Characterization of a vector measure: application in the \(GL(2;\mathbb{R})\) group

Author(s): Abalo Douhadji1, Yaovi Awussi2
1Department of Mathematics, University of Lomé, PObox 1515, Lomé, Togo.
2Department of Mathematics, Mathematics and Applications Laboratory, University of Lomé, PObox 1515, Lomé, Togo.
Copyright © Abalo Douhadji, Yaovi Awussi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper we characterize a bounded vector measure on Lie compact group \(G = GL(2;\mathbb{R})\). It is a question of considering a bounded vector measure \(m\) defined from \(K(G;E)\) the space of \(E\) valued functions with compact support on \(G\) and giving its integral form.

Keywords: Vector measure, Haar measure, Lie compact group, absolute continuity.

1. Introduction

In 1989, Assiamoua [1] worked on the properties of vector measure, introduced by Diestel in [2]. In 2013, Awussi [3] proved that any bounded vector measure is absolutely continue with respect to Haar measure. In 2013, Mensah [4] worked on Fourier-Stieljes transform of vector measures on compact groups. In this paper, we treat with a special case by giving a form to a bounded vector measure on \(G=GL(2;\mathbb{R})\). We consider a vector measure which is absolutely continue with respect to Haar measure and then give the integral form [5] to this vector measure. The first essential part of our work is to establish the form of Haar measure on \(G=GL(2;\mathbb{R})\). We prove that \begin{equation*} \mu(f)=\int_{\mathbb{R}^{4}}\dfrac{f(x_{11}; x_{12};x_{21}; x_{22})}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}; \end{equation*} \(\forall\, f\in K(G;E)\) and \(x=\left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array}\right)\in GL(2;\mathbb{R})\); is a Haar measure on \( G=GL(2;\mathbb{R})\). Once this demonstration achieved we go straight to generalize the form of a vector measure on \( K(GL(2;\mathbb{R});\mathbb{R}^{4})\). This paper is organized as follows: in Section 2, we give some definitions related to vector measure and matrices and prove the fundamental theorem which will help us in proving our main result and in Section 3, we present our main result.

2. Preliminaries

In this section, we give basic definitions and concepts concerning with vector measure and Lie groups.

Definition 1.[2] Let \(G\) be a locally compact group and \(K(G;E)\) be the space of \(E\) valued functions with compact support on \(G.\) A vector measure on \(G\) with respect to Banach spaces \(E\) and \(F\) is a linear map: \begin{eqnarray*} m:& K(G;E)&\to F \\ &f& \mapsto m(f) \end{eqnarray*} such as \(\forall\,\, K \) compact in \(G\) \(\exists \,\,a_{K} > 0, \| m(f) \|_{F} \leq a_{K}\|f \|_{\infty}.\) Where \(\| . \|_{F}\) is the norm on Banach spaces \(F\) and \(\|f \|_{\infty}= sup \{ \|f(t)\|_{E}, t \in G \}\) is the norm on \(K(G;E).\)

The value \( m(f)\) of \(m\) in \(f \in K(G;E)\) is called integral of \(f\) with respect to \(m\) and can be written as [5, 6]: \begin{equation*} \int_{G} f(t)dm(t)= m(f). \end{equation*} We consider \(GL(2;\mathbb{R})\) is the set of matrices of order two with real coefficients whose determinant is not equal to zero, i.e., $$GL(2;\mathbb{R})=\left\lbrace g = \left( \begin{array}{cc} g_{11} & g_{12} \\ g_{21} & g_{22} \end{array}\right); g_{ij}\in \mathbb{R};1\leqslant i,j\leqslant 2 | detg \neq 0 \right\rbrace. $$ \(G= GL(2;\mathbb{R})\) is a Lie group. Also \(G\) is a manifold such that, at any point \(g\in G\), there exists an open \(V_{g}\) of \(G\), an open \(U_{g}\) of \(\mathbb{R}^{4}\) and \(\varphi_{g}\) a diffeomorphism of \(V_{g}\) in \(U_{g}\). So each \(x=\left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array}\right)\in GL(2;\mathbb{R})\) is assimilated to \((x_{11}; x_{12};x_{21}; x_{22})\) of \(\mathbb{R}^{4}.\) In order to prove our main result, first we prove following fundamental theorem which allow us to get our final result. The following theorem gives us Haar’s measure on \(GL(2;\mathbb{R}).\)

