# Does a closed right ideal of a C$^*$-algebra have a C$^*$-algebra?

$$A$$ is an infinite dimensional C$$^*$$-algebra and $$J\subset A$$ is a closed right ideal. $$A$$ and $$J$$ are infinite dimensional(as a vector space). I want to find an infinite dimensional C$$^*$$-algebra subset of $$J$$. How can I find it?

I know an infinite dimensional C$$^*$$-algebra has an infinite dimensional commutative C$$^*$$-subalgebra. So if $$A_1$$ is infinite dimensional commutative C$$^*$$-subalgebra of $$A$$, Is the set $$A_1\cap J$$ an infinite dimensional C$$^*$$-algebra? If no, so what can I do?

• Small observation: every $C^*$-subalgebra of $J$ is contained in $J \cap J^*$, so your question is equivalent to asking if $J\cap J^*$ is infinite-dimensional for every infinite-dimensional closed right ideal $J\subset A$. – Yemon Choi Nov 16 at 12:26
It need not exist. Take $$A = B(l^2)$$ (or the compacts if you want $$A$$ to be separable) and let $$J$$ be the set of operators of the form $$u \mapsto \langle u,v\rangle e_1$$ for $$v \in l^2$$, where $$e_1$$ is the first standard basis vector. It is a right ideal because for any $$T \in B(l^2)$$ we have $$\langle Tu,v\rangle e_1 = \langle u,T^*v\rangle e_1$$, and it is closed because $$T \in J$$ if and only if $$\langle Tu,v\rangle = 0$$ for all $$u,v \in l^2$$ with $$v \perp e_1$$.
The adjoint of $$J$$ is the set of operators of the form $$u \mapsto \langle u,e_1\rangle v$$ for $$v \in l^2$$, and the intersection $$J \cap J^*$$ is one-dimensional, so we are done by Yemon's comment.