An ultrapower construction of the multiplier algebra of a $C^{\ast}$-algebra and an application to boundary amenability of groups

TL;DR

This paper introduces a novel ultrapower-based method for constructing the multiplier algebra of any $C^{st}$-algebra, extending previous work to more general cases and applying it to prove boundary amenability of certain groups.

Contribution

It provides a new ultrapower construction of the multiplier algebra applicable to noncommutative and nonseparable $C^{st}$-algebras, and offers a new proof of boundary amenability for specific group actions.

Findings

Extended the construction of multiplier algebras to noncommutative, nonseparable cases.

Provided a new proof of boundary amenability for groups acting on trees.

Generalized previous results to broader classes of $C^{st}$-algebras.

Abstract

Using ultrapowers of $C^{\ast}$-algebras we provide a new construction of the multiplier algebra of a $C^{\ast}$-algebra. This extends the work of Avsec and Goldbring [Houston J. Math., to appear, arXiv:1610.09276.] to the setting of noncommutative and nonseparable $C^{\ast}$-algebras. We also extend their work to give a new proof of the fact that groups that act transitively on locally finite trees with boundary amenable stabilizers are boundary amenable.

Equations

$$A_x(\epsilon ):=\{i\in \mathcal{I}:\Vert x(a_i-a_{\mathcal{U}})\Vert ,\Vert (a_i-a_{\mathcal{U}})x\Vert <\epsilon \}\in \mathcal{U},$$

$$B_x(\epsilon ):=\{i\in \mathcal{I}:\Vert x(b_i-b_{\mathcal{U}})\Vert ,\Vert (b_i-b_{\mathcal{U}})x\Vert <\epsilon \}\in \mathcal{U}.$$

$$|⟨a_{\mathcal{U}\text{-}WOT}\xi ,\eta ⟩-⟨b_i\xi ,\eta ⟩|\le |⟨(a_{\mathcal{U}\text{-}WOT}-a_i)\xi ,\eta ⟩|+|⟨(a_i-b_i)\xi ,\eta ⟩|<\epsilon .$$

$$\Vert x(b_i-a_{\mathcal{U}})\Vert \le \Vert x(b_i-a_i)\Vert +\Vert x(a_i-a_{\mathcal{U}})\Vert <\epsilon ,$$

$$\Vert (b_i-a_{\mathcal{U}})x\Vert \le \Vert (b_i-a_i)x\Vert +\Vert (a_i-a_{\mathcal{U}})x\Vert <\epsilon .$$

$${\mathcal{A}}^{s\mathcal{U}}:=\{(a_i{)}_{i\in \mathcal{I}}\in {\mathcal{A}}^{\mathcal{U}}:\text{there exists }{a}_{\mathcal{U}}\in B(\mathcal{H}):(a_i{)}_{i\in \mathcal{I}}\text{ is }\mathcal{U}\text{-strict convergent to }{a}_{\mathcal{U}}\}$$

$$\Vert (a_i+\lambda b_i-a_{\mathcal{U}}-\lambda b_{\mathcal{U}})x\Vert \le \Vert (a_i-a_{\mathcal{U}})x\Vert +|\lambda |\Vert (b_i-b_{\mathcal{U}})x\Vert <\epsilon ,$$

$$\Vert x(a_i+\lambda b_i-a_{\mathcal{U}}-\lambda b_{\mathcal{U}})\Vert \le \Vert x(a_i-a_{\mathcal{U}})\Vert +|\lambda |\Vert x(b_i-b_{\mathcal{U}})\Vert <\epsilon .$$

$$A_x\left(\frac{\epsilon }{2}\right)\cap B_x\left(\frac{\epsilon }{2|\lambda |}\right)\subset \{i\in \mathcal{I}:\Vert (a_i+\lambda b_i-a_{\mathcal{U}}-\lambda b_{\mathcal{U}})x\Vert <\epsilon ,\Vert x(a_i+\lambda b_i-a_{\mathcal{U}}-\lambda b_{\mathcal{U}})\Vert <\epsilon \},$$

$$\Vert (a_i{b}_i-a_{\mathcal{U}}{b}_{\mathcal{U}})x\Vert \le \Vert (a_i{b}_i-a_i{b}_{\mathcal{U}})x\Vert +\Vert (a_i{b}_{\mathcal{U}}-a_{\mathcal{U}}{b}_{\mathcal{U}})x\Vert \le \epsilon$$

$$\Vert x(a_i{b}_i-a_{\mathcal{U}}{b}_{\mathcal{U}})\Vert \le \Vert x(a_i{b}_i-a_{\mathcal{U}}{b}_i)\Vert +\Vert x(a_{\mathcal{U}}{b}_i-a_{\mathcal{U}}{b}_{\mathcal{U}})\Vert \le \epsilon ,$$

$$\{i\in\mathcal{I}:\lVert a_{i}^{n}x-a^{n}_{\mathcal{U}}x\rVert<\frac{\varepsilon}{4}\}\cap\{i\in\mathcal{I}:\lVert a_{i}^{m}x-a^{m}_{\mathcal{U}}x\rVert<\frac{\varepsilon}{4}\}\cap\Omega_{n}\ \big{(}\frac{\varepsilon}{4\lVert x\rVert}\big{)}\cap\Omega_{m}\big{(}\frac{\varepsilon}{4\lVert x\rVert}\big{)}$$

$$\Vert a_{\mathcal{U}}^{n}x-a_{\mathcal{U}}^{m}x\Vert$$

$$\le \frac{\epsilon }{4}+\Vert a_i^n-{\alpha }_i\Vert \Vert x\Vert +\Vert {\alpha }_i-a_i^m\Vert \Vert x\Vert +\frac{\epsilon }{4}<\epsilon .$$

$$\Vert \rho (x)-a_{\mathcal{U}}^{n}x\Vert <\frac{\epsilon }{3}\text{ and }{\Omega }_n\left(\frac{\epsilon }{3\Vert x\Vert }\right)\in \mathcal{U}.$$

$$\Vert \rho (x)-{\alpha }_{i}x\Vert \le \Vert \rho (x)-a_{\mathcal{U}}^{n}x\Vert +\Vert a_{\mathcal{U}}^{n}x-a_i^{n}x\Vert +\Vert a_i^{n}x-{\alpha }_{i}x\Vert \le \epsilon .$$

$$i\in \{i\in \mathcal{I}:|⟨({\alpha }_i-{\alpha }_{\mathcal{U}\text{-}WOT})x\xi ,\eta ⟩|<\frac{\epsilon }{2}\}\cap \{i\in \mathcal{I}:\Vert \rho (x)-{\alpha }_{i}x\Vert <\frac{\epsilon }{2}\}\in \mathcal{U}.$$

$$|⟨(\rho (x)-{\alpha }_{\mathcal{U}\text{-}WOT}x)\xi ,\eta ⟩|\le |⟨(\rho (x)-{\alpha }_{i}x)\xi ,\eta ⟩|+|⟨({\alpha }_{i}x-{\alpha }_{\mathcal{U}\text{-}WOT}x)\xi ,\eta ⟩|<\epsilon ,$$

$$J:=\{(a_i{)}_{i\in \mathcal{I}}\in {\mathcal{A}}^{s\mathcal{U}}:a_i\text{ is }\mathcal{U}\text{-strict convergent to }0\}$$

$$\{i\in \mathcal{I}:\Vert e_{i}x-x\Vert <\frac{\epsilon }{3M}\}\cap \{i\in \mathcal{I}:\Vert x-e_{i}x\Vert <\frac{\epsilon }{3M}\}\cap \{i\in \mathcal{I}:\Vert e_i{b}^{\prime }x-b^{\prime }x\Vert <\frac{\epsilon }{3M}\},$$

