$A$ 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?

  • 2
    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
  • So what is the answer of your question? – Dadrahm Nov 16 at 14:43
up vote 6 down vote accepted

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.

  • Thank you very much. – Dadrahm Nov 16 at 16:37
  • You are welcome! – Nik Weaver Nov 16 at 17:03

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