Question

Can anyone give me a relatively simple proof or Some reference for the following fact.(I know that there is a proof of this theorem in Gerard J. Murphy'book: "$C^*$-Algebras and Operator Theory", but I'm sure that there should be a simple proof of this.

Every hereditary C*-subalgebra of a simple $C^*$-algebra is also simple!

Maybe this is easy for someone, but it makes me confused for a long time. I am a novice!

Answer 1

I don't know much about C* algebras, but I would guess that there shouldn't be a simple proof of the theorem you state. Here's an algebraic example which seems to be a counterexample to the theorem once the modifier C* is removed from the statement. Let A be the Weyl algebra, i.e. the algebra generated by x,y with the relation xy-yx=1. It isn't hard to show this algebra is simple and that it has global dimension 1. The subalgebra generated by x also has global dimension 1, but of course it is far from simple.

Answer 2

Here is a direct argument which may not differ much in its essence from the one in the book that you mention:

Say $A$ is simple and $B$ is a closed hereditary subalgebra of $A$. This means that if $a\in A$ and $b_1,b_2\in B$ then $b_1ab_2\in B$. Let $x\in B$ be non-zero and let us show that it generates $B$ as a closed two sided ideal. Since $x$ generates $A$ as a closed two-sided ideal, the finite sums of elements of the form $axa'$, with $a,a'\in A$, form a dense subset in $A$. In particular, if $y\in B$ and $\epsilon>0$ then there exists an element of the form \[ \sum_{i=1}^{n} a_ixa'_i \] within a distance $\epsilon$ of $y$. The problem with this is that the $a_i$s and $a'_i$s are not in $B$. They are only in $A$. This is fixed using that in a C*-algebra one always has that $|c^*|^{1/n}c|c|^{1/n}\to c$ for any $c$ (alternatively, you can use an approximate unit for $B$). Then for $k$ large enough the element \[ \sum_{i=1}^{n} (|y^*|^{1/k}a_i|x^*|^{1/k}) x (|x|^{1/k}a'_i |y|^{1/k} ) \] is also within a distance of $\epsilon$ of $y$.