$$\Vert (b{e}_i{b}^{\prime }{e}_i-b{b}^{\prime }{e}_i)x\Vert$$

$$\le \Vert e_i{b}^{\prime }\Vert \Vert e_{i}x-x\Vert +\Vert e_i{b}^{\prime }x-b^{\prime }x\Vert +\Vert b^{\prime }\Vert \Vert x-e_{i}x\Vert <\epsilon .$$

$${\phi }^{\prime }(b)a={\phi }^{\prime }(b){\phi }^{\prime }(a)=ba=\phi (b)\phi (a)=\phi (b)a.$$

$$\mathcal{A}$$

$$\begin{gathered} |(f_n(x)-f_{\mathcal{U}}(x))g(x)|\le |f_n(x)-f_n(x_j)|\Vert g\Vert +|f_n(x_j)-f_{\mathcal{U}}(x_j)|\Vert g\Vert + \\ |f_{\mathcal{U}}(x_j)-f_{\mathcal{U}}(x)|\Vert g\Vert <\epsilon . \end{gathered}$$

$$|f_n(x)-f_n(y)|\le |f_n(x)-f_{\mathcal{U}}(x)|+|f_{\mathcal{U}}(x)-f_{\mathcal{U}}(y)|+|f_n(y)-f_{\mathcal{U}}(y)|\le \epsilon ,$$

$$T\ast S(\gamma )=\sum _{{\gamma }_1{\gamma }_2=\gamma }T({\gamma }_1)({\gamma }_1\cdot S({\gamma }_2))$$

$$T^{\ast }(\gamma )=\gamma \cdot T({\gamma }^{-1}{)}^{\ast }.$$

$${\Lambda }_i:=\bigcup _{v\in B(i)}\bigcup _{w\in B(i)}\left(g_v^{-1}{\Gamma }_i{g}_w\cap \Lambda \right).$$

$$\kappa (i):=\min \{k\in \mathbb{N}:||S_l-\lambda \ast S_l\Vert <\frac{1}{i}\text{ for all }l\ge k\text{ and for all }\lambda \in {\Lambda }_i\}.$$

$$T_{i}^{n}(\gamma)(t,y)=\frac{1}{\sqrt{\Big{\lvert}[o,t]\cap B(i)\Big{\rvert}}}\chi_{B(n)}(t)\sum_{v\in B(i)}{\chi_{[o,t]}(v)S_{\kappa(i)}(g_{v}^{-1}\gamma)(g_{v}^{-1}y)},$$

Full text

1. introduction

The multiplier algebra $\mathcal{M}(\mathcal{A})$ of a $C^{\ast}$-algebra $\mathcal{A}$ is a $C^{\ast}$-algebra that contains $\mathcal{A}$ as an essential ideal and satisfies the following universal property: for every $C^{\ast}$-algebra $\mathcal{B}$ containing $\mathcal{A}$ as an ideal, there exists a unique $\ast$-homomorphism $\varphi:\mathcal{B}\to\mathcal{M}(\mathcal{A})$ such that $\varphi$ is the identity on $\mathcal{A}$. If $\mathcal{A}$ is abelian, thus of the form $C_{0}(X)$ for some locally compact Hausdorff space $X$, then $\mathcal{M}(\mathcal{A})$ is isomorphic to $C_{b}(X)$ and this in turn can be identified with $C(\beta X)$, where $\beta X$ is the Stone-C̆ech compactification of $X$ (for more about multiplier algebras, see, for instance [4, 6]).

In the article [1], Avsec and Goldbring provided a new construction of the multiplier algebra of the abelian $C^{\ast}$-algebra $C_{0}(X)$ using ultraproducts of $C^{\ast}$-algebras, in the case when $X$ is a second countable locally compact Hausdorff space. From there, they inferred a new construction of the Stone-C̆ech compactification of $X$, and they used it to give a new proof of the fact that groups that act properly and transitively on trees are boundary amenable. In section 2 of this note, we extend their work providing a construction of the multiplier algebra of any $C^{\ast}$-algebra $\mathcal{A}$ by means of ultraproducts of $C^{\ast}$-algebras. In section 3, we focus on the case of commutative and separable $C^{\ast}$-algebras, and compare our main technical tool with the main technical tool used in [1] to explain why the work done here is indeed a generalization of [1]. Finally, in section 4, we extend the techniques of [1] to show that groups that act transitively on locally finite trees having boundary amenable stabilizers are boundary amenable.

2. Ultraproducts and Multipliers

Let $\mathcal{I}$ be a set. An ultrafilter over $\mathcal{I}$ is a nonempty collection $\mathcal{U}$ of subsets of $\mathcal{I}$ with the following properties:

1. (1)

finite intersection property: for every $\mathcal{I}_{0},\mathcal{I}_{1}\in\mathcal{U}$, then $\mathcal{I}_{0}\cap\mathcal{I}_{1}\in\mathcal{U}$; 2. (2)

directness: for every $\mathcal{I}_{o}\subset\mathcal{I}_{1}$, where $\mathcal{I}_{o}$ belongs to $\mathcal{U}$, then $\mathcal{I}_{1}\in\mathcal{U}$; 3. (3)

maximality: for every $\mathcal{I}_{0}\subset\mathcal{I}$, either $\mathcal{I}_{0}\in\mathcal{U}$ or $\mathcal{I}\setminus\mathcal{I}_{0}\in\mathcal{U}$.

An ultrafilter is principal if there exists $i\in\mathcal{I}$ such that the subsets of $\mathcal{I}$ that contains $i$ are in the ultrafilter. Ultrafilters not of this form are called nonprincipal or free. It is easy to show that an ultrafilter is nonprincipal exactly when it contains no finite sets. An ultrafilter is cofinal when the index set is a directed set, and the sets $\{i\in\mathcal{I}:i\geq i_{0}\}$ are in $\mathcal{U}$ for every $i_{0}\in\mathcal{I}$.

When dealing with directed sets with the property that there is no maximal element, every cofinal ultrafilter is nonprincipal. Moreover, when the ultrafilter is over $\mathbb{N}$, being cofinal is the same as being nonprincipal. If a directed set has a maximal element, then every cofinal ultrafilter is principal.

Definition 2.1**.**

Let $\mathcal{U}$ be an ultrafilter over $\mathcal{I}$. Let $(X,d)$ be a metric space and let $(a_{i})_{i\in\mathcal{I}}\subset X$. We say that $(a_{i})_{i\in\mathcal{I}}$ is convergent along $\mathcal{U}$, if there exists an element $a\in X$ such that, for every $\varepsilon>0$, the set $\{i\in\mathcal{I}:d(a_{i},a)<\varepsilon\}$ belongs to $\mathcal{U}$. The element $a$ is called the $\mathcal{U}$-limit of $(a_{i})_{i\in\mathcal{I}}$ and it is denoted by $\displaystyle\lim_{\mathcal{U}}{a_{i}}$.

2.1. Ultraproducts of $C^{\ast}$-algebras

Let $\mathcal{U}$ be an ultrafilter over a set $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. Denote by $\prod_{\mathcal{I}}{\mathcal{A}}$ the set $\{(a_{i})_{i\in\mathcal{I}}:\sup_{i\in\mathcal{I}}{\lVert a_{i}\rVert}<\infty\}$ and let $\mathcal{N}_{\mathcal{U}}$ be the subspace generated by those $(a_{i})_{i\in\mathcal{I}}\in\prod_{\mathcal{I}}{\mathcal{A}}$ such that $\displaystyle\lim_{\mathcal{U}}{\lVert a_{i}\rVert}=0$. Denote by $\mathcal{A}^{\mathcal{U}}$ the quotient $\prod_{\mathcal{I}}{\mathcal{A}}/\mathcal{N}_{\mathcal{U}}$. This is a vector space, and with the norm defined by $\lVert(a_{i})_{i\in\mathcal{I}}\rVert_{\mathcal{U}}:=\displaystyle\lim_{\mathcal{U}}{\lVert a_{i}\rVert}$, and the involution defined by $(a_{i})_{i\in\mathcal{I}}^{*}:=(a_{i}^{*})_{i\in\mathcal{I}}$, so $\mathcal{A}^{\mathcal{U}}$ becomes a $C^{\ast}$-algebra.

