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

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).

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.

• Nice one, +1 from me! Note that in general, to prove (2) one requires the fact that nuclearity is preserved under quotients which is hard to prove. In the case that the connecting maps are injective, (2) is easy to prove though. I think that in (1) the connecting maps are indeed injective so no "overkill" occurs here (but I am not 100% sure). Mar 30, 2021 at 18:53
• In order for (1) to hold, one first needs to prove that the proper subalgebras form an inductive (i.e. directed) set, and this is not at all clear to me.
– Ruy
Mar 30, 2021 at 18:54
• @Ruy Thanks for pointing that out. It is true that every $C^*$-algebra is the inductive limit of it's separable $C^*$-subalgebras, so this reasoning works if we assume the original algebra is non-separable. Mar 30, 2021 at 19:02
• @Aweygan Actually I was thinking of alternate answers and apparently things are easy for the non-separable world: a $C^*$-algebra is nuclear iff every separable $C^*$-subalgebra is contained in a separable, nuclear $C^*$-algebra. Thus the non-separable case easily follows. But what about the separable case? Mar 30, 2021 at 19:03
• I suppose the same trick works for C*-algebras that are not finitely generated. However most interesting C*-algebras are finitely generated, and often even singly generated!
– Ruy
Mar 30, 2021 at 19:20

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

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 $$c\to 0_+$$.

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.

• Thank you very much for the proof in the separable case: this was what I was more interested in, actually. Any ideas regarding how to address the unital case? It is straightforward to unitize a C* algebra, but I don't know of any technique to go the other way round (that is, to find a nonunital C* subalgebra of a unital C* algebra: it can't actually be done in finite dimensions, that much I can see). Somehow if we are able to argue that a unital separable non-nuclear C* algebra has a nonunital separable non-nuclear C* subalgebra, then we are done. But I have no idea if that's even true... Apr 4, 2021 at 19:25