2 Preliminaries

Group algebras and group actions on manifolds are two major sources for the construction of operator algebras. In applications, generalizations of groups (group algebras), such as semigroups (semigroup algebras), are also used. A brief description of semigroup algebras follows.

The Hilbert space $\mathcal {H}$ is $l^{2}(S)$ , the space of all square-summable complex valued functions on a cancellative semigroup S. The semigroup S (with unit e) is assumed to be discrete and countable. Hence $\mathcal {H}$ is separable. The family of functions $(\delta _s)_{s\in S}$ forms an orthonormal basis of $\mathcal {H}$ , where $\delta _s(s)=1$ and $\delta _s(t)=0$ for any $t\in S$ , $t\neq s$ . For each f and g in $\mathcal {H}$ , let $L_f$ be the left convolution operator defined as $L_fg= f\ast g$ , where $ f\ast g(s)=\sum _{uv=s}f(u)g(v)$ for each s in S. Note that the convolution operators may be unbounded. We denote by $\mathcal {L}(S)$ the set of all bounded left convolution operators on $\mathcal {H}$ . Then $\mathcal {L}(S)$ is a subalgebra of $B(\mathcal {H})$ . In general, $\mathcal {L}(S)$ is not a $*$ -algebra. Similarly, we denote by $\mathcal {R}(S)$ the subalgebra of $B(\mathcal {H})$ consisting of all bounded right convolution operators. Then $\mathcal {L}(S)'=\mathcal {R}(S)$ and $\mathcal {R}(S)'=\mathcal {L}(S)$ , which implies that $\mathcal {L}(S)$ and $\mathcal {R}(S)$ are both weak-operator closed algebras. For each s in S, the operator $L_{\delta _s}$ is an isometry on $\mathcal {H}$ and is denoted by $L_s$ in this paper for convenience. We denote by $\mathfrak {B}(S)$ the Banach algebra generated by $\{L_s:\ s\in S \}$ in norm topology in $B(\mathcal {H})$ . Then $\mathfrak {B}(S)$ is a Banach subalgebra of $\mathcal {L}(S)$ .

Specific examples for such Banach algebras result from choosing for S any of the free semigroup $\mathcal {F}_n$ on n ( $\geq 2$ ) generators, Thompson’s semigroup $\mathcal {S}$ , or the multiplicative semigroup of natural numbers $(\mathbb {N},\star )$ . The algebraic structure of $\mathfrak {B}(S)$ can reflect the structure of S. In [Reference Dong, Huang and Xue5], Dong, Huang, and Xue proved that the maximal ideal space of the commutative Banach algebra $\mathfrak {B}(\mathbb {N})$ is homeomorphic to the Cartesian product of unit closed disk indexed by primes (see [Reference Dong, Huang and Xue5, Theorem 1.1]). They pointed out that this result implies the fundamental theorem of arithmetic. Analogously, studying the cohomology of the Banach algebras $\mathfrak {B}(\mathcal {S})$ , $\mathfrak {B}(\mathcal {F}_n)$ , and $\mathfrak {B}(\mathcal {T})$ can help us to understand the properties of the corresponding semigroups.

In this paper, we will prove that derivations on $\mathfrak {B}(\mathcal {S})$ , $\mathfrak {B}(\mathcal {F}_n),$ and $\mathfrak {B}(\mathcal {T})$ are automatically continuous, and every derivation on $\mathfrak {B}(\mathcal {S})$ is spatial and induced by an operator in $\mathcal {L}(\mathcal {S})$ . Comparing with a result of Kadison [Reference Kadison18, Theorem 4] that every derivation on a $C^*$ -algebra is spatial, we give a nontrivial example in the case of Banach algebras. Moreover, we prove that the first cohomology group of $\mathfrak {B}(\mathcal {S})$ with coefficients in $\mathcal {L}(\mathcal {S})$ is zero, which gives a positive indication for the left amenability of Thompson’s semigroup.

In the following, we recall several concepts and results about amenable semigroups. We say that a discrete cancellative semigroup S is left (resp. right) amenable if there exists a left (resp. right) invariant mean on $l^{\infty }(S)$ . For example, the additive semigroup of natural numbers is amenable while the free semigroup on n ( $\geq 2$ ) generators is not. A left (resp. right) Følner net of S is a net of non-empty finite subsets $\{F_{\alpha }\}$ in S such that for any $s\in S$ ,

$$\begin{align*}\lim_{\alpha}\frac{|sF_{\alpha}\Delta F_{\alpha}|}{|F_{\alpha}|}=0\ \left( \text{resp}.\ \lim_{\alpha}\frac{|F_{\alpha}s\Delta F_{\alpha}|}{|F_{\alpha}|}=0\right). \end{align*}$$

It was proved for groups by Følner, and then generalized to discrete cancellative semigroups by Frey [Reference Frey7] that S is left (resp. right) amenable if and only if S has a left (resp. right) Følner net. In [Reference Følner6], Følner proved that every subgroup of an amenable group is still amenable. For semigroups, it is not always true. In [Reference Frey7], Frey gave an example of a left amenable semigroup which contains a non-amenable semigroup.

3 Unique factorization semigroup

For example, the multiplicative semigroup of natural numbers is a unique factorization semigroup and the set of all primes is the unique basis up to reorder. It is also clear that Thompson’s semigroup $\mathcal {S}=\langle X_0, X_1,\ldots \ |\ X_jX_i=X_iX_{j+1},\ i<j \rangle ^+$ is a unique factorization semigroup with the basis $\{X_n\in \mathcal {S}:\ n\in \mathbb {N}\}$ .

Next, we introduce some properties of $\mathcal {S}$ that will be frequently used in Sections 4 and 5.

It is clear that $\mathsf {ind}_{0}(uv)=\mathsf {ind}_{0}(u)+\mathsf {ind}_{0}(v)=\mathsf {ind}_0(vu)$ and $\mathsf {ind}(uv)=\mathsf {ind}(u)+\mathsf {ind}(v)=\mathsf {ind}(vu)$ for all u and v in $\mathcal {S}$ . In general, $\mathsf {ind}_i(uv)\neq \mathsf {ind}_i(u)+\mathsf {ind}_i(v)$ for $i{\kern-1pt}\geq{\kern-1pt} 1$ . For example, $\mathsf {ind}_2(X_0X_2X_1){\kern-1pt}={\kern-1pt}\mathsf {ind}_2(X_0X_1X_3){\kern-1pt}={\kern-1pt}0{\kern-1pt}\neq{\kern-1pt} 1{\kern-1pt}={\kern-1pt} \mathsf {ind}_2(X_0X_2)+\mathsf {ind}_2(X_1)$ .

For example, $X_1|X_0X_2=X_1X_0$ while $X_2\nmid X_0X_2$ . It is clear that if $u|v$ then $\mathsf {ind}(u)\leq \mathsf {ind}(v)$ .

With the aid of the index function we introduce a total order on $\mathcal {S}$ which plays an important role in Sections 4 and 5.

For example, $X_0\prec X_1\prec X_0X_1\prec X_0X_2 \prec X_1X_2$ . The relation “ $\preceq $ ” is a total order on $\mathcal {S}$ with the well-ordered properties.

Lemma 3.7 We have the following statements:

(i) There exists a unique minimal element in every non-empty subset of $\mathcal {S}$ under the total order.

$(\mathrm{ii})$ Let $u_i$ and $v_i$ ( $1\leq i\leq n$ ) be $2n$ elements in $\mathcal {S}$ . If $u_{i}\preceq v_{i}$ for each $1\leq i\leq n$ , then $\Pi _{i=1}^{n}u_{i}\preceq \Pi _{i=1}^{n}v_{i}$ . The equality holds if and only if $u_{i}=v_{i}$ for each $1\leq i\leq n$ .

Proof It is clear that (i) holds. We now give the proof of ( $ii$ ). First, we consider the case when $n=2$ . If $\mathsf {ind}(u_{1})<\mathsf {ind}(v_{1})$ or $\mathsf {ind}(u_{2})<\mathsf {ind}(v_{2})$ , then $\mathsf {ind}(u_{1}u_{2})=\mathsf {ind}(u_{1})+\mathsf {ind}(u_{2})<\mathsf {ind}(v_{1})+\mathsf {ind}(v_{2})=\mathsf {ind}(v_{1}v_{2})$ . This implies $u_{1}u_{2}\prec v_{1}v_{2}$ . In the case that $\mathsf {ind}(u_{1})=\mathsf {ind}(v_{1})$ and $\mathsf {ind}(u_{2})=\mathsf {ind}(v_{2})$ , if $\mathsf {ind}_{0}(u_{1})>\mathsf {ind}_{0}(v_{1})$ or $\mathsf {ind}_{0}(u_{2})>\mathsf {ind}_{0}(v_{2})$ , then $\mathsf {ind}_{0}(u_{1}u_{2})=\mathsf {ind}_{0}(u_{1})+\mathsf {ind}_{0}(u_{2})>\mathsf {ind}_{0}(v_{1})+\mathsf {ind}_{0}(v_{2})=\mathsf {ind}_{0}(v_{1}v_{2})$ . This also implies $u_{1}u_{2}\prec v_{1}v_{2}$ . Hence, we assume that $\mathsf {ind}_{0}(u_{1})=\mathsf {ind}_{0}(v_{1})$ and $\mathsf {ind}_{0}(u_{2})=\mathsf {ind}_{0}(v_{2})$ . If either $\mathsf {ind}(u_{1})$ or $\mathsf {ind}(u_{2})$ is zero, then it is trivial. Thus we may further assume that $\mathsf {ind}(u_{1})=\mathsf {ind}(v_{1})\geq 1$ and $\mathsf {ind}(u_{2})=\mathsf {ind}(v_{2})\geq 1$ .