Remark 2.2*.*

Let $((a_{i}^{n})_{i\in\mathcal{I}})_{n\in\mathbb{N}}\subset\mathcal{A}^{\mathcal{U}}$ be a sequence that converges to $(a_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{\mathcal{U}}$. Then, for all $\varepsilon>0$, there exists $n_{0}\in\mathbb{N}$ such that if $n\geq n_{0}$, then ${\lVert(a^{n}_{i})_{i\in\mathcal{I}}-(a_{i})_{i\in\mathcal{I}}\rVert}_{\mathcal{U}}<\varepsilon$. We claim that if $n\geq n_{0}$, then the set $\Omega_{n}(\varepsilon):=\{i\in\mathcal{I}:\lVert a_{i}^{n}-a_{i}\rVert<\varepsilon\}$ belongs to $\mathcal{U}.$ To show this, set $\alpha_{n}:={\lVert(a^{n}_{i})_{i\in\mathcal{I}}-(a_{i})_{i\in\mathcal{I}}\rVert}_{\mathcal{U}}$. For every $\delta>0$, we have $\{i\in\mathcal{I}:\lvert\lVert a_{i}^{n}-a_{i}\rVert-\alpha_{n}\rvert<\delta\}\in\mathcal{U}$. Taking $\delta=\varepsilon-\alpha_{n}>0$ we get $\varepsilon-\alpha_{n}>\lvert\lVert a_{i}^{n}-a_{i}\rVert-\alpha_{n}\rvert\geq\lVert a_{i}^{n}-a_{i}\rVert-\alpha_{n}$. It follows that $\{i\in\mathcal{I}:\lvert\lVert a_{i}^{n}-a_{i}\rVert-\alpha_{n}\rvert<\delta\}\subset\Omega_{n}(\varepsilon)$. By directness, this implies that $\Omega_{n}(\varepsilon)\in\mathcal{U}$.

From now on we fix a faithful and non-degenerate representation of $\mathcal{A}$ on $B(\mathcal{H})$.

Lemma 2.3**.**

Let $\mathcal{U}$ be an ultrafilter defined over $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. For each $(a_{i})_{i\in\mathcal{I}}\in\prod_{\mathcal{I}}{\mathcal{A}}$, there exists a unique element $a_{\mathcal{U}\text{-}WOT}\in B(\mathcal{H})$ such that for every $\xi,\eta\in\mathcal{H}$, it holds that $\langle a_{\mathcal{U}\text{-}WOT}\xi,\eta\rangle=\displaystyle\lim_{\mathcal{U}}{\langle a_{i}\xi,\eta\rangle}$. The operator $a_{\mathcal{U}\text{-}WOT}$ is called the $\mathcal{U}$-WOT-limit of $(a_{i})_{i\in\mathcal{I}}$.

Proof.

Let $(a_{i})_{i\in\mathcal{I}}\in\prod_{\mathcal{I}}{\mathcal{A}}$ and let $\xi,\eta\in\mathcal{H}$. Then $(\langle a_{i}\xi,\eta\rangle)_{i\in\mathcal{I}}\subset\mathbb{C}$ is bounded, hence it has a $\mathcal{U}$-limit, which is denoted by $b_{\xi,\eta}$. It is easy to see that the map $(\xi,\eta)\mapsto b_{\xi,\eta}$ is a bounded sesquilinear form on $\mathcal{H}\times\mathcal{H}$. Take $a_{\mathcal{U}\text{-}WOT}\in B(\mathcal{H})$ the unique operator associated to it. ∎

Definition 2.4**.**

Let $\mathcal{U}$ be an ultrafilter defined over $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. An element $(a_{i})_{i\in\mathcal{I}}\in\prod_{\mathcal{I}}{\mathcal{A}}$ is $\mathcal{U}$-strict convergent if there exists an operator $a_{\mathcal{U}}\in B(\mathcal{H})$ such that, for every $x\in\mathcal{A}$, and every $\varepsilon>0$ we have $\{i\in\mathcal{I}:\lVert a_{i}x-a_{\mathcal{U}}x\rVert<\varepsilon,\lVert xa_{i}-xa_{\mathcal{U}}\rVert<\varepsilon\}\in\mathcal{U}$. The operator $a_{\mathcal{U}}$ is called the $\mathcal{U}$-strict limit of $(a_{i})_{i\in\mathcal{I}}$. Observe that $a_{\mathcal{U}}x$ and $xa_{\mathcal{U}}$ are elements of $\mathcal{A}$ for every $x\in\mathcal{A}$.

In what follows, it will be convenient to have the following notation at hand.

Notation 2.5*.*

Let $(a_{i})_{i\in\mathcal{I}},(b_{i})_{i\in\mathcal{I}}\in\displaystyle\Pi_{\mathcal{I}}\mathcal{A}$ that are $\mathcal{U}$-strict convergent to $a_{\mathcal{U}}$ and $b_{\mathcal{U}}$, respectively. For every $x\in\mathcal{A}$ and every $\varepsilon>0$, put

[TABLE]

[TABLE]

Proposition 2.6**.**

Let $\mathcal{U}$ be an ultrafilter defined over $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. If $(a_{i})_{i\in\mathcal{I}},(b_{i})_{i\in\mathcal{I}}\in\displaystyle\Pi_{\mathcal{I}}\mathcal{A}$ define the same element in $\mathcal{A}^{\mathcal{U}}$, then

1. (1)

if $(a_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-WOT convergent to $a_{\mathcal{U}\text{-}WOT}$, then $(b_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-WOT convergent to $a_{\mathcal{U}\text{-}WOT}$; 2. (2)

if $(a_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $a_{\mathcal{U}}$, then $(b_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $a_{\mathcal{U}}$.

Proof.

To prove (1), take $\xi,\eta\in\mathcal{H}$ of norm 1, let $\varepsilon>0$, and take $i$ in the set $\{i\in\mathcal{I}:\lVert a_{i}-b_{i}\rVert<\frac{\varepsilon}{2}\}\cap\{i\in\mathcal{I}:\lvert\langle a_{\mathcal{U}\text{-}WOT}\xi,\eta\rangle-\langle a_{i}\xi,\eta\rangle\rvert<\frac{\varepsilon}{2}\}\in\mathcal{U}$. Then

[TABLE]

It follows that the set $\{i\in\mathcal{I}:\lvert\langle a_{\mathcal{U}\text{-}WOT}\xi,\eta\rangle-\langle b_{i}\xi,\eta\rangle\rvert<\varepsilon\}$ belongs to $\mathcal{U}$.

To prove (2), take $\varepsilon>0$, $x\in\mathcal{A}$ and $i\in\{i\in\mathcal{I}:\lVert a_{i}-b_{i}\rVert<\frac{\varepsilon}{2\lVert x\rVert}\}\cap A_{x}(\frac{\varepsilon}{2})\in\mathcal{U}$. Then

[TABLE]

[TABLE]

It follows that the set $\{i\in\mathcal{I}:\lVert x(b_{i}-a_{\mathcal{U}})\rVert,\lVert(b_{i}-a_{\mathcal{U}})x\rVert<\varepsilon\}$ belongs to $\mathcal{U}$. ∎

Proposition 2.7**.**

Let $\mathcal{U}$ be an ultrafilter defined over $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. The set

[TABLE]

is a $C^{\ast}$-algebra.

