Biflat-like Banach algebras

TL;DR

This paper introduces the concept of $\sigma$-biflatness in Banach algebras, explores its differences from biflatness, and examines its relationships with other properties like $\sigma$-biprojectivity and $\sigma$-amenability, including tensor products.

Contribution

It defines $\sigma$-biflatness for Banach algebras, distinguishes it from biflatness, and investigates its connections with related concepts and tensor product behavior.

Findings

$\sigma$-biflatness is a proper generalization of biflatness.

$\sigma$-biflatness and biflatness are shown to be distinct properties.

Relations between $\sigma$-biflatness, $\sigma$-biprojectivity, and $\sigma$-amenability are established.

Abstract

Given a Banach algebra $ \mathcal{A} $ and a continuous homomorphism $\sigma$ on it, the notion of $\sigma$-biflatness for $ \mathcal{A} $ is introduced. This is a generalization of biflatness and it is shown that they are distinct. The relations between $\sigma$-biflatness and some other close concepts such as $\sigma$-biprojectivity and $\sigma$-amenability are studied. The $\sigma$-biflatness of tensor product of Banach algebras are also discussed.

Equations

$$a\cdot (b\otimes c)=ab\otimes c\text{and}(b\otimes c)\cdot a=b\otimes ca(a,b,c\in \mathcal{A}).$$

$$\rho (\sigma (a)\cdot \lambda )=a\cdot \rho (\lambda )\text{and}\rho (\lambda \cdot \sigma (a))=\rho (\lambda )\cdot a(1)$$

$$⟨T^{\ast }(\sigma (a)\cdot \lambda ),x⟩$$

$$=⟨\lambda ,T(x)\cdot \sigma (a)⟩=⟨\lambda ,T(x\cdot a)⟩$$

$$=⟨T^{\ast }(\lambda ),x\cdot a⟩=⟨a\cdot T^{\ast }(\lambda ),x⟩,$$

$$⟨{\pi }^{\ast \ast }\circ \rho (a),\lambda ⟩$$

$$=⟨a,\theta \circ {\pi }^{\ast }(\lambda )⟩=⟨a,{\sigma }^{\ast }(\lambda )⟩$$

$$=⟨\sigma (a),\lambda ⟩=⟨{\kappa }_{\mathcal{A}}\circ \sigma (a),\lambda ⟩.$$

$$⟨\rho (ab),\xi ⟩$$

$$=⟨b,\theta (\xi \cdot \sigma (a))⟩=⟨\rho (b),\xi \cdot \sigma (a)⟩=⟨\sigma (a)\cdot \rho (b),\xi ⟩,$$

$$⟨\tilde{\rho }\circ {\pi }^{\ast }(\lambda ),a⟩$$

$$=⟨\lambda ,{\kappa }_{\mathcal{A}}\circ \sigma (a)⟩=⟨\lambda ,\sigma (a)⟩=⟨{\sigma }^{\ast }(\lambda ),a⟩.$$

$$⟨{\rho }^{\ast }\circ {\pi }^{\ast }(\lambda ),a⟩=⟨{\pi }^{\ast }(\lambda ),\rho (a)⟩=⟨\lambda ,\pi \circ \rho (a)⟩=⟨\lambda ,\sigma (a)⟩=⟨{\sigma }^{\ast }(\lambda ),a⟩.$$

$$⟨b,\acute{\rho }(\sigma (a)\cdot \lambda )⟩$$

$$=⟨\sigma (b),\sigma (a)\cdot \rho (\lambda )⟩=⟨\sigma (ba),\rho (\lambda )⟩$$

$$=⟨ba,{\sigma }^{\ast }\rho (\lambda )⟩=⟨b,a\cdot \acute{\rho }(\lambda )⟩$$

$$D(ab)=\sigma (a)\cdot D(b)+D(a)\cdot \sigma (b)(a,b\in \mathcal{A}),$$

$$⟨\sigma (a)\cdot M,\lambda ⟩$$

$$=\underset{\alpha }{\lim }⟨e_{\alpha },\rho (\lambda \cdot \sigma (a))⟩=\underset{\alpha }{\lim }⟨e_{\alpha },\rho (\lambda )\cdot a⟩$$

$$=\underset{\alpha }{\lim }⟨a{e}_{\alpha },\rho (\lambda )⟩=⟨a,\rho (\lambda )⟩$$

$${\pi }_{\mathcal{A}}^{\ast \ast }(M)\cdot \sigma (a)$$

$${\pi }^{\ast \ast }\circ \rho (a)={\pi }^{\ast \ast }(\sigma (a)\cdot M)={\pi }^{\ast \ast }(M)\cdot \sigma (a)=\sigma (a)={\kappa }_{\mathcal{A}}(\sigma (a)).$$

$$\Sigma :0⟶ker\pi \stackrel{ı}{⟶}\mathcal{A}\hat{\otimes }\mathcal{A}\stackrel{\pi }{⟶}\mathcal{A}⟶0,$$

