebras: a $k$-module with a bilinear product and no further hypothesis. The product need not be associative, commutative or unital. Where a dimension is counted, $n = \dim_k A$ and $(e_i)$ is a basis.

The symbols are the corpus's. The commutator is $[x,y] = xy - yx$, without a factor $\tfrac{1}{2}$; the outer product is $x \wedge y = \tfrac{1}{2}(xy - yx)$, so that $[x,y] = 2(x \wedge y)$, and the wedge is the notation of Symmetric and Antisymmetric Decomposition of the Four General Products. The symmetrised product is $x \bullet y = \tfrac{1}{2}(xy + yx)$, the corpus's bullet, as in Jordan Algebras, where the same symbol stands for the unhalved product $xy + yx$ over a ring in which $2$ is not assumed invertible; the two products differ by the unit $2$, so over the field $k$ the article follows the halved form. The associator is the ternary bracket

$$ [x,y,z] = (xy)z - x(yz) $$

of Non-Associative Algebras and the Property Ladder, and it is not the binary commutator $[x,y]$. The two algebras that the split produces are written $A^{\bullet}$ and $A^{\wedge}$. Jordan Algebras writes the first $A^{+}$ and Unital Algebras uses $A^{+}$ for the unitisation of a non-unital algebra, so the bullet is used here to keep the two apart; the second is written $A^{(-)}$ in The Commutator Operator and $A^-$ in Associative Algebras, and all these names are recorded in the notation table at the end.

The theories the two halves lead to are owned elsewhere and are cited rather than restated. The Lie theory is Lie Algebras, whose construction from an associative algebra is also in Associative Algebras and The Commutator Operator; the Jordan theory is Jordan Algebras; the operator layer is The Operators on an Algebra, Left and Right Multiplication, The Sandwich Operator on an Algebra and The Commutator Operator; the ternary product generated by the symmetrisation is The Lie–Jordan Decomposition of a Bilinear Product and the Jordan Triple System; the biquaternion instances are Symmetric and Antisymmetric Decomposition of the Four General Products; the case of a product conjugate-linear in one argument is The Symmetric and Antisymmetric Parts of a Sesqualgebra Product, with The Conjugate-Symmetric and Skew-Conjugate-Symmetric Parts of a Sesquilinear Product for the exchange that keeps the class; and the identities below associativity are Non-Associative Algebras and the Property Ladder. The article is algebraic throughout and uses no distance, no norm and no limit.

The Two Halves of a Product

The Symmetrisation and the Antisymmetrisation

Definition. For $x, y \in A$ put

$$ x \bullet y = \tfrac{1}{2}(xy + yx), \qquad x \wedge y = \tfrac{1}{2}(xy - yx). $$

The symmetrisation $x \bullet y$ is the symmetric part of the product and the antisymmetrisation $x \wedge y$ its antisymmetric part. The algebra $A$ with the product $\bullet$ is written $A^{\bullet}$, and with the product $\wedge$ is written $A^{\wedge}$.

Proposition. Both operations are $k$-bilinear, and for all $x, y \in A$

$$ x \bullet y = y \bullet x, \qquad x \wedge y = -y \wedge x, \qquad x \wedge x = 0, \qquad x \bullet x = x^2, $$

together with the three reconstructing identities

$$ \boxed{\;xy = x \bullet y + x \wedge y, \qquad yx = x \bullet y - x \wedge y, \qquad [x,y] = 2(x \wedge y)\;}. $$

Proof. Bilinearity is inherited from the product, coefficient by coefficient. The first two identities restate the definitions after the exchange of $x$ and $y$, and the third is $x \wedge x = \tfrac{1}{2}(x^2 - x^2) = 0$. The fourth is $x \bullet x = \tfrac{1}{2}(x^2 + x^2) = x^2$. Adding the two definitions gives the product, subtracting them gives the product in the other order, and doubling the antisymmetrisation gives the commutator.

The third and fourth identities are the reason the split is worth making: it separates the part of the product that the squares already carry from the part they cannot carry.

Theorem (uniqueness of the split). Suppose $xy = s(x,y) + t(x,y)$ for all $x, y$, with $s$ symmetric and $t$ antisymmetric. Then $s(x,y) = x \bullet y$ and $t(x,y) = x \wedge y$.

Proof. Interchanging the arguments in the assumed decomposition gives $yx = s(x,y) - t(x,y)$, because $s$ is symmetric and $t$ antisymmetric. Adding the two relations, $xy + yx = 2s(x,y)$, and subtracting them, $xy - yx = 2t(x,y)$; the two quotients by $2$ are the symmetrisation and the antisymmetrisation. The division by $2$ is the only place where the characteristic is used.

