\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \fancyhead[CE]{Mohammad Hossein Sattari, Vahideh Yousefiazar} \fancyhead[CO]{HOMOLOGICAL PROPERTIES} %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewenvironment{proof}{{\bfseries \noindent Proof.}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\thm@headfont{\bfseries}} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm %\setcounter{page}{} \thispagestyle{plain} {\noindent Journal of Mathematical Extension \\ Vol. XX, No. XX, (2014), pp-pp (Will be inserted by layout editor)}\\ ISSN: 1735-8299\\ URL: http://www.ijmex.com\\ \vspace*{9mm} \begin{center} {\Large \bf HOMOLOGICAL PROPERTIES OF BANACH MODULES ON HOMOGENEOUS SPACES\\} %{\bf HOMOLOGICAL PROPERTIES} \let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)} {\bf Mohammad Hossein Sattari }\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics\\ Azarbaijan Shahid Madani University} \vspace{2mm} {\bf Vahideh Yousefiazar}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics\\ Azarbaijan Shahid Madani University} \vspace{2mm} \end{center} \vspace{4mm} {\footnotesize \begin{quotation} {\noindent \bf Abstract.} Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$. The aim of this paper is to characterize some homological properties of $L^{1}(G/H)$, $ C_{0}(G/H)$ and $M(G/H)$ as left Banach $L^{1}(G)$-modules such as flatness, injectivity and projectivity. Also, we study the projectivity of $C_{0}(G/H)$ and $M(G/H)$ as Banach left $L^{1}(G/H)$-modules and $M(G)$-modules. \end{quotation} \begin{quotation} \noindent{\bf AMS Subject Classification:} 43A15; 46H25. \noindent{\bf Keywords and Phrases:} Banach module, flatness, homogeneous spaces, injectivity, locally compact group, projectivity. \end{quotation}} \section{Introduction} \label{intro} Homological properties of certain left Banach $L^{1}(G)$-modules have been studied by Dales and Polyakov in 2004 \cite{Dal}, and in 2008 some of those results were investigated by Ramsden for semigroup algebras \cite{Ram}. However, homological properties of left Banach $L^{1}(G)$-modules constructed on homogeneous spaces have not been investigated so far. Throughout this paper $G$ is a locally compact group and $H$ is a closed subgroup of $G$ with left Haar measures $\lambda_{G}$ and $\lambda_{H}$, respectively. Also, $\Delta_{G}$ and $ \Delta_{H}$ are the modular functions of $G$ and $H$, respectively. Let $q: G \longrightarrow G/H $ be the natural quotient map. Consider $G$-space $G/H$ as a homogeneous space that $G$ acts on which, by $x(yH)=(xy)H$. Let $ \mu$ be a Radon measure on $G/H $. For $x \in G$, $\mu_{x}$ is defined by $\mu_{x}(E)=\mu(xE)$, where $E \subset G/H $ is a Borel set. The measure $\mu$ is said to be $G$-invariant, if $\mu_{x}=\mu$ for all $x \in G$. The Radon measure $\mu$ is said to be strongly quasi-invariant if there is a continuous function $\theta:G\times G/H \longrightarrow(0,\infty)$ such that $ d\mu_{x} (yH) =\theta (x,yH) d\mu (yH) ,$ $(x, y \in G).$ A continuous function $\rho: G \longrightarrow(0,\infty)$ is called rho-function for the pair $(G,H),$ when $ \rho(x\xi)=(\frac{\Delta_{H}(\xi)}{\Delta_{G}(\xi)})\rho(x), (x \in G, \xi \in H).$ For every rho-function $\rho$ there exits a strongly quasi-invariant measure $\mu$ on $G/H $ such that the Weil's formula holds: \begin{eqnarray*} \int_{G} f(x)\rho(x)d\lambda_{G}(x)=\int_{G/H} \int_{H}f(x\xi)d \lambda_{H}(\xi)d\mu(xH)\quad(f \in C_{c}(G)), \end{eqnarray*} also the measure $\mu$ satisfies $\frac{d\mu_{x}}{d\mu} (yH)=\frac{\rho(xy)}{\rho(x)},$ $(x,y \in G)$.\\ The map $T_{\rho} :L^{1}(G) \longrightarrow L^{1}(G/H )$ is defined by Tavallei and Ramezanpour in \cite{Tav} as follows \begin{eqnarray*} T_{\rho}f(xH)=\int_{H}\frac{f(x\xi)}{\rho(x\xi)}d\lambda_{H}(\xi) \quad (xH \in G/H , \xi \in H), \end{eqnarray*} where $\mu$ is a strongly quasi-invariant measure on $G/H$ which arises from a rho-function $\rho$ and $L^{1}(G/H)=L^{1}((G/H), \mu )$. The map $T_{\rho}$ is a linear, bounded, and surjective map with $ \Vert T_{\rho} \Vert \leq 1$ satisfying \begin{eqnarray*} \int_{G/H} T_{\rho}f(xH)d\mu(xH)= \int_{G} f(x)d\lambda_{G}(x) \end{eqnarray*} for all $f \in L^{1}(G).