Closed ideals and Lie ideals of minimal tensor product of certain C*-algebras

TL;DR

This paper characterizes closed ideals and Lie ideals of minimal tensor products of certain C*-algebras, linking their structure to properties of the component algebras and the underlying space.

Contribution

It provides a characterization of closed ideals in $C_0(X,A)$ for algebras with finitely many ideals and describes the structure of Lie ideals in these tensor products.

Findings

Closed ideals of $C_0(X,A)$ are characterized via ideals of $A$ and subspaces of $X$.

Closed ideals of $C_0(X) ensor^{ ext{min}} A$ are finite sums of product ideals.

The centre-quotient property of $C_0(X,A)$ is equivalent to that of $A$.

Abstract

For a locally compact Hausdorff space $X$ and a $C^*$-algebra $A$ with only finitely many closed ideals, we discuss a characterization of closed ideals of $C_0(X,A) $ in terms of closed ideals of $A$ and certain (compatible) closed subspaces of $X$. We further use this result to prove that a closed ideal of $C_0(X) \otimes^{\min} A$ is a finite sum of product ideals. We also establish that for a unital $C^*$-algebra $A$, $C_0(X,A)$ has centre-quotient property if and only if $A$ has centre-quotient property. As an application, we characterize the closed Lie ideals of $C_0(X,A)$ and identify all closed Lie ideals of $ C_0(X) \otimes^{\min} B(H) $, $H$ being a separable Hilbert space.

Equations

$$S_i={\cap }_{f\in J}{f}^{-1}(I_i)={\cap }_{f\in J}{f}^{-1}({\cap }_{j\in \gamma }{I}_j)={\cap }_{f\in J}{\cap }_{j\in \gamma }{f}^{-1}(I_j)={\cap }_{j\in \gamma }{\cap }_{f\in J}{f}^{-1}(I_j)={\cap }_{j\in \gamma }{S}_j,$$

$$\Vert \sum _{i=0}^k{\widehat{g_i}}_{{|}_X}(y)h_{x_i}(y)-f(y)\Vert$$

$${\overline{J(Y)\otimes C}}^{\min }\cong \{f\in C_0(X,A):f(Y)=\{0\},f(X)\subseteq C\}.$$

$$\sum _{i=0}^n\widetilde{f_i}(x)=\sum _{i=0}^n\tilde{f}(x)f_i(x)=\sum _{i=0}^n{f}_i(x)=1.$$

$$\Vert h(x)-\sum _{i=1}^n\widetilde{f_i}(x)h(x_i)\Vert$$

$$C_0(X){\otimes }^{\min }I+J(Y){\otimes }^{\min }A=\{f\in C_0(X,A):f(Y)\subseteq I\}$$

$$K=\sum _{j=2}^{n}J({\cup }_{k\in {\gamma }_j}{S}_k){\otimes }^{\min }{I}_j,$$

$$\bigcap _{r\in {\alpha }_i}({\cup }_{k\in {\gamma }_r}{S}_k)=\bigcup _{1\le t_{r_j}\le p_{r_j}}(S_{r_{1,t_{r_1}}}\cap S_{r_{2,t_{r_2}}}\cap \cdots \cap S_{r_{q,t_{r_q}}}).$$

$$B(H)\oplus B(H)(=I_9)$$

$$S_9$$

$$\{f\in C_0(X,A):f(S_i)\subseteq \overline{[I_i,A]},\forall i\in \Delta \}\subseteq L\subseteq \{f\in C_0(X,A):f(S_i)\subseteq N(I_i),\forall i\in \Delta \}.$$

$$N(J)$$

$$\overline{[J,C_0(X,A)]}$$

$$fg-gf$$

$$L=\overline{J(S_1)\otimes K(H)+J(S_2)\otimes B(H)+K}.$$

Full text

The first named author was supported by a Junior Research Fellowship of CSIR with file number 09/045(1442)/2016-EMR-I

1. Introduction

Let $A\otimes^{\alpha}B$ denote the completion of the algebraic tensor product of two C∗-algebras $A$ and $B$ under an algebra cross norm $\|\cdot\|_{\alpha}$. A natural question arises whether the closed ideals of $A\otimes^{\alpha}B$ can be identified in terms of the closed ideals of $A$ and $B$ or not. It is known that the closed ideals of the Banach algebras $A\otimes^{h}B$, $A\otimes^{\gamma}B$ and $A\widehat{\otimes}B$ are directly related to the closed ideals of $A$ and $B$, where $\otimes^{h},\otimes^{\gamma}$ and $\widehat{\otimes}$ are the Haagerup tensor product, Banach space projective tensor product and operator space projective tensor product, respectively. Interestingly, if either $A$ or $B$ possesses finitely many closed ideals then every closed ideal of these spaces is a finite sum of product ideals (see, for instance, [1, 5, 7]). However, in 1978, Wassermann [16] established an astonishing result that not every closed ideal of $B(H)\otimes^{\min}B(H)$ is a finite sum of product ideals, $\otimes^{\min}$ being the minimal $C^{*}$-tensor norm. Recall that for every $x\in A\otimes B$, $\|x\|_{\text{min}}=\text{sup}\{\|(\pi_{1}\otimes\pi_{2})(x)\|\}$, where $\pi_{1}$ and $\pi_{2}$ run over all representations of $A$ and $B$ respectively (see [14] for details). In the present article, we prove that this anamoly can be removed by assuming one of the $C^{*}$-algebras to be commutative with the help of a totally different technique than those used for $\otimes^{h}$, $\otimes^{\gamma}$ and $\widehat{\otimes}$.

A Banach algebra $B$ naturally imbibes a Lie algebra structure with the Lie bracket given by $[a,b]=ab-ba$ for every $a,b\in B$. A closed subspace $L$ of $B$ is said to be a Lie ideal if $[B,L]\subseteq L$ where $[B,L]=\text{span}\{[b,l]:b\in B,l\in L\}$. The closed Lie ideals for $C^{*}$-algebras are extensively studied, one may refer to the expository article [11] for details. Recently some research has been done to identify the closed Lie ideals for the various tensor products of $C^{*}$-algebras. In [4, Section 5], [6, Section 4]) the closed Lie ideals of $C_{0}(X)\otimes^{\min}A$ have been characterized in terms of closed subspaces of $X$, $X$ being a locally compact Hausdorff space and $A$ being a simple $C^{*}$-algebra with at most one tracial state. However, if $A$ is not simple nothing is known about the closed Lie ideals of such spaces. In this article, we discuss a characterization of closed Lie ideals of $C_{0}(X,A)$, for any $C^{\ast}$-algebra $A$.

