# “Definitive” Noncommutative Space

Let $Y$ be a (locally compact) non-Hausdorff topological space. I want to know if there is a necessary and/or sufficient condition for $Y=X/G$, that is, $Y$ is the orbit space of a locally compact Hausdorff $G$-space $X$. I'd then indeed be able to call $Y$ an NC Space, with associated C*-algebra $C_0(X)\rtimes G$.

I couldn't find this "inverse" question in Connes' NCG or in Williams' Crossed Products of C*-algebras, where I might expect to find it (nor in Bourbaki, although there were similar things), though on pg. 112 Connes does state that when we have this situation (X/G being non-Hausdorff when X is LCH), the appropriate C*-algebra to assign is $C_0(X)\rtimes G$.

It seems like a natural question since many examples of nc spaces arise as orbit spaces.

• I don't see any reason why this construction wouldn't depend on the choice of $X$ and $G$. – Qiaochu Yuan Nov 25 '13 at 20:38
• Well in the "forward" direction (that is, given $X$ and $G$), if $G$ acts freely but not properly (in particular if $X$ is compact and $G$ is infinite), then $X/G$ is not Hausdorff (Connes pg 112). But I'm wondering if just knowing something about $Y$ is enough, or what must follow regarding $X,G$ when $Y$ is not Hausdorff. – Andrew Nov 26 '13 at 10:52
• In the question mathoverflow.net/questions/44109/… , and as a comment to the third answer, @david-carchedi stated "In fact, ALL compactly generated spaces (with no separation axiom) arise as quotients of locally compact Hausdorff spaces. In fact, this is if and only if." This is pretty close to the posed situation, modulo the group action. – Andrew Nov 26 '13 at 11:13