Proposition (what each half detects). The symmetrisation vanishes identically exactly when the product is anticommutative, and the antisymmetrisation vanishes identically exactly when the product is commutative:

$$ x \bullet y = 0 \ \text{ for all } x, y \iff xy = -yx \ \text{ for all } x, y, $$

$$ x \wedge y = 0 \ \text{ for all } x, y \iff xy = yx \ \text{ for all } x, y. $$

Proof. The symmetrisation is zero if and only if $xy + yx = 0$ for every pair, which is the anticommutative law, and the antisymmetrisation is zero if and only if $xy - yx = 0$ for every pair, which is the commutative law.

Remark. The two extremes are therefore the two degenerate cases of the construction. The antisymmetric half measures nothing but the failure of commutativity, and the symmetric half nothing but the failure of anticommutativity. Nothing else about the product enters either half.

The Exchange and the Structure Constants

Remark (the split is the exchange). The tensor square $A \otimes A$ carries the exchange

$$ \tau : A \otimes A \longrightarrow A \otimes A, \qquad \tau(x \otimes y) = y \otimes x, $$

an involution. Over a field in which $2$ is invertible its fixed part and its anti-fixed part are the symmetric square and the exterior square,

$$ S^2 A = \{\xi \in A \otimes A : \tau(\xi) = \xi\}, \qquad \Lambda^2 A = \{\xi \in A \otimes A : \tau(\xi) = -\xi\}, \qquad A \otimes A = S^2 A \oplus \Lambda^2 A, $$

the two projections onto the summands being $\tfrac{1}{2}(1 \pm \tau)$. A product is a linear map $m : A \otimes A \to A$, and the two halves of the product are its values on the two components:

$$ x \bullet y = m\Bigl(\tfrac{1}{2}(x \otimes y + y \otimes x)\Bigr), \qquad x \wedge y = m\Bigl(\tfrac{1}{2}(x \otimes y - y \otimes x)\Bigr). $$

The split of the product is the split of its domain, and over the field $k$ the exchange is diagonalisable exactly when the characteristic is not $2$, which is the hypothesis of the whole article.

Remark (the structure constants). Let $A$ be free of finite rank $n$ with basis $(e_i)$ and write the product of the basis vectors as

$$ e_i e_j = \sum_k c^k{}_{ij}\, e_k . $$

The symmetrisation and the antisymmetrisation have the structure constants obtained from the pair of lower indices alone,

$$ e_i \bullet e_j = \sum_k \tfrac{1}{2}\bigl(c^k{}_{ij} + c^k{}_{ji}\bigr) e_k, \qquad e_i \wedge e_j = \sum_k \tfrac{1}{2}\bigl(c^k{}_{ij} - c^k{}_{ji}\bigr) e_k . $$

Counting the constants, a product on $A$ is an element of $A^* \otimes A^* \otimes A$, of dimension $n^3$, its symmetric half is a symmetric bilinear map $A \times A \to A$, with $\binom{n+1}{2} n$ constants, and its antisymmetric half is an antisymmetric one, with $\binom{n}{2} n$. The two counts add to $n^3$:

$n$	products, $n^3$	symmetric half, $\binom{n+1}{2} n$	antisymmetric half, $\binom{n}{2} n$
2	8	6	2
3	27	18	9
4	64	40	24

Remark. Every row of the table is the ordinary identity $\binom{n+1}{2} + \binom{n}{2} = n^2$ multiplied by $n$, and it is the numerical shadow of the decomposition $S^2 A \oplus \Lambda^2 A = A \otimes A$ of the preceding remark.

The Space of Products and the Two Projections

Definition. On the space of bilinear products on $A$ define the two operators

$$ n(m)(x,y) = \tfrac{1}{2}\bigl(m(x,y) + m(y,x)\bigr), \qquad q(m)(x,y) = \tfrac{1}{2}\bigl(m(x,y) - m(y,x)\bigr). $$

Proposition. The operators $n$ and $m$ are $k$-linear and satisfy

$$ n^2 = n, \qquad m^2 = m, \qquad pq = qp = 0, \qquad n+m = \mathrm{id}, $$

so that the space of products is the direct sum of the symmetric products, the image of $n$, and the antisymmetric products, the image of $m$.

Proof. Linearity is immediate from the definition. For $n^2 = n$, apply $n$ twice:

$$ n(n(m))(x,y) = \tfrac{1}{2}\bigl(n(m)(x,y) + n(m)(y,x)\bigr) = \tfrac{1}{4}\bigl(m(x,y) + m(y,x) + m(y,x) + m(x,y)\bigr), $$

which is $\tfrac{1}{2}(m(x,y) + m(y,x)) = n(m)(x,y)$, and the computation for $m$ is the same with the two signs. For $pq$,

$$ n(q(m))(x,y) = \tfrac{1}{4}\bigl(m(x,y) - m(y,x) + m(y,x) - m(x,y)\bigr) = 0, $$

and $qp = 0$ likewise. Finally $n(m) + q(m)$ is $\tfrac{1}{2}(m(x,y) + m(y,x)) + \tfrac{1}{2}(m(x,y) - m(y,x)) = m(x,y)$, which is $n+m = \mathrm{id}$. A direct sum decomposition is a family of idempotents that sum to the identity and annihilate one another, so the space of products is the sum of the two images, and the sum is direct.