We present a brief summary of the main results of the article. In Section 2, we first establish an appropriate (surjective) correspondence between a class of closed subspaces of $X$ and the closed ideals of $C_{0}(X,A)$, for any $C^{\ast}$-algebra $A$. Interestingly, this correspondence turns out to be bijective if $A$ has finitely many closed ideals. We use this correspondence to identify the image of a closed ideal of $C_{0}(X,A)$ in $C_{0}(X)\otimes^{\min}A$ under the canonical isomorphism. This will pave our way to establish that every closed ideal of $C_{0}(X)\otimes^{\min}A$ is a finite sum of product ideals, where $X$ is a locally compact Hausdorff space and $A$ is a $C^{\ast}$-algebra with finitely many closed ideals. In Section 3, we characterize the closed Lie ideals of $C_{0}(X,A)$ in terms of some closed subspaces of $X$. To obtain a better picture of closed Lie ideals of $C_{0}(X)\otimes^{\min}A$, we prove that if $A$ is unital then the centre-quotient property of $A$ passes to $C_{0}(X,A)$ and vice-versa. As an application we establish an interesting result that a closed subspace $L$ of $C_{0}(X)\otimes^{\min}B(H)$ is a Lie ideal if and only if there exist closed subspaces $S_{1},S_{2}$ of $X$ with $S_{1}\subseteq S_{2}$ and a closed subspace $K$ of $C_{0}(X)\otimes\mathbb{C}1$ such that $L=\overline{J(S_{1})\otimes K(H)+J(S_{2})\otimes B(H)+K}$, where $H$ is a separable Hilbert space and for $F\subseteq X$, $J(F):=\{f\in C_{0}(X):f(F)\subseteq\{0\}\}$.

2. Closed ideals of $C_{0}(X)\otimes^{\min}A$

Let $X$ be a locally compact Hausdorff space and $A$ be any $C^{*}$-algebra. It is a well known fact that there is a bijective correspondence between the closed subspaces of $X$ and the closed ideals of $C_{0}(X)$ given by $F\leftrightarrow J(F)$. However, if we move from complex valued functions to the vector valued functions, such a correspondence is not known. Although, in the literature, it is established that every closed ideal of $C_{0}(X,A)$ is of the form $\{f\in C_{0}(X,A):f(x)\in I_{x},\,\forall x\in X\}$ where for every $x\in X$, $I_{x}$ is a closed ideal of $A$ [12, V.26.2.1], but this description fails to be fruitful while moving from $C_{0}(X,A)$ to $C_{0}(X)\otimes^{\min}A$ in order to determine the closed ideals.

We first generalize the former notion to the continuous vector valued functions by establishing a correspondence between the closed subspaces of X and closed ideals of $C_{0}(X,A)$. This correspondence will further enable us to characterize closed ideals of $C_{0}(X)\otimes^{\min}A$ in terms of closed ideals of $A$ and subspaces of $X$, when $A$ has finitely many closed ideals. This is due to the fact that there exists an isometric $\ast$-isomorphism $\tilde{\varphi}:C_{0}(X)\otimes^{\min}A\rightarrow C_{0}(X,A)$, which takes $f\otimes a$ to $af$ for every $f\in C_{0}(X)$ and $a\in A$, where $(af)(x)=f(x)a$ (see, [14, Theorem 4.14 (iii)], [9, Proposition 1.5.6]).

Let us first fix some notations for further use. For any $t\in\mathbb{N}$, the set $\{1,2,3,\dots,t\}$ is be denoted by $\mathbb{N}_{t}$. The spaces $C_{b}(X,A)$ and $C_{c}(X,A)$, as usual, denote the css of all bounded continuous functions and compactly supported continuous functions, respectively, from $X$ to $A$ endowed with sup norm. For a non-unital $C^{\ast}$-algebra $A$, $\tilde{A}$ will denote its unitization. For a locally compact Hausdorff space $X$ and any function $g\in C_{0}(X)$, we define $\hat{g}\in C_{0}(X,\tilde{A})$ (resp., $\hat{g}\in C_{0}(X,A)$) by $\hat{g}(x)=g(x)1$, where $1$ is the unit of $\tilde{A}$ (resp., of $A$) if $A$ is non unital (resp., if $A$ is unital). For any $a\in A$, we denote by $a^{\prime}$ the constant function in $C_{b}(X,A)$ such that $a^{\prime}(x)=a$ for every $x\in X$.

For an indexing set $\Delta$, let $S=\{S_{i}\}_{i\in\Delta}$ and $T=\{T_{i}\}_{i\in\Delta}$ be collections of subspaces of some sets $Y$ and $Z$. We define $S$* to be compatible with $T$* if whenever for some subset $\gamma$ of $\Delta$, $\cap_{j\in\gamma}T_{j}=T_{i}$ for some $i\in\Delta$, then $\cap_{j\in\gamma}S_{j}=S_{i}$. For a locally compact Hausdorff space $X$, $\alpha\in\Delta$ and $T$ as above, define a collection $\mathcal{T}_{T}^{\alpha}:=\{S=\{S_{i}\}_{i\in\Delta}:S_{i}$ is a closed subspace of $X$ for every $i\in\Delta,\ S\,\text{is compatible with}\ T\ \text{and}\ S_{\alpha}=X\}$.

Theorem 2.1**.**

Let $X$ be a locally compact Hausdorff space, $A$ be a $C^{\ast}$-algebra  and $\mathcal{I}=\{I_{i}\}_{i\in\Delta}$ be the collection of all closed ideals of $A$ with $I_{\beta}=A$. Then there exists a surjection $\theta$ from $\mathcal{T}_{\mathcal{I}}^{\beta}$ into $\mathcal{K}$, the set of all closed ideals of $C_{0}(X,A)$.

Proof.

Define $\theta:\mathcal{T}_{\mathcal{I}}^{\beta}\rightarrow\mathcal{K}$ by $\theta(S)=J(S)$ where $J(S)=\{f\in C_{0}(X,A):f(S_{i})\subseteq I_{i},\ \forall i\in\Delta\}$. Clearly $\theta$ is well defined. To see that $\theta$ is onto, consider $J\in\mathcal{K}$. For each $i\in\Delta$, set $S_{i}=\cap_{f\in J}{f^{-1}(I_{i})}$. If $I_{i}=\cap_{j\in\gamma}I_{j}$ for some subset $\gamma$ of $\Delta$, then

[TABLE]

so that $S=\{S_{i}\}_{i\in\Delta}$ is compatible with $\mathcal{I}$. Since, $S_{\beta}=\cap_{f\in J}f^{-1}(I_{\beta})=\cap_{f\in J}f^{-1}(A)=X$, we obtain $S\in\mathcal{T}_{\mathcal{I}}^{\beta}$. Clearly $J\subseteq J(S)$, so it is sufficient to prove that $J$ is dense in $J(S)$. For this, let $f\in J(S)$ be non-zero and $\epsilon>0$ be arbitrary.

We first claim that for any $x\in X$, there exists an element $h_{x}\in J$ such that $\|h_{x}(x)-f(x)\|\leq\frac{\epsilon}{3}$. Indeed, if $\gamma^{\prime}=\{i\in\Delta:x\in S_{i}\}$, then $x\in\cap_{i\in\gamma^{\prime}}S_{i}$ and $f(x)\in\cap_{i\in\gamma^{\prime}}I_{i}=I_{j}$ for some $j\in\Delta$. Let $I_{r}$ denote the closed ideal of $A$ generated by the set $\{g(x):g\in J\}$. Then $x\in S_{r}$ which implies that $f(x)\in I_{r}$. Hence $\|\sum_{i=1}^{k}a_{i}g_{i}(x)b_{i}-f(x)\|\leq\frac{1}{3}\epsilon$ for some $a_{1},a_{2},\dots a_{k},b_{1},b_{2},\dots b_{k}$ in $A$ (resp., in $\tilde{A}$) if $A$ is unital (resp., if $A$ is non unital). From [8, Corollary 4.2.10], we know that $J$ is a closed ideal of $C_{0}(X,\tilde{A})$, thus we have a function $h_{x}=\sum_{i=1}^{n}a_{i}^{\prime}g_{i}b_{i}^{\prime}$ in $J$ which satisfies the required condition.

