# Totally non hereditary $C^{*}$-subalgebras

Assume that $B$ is a $C^{*}$ subalgebra of $A$. We say $B$ is totally non hereditary subalgebra of $A$ if not only $B$ is not a hereditary subalgebra but also it is not isomorphic to any hereditary subalgebra of $A$.

Example: For every $C^{*}$ algebra $A$, $c(A)$ is a totally non hereditary subalgebra of $\ell^{\infty}(A)$. Here $c(A)$ and $\ell^{\infty}(A)$ are the algebra of all convergent and bounded sequence with $A$-elements, respectively.

There are some $C^{*}$ algebras which does not have such subalgebras. Example $\mathbb{C}^{n}$

To what extent all $C^{*}$ algebras without totally nonhereditary subalgebras are classified? What are some more examples(other than $\mathbb{C}^{n}$)? In the topological language, somehow, we are interested in some non trvial example of compact Hausdorff space $X$ such that each quotient of $X$ is homeomorphic to an open subset of $X$.

What is a totally nonhereditary subalgebra of $B(H)$ where $H$ is an infinite dimensional Hilbert space? What is an example of a simple $C^{*}$ algebra which contains a totally non hereditary sub $C^{*}$ algebra? Is there any reference about this concept?

• Let $X$ be a countable successor ordinal. Then each quotient of $X$ is homeomorphic to a clopen subset of $X$. Feb 8, 2015 at 21:21

So you want to consider a C*-algebra $A$ with the following property: Every sub-C*-algebra of $A$ is isomorphic to a hereditary sub-C*-algebra of $A$.

We can distinguish two cases:

1. If $A$ is finite-dimensional, then $A$ has to be commutative. Indeed, assume $A\cong M_{k_1}\oplus\ldots\oplus M_{k_l}\oplus\mathbb{C}^n$ for $k_j\geq 2$ (and $j\geq 1$). Then $A$ has a commutative sub-C*-algebra of the form $\mathbb{C}^{k_1}\oplus\ldots\oplus \mathbb{C}^{k_l}\oplus\mathbb{C}^n$, but a commutative hereditary sub-C*-algebra of $A$ has at most $l+n$ points in its spectrum.

2. If $A$ is infinite-dimensional, then it contains commutative sub-C*-algebras $C(X)$ where $X$ is any Peano continuum. Such a C*-algebra has to have `huge' primitive ideal space.

I don't think this property has been considered in the literature before.

Another remark: Every simple C*-algebra (not equal to $0$ or $\mathbb{C}$) contains a non-simple sub-C*-algebra. Every hereditary sub-C*-algebra of a simple C*-algebra is simple again. Therefore, every (nontrivial) simple C*-algebra contains a sub-C*-algebra with the condition you consider.

Similarly, $B(H)$ contains many sub-C*-algebras that are not isomorphic to a hereditary sub-C*-algebras of $B(H)$. This is because the primitive ideal space of $B(H)$ has only two points. Thus, any sub-C*-algebra $A$ of $B(H)$ such that the primitive ideal space of $A$ has at least three points will do.

• I learn from your answer that "simplicity" is a heritage of a $C^{*}$ algebra for its hereditary subalgebras. what type of other properties could be a heritage? Could you please give a reference which contains about these type of heritages? Feb 27, 2015 at 13:11
• Yes, given a C*-algebra A and a hereditary sub-C*-algebra B of A, the primitive ideal space of B, Prim(B) is (homeomorphic to) an open subset of Prim(A). Let's assume that all C*-algebras are nonzero. Then A is simple if and only if Prim(A) is the one-point space. An open subset of the one-point space contains (at most) one point, therefore B is simple. Feb 28, 2015 at 11:21
• Many properties pass to hereditary sub-C*-algebras: simplicity, real rank zero, stable rank one, being an AF-algebra, purely infinite ... A useful result in this context is Brown's stabilization theorem, in Brown (1977) # Stable isomorphism of hereditary subalgebras of C-algebras [Pac. J. Math. 71]. It says that in a (say separable) C*-algebra, every hereditary sub-C*-algebra is stably isomorphic to the ideal it generates. Thus, every property of C*-algebras that passes to ideals and is preserved under stable isomorphism will pass to hereditary sub-C*-algebras of separable C*-algebras. Feb 28, 2015 at 11:26
• Among these properties, is there a property to shows "$c(A)$ is totally non hereditary subalgebra of $\ell^{\infty}(A)$ for arbitrary A? Jul 5, 2015 at 12:57
• To be honest, I would like to realize whetherre the generalization of corollary 4 at page 7 of this note is a trivial obervation arxiv.org/abs/1301.3129: Jul 5, 2015 at 13:33