TITLE: Maximal ideal space of $c_{\mathcal{U}}$
QUESTION [11 upvotes]: Let $\mathcal{U}$ be an filter over $\mathbb{N}$. Define 
$$c_{\mathcal{U}} = \{{(x_n)\in \ell_\infty\colon \lim_{\mathcal{U}, n}x_n =0\}},$$
which is a C*-algebra. Is there an accessible topological description of the maximal ideal space of $c_{\mathcal{U}}$? At least for ultrafilters?

REPLY [3 votes]: (I will use $\omega$ for the ultrafilter since using a lowercase letter will improve readability)
The way I see it, your algebra $c_\omega$ is simply 
$$
c_\omega=\{f:\ f(\omega)=0\}\subset C(\beta \mathbb N).
$$
So you can make the identification $c_\omega=C_0(\beta\mathbb N\setminus\{\omega\})$. 
Note that $c_\omega$ is an ideal in $C(\beta\mathbb N)$, so the ideals in $c_\omega$ are ideals in $C(\beta\mathbb N)$. This is important because, with $\beta\mathbb N$ being compact, the ideals of $C(\beta\mathbb N)$ are precisely the sets of functions that annihilate a fixed closed subset. 
Thus,  the ideals of $c_\omega$ are the sets of the form 
$$
\{f\in C_0(\beta\mathbb N\setminus\{\omega\}):\ f=0\ \mbox{ on }\{\omega\}\cup K\}
$$
for a fixed closed $K\subset\mathbb N\setminus\{\omega\}$.
We conclude that maximal ideals of $c_\omega$ are of the form 
$$
\{f\in C_0(\beta\mathbb N\setminus\{\omega\}):\ f(\omega)=f(\eta)=0\}
$$
for some $\eta\in\beta\mathbb N\setminus\{\omega\}$.