Denote by $\tilde{X}=X\cup\{\infty\}$, the one point compactification of $X$. For each $x\in X$, let $\tilde{h}_{x}$ and $\tilde{f}$ be the continuous extensions of $h_{x}$ and $f$ to $\tilde{X}$ which take $\infty$ to [math], and let $\tilde{h}_{x_{0}}$ be the continuous extension of $h_{x_{0}}$, the zero function, to $\tilde{X}$ such that $\tilde{h}_{x_{0}}(\infty)=0$. As $\tilde{h}_{x_{0}}(\infty)=\tilde{f}(\infty)=0$, there exists a neighbourhood $V_{x_{0}}$ of $\infty$ in $\tilde{X}$ such that $\|\tilde{h}_{x_{0}}(y)-f(y)\|\leq\frac{1}{3}\epsilon$ for every $y\in V_{x_{0}}$. Further $\tilde{f}$ and $\tilde{h}_{x}$ are continuous at $x\in X$, so there exists a neighbourhood $V_{x}$ of $x$ such that $\|h_{x}(y)-f(y)\|\leq\epsilon$ for every $y\in V_{x}$.

The collection $\{V_{x_{0}},V_{x}\}_{x\in\tilde{X}}$ is an open cover for $\tilde{X}$. As $\tilde{X}$ is compact, there is a finite subcover $\{V_{x_{0}},V_{x_{1}},V_{x_{2}},\dots,V_{x_{k}}\}$ of $\tilde{X}$. Then there exists a partition of unity $\{g_{0},g_{1},g_{2},\dots g_{k}\}$ subordinate to this subcover such that for every $i\in\{0,1,2,\dots,k\}$, $g_{i}\in C(\tilde{X})$, $0\leq g_{i}\leq 1$, $\text{supp}(g_{i})\subseteq V_{x_{i}}$ and for each $x\in\tilde{X}$, $\sum_{i=0}^{k}g_{i}(x)=1$. Now for $\tilde{h}=\sum_{i=0}^{k}\hat{g_{i}}\tilde{h}_{x_{i}}\in C(\tilde{X},A)$, its restriction to $X$, given by $h=\sum_{i=0}^{k}\hat{g_{i}}_{|_{X}}h_{x_{i}}$, is an element in $J$. Hence for any $y\in X$, we have

[TABLE]

thus $\|h-f\|\leq\epsilon$. ∎

Remark 2.2**.**

Note that the notation $J(S)$ given in the above theorem coincides with the previous notation of $J(F)$ by taking $S=\{F,X\}$ and $\mathcal{I}=\{\{0\},A=\mathbb{C}\}$.

It is interesting to note that if $A$ possesses only finitely many closed ideals, then the above defined mapping $\theta$ turns out to be injective, hence providing a nice characterization of the closed ideals of $C_{0}(X,A)$. We would like to mention that the case when $A$ is a simple $C^{\ast}$-algebra is discussed in [6, Proposition 4.1], wherein we proved that every closed ideal of $C_{0}(X,A)$ is of the form ̃$\{f\in C_{0}(X,A):f(x)=0,\,\forall x\in F\}$, for some closed subspace F of X.

Theorem 2.3**.**

Let $X$ be a locally compact Hausdorff space, $A$ be a $C^{\ast}$-algebra and $\mathcal{I}=\{I_{1},I_{2},\dots,I_{n}=A\}$, $n\in\mathbb{N}$, be the set of all closed ideals of $A$. Then the map $\theta$ from $\mathcal{T}_{\mathcal{I}}^{n}$ to the set of all closed ideals of $C_{0}(X,A)$ is a bijection. In particular, every closed ideal of $C_{0}(X,A)$ is precisely of the form $J(S)=\{f\in C_{0}(X,A):f(S_{i})\subseteq I_{i},\ \forall i\in\mathbb{N}_{n}\}$, for a unique $S=\{S_{i}\}$ compatible with $\mathcal{I}$.

Proof.

Let $S=\{S_{i}\}_{i\in\mathbb{N}_{n}}$ and $S^{\prime}=\{S_{i}^{\prime}\}_{i\in\mathbb{N}_{n}}$ be two distinct elements of $\mathcal{T}_{\mathcal{I}}^{n}$ so that $S_{i}\neq S_{i}^{\prime}$ for some $i\in\mathbb{N}_{n}\setminus\{n\}$. Without loss of generality, we may assume that there exists $x\in S_{i}\setminus S_{i}^{\prime}$. Define a non-empty subset $\gamma=\{j\in\mathbb{N}_{n}:x\notin S_{j}^{\prime}\}$ of $\mathbb{N}_{n}$. By Urysohn’s Lemma [13, Theorem 2.12] , as $V_{x}=(\cup_{i\in\gamma}S_{i}^{\prime})^{c}$ is an open set containing $x$, there exists $g\in C_{c}(X)$ such that $g(x)=1$, $g(X)\subseteq[0,1]$ and $supp(g)\subseteq V_{x}$. It is easy to see that $(\cap_{j\in\gamma^{c}}I_{j})\setminus I_{i}\neq\emptyset$, because if $\cap_{j\in\gamma^{c}}I_{j}\subseteq I_{i}$ then $\cap_{j\in\gamma^{c}}S_{j}^{\prime}\subseteq S_{i}^{\prime}$, which is not true as $x\in\cap_{j\in\gamma^{c}}S_{j}^{\prime}$ but $x\notin S_{i}^{\prime}$. Let $a\in(\cap_{j\in\gamma^{c}}I_{j})\setminus I_{i}$. Consider the function $h=a^{\prime}\hat{g}\in C_{0}(X,A)$. Observe that $h\notin J(S)$, since $x\in S_{i}$ but $h(x)=a^{\prime}(x)\hat{g}(x)=a\notin I_{i}$. However, we assert that $h\in J(S^{\prime})$ which proves $J(S)\neq J(S^{\prime})$. For $k\in\gamma$, $y\in S_{k}^{\prime}$ implies $y\in V_{x}^{c}$, so that $h(y)=a\hat{g}(y)=0\in I_{k}$. Also, if $k\in\gamma^{c}$, then $h(y)=a\hat{g}(y)\in(\cap_{j\in\gamma^{c}}I_{j})\subseteq I_{k}$ for every $y\in S_{k}^{\prime}$. ∎

In the quest of proving the main result regarding the characterization of closed ideals of $C_{0}(X)\otimes^{\min}A$, we require few more ingredients.

Lemma 2.4**.**

Let $X$ be a locally compact Hausdorff space and $A$ be a $C^{\ast}$-algebra. Then for a closed subspace $C$ of $A$ and a closed ideal $J(Y)$ of $C_{0}(X)$, $Y\subseteq X$ being closed, there is an isometric isomorphism of Banach spaces

[TABLE]

Proof.

