text:
\begin{document}

\title[A simplified and unified generalization of some majorization results]{A simplified and unified generalization of some majorization results}

\author[Shirin Moein]{Shirin Moein\textsuperscript{1,2}}

\author[Rajesh Pereira]{Rajesh Pereira\textsuperscript{2}}

\author[Sarah Plosker]{Sarah Plosker\textsuperscript{3,2}}

\thanks{\textsuperscript{1}Department of Mathematical Sciences, Isfahan University of Technology, Isfahan 84156-83111, Iran}

\thanks{\textsuperscript{2}Department of Mathematics and Statistics, University of Guelph, Guelph, ON N1G 2W1, Canada}

\thanks{\textsuperscript{3}Department of Mathematics and Computer Science, Brandon University,
Brandon, MB R7A 6A9, Canada}

\keywords{matrix majorization;  multivariate majorization; sublinear functionals; convex functionals; stochastic operators; Markov operators;

}
\subjclass[2010]{
 	15B51,
 	26B25,
 	26D15,

 	47B65
}

\begin{abstract}
We consider positive, integral-preserving linear operators acting on $L^1$ space, known as stochastic operators or Markov operators. We show that, on finite-dimensional spaces, any stochastic operator can be approximated by a sequence of stochastic integral operators (such operators arise naturally when considering matrix majorization in $L^1$). We collect a number of results for vector-valued functions on $L^1$, simplifying some proofs found in the literature. In particular, matrix majorization and multivariate majorization are related in $\mathbb{R}^n$. In $\mathbb{R}$, these are also equivalent to convex function inequalities.
\end{abstract}

\maketitle

\section{Introduction}\label{sec:Intro}

In this work, we connect  several generalizations of majorization  in reference to vector-valued measurable functions; notably, matrix majorization, multivariate majorization, mixing distance, $f$-divergence, and coarse graining.
While some results are known, they appear rather obscure in the literature; we also simplify arguments when possible.

We first recall the definition of (vector) majorization: if $x, y\in \mathbb{R}^n$, we say $x$ is \emph{majorized} by $y$, denoted $x\prec y$, if \begin{eqnarray*}
\sum_{j=1}^{k}x^{\downarrow}_{j}\leq \sum_{j=1}^{k}y^{\downarrow}_{j}\quad \forall k\in \{1,\dots,n-1\}
\end{eqnarray*}
with equality when $k=n$, where $x$ has been reordered so that $x^{\downarrow}_{1}\geq x^{\downarrow}_{2}\geq \cdots \geq x^{\downarrow}_{n}$ (and similarly for $y$). A well-known theorem of Hardy, Littlewood, and P\'{o}lya states that $x\prec y$ is equivalent to the existence of a doubly stochastic matrix $S$ such that $x=Sy$ \cite[Theorem 8]{HLP}.

Consider now two matrices $R\in M_{m\times n}(\mathbb{R})$ and $T\in M_{p\times n}(\mathbb{R})$. We say $R$ is \emph{majorized} by $T$, denoted $R\prec T$ (where it is clear from context that this is \emph{matrix} majorization rather than vector majorization, although matrix majorization  is sometimes denoted $\prec_d$ or $\prec_S$ to distinguish it from vector majorization) if there exists a column stochastic matrix $S\in M_{m\times p}(\mathbb{R})$ such that $R= ST$. For more information on matrix majorization, see \cite{dahl1999}; when we restrict ourselves to the special case where $m=p$ and $S$ is doubly stochastic we get a more restrictive ordering called multivariate majorization, see \cite[Chapter 15]{MOA}. Matrix majorization has recently been generalized to quantum majorization between bipartite states \cite{Gour2017}.

  We denote by $L^1(X, \mu)$, or simply $L^1(X)$ if the measure $\mu$ is clear from context, the set of all functionals $f$ satisfying $\int_X|f|d\mu<\infty$.
 If $f\in L^1(X)$, the distribution function of $f$   is defined by $d_f(s)= \mu (\{x: f(x)>s\} )$ for all real $s$,
and the decreasing rearrangement of $f$ is defined by
\begin{eqnarray*}
f^\downarrow (t)&=& \inf \{s: d_f(s)\leq  t\},\quad \quad\quad 0 \leq  t \leq \mu( X)\\

&=& \sup \{s: d_f(s)> t\}, \quad \quad\quad 0 \leq  t \leq \mu(X).
\end{eqnarray*}

We are now in the position to  define continuous majorization. Typically the word ``continuous'' is dropped as it is clear from context.

\begin{definition}\label{def:cont} Let $(X,  \mu)$ and $(Y,  \nu)$ be finite measure spaces for which $a=\mu( X)=\nu(Y)$.
If $f\in L^1(X,\mu)$ and $g\in L^1( Y, \nu)$ satisfy
\begin{eqnarray*}
\int_0^t f^\downarrow d x&\leq & \int_0^t g^\downarrow d x\quad \forall t:\, 0\leq  t\leq  a\\
\textnormal{and }\int_0^a f^\downarrow d x&=& \int_0^a g^\downarrow d x,
\end{eqnarray*}
where the integration is with respect to Lebesgue measure, then we say that  $f$ is majorized by $g$, denoted $f\prec g$.
\end{definition}

Following \cite{dahl1999}, we define the positive homogeneous subadditive functionals on $\mathbb{R}^n$, also called \emph{sublinear} functionals, to be all functionals $\psi$ satisfying $\psi(\lambda x)=\lambda\psi(x)$ and $\psi(x+y)\leq \psi(x)+\psi(y)\ $ for all $x, y\in \mathbb{R}^n$, and $\lambda\geq 0$.

Part of  \cite[Theorem 3.3]{dahl1999} shows that if $R\in M_{m\times n}(\mathbb{R})$ and $T\in M_{p\times n}(\mathbb{R})$, then $R\prec T$ is equivalent to $\sum_{j=1}^m\psi(r_j)\leq \sum_{j=1}^p\psi(t_j)$  for all sublinear functionals $\psi$, where $r_j$ is the $j$th row of the matrix $R$, and similarly for $t_j$.

Given a measure space $(X,\mu)$, let $L^1(X, \mu, \mathbb{R}^n)$, or simply $L^1(X, \mathbb{R}^n)$, denote the set of all measurable functions $f$ from   $(X,\mu)$ to $\mathbb{R}^n$ that satisfy  $\int_X \vert f\vert d\mu <\infty$, where $\vert f \vert(x)=\sum_{k=1}^n |f_k(x)|$.

The notion of a stochastic matrix was generalized to a  stochastic operator on $L^1([0,1])$ in \cite{RSS1980}; we provide the corresponding definition for a stochastic operator from $L^1(Y, \nu)$ to  $L^1(X, \mu)$. Such an operator is sometimes referred to a   \emph{Markov operator} in the literature \cite{LMbook}.

\begin{definition}
Let $(X,\mu)$ and $(Y,\nu)$ be $\sigma$-finite measure spaces. A linear operator $S:L^1(Y)\to L^1(X)$ is called a {\em stochastic operator} if
\begin{enumerate}
    \item $S$ is positive (that is, $S$ takes positive elements to positive elements), and
    \item $\int_X Sf d\mu = \int_Y f d\nu, \quad \forall f\in L^1(Y)$.
\end{enumerate}
Moreover, if in addition to the two conditions above, $\mu(X)=\nu(Y)<\infty$ and $S1=1$, then $S$ is called a {\em doubly stochastic operator}.
\end{definition}

We have the following lemma which will be useful later on.

\begin{lemma} \label{absineq} Let $(X,\mu )$ and $(Y,\nu)$ be $\sigma$-finite measure spaces.  Let $f\in L^1(Y)$ and $S:L^{1}(Y)\to L^{1}(X)$ be a stochastic operator, then $\int_{X}\vert (Sf)(t)\vert d\mu(t)\le \int_{Y}\vert f(y)\vert d\nu(y)$. \end{lemma}

\begin{proof} Let $f_+(y)=\max(f(y),0)$ and $f_-(y)=\max(-f(y),0)$.  Then \begin{eqnarray*} \int_{X}\vert Sf(x) \vert d\mu(x) &\le &\int_{X} S(f_+)(x)d\mu(x) +  \int_{X} S(f_-)(x)d\mu(x)\\ &=& \int_{Y}f_+(y)d\nu(y)+\int_{Y}f_-(y)d\nu(y)\\   & =& \int_{Y}\vert f(y)\vert d\nu(y).\end{eqnarray*} \end{proof}

Note that the absolute value function is a nonnegative sublinear functional on $\mathbb{R}$; we will later show that a similar inequality holds for all nonnegative sublinear functionals on $\mathbb{R}^n$.  We first need to describe how $S$ acts on an element of $ L^1(Y,\mathbb{R}^n)$.
If $f=(f_1,f_2,...,f_n)\in L^1(Y,\mathbb{R}^n)$, then $S$ acts componentwise on $f$; that is,  $Sf=(Sf_1,Sf_2,...,Sf_n)$.

\begin{definition} \label{exp}  Let $(X,\mu)$ be a $\sigma$-finite measure space and let $\mathcal{P}=\{E_i\}_{i\in \mathbb{N}}$ be a partition of $X$ into disjoint measurable sets of finite measure.  We define $M_{\mathcal{P}}$ to be the operator which maps every $f\in L^1(X)$ to $\sum_{i\in \mathbb{N}} a_i \chi_{E_i}$ where $a_i=\frac{1}{\mu(E_i)}\int_{E_i}f(x)d\mu(x)$ if $E_i$ has positive measure and $a_i=0$ if $E_i$ is measure zero.
\end{definition}

 We note that it is easy to verify that $M_{\mathcal{P}}$ in Definition \ref{exp} is a  stochastic operator on $L^1(X)$; in fact, it maps $1\mapsto 1$ and is therefore a doubly stochastic operator.

Let $K$ be a convex set. A function $f:K\rightarrow K$ is \emph{affine} if $f(\lambda x +(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$ for all $x, y\in K$ and all $\lambda\in (0,1)$. We note that affine functions on the nonnegative face of the unit ball of $L^1$ are exactly the stochastic operators. Affine transformations on measure spaces are used to define coarse graining, a relation on the measurement statistics coming from two positive operator valued measures \cite{BQ1990, BQ1993, Heinonen, HZbook, Zan}.
A stochastic operator is an affine
 transformation between nonnegative faces of the unit balls in the respective measure spaces. Note that the $L^1$ norm is one of the few norms where the nonnegative elements of the unit ball form a face.

The following theorem is a combination of two well-known results in the literature.
\begin{theorem}\label{HLP}
If $f\in L^1(X,\mu)$, $g\in L^1(Y,\nu)$ where $\mu(X)=\nu(Y)< \infty$, then the following are equivalent:
\begin{enumerate}
    \item
$f\prec g$, as in Definition \ref{def:cont}.
\item For all convex functions $\phi:\mathbb{R}\to \mathbb{R}$,
$$\int_X \phi(f)d\mu\leq \int_Y \phi(g)d\nu.$$
\item
There exists a doubly stochastic operator $D:L^1(Y)\to L^1(X)$ such that $f=Dg$.
\end{enumerate}
\end{theorem}
\begin{proof}
Chong \cite{Chong} in Theorem 2.5 proved the equivalence of 1 and 2 and Day \cite{Day} in Theorem 4.9 proved the equivalence of 2 and 3.
\end{proof}

Of particular interest are the integral operators which are stochastic or doubly stochastic.

\begin{definition}
A \emph{stochastic kernel} is a measurable function $S:X\times Y\to[0,\infty)$ such that $\int_X S(x,y)d\mu(x)=1$ for almost all $y\in Y$.
A \emph{doubly stochastic kernel} is a stochastic kernel with the additional property that $\int_Y S(x,y)d\nu(y)=1$ for almost all $x\in X$.
\end{definition}

\begin{definition} An integral operator $M$ from $L^1(Y)$ to $L^1(X)$ given by $Mg=\int_{Y}S(x,y)g(y)d\nu(y)$ is said to be a \emph{stochastic integral operator} (resp.\ \emph{doubly stochastic integral operator}) if $S(x,y)$ is stochastic kernel (resp.\ doubly stochastic kernel).
\end{definition}

All stochastic integral operators are  stochastic operators, and all doubly stochastic integral operators are doubly stochastic operators. However, the converse of either statement is false. Indeed, consider the identity operator which is a doubly stochastic operator but is not a doubly stochastic integral operator nor a stochastic integral operator.