Case I: $u_{2}=v_{2}$ . If $u_{1}=v_{1}$ , then it is obvious. Otherwise, let $u_{1}=X_{0}^{\alpha _{0}}\dots X_{n}^{\alpha _{n}}$ and $v_{1}=X_{0}^{\beta _{0}}\dots X_{m}^{\beta _{m}}$ . By the definition of the total order, there exists an integer $i>0$ such that $\alpha _{j}=\beta _{j}$ whenever $j<i$ and $\alpha _{i}>\beta _{i}$ . Since $\{X_l\in \mathcal {S}:\ l\in \mathbb {N}\}$ is a basis of $\mathcal {S}$ , it only needs to prove $u_{1}X_{l}\prec v_{1}X_{l}$ for each $X_{l}\in \mathcal {S}$ . In fact, we have

(1) $$ \begin{align} u_{1}X_{l}= \begin{cases} X_{0}^{\alpha_{0}}\dots X_{l-1}^{\alpha_{l-1}}X_{l}^{\alpha_{l}+1} X_{l+2}^{\alpha_{l+1}}\dots X_{i+1}^{\alpha_{i}} \dots X_{n+1}^{\alpha_{n}},&l<i, \cr X_{0}^{\alpha_{0}}\dots X_{i-1}^{\alpha_{i-1}}X_{i}^{\alpha_{i}+1} X_{i+2}^{\alpha_{i+1}}\dots X_{n+1}^{\alpha_{n}},&l=i, \cr X_{0}^{\alpha_{0}}\dots X_{i-1}^{\alpha_{i-1}}X_{i}^{\alpha_{i}} X_{i_{1}}^{\alpha^{\prime}_{i_{1}}}\dots X_{i_{t}}^{\alpha^{\prime}_{i_{t}}},&l>i, \end{cases} \end{align} $$

and

(2) $$ \begin{align} v_{1}X_{l}= \begin{cases} X_{0}^{\beta_{0}}\dots X_{l-1}^{\beta_{l-1}}X_{l}^{\beta_{l}+1} X_{l+2}^{\beta_{l+1}}\dots X_{i+1}^{\beta_{i}} \dots X_{m+1}^{\beta_{m}},&l<i, \cr X_{0}^{\beta_{0}}\dots X_{i-1}^{\beta_{i-1}}X_{i}^{\beta_{i}+1} X_{i+2}^{\beta_{i+1}}\dots X_{m+1}^{\beta_{m}},&l=i, \cr X_{0}^{\beta_{0}}\dots X_{i-1}^{\beta_{i-1}}X_{i}^{\beta_{i}} X_{j_{1}}^{\beta^{\prime}_{j_{1}}}\dots X_{j_{s}}^{\beta^{\prime}_{j_{s}}},&l>i, \end{cases} \end{align} $$

where $i<i_{1}<\cdots <i_{t}$ and $i<j_{1}<\cdots <j_{s}$ . By comparing equation (1) with (2), we have $u_{1}X_{l}\prec u_{2}X_{l}$ .

Case II: $u_{2}\prec v_{2}$ . Let $u_{2}=X_{0}^{\gamma _{0}}\dots X_{k}^{\gamma _{k}}$ and $v_{2}=X_{0}^{\omega _{0}}\dots X_{l}^{\omega _{l}}$ , then there exists an integer $h>0$ such that $\gamma _{j}=\omega _{j}$ whenever $j<h$ and $\gamma _{h}>\omega _{h}$ . By the first case, it can be reduced to the case when $u_{2}=X_{h}^{\gamma _{h}}\dots X_{k}^{\gamma _{k}}$ and $v_{2}=X_{h}^{\omega _{h}}\dots X_{l}^{\omega _{l}}$ . If $u_{1}=v_{1}=X_{0}^{\alpha _{0}}\dots X_{n}^{\alpha _{n}}$ , then

(3) $$ \begin{align} u_{1}u_{2}= \begin{cases} X_{0}^{\alpha_{0}}\dots X_{n}^{\alpha_{n}}X_{h}^{\gamma_{h}}\dots X_{l}^{\gamma_{l}},&h>n, \cr X_{0}^{\alpha_{0}}\dots X_{h-1}^{\alpha_{h-1}}X_{h}^{\alpha_{h}+\gamma_{h}} X_{h_{1}}^{\alpha^{\prime}_{h_{1}}}\dots X_{h_{t}}^{\alpha^{\prime}_{h_{t}}},&h\leq n, \end{cases} \end{align} $$

and

(4) $$ \begin{align} v_{1}v_{2}= \begin{cases} X_{0}^{\alpha_{0}}\dots X_{n}^{\alpha_{n}}X_{h}^{\omega_{h}}\dots X_{l}^{\omega_{l}},&h>n, \cr X_{0}^{\alpha_{0}}\dots X_{h-1}^{\alpha_{h-1}}X_{h}^{\alpha_{h}+\omega_{h}} X_{h_{1}}^{\alpha^{\prime\prime}_{h_{1}}}\dots X_{h_{s}}^{\alpha^{\prime\prime}_{h_{s}}},&h\leq n, \end{cases} \end{align} $$

where $h<h_{1}<\cdots <h_{\max \{t,s\}}$ . Comparing equation (3) with (4), we have $u_{1}u_{2}\prec v_{1}v_{2}$ . The proof of the case that $u_{1}\prec v_{1}$ is similar, we omit it here. Moreover, we can obtain from the above process directly that $u_1u_2=v_1v_2$ if and only if $u_1=v_1$ and $u_2=v_2$ .

The general case when $n>2$ can be obtained by induction.

We now turn to the free semigroup $\mathcal {F}_n$ on n ( $\geq 2$ ) generators $a_i$ ( $1\leq i \leq n$ ). The following two definitions on $\mathcal {F}_n$ are parallel to Definitions 3.2 and 3.6, and Lemma 3.10 is parallel to Lemma 3.7.

It is clear that $\mathsf {ind}(gh)=\mathsf {ind}(g)+\mathsf {ind}(h)$ for all g and h in $\mathcal {F}_n$ .

The relation “ $\preceq $ ” is a total order on $\mathcal {F}_n$ .

Proposition 3.11 For $n\geq 2$ , the free semigroup $\mathcal {F}_n$ on n generators $a_i$ ( $1\leq i \leq n$ ) is a unique factorization semigroup.Footnote ¹

Proof Let $X_i=a_i$ ( $1\leq i\leq n$ ), then $X_1\prec X_2 \prec \cdots \prec X_n$ . Let $X_{n+1}$ be the minimal element of $\mathcal {F}$ such that $X_{n+1}$ cannot be represented as $X_1^{i_1}\dots X_n^{i_n}$ for any nonnegative integers $i_j$ ( $1\leq j\leq n$ ). Then $X_n\prec X_{n+1}$ . Let $X_{n+2}$ be the minimal element of $\mathcal {F}$ such that $X_{n+2}$ cannot be represented as $X_1^{i_1}\dots X_{n+1}^{i_{n+1}}$ for any nonnegative integers $i_j$ ( $1\leq j\leq n+1$ ). Then $X_{n+1}\prec X_{n+2}$ . Continuing this process, we can obtain a subset $\{X_1,X_2,\ldots \}$ of $\mathcal {F}_n$ such that $X_i\prec X_{i+1}$ for each i ( $\geq 1$ ). Then every element can be written as the product $X_1^{i_1}\dots X_n^{i_n}$ for some nonnegative integers $i_j$ ( $1\leq j\leq n$ ). We next show the uniqueness by induction on the index. It is clear that the uniqueness holds for index $\leq 1$ . Assume that the uniqueness holds for index $\leq k$ ( $k\geq 1$ ). Let $g\in \mathcal {F}_n$ and $\mathsf {ind}(g)= k+1$ . Suppose that g has two different forms:

$$\begin{align*}g= X_{i_1}^{\alpha_1}\dots X_{i_n}^{\alpha_n}= X_{j_1}^{\beta_1}\dots X_{j_m}^{\beta_m},\end{align*}$$

where $\alpha _i$ ( $1\leq i\leq n$ ) and $\beta _j$ ( $1\leq j\leq m$ ) are positive integers, $i_1<\cdots <i_n$ and $j_1<\cdots <j_m$ . If $i_1=j_1$ , then

$$\begin{align*}X_{i_1}^{\alpha_1-1}\dots X_{i_n}^{\alpha_n}= X_{j_1}^{\beta_1-1}\dots X_{j_m}^{\beta_m},\end{align*}$$

which implies that $n=m$ , $i_t=j_t$ and $\alpha _t=\beta _t$ ( $1\leq t\leq n$ ). This leads to a contradiction.