$$\theta (\sigma (a)\cdot f)=a\cdot \theta (f)\text{and}\theta (f\cdot \sigma (a))=\theta (f)\cdot a(a\in \mathcal{A},f\in (\mathcal{A}\hat{\otimes }\mathcal{A}{)}^{\ast }).$$

$$⟨a,\theta (f)⟩:=⟨f\cdot \sigma (a),M⟩(a\in \mathcal{A},f\in (\mathcal{A}\hat{\otimes }\mathcal{A}{)}^{\ast }).$$

$$⟨a,\theta {\pi }^{\ast }\lambda ⟩$$

$$=⟨\lambda ,\sigma (a){\pi }^{\ast \ast }M⟩=⟨\lambda ,\sigma (a)⟩=⟨a,{\sigma }^{\ast }\lambda ⟩$$

$$⟨b,\theta (\sigma (a).f)⟩$$

$$=⟨f.\sigma (b),\sigma (a).M⟩=⟨f.\sigma (ba),M⟩$$

$$=⟨ba,\theta (f)⟩=⟨b,a.\theta (f)⟩$$

$$\bar{\rho }(axb)$$

$$=w^{\ast }-\underset{\alpha }{\lim }\sigma (axb)\cdot \tilde{\rho }(e_{\alpha })(\text{since}axb\in \mathcal{B})$$

$$=w^{\ast }-\underset{\alpha }{\lim }\sigma (a)\sigma (xb)\cdot \tilde{\rho }(e_{\alpha })$$

$$=w^{\ast }-\underset{\alpha }{\lim }\sigma (a)\cdot \tilde{\rho }(e_{\alpha })\cdot \sigma (xb)(\text{since}xb\in \mathcal{B})$$

$$=\sigma (a)\cdot \bar{\rho }(x)\cdot \sigma (b)(a,x\in \mathcal{A},b\in \mathcal{B}).$$

$$\bar{\rho }(axb)$$

$$=\sigma (a)\cdot \bar{\rho }(x)\cdot \sigma (cd)=\sigma (a)\cdot \bar{\rho }(x)\cdot \sigma (b).$$

$$\bar{\rho }(ax)=\bar{\rho }(ayz)=\sigma (a)\cdot \bar{\rho }(y)\cdot \sigma (z)=\sigma (a)\cdot \bar{\rho }(yz)=\sigma (a)\cdot \bar{\rho }(x).$$

$${\pi }_{\mathcal{A}}^{\ast \ast }(\bar{\rho }(a))$$

Full text

Keywords: $\sigma$-amenable, $\sigma$-biflat, $\sigma$-virtual diagonal, $\sigma$-biprojective, $\sigma$-derivation.

MSC 2010: Primary: 46H25; Secondary: 16E40, 43A20.

1. Introduction

Biprojectivity and biflatness for Banach algebras, as introduced and studied in the works of Helemskii (see for instance [5]), have proved to be important and fertile notions. There are close relationship between these notions and some other concepts of Banach algebras such as amenability. It is known that every biprojective Banach algebra with a bounded approximate identity is amenable and in the presence of a bounded approximate identity, biflatness and amenability are the same notions [5].

Before preceding further, we recall some preliminaries. Let $\mathcal{A}$ be a Banach algebra. Then its projective tensor product $\mathcal{A}\widehat{\otimes}{\mathcal{A}}$ is a Banach $\mathcal{A}$-bimodule through

[TABLE]

For a Banach $\mathcal{A}$-bimodule $X$, the dual space $X^{*}$ becomes a Banach $\mathcal{A}$-bimodule in a natural manner. Let $X$ and $Y$ be Banach $\mathcal{A}$-bimodules. A bounded linear map $T:X\longrightarrow Y$ is an $\mathcal{A}$-bimodule homomorphism if $T(a\cdot x)=a\cdot T(x)$ and $T(x\cdot a)=T(x)\cdot a$ , for $a\in\mathcal{A}$, $x\in X$. It is obvious that the diagonal operator $\pi_{\mathcal{A}}:\mathcal{A}\widehat{\otimes}\mathcal{A}\rightarrow\mathcal{A}$ given by $\pi(a\otimes b)=ab$, is an $\mathcal{A}$-bimodule homomorphism. If it is clear to which algebra $\mathcal{A}$ we refer, we simply write $\pi$.

A Banach algebra $\mathcal{A}$ is biprojective if $\pi$ has a right inverse which is an $\mathcal{A}$-bimodule homomorphism. If there is an $\mathcal{A}$-bimodule homomorphism which is a left inverse of $\pi^{*}$, then we say that $\mathcal{A}$ is biflat.