Corollary (each half can be prescribed). A bilinear map on $A$ is a symmetrisation if and only if it is symmetric, and an antisymmetrisation if and only if it is antisymmetric. Moreover, for any symmetric $s$ and any antisymmetric $t$ the product $m = s + t$ satisfies $n(m) = s$ and $q(m) = t$.

Proof. A symmetric $s$ satisfies $n(s) = s$, and an antisymmetric $t$ satisfies $n(t) = 0$ and $q(t) = t$; this gives both claims of the first sentence. For the second, $n(s+t) = n(s) + n(t) = s$ and $q(s+t) = q(s) + q(t) = t$.

Remark. The corollary is the precise sense in which the two halves are independent data. The symmetric half of a product places no constraint on its antisymmetric half and conversely, and a condition imposed on one half can be studied with no hypothesis at all on the other.

What the Squares See

Proposition (polarisation of the square). For all $x, y \in A$,

$$ (x + y)^2 - x^2 - y^2 = xy + yx = 2\,(x \bullet y). $$

Proof. Expand the square by bilinearity and cancel the two squares:

$$ (x+y)^2 = x^2 + xy + yx + y^2 . $$

Remark. The identity is the reason the symmetrisation is the only half the square map carries. The symmetrisation determines the squares, by the fourth identity of the first proposition, and the squares determine the symmetrisation, by the polarisation identity; the two pieces of data are therefore equivalent, and the antisymmetric half is invisible to the square map. A product and its opposite have the same squares and the same symmetric half, and differ exactly in the sign of the antisymmetric half. The identity in this place is the degree-two case of the general reconstruction of a symmetric multilinear map from its diagonal, which is The Polarisation Operator.

The Two Admissibility Conditions

Lie-Admissible Algebras

Definition. The algebra $A$ is Lie-admissible if the antisymmetrisation is a Lie bracket, that is if

$$ x \wedge x = 0, \qquad (x \wedge y) \wedge z + (y \wedge z) \wedge x + (z \wedge x) \wedge y = 0 $$

for all $x, y, z$; the second is the Jacobi identity. The algebra $A^{\wedge}$ is then a Lie algebra, the antisymmetrised algebra of $A$.

The first condition is automatic, by the third identity of the first proposition. The second is not, and the whole content of the definition lies in it.

Proposition. The algebra $A$ is Lie-admissible if and only if the commutator satisfies the Jacobi identity,

$$ \boxed{\;[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0\;} $$

for all $x, y, z$.

Proof. The commutator is twice the outer product, $[x,y] = 2(x \wedge y)$. In each term of the Jacobi sum of the commutator the inner bracket contributes a factor $2$ and the outer bracket another, so

$$ [[x,y],z] + [[y,z],x] + [[z,x],y] = 4\Bigl((x \wedge y) \wedge z + (y \wedge z) \wedge x + (z \wedge x) \wedge y\Bigr), $$

and a scalar multiple of a quantity vanishes exactly when the quantity does, with $4 \neq 0$ because the characteristic is not $2$.

Proposition (the spanning-set criterion). The Jacobi identity is a trilinear condition: if it holds for all triples from a subset $S \subseteq A$, it holds for all triples from the span of $S$. In particular, if $A$ is free with basis $(e_i)$, Lie-admissibility is decided by the triples of basis elements.

Proof. The Jacobi sum is linear in each of its three arguments, so writing $x = \sum_i \alpha_i x_i$ and similarly for $y$ and $z$ expands the sum as a combination of Jacobi sums of the $x_i, y_j, z_k$. If those vanish, so does every combination.

The Associator Criterion

Theorem (the associator criterion of Lie-admissibility). The antisymmetrised algebra $A^{\wedge}$ satisfies the Jacobi identity if and only if the associator of $A$ satisfies the identity

$$ \boxed{\;[x,y,z] + [y,z,x] + [z,x,y] = [y,x,z] + [z,y,x] + [x,z,y]\;} $$

for all $x, y, z$. Equivalently, the Jacobi sum of the commutator is the signed sum of the six associators of the three arguments,

$$ [[x,y],z] + [[y,z],x] + [[z,x],y] = \sum_{\sigma \in S_3} \operatorname{sgn}(\sigma)\, [x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}]. $$

Proof. Expand the three terms of the Jacobi sum. The first is

$$ [[x,y],z] = (xy)z - (yx)z - z(xy) + z(yx), $$

and the second and the third are the same expression with the arguments cycled,

$$ [[y,z],x] = (yz)x - (zy)x - x(yz) + x(zy), $$

$$ [[z,x],y] = (zx)y - (xz)y - y(zx) + y(xz). $$

The right-hand side of the first display, expanded over the six permutations of $x, y, z$, is

$$ [x,y,z] - [y,x,z] + [y,z,x] - [z,y,x] + [z,x,y] - [x,z,y], $$

which is the signed sum of the six associators. Collecting the twelve monomials of the three expanded terms gives exactly this expression, monomial by monomial; the two cyclic sums of three associators are the positive and the negative part of the signed sum. The Jacobi sum therefore vanishes identically exactly when the two cyclic sums agree.