Denote by $J$ the closed subspace $\{f\in C_{0}(X,A):f(Y)=\{0\},f(X)\subseteq C\}$ of $C_{0}(X,A)$, and $I=J(Y)$. Let $\varphi$ denote the restriction of $\tilde{\varphi}$ to $\overline{I\otimes C}^{\text{min}}$, where $\tilde{\phi}:C_{0}(X)\otimes^{\min}A\to C_{0}(X,A)$ is the isometric $*$-isomorphism as discussed earlier. Then for $\sum_{j=1}^{n}f_{j}\otimes c_{j}\in I\otimes C$, we have $\varphi(\sum_{j\in\mathbb{N}_{n}}f_{j}\otimes c_{j})(Y)=\{0\}$, so that $\varphi(I\otimes C)\subseteq J$. Since $\varphi$ is an isometry, it is sufficient to prove that $\varphi(I\otimes C)$ is dense in $J$.

Let $g\in J$ and $\epsilon>0$ be arbitrary. Since $J$ is also a closed subspace of $C_{0}(X,C)$ and $C_{c}(X,C)$ is dense in $C_{0}(X,C)$, there exists a function $h\in C_{c}(X,C)$ such that $\|{g-h}\|<\epsilon/2$. Let $K:=\text{supp}(h)$, $B_{r}(b):=\{c\in C:\|c-b\|<r\}$ and $B_{r}^{\times}(b):=B_{r}(b)\setminus\{0\}$, where $b\in C$ and $r>0$. Since $\|h(y)\|=\|g(y)-h(y)\|<\epsilon/2$ for every $y\in Y$, the collection $\{h^{-1}(B^{\times}_{\frac{\epsilon}{2}}(h(x))\setminus\overline{h(Y)}):x\in K\setminus Y\}$ $\cup h^{-1}({B_{\epsilon}(0))}$ forms an open cover of the compact set $K$. Fix a finite subcover, say, $h^{-1}({B_{\epsilon}(0))}$ $\cup\{h^{-1}(B^{\times}_{\frac{\epsilon}{2}}(h(x_{i}))\setminus\overline{h(Y)})$ $:1\leq i\leq n\}$. Since $K$ is a compact subspace of a locally compact Hausdorff space $X$, there exists a partition of unity subordinate to this finite subcover, i.e. there exist functions $f_{0},f_{1},\dots,,f_{n}$ in $C_{c}(X)$ such that $0\leq f_{i}\leq 1$ for all $0\leq i\leq n$, $\text{supp}(f_{0})\subseteq U_{0}:=h^{-1}(B_{\epsilon}(0))$, $\text{supp}(f_{i})\subseteq U_{i}:=h^{-1}(B^{\times}_{\frac{\epsilon}{2}}(h(x_{i}))\setminus\overline{h(Y)})$ for all $1\leq i\leq n$ and $\sum_{i=0}^{n}f_{i}(x)=1$ for $x\in K$(see [13, Theorem 2.13]).

Let $V=(\sum_{i=0}^{n}f_{i})^{-1}(0,3/2)$. Then $V\cap(\cup_{i=0}^{n}U_{i})$ is an open set containing $K$. Pick $\tilde{f}\in C_{c}(X)$ such that $\tilde{f}$ is 1 on $K$, $\text{supp}(\tilde{f})\subseteq V\cap(\cup_{i=0}^{n}U_{i})$ and $0\leq\tilde{f}\leq 1$. Then for $\tilde{f_{i}}=\tilde{f}f_{i}$, $\text{supp}(\tilde{f_{i}})\subseteq V\cap U_{i}$ because $\text{supp}(f_{i})\subseteq U_{i}$ and $\text{supp}(\tilde{f})\subseteq V$. Now for $x\in K$, we have

[TABLE]

Also notice that $0\leq\sum_{i=0}^{n}\tilde{f_{i}}\leq 3/2$ because for $x\in V\cap(\cup_{i=0}^{n}U_{i})$, $\sum_{i=0}^{n}\tilde{f_{i}}(x)=\sum_{i=0}^{n}\tilde{f}(x)f_{i}(x)\leq\sum_{i=0}^{n}f_{i}(x)=3/2$ and for $x\in(V\cap(\cup_{i=0}^{n}U_{i}))^{c}$, we have $\sum_{i=0}^{n}\tilde{f_{i}}(x)=0$ .

Now for $1\leq i\leq n$, the open set $U_{i}$, and thus $V\cap U_{i}$ is disjoint from $Y$ so that $\sum_{i=1}^{n}\tilde{f_{i}}\otimes h(x_{i})\in I\otimes C$. Fix $x_{0}\in K^{c}$, then for each $x\in X$

[TABLE]

Hence we obtain $\|g-\varphi(\sum_{i=1}^{n}\tilde{f_{i}}\otimes h(x_{i}))\|<\frac{3\epsilon}{2}$, proving that $\varphi(I\otimes C)$ is dense in $J$. ∎

As a consequence of the above result, we have an interesting observation which identifies certain closed ideals of $C_{0}(X,A)$ with some closed ideals of $C_{0}(X)\otimes^{\min}A$.

Corollary 2.5**.**

Let $Y$ be a closed subspace of a locally compact Hausdorff space $X$. For any closed ideal $I$ of a $C^{\ast}$-algebra $A$, we have

[TABLE]

Proof.

Let $\mathcal{I}=\{I_{i}\}_{i\in\Delta}$ be the set of all closed ideals of $A$ with $I_{\beta}=A$, for some $\beta\in\Delta$, and set $I=I_{t}$. If $t=\beta$ then the result is trivial. Otherwise, let $J_{1}=C_{0}(X)\otimes^{\min}I_{t}+J(Y)\otimes^{\min}A$ and $J_{2}=\{f\in C_{0}(X,A):f(Y)\subseteq I_{t}\}$. Then, by Theorem 2.1, there exist elements $S=\{S_{i}\}_{i\in\Delta}$ and $S^{\prime}=\{S_{i}^{\prime}\}_{i\in\Delta}$ in $\mathcal{T}_{\mathcal{I}}^{\beta}$ such that $J_{1}=J(S)$ and $J_{2}=J(S^{\prime})$. It is sufficient to prove that $S=S^{\prime}$.

We first mention a common trick used in the proof. For any $x\in X$ and a closed subspace $F$ of $X$ with $x\notin F$, Urysohn’s Lemma implies that there exists $f\in C_{c}(X)$ such that $f(x)=1$ and $f(F)=0$. Then for any fixed $a\in A$ and any $y\in X$, there exists a function $g(y):=f(y)a$ in $C_{0}(X,A)$ such that $g(x)=a$ and $g$ vanishes on $F$.

We now claim that $S_{t}=\cap_{f\in J_{1}}f^{-1}(I_{t})=Y$. For $f\in J_{1}$, $f=f_{1}+f_{2}$ for some $f_{1}\in C_{0}(X)\otimes^{\min}I_{t}$ and $f_{2}\in J(Y)\otimes^{\min}A$. Thus, for any $y\in Y$, $f(y)=f_{1}(y)+f_{2}(y)\in I_{t}$ as $f_{2}(y)=0$ by 2.4, so that $Y\subseteq S_{t}$. For the reverse containment assume that $Y\subsetneq S_{t}$. Pick $a\in A\setminus I_{t}$ and $x\in S_{t}\setminus Y$, then there exists a function in $C_{0}(X,A)$ which vanishes on $Y$ and maps $x$ to $a$ which is a contradiction to the definition of $S_{t}$. On the similar lines, using the fact that $S_{t}^{\prime}=\cap_{g\in J_{2}}g^{-1}(I_{t})$, one can easily deduce that $S_{t}=Y=S_{t}^{\prime}$.