We assume that $i_1<j_1$ . Then there exists some l ( $1\leq l\leq n-1$ ) such that $(X_{i_1}^{\alpha _1}\dots X_{i_l}^{\alpha _l})^{-1}X_{j_1}$ belongs to $\mathcal {F}_n$ and $1\leq \mathsf {ind}((X_{i_1}^{\alpha _1}\dots X_{i_l}^{\alpha _l})^{-1}X_{j_1})< \mathsf {ind}(X_{i_{l+1}})$ , or $(X_{i_1}^{\alpha _1}\dots X_{i_{l-1}}^{\alpha _{l-1}}$ $X_{i_l}^{\alpha ^{\prime }_l})^{-1}X_{j_1}$ ( $1\leq l\leq n$ ) belongs to $\mathcal {F}_n$ for some $\alpha ^{\prime }_l$ ( $1\leq \alpha ^{\prime }_l< \alpha _l$ ) and $1\leq \mathsf {ind}((X_{i_1}^{\alpha _1}\dots X_{i_{l-1}}^{\alpha _{l-1}}X_{i_l}^{\alpha ^{\prime }_l})^{-1}$ $X_{j_1})< \mathsf {ind}(X_{i_l})$ . In the first case, we have $(X_{i_1}^{\alpha _1}\dots X_{i_l}^{\alpha _l})^{-1}X_{j_1}=X_{s_1}^{\gamma _1}\dots X_{s_t}^{\gamma _t}$ for some positive integers $\gamma _1,\ldots ,\gamma _t$ and $s_1<\cdots <s_t<\min \{ i_{l+1},j_1\}$ . Then

$$\begin{align*}X_{i_{l+1}}^{\alpha_{l+1}}\dots X_{i_n}^{\alpha_n}= X_{s_1}^{\gamma_1}\dots X_{s_t}^{\gamma_t}X_{j_1}^{\beta_1-1}\dots X_{j_m}^{\beta_m},\end{align*}$$

which leads to a contradiction. In the second case, we have $(X_{i_1}^{\alpha _1}\dots X_{i_l}^{\alpha ^{\prime }_l})^{-1}X_{j_1}=X_{s_1}^{\gamma _1}\dots X_{s_t}^{\gamma _t}$ for some positive integers $\gamma _1,\ldots ,\gamma _t$ and $s_1<\cdots <s_t<\min \{ i_{l},j_1\}$ . Then

$$\begin{align*}X_{i_l}^{\alpha_i-\alpha^{\prime}_i}X_{i_{l+1}}^{\alpha_{l+1}}\dots X_{i_n}^{\alpha_n}= X_{s_1}^{\gamma_1}\dots X_{s_t}^{\gamma_t}X_{j_1}^{\beta_1-1}\dots X_{j_m}^{\beta_m},\end{align*}$$

which also leads to a contradiction. Above all, we complete the proof.

Free semigroups $\mathcal {F}_n$ ( $n\geq 2$ ) are neither left nor right amenable. Thompson’s semigroup is not right amenable and whether it is left amenable is still unknown. For completeness, we construct the following both left and right amenable semigroup.

Proposition 3.12 Let $\mathcal {T}$ be the semigroup generated by

$$ \begin{align*} A= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix},\ B= \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix},\ C= \begin{pmatrix} 0 & 2 \\ 3 & 0 \end{pmatrix} \end{align*} $$

in $GL_{2}(\mathbb {Z})$ . Then $\mathcal {T}$ is a non-commutative unique factorization semigroup. Moreover, $\mathcal {T}$ is both left and right amenable.

Proof It can be verified directly that $AB=BA$ , $AC=CB$ , $BC=CA$ . Hence, $\mathcal {T}$ is non-commutative and every element in $\mathcal {T}$ can be written as $A^{\alpha _1}B^{\alpha _2}C^{\alpha _3}$ for some natural numbers $\alpha _1$ , $\alpha _2$ , and $\alpha _3$ . Let X be a matrix in $\mathcal {T}$ with two different representations: $X=A^{\alpha _{1}}B^{\alpha _{2}}C^{\alpha _{3}}=A^{\beta _{1}}B^{\beta _{2}}C^{\beta _{3}}$ , where $\alpha _i$ and $\beta _i$ ( $1\leq i\leq 3$ ) are natural numbers. Taking the determinant at the both sides, we have $2^{\alpha _{1}}\dot 2^{\alpha _{2}}\dot (-6)^{\alpha _{3}}=2^{\beta _{1}}\dot 2^{\beta _{2}}\dot (-6)^{\beta _{3}}$ . Hence $\alpha _{3}=\beta _{3}$ and $\alpha _{1}+\alpha _{2}=\beta _{1}+\beta _{2}$ . Then we have $A^{\alpha _{1}-\beta _{1}}=B^{\beta _{2}-\alpha _{2}}$ , which implies $\alpha _{1}=\beta _{1}$ and $\alpha _{2}=\beta _{2}$ . Thus $\mathcal {T}$ is a unique factorization semigroup. For each positive integer N ( $\geq 2$ ), let $F_{N}=\{A^{\alpha _{1}}B^{\alpha _{2}}C^{\alpha _{3}}\in \mathcal {T}\ |\ 0\leq \alpha _{i}\leq N,\ i=1,2,3\}$ , we have

$$ \begin{align*} AF_{N}\cap F_{N}=\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{1}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=2,3\},\\ BF_{N}\cap F_{N}=\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{2}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=1,3\},\\ CF_{N}\cap F_{N}=\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{3}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=1,2\} \end{align*} $$

and $|AF_{N}\cap F_{N}|=|BF_{N}\cap F_{N}|=|CF_{N}\cap F_{N}|=N(N+1)^2$ . Then

$$\begin{align*}\lim_{N\rightarrow\infty}\frac{|AF_{N}\cap F_{N}|}{|F_{N}|}=\lim_{N\rightarrow\infty}\frac{|BF_{N}\cap F_{N}|}{|F_{N}|}=\lim_{N\rightarrow\infty}\frac{|CF_{N}\cap F_{N}|}{|F_{N}|}=\lim_{N\rightarrow\infty}\frac{N(N+1)^2}{(N+1)^{3}}=1.\end{align*}$$

Since A, B, and C are generators of $\mathcal {T}$ , $(F_N)_{N\in \mathbb {N}}$ is a left Følner sequence of $\mathcal {T}$ . Thus $\mathcal {T}$ is left amenable. On the other hand, we have

$$ \begin{align*} F_{N}A\cap F_{N}=&\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{1}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=2,3,\ \alpha_{3}\ \text{is even}\}\\ \cup&\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{2}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=1,3,\ \alpha_{3}\ \text{is odd}\},\\ F_{N}B\cap F_{N}=&\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{2}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=1,3,\ \alpha_{3}\ \text{is even}\}\\ \cup&\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{1}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=2,3,\ \alpha_{3}\ \text{is odd}\},\\ F_{N}C\cap F_{N}=&\{A^{\alpha_{1}}B^{\alpha_{2}}C^{\alpha_{3}}\in \mathcal{T}\ |\ 1\leq \alpha_{3}\leq N,\ 0\leq \alpha_{i}\leq N,\ i=1,2\}. \end{align*} $$

Similarly, $(F_N)_{N\in \mathbb {N}}$ is also a right Følner sequence of $\mathcal {T}$ . Thus $\mathcal {T}$ is both left and right amenable. We complete the proof.

4 The continuity of derivations

In this section, we will prove the following theorem.

Before proving Theorem 4.1, we introduce the following definition.

By Lemmas 3.7 and 3.10, Thompson’s semigroup $\mathcal {S}$ and free semigroups $\mathcal {F}_n$ ( $n\geq 2$ ) are both lower stable.

Definition 4.3 A Banach algebra $\mathcal {A}$ is said to be semisimple if its Jacobson radical $\mathcal {J}$ equals zero, where

$$\begin{align*}\mathcal{J}=\bigcap_{\text{maximal left ideals of } \mathcal{A}}\mathcal{I}_{l}= \bigcap_{\text{maximal right ideals of } \mathcal{A}}\mathcal{I}_{r}.\end{align*}$$

The following lemma is a characterization of semisimple Banach algebras.

Lemma 4.5 Let S be a lower stable discrete semigroup and $\mathfrak {B}(S)$ the Banach algebra generated by $\{L_s:\ s\in S\}$ in $B(l^2(S))$ . Then $\mathfrak {B}(S)$ is semisimple.

Proof Let $L_f$ be a nonzero element in $\mathfrak {B}(S)$ . We claim that the spectral radius $r(L_f)>0$ . By the definition of lower stable semigroup, there is a unique minimal element X of the subset $\{X\in S:f(X)\neq 0\}$ under the total order. Then we have

$$\begin{align*}\underbrace{f\ast\cdots \ast f}_{n}(X^{n})=f(X)^{n}\end{align*}$$

for each $n\geq 1$ . Therefore,

$$ \begin{align*} r(L_{f})=\lim_{n\rightarrow\infty}\|L_{f}^{n}\|^{1/n}&\geq \lim_{n\rightarrow\infty}\|\underbrace{f\ast\cdots \ast f}_{n}\|_{2}^{1/n}\\ &\geq \lim_{n\rightarrow\infty}|\underbrace{f\ast\cdots \ast f}_{n}(X^{n})|^{1/n}=|f(X)|>0. \end{align*} $$

By Lemma 4.4, we obtain that $\mathfrak {B}(S)$ is semisimple.

In the semigroup $\mathcal {T}$ , if $A \prec B$ , then $AC \prec BC = CA \prec CB = AC.$ Hence, we cannot have $A \prec B$ . Similarly, $B \prec A$ is not allowed. Therefore, the semigroup $\mathcal {T}$ in Example 3.12 is not lower stable. The Banach algebra $\mathfrak {B}(\mathcal {T})$ is also semisimple. The proof is similar to that of Lemma 4.5, we omit it here. Now, we prove Theorem 4.1.

