I would like to know which $C^*$-algebras are ideals in their second duals?

There is a paper by S. Watanabe that claims in introduction that it is well known that a $C^*$-algebra is an ideal in its second dual iff it is a dual $C^*$-algebra. But I do not know what does he mean by term "dual $C^*$-algebra".

For the general case of normed algebras there is a criterion (see Banach algebras and the general theory of *-algebras. Volume I. Algebras and Banch algebras. Theodore W. Palmer, theorem 1.4.13): $A$ is a two-sided ideal in $A^{**}$ iff the maps $L_a:A\to A$, $b\mapsto ab$ and $R_a:A\to A$, $b\mapsto ab$ are weakly compact for all $a\in A$.

May be this criterion could be improved for the case of $C^*$-algebras. You may even assume that $A$ is commutative.

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    $\begingroup$ What about examples such as $c_0$ or $K(H)$? $\endgroup$
    – Yemon Choi
    Dec 5, 2020 at 22:02
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    $\begingroup$ I think that for any unital Banach algebra $A$ the natural embedding $A\to A^{**}$ is a unital homomorphism (for both Arens products). It follows that $A$ cannot be an ideal in $A^{**}$ unless $A$ is reflexive as a Banach space. $\endgroup$
    – Yemon Choi
    Dec 5, 2020 at 22:04
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    $\begingroup$ @MatthewDaws: The second of Yemon Choi's comments only refers to unital algebras. $\endgroup$ Dec 6, 2020 at 12:56
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    $\begingroup$ @JochenGlueck: Ah, of course!! I had somehow failed to see "unital". $\endgroup$ Dec 6, 2020 at 14:01
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    $\begingroup$ @MatthewDaws It's OK, I haven't suddenly gained the impression that $c_0=\ell_\infty$ ... :D $\endgroup$
    – Yemon Choi
    Dec 6, 2020 at 17:19

1 Answer 1


Warning: the following is just what I found from some work on MathSciNet, following Yemon's hint in the comments. It's not meant to be accurate historical notes.

A "dual" $C^\ast$-algebra is defined as follows. Let $A$ be an algebra and for a subset $M\subseteq A$ let $R(M) = \{ x\in A : Mx=\{0\}\}$; similarly define $L(M) = \{ x\in A : xM=\{0\}\}$. Then a $C^\ast$-algebra is dual if for each closed left ideal $I$ we have that $L(R(I))=I$. (The involution can be used to show that the analogous definition with right ideals gives the same notion).

An early paper which studied these is Kaplansky, The structure of certain operator algebras, see section 2. It seems that Berglund, Ideal $C^\ast$-algebras was the first to obtain the equivalence you seek (why is the Duke journal archive behind a paywall??) A short proof is in McCharen, A characterization of dual $B^{\ast}$-algebras These give the following:

Claim: A $C^\ast$-algebra $A$ is an ideal in its bidual if and only if $A$ is dual.

We now combine this with known characterisations of dual $C^\ast$-algebras:

Claim: A $C^\ast$-algebra $A$ is dual if and only if $A$ is isomorphic to a $C^\ast$-subalgebra of $K(H)$ for some $H$ if and only if $A$ is the $c_0$-direct sum of algebras of the form $K(H)$ for some (finite or infinite dimensional) $H$.

Thus $K(H)$ and $c_0$ really are the archetypal examples. These results are quoted in the paper of Kaplansky I linked above, and in Dixmier's book (English edition, section 4.7.20) though in both cases it seems only further references are given, not proofs. I am afraid that I don't know of a modern, self-contained treatment.

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    $\begingroup$ Here is an interesting example of how "well-known" is a dangerous phrase! Back in the day, such results probably were very well-known, when everyone knew lots about liminal $C^\ast$-algebras. 45+ years later, we still want to read the papers, but the people reading the papers are a new generation, and that folk knowledge has disappeared... $\endgroup$ Dec 6, 2020 at 10:57
  • $\begingroup$ Thank you very much! $\endgroup$
    – Norbert
    Dec 6, 2020 at 12:49
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    $\begingroup$ It feels like there should be a more direct way to see that the assumption that each left and right multiplication operator on $A$ is weakly compact (Ylinen's condition) forces $A$ to somehow be generated by projections $p$ such that $pA$ is finite-dimensional ... but I admit I don't see right now how to construct a proof. $\endgroup$
    – Yemon Choi
    Dec 6, 2020 at 17:23

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