Non-nuclear C* subalgebras of a non-nuclear C* algebra

Question

We know that a nuclear $C^\star$ algebra might have a $C^\star$ subalgebra that is not nuclear. Consider, for example, the construction in Man-Duen Choi's paper "A Simple $C^\star$ Algebra generated by Two Finite-Order Unitaries", Can. J. Math., Vol. XXXI, No. 4, 1979, pp. 867-880, where he defines two unitaries $u,v$ on an infinite-dimensional Hilbert space with $u^2=1$ and $v^3=1$ such that the $C^\star$ algebra generated by them, $C^\star(u,v)$, is not nuclear, but is a $C^\star$ subalgebra of the Cuntz algebra $\mathcal{O}_2$, which is nuclear.

It is also direct to see that any non-nuclear $C^\star$ algebra has a nontrivial nuclear $C^\star$subalgebra (simply consider any normal element $N$ not equal to $0$ or $1$, and the closure of the $\star$-subalgebra generated by $1$ and $N$ is abelian, and hence nuclear).

Now, my question is the following: Does every non-nuclear $C^\star$ algebra have a (proper) $C^\star$ subalgebra that is also not nuclear?

I am primarily trying to understand this for unital $C^\star$ algebras...

In case of a negative answer, this might imply some notion of "minimal" non-nuclear $C^\star$ algebras (in the sense that they have every $C^\star$ subalgebra nuclear).

Answer 1

Taking into account point (2) of Aweygan's answer, the problem actually boils down to proving that every C*-algebra is the inductive limit of a family of proper subalgebras.

As observed in the comments, this is true for every C*-algebra that is not finitely generated (including all non-separable algebras), as one could take the inductive system formed by the finitely generated subalgebras.

Here is a method for proving this for NON UNITAL separable algebras (including of course the finitely generated ones).

So let $A$ be a non-unital separable algebra. It is well known that $A$ contains a strictly positive element $h$, so let us fix one such element.

For each $c>0$, consider the real function $f_c$ of one real variable given by $$ f_c(x) = \left\{\matrix{ 0, & \text { if } x<c,\cr x-c, & \text { if } x\geq c. } \right. $$

Denoting by $A_c$ the hereditary subalgebra of $A$ given by $$ A_c = \overline{ f_c(h)Af_c(h)}, $$ I claim that

1. $A_c$ is a proper subalgebra,

2. if $c_1\leq c_2$, then $A_{c_1}\supseteq A_{c_2}$,

3. the union of the $A_c$ is dense in $A$.

To prove (1), first notice that zero cannot be an isolated point in the the spectrum of $h$, or otherwise the characteristic function $1_{(0, \infty )}$ would be continuous on the spectrum of $h$ and then $A$ would be unital with unit $1_{(0, \infty )}(h)$.

Letting $g$ be any continuous function on ${\mathbb R}$ with support $[0,c]$, we then have that $$ g(h)\neq 0, \quad \text{and}\quad g(h)f_c(h)=0, $$ so $g(h)\not \in A_c$. This proves that $A_c$ is proper.

To prove (2), observing that that $f_c(f_d(x)) = f_{c+d}(x)$, we have that $$ f_{c_2-c_1}\big (f_{c_1}(h)\big ) = f_{c_2}(h), $$ so $f_{c_2}(h)\in C^*\big (f_{c_1}(h)\big )\subseteq A_{c_1}$, and hence $A_{c_2}\subseteq A_{c_1}$.

Finally (3) follows because $f_c(h)\to h$, as $c\to 0_+$.

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This said, we have that $A$ is the inductive limit of the $A_c$, and hence, should $A$ fail to be nuclear, one of the $A_c$ would do too.

Answer 2

Yes, every non-nuclear $C^*$-algebra contains a (proper) non-nuclear $C^*$-subalgebra. This follows from the following two facts:

1. Every $C^*$-algebra is the inductive limit of its proper $C^*$-subalgebras (easy to see), and
2. Nuclearity is preserved under direct limits (an exercise in Brown and Ozawa's book "$C^*$-Algebras and Finite-Dimensional Apprixmations").

EDIT As pointed out in the comments, the above reasoning isn't quite correct, and can be modified to show that the result holds if we assume the original algebra is non-separable. I will continue thinking about this, and will keep this post up to date.