5 The first cohomology group of $\mathfrak {B}(\mathcal {S})$

A derivation D on a Banach algebra $\mathfrak {B}\subseteq B(\mathcal {H})$ is said to be spatial if there exists a bounded operator T in $B(\mathcal {H})$ such that $D(A)=TA-AT$ for each A in $\mathfrak {B}$ . In [Reference Kadison18], Kadison proved that all derivations on a $C^{*}$ -algebra are spatial. In this section, we will prove that all derivations on the Banach algebra $\mathfrak {B}(\mathcal {S})$ are spatial and induced by bounded operators in $\mathcal {L}(\mathcal {S})$ . Moreover, we prove that the first continuous cohomology group of $\mathfrak {B}(\mathcal {S})$ with coefficients in $\mathcal {L}(\mathcal {S})$ is zero (see Theorem 5.8).

The Hilbert space $\mathcal {H}$ is $l^2(\mathcal {S})$ . Recall that $\mathfrak {B}(\mathcal {S})$ is the Banach algebra generated by $\{L_s:s\in \mathcal {S} \}$ in $B(\mathcal {H})$ and $\mathcal {L}(\mathcal {S})=\left \{L_{f}\in B(\mathcal {H}):f\in \mathcal {H}\right \}$ . We use $\sum _{X\in \mathcal {S}}f(X)X$ and $\sum _{X\in \mathcal {S}}\overline {f(X)}X^{*}$ to denote $L_{f}$ and its adjoint operator $L_f^*$ for convenience. It is clear that $XX^{*}$ is the projection from $\mathcal {H}$ onto the closure of the subspace span $\{\delta _Y:\ Y\in \mathcal {S},\ X|Y\}$ and $X^{*}X=I$ (the identity map). Recall that “ $|$ ” is the partial order of Thompson’s semigroup introduced in Section 3 and “ $X|Y$ ” means that there exists an element Z in $\mathcal {S}$ such that $Y=XZ$ .

Let $(\mathbb {N},+)$ be the additive semigroup of natural numbers. The notation $\beta \mathbb {N}$ denotes the maximal ideal space of the commutative $C^*$ -algebra $l^{\infty }(\mathbb {N})$ . The elements in $\beta \mathbb {N} \setminus \mathbb {N}$ are called free ultrafilters. Let $\omega \in \beta \mathbb {N} \setminus \mathbb {N}$ be a free ultrafilter. For any n in $\mathbb {N} $ and any f in $l^{\infty }(\mathbb {N})$ , we define $E_n(f) =\frac {1}{n} \sum _{i=0}^{n-1} f(i)$ . Then for each f, the function $n\mapsto E_n(f)$ is in $l^{\infty }(\mathbb {N})$ . By Gelfand–Naimark theorem, we have $l^{\infty }(\mathbb {N})\cong C(\beta \mathbb {N})$ . Thus $E_n(f)$ is a continuous function on $\beta \mathbb {N}$ . We use $E_{\omega }(f)$ or the integral $\int _{\mathbb {N}} f(n) dE_{\omega }(n)$ to denote the limit of $E_n(f)$ at $\omega $ . Then $E_{\omega }$ is an invariant mean on $l^{\infty }(\mathbb {N})$ , i.e.,

$$\begin{align*}\int_{\mathbb{N}} f(n) dE_{\omega}(n)=\int_{\mathbb{N}} f(n+m) dE_{\omega}(n)\end{align*}$$

for each $m\geq 1$ . Moreover, $E_{\omega }$ satisfies that $E_{\omega }(f)=\lim _{n\rightarrow \infty }f(n)$ if the limit exists. The invariant mean $E_{\omega }$ is also called a Banach limit.

Let D be a continuous derivation from the Banach algebra $\mathfrak {B}(\mathcal {S})$ into $\mathcal {L}(\mathcal {S})$ . For each $\xi ,\eta \in \mathcal {H}$ , we define

(5) $$ \begin{align} \langle A\xi,\eta\rangle:=\int_{\mathbb{N}} \langle (X_{0}^{*})^{n}D(X_{0}^{n+1})\xi,\eta \rangle dE_{\omega}(n). \end{align} $$

Then A is a bounded linear operator on $\mathcal {H}$ . Moreover, we have

$$ \begin{align*} \langle A\xi,\eta\rangle&=\int_{\mathbb{N}} \langle (X_{0}^{*})^{n+1}D(X_{0}^{n+2})\xi,\eta \rangle dE_{\omega}(n)\\ &=\int_{\mathbb{N}} \langle D(X_{0})\xi,\eta \rangle dE_{\omega}(n)+ \int_{\mathbb{N}} \langle (X_{0}^{*})^{n+1}D(X_{0}^{n+1})X_0\xi,\eta \rangle dE_{\omega}(n)\\ &=\langle D(X_{0})\xi,\eta \rangle+ \langle X_{0}^{*}AX_0\xi,\eta \rangle, \end{align*} $$

which follows that

(6) $$ \begin{align} D(X_{0})=A-X_{0}^{*}AX_0. \end{align} $$

Similarly, we define

(7) $$ \begin{align} \langle B\xi,\eta\rangle:=\int_{\mathbb{N}} \langle (X_{1}^{*})^{n}D(X_{1}^{n+1})\xi,\eta \rangle dE_{\omega}(n). \end{align} $$

We have $B \in B(\mathcal {H})$ and

(8) $$ \begin{align} D(X_{1})=B-X_{1}^{*}BX_1. \end{align} $$

5.1 The case of $X_0$

We prove the following local inner derivative property for $X_0$ in this subsection. A similar result for $X_1$ will be given later.

Lemma 5.1 There exists some $\widehat {A}$ in $\mathcal {L}(\mathcal {S})$ such that $D(X_0)=\widehat {A}X_0-X_0\widehat {A}$ .

Proof Let

$$\begin{align*}D(X_{0})=\sum_{X_{0}|X}f(X)X+\sum_{X_{0}\nmid X}f(X)X.\end{align*}$$

By the fact that $X_{1}X_{0}=X_{0}X_{2}$ , we have

$$\begin{align*}D(X_{1})X_{0}+X_{1}D(X_{0})=X_{0}D(X_{2})+D(X_{0})X_{2}.\end{align*}$$

It follows that

$$\begin{align*}X_{1}\sum_{X_{0}\nmid X}f(X)X=\sum_{X_{0}\nmid X}f(X)XX_{2}.\end{align*}$$

Since $X_{1}X\prec XX_{2}$ when $X_{0}\nmid X$ , hence $f(X)=0$ in this case. Therefore,

$$\begin{align*}D(X_0)=\sum_{X_{0}|X}f(X)X=X_{0}L_{f_1}\end{align*}$$

for some $L_{f_1}$ in $\mathcal {L}(\mathcal {S})$ . By the Leibniz rule and induction, we can obtain that

(9) $$ \begin{align} D(X_0^n)=X_0^nL_{f_n} \end{align} $$

for some $L_{f_n}$ in $\mathcal {L}(\mathcal {S})$ . We define

$$ \begin{align*} \langle \widehat{A}\xi,\eta\rangle&:=-\int_{\mathbb{N}} \langle (X_{0}^{*})^{n+1}D(X_{0}^{n+1})\xi,\eta \rangle dE_{\omega}(n). \end{align*} $$

Then $\widehat {A}\in B(\mathcal {H})$ . For each T in $\mathcal {R}(\mathcal {S})$ , we have

$$ \begin{align*} \langle T\widehat{A}\xi,\eta\rangle&=-\int_{\mathbb{N}} \langle (X_{0}^{*})^{n+1}D(X_{0}^{n+1})\xi,T^*\eta \rangle dE_{\omega}(n)\\ &=-\int_{\mathbb{N}} \langle (X_{0}^{*})^{n+1}D(X_{0}^{n+1})T\xi, \eta \rangle dE_{\omega}(n)\\ &=\langle \widehat{A}T\xi,\eta\rangle. \end{align*} $$

The second equality is due to equation (9). It follows that $\widehat {A}\in \mathcal {R}(\mathcal {S})'=\mathcal {L}(\mathcal {S})$ . By equation (5), we have

$$ \begin{align*} \langle A\xi,\eta\rangle&=\int_{\mathbb{N}} \langle (X_{0}^{*})^{n}D(X_{0}^{n+1})\xi,\eta \rangle dE_{\omega}(n)\\ &=\int_{\mathbb{N}} \langle (X_{0}^{*})^{n+1}D(X_{0}^{n+1})\xi,X_0^*\eta \rangle dE_{\omega}(n)\\ &= \langle -\widehat{A}\xi,X_0^*\eta \rangle, \end{align*} $$

which implies $A=-X_0\widehat {A}$ . The second equality is due to equation (9). Then by equation (6), we have $D(X_0)=\widehat {A}X_0-X_0\widehat {A}$ . This completes the proof.

5.2 The case of $X_1$

The following lemma is the main conclusion of this subsection.

Lemma 5.2 There exists some $\widehat {B}$ in $\mathcal {L}(\mathcal {S})$ such that $D(X_{1})=\widehat {B}X_{1}-X_{1}\widehat {B}$ .

We need Lemmas 5.3 and 5.4 to obtain Lemma 5.2.

Lemma 5.3 There exists some B in $\mathcal {L}(\mathcal {S})$ such that $D(X_{1})=B-X_{1}^{*}BX_{1}$ .