$ Also, for every $ \varphi \in L^ {1}(G/H), $ \begin{eqnarray} \Vert\varphi\Vert_{1}=\inf\lbrace\Vert f \Vert_{1}: f \in L^{1}(G), \varphi = T_{\rho}f \rbrace. \end{eqnarray} The Banach space $L^ {1}(G/H) $ isometrically isomorphic to the quotient space $ \frac{L^{1}(G)}{Ker(T_{\rho})} $ equipped with the usual quotient norm. For a compact subgroup $H$ of $G$, let \begin{align*} L^{1}(G:H) = \lbrace f \in L^{1}(G) : R_{\xi} f=f \rbrace. \end{align*} Then $L^{1}(G:H)$ is a left ideal of $L^{1}(G)$ which is isometrically isomorphic to $L^{1}(G/H)$, and so $L^{1}(G/H)$ is a Banach algebra. It was shown in \cite{Tav} that \begin{align}\label{e:1} L^{1}(G:H) = \lbrace \psi \circ q: \psi \in L^{1}(G/H) \rbrace. \end{align} For more detailes see \cite{Tav}. \begin{lemma} Let $H$ be a closed subgroup of $G$. Then $kerT_{\rho}$ is a left ideal of $L^{1}(G)$. \end{lemma} \begin{proof} Let $f \in L^{1}(G)$ and $g \in kerT_{\rho}.$ Then one has \begin{eqnarray*} f.g(xH)=T_{\rho}(f \star g)(xH) &= \int_{G}f(y)\int_{H} \frac{g(y^{-1}x\xi)}{\rho(x)\frac{\Delta_{H}(\xi)}{\Delta_{G}(\xi)}}d\lambda_{H}(\xi)d\lambda_{G}(x)=0. \end{eqnarray*} Since $T_{\rho}(g)(y^{-1}xH) =0$, we have $\int_{H}\frac{g(y^{-1}x\xi )}{\frac{\Delta_{H}(\xi)}{\Delta_{G}(\xi )}}d\lambda_{H}(\xi) =0$. \\ Thus $f \star g \in kerT_{\rho} $. \end{proof} \begin{lemma} Let $G$ be a locally compact group and $H$ be a closed subgroup of $G$. Then $L^{1}(G/H)$ is a left $L^{1}(G)$-module. \end{lemma} \begin{proof} For $\varphi\in L^{1}(G/H)$ there is $g_{\varphi}\in L^{1}(G)$ such that $T_{\rho}(g_{\varphi})=\varphi $. Define a left module operation of $ L^{1}(G) $ on $ L^{1}(G/H) $ by \begin{align*} f.\varphi=T_{\rho} (f \star g_{\varphi}), (f \in L^{1}(G)). \end{align*} \begin{align*} \Vert f.\varphi \Vert_{L^{1}(G/H)} &\leq \int_{G/H}\int_{H} \vert \frac{f\star g_{\varphi}(x\xi)}{\rho(x\xi)} \vert d\lambda_{H}(\xi) d\mu(xH)\\ &=\int_{G}\vert f \star g_{\varphi} \vert d\lambda_{G}\\ %&= \Vert f \star g_{\varphi} \Vert_{1}\\ &\leq \Vert f \Vert_{1} \Vert g_{\varphi} \Vert_{1} < \infty. \end{align*} From equality (\ref{e:1}) we have $ \Vert f.\varphi \Vert_{L^{1}(G/H)} \leq \Vert f\Vert_{1}\Vert \varphi\Vert_{L^{1}(G/H)}$. It is easily seen that this operation, converts $ L^{1}(G/H) $ to a left Banach $ L^{1}(G)$-module and $ L^{1}(G/H) $ is essential as a left $L^{1}(G)$-module. \end{proof} we conclude this section with some examples of homogeneous spaces. \begin{example} Let $M_{n}(\mathbb{C})$ be the space of $n\times n$ matrices. If $G=GL_{n}(\mathbb{C})$ and $H=U(n)=\lbrace T\in M_{n}(\mathbb{C}): T^{*}T=I \rbrace $, then $H$ is a compact subgroup of $G.$ The $ SU(n)=\lbrace T\in U(n) : det T=1 \rbrace $ is a compact subgroup of $ U(n) $. \end{example} \begin{example} Let $G=\lbrace f : \mathbb{N} \longrightarrow \mathbb{N} $; $ f$ is a bijection $ \rbrace $ with discrete topology and $H_{k}=\lbrace f \in G: f(n)=n $, $ n >k \rbrace. $ Then $H_{k}$ is a compact subgroup of $G$. %=S_{k}$ ( the symmetric group) is a compact subgroup of $G$. \end{example} \section{The module $L^{1}(G/H)$} Let $E$ and $F$ be two Banach spaces, and $B(E,F)$ denote the space of all bounded linear operators from $E$ into $F$. An operator $T \in B(E,F) $ is called admissible, if there exists $S \in B(F,E)$ such that $T\circ S \circ T =T$. Let $A$ be a Banach algebra and let $E,F$ be left Banach $A$-modules. The linear space of all left $A$-module morphisms is denoted by ${}_{A}B(E,F).$ An operator $T\in {}_{A}B(E,F) $ is called retraction if there exists $S\in {}_{A}B(F,E)$ such that $T\circ S = I_{F}$. A left Banach $A$-module $P$ is called projective if for every admissible epimorphism $T \in {}_{A}B(E, F)$, and for every $S \in {}_{A}B(P, F)$, there exists $ R \in {}_{A}B(P,E)$ such that $T\circ R =S. $ A left Banach $A$-module $J$ is called injective if for every admissible monomorphism $T \in {}_{A}B(E, F)$, and for every $S \in {}_{A}B(E, J)$, there exists $ R \in {}_{A}B(F, J)$ such that $R\circ T =S. $ If a left (right) Banach $A$-module $E$ is projective, then the right (left) Banach $A$-module $E^{\prime}$ is injective. A left (right) Banach $A$-module $E$ is called flat if $E^{\prime}$ is injective as a right (left) $A$-module. If $A$ is a Banach algebra, then $A^{\flat}$ denotes the algebra formed by adjoining an identity to $A$. The morphism $ \Pi \in {}_{A} B(A^{\flat}\hat{\otimes}E, E)$ is defined by \begin{eqnarray*} \Pi (a\otimes x)=a.x \quad\quad (a \in A^{\flat}, x \in E ). \end{eqnarray*} We shall use the following theorem from ~\cite [IV.1.1, IV.1.2 ]{Hel}. \begin{theorem}\label{33} Let $A$ be a Banach algebra and $E$ be a left $A$-module. Then $E$ is projective if and only if the morphism $ \Pi \in {}_{A} B(A^{\flat}\hat{\otimes}E, E)$ is retraction. In the case that $E$ is essential, $E$ is projective if and only if the morphism $ \Pi \in {}_{A} B(A\hat{\otimes}E, E)$ is retraction \end{theorem} \begin{remark} We know that $ L^{1}(G)\hat{\otimes}L^{1}(G/H)$ is a left $L^{1}(G)$-module. If $f$, $f_{1}\in L^{1}(G) $ and $ \varphi \in L^{1}(G/H)$, then for $F= f_{1}\otimes \varphi \in L^{1}(G \times G/H)$ we have \begin{eqnarray*} f.F(x,zH) & = &(f \star f_{1})(x)\varphi(zH) \\ & =& \int_{G} f(y)(f_{1}\otimes \varphi )(y^{-1}x,zH) d\lambda_{G}(y)\\ & =&\int_{G}f(y)F(y^{-1}x,zH)d\lambda_{G}(y)\quad (x, z \in G), \end{eqnarray*} and so for any $F \in L^{1}(G \times G/H)$ this formula holds. \end{remark} \begin{theorem}\label{3.3} Let $H$ be a compact subgroup of $G$, and let $\mu$ be the $G$-invariant measure on $G/H$ arising from the constant rho-function $\rho=1$. Then $ L^{1}(G/H ,\mu) $ is projective as a left $L^{1}(G)$-module ($M(G)$-module), and hence flat as a left $L^{1}(G)$-module ($M(G)$-module). \end{theorem} \begin{proof} Let \begin{eqnarray*} \Pi : L^{1}(G)\hat{\otimes}L^{1}(G/H) \longrightarrow L^{1}(G/H), \end{eqnarray*} \begin{eqnarray*} f\otimes \varphi \longrightarrow f.\varphi , \end{eqnarray*} \begin{eqnarray*} f \cdot \varphi(xH)=\int _{G} f(y)\varphi(y^{-1}xH)d\lambda_{G}(y), \end{eqnarray*} and let $E$ be a compact and symmetric subset of $G$ such that $ \lambda_{G}(HE) >0$. We can choose a Haar measure on $G$ such that $ \lambda_{G}(HE)=1$. If $K=q(E) \subset G/H$, then $K$ is compact. Define \begin{eqnarray*} \rho : L^{1}(G/H) \longrightarrow L^{1}(G)\hat{\otimes}L^{1}(G/H) = L^{1}(G \times G/H), \end{eqnarray*} \begin{eqnarray*} \rho(\varphi)(s,tH)=\chi_{K}(tH)\varphi(stH). \end{eqnarray*} Therefore \begin{eqnarray*} \Vert\rho(\varphi)\Vert_{1} & = & \int_{G} \int_{G/H} \vert \chi_{K}(tH)\varphi(stH) \vert d\lambda_{G}(s)d\mu(tH) \\ &=& \int_{G} \int_{G/H} \vert \chi_{K}(zH)\varphi(tH) \vert d\lambda_{G}(s)d\mu(tH)\quad(z=\xi s^{-1}t, \xi \in H) \\ &=& \lambda_{G}(HE) \int_{G/H}\vert \varphi (tH) \vert d\mu(tH) < \infty, \end{eqnarray*} For $g \in L^{1}(G/H) $ and $x \in G $, we have \begin{eqnarray*} \Pi (\rho (\varphi))(xH) &=& \int _{G} \rho( \varphi ) (y,y^{-1}xH)d\lambda_{G}(y)\\ &=&\varphi(xH) \int_{G} \chi_{K}(y^{-1}H) d\lambda_{G}(y)=\varphi(xH)\lambda_{G}(HE)=\varphi(xH). \end{eqnarray*} Take elements $ f \in L^{1}(G)$, $ \varphi \in L^{1}(G/H)$ and $ s$, $t \in G$. Then \begin{eqnarray*} \rho(f.\varphi)(s,tH) &=& \chi _{K} (tH) \int_{G} f(y)\varphi(y^{-1}stH)d\lambda_{G}(y)\\ &= & f.\rho(\varphi)(s,tH). \end{eqnarray*} Since $L^{1}(G/H)$ is unital as a left Banach $M(G)$-module, the result follows by Theorem \ref{33} and~\cite[Theorem 3.1.1]{Ram}. \end{proof} \begin{lemma}\label{3.5} Let $G$ be a locally compact group, and $H$ be a closed subgroup of $G$ and $\mu_{1}$, $\mu_{2} $ be two strongly quasi-invariant measures on $G/H$ that they arise respectively from rho-functions $\rho_{1}$, $\rho_{2}$. Then $L^{1}(G/H ,\mu_{1})$ is isometrically isomorphic to $L^{1}(G/H ,\mu_{2})$ as left Banach $L^{1}(G)$-modules. \end{lemma} \begin{proof} If the function $\eta : G/H \longrightarrow (0 , \infty) $ is defined by $\eta (xH) =\frac{\rho_{1}(x)}{\rho_{2}(x)}$, then according to ~\cite[Theorem $2.59$]{Fol}, $d\mu_{1}=\eta d\mu_{2}$. Thus $T$ can be defined as follows: \begin{eqnarray*} T : L^{1}(G/H ,\mu_{1}) \longrightarrow L^{1}(G/H ,\mu_{2}), \end{eqnarray*} \begin{eqnarray*} \varphi \longrightarrow \eta \varphi. \end{eqnarray*} In this case $\int_{G/H} \vert \varphi \vert d\mu_{1} = \int_{G/H} \vert \varphi \vert \eta d\mu_{2}=\int_{G/H} \vert T( \varphi) \vert d\mu_{2} <\infty$, and if $ \varphi \in L^{1}(G/H, \mu_{2})$, then $T(\frac{1}{\eta}\varphi)= \varphi $. Therefore, $T$ is a surjective isometry linear map. For $ \varphi \in L^{1}(G/H ,\mu_{1})$, there exists $g_{\varphi}\in L^{1}(G)$ such that $T_{\rho_{1}}(g_{\varphi}) =\varphi $, so $T_{\rho_{1}}(g_{\varphi}) = \frac{1}{\eta} T_{\rho_{2}}(g_{\varphi}). $ Finally for $f \in L^{1}(G)$ and $ \varphi \in L^{1}(G/H, \mu_{1})$, we have \begin{eqnarray*} T(f.