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In a Jordan algebra elements $a$ and $b$ are said to operator-commute, whenever $a \circ (b \circ x) = b \circ (a \circ x)$ for every other element $x$. (That is: $T_aT_b = T_bT_a$, writing $T_x(y) = x \circ y$.) In a JB-algebra elements $a$ and $b$ operator-commute if and only if they generate an associative subalgebra. (See e.g. p.44 "Jordan operator algebras" by Hanche-Olsen and Størmer.) Does this generalise from pairs to arbitrary subsets?

Question: Assume $A$ is a JB-algebra. $S \subseteq A$ is a subset of pairwise operator-commuting elements. Is the algebra generated by $S$ in $A$ associative?

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Yes, this is true. I couldn't find any proof of the statement you quote in the article, and even after emailing the authors I didn't get any wiser, so I decided to work out the details myself, see my paper Commutativity in Jordan Operator Algebras. My main result is that if $a$ and $b$ operator commute then they indeed generate an associative JB-algebra of mutually operator commuting elements. In particular, $a^2$, $b^2$ and $a*b$ also commute with each other and with $a$ and $b$.

Now suppose $S$ contains more than two elements. Consider $J(S)$ the Jordan algebra generated by $S$ which consists of "polynomials" in the elements of $S$. For such an element $p(s_1,s_2,\ldots,s_n)$ we can always decompose the operator $T_p$ into simple terms $T_{s_i}$, $T_{s_i*s_j}$ and $T_{s_i^2}$ by repeatedly applying the "normalisation equation" (Eq. 1 in my paper). By the two-element case we know that these all operator commute, hence any two elements of $J(S)$ operator commute. Now we just need to close $J(S)$ in the norm to get the JB-algebra generated by $S$. As the Jordan product is continuous, this algebra also consists of operator commuting elements.

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