# Motivation for primitive ideals of a C*-algebra

I have recently learned about the primitive ideals and prime spectrum of a C*-algebra. I am looking for a 'reason' for why they are useful. I mean this in the sense that if I was a mathematician reseraching this area, why would it even cross my mind to define the primitive spectrum?

It seems to me, that the theory of irreducible representations of a C*-algebra are perfectly happy without the notion of a primitive ideal. In the commutative case, the correspondance between maximal modular ideals and characters is motivated by the fact that we are trying to prove results which require the exsitence of cahracters with certain properties (e.g. when proving the norm of the Gelfand transform, we want to construct a character). I think about this as 'certain accessible algebraic property (every modular ideal is contained in a maximal modular ideal) allows us to deduce some property about the character space. However I have not come across any similar result, where the algebra concerning primitive ideals would have allowed some results to be proven about representations. In fact I feel that the algebraic definition connecting them to maximal modular left ideals is really rigid.

## 1 Answer

Primitive ideals are mainly useful for studying the entire closed ideal structure of a $$C^*$$-algebra $$A$$. Let $$I(A)$$ be the lattice of closed ideals in $$A$$, and $$O(Prim(A))$$ the lattice of open subsets of the primitive spectrum $$Prim(A)$$. There is a lattice isomorphism $$O(Prim(A)) \to I(A)$$ given by $$$$U \mapsto \bigcap_{J \in Prim(A) \setminus U} J.$$$$ This is for instance proved in Gert Pedersen's book on $$C^\ast$$-algebras and their automorphism groups (but I don't have a precise reference at hand, I believe it is in chapter 4).

Being able to answer structural questions using representation theory is very useful, for instance, it follows that a $$C^*$$-algebra $$A$$ is simple if and only if $$0$$ is the only primitive ideal in $$A$$ if and only if every irreducible representation $$\pi \colon A \to \mathcal B(\mathcal H)$$ is faithful.

Here's an example where it gets used. If $$B$$ and $$C$$ are simple $$C^*$$-algebras then the minimal tensor product $$A = B \otimes_{\min{}} C$$ is also simple. Here is a sketch: Let $$\pi \colon A \to \mathcal B(\mathcal H)$$ be an irreducible representation. We should prove that $$\pi$$ is faithful, and it suffices to prove that it is faithful on the algebraic tensor product $$B \otimes_{alg} C$$. Assume for simplicity (no pun intended) that $$B$$ and $$C$$ are unital so that $$\pi$$ splits as two representations $$\pi_B : B \to \mathcal B(\mathcal H)$$ and $$\pi_C \colon C \to \mathcal B(\mathcal H)$$ by $$\pi_B(b) = \pi(b\otimes 1_C)$$ and $$\pi_C(c) = \pi(1_B \otimes c)$$ (this can be done without units, see the book of Brown-Ozawa). Since $$\pi$$ is irreducible, it is easy to see that $$M := \pi_B(B)''$$ and $$\pi_C(C)'' \subseteq M'$$ are factors. Hence $$\pi$$ restricted to the algebraic tensor product factors as $$B \otimes_{alg} C \to M \otimes_{alg} M' \to \mathcal B(\mathcal H)$$ (the second map is the product map). The first map is easily seen to be faithful since $$B$$ and $$C$$ are simple (so that $$\pi_B$$ and $$\pi_C$$ are faithful), and the second map is faithful since $$M$$ is a factor (this goes back to Murray and von Neumann, a proof can for instance be found in Takesaki's first book).