Now fix $i\in\Delta$ with $i\neq\beta,i\neq t$. Note that $J_{2}=\cap_{i\in\Delta}\{f\in C_{0}(X,A):f(S_{i}^{\prime})\subseteq I_{i}\}$, so that $G_{t}:=\{f\in C_{0}(X,A):f(S_{t}^{\prime})\subseteq I_{t}\}\subseteq\{f\in C_{0}(X,A):f(S_{i}^{\prime})\subseteq I_{i}\}=G_{i}$ (say), for every $i\in\Delta$.

Case(i): $I_{i}\subsetneq I_{t}$, then $S_{i}^{\prime}=\emptyset=S_{i}$. Because for $y\in S_{i}^{\prime}\subseteq S_{t}^{\prime}$ and $a\in I_{t}\setminus I_{i}$, there exists a function in $C_{0}(X,A)$ which takes $y$ to $a$. Then such a function is in $G_{t}$ but not in $G_{i}$. Also, if there exists an $x\in S_{i}\subseteq S_{t}=Y$, then $a^{\prime}g$ as defined above will be a function in $C_{0}(X)\otimes^{\min}I_{t}\subset J_{1}$ which takes an element $x$ of $S_{i}$ to $a$ which does not belong to $I_{i}$, which is a contradiction to the definition of $S_{i}$.

Case(ii): $I_{t}\subsetneq I_{i}$, then $S_{i}=S_{t}=S_{t}^{\prime}=S_{i}^{\prime}$. To see this, if $S_{t}^{\prime}$ is properly contained in $S_{i}^{\prime}$, then for $x\in S_{i}^{\prime}\setminus S_{t}^{\prime}$ and $a\notin I_{i}$, there exists a function in $C_{0}(X,A)$ which takes $x$ to $a$ and $S_{t}^{\prime}$ to 0. This function belongs to $G_{t}$ but does not belong to $G_{i}$, which is a contradiction. Similarly, if $S_{t}$ is properly contained in $S_{i}$, then for $x\in S_{i}\setminus S_{t}$ and $a\notin I_{i}$, there is a function in $J(Y)\otimes^{\min}A\subset J_{1}$ which takes $x$ outside $I_{i}$ which contradicts the definition of $S_{i}$.

Case(iii): $I_{i}$ is neither a subset nor a superset of $I_{t}$, then we claim that $S_{i}=S_{i}^{\prime}=\emptyset$. If $I_{j}=I_{t}\cap I_{i}$, then $I_{j}\subsetneq I_{t}$ so that by Case(i), $S_{t}\cap S_{i}=S_{j}=\emptyset=S_{j}^{\prime}=S_{t}^{\prime}\cap S_{i}^{\prime}$. Now, for $x\in S_{i}^{\prime}$, $x\notin S_{t}^{\prime}$, as argued in Case (ii), we obtain that $G_{t}$ is not contained in $G_{i}$, which is a contradiction, thus $S_{i}^{\prime}=\emptyset$. Similarly, for $x\in S_{i}$, $x$ is not a member of $Y=S_{t}$. So for any $a\in I_{t}\setminus I_{i}$, applying the technique mentioned in the beginning, we get a function $g$ in $J(Y)\otimes^{\min}A\subseteq J_{1}$ such that $g(x)\notin I_{i}$, a contradiction.

This proves that $S=S^{\prime}$ and hence $J_{1}=J_{2}$. ∎

We are now ready to prove the main result of this section. Note that a product ideal is a closed ideal of the form $I\otimes^{\min}J$, where $I$ and $J$ are closed ideals of $A$ and $B$, respectively.

Theorem 2.6**.**

Let $X$ be a locally compact Hausdorff space and $A$ be a $C^{\ast}$-algebra  with finitely many closed ideals, say, $I_{1},I_{2},\dots,I_{n}$ with $I_{1}=\{0\}$ and $I_{n}=A$. Then for any closed ideal $K$ of $C_{0}(X)\otimes^{\min}A$, there exists $S=\{S_{i}\}_{i\in\mathbb{N}_{n}}\in\mathcal{T}_{\mathcal{I}}^{n}$, where $\mathcal{I}=\{I_{i}\}_{i\in\mathbb{N}_{n}}$, such that

[TABLE]

where $\gamma_{j}=\{i\in\mathbb{N}_{n}:I_{j}\nsubseteq\ I_{i}\}$, for every $j\in\{2,3,\dots,n\}$.

In particular, every closed ideal of $C_{0}(X)\otimes^{\min}A$ is a finite sum of product ideals.

Proof.

By Theorem 2.3, there exists $S=\{S_{i}\}_{i\in\mathbb{N}_{n}}\in\mathcal{T}_{\mathcal{I}}^{n}$ such that $K=J(S)=\{f\in C_{0}(X,A):f(S_{i})\subseteq I_{i},i\in\mathbb{N}_{n}\}$. Set $K^{\prime}=\sum_{j=2}^{n}J(\cup_{k\in\gamma_{j}}S_{k})\otimes^{\min}I_{j}$, then by 2.4, $K^{\prime}$ can be considered as a closed ideal of $C_{0}(X,A)$. By virtue of Theorem 2.1, it is sufficient to prove that $S_{i}=\cap_{f\in K^{\prime}}f^{-1}(I_{i})$ for every $i\in\mathbb{N}_{n}$.

It is clear that $S_{n}=X=\cap_{f\in K^{\prime}}f^{-1}(A)$. Fix $i\in\mathbb{N}_{n-1}$ and consider any $x\in S_{i}$. For $f\in K^{\prime}$, $f=f_{2}+f_{3}+\dots+f_{n}$, where $f_{r}\in J(\cup_{k\in\gamma_{r}}S_{k})\otimes^{\min}I_{r}$ for every $r\in\{2,3,\dots,n\}$. Then for any such $r$, either $i\in\gamma_{r}$ or $i\in\gamma_{r}^{c}$. If $i\in\gamma_{r}$ then $f_{r}(x)=0\in I_{i}$. If $i\in\gamma_{r}^{c}$ then $f_{r}(x)\in I_{r}\subseteq I_{i}$. These two conclusions together imply that $S_{i}\subseteq\cap_{f\in K^{\prime}}f^{-1}(I_{i})$.

Next, pick $x\notin S_{i}$ and define $\alpha_{i}=\{j\in\mathbb{N}_{n}:I_{j}\nsubseteq\ I_{i}\}$. Note that $\alpha_{i}$ is non empty as $n\in\alpha_{i}$. It is sufficient to prove the existence of a function $f\in K^{\prime}$ such that $f(x)\notin I_{i}$. We shall actually prove that such a function exists in the subset $\sum_{r\in\alpha_{i}}J(\cup_{k\in\gamma_{r}}S_{k})\otimes^{\min}I_{r}$ of $K^{\prime}$. It is further enough to prove that there exists an $r\in\alpha_{i}$ such that $x\notin\cup_{k\in\gamma_{r}}S_{k}$, so that the required function $f$ exists in $J(\cup_{k\in\gamma_{r}}S_{k})\otimes^{\min}I_{r}$. In fact, by Urysohn’s Lemma there exists a function $g\in C_{c}(X)$ such that $0\leq g\leq 1$, $g(x)=1$ and $g(\cup_{k\in\gamma_{r}}S_{k})=\{0\}$. Then by 2.4, for $a\in I_{r}\setminus I_{i}$ (since $I_{r}\nsubseteq I_{i})$, the function $a^{\prime}\hat{g}$ serves the purpose. We claim that $\cap_{r\in\alpha_{i}}(\cup_{k\in\gamma_{r}}S_{k})=S_{i}$, which will ensure the existence of such an $r$.