\section{Convex function inequalities}
We now discuss some properties of convex and sublinear functionals.

\begin{proposition}  \label{prop:sublinHLP} Let $K$ be a convex cone  of $\mathbb{R}^n$.  Let $\phi:K \to \mathbb{R}$ be a convex functional.  Then $\psi (v,x):=x\phi(\frac{v}{x})$ is a sublinear functional on the cone $K\times (0,\infty)$.
\end{proposition}

\begin{proof}
For $\lambda>0$, we have $\psi(\lambda(v,x))=\lambda x \phi (\frac{\lambda v}{\lambda x})=\lambda x \phi (\frac{v}{x})=\lambda \psi(v,x)$.
Next, let $v_1,v_2\in K$ and  $x_1,x_2\in (0,\infty)$. Then by convexity of $\phi$ we have
$$\phi\Big(\frac{v_1+v_2}{x_1+x_2}\Big)=\phi\Big(\frac{x_1v_1 }{x_1(x_1+x_2)}+ \frac{x_2v_2}{x_2(x_1+x_2)}\Big)\leq \frac{x_1}{x_1+x_2}\phi\Big(\frac{v_1}{x_1}\Big)+\frac{x_2}{x_1+x_2}\phi\Big(\frac{v_2}{x_2}\Big).$$

It follows that $\psi$ is a sublinear functional on $K\times  (0,\infty)$.
\end{proof}

Due to issues with convergence, we have avoided considering $x=0$ in the above proposition, whence the restriction to $ (0,\infty)$. However, one can consider $x\rightarrow 0^+$ to obtain the recession function $\phi_\infty$. Note that the converse of the above proposition is immediate; any sublinear functional on a convex set is automatically convex on that set.

\begin{proposition}\label{thm:Lipseq}
 Let $K$ be a convex subset of $\mathbb{R}^n$. For any continuous convex  nonnegative functional $\phi$ on $K$, there exists an increasing sequence of Lipschitz convex nonnegative functionals $\{\phi_k\}_{k=1}^\infty$ that converges pointwise to it on $K$. If further, $K$ is a convex cone and $\phi$ is sublinear, then $\{\phi_k\}_{k=1}^\infty$ can be taken to be sublinear.
\end{proposition}

If $K$ is a closed convex cone in $\mathbb{R}^n$, then in fact every continuous sublinear functional $\phi:K\to \mathbb{R}$
is Lipschitz.

To prove Theorem \ref{contmaj}, we require the generalization of Jensen's inequality to the multivariate case; see \cite[Proposition 16.C.1]{MOA}.

\begin{theorem}\label{thm:Jensen}(multivariate Jensen's inequality)   Let $(X, \mu)$ be a probability measure space. Let $\phi:\mathbb{R}^n\to \mathbb{R}$ be a convex function and let $f\in L^1(X, \mathbb{R}^n)$. Then   $\phi(\int_{X}f(x) d\mu(x) )\le \int_{X}\phi(f(x)) d\mu(x)$.
\end{theorem}

If $\phi$ is a sublinear function, we no longer require the measure to be a probability measure.

\begin{theorem}[Roselli-Willem inequality]\label{RW}\cite[Theorem 6]{RW} Let $(X, \mu)$ be an arbitrary measure space. Let $\phi:\mathbb{R}^n\to \mathbb{R}$ be a sublinear function and let $f\in L^1(X, \mathbb{R}^n)$.   Then   $\phi(\int_{X}f(x) d\mu(x) )\le \int_{X}\phi(f(x)) d\mu(x)$.
\end{theorem}

The related concept of $f$-divergence (which we call $\phi$-divergence since $\phi$ is convex) was introduced by Csisz{\'a}r \cite{Csiszar} and was studied extensively by statisticians \cite[Chapter 2]{CohenBook} and \cite{CD05,  JC15, LV06, MM13, SV}.  Similar concepts with different names have also appeared in the Physics literature  \cite[Chapter 6]{QIStext} and \cite{Morimoto, Zan}.

\begin{definition} Let $(X,\mu)$ be a measure space and $V$ be a real vector space.  Let $\phi$ be a real valued convex function on $V$.  Let $f:X \to V$ and let $h:X\to (0,\infty)$ with all functions being measurable.  Then the $\phi$-divergence of $f$ with respect to $h$ is $\int_X h\phi (\frac{f}{h}) \,d\mu$.
\end{definition}

The following result is useful in relating $\phi$-divergence inequalities with sublinear functional integral inequalities:

\begin{theorem} \label{thm:all} Let $(X,\mu)$ and $(Y,\nu)$ be measure spaces, $f\in L^1(X,K)$, $g\in L^1(Y,K)$, $h\in  L^1(X,(0,\infty))$ and  $k\in L^1(Y,(0,\infty))$, where $K$ is a convex cone. Define  $F:X \to K \times \mathbb{R}$ and $G:Y \to K \times \mathbb{R}$  as $F(x)=(f(x),h(x))$ and $G(y)=(g(y),k(y))$.      The following are equivalent.
\begin{enumerate}
\item\label{i:fdiv} The $\phi$-divergence of $f$ with respect to $h$ is less than or equal to that of $g$ with respect to $k$; that is,  $\int_X \phi (\frac{f}{h})h d\mu\leq  \int_Y \phi (\frac{g}{k})k d\nu$, for all convex functions $\phi:K\to \mathbb{R}$.
\item\label{i:sublin} $\int_X \psi (F(x))d\mu(x)\leq \int_Y \psi (G(y))d\nu(y)$ for all sublinear functionals $\psi:K\times (0,\infty)\to\mathbb{R}$.
\end{enumerate}  \end{theorem}

\begin{proof}

 (\ref{i:fdiv})$\Rightarrow $ (\ref{i:sublin}):

 Let $\psi(v,t) :K \times (0,\infty) \to \mathbb{R}$ be a sublinear functional. Let $\phi(v)=\psi(v,1)$ for all $v\in K$.  Then $\phi$ is  a real valued convex function on $K$. We can then use the sublinearity of $\psi$ to prove the following:

$\begin{array}{rcl}
\int_X \psi (F(x))d\mu(x)

                         & = & \int_X \phi (\frac{f(x)}{h(x)})h(x)d\mu(x)\\
                         & \le & \int_Y \phi (\frac{g(x)}{k(x)})k(x)d\nu(x)\\

                         & = & \int_Y \psi (G(x))d\nu(x).
 \end{array}$

  (\ref{i:sublin})$\Rightarrow $ (\ref{i:fdiv}): This implication is a straightforward application of Proposition \ref{prop:sublinHLP}. Let $\phi $ be any real-valued convex functional on $K$.  Then $\psi (v,x)=x\phi(\frac{v}{x})$ is a sublinear functional on $K\times (0,\infty)$, and we have

  $\begin{array}{rcl}

  \int_X \phi (\frac{f(x)}{h(x)})h(x)d\mu(x) & = &   \int_X \psi(f(x),h(x))d\mu(x)\\
                                              & \le & \int_Y \psi (g(y),k(y))d\nu(y)\\
                                               & = & \int_Y \phi (\frac{g(y)}{k(y)})k(y)d\nu(y).
   \end{array}$
   \end{proof}

\section{A generalization of matrix majorization}

With the notion of a stochastic operator from $L^1(Y, \nu)$ to  $L^1(X, \mu)$, we can generalize the definition of matrix majorization:

\begin{definition}\label{defn:matrixmaj}
Let $(X,\mu)$ and $(Y,\nu)$ be measure spaces.  Let $f=(f_1, f_2, \dots, f_n)\in L^1(X,\mathbb{R}^n)$ and $g=(g_1, g_2, \dots, g_n)\in L^1(Y,\mathbb{R}^n)$.  Then we say that $f$ is     \emph{matrix majorized} by $g$, denoted $f\prec_M g$, if there exists a stochastic operator $S$ such that $f=S(g)$; i.e., $f_k=Sg_k$ for all $k=1, \dots, n$.
\end{definition}

It is straightforward to check that  matrix majorization between measurable functions in $L^1$ is a reflexive, transitive relation and therefore is a preorder, which generalizes the same result in  \cite[Theorem 3.3]{dahl1999} for matrix majorization on matrices.

The term matrix majorization was coined by Dahl \cite{dahl1999}.  To see that our formulation is a generalization of Dahl's, we now restrict ourselves to the special case where $X=\{ 1,2,...,m\}$ and $Y=\{ 1,2,...,p\}$ are finite sets, $\mu$ and $\nu$ are counting measures.  We can represent each function $f=(f_1,f_2,...,f_n)\in L^1(X,\mathbb{R}^n)$ as an $n$ by $m$
matrix $A_{f}$ whose $k$th column is $f(k)$.   We can see that $f$ is matrix majorized by $g$ if there exists a row stochastic matrix $S$ such that $A_{f}=A_{g}S$; the later is Dahl's formulation of matrix majorization on matrices.

We now note that part of \cite[Theorem 3.3]{dahl1999} can now be rephrased as follows:

\begin{theorem} Let $X=\{ 1,2,...,m\}$, $Y=\{ 1,2,...,p\}$, $f\in L^1(X,\mathbb{R}^n)$ and $g\in L^1(Y,\mathbb{R}^n)$. Then $f$ is matrix  majorized by $g$ if and only if $\sum_{k=1}^m \phi (f(k)) \le \sum_{k=1}^p \phi (g(k))$ for all sublinear functionals  $\phi: \mathbb{R}^n\to \mathbb{R} $.    \end{theorem}

This suggests the following one-sided extension to the general case.

\begin{theorem} \label{contmaj} Let $(X,\mu)$ and $(Y,\nu)$ be measure spaces, $f\in L^1(X,\mathbb{R}^n)$ and $g\in L^1(Y,\mathbb{R}^n)$. If there exists a stochastic kernel $S(x,y): X\times Y\rightarrow [0,\infty)$ such that $f(x)=\int_Y S(x,y)g(y)d\nu(y)$ then $\int_X \phi (f(x))d\mu(x) \le \int_Y \phi (g(y))d\nu(y)$ for all sublinear functionals $\phi: \mathbb{R}^n\to \mathbb{R}$.    \end{theorem}

\begin{proof}
 Suppose  there exists a stochastic kernel $S(x,y): X\times Y\rightarrow [0,\infty)$ such that $f(x)=\int_Y S(x,y)g(y)d\mu(y)$.  Let $\phi :\mathbb{R}^n\to \mathbb{R}$ be sublinear.
 Hence by using Theorem \ref{RW} and Fubini's Theorem

$\begin{array}{rcl}
\int_X \phi (f(x))d\mu(x)& = &\int_X \phi (\int_Y S(x,y)g(y)d\nu(y))d\mu(x) \\
   &\le & \int_X \int_Y \phi (S(x,y)g(y))d\nu(y)d\mu(x)\\
   & = &  \int_X \int_Y S(x,y)\phi(g(y))d\nu(y)d\mu(x)\\
   & = & \int_Y (\int_X S(x,y)d\mu(x))\phi(g(y))d\nu(y)\\
   & = & \int_Y \phi(g(y))d\nu(y),
   \end{array}$

as desired.
\end{proof}

Note that, if we take $X=Y=[0,1]$ in Theorem \ref{contmaj}, then this is nearly the definition of mixing distance \cite[Definition 1a]{RSS1978}, except that the authors of \cite{RSS1978} take $\phi$ to be any convex functions, or certain subsets thereof, whereas we are working with sublinear (positively homogeneous convex) functionals in accordance with \cite{dahl1999}. This similarity hints at a connection between matrix majorization and the mixing distance.