Corollary. An associative algebra is Lie-admissible, because its associator vanishes; and a commutative algebra is Lie-admissible, because its commutator vanishes. The identity of the theorem is thus the exact measure of what associativity contributes to the Jacobi identity, and the case of the commutative algebras shows that the criterion is not a disguised form of associativity.

The first clause of the corollary is the standard construction of a Lie algebra from an associative one, the passage from $A$ to its commutator algebra. The commutator algebra is defined in Associative Algebras, §The Universal Enveloping Algebra, and it is exhibited there as one of the algebras that fail associativity, in §Algebras That Fail Associativity; the bracket with its Jacobi identity is The Commutator Operator, §The Commutator Bracket, where the identity is proved and not repeated here.

Jordan-Admissible Algebras

Definition. The algebra $A$ is Jordan-admissible if the symmetrisation is a Jordan product, that is if it is commutative, which it always is, and satisfies the Jordan identity

$$ x^2 \bullet (y \bullet x) = (x^2 \bullet y) \bullet x, \qquad x^2 = x \bullet x, $$

for all $x, y \in A$. The algebra $A^{\bullet}$ is then a Jordan algebra, the symmetrised algebra of $A$. The identity as written is the form of Jordan Algebras, §The Jordan Product, and its nested form is $x \bullet (y \bullet (x \bullet x)) = (x \bullet y) \bullet (x \bullet x)$, the same statement with the arguments of the two outer products exchanged, which is legitimate because the product is commutative.

Theorem. Every associative algebra is Jordan-admissible.

Proof. This is the theorem that the symmetrisation of an associative algebra is a Jordan algebra, with the computation of the Jordan identity in Jordan Algebras, §The Symmetrisation of an Associative Algebra; the identity is not reproved here.

Proposition. A commutative algebra is Jordan-admissible if and only if it is a Jordan algebra, and its symmetrisation is itself: $A^{\bullet} = A$.

Proof. If the product is commutative then $x \bullet y = \tfrac{1}{2}(xy + yx) = xy$, so the two products coincide and the Jordan identity for the bullet product is the Jordan identity for the original product.

Remark. The two theorems give the two sufficient conditions that are used in the corpus. Associativity gives both halves their identity at once, and the two are then the commutator algebra and the special Jordan algebra of the same associative algebra. Jordan-admissibility is not automatic even for a commutative algebra, and the next section exhibits a commutative algebra in which it fails.

The Two Conditions Are Independent

Theorem. Neither admissibility condition implies the other. There is a Lie-admissible algebra that is not Jordan-admissible, and a Jordan-admissible algebra that is not Lie-admissible.

Proof. By the corollary of §The Space of Products and the Two Projections, the two halves of a product on a fixed space are independent data, so it is enough to exhibit a symmetric bilinear map that is not Jordan and an antisymmetric bilinear map that is not Lie, and to pair each with the other's good instance. The space is $k^3$, with basis $u, v, w$, in all four instances.

The symmetric part $s$ that is not Jordan. Define

$$ u \cdot u = v, \qquad u \cdot v = v \cdot u = v, \qquad v \cdot v = 0, $$

and let every product involving $w$ vanish. The product is commutative and not Jordan: with $x = y = u$, so that $x^2 = v$, the two sides of the Jordan identity are

$$ x^2 \bullet (y \bullet x) = v \bullet (u \bullet u) = v \bullet v = 0, \qquad (x^2 \bullet y) \bullet x = (v \bullet u) \bullet u = v \bullet u = v, $$

and $v \neq 0$.

The antisymmetric part $t$ that is not Lie. Define

$$ u \wedge v = u, \qquad v \wedge w = v, \qquad w \wedge u = w $$

and the reversed products by antisymmetry. The Jacobi sum at $(u, v, w)$ is

$$ (u \wedge v) \wedge w + (v \wedge w) \wedge u + (w \wedge u) \wedge v = u \wedge w + v \wedge u + w \wedge v = -w - u - v, $$

which is nonzero.

The two good instances. Let $t_0$ be the cross product on $k^3$, with $u \wedge v = w$, $v \wedge w = u$ and $w \wedge u = v$. It is a Lie bracket: for an alternating bracket the Jacobi sum vanishes whenever two arguments coincide, so on a basis only the orderings of the three distinct basis vectors need be examined, and there

$$ [u,[v,w]] + [v,[w,u]] + [w,[u,v]] = [u,u] + [v,v] + [w,w] = 0 , $$

the spanning-set criterion of §Lie-Admissible Algebras then giving the identity for all triples; the cross product on $\mathbb{R}^3$ is the standard three-dimensional example of Lie Algebras. Let $s_0$ be the componentwise product, $u \cdot u = u$, $v \cdot v = v$, $w \cdot w = w$ and every mixed product zero: it is commutative and associative, hence Jordan by the theorem of §Jordan-Admissible Algebras.