\varphi) =T( T_{\rho_{1}} ( f \star g_{\varphi})) &=&T(\frac{1}{\eta} T_{\rho_{2}} ( f \star g_{\varphi}))\\ &=&f. T_{\rho_{2}} ( g_{\varphi})\\ &=& f. T(\varphi). \end{eqnarray*} \end{proof} \begin{corollary} Let $G$, $H$, $\mu_{1}$ and $\mu_{2} $ be as in Lemma ~\ref{3.5}. If $L^{1}(G/H)$, $\mu_{1})$ is projective as a left $L^{1}(G)$-module, then $L^{1}(G/H ,\mu_{2})$ is projective as a left $L^{1}(G)$-module.\end{corollary} \begin{corollary} Let $H$ be a compact subgroup of $G$. Then $ L^{1}(G/H) $ is projective as a left Banach $L^{1}(G/H)$-module. \end{corollary} \begin{proof} We know $ L^{1}(G/H) $ is $ L^{1}(G/H)$-bimodule. Since $T_{\rho} $ is a surjective left $ L^{1}(G)$-module morphism, the result follows by Theorem \ref{3.3} and the proof of \cite[IV Proposition1.7]{Hel}. \end{proof} \begin{definition} Let $G$ be a locally compact group and let $E$ be a left Banach $L^{1}(G)$-module and $\varphi_{G}$ be the augmentation character defined by $ \varphi_{G}(f)=\int_{G}f d\lambda_{G}$. The module $E$ is augmentation-invariant, if there exists a non-zero $\lambda \in E^{\prime}$ such that \begin{eqnarray*} =\varphi_{G}(f)\quad (f \in L^{1}(G), x \in E). \end{eqnarray*} \end{definition} \begin{remark}\label{3.10}Let $E$ be a left Banach $L^{1}(G)$-module. If $E$ is augmentation-invariant, then $E^{\prime\prime}$ will also be augmentation-invariant.\\ Let $\lambda \in E^{\prime}$ such that $=\varphi_{G}(f),$ $(f \in L^{1}(G)$, $ x \in E$), we can consider $\lambda\in E^{\prime\prime\prime}$. For each $\varphi \in E^{\prime\prime},$ there exists a net $(x_{\alpha})_{\alpha}\in E$ with $x_{\alpha}\longrightarrow \varphi $, in $\sigma (E^{\prime\prime}, E^{\prime})$-topology so if $f \in L^{1}(G)$ and $\varphi \in E^{\prime\prime},$ we have \begin{eqnarray*} &= &<\varphi, \lambda.f>\\ &=&lim_{\alpha}\\ &=& lim_{\alpha}\\ &=& lim_{\alpha}\varphi_{G}(f)\\ &=&\varphi_{G}(f)<\varphi, \lambda>. \end{eqnarray*} \end{remark} \begin{lemma}\label{2.11} Let $E$ and $F$ be left Banach $L^{1}(G)$-modules and $T \in {}_{L^{1}(G)}B(E,F)$ be an isometry isomorphism of Banach space. If $E$ is an augmentation-invariant, then $F$ is also augmentation-invariant. \end{lemma} \begin{proof} Let $\lambda \in E^{\prime}$ such that \begin{eqnarray*} < f.x, \lambda > = \varphi_{G}(f)\quad (f \in L^{1}(G), x \in E). \end{eqnarray*} Since $\lambda \circ T^{-1} \in F^{\prime}$, for any $y\in F$ we have \begin{eqnarray*} = \lambda \circ T^{-1}(f.y) = \lambda (f(T^{-1}(y))) = \varphi_{G}(f). \end{eqnarray*} \end{proof} \begin{corollary}\label{2.14} Let $H$ be a closed subgroup of $G$. Then the left $L^{1}(G)$-module $L^{1}(G/H)$ is augmentation-invariant. \end{corollary} \begin{proof} For $f \in kerT_{\rho} $, we have \begin{eqnarray*} 0=\int _{G/H} T_{\rho}f(xH) d\mu (xH)= \int_{G} f(x) d\lambda_{G}(x) =\varphi_{G}(f). \end{eqnarray*} Let $ \lambda: L^{1}(G)/ kerT_{\rho} \longrightarrow \mathbb{C}$ by $\lambda (f+kerT_{\rho})=\varphi_{G} (f) $. Then \begin{eqnarray*} \lambda (f.(g + kerT_{\rho} & =\varphi_{G}(f)\lambda (g+kerT_{\rho}). \end{eqnarray*} Thus $ L^{1}(G)/ kerT_{\rho} $ as a left $L^{1}(G)$-module is augmentation-invariant, %Since $ L^{1}(G/H)$ and $ and so $ L^{1}(G/H)$ is augmentation-invariant by Lemma~\ref{2.11}. \end{proof} \begin{definition} Let $A$ be an algebra, and $E$ be a left $A$-module. Then $E$ is said to be faithful, if for each $x \in E\setminus\lbrace 0\rbrace $ there exists $a \in A$ such that $a.x \neq 0$%\end{defn} \end{definition} \begin{lemma} Let $E$ and $F$ be left Banach $A$-modules and $T : E \longrightarrow F$ be an isomorphism of left Banach $A$-modules. If $E$ is faithful, then $F$ is also faithful. \end{lemma} \begin{lemma}\label{2.13} The Banach space $L^{1}(G)/ kerT_{\rho} $ as a left $L^{1}(G)$-module