When $r\in\alpha_{i}$, we have $i\in\gamma_{r}$ and hence $S_{i}\subseteq\cap_{r\in\alpha_{i}}(\cup_{k\in\gamma_{r}}S_{k})$. We now prove the reverse inclusion. Set $\alpha_{i}=\{r_{1},r_{2},\dots,r_{q}\}$ and for each $r_{j}\in\alpha_{i}$, let there be $p_{r_{j}}$ number of elements in $\gamma_{r_{j}}$, say $\gamma_{r_{j}}=\{r_{j,t_{r_{j}}}:1\leq t_{r_{j}}\leq p_{r_{j}}\}$. So

[TABLE]

We have obtained that $\cap_{r\in\alpha_{j}}(\cup_{k\in\gamma_{r}}S_{k})$ is a union of $\Pi_{j=1}^{q}p_{r_{j}}$ objects, each of which is an intersection of $q$ objects which looks like $S_{r_{1,t_{r_{1}}}}\cap S_{r_{2,t_{r_{2}}}}\cap\dots\cap S_{r_{q,t_{r_{q}}}}$. Pick an ideal $I_{j_{r_{1},t_{r_{1}}}}\cap I_{j_{r_{2},t_{r_{2}}}}\cap\dots\cap I_{j_{r_{q},t_{r_{q}}}}$. Then there exists an $m\in\mathbb{N}_{n}$ such that $I_{m}=I_{r_{1,t_{r_{1}}}}\cap I_{r_{2,t_{r_{2}}}}\cap\dots\cap I_{r_{q,t_{r_{q}}}}$, and hence $S_{m}=S_{r_{1,t_{r_{1}}}}\cap S_{r_{2,t_{r_{2}}}}\cap\dots\cap S_{r_{q,t_{r_{q}}}}$. If $S_{m}\subseteq S_{i}$, we are done. Otherwise, $S_{m}\nsubseteq S_{i}$ will imply $I_{m}\nsubseteq I_{i}$ and hence $m\in\alpha_{i}$. Thus $m=r_{l}$ for some $l\in\{1,2,\dots k\}$. Then $r_{l,t_{r_{l}}}\in\gamma_{r_{l}}$ which implies $(I_{m}=)I_{r_{l}}\nsubseteq I_{r_{l,t_{r_{l}}}}$, which is a contradiction to the fact that $I_{m}=I_{r_{1,t_{r_{1}}}}\cap I_{r_{2,t_{r_{2}}}}\cap\dots\cap I_{r_{q,t_{r_{q}}}}$. ∎

Let us demonstrate the above theorem with the help of few examples.

Example 2.7**.**

Let $H$ be a separable Hilbert space and $A=B(H)\oplus B(H)$. For $\mathcal{I}=\{I_{1},\dots,I_{9}=A\}$, the lattice of closed ideals of $A$ is given in the diagram below.

[TABLE]

Looking at the diagram, we obtain that $\gamma_{2}=\{1,3,6\}$, $\gamma_{3}=\{1,2,4\}$, $\gamma_{4}=\{1,2,3,5,6,8\}$, $\gamma_{5}=\{1,2,3,4,6\}$, $\gamma_{6}=\{1,2,3,4,5,7\}$, $\gamma_{7}=\{1,2,3,4,5,6,8\}$, $\gamma_{8}=\{1,2,3,4,5,6,7\}$ and $\gamma_{9}=\{1,2,3,4,5,6,7,8\}$. Hence, every closed ideal of $C_{0}(X)\otimes^{\min}A$ is of the form:

$J(S_{6})\otimes^{\min}I_{2}+J(S_{4})\otimes^{\min}I_{3}+J(S_{8})\otimes^{\min}I_{4}+J(S_{4}\cup S_{6})\otimes^{\min}I_{5}+J(S_{7})\otimes^{\min}I_{6}+J(S_{8})\otimes^{\min}I_{7}+\\ J(S_{7})\otimes^{\min}I_{8}+J(S_{7}\cup S_{8})\otimes^{\min}I_{9}$, for a unique $S=(S_{1},S_{2},\dots,S_{9})\in\mathcal{T}_{\mathcal{I}}^{9}$, where a pictorial representation of $S$ is the following (here an arrow from $S_{j}$ to $S_{k}$ means that $S_{j}\subseteq S_{k}$, for $j,k\in\mathbb{N}_{9}$):

[TABLE]

We next discuss the precise form of closed ideals of $C_{0}(X)\otimes^{\min}B(H)$, in terms of product ideals. Note that for a Hilbert space $H$, the set of all closed ideals forms a chain (see, [10, Corollary 6.2]). In the following, $w_{0}$ denotes the cardinality of the set of all natural numbers and for every $i\in\mathbb{N}$, let $w_{i}=2^{w_{i-1}}$ so that $w_{1}$ is continuum.

Example 2.8**.**

Let $X$ be a locally compact Hausdorff space and $H$ be a Hilbert space with $w_{n}$ ($n\in\mathbb{N}$) as its Hilbert dimension. Then the closed ideals of $C_{0}(X)\otimes^{\min}B(H)$ are of the form $\sum_{j=2}^{n+3}J(S_{j-1})\otimes^{\min}I_{j}$ , where $I_{j}$’s are closed ideals of $B(H)$ and $S_{j}$’s are some closed subspaces of $X$.

To see this, let $\{0\}=I_{1}\subsetneq I_{2}\subsetneq I_{3}\subsetneq\dots\subsetneq I_{n+3}=B(H)$ be the chain of closed ideals of $B(H)$ and $J$ be a closed ideal of $C_{0}(X)\otimes^{\min}B(H)$. By Theorem 2.1, there exists $n+3$ closed subspaces $S_{1}\subseteq S_{2}\subseteq S_{3}\subseteq\dots\subseteq S_{n+3}=X$ such that $J=J(S)$ where $S=\{S_{i}\}_{i=1}^{n+3}$. As in Theorem 2.6, for $j\in\mathbb{N}_{n+3}$, $\cup_{k\in\gamma_{j}}S_{k}=S_{j-1}$ and hence $J=\sum_{j=2}^{n+3}J(S_{j-1})\otimes^{\min}I_{j}$.

3. Closed Lie ideals of $C_{0}(X,A)$

The Lie normalizer of a subspace $S$ of a Lie algebra $A$ is defined by $N(S):=\{a\in A:[a,A]\subseteq S\}$. It can be easily verified that $N(I)$ is a closed subalgebra of $A$ for a closed ideal $I$ in $A$. The Lie normalizer plays an important role in determining the Lie ideals of $A$ (for instance, see [4],[6]). We identify the Lie normalizer of ideals of $C_{0}(X,A)$ and use this identification to characterize its closed Lie ideals.

