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In [1, 5.6.3] Pedersen states without proof or reference that there are non-unital C*-algebras whose Pedersen ideal is the whole algebra.

  • Does anyone know where can I find such an example?
  • Is it possible to characterize algebras with this property?

[1] Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London - New York -San Francisco: Academic Press. X, 416 p. $ 60.00 (1979). ZBL0416.46043.

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Examples include all non-unital algebraically simple $C^\ast$-algebras. By [Blackadar, Bruce E.; Cuntz, Joachim The structure of stable algebraically simple C∗-algebras. Amer. J. Math. 104 (1982), no. 4, 813–822.] a simple, stable $C^\ast$-algebra is algebraically simple if and only if it contains an infinite projection. So an explicit example would be a stabilised Cuntz algebra $\mathcal O_n \otimes \mathcal K$.

Addon: in his original paper [Pedersen, Gert Kjaergȧrd, Measure theory for C∗ algebras. Math. Scand. 19 (1966), 131–145], Pedersen gives the following example of a non-unital $C^\ast$-algebra $A$ with $\mathrm{Ped}(A) = A$: Let $H$ be a non-separable Hilbert space, and let $A$ be the $C^\ast$-subalgebra of $B(H)$ of operators whos range projection project onto a separable subspace of $H$. This $C^\ast$-algebra is non-unital and $\mathrm{Ped}(A) = A$. To see this, let $a\in A$ and $p$ be the range projection of $a$. Then $p\in \mathrm{Ped}(A)$ and $a = pa \in \mathrm{Ped}(A)$ since $\mathrm{Ped}(A)$ is an ideal.

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  • $\begingroup$ Great answer, as always! $\endgroup$
    – Nik Weaver
    Nov 20, 2020 at 15:40
  • $\begingroup$ Thanks Jamie! I guess I edited my question at the exact same time you posted your answer. So I wonder if you have an answer to the second part? $\endgroup$
    – Black
    Nov 20, 2020 at 15:41
  • $\begingroup$ Another minor point: Pedersen's book predates the reference you mentioned by 3 years, so it might be interesting to know what was the example he had in mind. $\endgroup$
    – Black
    Nov 20, 2020 at 15:48
  • $\begingroup$ I don't know how you would characterise this property. If $A$ is stable, has a strictly positive element (e.g. $A$ separable), and the primitive ideal space $Prim(A)$ is compact, the following are equivalent: (i) $Ped(A) = A$, (ii) $A$ contains a properly infinite, full projection, (iii) the only (possibly unbounded) tracial functional on $Ped(A)$ is the zero functional. The proof is essentially the same as the Blackadar-Cuntz result is cited above. Otherwise I don't know how one could characterise this property. $\endgroup$
    – Jamie Gabe
    Nov 20, 2020 at 17:18
  • $\begingroup$ Thanks for the addon. It is a lot easier to understand than the stabilization of the Cuntz algebra! $\endgroup$
    – Black
    Nov 20, 2020 at 18:38

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