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## Is the generated subalgebra of a subset of pairwise operator-commuting part in a JB-algebra associative?

Yes, that is correct. I could not discover any proof of the assertion you quote within the article, and airplane after emailing the authors I did not purchase any wiser, so I made a decision to labor out the main points myself, graze my paper Commutativity in Jordan Operator Algebras.

My primary end result is that if $a$ and $b$ operator commute then they certainly generate an associative JB-algebra of mutually operator commuting components. In explicit, $a^2$, $b^2$ and $a*b$ too commute with one another and with $a$ and $b$.

Now suppose $S$ accommodates greater than two components. assume $J(S)$ the Jordan algebra generated by $S$ which consists of “polynomials” within the components of $S$. For such an part $p(s_1,s_2,ldots,s_n)$ we will at all times decompose the operator $T_p$ into unostentatious phrases $T_{s_i}$, $T_{s_i*s_j}$ and $T_{s_i^2}$ by repeatedly making use of the “normalisation equation” (Eq. 1 in my paper). By the two-component illustration we all know that these all operator commute, therefore any two components of $J(S)$ operator commute. Now we simply necessity to instant $J(S)$ within the norm to amass the JB-algebra generated by $S$. As the Jordan product is steady, this algebra too consists of operator commuting components.

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