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

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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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