Witness 1: Lie-admissible and not Jordan-admissible. Take the product $m_1 = s + t_0$, whose table is

$m_1$	$u$	$v$	$w$
$u$	$v$	$v + w$	$-v$
$v$	$v - w$	$0$	$u$
$w$	$v$	$-u$	$0$

Its antisymmetric part is $t_0$, so $A^{\wedge}$ is the cross-product Lie algebra and the algebra is Lie-admissible; its symmetric part is $s$, which fails the Jordan identity, so it is not Jordan-admissible.

Witness 2: Jordan-admissible and not Lie-admissible. Take the product $m_2 = s_0 + t$, whose table is

$m_2$	$u$	$v$	$w$
$u$	$u$	$u$	$-w$
$v$	$-u$	$v$	$v$
$w$	$w$	$-v$	$w$

Its symmetric part is $s_0$, so $A^{\bullet} = (k^3, s_0)$ is a Jordan algebra and the algebra is Jordan-admissible; its antisymmetric part is $t$, which fails the Jacobi identity, so it is not Lie-admissible.

The two witnesses live on the same space and differ only in the choice of the two halves.

Remark. The proof uses the independence of the halves in its strongest form: not only does admissibility of one half not imply admissibility of the other, but the condition on each half is a condition on that half alone, so an algebra may fail either condition without touching the other. The classes of Lie-admissible and of Jordan-admissible algebras therefore overlap without containing one another.

Remark. Admissibility of both halves is still weaker than associativity. The product $s_0 + t_0$ of the componentwise product and the cross product is Lie-admissible and Jordan-admissible, and it is not associative: on the basis triple $(u, u, v)$ the two bracketings differ, since

$$ (uu)v = uv = s_0(u,v) + u \times v = w, \qquad u(uv) = uw = s_0(u,w) + u \times w = -v, $$

and the associator is $w - (-v) = v + w \neq 0$. Something more than the two admissibility conditions is needed to reach associativity.

The Opposite Algebra and the Relatives of a Product

Definition. The opposite algebra $A^{\mathrm{opp}}$ is the same $k$-module with the reversed product $x \cdot^{\mathrm{opp}} y = yx$, as in Opposite Algebras and Anti-Isomorphisms. More generally, for $\lambda \in k$ let $A_\lambda$ be the same module with the product

$$ x \cdot_\lambda y = x \bullet y + \lambda\,(x \wedge y) = \tfrac{1}{2}(1+\lambda)\,xy + \tfrac{1}{2}(1-\lambda)\,yx . $$

The algebra $A_1$ is $A$, the algebra $A_{-1}$ is $A^{\mathrm{opp}}$, and $A_0$ is the commutative algebra $A^{\bullet}$.

Proposition. For every $\lambda \in k$ the symmetrisation of $A_\lambda$ is that of $A$, and the antisymmetrisation is the multiple by $\lambda$:

$$ (A_\lambda)^{\bullet} = A^{\bullet}, \qquad (A_\lambda)^{\wedge} = \lambda\, A^{\wedge}. $$

Proof. The symmetrisation of $x \cdot_\lambda y$ is computed from the two definitions: adding $x \cdot_\lambda y$ and $y \cdot_\lambda x$ cancels the two occurrences of $\wedge$ and leaves $2(x \bullet y)$, and subtracting them cancels the two occurrences of $\bullet$ and leaves $2\lambda(x \wedge y)$.

Corollary. The passage to the opposite algebra leaves the symmetric half fixed and reverses the sign of the antisymmetric half. Lie-admissibility and Jordan-admissibility are properties of $A$ and of $A^{\mathrm{opp}}$ together: an algebra is Lie-admissible if and only if its opposite is, because the Jacobi identity is homogeneous and a reversal of sign does not affect its vanishing, and it is Jordan-admissible if and only if its opposite is, because the symmetric half is unchanged.

Remark. The family $A_\lambda$ is the full family of algebras on the module with the prescribed symmetric half and an antisymmetric half proportional to the given one, so it shows exactly which features of $A$ the antisymmetric half can change. The symmetric half is untouched by $\lambda$, so Jordan-admissibility is the same for every member of the family; and the antisymmetric half is a nonzero multiple of that of $A$ for $\lambda \neq 0$ and vanishes at $\lambda = 0$ alone, so the algebras $A_\lambda$ with $\lambda \neq 0$ are Lie-admissible exactly when $A$ is, and $A_0$ is Lie-admissible always, its antisymmetrisation being zero. The vanishing of the antisymmetric half is the only qualitative change the parameter can make.