is faithful. Consequently $L^{1}(G/H)$ as a left $L^{1}(G) $-module is faithful. \end{lemma} \begin{proof} Let $g +kerT_{\rho}\neq0 $ and let $ \lbrace f_\upsilon \rbrace_\upsilon $ be a bounded approximate identity for $ L^{1}(G) $. Since $ f_{\upsilon}\star g +kerT_{\rho} \longrightarrow g+kerT_{\rho}\neq0$, there exists $\upsilon$ such that $ f_{\upsilon}\star g +kerT_{\rho} \neq 0$. \end{proof} \begin{remark} Since $L^{1}(G/H)$ is a left Banach $L^{1}(G)$-module, $L^{\infty}(G/H, \mu)=L^{1}(G/H,\mu)^{\prime}$ and $C_{0}(G/H)$ become right Banach $L^{1}(G)$-modules. For $\psi\in C_{0}(G/H)$ and $f \in L^{1}(G) $, take $g_{\varphi}\in L^{1}(G) $ with $T_{\rho}(g_{\varphi})=\varphi$, we have \begin{align*} <\varphi, \psi. f> &=\\ &= \int_{G/H} f.\varphi (xH) \psi(xH) d\mu (xH)\\ &=\int_{G}\int_{G/H} \int_{H} \frac{f(y) g_{\varphi} (y^{-1} x \xi)}{\rho (x \xi)} \psi (xH)\frac{\rho(y^{-1}x\xi)}{\rho(y^{-1}x\xi)}d\lambda_{H}(\xi)d\lambda_{G}(y) d\mu(xH)\\ &= \int_{G/H}\int_{G}f(y)\psi(xH)\varphi(y^{-1}xH) \theta(y^{-1}, xH)d\lambda_{G}(y)d\mu(xH) \\ &= \int_{G/H}\int_{G}f(y)\psi(yxH)\varphi(xH)d\lambda_{G}(y)d\mu(xH). \end{align*} Thus $\psi.f(xH) = \int_{G} f(y)R_{x}(\psi \circ q)(y) d\lambda_{G}(y)$ and so $ \psi.f \in C_{0}(G/H).$ \end{remark} \begin{theorem} Let $G/H$ be a discrete spase. Then $\ell^{1}(G/H)$ is injective as a left Banach $L^{1}(G)$-module if and if $G$ is amenable.\end{theorem} \begin{proof} According to Lemma ~\ref{2.13}, Corollary~\ref{2.14} and above remark, the result is obtained by \cite [Proposition4.6]{Dal}. \end{proof} \section{ The modules $C_{0}(G/H) $ and $L^{\infty}(G/H)$} Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$. The surjective linear map $T_{\infty}: L^{\infty}(G) \longrightarrow L^{\infty}(G/H)$ with $T_{\infty}f(xH)= \int_{H} f(x \xi) d\lambda_{H}(\xi)$ for $x \in G$ and $f \in L^{\infty}(G)$ is defined in ~\cite{Tav}. It has been proved that $ \Vert T_{\infty} \Vert \leq 1$ and $T_{\infty}(C_{c}(G)) \subset C_{c}(G/H)$. If $\varphi \in L^{\infty}(G/H) $, then \begin{eqnarray*} \Vert \varphi \Vert_{\infty} =\inf\lbrace \Vert f \Vert_{\infty} : f \in L^{\infty}(G) ,\varphi = T_{\infty}(f)\rbrace. \end{eqnarray*} Set $L^{\infty} (G:H) = \lbrace f \in L^{\infty}(G) :R_{\xi} f=f \quad \xi \in H \rbrace ,$ then the restriction of $T_{\infty}$ on $L^{\infty}(G:H) $ is an isometry isomorphism. The left module action of $L^{1}(G) $ on $ L^{\infty}(G/H) $ is defind by \begin{eqnarray*} f .\varphi =T_{\infty}(f \star g) \quad( \varphi \in L^{\infty}(G/H), f \in L^{1}(G), g \in L^{\infty}(G) ), \end{eqnarray*} where $T_{\infty}(g) =\varphi$. The space $C_{0}(G/H)$ is a closed $L^{1}(G)$-submodule of the left Banach $L^{1}(G)$-module $ L^{\infty}(G/H) $, and $C_{0}(G/H)$ is essential. \begin{remark} Since $L^{\infty}(G/H)$ is a left $L^{1}(G)$-module, $L^{\infty}(G/H)^{\prime}$ will be a right $L^{1}(G)$-module. Now, we show that $L^{1}(G/H)$ is also a right $L^{1}(G)$-module. \\ For $f \in L^{1}(G)$, $\varphi \in L^{1}(G/H)$, $\psi \in L^{\infty}(G/H)$ and $g_{\psi} \in L^{\infty}(G)$ such that $T_{\infty}(g_{\psi})=\psi$ we have \begin{eqnarray*} <\psi, \varphi.f>&=&\\ &=&\int_{G/H}\int_{H}\int_{G}f(y)g_{\psi}(y^{-1}x\xi) \varphi(xH) d\lambda_{H}(\xi) d\mu(xH) d\lambda_{G}(y)\\ &=&\int_{G/H}\int_{G}f(y)\psi(xH)\varphi(yxH)\theta(y,xH)d\lambda_{G}(y) d\mu(xH). \end{eqnarray*} Then $\varphi.f(xH) =\int_{G}f(y)\varphi(yxH) \theta(y,xH) d\lambda_{G}(y)$, and so $\varphi.f \in L^{1}(G/H).$ \begin{eqnarray*} \Vert\varphi.f \Vert_{1} & \leq &\int_{G/H}\int_{G}\vert f(y)\vert \vert\varphi(yxH) \vert \theta(y, xH) d\lambda_{G}(y) d\mu(xH)\\ &=& \int_{G/H}\int_{G}\vert f(y)\vert \vert\varphi(xH)\vert \theta(y^{-1}, xH)\theta(y, y^{-1}xH) d\lambda_{G}(y) d\mu(xH)\\ &=&\Vert f \Vert_{1} \Vert \varphi \Vert_{1}. \end{eqnarray*} It is known that $ L^{1}(G/H \times G)\cong L^{1}(G/H)\hat{\otimes}L^{1}(G)$ is a right $L^{1}(G)$-module, so for $F \in L^{1}(G/H \times G)$ and $ f\in L^{1}(G)$ we can write \begin{eqnarray*} F.f (tH, s)= \int_{G}F(tH, sy^{-1}) f(y) \Delta_{G}(y^{-1})d\lambda_{G}(y). \end{eqnarray*} If $G$ is compact and rho-function $\rho=1$, $L^{1}(G/H)$ is essential as a right $L^{1}(G)$-module. \end{remark} \begin{theorem} Let $G$ be a compact group and $H$ be a closed subgroup of $G$. Then $L^{1}(G/H)$ is projective as a right $L^{1}(G)$-module and flat as a right $L^{1}(G)$-module. \end{theorem} \begin{proof} Let $ \Pi : L^{1}(G/H)\hat{\otimes}L^{1}(G) \longrightarrow L^{1}(G/H)$ be given by \begin{eqnarray*} \Pi(\varphi \otimes f)= \varphi.f \quad (\varphi \in L^{1}(G/H), f\in L^{1}(G) ), \end{eqnarray*} \begin{eqnarray*} \varphi \cdot f(xH)=\int _{G} f(y)\varphi (yxH) \theta(y,xH)d\lambda_{G}(y). \end{eqnarray*} Define \begin{eqnarray*} \rho : L^{1}(G/H) \longrightarrow L^{1}(G/H)\hat{\otimes}L^{1}(G) \cong L^{1}((G/H)\times G), \end{eqnarray*} \begin{eqnarray*} \rho(\varphi)(tH, s))=\varphi (s^{-1}tH) \theta(s^{-1}, tH). \end{eqnarray*} \begin{eqnarray*} \Pi(\rho)(\varphi)(xH) &=& \int_{G}\varphi(xH)\theta(y^{-1},yxH)\theta(y,xH) d\lambda_{G}(y)\\ &=& \lambda_{G}(G)\varphi(xH)=\varphi(xH). \end{eqnarray*} \begin{eqnarray*} \rho(\varphi.f)(tH,s) &=& \int_{G}f(y) \varphi(ys^{-1}tH)\theta(y, s^{-1}tH) \theta(s^{-1}, tH)d\lambda_{G}(y)\\ &=& (\rho(\varphi).f)(tH,s). \end{eqnarray*} \end{proof} The following theorem is proved with a similar argument as in \cite[Theorem 3.1]{Dal}. \begin{theorem}\label{4.3} Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$, and let $E$ be a closed, left $L^{1}(G)$-submodule of $L^{\infty}(G/H)$ such that $C_{c}(G/H) \subset E\subset C^{b}(G/H)$. If $E$ is projective, then $G/H$ is a compact space. \end{theorem} \begin{proof} Suppose that $G/H$ is not compact. Then $G$ is not compact. There exists a compact and symmetric neighbourhoods of $e_{G}$ such as $V,W$ such that $V^{2} \subset W$ and there exists $0 \leq f_{1} \leq 1, f_{1} \in C_{c}(G)$ with $f_{1}(e_{G}) =1 , supp f_{1} \subset V$ and $ \Vert f_{1} \Vert_{1}\leq 1 $. Then $supp ( f_{1}\star f_{1}) \subset V^{2} \subset W , f_{1} \star f_{1}(e_{G}) \neq 0$ and $ f_{1} \star f_{1}\geq 0$. Put $T_{\rho}(f_{1})=f $. Since $f_{1} \geq 0$ and $f_{1}$ is continous, so $ f_{1}.f = T_{\rho}( f_{1} \star f_{1}) \neq 0$ and $ T_{\rho}( f_{1} \star f_{1}) \in C_{c}(G/H)$. Set $A=L^{1}(G)$, because $E$ is projective, there exists $T \in {}_A B(E,A^{\flat})$ such that $T(f_{1}.f )\neq 0$ by \cite [Proposition1.2]{Dal}.\\ We may suppose $T(E) \subset A$. Let $n\in\mathbb{N}$ and $F$ be a compact subset of $G$. Since $G$ is not compact, there exist $ s_{1} , . . . , s_{n} \in G$ such that $s_{i}F \cap s_{j}F =\emptyset$ where $ i\neq j.$\\ Set $\eta =\frac{\Vert f_{1}\star T(f) \Vert_{1}}{2}>0 $ and pick $k\in \mathbb{N} $ with $\frac{1}{k}<\eta $. There exists $g \in C_{c}(G) $ such that $\Vert T(f)-g \Vert_{1}<\dfrac{1}{k},$ and so \begin{eqnarray*} \Vert f_{1}\star(T(f)- g)\Vert_{1} \leq \Vert f_{1} \Vert_{1} \Vert T(f)-g \Vert_{1}\leq\frac{1}{k} <\eta. \end{eqnarray*} Therefore we have \begin{eqnarray*} \Vert f_{1} \star g \Vert_{1} &=&\Vert f_{1}\star g-f_{1} \star T(f) +f_{1} \star T(f) \Vert_{1}\\ & \geq & \Vert f_{1} \star T(f)\Vert_{1} - \Vert f_{1}\star g-f_{1} \star T(f) \Vert_{1}\\ & > &2\eta - \eta = \eta. \end{eqnarray*} Set $ K=supp (g)$ and observe that $(K \cup V )^{2} $ is compact. There exist $s_{1}, . . . , s_{k} \in G $ such that the sets $s_{i}(K \cup V)^{2}$ are pairwise disjoint for $i=1, . . . ,k $. Set $f_{1j}=s_{j} \star f_{1}=L_{s_{j}}f $ then $suppf_{1j} \subset s_{j}V$, $supp(f_{1j}\star f) \subset s_{j}V.V$ and $\vert\Sigma_{j=1}^{k}f_{1j} \vert_{G}=1$. \\ Therefore $\Vert \Sigma_{1}^{k} f_{1j}\star g \Vert_{1}=\Sigma_{1}^{k} \int_{s_{j}(K \cup V)^{2} } \vert f_{1j}\star g \vert =k \Vert f_{1} \star g \Vert_{1}.