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\fancyhead[CE]{Mohammad Hossein Sattari, Vahideh Yousefiazar}
\fancyhead[CO]{HOMOLOGICAL PROPERTIES}



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{\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*}
<f.x , \lambda >=\varphi_{G}(f)<x, \lambda>\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 $<f.x , \lambda >=\varphi_{G}(f)<x, \lambda>,$ $(f \in L^{1}(G)$, $ 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*}
 <f.\varphi, \lambda>&= &<\varphi, \lambda.f>\\
 &=&lim_{\alpha}<x_{\alpha}, \lambda.f>\\
 &=& lim_{\alpha}<f.x_{\alpha},  \lambda>\\
 &=& lim_{\alpha}\varphi_{G}(f)<x_{\alpha},  \lambda>\\
 &=&\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)<x,\lambda>\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*}
<f.y, \lambda \circ T^{-1}> =  \lambda \circ T^{-1}(f.y) = \lambda (f(T^{-1}(y))) = \varphi_{G}(f)<y, \lambda \circ T^{-1}>.
\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> &=<f.\varphi , \psi>\\
&= \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>&=&<f.\psi, \varphi>\\
&=&\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 $ <f\cdot m, \lambda >=\varphi_{G}(f)<m , \lambda>$ 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} 