When 2 Is Not Invertible

The Collapse in Characteristic 2

Remark. Over a field of characteristic $2$ the sign $-1$ is $1$, and the exchange has only one eigenvalue. The antisymmetric tensors and the symmetric tensors coincide, because $x \otimes y - y \otimes x = x \otimes y + y \otimes x$, so $\Lambda^2 A = S^2 A$ is the fixed space of the exchange, of dimension $\binom{n+1}{2}$, which does not fill $A \otimes A$ for $n \geq 2$. The projections $\tfrac{1}{2}(1 \pm \tau)$ and the operators $n$ and $m$ of §The Space of Products and the Two Projections are not available, and the two integral combinations that remain, $xy + yx$ and $xy - yx$, coincide. Writing $\Sigma(m)(x,y) = m(x,y) + m(y,x)$ for the unhalved symmetrisation, the operation is not a projection but is nilpotent:

$$ \Sigma(\Sigma(m))(x,y) = \Sigma(m)(x,y) + \Sigma(m)(y,x) = 2\,\Sigma(m)(x,y), $$

an identity over every commutative ring, which in characteristic $2$ reads $\Sigma^2 = 0$. The symmetrisation is therefore square-zero: it carries every product to a product whose symmetrisation vanishes.

Example. In characteristic $2$ the symmetrisation of a commutative product vanishes identically, because $xy + yx = 2xy = 0$, and its antisymmetrisation vanishes identically too, because the product is commutative. Two distinct commutative products on the same space are therefore indistinguishable by their two halves. On the two-dimensional space over $\mathbb{F}_2$ with basis $u, v$, let $A$ have the product $uu = u$ with $uv = vu = vv = 0$, and let $B$ have the product

$$ uu = u + v, \qquad uv = vu = u, \qquad vv = 0 . $$

Both products are commutative, so the symmetrisation and the antisymmetrisation of both vanish, and the two products are distinct: $uu$ is $u$ in the first and $u+v$ in the second. In characteristic $2$ the product is not determined by its two halves, and the split of this article does not exist.

Remark. In characteristic $2$ the two laws $xy = yx$ and $xy = -yx$ are one law, so commutativity and anticommutativity coincide, and the anticommutative law no longer forces the square of an element to vanish: the product of $A$ above satisfies it and has $uu = u \neq 0$.

Remark. The same collapse affects the notion of a Lie algebra, and for the same reason: in characteristic $2$ the antisymmetry of a bracket is vacuous, since $-[x,y] = [x,y]$, and a Lie bracket must be defined by the explicit alternation $[x,x] = 0$ together with the Jacobi identity. The corpus states the hypothesis exactly where it is needed and no more: Jordan Algebras works over a commutative ring and names the invertibility of $2$ at the point where the symmetrisation is defined, while the articles that carry the bracket alone, such as The Commutator Operator, need no hypothesis, the expression $xy - yx$ being integral. The present article needs it throughout, and the two sections above record what fails without it.

The General Commutative Ring

Remark. The construction needs $\tfrac{1}{2}$, not a field: over a commutative ring $R$ in which $2$ is invertible the whole of the article goes through with $k$ replaced by $R$. If $2$ is not invertible, the two integral combinations $xy + yx$ and $xy - yx$ still exist, and they still determine the product when $2$ is not a zero divisor, since $xy$ is recovered from $xy + yx$ and $xy - yx$ by halving a relation that is then unique; but the two combinations are no longer the two summands of a direct sum, and the idempotents $n$ and $m$ of §The Space of Products and the Two Projections are not available. The field of characteristic not $2$ is the setting in which the decomposition itself, and not only the two operations, exists.

Where the Corpus Uses the Split

The split is used at several places in the corpus, and each use is recorded in the article that owns it.

The commutator algebra and the special Jordan algebra. For an associative algebra the two halves are the source of the corpus's Lie and Jordan examples: the commutator algebra $A^{(-)}$ is a Lie algebra, by the theorem of §The Associator Criterion, and the symmetrised algebra $A^{+}$ is the special Jordan algebra of Jordan Algebras, §The Symmetrisation of an Associative Algebra. The names are the corpus's: $A^{+}$ from Jordan Algebras, and $A^{(-)}$ from The Commutator Operator, with $A^-$ the variant of Associative Algebras; they denote the algebras written $A^{\wedge}$ and $A^{\bullet}$ above. The dichotomy between the special and the exceptional Jordan algebras is stated in Special and Exceptional Jordan Algebras, not here.

The operator layer. The two halves are the symmetrisation and the antisymmetrisation of the one-sided multiplications. With $L(x)(y) = xy$ and $R(x)(y) = yx$ the left and right multiplications of Left and Right Multiplication,

$$ x \bullet y = \tfrac{1}{2}(L(x) + R(x))(y), \qquad x \wedge y = \tfrac{1}{2}(L(x) - R(x))(y), $$

so the symmetric half is the balanced multiplication and the antisymmetric half is half the commutator operator $\mathrm{ad}_x = L(x) - R(x)$ of The Commutator Operator. That operator is the inner derivation of the algebra when the product is associative, and it is the reason the bracket belongs to the operator layer rather than to the product. In a unital algebra the sandwich operator of The Sandwich Operator on an Algebra recovers the commutator as $T_{a,b}(1) - T_{b,a}(1) = [a,b]$.

