Dear Carlo Pandiscia For the Lie bracket [a,b]=ab-ba and the commutator

[J_a,J_b]:=J_aJ_b-J_bJ_a you can check that, for any x in A:

[J_a,J_b](x)=1/4[[a,b],x] so A^c is contained in Z(A^+) (as you already mentioned). For the converse, if the Lie algebra (A,[,]) is 2-nilpotent (as the Heisenberg algebra) then J_aJ_b=J_bJ_a for any b without a being in the center!

Dear Mohamed amine Bahayou

Thanks for the reply

Summarize,

denoted with $Ad_a (b)=[b,a]$, if $a\in Z(A^+)$ then $ [Ad_a (b), x]=0 \forall b,x\in\A $

therefore I have to prove that it follows $Ad_a (b)=0 , \forall b\in A$.

I am work with infinite dimensional associative algebras.

Example, if $A=B(H)$ bounded linear operator on real Hilbert space H,

we have $Z(A^+)= \mathbb R I $.

Indeed, $ [Ad_a (b), x]=0 \forall x\in\B(H) $ then $Ad_a (b)\in \mathbb R I , \forall b in B(H)$ then $[b,a]=0 \forall b\in B(H)$ it follows again that $ Z(A^+)=\mathbb R I = B(H)^c$

Thanks Carlo for the precision. I think it's also true for infinite dimensional algebras (the infinite dimensional Heisenberg algebra).

So what are the 'extra' conditions, in order to have $Z(A^+)\subset A^c$, apart the case of linear bounded operators on a Hilbert space?

Dear Mohamed,

maybe I came to a partial conclusion

if J is a JBW algebra, for Shultz 1979(normed Jordan algebra) theorem 3.9, it can be decomposed in direct sum $ J = J_1 \ oplus J_2 $ where is it

$ J_1 $ is isomorphic to JW algebra while $ J_2 $ is (purely) exceptional

It follows that if A is a W* algebra then $A^+$ is isomorfich to JW algebra and for Topping's theorem the centre and its commutant they should coincide.

the problem is JB algebras , because the JB-algebras cannot in general be represented as an algebra of operators on a Hilbert space (Shultz 79).

Is there another procedure?