Let $\mathcal{A}$ be a Banach algebra. We write $Hom(\mathcal{A})$ for the set of all continuous homomorphisms on $\mathcal{A}$. Let $X$ and $Y$ be Banach $\mathcal{A}$-bimodule, and let $\sigma\in Hom(\mathcal{A})$. A bounded linear map $T:X\rightarrow Y$ is a $\sigma$-$\mathcal{A}$-bimodule homomorphism if $T(a\cdot x)=\sigma(a)\cdot T(x)$ and $T(x\cdot a)=T(x)\cdot\sigma(a)$ where $x\in X$ and $a\in\mathcal{A}$. A Banach algebra $\mathcal{A}$ is $\sigma$-biprojective if there exists a $\sigma$-$\mathcal{A}$-bimodule homomorphism $\rho:\mathcal{A}\longrightarrow\mathcal{A}\widehat{\otimes}\mathcal{A}$ such that $\pi\circ\rho=\sigma$ [13].

The purpose of this paper is to study the concept of $\sigma$-biflatness for Banach algebras. We have to stress that our definition is completely different from what have introduced in [4, Definition 2.11]. For comparison, unlike our definition, $\sigma$-biflatness in [4] is not a generalization of the notion of biflatness [4, Remark 2.13].

The organization of the paper is as follows. Firstly, in section 2 we investigate some basic properties of $\sigma$-biflat Banach algebras. We find an equivalent condition to $\sigma$-biflatness (Theorem 2.3). We prove that every $\sigma$-biprojective Banach algebra is $\sigma$-biflat (Proposition 2.4). However biflat Banach algebras are $\sigma$-biflat (Proposition 2.5), we give an example to show that the class of $\sigma$-biflat Banach algebras is larger than that for biflat Banach algebras (Example 2.1).

In section 3, we find the relations between $\sigma$-biflatness and both $\sigma$-amenability and the existence of some certain $\sigma$-diagonals. There are examples of $\sigma$-biflat Banach algebras which are not $\sigma$-amenable (Examples 3.1 and 3.2).

In section 4, we deal with the short exact sequence $\Sigma:0\longrightarrow ker\pi\stackrel{{\scriptstyle\imath}}{{\longrightarrow}}\mathcal{A}\widehat{\otimes}\mathcal{A}\stackrel{{\scriptstyle\pi}}{{\longrightarrow}}\mathcal{A}\longrightarrow 0$, where $\mathcal{A}$ is a Banach algebra. We prove that if $\mathcal{A}$ is $\sigma$-amenable, then $\Sigma^{*}$ and $\Sigma^{**}$ behave like splitting sequences (Proposition 4.2).

Finally in section 5, we generalize Ramsden’s theorem [11, Proposition 2.5] related to biflatness of tensor product of Banach algebras to the $\sigma$-case (Theorem 5.1).

2. Basic properties

Definition 2.1**.**

Let $\mathcal{A}$ be a Banach algebra and let $\sigma\in Hom(\mathcal{A})$. Then $\mathcal{A}$ is $\sigma$-biflat if there exists a bounded linear map $\rho:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{*}\longrightarrow\mathcal{A}^{*}$ satisfying

[TABLE]

for $a\in\mathcal{A},\lambda\in\mathcal{A}^{\ast}$ such that $\rho\circ\pi^{*}=\sigma^{*}$.

Lemma 2.2**.**

Let $\mathcal{A}$ be a Banach algebra, let $\sigma\in Hom(\mathcal{A})$ and let $X$ and $Y$ be Banach $\mathcal{A}$-bimodules. If $T:X\rightarrow Y$ is a $\sigma$-$\mathcal{A}$-bimodule homomorphism, then $T^{\ast}$ satisfies $(1)$, that is, $T^{\ast}(\sigma(a)\cdot\lambda)=a\cdot T^{\ast}(\lambda)$ and $T^{\ast}(\lambda\cdot\sigma(a))=T^{\ast}(\lambda)\cdot a$, for $a\in\mathcal{A},\lambda\in Y^{\ast}$.

Proof.

For every $x\in X$, $a\in A$ and $\lambda\in Y^{\ast}$, we have

[TABLE]

so $T^{\ast}(\sigma(a)\cdot\lambda)=a\cdot T^{\ast}(\lambda)$. Similarly $T^{\ast}(\lambda\cdot\sigma(a))=T^{\ast}(\lambda)\cdot a$. ∎

The following characterization is useful.

Theorem 2.3**.**

Let $\mathcal{A}$ be a Banach algebra and let $\sigma\in Hom(\mathcal{A})$. Then, the following are equivalent:

1. (i)

$\mathcal{A}$ is $\sigma$-biflat; 2. (ii)

There is a $\sigma$-$\mathcal{A}$-bimodule homomorphism $\rho:\mathcal{A}\rightarrow(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast\ast}$ such that $\pi^{\ast\ast}\circ\rho=\kappa_{\mathcal{A}}\circ\sigma$, where $\kappa_{\mathcal{A}}$ is the canonical embedding of $\mathcal{A}$ into $\mathcal{A}^{\ast\ast}$.

Proof.