Proof.

Let $(a_{i})_{\in\mathcal{I}},(b_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{s\mathcal{U}}$ that are $\mathcal{U}$-strict convergent to $a_{\mathcal{U}}$ and $b_{\mathcal{U}}$, respectively, let $\lambda\neq 0$ be a complex number, and let $x\in\mathcal{A}$. By following Notation 2.5, if $i\in A_{x}\left(\displaystyle\frac{\varepsilon}{2}\right)\cap B_{x}\left(\displaystyle\frac{\varepsilon}{2\lvert\lambda\rvert}\right)\in\mathcal{U}$, then

[TABLE]

and

[TABLE]

It follows that

[TABLE]

so $(a_{i}+\lambda b_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $a_{\mathcal{U}}+\lambda b_{\mathcal{U}}$.

It is clear that $\mathcal{A}^{s\mathcal{U}}$ is $*$-closed. To show that $\mathcal{A}^{s\mathcal{U}}$ is closed under taking products, set $M=\sup_{i\in\mathcal{I}}\{\lVert a_{i}\rVert,\lVert b_{i}\rVert\}$ and take i\in A_{x}\big{(}\displaystyle\frac{\varepsilon}{2M}\big{)}\cap A_{b_{\mathcal{U}}x}\big{(}\displaystyle\frac{\varepsilon}{2}\big{)}\cap B_{x}\big{(}\displaystyle\frac{\varepsilon}{2M}\big{)}\cap B_{xa_{\mathcal{U}}}\big{(}\displaystyle\frac{\varepsilon}{2}\big{)}\in\mathcal{U}. Then

[TABLE]

and

[TABLE]

which means that $(a_{i}b_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $a_{\mathcal{U}}b_{\mathcal{U}}$.

It is left to show that $\mathcal{A}^{s\mathcal{U}}$ is norm closed. Let $((a_{i}^{n})_{i\in\mathcal{I}})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{A}^{s\mathcal{U}}$ that converges to $(\alpha_{i})_{i\in\mathcal{I}}$ in $\mathcal{A}^{\mathcal{U}}$. We need to see that $(\alpha_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent.

As a first step, we will show that, for a fixed element $x\in\mathcal{A}$, $(\alpha_{i}x)_{i\in\mathcal{I}}$ and $(x\alpha_{i})_{i\in\mathcal{I}}$ have $\mathcal{U}$-limit in $\mathcal{A}$ (in the sense of Definition 2.1).

Let $x\in\mathcal{A}$ be fixed and $x\neq 0$. For each $n\in\mathbb{N}$ let $a^{n}_{\mathcal{U}}$ be the $\mathcal{U}$-strict limit of $(a_{i}^{n})_{i\in\mathcal{I}}\in\mathcal{A}^{s\mathcal{U}}$. We proceed to show that $(a^{n}_{\mathcal{U}}x)_{n\in\mathbb{N}}$ is a Cauchy sequence in $\mathcal{A}$. By Remark 2.2, for every $\varepsilon>0$, there exists $n_{0}\in\mathbb{N}$ such that, for all $n\geq n_{0}$, the sets \Omega_{n}\big{(}\frac{\varepsilon}{4\lVert x\rVert}\big{)} are elements of $\mathcal{U}$. It follows that the set

[TABLE]

is an element of $\mathcal{U}$ for all $n,m\geq n_{0}$. Take $i$ in this set. Then

[TABLE]

Let $\rho(x):=\displaystyle\lim_{n\in\mathbb{N}}{a^{n}_{\mathcal{U}}x}\in\mathcal{A}$. We will show that $\displaystyle\lim_{\mathcal{U}}{\alpha_{i}x}=\rho(x)$, that is,

$\{i\in\mathcal{I}:\lVert\rho(x)-\alpha_{i}x\rVert<\varepsilon\}\in\mathcal{U}$ for each $\varepsilon>0$. Let $n\in\mathbb{N}$ large enough such that

[TABLE]

For such $n\in\mathbb{N}$, take i\in\{i\in\mathcal{I}:\lVert a^{n}_{\mathcal{U}}x-a_{i}^{n}x\rVert<\displaystyle\frac{\varepsilon}{3}\}\cap\Omega_{n}\big{(}\displaystyle\frac{\varepsilon}{3\lVert x\rVert}\big{)}\in\mathcal{U}. Then

[TABLE]

Repeating this with $(x\alpha_{i})_{i\in\mathcal{I}}$ concludes the first step.

Let $\alpha_{\mathcal{U}\text{-}WOT}\in B(\mathcal{H})$ be the $\mathcal{U}$-WOT-limit of $(\alpha_{i})_{i\in\mathcal{I}}$. We will show that $\alpha_{\mathcal{U}\text{-}WOT}x=\rho(x)$. Take $\eta,\xi\in\mathcal{H}$ of norm 1, $\varepsilon>0$, and

[TABLE]

We then have that

[TABLE]

which implies that $\alpha_{\mathcal{U}\text{-}WOT}x=\rho(x)=\displaystyle\lim_{\mathcal{U}}{\alpha_{i}x}$. Therefore, for all $x\in\mathcal{A}$ and $\varepsilon>0$, we have $\{i\in\mathcal{I}:\lVert\alpha_{\mathcal{U}\text{-}WOT}x-\alpha_{i}x\rVert<\varepsilon\}\in\mathcal{U}$. In a similar manner, one shows that $\{i\in\mathcal{I}:\lVert x\alpha_{\mathcal{U}\text{-}WOT}-x\alpha_{i}\rVert<\varepsilon\}\in\mathcal{U}$. It follows that $(\alpha_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $\alpha_{\mathcal{U}\text{-}WOT}$. ∎

Proposition 2.8**.**

Let $\mathcal{U}$ be an ultrafilter defined over $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. The set

[TABLE]

is an ideal of $\mathcal{A}^{s\mathcal{U}}$.

Proof.

We only have to show that $J$ is norm closed. Consider $((a_{i}^{n})_{i\in\mathcal{I}})_{n\in\mathbb{N}}\subset J$ a sequence that converges in norm to $(\alpha_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{s\mathcal{U}}$. Let $\alpha_{\mathcal{U}}$ be the $\mathcal{U}$-strict limit of $(\alpha_{i})_{i\in\mathcal{I}}$. Let $\varepsilon>0$. Take $n\in\mathbb{N}$ large enough such that \Omega_{n}\Big{(}\displaystyle\frac{\varepsilon}{3\lVert x\rVert}\Big{)}\in\mathcal{U}, and i\in\{i\in\mathcal{I}:\lVert(\alpha_{i}-\alpha_{\mathcal{U}})x\rVert<\frac{\varepsilon}{3}\}\cap\{i\in\mathcal{I}:\lVert a_{i}^{n}x\rVert<\frac{\varepsilon}{3}\}\cap\Omega_{n}\Big{(}\displaystyle\frac{\varepsilon}{3\lVert x\rVert}\Big{)}\in\mathcal{U}. We then have that $\lVert\alpha_{\mathcal{U}}x\rVert\leq\lVert(\alpha_{\mathcal{U}}-\alpha_{i})x\rVert+\lVert\alpha_{i}x\rVert\leq\frac{\varepsilon}{3}+\lVert(\alpha_{i}-a_{i}^{n})x\rVert+\lVert a_{i}^{n}x\rVert\leq\varepsilon.$ Since the action of $\mathcal{A}$ on $\mathcal{H}$ is nondegenerate, $\alpha_{\mathcal{U}}=0$. ∎

There exists a natural embedding of $\mathcal{A}$ in $\mathcal{A}^{s\mathcal{U}}$, via the constant sequences $a\mapsto(a)_{i\in\mathcal{I}}$. It is clear that this element is $\mathcal{U}$-strict convergent to $a$. Moreover, since the representation of $\mathcal{A}$ in $B(\mathcal{H})$ is faithful and nondegenerate, it follows that there exists a natural embedding of $\mathcal{A}$ in $\mathcal{A}^{s\mathcal{U}}/J$.