$\\ Now set $\lambda = \Sigma_{j=1}^{k}( f_{1j} . f )$. Since $s_{j}V.V \cap s_{i}V.V = \emptyset $ for $ i\neq j$ and $f_{1j}\star f_{1} \geq 0$, we have\\ \begin{eqnarray*} \vert \lambda \vert_{G/H} & = & sup_{t \in G} \vert \lambda (tH) \vert\\ %& =sup_{t \in G} \vert \sum_{1}^{k} f_{1j}.f (tH) \vert \\ &= & sup _{t \in G} \vert\sum_{1}^{k} T_{\rho}(f_{1j} \star f_{1})(tH) \vert\\ %& =sup_{t \in G} \vert \sum_{1}^{k} \int_{H} \frac{f_{1j} \star f_{1}(t\xi)}{\rho (t \xi)} d\lambda_{H}(\xi) \vert \\ %&\leq M \sum_{1}^{k} \vert f_{1j} \star f_{1}\vert_{G} \\ & =& M \vert f_{1} \star f_{1} \vert_{G}, \\ \end{eqnarray*} in which $M=sup( \frac{1}{\rho})$ on $\cup_{j=1}^{k}( s_{j}V.V). $\\ Since \begin{eqnarray*} \Vert f_{1j} \star (T(f) -g) \Vert _{1} \leq \Vert f_{1j} \Vert_{1}\Vert T(f)-g \Vert_{1} \leq \frac{1}{k}, \end{eqnarray*} thus we have \begin{eqnarray*} \Vert T(\lambda) \Vert_{1} =\Vert \sum_{1}^{k} f_{1j} \star T(f) \Vert_{1} \geq\Vert \sum_{1}^{k} f_{1j} \star g \Vert_{1} -1= k\Vert f_{1} \star g \Vert_{1}-1 \geq k \eta -1. \end{eqnarray*} Therefore \begin{eqnarray*} k \eta -1 \leq \Vert T(\lambda) \Vert_{1} \leq \Vert T \Vert \vert\lambda\vert_{G/H} \leq \vert f_{1} \star f_{1}\vert_{G}M\Vert T \Vert. \end{eqnarray*} This holds for each $k\in \mathbb{N}$, and this is a contradiction with boundedness of $T$. \end{proof} \begin{theorem}\label{4.6} Let $H$ be a compact subgroup of $G$ and $G/H$ is a compact space. Then $C(G/H)$ as a left $ L^{1}(G)$-module, is projective. \end{theorem} \begin{proof} Since $H$ and $G/H$ are compact, $G$ is compact and so $ L^{1}(G)$ is biprojective. If $C_{0}(G:H)=\lbrace \varphi \in C_{0}(G) : \varphi(x\xi)=\varphi(x)$, $x\in G , \xi \in H\rbrace$, then $C_{0}(G:H)$ is a left $ L^{1}(G)$-module by the module action of $f.\varphi(x)=f\star\varphi(x)= \int_{G}f(y)\varphi (y^{-1}x) d \lambda_{G}(y)$, $(f\in L^{1}(G)$, $\varphi \in C_{0}(G:H))$. Take $(f_{\alpha})_{\alpha}$ to be a bounded approximate identity for $ L^{1}(G)$. Then $f_{\alpha} .\varphi=f_{\alpha} \star\varphi\longrightarrow\varphi$, therefore $ L^{1}(G).C_{0}(G:H)=L^{1}(G) \star C_{0}(G:H)= C_{0}(G:H)$. By ~\cite[IV, Proposition 5.3]{Hel} $ C_{0}(G:H)$ is projective as a left $ L^{1}(G)$-module. Since $T_{\infty}$ is an isometry isomorphism of $C_{0}(G:H)$ onto $C_{0}(G/H)=C(G/H)$ as a left $ L^{1}(G)$-module, $C(G/H)$ is projective. \end{proof} \begin{corollary} Let $H$ be a compact subgroup of $G $. Then $C_{0}(G/H)$ as a left $ L^{1}(G)$-module ($ M(G)$-module) is projective if and only if $G/H$ is a compact space. \end{corollary} \begin{remark} If $G$ is a locally compact group, and if $H$ is a compact subgroup of $G$, then by the theorem \ref{3.3} $L^{\infty}(G/H)$ is an injective right $L^{1}(G)$-module. If $G/H$ is finite, by Theorem \ref{4.6} and the proof of the ~\cite[IV Proposition1.7]{Hel} $ L^{\infty}(G/H) $ is projective as a left $L^{1}(G/H)$-module. \end{remark} Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$. According to \cite{Rei}, we can define a norm decreasing linear map $\tilde{T} : M(G) \longrightarrow M(G/H) $ by %\begin{eqnarray*} $ \tilde{T}m(E)=m(q^{-1}(E))$ where $ E$ is a Borel subset of $ G/H,$ %\end{eqnarray*} that satisfies \begin{eqnarray*} \int_{G/H}\varphi(xH)d\tilde{T}m(xH) = \int_{G}\varphi (xH) dm(x) \quad (\varphi \in C_{0}(G/H) ). \end{eqnarray*} Set $M(G:H)= \lbrace m\in M(G):m(Ah)= m(A)$, $A \in B_{G}$ , $h\in H\rbrace$, where $ B_{G}$ is the $\sigma$-algebra of Borel sets. Then, $M(G:H)$ is a closed left ideal of $M(G)$ and $ \tilde{T} : M(G:H) \longrightarrow M(G/H) $ is an isometry isomorphism of Banach spaces, ~\cite{Jav}. \\ For $ m \in M(G)$ and $ \nu \in M(G/H)$ define a left $M(G)$-module action on $M(G/H) $ by \begin{eqnarray*} m . \nu(\varphi) = \int_{ G/H} \int_{G} \varphi(yxH) dm (y) d \nu(xH)\quad (\varphi \in C_{0}(G/H). \end{eqnarray*} Therefore $M(G/H) $ is a left Banach $L^{1}(G)$-module and $ \tilde{T}(m_{1}\star m_{2})=m_{1} . \tilde{T}(m_{2})$ for all $m_{1}$, $m_{2} \in M(G)$. Also, if $\omega,\nu \in M(G/H)$, then $\omega\star \nu= \tilde{T}(\omega_{v}\star\nu_{v}),$ where $\omega_{v},\nu_{v} \in M(G:H)$ such that $\tilde{T}(\omega_{v})=\omega$, $\tilde{T}(\nu_{v})=\nu,$ and $M(G/H) $ with this convolution is a Banach algebra and $L^{1}(G/H)$ is an ideal of $M(G/H)$. See ~\cite{Jav} for more details. \begin{remark} If $G/H$ is discrete, then $M(G/H) = \ell^{1}(G/H)$ is projective as a left $L^{1}(G)$-module ( $M(G)$-module and $L^{1}(G/H)$-module). \end {remark} \begin{lemma}\label{4.18} Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$. Then $M(G/H)$ is faithful and augmentation-invariant as a left $L^{1}(G)$-module. \end{lemma}\begin{proof} Since $M(G:H)$ is a submodule of the faithful Banach module $M(G)$ as a left $L^{1}(G)$-module, $M(G/H)$ is faithful. %The map $ \tilde{T} : M(G:H) \longrightarrow M(G/H) $ is an isometry isomorphism of left Banach $L^{1}(G)$-modules, and $M(G:H)$ is a submodule of the faithful Banach module $M(G)$, as a left $L^{1}(G)$-module. So $M(G/H)$ is faithful.