(i)$\Longrightarrow$(ii) Since $\mathcal{A}$ is $\sigma$-biflat, there exists a bounded linear map $\theta:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast}\rightarrow\mathcal{A}^{\ast}$ satisfying ($1$) and $\theta\circ\pi^{\ast}=\sigma^{\ast}$. We set $\rho$ to be the restriction of $\theta^{\ast}$ to $\mathcal{A}$. Then, for every $a\in\mathcal{A}$ and $\lambda\in\mathcal{A}^{\ast}$ we have

[TABLE]

Next, for every $a,b\in\mathcal{A}$ and $\xi\in(\mathcal{A}\otimes\mathcal{A})^{\ast}$

[TABLE]

so $\rho(ab)=\sigma(a)\cdot\rho(b)$. A Similar argument shows that $\rho(ba)=\rho(b)\cdot\sigma(a)$.

(ii)$\Longrightarrow$(i) Let $\rho$ be as specified in the clause (ii). Suppose that $\tilde{\rho}:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast}\rightarrow\mathcal{A}^{\ast}$ is the restriction of $\rho^{\ast}$ into $(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast}$. Clearly $\tilde{\rho}$ is a bounded linear map and satisfies $(1)$, by Lemma 2.2. For every $a\in\mathcal{A}$ and $\lambda\in\mathcal{A^{\ast}}$

[TABLE]

showing that $\tilde{\rho}\circ\pi^{\ast}=\sigma^{\ast}$. ∎

It is well-known that every biprojective Banach algebra is biflat. The next proposition gives a generalization of this fact.

Proposition 2.4**.**

Let $\mathcal{A}$ be a Banach algebra and let $\sigma\in Hom(\mathcal{A})$. If $\mathcal{A}$ is $\sigma$-biprojective, then $\mathcal{A}$ is $\sigma$-biflat.

Proof.

If $\mathcal{A}$ is a $\sigma$-biprojective Banach algebra, then there exists a bounded $\sigma$-$\mathcal{A}$-bimodule homomorphism $\rho:\mathcal{A}\rightarrow\mathcal{A}\widehat{\otimes}\mathcal{A}$ such that $\pi\circ\rho=\sigma$. For every $\lambda\in\mathcal{A}^{\ast}$ and $a\in\mathcal{A}$ we have

[TABLE]

Since $\rho$ is a $\sigma$-$\mathcal{A}$-bimodule homomorphism, $\rho^{\ast}$ satisfies $(1)$ by Lemma 2.2, so $\mathcal{A}$ is $\sigma$-biflat. ∎

The relation between biflatness and $\sigma$-biflatness appears as follows.

Proposition 2.5**.**

Let $\mathcal{A}$ be a Banach algebra and let $\sigma\in Hom(\mathcal{A})$. Then:

1. (i)

If $\mathcal{A}$ is biflat, then $\mathcal{A}$ is $\sigma$-biflat. 2. (ii)

If $\mathcal{A}$ is $\sigma$-biflat and has a bounded approximate identity, and if $\sigma$ has a dense range, then $\mathcal{A}$ is biflat.

Proof.

(i) There exists a bounded $\mathcal{A}$-bimodule homomorphism $\rho:(\mathcal{A}\hat{\otimes}\mathcal{A})^{*}\longrightarrow\mathcal{A}^{\ast}$ such that $\rho\pi^{\ast}=i_{\mathcal{A}^{\ast}}$. Define $\acute{\rho}=\sigma^{\ast}\rho:(\mathcal{A}\hat{\otimes}\mathcal{A})^{\ast}\longrightarrow\mathcal{A}^{\ast}$. Then, for every $\lambda\in(\mathcal{A}\hat{\otimes}\mathcal{A})^{\ast}$ and $a,b\in\mathcal{A}$, we have

[TABLE]

and analogously, $\acute{\rho}(\lambda\cdot\sigma(a))=\acute{\rho}(\lambda)\cdot a$. It is clear that $\acute{\rho}\pi^{\ast}=\sigma^{\ast}$, and hence $\mathcal{A}$ is $\sigma$-biflat.

(ii) By Proposition 3.3 below, $\mathcal{A}$ is amenable and then by [3, Theorem 2.9.65], $\mathcal{A}$ is biflat. ∎

Now, we give a $\sigma$-biflat Banach algebra which is not biflat.

Example 2.1**.**

It is known that $\mathcal{A}=l^{1}(\mathbb{Z}^{+})$ is not amenable, and so $\mathcal{A}^{\sharp}$ is not amenable by [12, Corollary 2.3.11]. Therefore, $\mathcal{A}^{\sharp}$ is not biflat by [5]. We consider the homomorphism $\sigma:\mathcal{A}^{\sharp}\rightarrow\mathcal{A}^{\sharp}$ defined by $\sigma(a+\lambda e)=\lambda$, $(a\in\mathcal{A},\lambda\in\mathbb{C})$. An argument similar to [4, Example 2.7], shows that $\mathcal{A}^{\sharp}$ is $\sigma$-amenable. Then by Remark 3.4 below, $\mathcal{A}^{\sharp}$ is $\sigma$-biflat.