\begin{lemma}\label{seqpart}  Let $(X,\mu)$ be a $\sigma$-finite measure space and let $f\in L^1(X)$.  Then there exists a sequence of partitions $\{\mathcal{P}_n\}_{n=1}^\infty$ of $X$ into disjoint sets of finite measure such that $\{M_{\mathcal{P}_n}f\}_{n=1}^\infty$ converges to $f$ in the $L^1$ norm. \end{lemma}

\begin{proof}  Let $P_n=\{E_k \}_{k=0}^{2n^2+1}$ where $E_0=\{ x\in X: f(x)<-n\}$  and $E_j=\{x\in X: f(x)\in [-n+\frac{j-1}{n},-n+\frac{j}{n})\} $ if $1\le j\le 2n^2$ and $E_{2n^2+1}=\{ x\in X: f(x)\ge n\}$.  If $E_k$ has infinite measure for some $k\in \{0, \dots, 2n^2+1\}$, since $X$ is $\sigma$-finite, $E_k$ is a countable disjoint union of sets of finite measure. Replace every such $E_k$ in the partition with the sets of finite measure.  It is then easy to verify that $\{M_{\mathcal{P}_n}f\}_{n=1}^\infty$ converges to $f$ in the $L^1$ norm. \end{proof}

We note that if  $\mathcal{P}=\{E_i\}_{i\in \mathbb{N}}$ and  $\mathcal{Q}=\{F_i\}_{j\in \mathbb{N}}$ are two partitions of $X$ into disjoint sets of finite measure, then we can form  the intersection partition $\mathcal{P}\cap \mathcal{Q}=\{E_i\cap F_j: i,j\in \mathbb{N}\}$.

\begin{lemma}\label{seqpart2}  Let $(X,\mu)$ be a $\sigma$-finite measure space and let $V$ be a finite dimensional subspace of $L^1(X)$.  Then there exists a sequence of partitions $\{\mathcal{P}_n\}_{n=1}^\infty$ of $X$  into disjoint sets of finite measure such that $\{M_{\mathcal{P}_n}f\}_{n=1}^\infty$ converges to $f$ in the $L^1$ norm for all $f\in V$. \end{lemma}

\begin{proof} The proof is by induction on the dimension of $V$ with the base case being Lemma \ref{seqpart}.  Now suppose the induction hypothesis holds for dimension $n$ and let $V$ be a subspace of dimension $n+1$.  Let $S$ be a subspace of $V$ of dimension $n$; by the induction hypothesis there exists a sequence of partitions $\{\mathcal{P}_n\}_{n=1}^\infty$ of $X$ into disjoint sets of finite measure such that $\{M_{\mathcal{P}_n}f\}_{n=1}^\infty$ converges to $f$ in the $L^1$ norm for all $f\in S$.  Now let $g\in V$ with $g\not \in S$, then by Lemma \ref{seqpart} there exists a sequence of partitions $\{\mathcal{Q}_n\}_{n=1}^\infty$ into disjoint sets of finite measure such that $\{M_{\mathcal{Q}_n}g\}_{n=1}^\infty$ converges to $g$ in the $L^1$ norm.  It is easy to see that sequence of intersection partitions  $\{\mathcal{R}_n=\mathcal{P}_n\cap \mathcal{Q}_n\}_{n=1}^\infty$ now satisfies the property that that $\{M_{\mathcal{R}_n}f\}_{n=1}^\infty$ converges to $f$ in the $L^1$ norm for all $f\in V$.  \end{proof}

\begin{theorem} \label{sto:markov} Let $(X,\mu )$ and $(Y,\nu)$ be $\sigma$-finite measure spaces.  Let $S:L^{1}(Y)\to L^{1}(X)$ be a stochastic operator and let $V$ be a finite dimensional subspace of $L^{1}(Y)$.  Then there exists a sequence of stochastic integral operators from $L^{1}(Y)$ to $L^{1}(X)$ which converges to $S$ on $V$. \end{theorem}

\begin{proof} Let $\mathcal{P}=\{E_i\}_{i\in \mathbb{N}}$ be a partition of $X$ into disjoint sets of finite measure.  Then $M_{\mathcal{P}}S$ is a stochastic operator from $L^{1}(Y)\to L^{1}(X)$; we will show that it is a stochastic integral operator. Fix $x\in X$. Then there exists a unique $k$ such that $x\in E_k$.  Define the functional $g_x(f)=(M_{\mathcal{P}}Sf)(x)$.  Then $$\begin{array}{rcl}  \vert g_x(f)\vert &=& \vert \frac{1}{\mu(E_k)}\int_{E_k}(Sf(t))d\mu(t)\vert \\
& \le&  \frac{1}{\mu(E_k)}\int_{X}\vert (Sf)(t)\vert d\mu(t) \\ & \le & \frac{1}{\mu(E_k)}\int_{Y}\vert f(y)\vert d\nu(y)\quad \textnormal{by Lemma \ref{absineq} }\\ & = &\frac{1}{\mu(E_k)}\Vert f\Vert_1.\end{array}$$  Hence $g_x$ is a bounded linear functional of $L^1(Y)$.  So by the Riesz representation theorem, there exist a nonnegative function $h_x\in L^{\infty}(Y)$ such that $g_x(f)=\int_Y f(y)h_x(y)d\nu(y)$.  Now let $K_{\mathcal{P}}(x,y)=h_x(y)$ for all $x\in X$ and all $y\in Y$. Since $K_{\mathcal{P}}(x,y)=\sum_{i\in \mathbb{N}}\chi_{E_i}(x)h_x(y)$,  $K_{\mathcal{P}}(x,y)$ is measurable. Then $(M_{\mathcal{P}}Sf)(x)=\int_Y K_{\mathcal{P}}(x,y)f(y)d\nu(y)$.

Since $M_{\mathcal{P}}S$ is a stochastic operator, we have
$$\int_X ((M_{\mathcal{P}}S)f)(x) d\mu(x)=\int_Y f(y)d\nu(y)\quad \forall f\in L^1(Y)$$ and by using Fubini's Theorem, we find

$$\begin{array}{rcl}
\int_Y f(y)d\nu(y) &=& \int_X ((M_{\mathcal{P}}S)f)(x) d\mu(x)\\ &=& \int_X\int_Y K_{\mathcal{P}}(x,y)f(y)d\nu(y)d\mu(x)\\
&=&\int_Y f(y) (\int_X K_{\mathcal{P}}(x,y)d\mu(x))d\nu(y).
\end{array}$$

Therefore $\int_X K_{\mathcal{P}}(x,y)d\mu(x)=1$ for almost all $y\in Y$.
Since $V$ is a finite dimensional subspace of $L^{1}(Y)$, the forward image $S(V)$ is a finite dimensional subspace of $L^{1}(X)$.  The result now follows from Lemma \ref{seqpart2}.
\end{proof}
We also have a doubly stochastic version of this theorem:
\begin{theorem}
Let $(X,\mu )$ and $(Y,\nu)$ be finite measure spaces. A doubly stochastic operator $D:L^1(Y)\to L^1(X)$ on a finite dimensional subspace $V$ of $L^1(Y)$ can be approximated by doubly stochastic integral operators.
\end{theorem}
\begin{proof}
 Let $\mathcal{P}=\{E_i\}_{i=1}^n$ be a partition of $X$ into disjoint sets of finite measure.
The  operator $M_\mathcal{P}:L^1(X)\to L^1(X)$ from Definition \ref{exp} is a doubly stochastic operator and since the composition of doubly stochastic operators is a doubly stochastic operator, $M_\mathcal{P}D$ is a doubly stochastic operator. By a proof similar to that of Theorem \ref{sto:markov}, for all $f\in L^1(Y)$ we have  $(M_{\mathcal{P}}Df)(x)=\int_Y K_{\mathcal{P}}(x,y)f(y)d\nu(y)$ such that  $\int_X K_{\mathcal{P}}(x,y)d\mu(x)=1$ for almost all $y\in Y$. Now suppose $f=1$. Then  $\int_Y K_{\mathcal{P}}(x,y)d\nu(y)=1$ for almost all $x\in X$ and hence $M_{\mathcal{P}}D$ is a doubly stochastic integral operator.
\end{proof}

   \begin{theorem}\label{main2}
   Let $(X,\mu)$ and $(Y,\nu)$ be two $\sigma$-finite measure spaces, $f\in L^1(X,\mathbb{R}^n)$ and $g\in L^1(Y,\mathbb{R}^n)$. If
  $f$ is matrix  majorized by $g$,

  then
   $$\int_X \phi (f(x))d\mu(x)\leq \int_Y \phi (g(y))d\nu(y)$$ for all nonnegative sublinear functionals
  $ \phi: \mathbb{R}^n\rightarrow [0,\infty)$.

   \end{theorem}
   \begin{proof}

Let $g=(g_1,...,g_n)$ and   $V=\operatorname{span}\{g_1,..., g_n\}$. Since $V$ is a finite dimensional subspace of $L^1(Y)$, by using Theorem \ref{sto:markov}, there exists a sequence of stochastic integral operators   $\{S_k\}_{k=1}^\infty$ which converges to  the stochastic operator $S$ coming from Definition \ref{defn:matrixmaj}.
Now by using Theorem \ref{contmaj}, for each $k\in\mathbb{N}$ we obtain $\int_X \phi(S_k g)d\mu(x)\leq \int_Y \phi(g)d\nu(y)$ for all  sublinear functionals $\phi$.

Since $\phi$ is sublinear on  $\mathbb{R}^n$, it is Lipschitz; denote its Lipschitz constant as $c$.  Then we have,
$$\int_{X}|\phi(S_kg)-\phi(Sg)|d\mu(x) \leq c\sum_{j=1}^n\int_X|S_kg_j-Sg_j|d\mu(x)\quad \forall k\in\mathbb{N}, \, \textnormal{ where } c\geq 0.$$
Since $\lim_{k\to \infty} S_kg_j=Sg_j$ in $L^1$ for all $j$, the left hand side must go to zero which means that $\lim_{k\to \infty }\int_{X}\phi(S_kg)d\mu(x)=\int_{X}\phi(Sg)d\mu(x) $.  Therefore we have
$$\int_X \phi(f)d\mu(x)=\int_X \phi(S g)d\mu(x)\leq \int_Y \phi(g)d\nu(y).$$

   \end{proof}

  We note that a special case of this result is a slight generalization of a theorem of Alberti; when $X=Y$ and $\mu=\nu$, Theorem \ref{main2} reduces to one direction of the following result which was proved using methods from the theory of von Neumann algebras.

  \begin{theorem}  \cite[Theorem 1]{Alberti}
   Let $(X,\mu)$  be  a $\sigma$-finite measure space and  $f,g\in L^1(X,\mathbb{R}^n)$.  Then
  $f$ is matrix  majorized by $g$,

  if and only if
   $$\int_X \phi (f(x))d\mu(x)\leq \int_X \phi (g(x))d\mu(x)$$ for all nonnegative sublinear functionals
  $ \phi: \mathbb{R}^n\rightarrow [0,\infty)$.
 \end{theorem}

 We do not know if the converse to Theorem \ref{main2} holds for arbitrary measures.

  We now consider a generalization of majorization known as multivariate majorization. In the setting of $\mathbb{R}^n$, we can show (Theorem \ref{thm:DSmultivar}) that  matrix majorization and multivariate majorization are strongly related.

  \begin{definition}
  Let $(X,\mu)$ and $(Y,\nu)$ be finite measure spaces, $f\in L^1(X,\mathbb{R}^n)$,  and $g\in L^1(Y,\mathbb{R}^n)$. Then $f$ is \emph{multivariate majorized} by $g$ if there exists a doubly stochastic operator $D:L^1(Y)\to L^1(X)$ such that  $f=Dg$.
  \end{definition}

 \begin{theorem}\label{thm:DSmultivar}
Let $(X,\mu)$ and $(Y,\nu)$ be finite measure spaces, $f\in L^1(X,\mu ,\mathbb{R}^n)$, $g\in L^1(Y,\nu ,\mathbb{R}^n)$, $h\in  L^1(X,\mu, (0,\infty))$, and  $k\in L^1(Y,\nu, (0,\infty))$. The following are equivalent:
\begin{enumerate}
    \item \label{i1} $(f_1, f_2, \dots, f_n, h)$ is matrix majorized by $(g_1, g_2, \dots, g_n, k)$; i.e.,
    there exists a stochastic operator $S:L^1(Y,\nu)\to L^1(X,\mu)$ such that $Sg_i=f_i$ for all $i=1,...,n$ and $Sk=h$,
    \item \label{i2}$\left(\frac{f_1}{h}, \frac{f_2}{h}, \dots, \frac{f_n}{h}\right)$ is multivariate majorized by $\left(\frac{g_1}{k}, \frac{g_2}{k}, \dots, \frac{g_n}{k}\right)$ with respect to measures $\alpha$ and $\beta$ where the measures $\alpha$ and $\beta$ are defined by
    $\alpha=h\, d\mu$ and $\beta=k\, d\nu$; i.e.,
    there exists a doubly stochastic operator $D:L^1(Y,\beta)\to L^1(X,\alpha)$ such that $D\frac{g_i}{k}=\frac{f_i}{h}$ for all $i=1,...,n$.
    \end{enumerate}
\end{theorem}

\begin{proof}
 Let  $T_s$ denote the multiplication operator which maps any function $f$ to the product $sf$.