Recall that an ideal $I$ in a $C^{\ast}$-algebra $\mathcal{A}$ is essential if $I\cap K$ is nontrivial for every ideal $K\neq\{0\}$, or equivalently, $aI=0$ implies $a=0$.

Lemma 2.9**.**

Let $\mathcal{U}$ be an ultrafilter defined over $\mathcal{I}$ and let $\mathcal{A}$ be a $C^{\ast}$-algebra. Consider the $C^{\ast}$-algebra $\mathcal{A}^{s\mathcal{U}}/J$. The image of $\mathcal{A}$ in $\mathcal{A}^{s\mathcal{U}}/J$ is an essential ideal.

Proof.

Take $(b_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{s\mathcal{U}}$, let $b_{\mathcal{U}}$ be its $\mathcal{U}$-strict limit, and take $a\in\mathcal{A}$. Then $(b_{i}a)_{i\in\mathcal{I}}$ and $(b_{\mathcal{U}}a)_{i\in\mathcal{I}}$ are both $\mathcal{U}$-strict convergent to $b_{\mathcal{U}}a$. It follows that $(b_{i}a)_{i\in\mathcal{I}}$ and $(b_{\mathcal{U}}a)_{i\in\mathcal{I}}$ are equal in $\mathcal{A}^{s\mathcal{U}}/J$. Analogously, $(ab_{i})_{i\in\mathcal{I}}$ is equal to $(ab_{\mathcal{U}})_{i\in\mathcal{I}}$ in $\mathcal{A}^{s\mathcal{U}}/J$.

Suppose that $J^{\prime}\subset\mathcal{A}^{s\mathcal{U}}/J$ is an ideal such that $J^{\prime}\cap\mathcal{A}=\{0\}$. If $(b_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{s\mathcal{U}}$ projects to $J^{\prime}$, then $(b_{i}x)_{i\in\mathcal{I}}\in J$,for each $x\in\mathcal{A}$. Let $b_{\mathcal{U}}$ be the $\mathcal{U}$-strict limit of $(b_{i})_{i\in\mathcal{I}}$. Hence $(b_{i}x)_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $b_{\mathcal{U}}x$. Then $b_{\mathcal{U}}x=0$ for all $x\in\mathcal{A}$. It follows that $b_{\mathcal{U}}=0$ and then $J^{\prime}=\{0\}$. ∎

2.2. Ultrafilters and approximate units

Every $C^{\ast}$-algebra $\mathcal{A}$ has an approximate unit, namely, there exist a directed set $\mathcal{I}$ and a net $(e_{i})_{i\in\mathcal{I}}\subset\mathcal{A}$ such that for every $x\in\mathcal{A}$, the nets $(xe_{i})_{i\in\mathcal{I}}$ and $(e_{i}x)_{i\in\mathcal{I}}$ converge to $x$ (see [6, Chapter I.4]). Approximate units can be taken to be positive and uniformly bounded, in which case they are elements of $\prod_{\mathcal{I}}{\mathcal{A}}$. In what follows, we will focus in the case where the ultrafilters are defined over this directed set $\mathcal{I}$. Observe that for a cofinal ultrafilter $\mathcal{U}$, the sets $\{i\in\mathcal{I}:\lVert xe_{i}-x\rVert<\varepsilon\}$ and $\{i\in\mathcal{I}:\lVert e_{i}x-x\rVert<\varepsilon\}$ belong to $\mathcal{U}$, for every $x\in\mathcal{A}$ and for every $\varepsilon>0$. Moreover, when $\mathcal{A}$ is unital, $\mathcal{I}$ can be taken to be the set with one element $\{1_{\mathcal{A}}\}$ and the approximate unit to be equal to $\{1_{\mathcal{A}}\}$. In this case, the only ultrafilter is the set $\{1_{\mathcal{A}}\}$, which is cofinal.

Theorem 2.10**.**

Let $\mathcal{A}$ be a $C^{\ast}$-algebra and let $(e_{i})_{i\in\mathcal{I}}\in\prod_{i\in\mathcal{I}}\mathcal{A}$ be an approximate unit for $\mathcal{A}$. Let $\mathcal{U}$ be a cofinal ultrafilter over $\mathcal{I}$. Then the $C^{\ast}$-algebra $\mathcal{A}^{s\mathcal{U}}/J$ is the multiplier algebra of $\mathcal{A}$.

Proof.

We saw in Lemma 2.9 that $\mathcal{A}$ is an essential ideal in $\mathcal{A}^{s\mathcal{U}}/J$. We are left to show that for any $C^{\ast}$-algebra $\mathcal{B}$ containing $\mathcal{A}$ as an ideal, there is a unique $C^{\ast}$-homomorphism $\varphi:\mathcal{B}\to\mathcal{A}^{s\mathcal{U}}/J$ such that $\varphi(a)=a$.

To this end, let $b\in\mathcal{B}$ and consider $\psi:\mathcal{B}\to\mathcal{A}^{s\mathcal{U}}$ defined by $\psi(b)=(be_{i})_{i\in\mathcal{I}}$. To see that $\psi$ is well defined, let $\varepsilon>0$ and let $x\in\mathcal{A}$. Let $i_{0}\in\mathcal{I}$ such that if $i\geq i_{0}$, then $\lVert xbe_{i}-xb\rVert<\varepsilon$. Let $i_{1}\in\mathcal{I}$ such that if $i\geq i_{1}$, then $\lVert e_{i}x-x\rVert<\displaystyle\frac{\varepsilon}{\lVert b\rVert}$. By the cofinality of $\mathcal{U}$ one obtains that $\{i\in\mathcal{I}:\lVert be_{i}x-bx\rVert<\varepsilon,\lVert xbe_{i}-xb\rVert<\varepsilon\}\in\mathcal{U}$. Observe that since $b\notin B(\mathcal{H})$, the last line does not imply that $(be_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{s\mathcal{U}}$. We must “represent” $b$ in $B(\mathcal{H})$. To this end, let $b_{\mathcal{U}\text{-}WOT}\in B(\mathcal{H})$ be the $\mathcal{U}$-WOT-limit of $(be_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{\mathcal{U}}$, an argument similar to one given in the proof of Proposition 2.7, shows that $b_{\mathcal{U}\text{-}WOT}x=bx$ and $xb_{\mathcal{U}\text{-}WOT}=xb$. Thus $(be_{i})_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $b_{\mathcal{U}\text{-}WOT}$. The same procedure shows that $(e_{i}b)_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to $b_{\mathcal{U}\text{-}WOT}$.

Call $\pi$ the quotient projection to $\mathcal{A}^{s\mathcal{U}}/J$, and let $\varphi=\pi\circ\psi$. It is clear that $\varphi$ is linear and bounded. Since $\psi(b^{*})=(b^{*}e_{i})_{i\in\mathcal{I}}$ and $\psi(b)^{*}=(e_{i}b^{*})_{i\in\mathcal{I}}$ and they are both $\mathcal{U}$-strict convergent to $(b^{*})_{\mathcal{U}\text{-}WOT}$, hence $\psi(b^{*})-\psi(b)^{*}$ is an element of $J$, so $\varphi$ is a $\ast$-preserving homomorphism.

