How to characterize the kernels of pure states on a $C^*$-algebra?

If $$A$$ is a commutative $$C^*$$ $$*$$-algebra, then the kernel of a nontrivial character is a maximal ideal.

If $$A$$ is an arbitrary $$C^*$$-algebra, what kind of subspace is the kernel of a pure state?

The motivation is the following: in the commutative case, we have the $$C^*$$-algebra version of the Stone-Weierstrass theorem, in which the concept of "character" (or, equivalently: maximal ideal) is central; if the algebra is no longer commutative, the space of characters (the spectrum of the algebra) is replaced by the pure state space. I would therefore like to understand what the correspondence between characters and maximal ideals changes into when the underlying algebra is no longer commutative.

• The set of pure states is in bijection with the set of all modular maximal left ideals of $A$. This is not an exact answer to the question you ask, but I thought it might interest you. Aug 25, 2021 at 9:27

$\tau$ is a pure state and $\tau(a)=0$. Then $a=b+c$ for some $b,c$ with $\tau (b^*b)=\tau(cc^*)=0$.
I believe you can conclude a state is pure iff its kernel is equal to $$I+I*$$ where $$I$$ is a left ideal. The first direction is obvious. To show the converse, according to the second post, we have that $$I$$ is the kernel of a state and hence, by the first post, we know that state is pure.