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?



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> **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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Is the generated subalgebra of a subset of pairwise operator-commuting element in a JB-algebra associative?
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](https://doi.org/10.1016/j.jpaa.2020.106407).
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.