The compatibility and the ternary product. In the associative case the bracket is not only a Lie bracket and the bullet product not only a Jordan product: the bracket is a derivation of the bullet product, $[x, y \bullet z] = [x,y] \bullet z + y \bullet [x,z]$, and the two halves satisfy the fundamental identity that accompanies it,

$$ [x, y \bullet z] + [y, z \bullet x] + [z, x \bullet y] = 0 . $$

The ternary product that the symmetrisation generates, its symmetry in the outer arguments and the five-linear identity it satisfies, together with the compatibility of the two halves, are the subject of The Lie–Jordan Decomposition of a Bilinear Product and the Jordan Triple System, for the biquaternion algebra. The compatibility is a third condition, not a consequence of the two admissibility conditions: the product $s_0 + t_0$ of the preceding section is Lie-admissible and Jordan-admissible, and it fails the derivation identity, the two sides differing by $2w$ at $x = u$ and $y = z = v$.

Deformation quantisation. The split is the first line of the theory of star products, in this category. There the symmetric half is a commutative associative product, a deformation of it is an associative product whose first-order antisymmetrisation is a Poisson bracket, and the antisymmetrisation of the deformed product is therefore a Lie bracket to first order in the deformation parameter, which is Lie-admissibility at first order. The Moyal product with a constant bracket is the closed-form example, and the theory is Deformation Quantization, §The Moyal star product; the geometric side, the symplectic and Poisson structures on which the bracket is defined, is Symplectic Forms and Poisson Brackets, in Part II.

The biquaternion instances. The four general products carried by the biquaternion algebra are split into their symmetric and antisymmetric halves in Symmetric and Antisymmetric Decomposition of the Four General Products, and the two halves are developed into Symmetric and Antisymmetric Decomposition of the Four General Products. In that algebra the two halves meet the Hermitian and anti-Hermitian subspaces in the way each structure requires: the antisymmetric half is a Lie subalgebra on the anti-Hermitian subspace $\mathbb{M}_-$, which it carries to itself, and the symmetric half is a Jordan product on the Hermitian subspace $\mathbb{M}_+$; the closure of the subspaces, half by half, is the commutator row and the symmetrised row of the tables of Remarkable Subspaces and the Four General Products, Forms and Norms.

The sesquilinear case. When the product is conjugate-linear in one argument the exchange is of a different kind, and it is read in two ways. Read plainly, $x\star y\mapsto y\star x$, the two halves are the symmetrised sesquilinear product and half the sesquilinear commutator; they are only $R^{\varsigma}$-bilinear, because the swapped product is $\varsigma$-semilinear in the first slot, and they split the scalar form into its $\varsigma$-even and $\varsigma$-odd parts rather than into a commutative and an anticommutative half. Read through a conjugation, $x\star^{c}y=c(y\star x)$, the exchange is an involution of the class and its two halves are the conjugate-symmetric and the skew-conjugate-symmetric parts, which in the derived case $x\star y=xy^{*}$ carry the scalar form in the conjugate-symmetric half and the vector part of the value in the skew half: the conjugate-symmetric half is $\tfrac12(xy^{*}+c(yx^{*}))=\varepsilon(xy^{*})e_t$, the whole scalar form of The Conjugate-Symmetric and Skew-Conjugate-Symmetric Parts of a Sesquilinear Product, §The Case of a Hermitian Form, with values in the centre and not necessarily in the Hermitian subspace, while the skew half is scalar-free with values in the vector subspace. That half is not Hermitian-valued in general — for $x=e_t$ and $y=ie_t$ it is $-ie_t$, which is skew-Hermitian — and it is confused with the Hermitian-valued symmetrisation of the plain split only in the bilinear case, where the two exchanges coincide. The two constructions are The Symmetric and Antisymmetric Parts of a Sesqualgebra Product and The Conjugate-Symmetric and Skew-Conjugate-Symmetric Parts of a Sesquilinear Product, and the plain derived operation with its Hermitian values is in The Sesquilinear Symmetrised Product and Hermitian and Skew-Hermitian Elements, all in the sesqualgebra category of this part.

The enveloping algebra. The commutator algebra of an associative algebra is a Lie algebra, so it has a universal enveloping algebra, and the construction $\mathrm{G} \mapsto \mathrm{U}(\mathrm{G})$ is the left adjoint of the functor $A \mapsto A_{\mathrm{Lie}}$ that attaches to an associative algebra its commutator Lie algebra; the universal property, the Poincaré–Birkhoff–Witt theorem and the injectivity of the canonical map are Universal Enveloping Algebras. The construction is also recorded in Associative Algebras, §The Universal Enveloping Algebra, where the adjunction with the commutator functor is stated.

