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The center $Z(A)$ of an algebra $A$ is the set of all those elements that commute with all other elements. If $A$ is the algebra of compact operators on a Hilbert space $H$, then $A^{**}$ is the algebra of all operators on $H$, and $Z(A^{**})$ is trivial: the multiples of the identity operator. I would like to know examples of $C^*$-algebras $A$ such that $Z(A^{**})$ is non-trivial.

The case $A$ commutative, say $A=C_0(L)$ with $L$ locally compact, is one example. And taking products we get additional examples. Are there other examples?

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    $\begingroup$ Group C*-algebras of finite or infinite groups give many examples. $\endgroup$
    – Francois Ziegler
    Commented Feb 28, 2017 at 13:07
  • $\begingroup$ I think that $Z(A^{\ast\ast})$ is nontrivial iff the $C^{\ast}$-algebra $A$ is not simple, i.e. admits a non-trivial closed ideal. $\endgroup$
    – Mateusz Wasilewski
    Commented Feb 28, 2017 at 17:21
  • $\begingroup$ In which conditions is $C_0(L)$ an ideal in $C_0(L)^{**}$? $\endgroup$
    – M.González
    Commented Feb 28, 2017 at 18:27
  • $\begingroup$ $Z(A^{**})$ will be non-trivial in almost every situation. Whenever $A$ has two non-unitarily equivalent irreducible representations, $Z(A^{**})$ will be non-trivial. The relevant facts can be found in any C*-algebra book that discusses representations, for example Brown and Ozawa Section 1.4. $\endgroup$
    – Caleb Eckhardt
    Commented Feb 28, 2017 at 22:54
  • $\begingroup$ one of you could write an answer $\endgroup$
    – user62639
    Commented May 3, 2017 at 2:02

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