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