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The self-adjoint part need not be an associative subalgebra

Statement

In $A = M_{2} (C)$ with conjugate-transpose involution, the self-adjoint part is closed under the symmetrized product but not under the original associative multiplication. For $X = E_{11}, Y = E_{12} + E_{21},$ both $X, Y$ are self-adjoint while $X Y = E_{12}$ is not.

Facts & Assumptions

Given: $A = M_{2} (C)$ , with ordinary matrix multiplication and $Z^{*} = {\overline{Z}}^{T}$ .

[F1]

A complex associative algebra with involution has a self-adjoint part $A_{s a} = {Z : Z^{*} = Z}$ , and the involution reverses products (Complex associative algebra with involution).

[F2]

For $a, b \in {1, 2}$ , $E_{a b}$ is the matrix with $1$ in position $(a, b)$ and zeros elsewhere; these units obey $E_{a b} E_{c d} = δ_{b c} E_{a d}$ . The usual product makes $M_{2} (C)$ a unital associative complex algebra, and $C$ has characteristic zero. Conjugate transpose is conjugate-linear, satisfies $(Z^{*})^{*} = Z$ , $(Z W)^{*} = W^{*} Z^{*}$ , and fixes $I_{2}$ , so it is an involution as required by [F1]. It sends each real matrix unit $E_{a b}$ to $E_{b a}$ ; in particular, it fixes $E_{11}$ and $E_{22}$ and interchanges $E_{12}$ with $E_{21}$ . The self-adjoint part is closed under the symmetrized Jordan product by The symmetrized product of an associative algebra is Jordan.

Proof

1.1givenF1F2algebra

Since $E_{11}^{*} = E_{11}$ and $(E_{12} + E_{21})^{*} = E_{21} + E_{12}$ , both $X$ and $Y$ belong to $A_{s a}$ . The matrix-unit rule gives $X Y = E_{12}$ .

2.1givenF1F2step 1.1algebra∎

But $(X Y)^{*} = E_{21} \neq E_{12} = X Y$ , so $X Y \notin A_{s a}$ . Hence the self-adjoint part is not an associative subalgebra, even though it is closed under the symmetrized Jordan product by [F2].

Depends on

Used by

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Sources

N. Jacobson, Lectures on Quadratic Jordan Algebras, §1 (standard reference, not scraped)
John C. Baez, The Octonions, PDF pp. 20–21 (standard reference, not scraped)