Consider a $C^*$ algebra $\mathcal A$. A closed two-sided hermitian ideal $I$ is called primitive if it is the kernel of an irreducible $*$ representation, alternatively if $\mathcal A/I$ admits a faithful irreducible $*$ representation. Let $\mathrm{Prim}(\mathcal A)$ be the set of primitive ideals of $\mathcal A$.

One topologises $\mathrm{Prim}(\mathcal A)$ by defining a closure operation: $$Z\subset\mathrm{Prim}(\mathcal A)\quad :\quad\overline Z:=\left\{\rho\in\mathrm{Prim}(\mathcal A)\ \middle|\ \ \rho\supseteq\bigcap_{\sigma\in Z}\sigma\right\}$$ this is called the "hull kernel topology".

If $\mathcal A$ is commutative, one can verify that $\mathrm{Prim}(\mathcal A)$ with this topology corresponds to the Gelfand transformation of $\mathcal A$ (one uses that the only commutative $C^*$ algebra that admits a faithful $*$ representation is $\Bbb C$, giving an identification of $\mathrm{Prim}(\mathcal A)$ with the character space of $\mathcal A$, definition pushing of the construction of the Gelfand transform shows that the topologies are the same).

Beside this fact I have no understanding of this space, so I'm interested in figuring out what the simplest non-commutative $\mathrm{Prim}(\mathcal A)$ spaces look like.

What is the space $\mathrm{Prim}(M_{n\times n}(\Bbb C))$?

My problem in understanding this space is that I don't know what the primitive ideals of $M_{n\times n}$ are. So I guess at this point I am treating that question to be equivalent to

What are the primitive ideals of $M_{n\times n}(\Bbb C)$?


This algebra is simple, so it doens't have any non trivial two-sided ideal.

  • $\begingroup$ Huh, I had forgotten that! Thats kind of weird and makes this construction seem a bit singular. $\endgroup$ – s.harp Jun 11 '17 at 11:57
  • $\begingroup$ @s.harp It looks like your definition of "primitive" matches the classical ring theory version. $\{0\}$ is the (one and only) primitive ideal of any full matrix ring over a field. $\endgroup$ – rschwieb Jun 11 '17 at 23:36

You will not get anything interesting out of the matrices.

For a non-trivial example, consider the Toeplitz algebra, $\mathcal A=K(H)+C(\mathbb T)$ (this is the C$^*$-algebra generated by the unilateral shift). The irreducible representations are $\pi_0:k+f\longmapsto k$ and $\pi_\lambda:k+f\longmapsto f(\lambda)$ for each $\lambda\in\mathbb T$.

So $$\text{Prim}(\mathcal A)=\{0\}\cup\mathbb T,$$ where $0$ represents $\ker\pi_0=0+C(\mathbb T)$ and each $\lambda$ represents $\ker\pi_\lambda=K(H)+I_\lambda$ (where $I_\lambda=\{f:\ f(\lambda)=0\}$.

The interesting part comes when you pay attention to the topology. For any $X\subset \mathbb T$, \begin{align} \overline X&=\{\mu\in\{0\}\cup\mathbb T:\ K(H)+I_\mu\supset\bigcap_{\lambda\in X}K(H)+I_\lambda\}\\ \ \\ &=\{K(H)+I_\mu:\ \mu\in \overline X\}. \end{align} No surprise here, the closure of $X$ is the closure in the usual topology in $\mathbb T$. So the circle inherits its usual topology.

But consider the point $\{0\}$. It corresponds to the irreducible representation that is nonzero on the compacts. It is not hard to check that this representation has to be the identity representation, and so its kernel is trivial. Then $$ \overline{\{0\}}=\{J\in\text{Prim}(\mathcal A):\ J\supset\{0\}=\text{Prim}(\mathcal A). $$ That is, the closure of the singleton $\{0\}$ is the whole uncountable space. Thus the space is not Hausdorff (for instance, the constant sequence $\{0\}$ converges to every point; not being Hausdorff is weird).

  • $\begingroup$ Thank you for the details! Here $K(H)$ is the algebra of compact operators on $\ell^2(\Bbb Z)$ which is identified with $L^2(\Bbb T)$ via the fourier transform? And $C(\Bbb T)$ is identified as the commutative sub-algebra of multiplications with continuous functions? $\endgroup$ – s.harp Jun 11 '17 at 11:54
  • $\begingroup$ No. The Toeplitz algebra is the C $^*$-algebra generated by the unilateral shift. $C (\mathbb T) $ is identified with the Toeplitz operators with continuous symbol. $\endgroup$ – Martin Argerami Jun 11 '17 at 14:40

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