Proof Let $D(X_{1})=L_{f}$ in $\mathcal {L}(\mathcal {S})$ . By the continuity of D, we have $f(X_{1}^{n})=0$ for each $n\in \mathbb {N}$ . For each $m\geq 1$ , we have

$$\begin{align*}(X_{1}^{*})^{m-1}D(X_{1}^{m})=\sum_{i=0}^{m-1}(X_1^*)^iL_fX_1^{i}\end{align*}$$

and

$$\begin{align*}\|(X_{1}^{*})^{m-1}D(X_{1}^{m})\|\leq\|D\|.\end{align*}$$

Let $h_{1}$ and $ h_{2}$ be two elements in $\mathcal {S}$ . If $h_{2}h_{1}^{-1}= X_{1}^{k}$ for some $k\in \mathbb {N}$ , then

$$\begin{align*}\lim_{n,m\rightarrow\infty}\left\langle\bigg((X_{1}^{*})^{m-1}D(X_{1}^{m})-(X_{1}^{*})^{n-1}D(X_{1}^{n})\bigg)\delta_{h_{1}},\delta_{h_{2}}\right\rangle=\lim_{n,m\rightarrow\infty}\sum_{i=n+1}^{m}f(X_{1}^{i}h_{2}h_{1}^{-1}X_{1}^{-i})=0.\end{align*}$$

If $h_{2}h_{1}^{-1}\neq X_{1}^{k}$ for any $k\in \mathbb {N}$ , then $X_{1}^{i}h_{2}h_{1}^{-1}X_{1}^{-i}\notin \mathcal {S}$ when i is sufficiently large. Therefore,

$$\begin{align*}\lim_{n,m\rightarrow\infty}\left\langle\bigg((X_{1}^{*})^{m-1}D(X_{1}^{m})-(X_{1}^{*})^{n-1}D(X_{1}^{n})\bigg)\delta_{h_{1}},\delta_{h_{2}}\right\rangle=\lim_{n,m\rightarrow\infty}\sum_{i=n+1}^{m}f(X_{1}^{i}h_{2}h_{1}^{-1}X_{1}^{-i})=0.\end{align*}$$

We define

(10) $$ \begin{align} \langle T\delta_{h_1},\delta_{h_2}\rangle:= \lim_{m\rightarrow\infty}\big\langle(X_{1}^{*})^{m-1}D(X_{1}^{m})\delta_{h_{1}},\delta_{h_{2}}\big\rangle. \end{align} $$

Then T is the weak-operator limit of $(X_{1}^{*})^{m-1}D(X_{1}^{m})$ in $B(\mathcal {H})$ .

Claim $T=L_{T\delta _{e}}\in \mathcal {L}(\mathcal {S})$ .

To prove this claim, we distinguish two cases:

Case I: $h_{2}h_{1}^{-1}\in \mathcal {S}$ . We have

$$\begin{align*}\langle T\delta_{h_{1}},\delta_{h_{2}}\rangle=\sum_{n=0}^{\infty}f(X_{1}^{n}h_{2}h_{1}^{-1}X_{1}^{-n})=\langle T\delta_{e},\delta_{h_{2}h_{1}^{-1}}\rangle=\langle T\delta_{e}\ast\delta_{h_{1}},\delta_{h_{2}}\rangle.\end{align*}$$

Case II: $h_{2}h_{1}^{-1}\notin \mathcal {S}$ . If $X_{1}^{n}h_{2}h_{1}^{-1}X_{1}^{-n}\notin \mathcal {S}$ for any $n\in \mathbb {N}$ , then

$$\begin{align*}\langle T\delta_{h_{1}},\delta_{h_{2}}\rangle=\langle T\delta_{e}\ast\delta_{h_{1}},\delta_{h_{2}}\rangle=0.\end{align*}$$

On the other hand, there exists a natural number n such that $X_{1}^{n+1}h_{2}h_{1}^{-1}X_{1}^{-n-1}\in \mathcal {S}$ and $X_{1}^{i}h_{2}h_{1}^{-1}X_{1}^{-i}\notin \mathcal {S}$ whenever $i\leq n$ . Let $X=X_{1}^{n+1}h_{2}h_{1}^{-1}X_{1}^{-n-1}$ , then $X_{0}|X$ . By Lemma 3.5, we have that $X_{1}^{-k}XX_{1}^{k}\notin \mathcal {S}$ for any $k\geq 1$ and $X_{1}^{k}XX_{1}^{-k}\notin \mathcal {S}$ when $k>\mathsf {ind}(X)$ . Let m be an even integer such that m is sufficiently lager than $\mathsf {ind}(X)$ . Then we have

$$ \begin{align*} \sum_{i=1}^{m}\left|\langle D(X_1^{2m})\delta_e,\delta_{X_1^{2m-i}XX_1^{i-1}}\rangle \right|^2= \sum_{i=1}^{m}\left|\sum_{j=-(i-1)}^{2m-i}f(X_1^{j}XX_1^{-j})\right|^2= \sum_{i=1}^{m}\left|\sum_{j=0}^{\mathsf{ind}(X)}f(X_1^{j}XX_1^{-j})\right|^2. \end{align*} $$

Note that $X_{1}^{2m-i}XX_{1}^{i-1}\neq X_{1}^{2m-j}XX_{1}^{j-1}$ when $1\leq i< j\leq m$ , then we have

$$\begin{align*}\sum_{i=1}^{m}\left|\sum_{j=0}^{\mathsf{ind}(X)}f(X_1^{j}XX_1^{-j})\right|^2 \leq \left\|D(X_{1}^{2m})\delta_{e}\right\|_{2}^{2} \leq\left\|D\right\|^{2} \end{align*}$$

for any sufficiently large even integer m. This implies that

$$\begin{align*}\sum_{j=0}^{\mathsf{ind}(X)}f(X_1^{j}XX_1^{-j})=0.\end{align*}$$

Therefore,

$$\begin{align*}\langle T\delta_{h_{1}},\delta_{h_{2}}\rangle=\sum_{j=0}^{\infty}f(X_{1}^{j}h_{2}h_{1}^{-1}X_{1}^{-j}) =\sum_{j=0}^{\mathsf{ind}(X)}f(X_{1}^{j}XX_{1}^{-j})=0.\end{align*}$$

From the above discussion, we have $\langle T\delta _{h_{1}},\delta _{h_{2}}\rangle =\langle T\delta _{e}\ast \delta _{h_{1}},\delta _{h_{2}}\rangle $ for all $h_{1}$ and $h_{2}$ in $\mathcal {S}$ . Thus the claim holds.

By equations (7) and (10), we have

$$ \begin{align*} \langle T\delta_{h_1},\delta_{h_2}\rangle&= \lim_{m\rightarrow\infty}\big\langle(X_{1}^{*})^{m-1}D(X_{1}^{m})\delta_{h_{1}},\delta_{h_{2}}\big\rangle\\ &=\int_{\mathbb{N}}\big\langle(X_{1}^{*})^{m-1}D(X_{1}^{m})\delta_{h_{1}},\delta_{h_{2}}\big\rangle dE_{\omega}(m)\\ &= \langle B\delta_{h_1},\delta_{h_2}\rangle, \end{align*} $$

which follows that $B=T$ . This completes the proof.

Lemma 5.4 Let $D(X_{1})=L_{f}\in \mathcal {L}(\mathcal {S})$ . Then

$$\begin{align*}D(X_{1})=\sum_{\ X_{0}|X}f(X)X+\sum_{ X_{0}\nmid X,X_{1}|X}f(X)X.\end{align*}$$

Proof Let

$$\begin{align*}D(X_{1})=\sum_{X_{0}|X}f(X)X+\sum_{X_{0}\nmid X,X_{1}|X}f(X)X+\sum_{X_{0}\nmid X,X_{1}\nmid X}f(X)X.\end{align*}$$

By the fact that $X_{1}X_{3}=X_{2}X_{1}$ , we have

$$\begin{align*}D(X_{1})X_{3}+X_{1}D(X_{3})=D(X_{2})X_{1}+X_{2}D(X_{1}).\end{align*}$$

Then

$$\begin{align*}\sum_{X_{0}\nmid X,X_{1}\nmid X}f(X)XX_{3}=X_{2}\sum_{X_{0}\nmid X,X_{1}\nmid X}f(X)X.\end{align*}$$

Let X be the minimal element of the set $\{X\in \mathcal {S}:\ X_{0}\nmid X,\ X_{1}\nmid X,\ f(X)\neq 0\}$ . Then $X_2X\prec XX_3$ . It follows that $f(X)=0$ when $X_{0}\nmid X$ and $X_{1}\nmid X$ . This completes the proof.

Now, we are ready to prove Lemma 5.2.

Proof of Lemma 5.2

By Lemma 5.3, we have $D(X_{1})=B-X_{1}^{*}BX_{1}$ for some $B=L_{g}\in \mathcal {L}(\mathcal {S})$ . We assume that $g(e)=0$ , where e is the unit element of $\mathcal {S}$ . Let

$$\begin{align*}L_{g}=\sum_{X_{1}|X}g(X)X+\sum_{X_{1}\nmid X,X_{0}|X}g(X)X+\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)X.\end{align*}$$

Then

$$ \begin{align*} D(X_{1})=&\sum_{X_{1}|X}g(X)X-X_{1}^{*}\sum_{X_{1}|X}g(X)XX_{1}+\!\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)X-X_{1}^{*}\!\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)XX_{1} \end{align*} $$

$$ \begin{align*} +&\sum_{X_{1}\nmid X,X_{0}|X}g(X)X-X_{1}^{*}\sum_{X_{1}\nmid X,X_{0}|X}g(X)XX_{1}. \end{align*} $$

Since $D(X_{1})\in \mathcal {L}(\mathcal {S})$ and $X_1^{-1}XX_1\notin \mathcal {S}$ when $X_{1}\nmid X$ and $X_{0}|X$ , we have

$$ \begin{align*}X_1^*\sum_{X_{1}\nmid X,X_{0}|X}g(X)XX_1=0.\end{align*} $$

For $X\in \mathcal {S}$ with $X_{1}\nmid X$ and $X_{0}|X$ , let $X_1^{-1}XX_1=UV^{-1}$ , where $U,V\in \mathcal {S}$ . Then

$$ \begin{align*} g(X)=\left\langle X_1^*\sum_{X_{1}\nmid X,X_{0}|X}g(X)XX_1\delta_V,\delta_U \right\rangle=0. \end{align*} $$