3. Relation to $\sigma$-amenability

Let $\mathcal{A}$ be a Banach algebra, let $\sigma\in Hom(\mathcal{A})$, and let $X$ be a Banach $\mathcal{A}$-bimodule. A bounded linear map ${D}:\mathcal{A}\rightarrow X$ is a $\sigma$-derivation if

[TABLE]

and it is $\sigma$-inner derivation if there is an element $x\in X$ such that $D(a)=\sigma(a)\cdot x-x\cdot\sigma(a)$ for all $a\in\mathcal{A}$. A Banach algebra $\mathcal{A}$ is $\sigma$-amenable if for every Banach $\mathcal{A}$-bimodule $X$, every $\sigma$-derivation ${D}:\mathcal{A}\rightarrow X^{*}$ is $\sigma$-inner [8, 9]. An element $M\in(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast\ast}$ is said to be a $\sigma$-virtual diagonal for $\mathcal{A}$, if $\sigma(a)\cdot M=M\cdot\sigma(a)$ and $\pi^{**}(\it M)\cdot\sigma(a)=\sigma(a)$ for all $a\in\mathcal{A}$. A bounded net $(m_{\alpha})\subseteq\mathcal{A}\widehat{\otimes}\mathcal{A}$ is said to be a $\sigma$-approximate diagonal for $\mathcal{A}$ if $\lim_{\alpha}m_{\alpha}\cdot\sigma(a)-\sigma(a)\cdot m_{\alpha}=0$ and $\lim_{\alpha}\pi(m_{\alpha})\cdot\sigma(a)=\sigma(a)$ for all $a\in\mathcal{A}$ [10]. In [4, Proposition 2.4] it is shown that if $\mathcal{A}$ has a $\sigma$-virtual diagonal, then it has a $\sigma$-approximate diagonal. An easy verification shows that the converse is also true.

In the following two propositions, we establish a connection between $\sigma$-biflatness and existence of $\sigma$-virtual diagonals.

Proposition 3.1**.**

Let $\mathcal{A}$ be a Banach algebra with a bounded approximate identity and $\sigma\in Hom(\mathcal{A})$. If $\mathcal{A}$ is $\sigma$-biflat, then $\mathcal{A}$ has a $\sigma$-virtual diagonal.

Proof.

Let $(e_{\alpha})$ be a bounded approximate identity for $\mathcal{A}$. Since $\mathcal{A}$ is $\sigma$-biflat there exists a bounded linear map $\rho:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast}\rightarrow\mathcal{A}^{\ast}$ satisfying $(1)$ such that $\rho\circ\pi^{\ast}=\sigma^{\ast}$. We may suppose that $\rho^{\ast}(e_{\alpha})$ converges in the weak* topology to an element of $(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast\ast}$, say $M$. Then for each $a\in\mathcal{A}$ and $\lambda\in(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast}$, we have

[TABLE]

and similarly $\langle M\cdot\sigma(a),\lambda\rangle=\langle a,\rho(\lambda)\rangle$. Thus $\sigma(a)\cdot M=M\cdot\sigma(a)$. An application of Theorem 2.3 shows that

[TABLE]

hence $M$ is a $\sigma$-virtual diagonal for $\mathcal{A}$. ∎

Proposition 3.2**.**

Let $\mathcal{A}$ be a Banach algebra and $\sigma\in Hom(\mathcal{A})$. If $\mathcal{A}$ has a $\sigma$-virtual diagonal, then $\mathcal{A}$ is $\sigma$-biflat.

Proof.

Let $M\in(\mathcal{A}\hat{\otimes}\mathcal{A})^{\ast\ast}$ be a $\sigma$-virtual diagonal. Define $\rho:\mathcal{A}\rightarrow(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast\ast}$ by $\rho(a)=\sigma(a)\cdot M$  $(a\in\mathcal{A})$. Clearly $\rho$ is bounded, linear and $\sigma$-$\mathcal{A}$-bimodule homomorphism. Also

[TABLE]

Thus $\mathcal{A}$ is $\sigma$-biflat. ∎

Now, we describe the relation between $\sigma$-biflatness and $\sigma$-amenability.

Proposition 3.3**.**

Let $\mathcal{A}$ be a $\sigma$-biflat Banach algebra with a bounded approximate identity. If $\sigma\in Hom(\mathcal{A})$ has a dense range, then $\mathcal{A}$ is amenable so is $\sigma$-amenable.

Proof.

