Closed ideal in the hereditary C*-subalgebra

Question

Let B is a hereditary C*-subalgebra of a C*-algebra A and J be a closed ideal of B, is AJA a closed ideal of A?

I do not find the definition of product of ideals in the C*-algebras, is it the same with the algebras?

Answer 1

In general, No. Take $A = \mathcal{K}(\ell^2)$, the C* algebra of compact operators on $\ell^2$, and take $B = \mathbb{C}e_1\otimes e_1$, where $$ e_1\otimes e_1(e_j) = \delta_{1j}e_1 $$ (where $e_j$ are the standard basis). It is easy to see that $B$ is a hereditary subalgebra of $A$.

Now take $J = B$, then $$ AJA = \text{span}\{a_1ja_2 : a_1,a_2\in A, j\in J \} $$ is the algebra of finite dimensional operators, and hence is not closed in $A$.

In answer to your second question, yes, I believe the definition is the same as usual : If $I_1$ and $I_2$ are two closed ideals, then $$ I_1I_2 = \text{span}\{a_1a_2 : a_j \in I_j\} $$ In fact, one can show using approximate units that $I_1I_2 = I_1\cap I_2$