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If $A=K(H)$, where $H$ is an infinite dimensional separable Hilbert space, then $A$ is simple and nuclear, and the multiplier algebra $M(A)$ of $A$ is not nuclear.

My question is: can we find a non-unital simple nuclear $\mathbb C$ $*$- algebra $B$ such that the multiplier algebra of $B$ is also nuclear?

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    $\begingroup$ My instinct is that multiplier algebras of any non-unital Cstar-algebra $B$ are usually "too big" to be nuclear, unless $B$ is close to being commutative e.g. $C_0(X; {\bf M}_2)$ which of course is very far from being simple. So I suspect the answer to your question is "no" but I don't see a proof at the moment $\endgroup$
    – Yemon Choi
    Commented Apr 3, 2019 at 17:45
  • $\begingroup$ One possible idea for an attempted proof that the answer is "no": since $\prod_n {\bf M}_n$ is not exact as a $C^*$-algebra, one could try to show that $B$ being non-unital and simple implies that $M(B)$ contains a subalgebra isomorphic to $\prod_n {\bf M}_n$... $\endgroup$
    – Yemon Choi
    Commented Apr 4, 2019 at 0:07

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If $A$ is a $\sigma$-unital, simple, non-unital $C^\ast$-algebra, then $M(A)$ is non-exact (in particular, it is non-nuclear). The following argument is modelled after Yemon's idea in the comments above. Also, it uses the very deep theorem of Kirchberg, that quotients of exact $C^\ast$-algebras are exact (this can probably be avoided).

If $A$ is type I, this is the fact that $B(H)$ is not exact, so we assume that $A$ is not type I. Assume for contradiction that $M(A)$ is exact.

Let $h\in A$ be strictly positive, $\| h \|=1$, let $t_1 > t_2 > \dots$ be in the spectrum of $h$, with $\lim t_n =0$, and let $f_n\colon [0,1] \to [0,1]$ be continuous functions of norm 1, supported only close to $t_n$ (so that all supports are mutually disjoint), such that $f_n(t_n) = 1$. Let $d_n = f_n(h)$. It is easily seen that the map $(a_n)_{n\in \mathbb N} \mapsto \sum a_n$ (strict convergence) induces an embedding $\prod_{n\in \mathbb N} \overline{d_n A d_n} \hookrightarrow M(A)$. As exactness passes to $C^\ast$-subalgebras, it follows by our assumption that $M(A)$ is exact, that $\prod \overline{d_n A d_n}$ is exact.

As each $\overline{d_n A d_n}$ is not type I, a theorem of Glimm (Cor. 6.7.4 in Pedersen's book "$C^\ast$-algebras and their automorphism groups") implies that there are $C^\ast$-subalgebras $B_n \subseteq \overline{d_n A d_n}$ such that $B_n$ has a quotient isomorphic to the UHF algebra $M_{n^\infty}$. Hence $\prod_{n\in \mathbb N} B_n$, which is exact since it embeds into $\prod \overline{d_n A d_n}$, has a quotient isomorphic to $\prod_{n\in \mathbb N} M_{n^\infty}$. As quotients of exact $C^\ast$-algebras are exact (by a deep theorem of Kirchberg), $\prod M_{n^\infty}$ is exact. However, this has a $C^\ast$-subalgebra $\prod_{n\in \mathbb N} M_n$ which is then exact, hence a contradiction.

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  • $\begingroup$ Is the conclusion also true in the non-$\sigma$-unital case? $\endgroup$
    – math112358
    Commented Apr 5, 2019 at 1:05
  • $\begingroup$ Nicely done! It seems like you only need simplicity to get each $d_nAd_n$ having non-Type I closure -- are there natural weakenings of the simplicity condition which allow the same argument to go through? $\endgroup$
    – Yemon Choi
    Commented Apr 5, 2019 at 1:51
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    $\begingroup$ I have no idea what happens in the non-$\sigma$-unital case. As for $\overline{d_n A d_n}$ not being type I, this is the only thing that is needed. So for instance, simplicity could be replaced by "$A$ has no $\sigma$-unital hereditary $C^\ast$-subalgebras of type I". In particular, this applies to any $C^\ast$-algebra that absorbs the Jiang-Su algebra tensorially (a normal criterion in the classification programme of Elliott). $\endgroup$
    – Jamie Gabe
    Commented Apr 5, 2019 at 2:15

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