Summary

Every bilinear product on a $k$-algebra over a field of characteristic not $2$ splits canonically into a commutative and an anticommutative half,

$$ xy = x \bullet y + x \wedge y, \qquad x \bullet y = \tfrac{1}{2}(xy + yx), \qquad x \wedge y = \tfrac{1}{2}(xy - yx), $$

and the split is unique. It is the decomposition of the product tensor under the exchange $x \otimes y \mapsto y \otimes x$, and the projections $n$ and $m$ of the space of products onto the symmetric products and onto the antisymmetric products return the two halves from the product; equivalently the space of products is the direct sum of those two subspaces, with the $n^3$ structure constants splitting as $\binom{n+1}{2} n$ and $\binom{n}{2} n$. The symmetrisation is the polarisation of the square, $(x+y)^2 - x^2 - y^2 = 2(x \bullet y)$, so the symmetric half is exactly what the square map carries and the antisymmetric half exactly what it cannot see; the antisymmetric half vanishes exactly for a commutative product and the symmetric half exactly for an anticommutative one.

Each half satisfies the identity it is named after only under a condition on the algebra. The algebra is Lie-admissible when the antisymmetrisation satisfies the Jacobi identity, and this holds exactly when the associator of the algebra satisfies the cyclic identity $[x,y,z] + [y,z,x] + [z,x,y] = [y,x,z] + [z,y,x] + [x,z,y]$; the Jacobi sum of the commutator is the signed sum of the six associators. The algebra is Jordan-admissible when the symmetrisation satisfies the Jordan identity. Associativity implies both, the commutative algebras show that Lie-admissibility is not a disguised associativity, and neither condition implies the other: on $k^3$ the product of a commutative non-Jordan symmetric part with the cross product is Lie-admissible and not Jordan-admissible, and the product of the componentwise Jordan product with a three-dimensional antisymmetric part failing Jacobi is Jordan-admissible and not Lie-admissible. Admissibility of both is still weaker than associativity, and it does not give the derivation identity that links the two halves, which is a third condition.

The passage to the opposite algebra leaves the symmetric half fixed and reverses the antisymmetric half, and so preserves both conditions; and on the family $A_\lambda$ with the product $x \bullet y + \lambda(x \wedge y)$ the symmetric half is fixed and the antisymmetric half is multiplied by $\lambda$. When $2$ is not invertible the decomposition fails: in characteristic $2$ the symmetric and the antisymmetric tensors coincide, the symmetrisation is square-zero, and distinct products have the same symmetrisation.

Summary of Notation

Symbol	Meaning
$k$, $A$, $n$	A field of characteristic not $2$; a $k$-algebra in the broad sense of Algebras; the dimension where finite
$xy$	The product, bilinear and with no further hypothesis
$x \bullet y = \tfrac{1}{2}(xy + yx)$	The symmetrisation, the symmetric part of the product, $x \bullet x = x^2$ and $x \bullet y = y \bullet x$
$x \wedge y = \tfrac{1}{2}(xy - yx)$	The antisymmetrisation, or outer product, the antisymmetric part, $x \wedge x = 0$ and $x \wedge y = -y \wedge x$
$[x,y] = xy - yx = 2(x \wedge y)$	The commutator, without a factor $\tfrac{1}{2}$
$[x,y,z] = (xy)z - x(yz)$	The associator, a ternary bracket, as in Non-Associative Algebras and the Property Ladder
$A^{\bullet}$, $A^{\wedge}$	The symmetrised and the antisymmetrised algebra of $A$
$A^{+}$, $A^{(-)}$, $A^-$	The same two algebras in the notation of the corpus: $A^{+}$ for the symmetrised one (Jordan Algebras), $A^{(-)}$ (The Commutator Operator) and $A^-$ (Associative Algebras) for the antisymmetrised one
$A^{\mathrm{opp}}$, $A_\lambda$	The opposite algebra, and the algebra with product $x \bullet y + \lambda(x \wedge y)$; $A_{-1} = A^{\mathrm{opp}}$, $A_0 = A^{\bullet}$
$S^2 A$, $\Lambda^2 A$, $\tau$	The symmetric and exterior squares, the fixed and anti-fixed parts of the exchange $\tau(x \otimes y) = y \otimes x$
$c^k{}_{ij}$	The structure constants, $e_i e_j = \sum_k c^k{}_{ij} e_k$
$n$, $m$	The projections of a product onto its symmetric and its antisymmetric part, $n^2 = n$, $m^2 = m$, $pq = 0$, $n+m = \mathrm{id}$
$\Sigma$	The unhalved symmetrisation $\Sigma(m)(x,y) = m(x,y) + m(y,x)$, which satisfies $\Sigma^2 = 2\Sigma$ and is used in §The Collapse in Characteristic 2
Lie-admissible, Jordan-admissible	$A^{\wedge}$ satisfies Jacobi; $A^{\bullet}$ satisfies the Jordan identity
$L(x)$, $R(x)$, $\mathrm{ad}_x$, $T_{a,b}$	Left and right multiplication, the commutator operator $L(x) - R(x)$, the sandwich $T_{a,b}(x) = axb$