Theorem 1. Let \(K(G;E)\) be the space of \(E\) valued functions with compact support on \(G\), where \(G= GL(2;\mathbb{R})\) and \( E=\mathbb{R}^{4}\). Then \( \mu:K(G;E)\to \mathbb{R}^{+}\) defined as

\begin{equation} \mu(f)=\int_{\mathbb{R}^{4}}\dfrac{f(x_{11}; x_{12};x_{21}; x_{22})}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}; \end{equation}
(1)
\(\forall\, f\in K(G;E)\) and \(x=\left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array}\right)\in GL(2;\mathbb{R})\) is a Haar measure on \(G= GL(2;\mathbb{R})\).

Proof. If \(\mu\) is a Haar measure on \(G= GL(2;\mathbb{R})\) then \(d\mu(x)=\dfrac{1}{(detx)^{2}}dx, \quad \forall\,\, x \in G \). As \begin{equation*} d\mu(x)=\dfrac{1}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}. \end{equation*} Since \(\mu(f)=\int_{G}f(x)d\mu(x),\,\forall\, f\in K(G;E)\), we get:

\begin{equation} \mu(f)=\int_{\mathbb{R}^{4}}\dfrac{f(x_{11}; x_{12};x_{21}; x_{22})}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}; \end{equation}
(2)
Conversely if \(\mu\) is a Haar measure on \(G= GL(2;\mathbb{R})\) then we have [6]:
\begin{equation}\label{Equa3} \mu(f)=\int_{G}f(x)|J(L_{x})|^{-1}dx\quad \forall\,f\in K(G;E) \end{equation}
(3)
The translation on the left \(L_{x}: y\mapsto xy \quad x;y \in G= GL(2;\mathbb{R})\) \begin{equation*} L_{x}(y)=\left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array}\right)\left( \begin{array}{cc} y_{11} & y_{12} \\ y_{21} & y_{22} \end{array}\right) =\left( \begin{array}{cc} x_{11}y_{11}+ x_{12}y_{21}& x_{11}y_{12}+x_{12}y_{22} \\ x_{21}y_{11}+x_{22}y_{21} & x_{21}y_{12}+x_{22}y_{22} \end{array}\right) \end{equation*} \begin{equation*} L_{x}(y)= (x_{11}y_{11}+ x_{12}y_{21}; x_{11}y_{12}+x_{12}y_{22}; x_{21}y_{11}+x_{22}y_{21}; x_{21}y_{12}+x_{22}y_{22}) \end{equation*} \begin{equation*} J(L_{x})=\dfrac{dL}{dy}= \left(\begin{array}{cccc} x_{11} & 0& x_{12} & 0 \\ 0 & x_{11} & 0 & x_{12} \\ x_{21} & 0 & x_{22} & 0 \\ 0 & x_{21} & 0 & x_{22} \end{array} \right) \end{equation*} The Jacobian gives: \begin{equation*} |J(L_{x})|=\left|\dfrac{dL}{dy}\right|=x_{11}\left|\begin{array}{ccc} x_{11} & 0 & x_{12} \\ 0 & x_{22} & 0 \\ x_{21} & 0 & x_{22} \end{array} \right|+ x_{12}\left|\begin{array}{ccc} 0 & x_{11} & x_{12} \\ x_{21} & 0 & 0 \\ 0& x_{21} & x_{22} \end{array} \right| \end{equation*} \begin{eqnarray*} |J(L_{x})|=\left|\dfrac{dL}{dy}\right|&=&(x_{11}x_{22})^{2}-x_{11}x_{12}x_{21}x_{22}-x_{11}x_{12}x_{21}x_{22}+ (x_{12}x_{21})^{2}\\ &=&(x_{11}x_{22})^{2}-2 x_{11}x_{12}x_{21}x_{22}+ (x_{12}x_{21})^{2}\\ &=&(x_{11}x_{22}-x_{21}x_{12})^{2}\\ &=& (det\, x)^{2} \end{eqnarray*} According to the relationship (3) we have: \begin{eqnarray*} \mu(f)&=&\int_{G}f(x)|J(L_{x})|^{-1}dx\quad \forall\,f\in K(G;E)\quad x\in G\\ &=&\int_{G}f(x)\left( (det\, x)^{2}\right) ^{-1}dx\\ &=&\int_{G}\dfrac{f(x)}{(det\, x)^{2}}dx\\ &=& \int_{\mathbb{R}^{4}}\dfrac{f(x_{11}; x_{12};x_{21}; x_{22})}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}; \end{eqnarray*} The last form of \(\mu\) is a \(4\) linear, alternating, positive, finite, left-invariant form so it is a Haar measure on \(GL(2;\mathbb{R}).\)

