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The symmetrized product need not be associative

Statement

The symmetrized product $A^{+}$ of an associative algebra need not be associative. In $M_{2} (R)^{+}$ , put $x = E_{12}$ , $y = E_{21}$ , and $z = E_{11}$ . Then $(x \circ y) \circ z = \frac{1}{2} E_{11}$ while $x \circ (y \circ z) = \frac{1}{4} I_{2}$ .

Facts & Assumptions

Given: $M_{2} (R)$ with usual matrix multiplication and $a \circ b = (a b + b a) / 2$ .

[F1]

The symmetrized product is $a \circ b = (a b + b a) / 2$ (Symmetrized product of an associative algebra).

[F2]

The usual matrix product on $M_{2} (R)$ is bilinear, associative, and unital with identity $I_{2}$ , and $R$ has characteristic zero. For $a, b \in {1, 2}$ , $E_{a b}$ is the matrix with $1$ in position $(a, b)$ and zeros elsewhere; these units satisfy $E_{a b} E_{c d} = δ_{b c} E_{a d}$ . The symmetrized product of an associative algebra is Jordan (The symmetrized product of an associative algebra is Jordan).

Proof

1.1givenF1F2algebra

The matrix-unit rule gives $x \circ y = I_{2} / 2$ and $y \circ z = E_{21} / 2$ . Therefore $(x \circ y) \circ z = (I_{2} / 2) \circ E_{11} = E_{11} / 2$ .

2.1givenF1F2step 1.1algebra∎

It also gives $x \circ (y \circ z) = E_{12} \circ (E_{21} / 2) = \frac{1}{4} (E_{11} + E_{22}) = I_{2} / 4.$ These differ, since their $(1, 1)$ entries are $1 / 2$ and $1 / 4$ . Thus the product is not associative; it is nevertheless a Jordan product by [F2].

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Sources

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