It follows that

$$\begin{align*}\sum_{X_{1}\nmid X,X_{0}|X}g(X)X=0.\end{align*}$$

Therefore,

$$\begin{align*}D(X_{1})=\sum_{X_{1}|X}g(X)X-X_{1}^{*}\!\sum_{X_{1}|X}g(X)XX_{1}+\!\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)X-X_{1}^{*}\!\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)XX_{1}.\end{align*}$$

By Lemma 5.4, we have

$$\begin{align*}\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)X-X_{1}^{*}\sum_{X_{1}\nmid X,X_{0}\nmid X}g(X)XX_{1}=0.\end{align*}$$

Let X be the minimal element of the set $\{X\in \mathcal {S}:\ X_{1}\nmid X,\ X_{0}\nmid X,\ g(X)\neq 0\}$ . Then $X\prec X_1^{-1}XX_1$ since $X\neq e$ . It follows that $g(X)=0$ when $X_{1}\nmid X$ and $X_{0}\nmid X$ . Thus $B=\sum _{X_{1}|X}g(X)X$ . Let $\widehat {B}=-X_{1}^{*}B$ , then $\widehat {B}\in \mathcal {L}(S)$ and

$$\begin{align*}D(X_{1})=B-X_{1}^{*}BX_{1}=X_{1}X_{1}^{*}B-X_{1}^{*}BX_{1}=\widehat{B} X_{1}-X_{1}\widehat{B}.\end{align*}$$

We complete the proof.

5.3 Conditional expectation

Definition 5.5 Let $\mathfrak {B}$ be a Banach algebra, and let $\mathcal {A}$ be a Banach subalgebra of $\mathfrak {B}$ . Let E: $\mathfrak {B}\rightarrow \mathcal {A}$ be a contraction such that:

(i) $E(I)=I$ ;

( $\mathrm{ii}$ ) $E(A_1BA_2)=A_1E(B)A_2$ whenever $A_1$ , $A_2$ $\in \mathcal {A}$ and B $\in \mathfrak {B}$ .

Then E is described as a conditional expectation from $\mathfrak {B}$ onto $\mathcal {A}$ .

Definition 5.6 We denote by $\mathcal {L}_0(\mathcal {S})$ the subset of $\mathcal {L}(\mathcal {S})$ such that for each $L_f$ in $\mathcal {L}_0(\mathcal {S})$ , $f(X)=0$ if X is not in the semigroup generated by $X_0$ .

It is not difficult to check that $\mathcal {L}_0(\mathcal {S})$ is a Banach subalgebra of $\mathcal {L}(\mathcal {S})$ . We have the following theorem.

Theorem 5.7 Let E be the map: $\mathcal {L}(\mathcal {S})$ $\rightarrow $ $\mathcal {L}_0(\mathcal {S})$ , $\sum _{X\in \mathcal {S}} f(X)X$ $\mapsto $ $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ . Then E is a well-defined conditional expectation from $\mathcal {L}(\mathcal {S})$ onto $\mathcal {L}_0(\mathcal {S})$ .

Proof For each $g\in l^2(\mathcal {S})$ such that $g(X)=0$ when $X\neq X_0^n$ , we have

$$ \begin{align*} \bigg\|\sum_{n=0}^{\infty} f(X_0^n)X_0^n \sum_{n=0}^{\infty}g(X_0^n)\delta_{X_0^n} \bigg\|_2^2\leq \|L_fg\|_2^2\leq \|L_f\|^2\|g\|_2^2. \end{align*} $$

This implies that $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ is bounded on the Hilbert subspace $l^2(\mathcal {S}_0)$ , where $\mathcal {S}_0$ is the subsemigroup of $\mathcal {S}$ generated by $X_0$ . The norm of $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ is bounded by $\|L_f\|$ . Let $\mathscr {F}$ be the Fourier transform: $\mathbb {Z}\rightarrow \mathbb {S}^1$ , $n\mapsto e^{2\pi i n\theta }$ , $\theta \in [0,1)$ . This induces the following isomorphisms [Reference Ge and Xi8]:

$$ \begin{align*} &\mathbb{Z}\quad \subseteq\quad\ \ \mathbb{C}\mathbb{Z}\quad\quad \subseteq \quad l^1(\mathbb{Z})\ \quad \subseteq \quad C^*(\mathbb{Z})\quad \subseteq \quad\ \mathcal{L}(\mathbb{Z})\quad\ \ \subseteq\quad l^2(\mathbb{Z})\quad\ \subseteq\quad \cdots\\ &\updownarrow \quad\quad\quad\quad\ \updownarrow \quad\quad\quad\quad\quad\ \updownarrow \quad\quad\quad\quad\quad\quad \updownarrow \quad\quad\quad\quad\quad\ \updownarrow \quad\quad\quad\quad\quad\ \ \updownarrow \\ &\mathbb{S}^1\quad \subseteq\quad \mathbb{C}[\mathbb{Z}]_{\mathbb{S}^1}\quad \subseteq \quad R(\mathbb{S}^1)\quad \subseteq \quad C(\mathbb{S}^1)\quad \subseteq \quad L^{\infty}(\mathbb{S}^1)\quad \subseteq\quad l^2(\mathbb{S}^1)\quad \subseteq\quad \cdots \end{align*} $$

Restricting the Fourier transform on $\mathbb {N}$ , we have

$$ \begin{align*} &\mathbb{C}\mathbb{N}\quad\quad \subseteq \quad\ l^1(\mathbb{N})\quad\ \subseteq \quad B(\mathbb{N})\quad\ \subseteq \quad\ \ \mathcal{L}(\mathbb{N})\quad\ \subseteq\quad\ l^2(\mathbb{N})\quad\ \subseteq\quad \cdots\\ &\ \updownarrow \quad\quad\quad\quad\quad\quad \updownarrow \quad\quad\quad\quad\quad\ \updownarrow \quad\quad\quad\quad\quad\quad \updownarrow \quad\quad\quad\quad\quad\ \ \updownarrow \\ &\mathbb{C}[\mathbb{Z}]_{\mathbb{D}}\quad \subseteq \quad H^1(\mathbb{D})\quad \subseteq \quad H_c(\mathbb{D})\quad \subseteq \quad H^{\infty}(\mathbb{D})\quad \subseteq\quad H^2(\mathbb{D})\quad \subseteq\quad \cdots \end{align*} $$

Since $\mathcal {S}_0$ is isomorphic to ( $\mathbb {N},+$ ), $\sum _{n=0}^{\infty } f(X_0^n)\delta _{n}$ belongs to $\mathcal {L}(\mathbb {N})$ and $\mathscr {F}(\sum _{n=0}^{\infty } f(X_0^n)\delta _{n})$ is in $H^{\infty }(\mathbb {D})$ . It follows that $\mathscr {F}(\sum _{n=0}^{\infty } f(X_0^n)\delta _{n})$ belongs to $L^{\infty }(\mathbb {S}^1)$ . Thus $\sum _{n=0}^{\infty } f(X_0^n)\delta _{n}$ is a bounded operator on $l^2(\mathbb {Z})$ and belongs to $\mathcal {L}(\mathbb {Z})$ . Since the subgroup $H=\left \langle X_0\right \rangle $ of Thompson’s group $\mathcal {F}$ is isomorphic to $\mathbb {Z}$ , we have that $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ is bounded on $l^2(H)$ . Let $\{H_n\}_{n=1}^{\infty }$ be all right cosets of H in Thompson’s group $\mathcal {F}$ such that $\mathcal {F}=\sqcup H_n$ . Then

$$ \begin{align*} l^2(\mathcal{F})=\bigoplus_{n=1}^{\infty} l^2(H_n). \end{align*} $$

Moreover, $l^2(H_n)$ are invariant subspaces of $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ and the operator norms on each subspace are same. Hence $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ is bounded on $l^2(\mathcal {F})$ . Thus $\sum _{n=0}^{\infty } f(X_0^n)X_0^n$ is in $\mathcal {L}_0(\mathcal {S})$ . Then E is well-defined. It is clear that conditions (i) and ( $ii$ ) in Definition 5.5 hold for E. We complete the proof.

5.4 Proof of main results

In this subsection, we will prove the main results of this paper (see Theorems 5.8 and 5.10). When the $\mathfrak {B}(\mathcal {S})$ -bimodule is $\mathcal {L}(\mathcal {S})$ , the first continuous cohomology group of $\mathfrak {B}(\mathcal {S})$ vanishes.

Theorem 5.8 The first continuous cohomology group $H^1(\mathfrak {B}(\mathcal {S}),\mathcal {L}(\mathcal {S}))=0$ .

The following result is an immediate corollary of Theorems 4.1 and 5.8.

Corollary 5.9 Derivations on the Banach algebra $\mathfrak {B}(\mathcal {S})$ are spatial and induced by operators in $\mathcal {L}(\mathcal {S})$ .

Let $\mathcal {M}(\mathcal {S})$ be the set of all $T\in \mathcal {L}(\mathcal {S})$ such that $[T,A]\in \mathfrak {B}(\mathcal {S})$ for each $A\in \mathfrak {B}(\mathcal {S})$ , where $[T,A]=TA-AT$ . Then $\mathcal {M}(\mathcal {S})$ is a norm closed linear subspace of $\mathcal {L}(\mathcal {S})$ containing $\mathfrak {B}(\mathcal {S})$ . The first cohomology group of $\mathfrak {B}(\mathcal {S})$ is characterized as the linear space $\mathcal {M}(\mathcal {S})$ module $\mathfrak {B}(\mathcal {S})$ .