Theorem 3.1**.**

Let $X$ be a locally compact Hausdorff space and $A$ be a $C^{\ast}$-algebra  with $\mathcal{I}=\{I_{i}\}_{i\in\Delta}$ as the collection of all closed ideals such that $I_{\beta}=A$. Then a closed subspace $L$ of $C_{0}(X,A)$ is a closed Lie ideal if and only if there is an element $S=\{S_{i}\}_{i\in\Delta}\in\mathcal{T}_{\mathcal{I}}^{\beta}$ such that

[TABLE]

Proof.

We know that a closed subspace $L$ of the $C^{\ast}$-algebra $C_{0}(X,A)$ is a Lie ideal if and only if there exists a closed ideal $J\subseteq C_{0}(X,A)$ such that $\overline{[J,C_{0}(X,A)]}\subseteq L\subseteq N(J)$ ([3, Proposition 5.25, Theorem 5.27]). By Theorem 2.1, $J=J(S)$ for some $S=\{S_{i}\}_{i\in\Delta}\in\mathcal{T}_{\mathcal{I}}^{\beta}$. Since for any fixed $a\in A$ and $x\in X$, there is an element in $C_{0}(X,A)$ which takes $x$ to $a$, we have

[TABLE]

We know from [3, Proposition 5.25] that $\overline{[I,B]}=I\cap\overline{[B,B]}$ for closed ideal $I$ of a $C^{\ast}$-algebra $B$. This fact, along with 2.4 gives

[TABLE]

Hence the result. ∎

From the last result one can observe that if $N(I)=I+\mathcal{Z}(A)$, $\mathcal{Z}(A)$ being the centre of $A$, for every closed ideal $I$ of $A$, then for a closed ideal $J$ of $C_{0}(X,A)$, $N(J)=\{f\in C_{0}(X,A):f(S_{i})\subseteq I_{i}+\mathcal{Z}(A),\ \forall i\in\Delta\}$ for some $\{S_{i}\}_{i\in\Delta}\in\mathcal{T}_{\mathcal{I}}^{\beta}$. The question that comes next is the following: Can we write every element of $N(J)$ as $g+h$ such that $g(S_{i})\subseteq I_{i}$ for every $i\in\Delta$ and $h$ is $\mathcal{Z}(A)$-valued? We shall observe in 3.6 that a positive answer of this question will help us to obtain a better representation of all closed Lie ideals of $C_{0}(X,A)$.

Recall that a $C^{*}$-algebra $A$ is said to have the centre-quotient property if $\mathcal{Z}(A/I)=(\mathcal{Z}(A)+I)/I$ for every closed ideal $I$ of $A$ (see [2] for details). Using the fact that for the natural quotient map $\pi:A\rightarrow A/I$, $N(I)=\pi^{-1}(\mathcal{Z}(A/I))$, we have a nice relation between Lie normalizer and the centre-quotient property.

Lemma 3.2**.**

A $C^{\ast}$-algebra $A$ has centre-quotient property if and only if $N(I)=I+\mathcal{Z}(A)$ for every closed ideal $I$ in $A$.

A unital $C^{*}$-algebra $A$ is called weakly central if the continuous surjection $\psi:\text{Max}(A)\rightarrow\text{Max}(\mathcal{Z}(A))$ given by $\psi(I)=I\cap\mathcal{Z}(A)$ is an injection, where $\text{Max}(B)$ denotes the space of all maximal ideals of a $C^{*}$-algebra $B$ endowed with the hull-kernel topology. It is well known that a unital $C^{\ast}$-algebra with unique maximal ideal must have one dimensional centre [2, Lemma 2.1]. Since weak centrality and centre-quotient property are equivalent in unital css (see, [15, Theorem 1 and 2]), presence of unique maximal ideal in a unital $C^{\ast}$-algebra $A$ implies that $A$ has centre-quotient property. In [6, Lemma 4.6] it was observed that for a simple unital $C^{\ast}$-algebra $A$ and a closed ideal $I\subseteq A$, $N(I)=I+C_{0}(X,\mathbb{C}1)$.

Theorem 3.3**.**

Let $X$ be a locally compact Hausdorff space and $A$ be a unital $C^{\ast}$-algebra with unique maximal ideal. Then $N(J)=J+C_{0}(X,\mathbb{C}1)$ for any closed ideal $J$ of $C_{0}(X,A)$.

Proof.

Note that $\mathcal{Z}(C_{0}(X,A))=C_{0}(X,\mathcal{Z}(A))=C_{0}(X,\mathbb{C}1)$ and hence $J+C_{0}(X,\mathbb{C}1)\subseteq N(J)$. Let $\mathcal{I}=\{I_{i}\}_{i\in\Delta}$ be the collection of all closed ideals of $A$ with $A=I_{\beta}$ and let $I_{\beta^{\prime}}$ be the unique maximal ideal of $A$, $\beta,\beta^{\prime}\in\Delta$. Then, Theorem 2.1, there exists an element $S=\{S_{i}\}_{i\in\Delta}\in\mathcal{T}_{\mathcal{I}}^{\beta}$ such that $J=J(S)$. Since $A$ has centre-quotient property, by 3.2, $N(I_{i})=I_{i}+\mathbb{C}1$ for every $i\in\Delta$. Let $f\in N(J)=\{g\in C_{0}(X,A):g(S_{i})\subseteq I_{i}+\mathbb{C}1,\ \forall i\in\Delta\}$ as noted in Theorem 3.1. Since $I_{\beta^{\prime}}$ is the unique maximal ideal of $A$ and $S\in\mathcal{T}_{\mathcal{I}}^{\beta}$, we have $S_{\alpha}\subseteq S_{\beta^{\prime}}$ for every $\alpha\in\Delta\setminus\{\beta\}$.

On $S_{\beta^{\prime}}$, write $f=g+h$ which satisfy $h(S_{\beta^{\prime}})\subseteq\mathbb{C}1$ and $g(S_{i})\subseteq I_{i}$ for every $i\in\Delta\setminus\{\beta\}$. This is possible as no proper ideal of $A$ can intersect $\mathbb{C}1$. Since $I_{\beta^{\prime}}\cap\ \mathbb{C}1=\{0\}$, by Hahn-Banach Theorem, there exists $T\in A^{\ast}$ such that $\|T\|=1$, $T(I_{\beta^{\prime}})=\{0\}$ and $T(\lambda 1)=\lambda$ for $\lambda\in\mathbb{C}$. Then $Tf=Th$ on $S_{\beta^{\prime}}$. Also $f$ vanishes at infinity and $\|T\|=1$, so we obtain that $Tf$ vanishes at infinity because $\|Tf\|\leq\|f\|$. Since $\|Th\|=\|h\|$, $Th$ and hence $h$ is a continuous function vanishing at infinity. So $g=f-h$ is continuous on $S_{\beta^{\prime}}$ and is vanishing at infinity. By [6, Theorem 4.5], there exists an $h^{\prime}\in C_{0}(X)$ such that $h^{\prime}_{|_{S_{\beta^{\prime}}}}=h$. For $x\in S_{\beta^{\prime}}^{c}$, define $g^{\prime}(x)=f(x)-h^{\prime}(x)$. Then $f=g^{\prime}+h^{\prime}$ with $g^{\prime}\in J(S)$ and $h^{\prime}\in C_{0}(X,\mathbb{C}1)$ and we are done. ∎

It is observed in the previous result that if $A$ is a unital $C^{\ast}$-algebra with a unique maximal ideal then $C_{0}(X,A)$ has the centre-quotient property. We now generalize this result and prove that the centre-quotient property of $A$ passes to $C_{0}(X,A)$. We would like to point out that the proof given above does not work when $A$ has more than one maximal ideal because in this case a closed ideal may intersect the centre non-trivially, that is, $I\cap\mathcal{Z}(A)\neq\{0\}$, and this is where the proof will fail. As an intermediate step towards this generalization, we provide the following result.