To see that $\varphi$ is multiplicative, fix $b,b^{\prime}\in\mathcal{B}$ of norm $1$ and take $x\in\mathcal{A}$, $\varepsilon>0$, $M=\sup_{i\in\mathcal{I}}\{\lVert e_{i}\rVert\}$ and $i$ in the set

[TABLE]

which is an element of $\mathcal{U}$. Then

[TABLE]

Take $i\in\{i\in\mathcal{I}:\lVert xbe_{i}-xb\rVert<\displaystyle\frac{\varepsilon}{M}\}\in\mathcal{U}$. Then $\lVert x(be_{i}b^{\prime}e_{i}-bb^{\prime}e_{i})\rVert\leq\lVert xbe_{i}-xb\rVert\lVert b^{\prime}e_{i}\rVert<\varepsilon.$ It follows that $\psi(b)\psi(b^{\prime})-\psi(bb^{\prime})=(be_{i}b^{\prime}e_{i}-bb^{\prime}e_{i})_{i\in\mathcal{I}}$ is an element of $J$.

Since $(ae_{i}-a)_{i\in\mathcal{I}}$ is $\mathcal{U}$-strict convergent to [math], for all $a\in\mathcal{A}$, so $\varphi(a)=a$ in $\mathcal{A}^{s\mathcal{U}}/J$.

Suppose that there exists another $C^{\ast}$-homomorphism $\varphi^{\prime}:\mathcal{B}\to\mathcal{A}^{s\mathcal{U}}/J$ such that $\varphi^{\prime}(a)=a$ for all $a\in\mathcal{A}$. Then

[TABLE]

By Lemma 2.9, $\varphi(b)=\varphi^{\prime}(b)$. ∎

Ultraproducts provide a new point of view for dealing with multiplier algebras. For instance, the identification of $\mathcal{M}(\mathcal{A})$ with $\mathcal{A}^{s\mathcal{U}}/J$ yields an easy proof of the next characterization of multipliers, without using double centralizers.

Corollary 2.11**.**

$\mathcal{M}(\mathcal{A})$* is isomorphic to $\mathcal{M}:=\{m\in B(\mathcal{H}):\text{ for all }a\in\mathcal{A},am\in\mathcal{A}\,,\,ma\in\mathcal{A}\}$. In particular, $\mathcal{M}(\mathcal{A})$ is unital.*

Proof.

Consider $\varphi:\mathcal{A}^{s\mathcal{U}}\to\mathcal{M}$ defined by $\varphi((a_{i})_{i\in\mathcal{I}})=\displaystyle\lim_{\mathcal{U}\text{-}strict}{a_{i}}$. This map is well defined, it is a $C^{\ast}$-homomorphism (Proposition 2.7), and $\ker(\varphi)=J$. To show that $\varphi$ is surjective, let $m\in\mathcal{M}$. Then $am\in\mathcal{A}$ and $ma\in\mathcal{A}$ for every $a\in\mathcal{A}$. Hence, for all $\varepsilon>0$, the sets $\{i\in\mathcal{I}:\lVert a(me_{i}-m)\rVert<\varepsilon\}$ and $\{i\in\mathcal{I}:\lVert(me_{i}-m)a\rVert<\varepsilon\}$ are elements of $\mathcal{U}$. Then $(me_{i})_{i\in\mathcal{I}}\in\mathcal{A}^{\mathcal{U}}$ is $\mathcal{U}$-strict convergent to $m$.

Taking $m=1$, it follows that the image of $(e_{i})_{i\in\mathcal{I}}$ in $\mathcal{A}^{s\mathcal{U}}/J$ is the unit of $\mathcal{A}^{s\mathcal{U}}/J$. ∎

For a second application, observe that every $C^{\ast}$-homomorphism $\phi:\mathcal{A}\rightarrow\mathcal{B}$ defines a natural $C^{\ast}$-homomorphism $\phi^{\prime}:\mathcal{A}^{\mathcal{U}}\rightarrow\mathcal{B}^{\mathcal{U}}$. When $\phi$ is surjective, a proof similar to one given in Proposition 2.7 shows that $\phi^{\prime}(\mathcal{A}^{s\mathcal{U}})\subset\mathcal{B}^{s\mathcal{U}}$. This together with Theorem 2.10 immediately gives the following known result.

Proposition 2.12**.**

Let $\mathcal{A},\mathcal{B}$ be $C^{\ast}$-algebras and let $\phi:\mathcal{A}\rightarrow\mathcal{B}$ be a surjective homomorphism. The natural extension $\phi^{\prime}:\mathcal{A}^{\mathcal{U}}\rightarrow\mathcal{B}^{\mathcal{U}}$ induces the following commutative diagram:

[TABLE]

3. the case of commutative and separable $C^{\ast}$-algebras

Recall that when a $C^{\ast}$-algebra $\mathcal{A}$ is separable, it is $\sigma$-unital, namely, there exists a countable approximate unit (see [6, Chapter I.4]). That entails that the index set $\mathcal{I}$ of the previous section can be taken to be equal to $\mathbb{N}$, in which case nonprincipal ultrafilters are cofinal and $\prod_{\mathcal{I}}\mathcal{A}$ is $\ell^{\infty}(\mathcal{A})$.

In [1], the authors built the multiplier algebra for commutative and separable $C^{\ast}$-algebras using ultraproducts of $C^{\ast}$-algebras. More precisely they considered $\mathcal{A}=C_{0}(X)$ where $X$ is a second countable, locally compact topological space and took $\mathcal{U}$ a nonprincipal ultrafilter defined over $\mathbb{N}$ to construct the multiplier algebra of $\mathcal{A}$, $C_{b}(X)$, identifying it with a quotient of a sub-$C^{\ast}$-algebra of $\prod_{\mathcal{I}}{\mathcal{A}}$. For that, the authors usee the key fact that the hypothesis on $X$ entails the existence of a proper metric compatible [8]. In what follows, we will then identify the second countable, locally compact topological space $X$ with the metric space $(X,d)$, where $d$ is a proper metric on $X$. The closed ball of radius $r>0$ centered at a fixed base-point $o\in X$, will be denoted by $B_{o}(r)$. The main technical tool of [1] is the following definition.

Definition 3.1**.**

[1, Section 3] Let $(X,d)$ be as in the preceding discussion. Let $\mathcal{A}=C_{0}(X)$ and let $\mathcal{U}$ be a nonprincipal ultrafilter over $\mathbb{N}$. For $(f_{n})_{n\in\mathbb{N}}\in\ell^{\infty}(\mathcal{A})$, we say that $(f_{n})_{n\in\mathbb{N}}$ is $\mathcal{U}$-equicontinuous on bounded sets if, for every $r,\varepsilon>0$, there is $\delta>0$ such that the set $\{n\in\mathbb{N}:\text{ for all }s,t\in B_{o}(r)\text{ with }d(s,t)<\delta\implies\lvert f_{n}(s)-f_{n}(t)\rvert<\varepsilon\}$ belongs to $\mathcal{U}$.

Given $(f_{n})_{n\in\mathbb{N}}\in\ell^{\infty}(\mathcal{A})$, and a fixed $x\in X$, the $\mathcal{U}$-limit of the sequence $(f_{n}(x))_{n\in\mathbb{N}}$ is well defined. We denote $f_{\mathcal{U}}:X\rightarrow\mathbb{C}$, $f_{\mathcal{U}}(x):=\displaystyle\lim_{\mathcal{U}}(f_{n}(x))$. The following fact was observed in [1].

