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The symmetrized product of an associative algebra is Jordan

Statement

Let $k$ be a field with $char (k) \neq 2$ and $A$ a unital associative $k$ -algebra. The product $a \circ b = (a b + b a) / 2$ makes $A^{+}$ of Symmetrized product of an associative algebra a Jordan algebra. If $A$ is a complex associative algebra with involution as in Complex associative algebra with involution, then its self-adjoint part $A_{s a}$ is a real vector subspace closed under $\circ$ , hence a real Jordan subalgebra of $A^{+}$ .

Facts & Assumptions

Given: A field $k$ of characteristic not $2$ , a unital associative $k$ -algebra $A$ , and, for the second claim, a complex algebra involution on $A$ . Write $a^{3} : = a a a$ in the associative multiplication.

[F1]

The product on $A^{+}$ is $a \circ b = (a b + b a) / 2$ (Symmetrized product of an associative algebra).

[F2]

Multiplication on $A$ is bilinear and associative (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[F3]

A Jordan algebra has a commutative bilinear product satisfying $x \circ (y \circ x^{2}) = (x \circ y) \circ x^{2}$ (Jordan algebra over a field of characteristic not two).

[F4]

In a complex algebra with involution, $*$ is conjugate-linear, $(a b)^{*} = b^{*} a^{*}$ , $(a^{*})^{*} = a$ , and $1^{*} = 1$ (Complex associative algebra with involution).

Proof

1.1givenF1F2algebra

By [F1] and the bilinearity of multiplication in [F2], $\circ$ is bilinear; swapping $a, b$ leaves $a b + b a$ unchanged, so it is commutative. Also $a_{\circ}^{2} : = a \circ a = a a$ because $2$ is invertible in $k$ .

2.1givenF1F2step 1.1algebra

Using associativity from [F2], expand $a \circ (b \circ a_{\circ}^{2}) = \frac{1}{4} (a b a^{2} + a^{3} b + b a^{3} + a^{2} b a) .$ Here $a^{3} = (a a) a = a (a a)$ .

2.2givenF1F4step 1.1algebra

In the complex involutive case, $A_{s a}$ is closed under addition and real scalar multiplication by conjugate-linearity in [F4], and it contains the unit because $1^{*} = 1$ . For $a, b \in A_{s a}$ , $(a \circ b)^{*} = \frac{1}{2} ((a b)^{*} + (b a)^{*}) = \frac{1}{2} (b a + a b) = a \circ b,$ so it is closed under $\circ$ .

3.1givenF1F2step 1.1step 2.1algebra

The other side expands, using [F2], as $(a \circ b) \circ a_{\circ}^{2} = \frac{1}{4} (a b a^{2} + b a^{3} + a^{3} b + a^{2} b a),$ which contains the same four terms as step 2.1.

4.1givenF1F2step 1.1step 2.1step 3.1F3

Since $A$ is unital, [F1] gives $1 \circ a = (1 a + a 1) / 2 = a$ for every $a \in A$ . Steps 1.1, 2.1, and 3.1 prove bilinearity, commutativity, and the Jordan identity in [F3]. Thus $A^{+}$ is a unital Jordan algebra.

5.1step 4.1step 2.2F3∎

The real subspace $A_{s a}$ inherits commutativity, bilinearity, and the Jordan identity from $A^{+}$ , so it is a real Jordan subalgebra.

Depends on

Used by

The self-adjoint part need not be an associative subalgebra Counterexample
The symmetrized product need not be associative Counterexample
The spin factor and the plus-quadratic Clifford convention Proposition

Cited to discharge well-definedness by Complex associative algebra with involution.

Dependency tree · two levels

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Sources

N. Jacobson, Lectures on Quadratic Jordan Algebras, §1 (standard reference, not scraped)
John C. Baez, The Octonions, §3.1, PDF p. 20 (standard reference, not scraped)