Let $\mathcal{A}$ be a $\sigma$-biflat Banach algebra. By Proposition 3.1 $\mathcal{A}$ has a $\sigma$-virtual diagonal, equivalently, $\mathcal{A}$ has a $\sigma$-approximate diagonal $m_{\alpha}\subseteq(\mathcal{A}\hat{\otimes}\mathcal{A})$. We show that $\mathcal{A}$ has an approximate diagonal, so it is amenable by [3, Theorem 2.9.65] and thus by [10, Corollay 2.2] it is $\sigma$-amenable, as required. Suppose that $a\in\mathcal{A}$ and $\varepsilon>0$. There exists $N$ such that for every $n\geq N$, $\lVert a-\sigma(a_{n})\rVert<\varepsilon$ then $\lVert a-\sigma(a_{N})\rVert<\varepsilon$. There is $\alpha_{0}$ such that for all $\alpha\geq\alpha_{0}$, $\lVert\sigma(a_{N})\cdot m_{\alpha}-m_{\alpha}\cdot\sigma(a_{N})\rVert<\varepsilon$. Hence for $\alpha\geq\alpha_{0}$, we have $\lVert a\cdot m_{\alpha}-m_{\alpha}\cdot a\rVert\leq\lVert a\cdot m_{\alpha}-\sigma(a_{N})\cdot m_{\alpha}\rVert+\lVert\sigma(a_{N})\cdot m_{\alpha}-m_{\alpha}\cdot\sigma(a_{N})\rVert+\lVert m_{\alpha}\cdot\sigma(a_{N})-m_{\alpha}\cdot a\rVert\leq(2M+1)\varepsilon$, where $M$ is a bounded of $(m_{\alpha})$. This shows that $a\cdot m_{\alpha}-m_{\alpha}\cdot a\rightarrow 0$ and similarly, $a\pi(m_{\alpha})\rightarrow a$. ∎

Remark 3.4**.**

There is a converse for Proposition 3.3. Indeed if $\mathcal{A}$ is $\sigma$-amenable with a bounded approximate identity, then it has $\sigma$-virtual diagonal [4, Theorem 2.2], and whence by Proposition 3.2, $\mathcal{A}$ is $\sigma$-biflat.

We conclude the current section with two examples of $\sigma$-biflat Banach algebras which are not $\sigma$-amenable. We refer the reader to [12, Definition 3.1.8 and Definition C.1.1] for the definitions of property ($\mathbb{A}$) and approximation property for Banach spaces. We also write $\mathfrak{A}(E)$ for the space of approximable operators on a Banach space $E$.

Example 3.1**.**

let $E$ be a Banach space with property ($\mathbb{A}$) such that $E^{**}$ does not have the bounded approximation property. Then $\mathfrak{A}(E^{*})$ is biflat while it is not amenable [12, Theorem 4.3.24]. Therefore, $\mathfrak{A}(E^{*})$ is $\sigma$-biflat for each $\sigma\in Hom(\mathfrak{A}(E^{*}))$. On the other hand, One may check that every $\sigma$-amenable Banach algebra for which $\sigma$ has a dense range, is amenable. Hence, choosing a homomorphism $\sigma\in Hom(\mathfrak{A}(E^{*}))$ with a dense range, it is readily seen that $\mathfrak{A}(E^{*})$ is not $\sigma$-amenable.

Example 3.2**.**

let $H$ be an infinite dimensional Hilbert space. It was shown in [12, Example 4.3.25] that $\mathfrak{A}(H\widehat{\otimes}H)$ is biflat but not amenable. Then, an argument similar to Example 3.1 shows that $\mathfrak{A}(H\widehat{\otimes}H)$ is $\sigma$-biflat which is not $\sigma$-amenable, whereas $\sigma$ is a homomorphism in $Hom(\mathfrak{A}(H\widehat{\otimes}H))$ with a dense range.

4. The role of sequences

We start with the following which is similar to Lemma 2.2.

Lemma 4.1**.**

Let $\mathcal{A}$ be a Banach algebra, $X$ and $Y$ be Banach $\mathcal{A}$-bimodule and let $\sigma\in Hom(\mathcal{A})$. If $T:X\longrightarrow Y$ is a bounded linear map satisfying $T(\sigma(a)\cdot x)=a\cdot Tx$ and $T(x\cdot\sigma(a))=Tx\cdot a$ ($a\in\mathcal{A},x\in X$), then $T^{*}$ is a $\sigma$-$\mathcal{A}$-bimodule homomorphism.

For a Banach algebra $\mathcal{A}$, we consider the short exact sequence

[TABLE]

and its duals $\Sigma^{*}$ and $\Sigma^{**}$, where $\imath$ is the natural injection. It is known that a Banach algebra $\mathcal{A}$ with a bounded approximate identity is amenable if and only if $\Sigma^{*}$ (and so $\Sigma^{**}$) split [2].

Proposition 4.2**.**

Let $\mathcal{A}$ be a Banach algebra with a bounded approximate identity and $\sigma\in Hom(\mathcal{A})$. Suppose that $\mathcal{A}$ is $\sigma$-amenable. Then we have the following statements:

$(i)$ Regarding the sequence $\Sigma^{*}$, there is a bounded linear map $\theta:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{*}\longrightarrow\mathcal{A}^{*}$ such that $\theta\pi^{*}=\sigma^{*}$, and

[TABLE]

$(ii)$ Regarding the sequence $\Sigma^{**}$, there is a $\sigma$-$\mathcal{A}$-bimodule homomorphism $\gamma:\mathcal{A}^{**}\longrightarrow(\mathcal{A}\widehat{\otimes}\mathcal{A})^{**}$ such that $\pi^{**}\gamma=\sigma^{**}$.