 (\ref{i1}) $\Rightarrow (\ref{i2})$:
 Suppose there exists a stochastic operator $S:L^1(Y,\nu)\rightarrow L^1(X,\mu)$ such that $Sg_i=f_i$ for all $i=1, \dots, n$ and $Sk=h$. We now show that \newline $\left(\frac{f_1}{h}, \frac{f_2}{h}, \dots, \frac{f_n}{h}\right) \in L^1(Y, \beta)$ is multivariate majorized by $\left(\frac{g_1}{k}, \frac{g_2}{k}, \dots, \frac{g_n}{k}\right) \in L^1(X, \alpha)$.

   The multiplication operator $T_{1/h}$ is a stochastic operator from $L^1(X,\mu)$ to $L^1(X,\alpha)$.  Note that for all $i=1, \dots, n$,  $f_i\in L^1(X,\mu)$ and $T_{1/h}(f_i)=\frac{f_i}{h}$. Similarly,  $T_{k}$ is a stochastic operator from $L^1(Y,\beta)$ to $L^1(Y,\nu)$ and   for all $i=1, \dots, n$, $g_i\in L^1(Y,\beta)$ and  $T_{k}(g_i)=kg_i$. Construct $D=T_{1/h}ST_{k}:L^1(Y,\beta)\to L^1(X,\alpha)$, which  is a stochastic operator since it is a product of stochastic operators. Furthermore, $D1=1$,   therefore $D$ is a doubly stochastic operator that maps $\frac{g_i}{k}$ to $\frac{f_i}{h}$.

 (\ref{i2}) $\Rightarrow (\ref{i1})$:   Assume there exists a  doubly stochastic $D:L^1(Y, \beta)\to L^1(X,\alpha)$ such that $\frac{f_i}{h}=D\frac{g_i}{k}$. We define a stochastic operator $T_{h}:L^1(X,\alpha)\to L^1(X,\mu)$ such that for all $i=1, \dots, n$, $f_i\in L^1(X,\mu )$, and $T_{h}(f_i)=f_ih$. Similarly, define $T_{1/k}:L^1(Y,\nu)\to L^1(Y,\beta)$ such that for all  $i=1, \dots, n$, $g_i\in L^1(Y,\nu)$ and $T_{1/k}(g_i)=\frac{g_i}{k}$. Construct  $S=T_{h}DT_{1/k}:L^1(Y,\nu)\to L^1(X,\mu)$. Since $D$ is a (doubly) stochastic operator and the product of stochastic operators is a stochastic operator, $S$ is a stochastic operator such that $Sg_i=f_i$ for all $i=1,...,k$ and $Sk=h$.
\end{proof}

In the setting of $\mathbb{R}$ we can show that  matrix majorization,  multivariate majorization, and the convex function inequalities are all strongly related. The following theorem can be viewed as a simplified version of the result on mixing distance in \cite{RSS1978}.

 \begin{theorem}
Let $(X,\mu)$ and $(Y,\nu)$ be finite measure spaces, $f\in L^1(X,\mathbb{R})$, $g\in L^1(Y,\mathbb{R})$, $h\in  L^1(X,(0,\infty))$,  $k\in L^1(Y,(0,\infty))$, with $\int_X h \,d\mu=\int_Y k \,d\nu$. The following are equivalent:
\begin{enumerate}
\item \label{i:S} There exists a stochastic operator $S:L^1(Y, \nu)\rightarrow L^1(X, \mu)$ such that $Sg=f$ and $Sk=h$.
    \item \label{i:C} For all real valued convex functions on $\mathbb{R}$,
$$\int_X \phi \left(\frac{f}{h}\right)h \,d\mu\leq  \int_Y \phi \left(\frac{g}{k}\right)k \,d\nu. $$
\item \label{i:DS}
There exists a  doubly stochastic $D:L^1(Y, \beta)\to L^1(X, \alpha)$ such that $D\left(\frac{g}{k}\right)=\frac{f}{h}$, where the measures $\alpha$ and $\beta$ are defined by
 $\alpha=h\, d\mu$ and $\beta=k\, d\nu$.
\end{enumerate}
\end{theorem}
\begin{proof}

Both (\ref{i:C}) and (\ref{i:DS}) are equivalent to $\frac{g}{k}\prec \frac{f}{h}$ by Theorem \ref{HLP}.

The equivalence of (\ref{i:S}) and (\ref{i:DS}) follows from Theorem \ref{thm:DSmultivar} with $n=1$.
\end{proof}

\section*{Acknowledgements}
S.M.\  was supported by a travel grant from the Iranian Ministry of Science Research and Technology. She gratefully acknowledges  the University of Guelph and Brandon University for the time she spent at the respective universities during the course of this work. R.P.\ was supported by NSERC Discovery Grant number 400550. S.P.\ was supported by NSERC Discovery Grant number 1174582, the Canada Foundation for Innovation (CFI) grant number 35711, and the Canada Research Chairs (CRC) Program grant number 231250.  The authors thank the anonymous referee for their helpful comments.


text:
\begin{document}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{defin}{Definition}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{corol}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{rem}{Remark}[section]
\newtheorem{example}{Example}[section]

\title{Soldered tensor fields of normalized submanifolds}
\author{{\small by}\vspace{2mm}\\Izu Vaisman}
\date{{\footnotesize Dedicated to the centenary of the Mathematical
Seminar ``Al. Myller", Jassy, Romania and to Acad. Prof.
Constantin Corduneanu on his 80-eth anniversary}}
\maketitle
{\def\thefootnote{*}\footnotetext[1]
{{\it 2000 Mathematics Subject Classification: 53C40} .
\newline\indent{\it Key words and phrases}: Normalized submanifold, Soldered
tensor field}}
\begin{center} \begin{minipage}{12cm}
A{\footnotesize BSTRACT. Soldered forms, multivector fields and
Riemannian metrics were studied in our earlier paper
\cite{V1}. In particular, it was shown that a Riemannian
submanifold is totally geodesic iff the metric is soldered to the
submanifold. In the present paper, we discuss general, soldered
tensor fields. In particular, we prove that the almost complex
structure of an almost K\"ahler manifold is soldered to a
submanifold iff the latter is an invariant, totally geodesic
submanifold.}
\end{minipage}
\end{center}
\vspace*{5mm}

\section{Introduction}
In the present paper, all the manifolds, mappings, bundles, tensor
fields, etc. are differentiable of class $C^\infty$ and we use the
standard notation of Differential Geometry, including the Einstein
summation convention. The reader may consult \cite{KN} for all the
differential geometric notion and results that are used in the
paper.

If $N^n$ is a submanifold of $M^m$ (indices denote dimension),
then a {\it normalization} of $N$ by a {\it normal bundle} $\nu N$
is a splitting
\begin{equation} \label{split} TM|_N=TN\oplus\nu N
\end{equation} ($T$ denotes tangent bundles). A submanifold
endowed with a normalization is called a {\it normalized
submanifold} and a vector field $X$ on $M$ is tangent or normal to
$N$ if $X|_N$ belongs to $TN,\nu N$, respectively. The best known
case is that of a {Riemannian normalization} $\nu N=T^{\perp_g}N$,
where $g$ is a Riemannian metric on $M$. In fact, given an
arbitrary normalization, it is easy to construct metrics $g$ such
that the normalization is $g$-Riemannian. Similarly, if $N$ is a
symplectic submanifold of a symplectic manifold $(M,\omega)$, $\nu
N=T^{\perp_\omega}N$ defines the {\it symplectic normalization}.
Another interesting example is that of a submanifold $N$ such
that, $\forall x\in N$, $T_xM=T_xN\oplus T_x\mathcal{F}$, where $
\mathcal{F}$ is a foliation  of $M$; then we may take $\nu
N=T\mathcal{F}|_N$.

In our earlier paper \cite{V1} we discussed differential forms,
multivector fields and Riemannian metrics that have a special kind
of contact with a normalized submanifold; these were said to be
{\it soldered} to the submanifold. In particular, it was shown
that a Riemannian submanifold is totally geodesic iff the metric
is soldered to the submanifold and that a submanifold of a Poisson
manifold is a (totally) Dirac submanifold iff there exists a
normalization such that the Poisson bivector field is soldered to
the submanifold with respect to this normalization.

In the present paper we give a general definition for the notion
of soldering of an arbitrary tensor field and we consider an
obstruction to soldering, which, essentially, is a generalization
of the second fundamental form of a Riemannian submanifold. We
establish some formulas for the calculation of this obstruction
and get corresponding applications. In particular, we prove that
the almost complex structure $J$ of an almost K\"ahler manifold is
soldered to a submanifold iff the latter is a $J$-invariant,
totally geodesic submanifold.
\section{Soldered tensor fields}
Let $(N^n,\nu N)$ be a normalized submanifold of $M^m$ and let
$\iota:N\subseteq M$ be the corresponding embedding.

First, we exhibit some adequate, local coordinates around the
points of $N$. Let $\sigma:W\rightarrow N$ be a tubular
neighborhood of $N$ such that $\forall x\in N$, $T_x(W_x)=\nu_x N$
($W_x$ is the fiber of $W$ and $\nu_xN$ is the fiber of $\nu N$ at
$x$). For every point $x\in N$ there exists a
$\sigma$-trivializing neighborhood $U$ with coordinates $(x^a)$
$(a,b,c,...= 1,...,m-n)$ around $x$ on the fibers of $\sigma$,
such that $x^a|_{N\cap U}=0$, and coordinates $(y^u)$
$(u,v,w,...=m-n+1,...,m)$ around $x$ on $N\cap U$. We say that
$(x^a,y^u)$ are {\it adapted local coordinates}.

Then,
\begin{equation} \label{adapted} TN|_{N\cap U}=span\left\{\left.\frac{\partial}
{\partial y^u}\right|_{x^a=0}\right\},\;\; \nu N|_{N\cap U}
=span\left\{\left.\frac{\partial} {\partial
x^a}\right|_{x^a=0}\right\}\end{equation} and the transition
functions between systems of adapted local coordinates have the
local form
\begin{equation}
\label{transition} \tilde x^a=\tilde x^a(x^b,y^v),\;\tilde
y^u=\tilde y^u(y^v),
\end{equation} where \begin{equation} \label{transcond}
\left.\frac{\partial\tilde x^a}{\partial y^v}\right|_{x^b=0}=0,
\;\frac{\partial\tilde y^u}{\partial x^b}\equiv0.
\end{equation}
Furthermore, (\ref{split}) implies
\begin{equation} \label{dualsplit}
T^*M|_N=T^*N\oplus\nu^*N,\end{equation} and
\begin{equation}\label{dualspan}
\begin{array}{c} T^*N=ann(\nu N)= span\{dy^u|_{x^a=0}\},\vspace{2mm}\\
\nu^*N=ann(TN)=span\{dx^a|_{x^a=0}\} \end{array}
\end{equation} ($ann$ denotes annihilator spaces).