Lemma 3.2**.**

If $(f_{n})_{n\in\mathbb{N}}\in\ell^{\infty}(\mathcal{A})$ is $\mathcal{U}$-equicontinuous on bounded sets, then $f_{\mathcal{U}}$ is uniformly continuous on bounded sets.

The next proposition shows that our notion of $\mathcal{U}$-strict convergence coincides with the notion of $\mathcal{U}$-equicontinuity on bounded sets in the case of $\mathcal{A}=C_{0}(X)$, where $X$ is a locally compact, second countable, topological space. This entails that the work done here in section 2 is indeed a generalization of the work done in [1, Section 3].

Proposition 3.3**.**

Take $(f_{n})\in\ell^{\infty}(\mathcal{A})$ and let $f_{\mathcal{U}}(x)=\displaystyle\lim_{\mathcal{U}}(f_{n}(x))$. The following two conditions are equivalent:

1. (1)

The sequence $(f_{n})_{n\in\mathbb{N}}$ is $\mathcal{U}$-equicontinuous on bounded sets. 2. (2)

The sequence $(f_{n})_{n\in\mathbb{N}}$ is $\mathcal{U}$-strict convergent to $f_{\mathcal{U}}$.

Proof.

To show that $\eqref{goldequi}$ implies $\eqref{strictnatural}$, let $(f_{n})_{n\in\mathbb{N}}\in\ell^{\infty}(\mathcal{A})$ be $\mathcal{U}$-equicontinuous on bounded sets. Let $\varepsilon>0$ and let $g\in C_{0}(X)$. Set $M=\sup_{n\in\mathbb{N}}\{\lVert f_{n}\rVert,\lVert g\rVert\}$, and take $K\subset X$ a compact set such that $\lvert g(x)\rvert<\displaystyle\frac{\varepsilon}{2M}$ if $x\notin K$. There exists $\delta_{1}$ such that for $x,y\in K$ with $d(x,y)<\delta_{1}$, $\{n\in\mathbb{N}:\lvert f_{n}(x)-f_{n}(y)\rvert\leq\displaystyle\frac{\varepsilon}{3M}\}\in\mathcal{U}$. By Lemma 3.2, there exists $\delta_{2}>0$ such that $\lvert f_{\mathcal{U}}(x)-f_{\mathcal{U}}(y)\rvert<\displaystyle\frac{\varepsilon}{3}$, for $x,y\in K$ with $d(x,y)<\delta_{2}$. Take $\delta=\min\{\delta_{1},\delta_{2}\}$ and cover $K$ with a finite number of balls $B_{x_{j}}(\delta)$ of radius $\delta$ centered at $x_{j}$, $j=1,\ldots,m$. Since $f_{\mathcal{U}}(x_{j})=\displaystyle\lim_{\mathcal{U}}f_{n}(x_{j})$, it follows that the sets $A_{j}=\left\{n\in\mathbb{N}:\lvert f_{n}(x_{j})-f_{\mathcal{U}}(x_{j})\rvert<\displaystyle\frac{\varepsilon}{3M}\right\}$ belong to $\mathcal{U}$. Therefore if $n\in\bigcap_{j=1}^{m}A_{j}\in\mathcal{U}$, and $x\in K$, taking $x_{j}$ with $d(x,x_{j})<\delta$ we get

[TABLE]

On the other hand, if $x\notin K$, then $\lvert(f_{n}-f_{\mathcal{U}})g(x)\rvert<\varepsilon$. This shows that $\{n\in\mathbb{N}:\|f_{n}g-f_{\mathcal{U}}g\|<\varepsilon\}\in\mathcal{U}$.

To show the converse, take $g\in C_{0}(X)$ such that $g=1$ in $B_{o}(r)$. By hypothesis, for all $\varepsilon>0$ the sets $\{n\in\mathbb{N}:\lVert(f_{n}-f_{\mathcal{U}})g\rVert<\varepsilon\}$ are in $\mathcal{U}$ and are infinite. So we can build a subsequence $(f_{n_{k}})_{k\in\mathbb{N}}$ such that $(f_{n_{k}})_{k\in\mathbb{N}}$ is uniformly convergent to $f_{\mathcal{U}}$ in $B_{0}(r)$. By hypothesis, closed balls are compact, so $f_{\mathcal{U}}$ is uniformly continuous on $B_{o}(r)$. Let $\delta>0$ such that $d(x,y)<\delta$ implies $\lvert f_{\mathcal{U}}(y)-f_{\mathcal{U}}(x)\rvert<\varepsilon$ in $B_{o}(r)$. Take $n\in\{n\in\mathbb{N}:\lVert(f_{n}-f_{\mathcal{U}})g\rVert<\displaystyle\frac{\varepsilon}{3}\}\in\mathcal{U}$. Then for $x,y\in B_{o}(r)$ such that $d(x,y)<\delta$ we have

[TABLE]

therefore $\{n\in\mathbb{N}:\lVert(f_{n}-f_{\mathcal{U}})g\rVert<\frac{\varepsilon}{3}\}$ is a subset of $\{n\in\mathbb{N}:\text{ for all }x,y\in B_{o}(r)\text{ with }d(x,y)<\delta\implies\lvert f_{n}(x)-f_{n}(y)\rvert<\varepsilon\}$. This shows that the last set belongs to $\mathcal{U}$. ∎

4. An application to boundary amenability of groups

Let $\Gamma$ be a countable group. Let $\mathcal{A}$ be a unital $C^{\ast}$-algebra endowed with a $\Gamma$-action by $\ast$-automorphisms. Let $C_{c}(\Gamma,\mathcal{A})$ be the space of finitely supported functions from $\Gamma$ to $\mathcal{A}$. This is a $\ast$-algebra with the product given by

[TABLE]

and the involution given by

[TABLE]

The space $C_{c}(\Gamma,\mathcal{A})$ has a pre-Hilbert $\mathcal{A}$-module structure via the inner product $\langle T,S\rangle_{\mathcal{A}}:=\sum\limits_{\gamma\in\Gamma}{T(\gamma)^{*}S(\gamma)}$, and the corresponding norm $\lVert T\rVert_{\mathcal{A}}:=\langle T,T\rangle^{1/2}$.

Definition 4.1**.**

A group $\Gamma$ is boundary amenable (or exact) if there exist a compact space $Y$ and a sequence $(S_{i})_{i\in\mathbb{N}}\subset C_{c}(\Gamma,C(Y))$ such that

1. (1)

$S_{i}$ is positive, that is, $S_{i}(\gamma)\geq 0$ for all $\gamma\in\Gamma$; 2. (2)

$\sum_{\gamma\in\Gamma}{S_{i}^{2}(\gamma)}=1$; 3. (3)

for every $\gamma_{1}\in\Gamma$, it follows that $\lVert S_{i}-\gamma_{1}\ast S_{i}\rVert$ goes to [math] when $i$ goes to infinity.

In [1, Section 4], the authors applied their ultraproduct construction of the multiplier algebra of a separable $C^{\ast}$ algebra to show that groups acting properly and transitively on a locally finite tree are boundary amenable. The assumption that the action is proper and transitive implies that the vertex stabilizers are all isomorphic finite groups of cardinal $m$. In the proof given in [1], this constant $m$ contains all the information that is needed about the stabilizers. Since it is known that groups acting on a locally finite tree with exact stabilizers are boundary amenable (see, for instance,[2, 3, 5, 7]), it is interesting to see if the strategy developed in [1] can be extended to prove this more general result. We partially achieve this in the proof of the next theorem.

Theorem 4.2**.**

If a countable group $\Gamma$ acts transitively on a locally finite tree $\mathcal{T}$ with boundary amenable stabilizers, then $\Gamma$ is boundary amenable.

Proof.