Proof.

$(i)$ Let $M\in(\mathcal{A}\widehat{\otimes}\mathcal{A})^{**}$ be a $\sigma$-virtual diagonal for $\mathcal{A}$. We define the map $\theta:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{*}\longrightarrow\mathcal{A}^{*}$ via

[TABLE]

Then, for $a\in\mathcal{A}$ and $\lambda\in\mathcal{A}^{*}$

[TABLE]

so that $\theta\pi^{*}=\sigma^{*}$. Next for $a,b\in\mathcal{A}$ and $f\in(\mathcal{A}\widehat{\otimes}\mathcal{A})^{*}$, we have

[TABLE]

whence $\theta(\sigma(a)\ .\ f)=a\ .\ \theta(f)$. The equality $\theta(f\ .\ \sigma(a))=\theta(f)\ .\ a$ is even easier.

$(ii)$ Take $\gamma:=\theta^{*}$, where $\theta$ is given by the clause $(i)$. Then, it is immediate by Lemma 4.1. ∎

Finally, we generalize [12, Proposition 4.3.23] where the proof reads somehow the same lines.

Theorem 4.3**.**

Let $\mathcal{A}$ be Banach algebra, let $\mathcal{B}$ be a closed subalgebra of $\mathcal{A}$ and $\sigma\in Hom(\mathcal{A})$ for which $\sigma(\mathcal{B})\subseteq\mathcal{B}$ with the followings properties:

1. (i)

$\mathcal{B}$ is $\sigma$-amenable; 2. (ii)

$\mathcal{B}$ is a left ideal of $\mathcal{A}$; 3. (iii)

$\mathcal{B}$ has a bounded approximate identity which is also a bounded left approximate identity for $\mathcal{A}$.

Then $\mathcal{A}$ is $\sigma$-biflat.

Proof.

Since $\mathcal{B}$ is $\sigma$-amenable, Proposition 4.2(ii) yields the existence of a $\sigma$-$\mathcal{B}$-bimodule homomorphism $\rho:\mathcal{B}^{**}\longrightarrow(\mathcal{B}\hat{\otimes}\mathcal{B})^{**}$ such that $\pi^{**}_{\mathcal{B}}\circ\rho=(\sigma\mid_{\mathcal{B}})^{**}$. Set $\tilde{\rho}=(\iota\hat{\otimes}\iota)^{**}\circ\rho$, where $\iota:\mathcal{B}\hookrightarrow\mathcal{A}$ is the canonical embedding. Let $(e_{\alpha})_{\alpha}$ be a bounded approximate identity for $\mathcal{B}$ which is a left bounded approximate identity for $\mathcal{A}$. For each $a\in\mathcal{A}$, $(\tilde{\rho}(e_{\alpha})\cdot\sigma(a))_{\alpha}$ is a bounded net in $(\mathcal{A}\hat{\otimes}\mathcal{A})^{**}$, and so without loss of generality we may suppose that it is convergence. Therefore we obtain a bounded linear map $\bar{\rho}:\mathcal{A}\longrightarrow(\mathcal{A}\hat{\otimes}\mathcal{A})^{**}$, defined by $\bar{\rho}(a)=w^{*}-\lim_{\alpha}\tilde{\rho}(e_{\alpha})\cdot\sigma(a)$. It is immediate that $\bar{\rho}$ is a right $\sigma$-$\mathcal{A}$-module homomorphism. To check that $\bar{\rho}$ is a left $\sigma$-$\mathcal{A}$-module homomorphism, we first notice that

[TABLE]

Let $a,x,b\in\mathcal{A}$. By Cohen’s factorization theorem, $b=cd$ for some $c\in\mathcal{B}$ and $d\in\mathcal{A}$. Then, it follows from the above observation that

[TABLE]

Now, let $a,x\in\mathcal{A}$. Again, by Cohen’s factorization theorem, there are $y,z\in\mathcal{A}$ such that $b=yz$. Then

[TABLE]

So $\bar{\rho}$ is $\sigma$-$\mathcal{A}$-bimodule homomorphism. It remains to prove that $\pi^{**}_{\mathcal{A}}\circ\bar{\rho}=\kappa_{\mathcal{A}}\circ\sigma$. For each $a\in\mathcal{A}$ we have

[TABLE]

So by Theorem 2.3, $\mathcal{A}$ is $\sigma$-biflat. ∎

5. Application for Tensor products

Let $\mathcal{A}$ be a Banach algebra, let $\sigma\in Hom(\mathcal{A})$ and let $C>0$. We say that $\mathcal{A}$ is $C$-$\sigma$-biflat, If there exists a map $\rho:(\mathcal{A}\otimes\mathcal{A})^{\ast}\longrightarrow\mathcal{A}^{\ast}$, satisfying $(1)$ such that $\rho\circ\pi^{\ast}=\sigma^{\ast}$ and $\lVert\rho\rVert<C$.