Proposition 3.4**.**

Let $X$ be a compact Hausdorff space and $A$ be a unital $C^{\ast}$-algebra having centre- quotient property. Then $C(X,A)$ has centre-quotient property.

Proof.

It is sufficient to prove that $C(X,A)$ is weakly central. Consider maximal ideals $J_{1}$ and $J_{2}$ of $C(X,A)$ such that $J_{1}\cap\mathcal{Z}(C(X,A))=J_{2}\cap\mathcal{Z}(C(X,A))$. For $i=1,2$, there exist maximal ideals $I_{i}$ of $A$ and $x_{i}\in X$ such that $J_{i}=\{f\in C(X,A):f(x_{i})\in I_{i}\}$ ([12, Corollary V.26.2.2]). With the help of Urysohn’s Lemma, one can easily verify that $I_{i}$ and $x_{i}$ are unique. Since $\mathcal{Z}(C(X,A))=C(X,\mathcal{Z}(A))$, we have $J_{i}\cap\mathcal{Z}(C(X,A))=\{f\in C(X,Z(A)):f(x_{i})\in I_{i}\cap\mathcal{Z}(A)\}$. By the uniqueness, $x_{1}=x_{2}$ and $I_{1}\cap\mathcal{Z}(A)=I_{2}\cap\mathcal{Z}(A)$. Since $A$ is weakly central, we obtain $I_{1}=I_{2}$ and hence $J_{1}=J_{2}$. ∎

Theorem 3.5**.**

If $X$ is a locally compact Hausdorff topological space and $A$ is a $C^{\ast}$-algebra. If $C_{0}(X,A)$ has centre-quotient property then $A$ has centre-quotient property. Converse is true if $A$ is unital.

Proof.

In order to prove that $A$ has centre-quotient property, it is sufficient to prove that for a closed ideal $I$ of $A$, $N(I)\subseteq I+\mathcal{Z}(A)$. For a fixed $x\in X$, consider a closed ideal $J$ of $C_{0}(X,A)$ given by $\{f\in C_{0}(X,A):f(x)\in I\}$. Now, for $a\in N(I)$, let $f\in C_{0}(X,A)$ such that $f(x)=a$. For any $g\in C_{0}(X,A)$, $(fg-gf)(x)=ag(x)-g(x)a\in I$, which implies that $f\in N(J)$. Since $C_{0}(X,A)$ has centre-quotient property, $f=g+h$, for some $g\in J$ and $h\in C_{0}(X,\mathcal{Z}(A))$. Thus $a=f(x)=g(x)+h(x)\in I+\mathcal{Z}(A)$.

Conversely, suppose that $A$ is unital and has centre-quotient property. If $\tilde{X}$ denotes the one point compactification of $X$, then we have a natural inclusion $C_{0}(X,A)\subseteq C(\tilde{X},A)$. For a closed ideal $J$ of $C_{0}(X,A)$, we claim that $N(J)_{C_{0}(X,A)}\subseteq N(J)_{C(\tilde{X},A)}$, where $N(J)_{B}$ represents the Lie normalizer in $B$. Let $f\in N(J)_{C_{0}(X,A)}$ and $\{f_{\mu}\}$ be a quasi central approximate identity of the closed ideal $C_{0}(X,A)$ of $C(\tilde{X},A)$. Then for any $g\in C(\tilde{X},A)$,

[TABLE]

since $\{f_{\mu}\}$ is quasi central approximate identity and $gf\in C_{0}(X,A)$. Note that $f_{\mu}g\in C_{0}(X,A)$ and $f\in N(J)_{C_{0}(X,A)}$ together imply that $ff_{\mu}g-f_{\mu}gf\in J$ for every $\mu$. Thus $fg-gf\in J$ which gives $f\in N(J)_{C(\tilde{X},A)}$.

Since $C(\tilde{X},A)$ has centre-quotient property (Theorem 3.5) and $J$ is also a closed ideal of $C(\tilde{X},A)$, we have $N(J)_{C(\tilde{X},A)}=J+C(\tilde{X},\mathcal{Z}(A))$. Now, for $f\in N(J)_{C_{0}(X,A)}$, there exist $g\in J$ and $h\in C(\tilde{X},\mathcal{Z}(A))$ such that $f=g+h$. If $\tilde{X}=X\cup\{\infty\}$, then $f(\infty)=0=g(\infty)$ which gives $h(\infty)=f(\infty)-g(\infty)=0$ so that $h\in C_{0}(X,\mathcal{Z}(A))$. Thus $N(J)_{C_{0}(X,A)}\subseteq J+C_{0}(X,\mathcal{Z}(A))$ and hence the result holds. ∎

We now characterize the Lie ideals of a class of $C^{*}$-algebras. Recall that a bounded linear functional $f$ on a $C^{\ast}$-algebra $A$ is said to be a tracial state if $f$ is positive of norm 1 and $f([a,b])=0$ for every $a,b\in A$.

Corollary 3.6**.**

Let $X$ be a locally compact Hausdorff space and $A$ be a unital $C^{\ast}$-algebra with centre-quotient property and no tracial states. Then a closed subspace $L$ of $C_{0}(X,A)$ is a Lie ideal if and only if it is of the form $\overline{J+K}$ for some closed ideal $J$ of $C_{0}(X,A)$ and a closed subspace $K$ of $C_{0}(X,\mathcal{Z}(A))$.

Proof.

Since $A$ has no tracial states, from [6, Lemma 2.4], $C_{0}(X,A)$ has no tracial states. Thus, by [3, Proposition 5.25], $\overline{[J,A]}=J$ for every closed ideal $J$ of $C_{0}(X,A)$. From [3, Theorem 5.27] and Theorem 3.5, a closed subspace $L$ of $C_{0}(X,A)$ is a Lie ideal if and only if there exists a closed ideal $J$ of $C_{0}(X,A)$ such that $J\subseteq L\subseteq J+C_{0}(X,\mathcal{Z}(A))$. Hence $L$ must be of the form $\overline{J+K}$ for some closed subspace $K$ of $C_{0}(X,\mathcal{Z}(A))$. ∎

As a consequence, we can now characterize all closed ideals of $C_{0}(X)\otimes^{\min}B(H)$.

Corollary 3.7**.**

For a separable Hilbert space $H$ and a locally compact Hausdorff space $X$, a closed subspace $L$ of $C_{0}(X)\otimes^{\min}B(H)$ is a Lie ideal if and only if there exist two closed subspaces $S_{1}\subseteq S_{2}$ of $X$ and a closed subspace $K$ of $C_{0}(X)\otimes\mathbb{C}1$ such that

[TABLE]

Proof.

Since $B(H)$ has no tracial states and has a unique maximal ideal, the result is an easy consequence of Example 2.8 and 3.6. ∎