Now we give the following general definition.
\begin{defin}\label{formsold} {\rm A tensor field
$A\in\mathcal{T}^p_q(M)$ (where $\mathcal{T}$ denotes a space of
tensor fields) is {\it soldered} to the normalized submanifold
$(N,\nu N)$ if for any normal vector field $X\in\Gamma TM$ of $N$
one has
\begin{equation}\label{eqsoldA} (L_XA)_x(Y_1,...,Y_q,\xi_1,...,\xi_p)=0,
\end{equation} for any $x\in N$ and any arguments
$Y_1,...,Y_q\in T_xN,\xi_1,...,\xi_p\in ann\,\nu_x N$.}
\end{defin}

In (\ref{eqsoldA}) $L$ denotes the Lie derivative and it turns out
that (\ref{eqsoldA}) is a combination of algebraic and
differential conditions. Indeed, we have
\begin{prop}\label{propalogcond} If the tensor field
$A\in\mathcal{T}^p_q(M)$ is soldered to the normalized submanifold
$(N,\nu N)$, then, for any fixed vectors $Y_1,...,Y_q\in T_xN$ and
covectors $\xi_1,...,\xi_p\in ann\,\nu_x N$, the following algebraic
conditions must hold:

1) the $1$-forms $\alpha_i\in T^*M|_N$, $i=1,...,q$, defined by
\begin{equation}\label{defalpha}\alpha_i(V)=A|_N(Y_1,...,Y_{i-1},V,
Y_{i+1},...,Y_q,\xi_1,...,\xi_p),\;\;V\in TM|_N,\end{equation}
belong to $ann\,\nu N$;\\

2) the vector fields $Z_j\in TM|_N$, $j=1,,,,p$, defined by
\begin{equation}\label{defZ}
Z_j(\gamma)=A|_N(Y_1,...Y_q,\xi_1,...,,\xi_{j-1},\gamma,
\xi_{j+1},...,\xi_p),\;\;\gamma\in T^*M|_N,\end{equation} are tangent to $N$.
\end{prop}
\begin{proof} Consider the general formula
\begin{equation}\label{Lgenform}(L_{\varphi V}A)(Y_1,...,Y_q,\xi_1,...,\xi_p)=
\varphi(L_{V}A)(Y_1,...,Y_q,\xi_1,...,\xi_p)\end{equation}
$$-\sum_{i=1}^q(Y_i\varphi)\alpha_i(V)-\sum_{j=1}^p\xi_j(V)(Z_j\varphi),$$
where $\alpha_i(V)$ and $Z_j\varphi=d\varphi(Z_j)$ are defined by
(\ref{defalpha}), (\ref{defZ}), respectively; the formula holds
for arbitrary arguments (not necessarily related to $N$) and for
any function $\varphi\in C^\infty(M)$. Conditions 1), 2), follow
from (\ref{Lgenform}) by taking $V=fX+lY$ where $l|_N=0$; this
vector field is normal to $N$ again, therefore, it also satisfies
(\ref{eqsoldA}). \end{proof}
\begin{prop}\label{condsoldloc} The tensor field $A\in\mathcal{T}^p_q(M)$
is soldered to the normalized submanifold $N$ iff the local
components of $A$ with respect to adapted coordinates satisfy the
conditions
\begin{equation}\label{12local}
A_{u_1,...,u_{i-1},a,u_{i+1},...,u_q}^{v_1,...,v_p}(0,y^w)=0,\;
A_{u_1,...,u_q}^{v_1,...,v_{j-1},a,v_{j+1},...,v_p}(0,y^w)=0
\end{equation} and
\begin{equation}\label{eqsoldloc}\left.
\frac{\partial A^{v_1,...,v_p}_{u_1,...,u_q}}{\partial
x^a}\right|_{x^b=0}=0.\end{equation}\end{prop}
\begin{proof} Using the bases (\ref{adapted}), (\ref{dualspan}),
we see that conditions 1), 2) of Proposition \ref{formsold} are
equivalent to (\ref{12local}) and (\ref{eqsoldloc}) is
(\ref{eqsoldA}) expressed for $X=\partial/\partial x^a$.
Conversely, using formula (\ref{Lgenform}), it is easy to derive
(\ref{eqsoldA}) from (\ref{eqsoldloc}) and the algebraic
conditions 1), 2).\end{proof}

In the case of either a differential form or a multivector field
formulas (\ref{12local}), (\ref{eqsoldloc}) reduce to the
conditions for soldering forms and multivector fields given in
\cite{V1}.
\begin{example}\label{exfol} {\rm Assume that there exists a
foliation $\mathcal{F}$ of $M$ such that $\nu N=T\mathcal{F}|_N$
is a normalization of $N$. A tensor field $A\in\mathcal{T}^p_q(M)$
is said to be {\it projectable} or {\it foliated} if for any local
quotient manifold $Q_U=U/\mathcal{F}\cap U$ ($U$ is an open
neighborhood in $M$ where $\mathcal{F}$ is simple) there exists a
tensor field $A'\in\mathcal{T}^p_q(Q_U)$ that is $\pi$-related to
A ($\pi$ is the natural projection $U\rightarrow Q_U$). Let
$(y^u,x^a)$ be local coordinates such that the local equations of
the leaves are $y^u=const.$ (In particular, around points $x\in N$
we may use $N$-adapted local coordinates.) Then, it is easy to see
that $A$ is projectable iff it has a local expression of the
following form
\begin{equation}\label{Aproj} A= dy^{u_1}\otimes...\otimes
dy^{u_q}\otimes
[A_{u_1...u_q}^{v_1...v_p}(y)\frac{\partial}{\partial y^{v_1}}
\otimes...\otimes\frac{\partial}{\partial y^{v_p}}\end{equation} $$+
A_{u_1...u_q}^{a_1v_2...v_p}(x,y)\frac{\partial}{\partial x^{a_1}}
\otimes\frac{\partial}{\partial
y^{v_2}}...\otimes\frac{\partial}{\partial y^{v_p}} + ... +
A_{u_1...u_q}^{a_1a_2...a_p}(x,y)\frac{\partial}{\partial x^{a_1}}
...\otimes\frac{\partial}{\partial x^{a_p}}].$$ Formula
(\ref{Aproj}) shows that the projectable tensor fields are
characterized by the following global properties:

\hspace*{1.5mm}(i)\hspace*{5mm}
$A\in[\otimes^q(ann\,T^*\mathcal{F})]\otimes[\otimes^pTM]$,

(ii)\hspace*{5mm} $\forall X\in T\mathcal{F}$ one has $L_XA\in
T\mathcal{F}\otimes[\otimes^q(ann\,T^*\mathcal{F})]\otimes[\otimes^{p-1}TM]$.\\
Accordingly, we see that an $ \mathcal{F}$-projectable tensor field
$A$ is soldered to the submanifold $N$ iff the algebraic condition
2) of Proposition \ref{formsold} is satisfied. In particular, a
totally covariant, foliated tensor field necessarily is soldered to
any local transversal submanifold $N^n$ of the foliation
$\mathcal{F}^{m-n}$.}\end{example}
\begin{defin}\label{algadapt} {\rm A tensor field
$A\in\mathcal{T}^p_q(M)$ that satisfies the algebraic conditions 1),
2) of Proposition \ref{formsold} (equivalently, satisfies
(\ref{12local})) will be called {\it algebraically adapted to
$N$}.}\end{defin}
\begin{prop}\label{form2} If $A\in\mathcal{T}^p_q(M)$ is
algebraically adapted to $N$, the morphism $w_A:\nu N\rightarrow
\mathcal{T}^p_q(N)$ defined by
\begin{equation}\label{defw} w_A(\bar
X)(Y_1,...,Y_q,\xi_1,...,\xi_p)=L_{
X}A(Y_1,...,Y_q,\xi_1,...,\xi_p)|_N,\end{equation} where
$Y_1,...,Y_q\in\Gamma TN$, $\xi_1,...,\xi_p\in\Gamma(ann\,\nu N)$,
$\bar X\in\Gamma\nu N$ and $X$ is a vector field on $M$ with the
restriction $\bar X$ to $N$, is independent of the choice of the
extension $X$ of $\bar X$.
\end{prop} \begin{proof} Since $A$ is algebraically $N$-adapted,
formula (\ref{Lgenform}) yields
\begin{equation}\label{invarianta}(L_{fX+lZ}A)(Y_1,...,Y_q,\xi_1,...,\xi_p)|_N=
f(L_{X}A)(Y_1,...,Y_q,\xi_1,...,\xi_p)|_N,\end{equation} for any
functions $f,l\in C^\infty(M)$ such that $l|_N=0$, any vector
field $X$ normal to $N$ and any vector field $Z$ on $M$. The case
$Z=0$ shows that $w_A$ is $C^\infty(N)$-linear in $\bar{X}$. The
case $f=1$ shows that $w_A(\bar{X})$ is independent of the choice
of the extension $X$ of $\bar{X}$, since, using coordinate
expressions, it easily follows that two extensions $X_1,X_2$ are
related by an equality of the form $X_2=X_1+\sum l_kZ_k$ where the
functions $l_k$ vanish on $N$.
\end{proof}
\begin{defin}\label{defobstr} {\rm The morphism $w_A$ will be called
the {\it soldering obstruction} of the algebraically $N$-adapted
tensor field $A$.}\end{defin}

The name is motivated by the fact that if $w_A=0$ then $A$ is
soldered to $N$. Notice that $w_A(\bar X)$ has the same symmetries
like $A$.
\section{Applications}
In this section we consider only Riemannian normalizations,
therefore, $M$ is endowed with a Riemannian metric $g$ and $\nu
N=T^{\perp_g}N$. All the vector fields denoted by $Y$ are tangent to
$N$ and all the vector fields denoted by $X$ are normal to $N$. The
reader is asked to pay attention to the situations where
calculations take place only along $N$. In \cite{V1}, as a
consequence of the Gauss-Weingarten formulas \cite{KN}, we proved
that the soldering obstruction of the metric $g$ is
\begin{equation}\label{invptg}
w_g(\bar X)(Y_1,Y_2)=(L_Xg)(Y_1,Y_2)|_N=-2g(\beta(Y_1,Y_2),\bar
X)\end{equation} where $X|_N=\bar X\in\Gamma\nu N$ and $\beta$ is
the second fundamental form of $N$. Accordingly, the metric is
soldered to a submanifold $N$ iff $N$ is a totally geodesic
submanifold of $M$.

For more applications we compute the soldering obstruction of a
tensor field $A$ of type $(1,1)$. From (\ref{12local}) it follows
that $A$ is algebraically adapted to $(N,\nu N)$ iff both $TN$ and
$\nu N$ are invariant by the endomorphism $A$ and we shall assume
that this condition holds. Together with the soldering invariant of
$A$ we define the bilinear {\it soldering form}
$\sigma_A(Y_1,Y_2)\in\nu N$ given by
\begin{equation}\label{soldform}
g(\sigma_A(Y_1,Y_2),\bar X)=w_A(\bar
X)(Y_1,\flat_gY_2).\end{equation} Of course, $A$ is soldered to
$(N,T^{\perp_g}N)$ iff $\sigma_A=0$.
\begin{prop}\label{symsold} Assume that the operator $A$ is either
symmetric or skew-symmetric with respect to $g$. Then, $\sigma_A$ is
symmetric, respectively, skew-symmetric iff $A$ is symmetric,
respectively, skew-symmetric with respect to the second fundamental
form $\beta$ of the submanifold $N$ of $(M,g)$.\end{prop}
\begin{proof} The assumed symmetry property is
\begin{equation}\label{symA}
g(AV_1,V_2)=\pm g(V_1,AV_2),\hspace{3mm} \forall V_1,V_2\in\Gamma
TM.\end{equation} If we take the Lie derivative $L_X$ of this
equality, modulo the equality itself, and use (\ref{invptg}), we
get
$$g(\sigma_A(Y_1,Y_2)\mp\sigma_A(Y_2,Y_1),\bar X)=
2g(\beta(AY_1,Y_2)\mp\beta(Y_1,AY_2),\bar X),$$ whence the
conclusion.\end{proof}