Let $\mathcal{A}$ and $\mathcal{B}$ be Banach algebras and $\sigma_{\mathcal{A}}\in Hom(\mathcal{A})$ and $\sigma_{\mathcal{B}}\in Hom(\mathcal{B})$. Let $E$ be a Banach $\mathcal{A}$-bimodule, and let $F$ be a Banach $\mathcal{B}$-bimodule. We regard $E\widehat{\otimes}F$ as a Banach $\mathcal{A}\widehat{\otimes}\mathcal{B}$-bimodule with the actions

[TABLE]

for every $a\in\mathcal{A},b\in\mathcal{B},x\in E$ and $y\in F$.

Using Ramsden’s notation [11], we construct $\tilde{\mathcal{B}}(E,F)$ and $\hat{\mathcal{B}}(F,E)$ as follows.

Let $\tilde{\mathcal{B}}(E,F)$ be the space $\mathcal{B}(E,F)$ with the following module actions:

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for every $T\in\mathcal{B}(E,F)$, $a\in\mathcal{A}$, $b\in\mathcal{B}$, $x\in E$. Consider the map $\tilde{T}:(E\widehat{\otimes}F)^{\ast}\rightarrow{\tilde{\mathcal{B}}}(E,F^{*});\lambda\rightarrow\tilde{T}(\lambda)$ defined by $\tilde{T}(\lambda)(x)(y)=\lambda(x\otimes y)$. By [1, 42. Proposition 13], $\tilde{T}$ is an isometric isomorphism of Banach spaces. We claim that $\tilde{T}$ satisfies (1). Indeed

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and on the other hand

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Let $\widehat{\mathcal{B}}(F,E)$ be $\mathcal{B}(F,E)$ with the following module actions

[TABLE]

for all $T\in\mathcal{B}(F,E)$, $a\in A$, $b\in B$, $y\in F$. Consider the map $\widehat{T}:(E\widehat{\otimes}F)^{\ast}\rightarrow{\hat{\mathcal{B}}}(F,E^{\ast});\lambda\mapsto\widehat{T}(\lambda)$ defined by $\widehat{T}(\lambda)(y)(x)=\lambda(x\otimes y)$. A similar argument as we use for $\tilde{T}$, shows that $\hat{T}$ is an isometric isomorphism of Banach spaces satisfying (1)

Now, we are ready to prove the main goal of the current section.

Theorem 5.1**.**

Let $\mathcal{A}$ be $C_{1}$-$\sigma_{\mathcal{A}}$-biflat Banach algebra and let $\mathcal{B}$ be $C_{2}$-$\sigma_{\mathcal{B}}$-biflat Banach algebra and if both $\sigma_{\mathcal{A}}$ and $\sigma_{\mathcal{B}}$ are idempotents. Then $(\mathcal{A}\widehat{\otimes}\mathcal{B})$ is $C_{1}C_{2}$-$\sigma_{\mathcal{A}}\otimes\sigma_{\mathcal{B}}$-biflat.

Proof.

There exists a bounded linear map $\rho_{\mathcal{A}}:(\mathcal{A}\widehat{\otimes}\mathcal{A})^{\ast}\rightarrow\mathcal{A}^{\ast}$ such that $\rho_{\mathcal{A}}\circ\pi^{\ast}_{\mathcal{A}}=\sigma^{\ast}_{\mathcal{A}}$, satisfying (1) and $\|\rho_{\mathcal{A}}\|\leq C_{1}$ and also a bounded linear map $\rho_{\mathcal{B}}:(\mathcal{B}\widehat{\otimes}\mathcal{B})^{\ast}\rightarrow\mathcal{B}^{\ast}$ such that $\rho_{\mathcal{B}}\circ\pi^{\ast}_{\mathcal{B}}=\sigma^{\ast}_{\mathcal{B}}$, satisfying (1) and $\|\rho_{\mathcal{B}}\|\leq C_{2}$. Consider the composition

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All the maps satisfy the equation (1), and $\|\rho_{\mathcal{A}\widehat{\otimes}\mathcal{B}}\|\leq C_{1}C_{2}$. For $\lambda\in(\mathcal{A}\widehat{\otimes}\mathcal{B})^{\ast}$, we follow $\pi^{\ast}_{\mathcal{A}\widehat{\otimes}\mathcal{B}}(\lambda)$ under the sequence of composition $\rho_{\mathcal{A}\widehat{\otimes}\mathcal{B}}$. We have

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So

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hence $\rho_{\mathcal{A}\widehat{\otimes}\mathcal{B}}\circ\pi^{\ast}_{\mathcal{A}\widehat{\otimes}\mathcal{B}}=\sigma^{\ast}_{\mathcal{A}\widehat{\otimes}\mathcal{B}}$.

Note that the last equality comes from the following isomorphisms

[TABLE]

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