Fix a vertex $o\in\mathcal{T}$ as a base-point. We will denote with $B(i)$ the closed ball of radius $r$ centered in $o$. The geodesic that connects $o$ with $t$ will be denoted by $[o,t]$. In what follows $Stab\{o\}$ will be denoted with $\Lambda$.

Since the action of $\Gamma$ on $\mathcal{T}$ is transitive, we choose a cross section for it, namely, for each $v\in\mathcal{T}$ we choose $g_{v}\in\Gamma$ such that $g_{v}o=v$.

Set \big{(}\Gamma_{i}\big{)}_{i\in\mathbb{N}} an increasing sequence of finite subsets of $\Gamma$ that covers $\Gamma$ and let

[TABLE]

Since $\mathcal{T}$ is locally finite, $\Lambda_{i}$ is finite.

Since $\Lambda$ is boundary amenable, there exist a compact set $Y$ and a sequence $(S_{i})_{i\in\mathbb{N}}\subset C_{c}((\Lambda,C(Y))$ satisfying the three conditions of Definition 4.1. The compact space $Y$ will be taken to be a $\Gamma$-space (see [3, pg. 178]). Moreover, we can assume that $S_{i}$ is defined over all $\Gamma$ by extending it by [math] in $\Gamma\setminus\Lambda$. For every $i\in\mathbb{N}$, let

[TABLE]

Consider the $C^{\ast}$-algebra $\mathcal{A}:=C_{0}(\mathcal{T}\times Y)$. Since $Y$ is compact and $\mathcal{T}$ is countable, $\mathcal{A}$ is a $\sigma$-unital $C^{\ast}$-algebra, hence we are in the case discussed at the beginning of Section 3. Let $\mathcal{U}$ be a non principal ultrafilter over $\mathbb{N}$ and consider the corona algebra $\mathcal{M}(\mathcal{A})/\mathcal{A}$, which we identify with \mathcal{A}^{s\mathcal{U}}/J\big{/}\mathcal{A}. This is a unital $C^{\ast}$-algebra.

For each $i\in\mathbb{N}$, set $T_{i}:\Gamma\rightarrow\mathcal{A}^{s\mathcal{U}}$, where $T_{i}(\gamma)=(T_{i}^{n}(\gamma))_{n\in\mathbb{N}}$ is defined by

[TABLE]

where \Big{\lvert}[o,t]\cap B(i)\Big{\rvert} denotes the number of points in the geodesic $[o,t]$ that lie inside the ball $B(i)$. Note that $(T_{i}^{n}(\gamma))_{n\in\mathbb{N}}$ is $\mathcal{U}$-strict convergent to

[TABLE]

We will denote by $T_{i}$ its projection to \mathcal{A}^{s\mathcal{U}}/J\big{/}\mathcal{A}. We remark that the definition of $T_{i}$ is inspired by the definition of $\mu_{x,y}$ given by Ozawa in [5].

Claim: The sequence (T_{i})_{i\in\mathbb{N}}\subseteq C(\Gamma;\mathcal{A}^{s\mathcal{U}}/J\big{/}\mathcal{A}) satisfies the conditions of Definition 4.1.

To show this, first observe that if $\Omega_{\kappa(i)}\subseteq\Lambda$ denotes the support of $S_{\kappa(i)}$, then the support of $T_{i}$ is contained in $\bigcup_{v\in B(i)}g_{v}\Omega_{\kappa(i)}$. Since $\mathcal{T}$ is locally finite, this is a finite set.

To show that condition (1) holds true, it is enough to notice that $T_{i}$ is sum and product of positive functions.

To show that condition (2) holds true, note that for each fixed $\gamma\in\Gamma$, there exists only one $g_{v}$ such that $g_{v}^{-1}\gamma\in\Lambda$. Then there exists at most one $g_{v}$ such that $S_{\kappa(i)}(g_{v}^{-1}\gamma)$ is nonzero. Then

[TABLE]

It remains to show that condition (3) holds true. To this end, fix $\gamma_{1}\in\Gamma$. Observe that if $(t,y)\in\mathcal{T}\times Y$, then $\gamma_{1}\ast T_{i}(\gamma)(t,y)=T_{i}(\gamma_{1}^{-1}\gamma)(\gamma_{1}^{-1}t,\gamma_{1}^{-1}y)$. Hence

[TABLE]

If we set \theta(i,t):=\Big{\lvert}[o,t]\cap B(i)\Big{\rvert}^{-1/2}\,\Big{\lvert}[o,\gamma_{1}^{-1}t]\cap B(i)\Big{\rvert}^{-1/2}, then

[TABLE]

where the last equality follows from the changes of variables $\lambda:=g_{v}^{-1}\gamma$ and $z:=g_{v}^{-1}y$, and by recalling that $S_{\kappa(i)}(\lambda)=0$ if $\lambda\notin\Lambda$.

In order to get that $S_{\kappa(i)}(g_{w}^{-1}\gamma_{1}^{-1}g_{v}\lambda)\neq 0$, it is necessary that $g_{w}^{-1}\gamma_{1}^{-1}g_{v}$ belongs to $\Lambda$. Note that for a fixed $v$, there exists only one $g_{w}$ such that $g_{w}^{-1}\gamma_{1}^{-1}g_{v}\in\Lambda$. For this $g_{w}$, we have that $w=\gamma_{1}^{-1}v$ and $\chi_{[o,\gamma_{1}^{-1}t]}(w)=\chi_{[o,\gamma_{1}^{-1}t]}(\gamma_{1}^{-1}v)=\chi_{[\gamma_{1}o,t]}(v)$. Therefore, (4) is equal to the following sum, only depending on $v$.

[TABLE]

Replacing this in (4), we get that $\lVert T_{i}-\gamma_{1}\ast T_{i}\rVert_{C_{c}(\Gamma,\mathcal{A}^{s\mathcal{U}}/J/\mathcal{A})}^{2}$ is equal to

[TABLE]

By the triangle inequality, to estimate (4.3), it is enough to estimate the following two quantities

[TABLE]

and

[TABLE]

In order to bound (4.4), first note that

[TABLE]

Then note that there exists $r\in\mathbb{N}$ such that $\gamma_{1}\in\Gamma_{i}$ for all $i\geq r$. It follows that for all $i\geq r$, $g_{v}^{-1}\gamma_{1}g_{w}\in\Lambda_{i}$, whenever $v,w\in B_{i}$. Then, by the definition of $\kappa(i)$, it follows that

[TABLE]

Then, for all $i\geq r$, (4.4) is bounded by

[TABLE]

In order to bound (4.5), for each $i\in\mathbb{N}$, we choose

[TABLE]

This choice of $a\in\mathcal{A}$ left us to bound (4.5) only when $t\notin B(i)\cup\gamma_{1}B(i)$. In this case $\theta(i,t)=\frac{1}{i}$. Moreover, if $i\geq d(o,\gamma_{1}o)$ then

[TABLE]

It follows that in (4.5) we have the bound

[TABLE]

All this combined entails that if $\varepsilon>0$, and $i\geq\max\{r;d(o,\gamma_{1}o),2/\varepsilon;2d(o,\gamma_{1}o)/\varepsilon\}$, then $\lVert T_{i}-\gamma_{1}\ast T_{i}\rVert_{C_{c}(\Gamma,\mathcal{A}^{s\mathcal{U}}/J/\mathcal{A})}^{2}<\varepsilon$. ∎

Acknowledgments. F. Poggi was supported in part by a CONICET Doctoral Fellowship. R. Sasyk was supported in part through the grant PIP-CONICET 11220130100073CO. We thank Prof. Isaac Goldbring for his comments on the first version of the article that helped us improve the exposition. We thank the referee for his detailed reading of the manuscript and for prompting us to add a section on boundary amenability of groups.