The following proposition expresses the soldering form of a
$g$-(skew)-symmetric $(1,1)$-tensor field $A$ in terms of the
Levi-Civita connection $\nabla$ of $g$ and the second fundamental
form $\beta$ of the submanifold $N$.
\begin{prop}\label{legAR} Assume that the algebraically
$N$-adapted tensor field $A\in\mathcal{T}_1^1(M)$ satisfies
{\rm(\ref{symA})}. Then, the following formula, where the sign in
the right hand side is opposite to the sign in {\rm(\ref{symA})},
holds:
\begin{equation}\label{legAR2}\begin{array}{l}
g(\sigma_A(Y_1,Y_2),\bar X)=g(\nabla_{\bar
X}A(Y_1),Y_2)\vspace*{2mm}\\ +g(\beta(AY_1,Y_2)\mp
\beta(Y_1,AY_2),\bar X).\end{array}\end{equation}
\end{prop}
\begin{proof} We prove the equality
at every fixed point $x\in N$. During the calculations, we extend
the vectors $Y_1(x),Y_2(x),\bar X(x)$ to vector fields $\tilde
Y_1,\tilde Y_2,X$ on $M$ that are tangent, respectively, normal to
$N$. Since the final result is independent of the choice of the
extension, we may use local, adapted coordinates and take
\begin{equation}\label{extensiiaux}\tilde
Y_1=\mu^u_1\frac{\partial}{\partial y^u}\,,\tilde
Y_2=\mu^u_2\frac{\partial}{\partial y^u}\, ,
X=\xi^a\frac{\partial}{\partial x^a},\end{equation} where
$\mu^u_1,\mu^u_2,\xi^a$ are constant (namely, the components of
$Y_1(x),Y_2(x),\bar X(x)$ at the fixed point $x$). From the
equality
$$L_X(A\tilde Y_1)=[X,A\tilde Y_1]=\nabla_X(A\tilde Y_1)-\nabla_{A\tilde
Y_1}X,$$ we get
$$[(L_XA)(Y_1)]_x=[A\nabla_{X}\tilde Y_1-\nabla_{AY_1}X
+(\nabla_{X}A)(Y_1)]_x.$$ In this result we may replace
$\nabla_{A_xY_1(x)} X=-W_{\bar X(x)}\tilde{Y}_1+D_{Y_1(x)}X$,
where $W$ is the Weingarten operator of $N$ and $D$ is the
connection induced by $\nabla$ in $\nu N$. Then, using also
(\ref{symA}), we get
$$[g(\sigma_{A}(Y_1,Y_2),\bar X)]_x=[g(\beta(AY_1,Y_2),\bar X)\pm
g(\nabla_{\bar X}\tilde Y_1,AY_2)+g((\nabla_{\bar
X}A)(Y_1),Y_2)]_x.$$ But, $\nabla_{\bar X}\tilde Y_1=
\nabla_{Y_1}X+[X,\tilde Y_1]$, and the last bracket
vanishes for the chosen extensions (\ref{extensiiaux}).
Accordingly, $$[g(\nabla_{\bar X}\tilde
Y_1,AY_2)]_x=[g(\nabla_{Y_1}X,AY_2)=g(-W_{\bar
X}Y_1+D_{Y_1}X,AY_2)]_x$$
$$=-[g(\beta_x(Y_1,AY_2),X)]_x$$ and (\ref{legAR2}) follows.
\end{proof}
\begin{corol}\label{corolAR} Assume that $A$ is parallel with respect
to the Levi-Civita connection of $g$. Then, if $A$ is $g$-symmetric
$\sigma_A=0$ and if $A$ is $g$-skew-symmetric
$\sigma_A(Y_1,Y_2)=2\beta(AY_1,Y_2)$, where $\beta$ is the
Riemannian, second fundamental form of $N$ in $M$.
\end{corol}
\begin{proof}
The Gauss equation $$\nabla_{Y_1}(A\tilde Y_2)=\nabla'_{Y_1}(A
Y_2)+\beta(AY_1,Y_2),$$ where $\nabla'$ is the induced Levi-Civita
connection on $N$, implies that, if $\nabla A=0$, then
$\beta(AY_1,Y_2)=A\beta(Y_1,Y_2)$. Similarly,
$\beta(Y_1,AY_2)=A\beta(Y_1,Y_2)$. Inserting $\nabla A=0$ and the
previous results for $\beta$ in (\ref{legAR2}) we get the
announced results.\end{proof}

Another nice formula is given by
\begin{prop}\label{AsiNij} Let $A$ be either a $g$-symmetric
or a $g$-skew-symmetric $(1,1)$-tensor field on $(M,g)$ that is
algebraically adapted to the submanifold $N$ and let
$\mathcal{N}_A$ be its Nijenhuis tensor. Then, one has
\begin{equation}\label{sigmaNij}
g(\sigma_A(Y_1,Y_2),A\bar X)=\pm g(\sigma_A(Y_1,AY_2),\bar
X)+g(\mathcal{N}_A(X,Y_1),Y_2).\end{equation}
\end{prop}
\begin{proof} With the notation in the proof of Proposition \ref{legAR},
if $x\in N\subseteq M$, one has the following expression of the
Nijenhuis tensor
\begin{equation}\label{NijcuL}
[(\mathcal{N}_A)(
X,\tilde{Y}_1)]_x=[(L_{AX}A)(\tilde{Y}_1)-A(L_{X}A)(\tilde{Y}_1)]_x.\end{equation}
The required result follows by taking the $g$-scalar product of
the previous equality by $Y_2$.\end{proof}

For a concrete application, let $(M,J,g)$ be an almost Hermitian
manifold, which means that the almost complex structure $J$ is
$g$-skew-symmetric. The tensor field $J$ is algebraically adapted
to the submanifold $N$ iff $N$ is $J$-invariant, which we shall
assume hereafter. If $(M,J,g)$ is a K\"ahler manifold and $N$ is a
complex submanifold, Corollary
\ref{legAR} gives $\sigma_J(Y_1,Y_2)=2\beta(JY_1,Y_2)$ and we see
that $J$ is soldered to $(N,T^{\perp_g}N)$ iff $N$ is a totally
geodesic submanifold.

We shall extend this result to almost K\"ahler manifolds. For any
almost Hermitian manifold $(M,J,g)$ one has the K\"ahler form
$\Omega(Y_1,Y_2)=g(JY_1,Y_2)$.
\begin{prop}\label{soldJOmega} The soldering form of the almost
complex structure $J$ is related to the K\"ahler form $\Omega$ by
means of the formula
\begin{equation}\label{leg3forme2}
g(\sigma_J(Y_1,Y_2),\bar X)=2g(\beta(JY_1,Y_2),\bar
X)+d\Omega(\bar X,Y_1,Y_2).\end{equation}
\end{prop}
\begin{proof} From the definition of $\Omega$ we
get
$$(L_X\Omega)(Y_1,Y_2)=(L_Xg)(JY_1,Y_2)+g(L_XJ(Y_1),Y_2),$$
which, for $\bar X\in\nu N$ for $Y_1,Y_2\in TN$ becomes
\begin{equation}\label{leg3forme}g(\sigma_J(Y_1,Y_2),\bar X)=(L_X\Omega)(Y_1,Y_2)
-g(\sigma_g(JY_1,Y_2),\bar X).\end{equation} Notice that $\Omega$
is algebraically compatible with $(N,T^{\perp_g})$, therefore,
$$(L_X\Omega)(Y_1,Y_2)=g(\sigma_\Omega(Y_1,Y_2),\bar X).$$
Since it is easy to check that for the involved arguments one has
$(di(X)\Omega)(Y_1,Y_2)=0$, by using $L_X=di(X)+i(X)d$ in
(\ref{leg3forme}), we get
\begin{equation}\label{leg3forme'}g(\sigma_J(Y_1,Y_2),\bar X)=d\Omega(\bar X,Y_1,Y_2)
-g(\sigma_g(JY_1,Y_2),\bar X).\end{equation} In view of
(\ref{invptg}), formula (\ref{leg3forme'}) is the same as the one
required by the proposition.\end{proof}
\begin{corol}\label{JsoldO} The almost complex structure $J$ is
soldered to the $J$-invariant submanifold $N$ iff the second
fundamental form of $N$ is given by the formula
\begin{equation}\label{f2ptJsold}
g(\beta(JY_1,Y_2),X)=-\frac{1}{2}d\Omega(X,Y_1,Y_2).\end{equation}
\end{corol}

Then, since an almost K\"ahler manifold is characterized by the
property $d\Omega=0$, we get the main application:
\begin{prop}\label{almKtgeod} If $(M,J,g)$ is an almost K\"ahler manifold,
the almost complex structure $J$ is soldered to the submanifold $N$
iff $N$ is a $J$-invariant, totally geodesic submanifold of
$M$.\end{prop}

config: arxiv


TITLE: Homogeneous restriction
QUESTION [1 upvotes]: If $G$ is a finite group and $N$ is a normal, abelian subgroup of $G$. What means a homogeneous restriction of an irreducible representation of $G$ to $N$?

REPLY [2 votes]: Let me explain it in characters. If $\chi \in Irr(G)$, and $N \unlhd G$ is abelian, then in general the restriction $\chi_N$ splits in several linear characters of $N$. But homogeneous means $\chi_N=e\lambda$ for some $\lambda \in Irr(N)$. Note that $e=\chi(1)$. It also holds that $\chi(1) \mid |G:N|$.


TITLE: Integration with multiple constants
QUESTION [1 upvotes]: Question:
$$\int \frac{ax^2-b}{x\sqrt{{c^2x^2-(ax^2+b)^2}}}\ \text dx$$
My approach:
I can't understand whether I should integrate it normally or use a trigonometric function.

REPLY [1 votes]: Let $$I =\int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx = \int\frac{ax^2-b}{x\cdot x \sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$
So $$I = \int\frac{a-\frac{b}{x^2}}{\sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$
Put $\displaystyle \left(ax+\frac{b}{x}\right) = t\;,$ Then $\displaystyle \left(a-\frac{b}{x^2}\right)dx = dt$
$$I = \int\frac{1}{\sqrt{c^2-t^2}}dt = \arcsin\left(\frac{t}{c}\right)+\mathcal{C}$$
So $$I =\arcsin\bigg(\frac{ax^2+b}{cx}\bigg)+\mathcal{C}$$


TITLE: Is there a way to solve a sum of powers with the same base?
QUESTION [1 upvotes]: If you define y as a function of x which looks like this:
$y(x) = x^2$
You can calculate the distance from the zero-point to any result of this function like this: $d = \sqrt{x^{2} + y^{2}}$ and since we know that y(x) = x2 we can now calculate the distance as following: $d = \sqrt{x^{2} + x^{4}}$ or $d^{2} = x^{2} + x^{4}$. Is there a way to  solve this eqation by $x$ in order to calculate $x$ using the distance? Am I just blind or is this not that simple or even impossible to solve?
I know some algebra, some trigonometry a tiny little bit of complex stuff but try not to simplify too much I'd love to learn more about things in math :)

REPLY [0 votes]: Since $d\ge 0$ we have
$$d = \sqrt{x^{2} + x^{4}} \iff x^4+x^2-d^2=0$$
which is a quadratic equation in $x^2$ therefore, keeping only the positive root
$$x^2=\frac{-1+ \sqrt{1+4d^2}}{2} \implies \ldots$$


TITLE: Complex exponential function
QUESTION [0 upvotes]: Just a basic question. I have $e^{-i9\pi/4}$, and I'm struggling to understand why this is equal to $e^{-i\pi/4}$.
Thanks in advance for any help.

REPLY [8 votes]: Note that $9\pi/4 = (1+8)\left(\pi/4\right) = \pi/4 + 2\pi$ and $e^{2\pi i}=1$.

REPLY [4 votes]: Because

$e^{i\theta}=\cos\theta + i\cdot \sin\theta$,
$e^{-i\theta}=\cos(-\theta)+i\cdot \sin(-\theta)=\cos\theta - i \sin(\theta)$.
$\sin(2\pi+t)=\sin(t)$, same for cosine also.
$\sin\left(\frac{9\pi}{4}\right)=\sin \left(2\pi + \frac{\pi}{4}\right)$.