Theorem 5.10 The first continuous cohomology group $H^{1}(\mathfrak {B}(\mathcal {S}),\mathfrak {B}(\mathcal {S}))=\dfrac {\mathcal {M}(\mathcal {S})}{\mathfrak {B}(\mathcal {S})}$ .

The following two lemmas are crucial to obtain the above results.

Lemma 5.11 Let D be a continuous derivation from the Banach algebra $\mathfrak {B}(\mathcal {S})$ into $\mathcal {L}(\mathcal {S})$ . If $D(X_0)=AX_{0}-X_{0}A$ and $D(X_{1})=AX_{1}-X_{1}A$ for some A in $B(l^{2}(\mathcal {S}))$ , then $D(T)=AT-TA$ for each T in $\mathfrak {B}(\mathcal {S})$ .

Proof For each $n\geq 1$ , if $D(X_{0}^{n})=AX_{0}^{n}-X_{0}^{n}A$ , then

$$\begin{align*}D(X_{0}^{n+1})=X_{0}D(X_{0}^{n})+D(X_{0})X_{0}^{n}=AX_{0}^{n+1}-X_{0}^{n+1}A.\end{align*}$$

Therefore, by induction, we have $D(X_{0}^{n})=AX_{0}^{n}-X_{0}^{n}A$ for any $n\geq 1$ . By the definition of Thompson’s semigroup, we have $X_{1}X_{0}^{m}=X_{0}^{m}X_{m+1}$ for any $m\geq 1$ . Then

$$\begin{align*}D(X_{1})X_{0}^{m}+X_{1}D(X_{0}^{m})=D(X_{0}^{m})X_{m+1}+X_{0}^{m}D(X_{m+1}).\end{align*}$$

We have

$$ \begin{align*} D(X_{m+1})&=X_0^{*m}(D(X_{1})X_{0}^{m}+X_{1}D(X_{0}^{m})-D(X_{0}^{m})X_{m+1})\\ &=X_0^{*m}(AX_1X_{0}^{m}-X_1AX_{0}^{m}+ X_1AX_0^m-X_1X_0^mA-AX_0^mX_{m+1}+ X_0^mAX_{m+1})\\ &=X_0^{*m}(X_0^mAX_{m+1}- X_1X_0^mA)\\ &=AX_{m+1}-X_{m+1}A. \end{align*} $$

Similarly, we can prove that $D(X)=AX-XA$ for any X in $\mathcal {S}$ . By linearity, we have $D(T)=AT-TA$ for each T in the semigroup algebra $\mathbb {C}[\mathcal {S}]$ . Since $\mathbb {C}[\mathcal {S}]$ is a dense subalgebra of $\mathfrak {B}(\mathcal {S})$ , we obtain that $D(T)=AT-TA$ for each T in $\mathfrak {B}(\mathcal {S})$ by the continuity of D.

The following lemma is a generalization of the above conclusion.

Lemma 5.12 Let D be a continuous derivation from the Banach algebra $\mathfrak {B}(\mathcal {S})$ into $\mathcal {L}(\mathcal {S})$ . If $D(X_0)=AX_{0}-X_{0}A$ and $D(X_{1})=BX_{1}-X_{1}B$ for some A and B in $\mathcal {L}(\mathcal {S})$ , then there exists an operator C in $\mathcal {L}(\mathcal {S})$ such that $D(T)=CT-TC$ for each T in $\mathfrak {B}(\mathcal {S})$ .

Proof By the definitions of Thompson’s semigroup and derivation, we have

$$\begin{align*}X_{1}X_{0}=X_{0}X_{2},\ X_{2}X_{0}=X_{0}X_{3},\ X_{2}X_{1}=X_{1}X_{3}\end{align*}$$

and

$$ \begin{align*} D(X_{1})X_{0}+X_{1}D(X_{0})&=D(X_{0})X_{2}+X_{0}D(X_{2}),\\ D(X_{2})X_{0}+X_{2}D(X_{0})&=D(X_{0})X_{3}+X_{0}D(X_{3}),\\ D(X_{2})X_{1}+X_{2}D(X_{1})&=D(X_{1})X_{3}+X_{1}D(X_{3}). \end{align*} $$

By the above equations, we have

$$ \begin{align*} D(X_{2})&=X_{0}^{*}X_{1}(A-B)X_{0}+X_{0}^{*}(B-A)X_{1}X_{0}+(AX_{2}-X_{2}A),\\D(X_{3})&=(X_{0}^{*})^{2}X_{1}(A-B)X_{0}^{2}+(X_{0}^{*})^{2}(B-A)X_{1}X_{0}^{2}+(AX_{3}-X_{3}A),\\D(X_{3})&=X_{1}^{*}X_{0}^{*}X_{1}(A-B)X_{0}X_{1}+X_{1}^{*}X_{0}^{*}(B-A)X_{1}X_{0}X_{1} +X_{1}^{*}(A-B)X_{2}X_{1}\\& \quad +X_{1}^{*}X_{2}(B-A)X_{1}+(BX_{3}-X_{3}B). \end{align*} $$

It follows that

$$ \begin{align*} &(X_{0}^{*})^{2}X_{1}(A-B)X_{0}^{2}+(X_{0}^{*})^{2}(B-A)X_{1}X_{0}^{2}+(A-B)X_{3}+X_{3}(B-A)\\& \quad = X_{1}^{*}X_{0}^{*}X_{1}(A-B)X_{0}X_{1}+X_{1}^{*}X_{0}^{*}(B-A)X_{1}X_{0}X_{1}+X_{1}^{*}(A-B)X_{2}X_{1}\\& \qquad +X_{1}^{*}X_{2}(B-A)X_{1}. \end{align*} $$

Let $A-B=L_{f}\in \mathcal {L}(\mathcal {S})$ and $L_{g}=L_{f}-f(e)I-\sum _{n\geq 0}\sum _{m\geq 1}f(X_{n}^{m})X_{n}^{m}$ . Then we have

(11) $$ \begin{align} &(X_{0}^{*})^{2}X_{1}L_{g}X_{0}^{2}-(X_{0}^{*})^{2}L_{g}X_{1}X_{0}^{2}+L_{g}X_{3} - X_{3}L_{g}=X_{1}^{*}X_{0}^{*}X_{1}L_{g}X_{0}X_{1}-X_{1}^{*}X_{0}^{*}L_{g}X_{1}X_{0}X_{1}\nonumber\\& \quad +X_{1}^{*}L_{g}X_{2}X_{1}-X_{1}^{*}X_{2}L_{g}X_{1}+\left(\sum_{n\geq2}\sum_{m\geq1} f(X_{n}^{m})X_{n+1}^{m}\right)X_{3}- X_{3}\left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n+1}^{m}\right)\nonumber \\& \quad +X_{3}\left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n}^{m}\right)-\left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n}^{m}\right)X_{3}. \end{align} $$

If $g\neq 0$ , then we take the minimal element X of the set $\{X\in \mathcal {S}\ |\ g(X)\neq 0\}$ under the total order. By the definition of $L_{g}$ , we have $X=X_{i_{1}}^{\alpha _{i_{1}}}\dots X_{i_{t}}^{\alpha _{i_{t}}}$ , where $i_{1}<i_{2}<\cdots <i_{t}$ and $\alpha _{i_{1}},\ldots ,\alpha _{i_{t}}\geq 1$ , $t\geq 2$ .

Case I: $i_{1}\geq 3$ . We have $X_{3}X\prec XX_{3}$ . Taking $\langle \cdot \ \delta _{e},\delta _{X_{3}X}\rangle $ on the both sides of equation (11), we obtain that $g(X)=0$ .

Case II: $i_{1}=1$ or $2$ . We have $ XX_{3}\prec X_{3}X$ . Analogously, taking $\langle \cdot \ \delta _{e},\delta _{XX_{3}}\rangle $ on the both sides of equation (11), we also have $g(X)=0$ .

Case III: $i_{1}=0$ . If $X_1|XX_{2}X_{1}$ , then taking $\langle \cdot \ \delta _{e},\delta _{X_{1}^{-1}XX_{2}X_{1}}\rangle $ , we can obtain that $g(X)=0$ . If $X_1\nmid XX_{2}X_{1}$ , by the normal form (of elements) in Thompson”s group $\mathcal {F}$ , there exist Y and $X_{k}$ in $\mathcal {S}$ such that $X_{1}^{-1}XX_{2}X_{1}=YX_{k}^{-1}$ , where $X_{1}^{-1}$ and $X_{k}^{-1}$ are in $\mathcal {F}$ . Then taking $\langle \cdot \ \delta _{X_{k}},\delta _{Y}\rangle $ , we can also obtain that $g(X)=0$ .