The following theorem is of a capital importance.

Theorem 2. Let \(m\) be a bounded vector measure on compact Lie group \(G\) and \(E\) and \(F\) two Banach spaces. If \(m\) is a continuous alternating linear form in \(L^{p}(G;E)\) then \(m\) is absolutely continuous.

The following theorems are important, because once we establish the form of the Haar measure on \(G \), it will be easy to establish a vector measure on \(K(G;E).\)

Theorem 3. [3] Let \(G\) be a compact group, \(m\) be a bounded vector measure on \(G\) and \(p\) and \(q\) are two conjugates numbers with \( p\geq 1 \). Then the following two assertions are equivalent:

  1. \( \forall \, h \in L^{p}(G;E),m\ast h \in C(G;E), \)
  2. \( \exists \, f \in L^{q}(G;E)\; such \; as \quad m = f\mu. \)

Theorem 4.[3] Let \(\mu \) be a Haar measure on compact Lie group \(G\), \(p \in [1,\infty[\) and \(q\) conjugate of \(p\). If \(\phi\) is a linear continuous form on \(L^{p}(G;E)\) then there exists a map \(f \in L^{q}(G,E)\) such as for any \(g \in L^{p}\), we have \(\phi(g)=\int_{G} gfd\mu. \)

We use the duality theorem for \(p= 1 \) and \( q=\infty.\)

3. Main result

In this section, we give our main result.

Theorem 5. Let \(K(G;E)\) be the space of \(E\) valued functions with compact support on \(G\), where \(G= GL(2;\mathbb{R})\) and \( E=\mathbb{R}^{4}\). If \(m\) is a vector measure on \(K(G;E)\), then

\begin{equation} m(f)= \int_{\mathbb{R}^{4}}\dfrac{g(x_{11}; x_{12};x_{21}; x_{22})f(x_{11}; x_{12};x_{21}; x_{22})}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}; \end{equation}
(4)
\begin{equation*}\forall \quad f\in K(GL(2;\mathbb{R});\mathbb{R}^{4}),\quad g\in L^{\infty}(GL(2;\mathbb{R});\mathbb{R}^{4}), \quad x \in GL(2;\mathbb{R}) \end{equation*}

Proof. Since \(m\) being a vector measure on \(K(GL(2;\mathbb{R});\mathbb{R}^{4})\), so \(m\) is continuous, alternating and bounded linear form. Using Theorem 2, we get \(m \ll \mu\). The Theorems 3 and 4 allow us to write \(dm=gd\mu;\quad g\in L^{\infty}(GL(2;\mathbb{R}))\), which implies \begin{eqnarray*} dm(x)&=& \dfrac{g(x)}{(detx)^{2}}dx; \\ m(f)&=& \int_{G}\dfrac{g(x)f(x)}{(detx)^{2}}dx \quad f\in K(GL(2;\mathbb{R});\mathbb{R}^{4}) \\ &=& \int_{\mathbb{R}^{4}}\dfrac{g(x_{11}; x_{12};x_{21}; x_{22})f(x_{11}; x_{12};x_{21}; x_{22})}{(x_{11} x_{22}-x_{12} x_{21})^{2}}dx_{11}dx_{12}dx_{21}dx_{22}. \end{eqnarray*}

Acknowledgments

The authors would like to express their thanks to the referees for their useful remarks and encouragements.

Author contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflict of interests

The authors declare no conflict of interest.

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