So that
\begin{align*}
e^{-i\frac{9\pi}{4}}
&=\cos\frac{-9\pi}{4}+i \sin\frac{-9\pi}{4}\\
&=\cos\frac{9\pi}{4} +i \sin\frac{-9\pi}{4}\\
&=\cos\frac{9\pi}{4} -i\sin\frac{9\pi}{4}\\
&= \cos\left(2\pi + \frac{\pi}{4}\right)-i\sin\left(2\pi +\frac{\pi}{4}\right)\\
&=\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\\
&=e^{-i\frac{\pi}{4}}
\end{align*}
Also while doing such problems, these formulae may come handy:

$\sin(\pi+\theta)= - \sin\theta\ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\sin(\pi-\theta) = \sin\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\sin\bigl(\frac{\pi}{2}+\theta\bigr) = \cos\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\sin\bigl(\frac{\pi}{2} -\theta\bigr) = \cos\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\cos(\pi+\theta)=-\cos\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\cos(\pi - \theta)=-\cos\theta\ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\cos\bigl(\frac{\pi}{2}-\theta\bigr)=\sin\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$
$\cos\bigl(\frac{\pi}{2} + \theta\bigr)= -\sin\theta\ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$


text:
\begin{document}
\title{Opportunistic Interference Alignment in MIMO Interference Channels}
\author{\IEEEauthorblockN{Samir Medina Perlaza$^{1}$, M\'{e}rouane Debbah$^{2}$, Samson Lasaulce$^{3}$ and Jean-Marie Chaufray$^{1}$}
\IEEEauthorblockA{\\$^{1}$ France Telecom R\&D - Orange Labs Paris. France\\
$\{$Samir.MedinaPerlaza, JeanMarie.Chaufray$\}$@orange-ftgroup.com\\
$^{2}$ Alcatel Lucent Chair in Flexible Radio - SUPELEC. France\\
Merouane.Debbah@supelec.fr\\
$^{3}$ Laboratoire des Signaux et Syst\`emes (LSS) - CNRS, SUPELEC, Univ. Paris Sud.  France\\
Samson.Lasaulce@lss.supelec.fr}\\
}
\maketitle

\begin{abstract}
\boldmath We present two interference alignment techniques such that an opportunistic point-to-point multiple input multiple output
(MIMO) link can reuse, without generating any additional interference, the same frequency band of a similar pre-existing primary
link. In this scenario, we exploit the fact that under power constraints, although each radio maximizes independently its rate by water-filling on their channel transfer matrix singular values, frequently, not all of them are used. Therefore, by aligning the interference of the
opportunistic radio it is possible to transmit at a significant rate while insuring zero-interference on the pre-existing link. We
propose a linear pre-coder for a perfect interference alignment and a power allocation scheme which maximizes the individual data rate
of the secondary link. Our numerical results show that significant data rates are achieved even for a reduced number of antennas.
\end{abstract}

\section{Introduction}\label{SecIntroduction}
We consider the case of radio devices attempting to
opportunistically exploit the same frequency band being utilized by
licensed networks under the constraint that no additional
interference must be generated. This can be implemented for example
by assuming that opportunistic users are cognitive
radios \cite{Haykin2005, fette-book-2006}. Typically, cognitive
radios temporally
 exploit the unused frequency bands, named white-spaces, to
  transmit their data. This clearly improves the spectral
   efficiency since more users are allowed to co-exist in the
   same bandwidth. However, a higher spectrum
   efficiency could be attained by simultaneously allowing the
    opportunistic radios to transmit with the licensees
    if no harmful interference is generated. In this case,
     interference alignment (IA) has been identified as a
      powerful tool to achieve such a goal. This technique, was initially introduced in \cite{Khandani2006Report, Khandani2006}. IA allows a
       given transmitter to partially or completely  ``align''
       its interference with unused dimensions of the primary
       terminals \cite{Jafar2008Alignment}. The concept of dimension could be associated
       with a specific spatial direction, frequency carrier or
       time slot \cite{Khandani2006}, \cite{Jafar2007}. An
       extensive study has been conducted \cite{Khandani2006, Jafar2007, JafarXChannel2008} to
       estimate the number of interference-free dimensions a given radio might
       find to transmit when several radio systems co-exist. In
       \cite{Khandani2006} and \cite{Jafar2007} several IA schemes to exploit such dimensions are proposed. Similarly, in \cite{Cardoso2008}, a linear pre-coder based on Vandermonde matrices allows an
       orthogonal frequency division multiplexing (OFDM) radio to
       co-exist with similar pre-existing terminals without generating any additional interference. The idea is to exploit the redundancy of the OFDM cyclic prefix and frequency selectivity of the channel.
\begin{figure}
\begin{center}
\includegraphics[width=.6\linewidth]{Figures/InterferenceChannel.eps}
\caption{\label{FigSystemModel} Two-user MIMO interference channel.}
\end{center}
\end{figure}

In this study, we propose a novel interference alignment technique
for the secondary users exploiting the fact that under a power-limitation, a primary user which maximizes its own rate
  by water-filling on its MIMO channel singular values, might leave some of them unused. \emph{i.e.} no transmission takes place along the corresponding spatial directions. These unused directions may be opportunistically utilized by a secondary transmitter, since its signal would not
  interfere with the signal sent by the primary transmitter. We present a linear
  pre-coder which perfectly aligns the interference generated by the
  secondary transmitter with such unused spatial directions. Similarly,
  we present a power allocation scheme based on the water-filling idea
  which maximizes the individual data rate of the opportunistic radio. Simulation
   results show that a significant data rate can be achieved by the secondary
   link following our approach.

\section{System Model}\label{SecSystemModel}
\emph{Notation:} In the following, matrices and vectors are denoted
by boldface upper case symbols and boldface lower case symbols, respectively. The
 $i^{th}$ entry of the vector $\boldsymbol{x}$ is denoted $\boldsymbol{x}(i)$. The
 entry corresponding to the $i^{th}$ row and $j^{th}$ column of the matrix $\boldsymbol{X}$ is
 denoted by  $\boldsymbol{X}(i,j)$. The $N$-dimension identity and null matrix are
 represented by $\boldsymbol{I}_N$ and $\boldsymbol{0}_N$, respectively. The
 Hermitian transpose is denoted $(\cdot)^H$, and the expected value is represented by the operator $\mathbb{E}\left[.\right]$.

We consider two point-to-point unidirectional links simultaneously
operating in the same frequency band and producing mutual interference as
shown in Fig. \ref{FigSystemModel} \cite{Sato1977}. Both transmitters are equipped
with $N_t$ antennas while both receivers use $N_r$ antennas. The first transmitter-receiver pair, i.e. $Tx_1$ and $Rx_1$, is a primary link licensed to
exclusively exploit a given frequency band. The pair, $Tx_2$ and $Rx_2$ is
an opportunistic link exploiting the same frequency band subject
to the constraint that no additional interference must be generated
over the primary system. Note that no cooperation between
terminals is allowed, \emph{i.e.} transmitters do not share or
exchange any signal before transmitting. Therefore, the multiple
access interference (MAI) is considered as additive white Gaussian
noise (AWGN).

The channel transfer matrix from transmitter $j  \in \left\lbrace 1,
2 \right\rbrace$ to receiver $i  \in \left\lbrace 1, 2
\right\rbrace$ is denoted  $\boldsymbol{H}_{i,j} \in
\mathbb{C}^{N_r\times N_t}$, where the entries of
$\boldsymbol{H}_{i,j}$  are independent and identically distributed
(i.i.d) complex Gaussian circularly symmetric random variables. The
channel matrices are supposed to be fixed for the whole transmission
duration. This correspond to assuming (static) Gaussian links. But
our analysis readily extends to the case of slow-fading channels by
assuming the channels to be constant over each data block. Regarding
the channel state information (CSI) conditions, we assume the
primary terminals (transmitter and receiver) to only have perfect
 knowledge of the matrix $\boldsymbol{H}_{1,1}$. On the other hand,
 the secondary terminals have perfect knowledge of all the channel
 transfer matrices $\boldsymbol{H}_{i,j}$, for every $i$
 and $j \in \left\lbrace 1, 2 \right\rbrace$. Although unrealistic,
 this condition provides us with an upper bound on the achievable
 rate of the secondary user. It can however be met in practice in the TDD (Time Division Duplex) mode if the secondary
 user exploits opportunistically the training sequences and signaling communication between the
 primary devices.

Following a matrix notation, the primary and secondary received
signals can be written as
\begin{equation}\label{EqReceivedSignal}
\left( \begin{array}{c} \boldsymbol{y}_1 \\
\boldsymbol{y}_2\end{array}\right) = \left( \begin{array}{cc}
\boldsymbol{H}_{1,1} & \boldsymbol{H}_{1,2} \\ \boldsymbol{H}_{2,1}
& \boldsymbol{H}_{2,2} \end{array}\right) \left(\begin{array}{c}
\boldsymbol{V}_1  \boldsymbol{s}_1 \\ \boldsymbol{V}_2
\boldsymbol{s}_2 \end{array}\right) + \left( \begin{array}{c}
\boldsymbol{n}_1 \\ \boldsymbol{n}_2 \end{array}\right),
\end{equation}
where the vectors $\boldsymbol{s}_i \in \mathbb{C}^{N_t \times 1}$ and
  $\boldsymbol{n}_i \in \mathbb{C}^{N_r \times 1}$ represent the transmitted
  symbols and an AWGN process with zero mean and covariance matrix $\sigma^2 \boldsymbol{I}_{N_r}$ for
  the $i^{th}$ link. For all $i \in \left\lbrace 1,2 \right\rbrace$ the
   matrices $\boldsymbol{V}_i \in \mathbb{C}^{N_t \times N_t}$ represent the linear pre-coders used
    for interference alignment. Furthermore, at each receiver, the
    input signal is linearly processed with the matrix
    $\boldsymbol{F}_i \in \mathbb{C}^{N_r \times N_r}$. The signal at the output of the linear filter $i$ is $\boldsymbol{r}_i = \boldsymbol{F}_i
    \boldsymbol{y}_i$. Both matrices $\boldsymbol{V}_i$ and $\boldsymbol{F}_i$ are
    described
    later on. The power allocation matrices are defined as the input covariance
matrices $\boldsymbol{P}_i = \mathbb{E}\left[\boldsymbol{s}_i
\boldsymbol{s}_i^H \right] \in \left(\mathbb{R}^+ \right)^{N_t
\times N_t}$, for the $i^{th}$ transmitter. The power constraints
are
\begin{equation}\label{EqPowerConstraints}
\forall i \in \left\lbrace 1, 2 \right\rbrace, \quad \text{Trace}\left( \boldsymbol{V}_i \boldsymbol{P}_{i} \boldsymbol{V}_i^H \right) \leqslant p_{i, \mathrm{max}},
\end{equation}
where $p_{i, \mathrm{max}}$ is the maximum transmit
 power level for the $i^{th}$ transmitter. Without loss of generality, we assume
 identical maximum transmit powers for
 all the terminals \emph{i.e.} $\forall i \in \left\lbrace 1,2 \right\rbrace$, $p_{i, \mathrm{max}} = p_{\mathrm{max}}$.

