# NonCommutative Baire theorem

The classical Baire theorem says that the intersection of a sequence of open dense subsets of $X$, is dense, if the space is compact Hausdorff. In the language of $C^{*}$ algebras this is equivalents to the following:

Assume that $I_{n},s$ are a sequence of essential ideals in a commutative unital $C^{*}$ algebra $A$. Assume that $J$ is an arbitrary nontrivial ideal.Then there is a maximal ideal $L$ which neither contain $J$, nor contains $I_{n}$. (no $I_{n}$ is contained in $L$).

This is a motivation to ask:

Is the above statement true for a commutative unital semisimple Banach algebra?

Is the above statement true for a unital non commutative $C^{*}$ algebra, provided we replace the "maximal ideals" by primitive Ideals?

• I believe your second question asks if the primitive ideal space of a C*-algebra is a Baire space. It is (by a more general theorem of Choquet), see Blackadar's operator algebra book II.6.5.14 for details. Apr 27 '14 at 19:35
• Two questions: (1) can you say a bit more about why your statement about ideals is equivalent to Baire category? Closed subsets of X correspond to zero sets of closed ideals in A, and Baire category says that the union of closed nowhere dense subsets is itself nowhere dense, but how does that translate into your statement about essential ideals? (2) What do you mean by an "essential ideal" in an arbitrary unital semisimple CBA? Apr 27 '14 at 19:41
• @CalebEckhardt thank you very much for the comment. I will look at the reference which you mentioned. Apr 28 '14 at 2:50
• @YemonChoi an essential ideal is an ideal with nontrivial intersection with any other ideal. This is the algebraic analogy of open-dense subsets. I considered the "complement" version of the Baire theorem:the intersection of a sequence of open dense sets is dense.Let $U_{n}$ be a sequence of open dense sets. for every open set $W$, there is a point $p\in W$ which belongs $\cap U_{n}$. But a point of a classical space corresponds to a maximal ideal. On the other hand,$p\in W$ means that $I_{W^{c}}$ is not contained in $I_{p}$. the later two ideals are the same as you mentioned. Apr 28 '14 at 3:00
• In fact an open set $U$ correspond to the ideal $I_{U^{c}}$. The fact that $\cap U_{n}$ is dense implies that for each open $W$,there is a $p$ with $p\in W$ ($I_{W^{c}}$ not contained in $I_{p}$), such that $p\in U_{n}$ for all $n$, that is no $I_{U_{n}^{c}}$ is contained in $I_{p}$ Apr 28 '14 at 3:13

This answer (should have been a comment) refers to the second question: The kinds of ideal lattices that one can get in a C*-algebra prevent this from happening in general. Take $\mathcal K^\sim$, the unitization of the compact operators (over a separable Hilbert space). The lattice of ideals of $\mathcal K^\sim$ isomorphic to {0,1,2}. The ideal corresponding to the compacts is essential and minimal among the non-zero ideals. So the property of the question cannot hold here. (One can even have the lattice of closed two-sided ideals of a C*-algebra isomorphic to the ordered set [0,1]. See "A purely infinite AH-algebra and an application to AF-embeddability", by Mikael Rordam.)