# Commutative multiplier algebra

In my course of spectral theory and operator algebras I came across the following exercise:

Let $\mathcal{A}=C_0(X)$ where $X$ is a locally compact Hausdorff space. Describe the multiplier algebra $M(\mathcal{A})$. Moreover, since the multiplier algebra is commutative, it is of the form $C(\beta X)$ where $\beta X$ is a compact Hausdorff space. Show that there exists an injection $j:X\rightarrow \beta X$ such that $j(X)$ is open in $\beta X$.

I was able to show that $M(\mathcal{A})\cong C_b(X)$, the bounded continuous functions on $X$, however the second part is annoying me. It seems very intuitive though.

I guess the more general statement is that if $\mathcal{A}\cong C_0(X)$ and $\mathcal{B}\cong C(Y)$ such that $\mathcal{A}$ is an essential two-sided ideal of $\mathcal{B}$, then $X$ injects into $Y$ such that $X$ is open in $Y$.

I'm aware that $\beta X$ is the Stone-Cech compactification, however I do not want to use that knowledge. How can I construct this map $j$ more or less constructively? Hints and tips are welcome, I tried certain things but I failed to prove the essential claims that I made.

You are almost there. The ideals of $$C (Y)$$ are given by its closed subsets, and the essential ideals are those that correspond to closed nowhere dense subsets. In other words, the essential ideals of $$C (Y)$$ are precisely $$C _0 (T)$$, where $$T\subset Y$$ is open and dense.

Edit: proof.

Let $$Y_0=\{t\in Y:\ f(t)=0,\ \text{ for all }f\in J\}$$. Note that $$Y_0=\bigcap_{f\in J} f^{-1}(\{0\}),$$ so $$Y_0$$ is closed and thus compact. Write $$J_0=\{f\in C_0(T):\ f|_{Y_0}=0\}$$. It is clear that $$J\subset J_0$$.

Let $$V\subset Y$$ be open with $$Y_0\subset V$$. For any $$g\in J$$, we have $$|g|^2=\overline g\,g\in J$$. Given $$t\in Y\setminus V$$, there exists $$g_1\in J$$ with $$g_1(t)\ne0$$; by replacing $$g_1$$ with $$|g_1|^2$$, we may assume that $$g_1(t)>0$$. As $$Y\setminus V$$ is compact, there exists $$c>0$$ with $$g_1(t) \geq c$$ for all $$t\in Y\setminus V$$. Define $$g_2:Y\setminus V\to\mathbb C$$ by $$g_2(t)=1/g_1(t)$$. Using again that $$Y\setminus V$$ is compact, we get from Tietze's Extension an extension, that we still call $$g_2$$, to all of $$Y$$. Let $$g_V=g_1g_2\in J$$ (since $$g_1\in J$$). We have $$g_V|_{Y\setminus V}=1$$, $$g_V|_{Y_0}=0$$, and $$\|g_V\|\leq1$$.

Now let $$f\in J_0$$, and fix $$\varepsilon>0$$. For each $$t\in Y_0$$ we have $$f(t)=0$$, so there exists an open set $$V_t$$ with $$t\in V_t$$ and $$|f|<\varepsilon$$ on $$V_t$$. As $$Y_0$$ is compact, it is covered by finitely many $$V_{t_1},\ldots,V_{t_n}$$. Let $$V=\bigcup_{j=1}^nV_{t_j}$$, open. We have $$Y_0\subset V$$ and $$|f|<\varepsilon$$ on $$V$$. Using again that $$J$$ is an ideal, $$fg_V\in J$$.

For any $$t\in Y\setminus V$$, since $$g_V(t)=1$$ we get $$f(t)-f(t)g_V(t)=0$$. For $$t\in V$$ we have $$|f(t)-f(t)g_V(t)|=|f(t)|\,|1-g_V(t)|\leq 2\varepsilon$$. Thus $$\|f-fg_V\|\leq2\varepsilon.$$ As $$\varepsilon$$ was arbitrary, we have shown that $$f\in \overline J=J$$. So $$J_0\subset J$$, and the equality follows.

• I don't think so. – Martin Argerami Nov 16 '18 at 12:55
• Edited. $\ \ \ \$ – Martin Argerami Jun 10 '19 at 21:36
• in where the condition "$T\subset Y$ is open and dense" be used? – math112358 Jun 12 '19 at 17:17
• The argument is a characterization of all the ideals. I'll let you deal with the essential ones. – Martin Argerami Jun 12 '19 at 17:18