\\ Since $M(G)$ is augmentation-invariant, there exists $\lambda \in M(G)^{\prime}$ such that $ =\varphi_{G}(f)$ for $f \in L^{1}(G)$ and $ m \in M(G)$. The map $\lambda \circ \iota\circ\tilde{T}^{-1}$ is an element of $M(G/H)^{\prime}$, that satisfies in definition of augmentation- invarintness of $M(G/H)$, where $\iota: M(G:H) \longrightarrow M(G)$ is the inclusion. \end{proof} %\begin{lem}\label{4.19}Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$. Then % $(L^{\infty}(G/H))^{\prime}$ is augmentation invariant as a left $L^{1}(G)$-module. % \end{lem} %\begin{proof} %Since $L^{1}(G/H)$ is augmentation-invariant, so $L^{1}(G/H)^{\prime\prime}=(L^{\infty}(G/H))^{\prime}$ is augmentation-invariant by Remark ~\ref{3.10}. %\end{proof} %The following theorem is a useful characterization of injective modules, given in \cite[Theorem 3.4.2]{Ram}.\\ %\begin{thm}\label{4.21} %Let $G$ be a locally compact group, and $E$ be an augmentation-invariant right (left) Banach $L^{1}(G)$-module. Suppose that $E$ is a dual space. Then $E$ is injective as a right (left) $L^{1}(G)$-module if and only if $G$ is amenable. %\end{thm} %$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad\quad \quad \quad \quad \quad\quad \quad \quad \quad \quad %\Box$ \begin{theorem} Let $G$ be a locally compact group and $H$ be a compact subgroup of $G$. The following conditions are equivalent.\\ (a) The group $G$ is amenable;\\ (b) $M(G/H)$ is injective as a left Banach $L^{1}(G)$-module;\\ (c) $L^{\infty}(G/H)$ is flat as a right Banach $L^{1}(G)$-module;\\ (d) $C_{0}(G/H)$ is flat as a right Banach $L^{1}(G)$-module. \end{theorem} \begin{proof} According to Lemma ~\ref{4.18}, $ M(G/H)$ is an augmentation-invariant as a left Banach $L^{1}(G)$-module, and also based on Remark ~\ref{3.10}, $(L^{\infty}(G/H))^{\prime}$ is augmentation-invariant as a left Banach $L^{1}(G)$-module. Therefore, according to \cite[Theorem 3.4.2]{Ram}, $(d) \Leftrightarrow (a) \Leftrightarrow (b) \Leftrightarrow (c)$. %Since %$M(G/H) = C_{0}(G/H)^{\prime}$ is augmentation-invariant as a left Banach $L^{1}(G)$-module, so by Theorem \cite[Theorem 3.4.2]{Ram} $ (a) \Leftrightarrow (d).$ \end{proof} \begin{center} \begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format. \bibitem{Dal} H. G. Dales and M. E. Polyakov, Homological properties of modules over group algebras, {\it Proc. London Math. Soc}, 3(89)(2004), 390-426. %F. Author, S. Author and T. Author, Article title should be written here, {\it Journal Name}, Volume (year), pp-pp. \bibitem{Jav} T. Derikvand, R. A. Kamyabi-Gol, M. Janfada, Banach algebra of complex Radon measures on homogeneous space, {\it Iran. J. Sci. Technol. Trans. Sci.} 44(2020) 1429-1437. %F. Author, {\it Book Title Should Be Written Here}, pp-pp, Publisher, place (year) \bibitem{Fol} G. B. Folland, {\it A course in abstract harmonic analysis}, CRC Press, Boca Raton (1995). \bibitem{Hel} A. YA. Helemskii, {\it The homology of Banach and topological algebras}, Kluwer Academic (1989). \bibitem{Ram} P. Ramsden, {\it Homological properties of semigroup algebra}, PhD thesis, University of Leeds (2008). \bibitem{Rei} H. Reiter and J.D. Stegeman, {\it Classical harmonic analysis and locally compact groups}, 2nd Ed Oxford Science Publications, Clarendon Press, New York (2000). \bibitem{Tav} N. TavallaeIi and M. Ramezanpour, Structural transition between $L^{p}(G)$ and $L^{p}(G/H)$, {\it Banach J. Math. Anal}, 3 (2015), 194-205. \end{thebibliography} \end{center} {\small \noindent{\bf Mohammad Hossein Sattari} \noindent Associate Professor of Mathematics \noindent Department of Mathematics \noindent Azarbaijan Shahid Madani University \noindent Tabriz, Iran \noindent E-mail: sattari@azaruniv.ac.ir}\\ {\small \noindent{\bf Vahideh Yousefiazar } \noindent Ph. D. Student \noindent Department of Mathematics \noindent Azarbaijan Shahid Madani University \noindent Tabriz, Iran \noindent E-mail: v.yousefiazar@azaruniv.ac.ir}\\ \end{document}