It follows from the above discussion that $g=0$ . This leads to

$$ \begin{align*} \left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n+1}^{m}\right)X_{3}-&X_{3}\left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n+1}^{m}\right)\\ +&X_{3}\left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n}^{m}\right)-\left(\sum_{n\geq2}\sum_{m\geq1}f(X_{n}^{m})X_{n}^{m}\right)X_{3}\\ =&\ 0. \end{align*} $$

It is not difficult to verify that $f(X_2^m)=0$ and $f(X_n^m)=f(X_3^m)$ for all $n \geq 4$ and $m\geq 1$ . Since $f\in l^2(\mathcal {S})$ , we have $f(X_{n}^{m})=0$ for any $n\geq 2$ and $m\geq 1$ . Therefore,

$$\begin{align*}A-B=\sum_{m=1}^{\infty}f(X_{0}^{m})X_{0}^{m}+\sum_{m=1}^{\infty}f(X_{1}^{m})X_{1}^{m}+f(e)I.\end{align*}$$

Let

$$\begin{align*}C=A-\sum_{m=1}^{\infty}f(X_{0}^{m})X_{0}^{m}=B+\sum_{m=1}^{\infty}f(X_{1}^{m})X_{1}^{m}+f(e)I.\end{align*}$$

Then by Theorem 5.7, we have $C\in \mathcal {L}(\mathcal {S})$ and

$$\begin{align*}D(X_{0})=CX_{0}-X_{0}C,\quad D(X_{1})=CX_{1}-X_{1}C.\end{align*}$$

By Lemma 5.11, we have $D(T)=CT-TC$ for each T in $\mathfrak {B}(\mathcal {S})$ .

Proof of Theorem 5.8

Let D be a continuous derivation from $\mathfrak {B}(\mathcal {S})$ into $\mathcal {L}(\mathcal {S})$ . By Lemma 5.1, Lemma 5.2 and Lemma 5.12, there exits an operator C in $\mathcal {L}(S)$ such that $D(T)=CT-TC$ for each T in $\mathfrak {B}(\mathcal {S})$ . Thus D is an inner derivation and then $H^1(\mathfrak {B}(\mathcal {S}),\mathcal {L}(\mathcal {S}))=0$ . We complete the proof.

Let us see the following lemma before proving Theorem 5.10.

Lemma 5.13 The intersection $\mathcal {L}(\mathcal {S})\cap \mathcal {R}(\mathcal {S})=\mathbb {C}I$ .

Proof For any $L_f\in \mathcal {L}(\mathcal {S})\cap \mathcal {R}(\mathcal {S}) $ , there exists an operator $R_g\in \mathcal {R}(\mathcal {S}) $ such that $L_f=R_g$ . Hence $f=L_f\delta _e=R_g\delta _e=g$ . Moreover,

$$ \begin{align*} f\ast \delta_{X_0}=L_f\delta_{X_0}=R_f\delta_{X_0}=\delta_{X_0}\ast f. \end{align*} $$

It is not hard to check that $f=\sum _{n=0}^{\infty }f(X_0^n)\delta _{X_0^n}$ using Lemma 3.7. Similarly,

$$ \begin{align*} f\ast \delta_{X_1}=\delta_{X_1}\ast f. \end{align*} $$

Hence $f=f(e)\delta _e$ . This completes the proof.

Proof of Theorem 5.10

Let D be a derivation on the Banach algebra $\mathfrak {B}(\mathcal {S})$ then D is continuous. By Lemmas 5.1, 5.2, and 5.12, there exists an operator C in $\mathcal {L}(\mathcal {S})$ such that $D(T)=CT-TC$ for each T in $\mathfrak {B}(\mathcal {S})$ . This induces the following map:

$$ \begin{align*} \pi:\quad H^{1}(\mathfrak{B}(\mathcal{S}),\mathfrak{B}(\mathcal{S}))\quad&\longrightarrow\qquad \frac{ \mathcal{M}(\mathcal{S})}{\mathfrak{B}(\mathcal{S})}\\ \overline{D}\qquad\qquad&\longrightarrow\quad\ C+\mathfrak{B}(\mathcal{S}). \end{align*} $$

We will show that $\pi $ is well-defined. If there exist two operators $C_1$ and $C_2$ in $\mathcal {L}(\mathcal {S})$ such that $D(T)=C_1T-TC_1=C_2T-TC_2$ for each T in $\mathfrak {B}(\mathcal {S})$ , then $C_1-C_2\in \mathcal {L}(\mathcal {S}) \cap \mathfrak {B}(\mathcal {S})'=\mathcal {L}(\mathcal {S}) \cap \mathcal {R}(\mathcal {S})=\mathbb {C}I$ from Lemma 5.13. Thus $C_1+\mathfrak {B}(\mathcal {S})=C_2+\mathfrak {B}(\mathcal {S})$ . Now, if $\overline {D_1}=\overline {D_2}$ , then $D_1-D_2$ is an inner derivation of $\mathfrak {B}(\mathcal {S})$ . There exists an operator $C_3$ in $\mathfrak {B}(\mathcal {S})$ such that $(D_1-D_2)(T)=C_3T-TC_3$ for each T in $\mathfrak {B}(\mathcal {S})$ . Assume that $D_1(T)=C_1^{\prime }T-TC_1^{\prime }$ and $D_2(T)=C_2^{\prime }T-TC_2^{\prime }$ , where $C_1^{\prime }$ and $C_2^{\prime }$ are in $\mathcal {L}(\mathcal {S})$ , then $(C_1^{\prime }-C_2^{\prime })T-T(C_1^{\prime }-C_2^{\prime })=C_3T-TC_3$ . It follows that $C_1^{\prime }-C_2^{\prime }-C_3$ belongs to $\mathcal {L}(\mathcal {S}) \cap \mathfrak {B}(\mathcal {S})'=\mathbb {C}I$ . Thus $C_1^{\prime }-C_2^{\prime }$ belongs to $ \mathfrak {B}(\mathcal {S})$ , that is $C_1^{\prime }+\mathfrak {B}(\mathcal {S})=C_2^{\prime }+\mathfrak {B}(\mathcal {S})$ . The map $\pi $ is a well-defined group homomorphism. If $\pi (\overline {D})=0$ , then there exists an operator $C'$ in $\mathfrak {B}(\mathcal {S})$ such that $D(T)=C'T-TC'$ , which means that D is an inner derivation. The map $\pi $ is injective. The surjectivity of $\pi $ is obvious. It follows from the above discussion that $\pi $ is a group isomorphism. We complete the proof.

6 Further discussions

We now recall the definition of higher order continuous Hochschild cohomology for Banach algebras. Let $\mathcal {M}$ be a Banach algebra, and let $\mathcal {X}$ be a Banach $\mathcal {M}$ -bimodule. The space of all n-linear (continuous) maps from n-fold Cartesian product $\mathcal {M}^n=\mathcal {M}\times \cdots \times \mathcal {M}$ into $\mathcal {X}$ is denoted by $L^{n}(\mathcal {M},\mathcal {X})$ for $n\geq 1$ , while $L^{0}(\mathcal {M},\mathcal {X})$ is defined to be $\mathcal {X}$ .

The coboundary operator $\partial ^n$ : $L^{n}(\mathcal {M},\mathcal {X})\rightarrow L^{n+1}(\mathcal {M},\mathcal {X})$ is defined, for $n\geq 1$ , by

$$ \begin{align*} \partial^n\phi(a_{1},a_{2},\ldots,a_{n+1})&=a_{1}\phi(a_{2},\ldots,a_{n+1})\\ &\quad+\sum_{i=1}^{n}(-1)^{i}\phi(a_1,\ldots,a_{i-1},a_ia_{i+1},a_{i+2},\ldots,a_{n+1})\\ &\quad+(-1)^{n+1}\phi(a_{1},\ldots,a_{n})a_{n+1}, \end{align*} $$

where $\phi \in L^{n}(\mathcal {M},\mathcal {X})$ and $a_1,\ a_2,\ldots ,\ a_{n+1}\in \mathcal {M}$ . When $n=0$ , we define $\partial ^0$ by

$$\begin{align*}\partial^0 x(m)=mx-xm\quad\quad\quad (x\in\mathcal{X},\ m\in\mathcal{M}).\end{align*}$$

It is routine to check that $\partial ^{n}\partial ^{n-1}$ : $L^{n-1}(\mathcal {M},\mathcal {X})$ $\rightarrow $ $L^{n+1}(\mathcal {M},\mathcal {X})$ is zero for all $n\geq 1$ , and so $\mathrm {Im}(\partial ^{n-1})$ is a linear subspace of $\mathrm {Ker}(\partial ^n)$ . The $n\mathrm {th}$ Hochschild cohomology group $H^{n}(\mathcal {M},\mathcal {X})$ is then defined to be the following quotient space:

$$\begin{align*}\frac{\mathrm{Ker}(\partial^n:\ L^{n}(\mathcal{M},\mathcal{X})\rightarrow L^{n+1}(\mathcal{M},\mathcal{X}))}{\mathrm{Im}(\partial^{n-1}:\ L^{n-1}(\mathcal{M},\mathcal{X})\rightarrow L^{n}(\mathcal{M},\mathcal{X}))}\end{align*}$$

for $n\geq 1$ . We end this paper by proposing some problems for future study:

• • What are the higher order cohomology groups $H^{n}(\mathfrak {B}(S),\mathfrak {B}(S))$ for $n\geq 2$ ?

• • When $n\geq 2$ , does $H^n(\mathfrak {B}(\mathcal {S}),\mathcal {L}(\mathcal {S}))=0$ ? The first step to calculate the high order cohomology groups should be the following. Given a 2-cocycle $\phi $ , we need to modify it by a 1-coboundary such that $\phi $ is $X_0$ -multimodular, i.e., $\phi (X_0A,B)= X_0\phi (A,B)$ , $\phi (AX_0,B)= \phi (A,X_0B)$ , and $\phi (A,BX_0)= \phi (A,B)X_0$ for all $A,B\in \mathfrak {B}(\mathcal {S})$ .

• • What are the cohomology groups of $\mathfrak {B}(\mathcal {F}_n)$ and $\mathfrak {B}(\mathcal {T})$ ?