\section{Interference Alignment Strategy} \label{SecStrategy}

In this section we focus on the study of the pre-coding
$\boldsymbol{V}_i$ and post-processing $\boldsymbol{F}_i$ matrices.
Suppose that  the primary terminals completely ignore the presence
of the opportunistic transmitter. Hence, in order to maximize its
own data rate, the primary transmitter follows a water-filling power
allocation as in the single-user case \cite{Cioffi2004}.
\subsection{Primary link design}
Under the assumption that the channel matrix
$\boldsymbol{H}_{1,1}$ is known at the receiver and transmitter, the
primary terminal chooses its pre-coding $\boldsymbol{V}_1$ and
post-processing $\boldsymbol{F}_1$ matrices in such a way that their
channel transfer matrix is diagonalized, \emph{i.e.}
$\boldsymbol{V}_1$ and $\boldsymbol{F}_1 = \boldsymbol{U}_1^H$
satisfy the singular value decomposition $\boldsymbol{H}_{1,1} =
\boldsymbol{U}_1 \boldsymbol{\Lambda}_1 \boldsymbol{V}^{H}_1$, where
$\boldsymbol{U}_1 \in \mathbb{C}^{N_r \times N_r}$ and
$\boldsymbol{V}_1 \in \mathbb{C}^{N_t \times N_t}$ are unitary
matrices and $\boldsymbol{\Lambda_1} \in
\left(\mathbb{R}^+\right)^{N_r \times N_t}$ is a diagonal matrix
which contains $\mathrm{min}(N_r,N_t)$ non-zero singular values, $
\lambda_1,\ldots,\lambda_{\mathrm{min}(N_r,N_t)}$. Thus, the received signal after
linear processing, $\boldsymbol{r}_1$, can be written as
\begin{equation}\label{EqWhitenedReceivedSignal}
\boldsymbol{r}_1 = \boldsymbol{F}_1 \boldsymbol{y}_1 = \boldsymbol{\Lambda}_1 \boldsymbol{s}_1  + \boldsymbol{n}_1',
\end{equation}
where $\boldsymbol{n}_1' = \boldsymbol{U}^H \boldsymbol{n}_1$ is an
AWGN process with zero mean and covariance matrix $\sigma^2 {\bf
I}_{N_r}$. Then, the achievable rate of the primary user is
maximized by the power allocation matrix $\boldsymbol{P}_1$ which is
a solution to the following optimization problem
\begin{equation}\label{EqOptimizationPrimaryUser}
\begin{array}{lc}
\text{maximize} &   \log_2 \left|\boldsymbol{I}_{N_t} + \frac{1}{\sigma^2} \boldsymbol{H}_{1,1}  \boldsymbol{V}_1 \boldsymbol{P}_1 \boldsymbol{V}_1^H \boldsymbol{H}_{1,1}^H  \right|\\
\text{subject to} & \text{Trace}\left( \boldsymbol{P}_1 \right) \leqslant p_{\mathrm{max}}.
\end{array}
\end{equation}
The solution to (\ref{EqOptimizationPrimaryUser}) is the classical water-filling algorithm
\cite{Cioffi2004}. Following this approach, the optimal power
allocation matrix is a diagonal matrix with entries
\begin{equation}\label{Eqwater-filling}
\forall n \in \left\lbrace 1,\ldots,N_t \right\rbrace, \quad
P_1(n,n) = \left[\beta - \frac{\sigma^2}{\lambda_n^2}\right]^+,
\end{equation}
with $\left[ p \right]^+ = \max\left(0,p\right)$. The constant $\beta$ is a Lagrangian multiplier that is determined to satisfy
\begin{equation*}
\displaystyle\sum_{j =1}^{N} \boldsymbol{P}_1(j,j) = p_{\mathrm{max}}.
\end{equation*}
\subsection{Secondary link design}
Depending on the channel singular values $\lambda_1, \ldots,
\lambda_{\mathrm{min}(N_r,N_t)}$, the power allocation matrix $\boldsymbol{P}_1$ might
contain  zeros in its main diagonal. A zero power allocation for a
given singular value means that no transmission takes place along
the corresponding spatial direction. This means that the secondary
terminal can align its transmitted signal with the unused singular
modes such that it does not interfere with the signal transmitted by
the primary user. If one converts the spatial problem into the
frequency one, the result is similar to the cognitive scenario where
the secondary would opportunistically use the unexploited frequency
modes. The main difference here lies in the fact that in the spatial
domain, there is no universal precoder which diagonalizes the basis
of all the devices whereas this is the case in the frequency domain
with the use of the FFT.

As a consequence, in the spatial domain, the corresponding
orthogonality condition (such that the secondary user generates no
interference on the primary link) is given by
\begin{equation}\label{EqOrthogonalityCondition}
\boldsymbol{U}_1^H \boldsymbol{H}_{1,2} \boldsymbol{V}_{2} = \alpha
\bar{\boldsymbol{P}_1},
\end{equation}
where the matrix $\bar{\boldsymbol{P}_1}$ is a diagonal matrix with entries
\begin{equation}\label{EqComplementaryPowerAllocation}
 \forall n \in \left\lbrace 1,\ldots,N_t\right\rbrace, \quad \boldsymbol{\bar{P_1}}(n,n) = \left[\frac{\sigma^2}{\lambda_n^2} - \beta \right]^+,
\end{equation}
such that the condition $\boldsymbol{P}_1 \bar{\boldsymbol{P}}_1 =
\boldsymbol{0}_{N_r}$ always holds. It can be easily verified since both matrices are diagonal.
Additionally, the constant $\alpha$ is chosen
to satisfy the power constraints (\ref{EqPowerConstraints}) with $i
= 2$.

Assuming that perfect estimates of $\boldsymbol{H}_{1,1}$ and
$\boldsymbol{H}_{1,2}$ are available at the secondary transmitter,
the secondary precoder (when the inverse of $\boldsymbol{H}_{1,2}$
exists) is given by:
\begin{equation}\label{EqPrecoder}
\boldsymbol{V}_2 = \alpha \boldsymbol{H}_{1,2}^{-1}
\boldsymbol{U}_{1} \bar{\boldsymbol{P}}_1.
\end{equation}
For the case where $N_r > N_t$, \emph{i.e.} the receiver has more
antennas than the transmitter, it is still possible to obtain the
pre-coding matrix by using the Moore-Penrose pseudo-inverse of
$\boldsymbol{H}_{1,2}$,
\begin{equation}\label{EqPrecoderIllConditioned}
\boldsymbol{V}_2 =  \alpha \left( \boldsymbol{H}_{1,2}^H
\boldsymbol{H}_{1,2} \right)^{-1}\boldsymbol{H}_{1,2}^H
\boldsymbol{U}_{1}  \bar{\boldsymbol{P}}_1.
\end{equation}

Once the pre-decoder $\boldsymbol{V}_2$ has been adapted to satisfy
(\ref{EqPrecoder}) or (\ref{EqPrecoderIllConditioned}) at the
secondary transmitter, no additional interference impairs the
primary user. However, the secondary receiver still undergoes the
interference from the primary transmitter. Typically, this effect is
a colored noise with covariance $\boldsymbol{Q} \in \mathbb{C}^{N_r
\times N_r}$ due to the channel $\boldsymbol{H}_{2,1}$ and the
pre-coder $\boldsymbol{V}_1$. Here,
\begin{equation}\label{EqMAINCovariance}
\boldsymbol{Q} = \boldsymbol{H}_{2,1} \boldsymbol{V}_1
\boldsymbol{P}_1 \boldsymbol{V}_1^H \boldsymbol{H}_{2,1}^H  +
\sigma^2 \boldsymbol{I}_{N_r}.
\end{equation}
Hence, the received signal $\boldsymbol{y}_2$ can be whitened  by
using the
 matrix $\boldsymbol{F}_2 = \boldsymbol{Q}^{-\frac{1}{2}}$, to
 obtain $\boldsymbol{r}_2 =\boldsymbol{F}_2 \boldsymbol{y}_2$, such that
\begin{equation}\label{EqFilteredSignal2}
\boldsymbol{r}_2 = \boldsymbol{Q}^{-\frac{1}{2}}
\boldsymbol{H}_{2,2} \boldsymbol{V}_2  \boldsymbol{s}_2 +
\boldsymbol{n}_2',
\end{equation}
where $\boldsymbol{n}_2' = \boldsymbol{Q}^{-\frac{1}{2}} \left(
\boldsymbol{H}_{2,1}  \boldsymbol{V}_1 \boldsymbol{s}_1 +
\boldsymbol{n}_2 \right)$ is an i.i.d. AWGN process with zero mean
and a covariance matrix proportional to the identity. Let $S$ be the
number of zeros on the main diagonal of $\boldsymbol{P}_1$. Then,
the matrix $\boldsymbol{V}_2$ contains $N_{r} - S$ zero columns.
Note that $S = 0$ implies that no transmission takes
  place in the secondary link. In the sequel, we always assume that $S > 0$ (which will be the case at low signal to noise ratio as shown in the simulations).

\section{Input Covariance Matrix Optimization}\label{SecInputCovarianceMatrix}
In the latter section the proposed pre-coding scheme does not generate
any interference on the primary user but the transmission rate for
the secondary user was not optimized. For this purpose, the choice
of the power allocation of the secondary transmitter, \emph{i.e.} the matrix ${\bf P}_2$, needs to be optimized. First, we present the most
simple case where uniform power allocation is performed. Second, we
introduce a power allocation which maximizes the individual
transmission rate. In both cases we assume that the pre-coder has
been previously adapted to satisfy the orthogonality conditions
(\ref{EqPrecoder}) or (\ref{EqPrecoderIllConditioned}).

\subsection{Uniform Power Allocation}

For the uniform power allocation scheme the input covariance matrix
is set to $\boldsymbol{P}_2 = \boldsymbol{I}_{N_t}$ and the constant
$\alpha$ from (\ref{EqOrthogonalityCondition}) is tuned in order to
meet the condition $\text{Trace}\left(\boldsymbol{V}_2
\boldsymbol{V}_2^H \right) = p_{\mathrm{max}}$. The rate achieved by the
secondary user while generating zero-interference to the primary
receiver is
\begin{equation}\label{EqSecondaryRate}
R_2 = \log_2 \left| \boldsymbol{I}_{N_r} + \boldsymbol{Q}^{-\frac{1}{2}} \boldsymbol{H}_{2,2} \boldsymbol{V}_2  \boldsymbol{V}_2^H  \boldsymbol{H}_{2,2}^H \boldsymbol{Q}^{-\frac{1}{2}} \right|.
\end{equation}

\subsection{Optimal Power Allocation}
The transmission rate for the secondary link is maximized by
adopting a power allocation matrix $\boldsymbol{P}_2$ which is a
solution of the following optimization problem,
\begin{equation}\label{EqOptimizationProblem2}
\begin{array}{lc}
\displaystyle\arg\max_{\boldsymbol{P}_2} & R_2(\boldsymbol{P}_2)\\
\text{s.t.} & \text{Trace}\left( \boldsymbol{V}_2 \boldsymbol{P}_2
 \boldsymbol{V}_2^H \right) \leqslant p_{\mathrm{max}},
\end{array}
\end{equation}
where
\begin{equation}
R_2(\boldsymbol{P}_2) = \log_2 \left|\boldsymbol{I}_N +
\boldsymbol{Q}^{-\frac{1}{2}} \boldsymbol{H}_{2,2}
 \boldsymbol{V}_2 \boldsymbol{P}_2 \boldsymbol{V}_2^H
 \boldsymbol{H}_{2,2}^H \boldsymbol{Q}^{-\frac{1}{2}}  \right|.
\end{equation}
Note that
  solving this optimization problem requires the knowledge of
  the covariance matrix $\boldsymbol{Q}$, which is calculated at
  the secondary receiver based on the knowledge of the
  channel $\boldsymbol{H}_{2,1}$. This can be done if  the
  secondary receiver estimates $\boldsymbol{Q}$ and feeds
  it back to the secondary transmitter. Here, we assume a perfect
      knowledge of $\boldsymbol{Q}$ is available at the secondary transmitter.

By definition (Eq. (\ref{EqOptimizationProblem2})), the matrix $
\boldsymbol{V}_2$ is not full rank. Therefore, the optimization
problem (\ref{EqOptimizationProblem2}) does not have a simple
solution. We propose a two-step optimization which leads to a
water-filling solution. First, we define a new input covariance
$\hat{\boldsymbol{P}}_2$ such that,
\begin{equation}\label{EqNewInputCovariance}
\hat{\boldsymbol{P}_2} =  \left(\boldsymbol{V}_2^H
 \boldsymbol{V}_2 \right)^{\frac{1}{2}} \boldsymbol{P}_2
  \left(\boldsymbol{V}_2^H \boldsymbol{V}_2 \right)^{\frac{1}{2}}.
\end{equation}
By replacing the expression (\ref{EqNewInputCovariance}) in
 (\ref{EqOptimizationProblem2}), the optimization problem becomes
\begin{equation}\label{}
\begin{array}{lc}
\displaystyle\arg\max_{\hat{\boldsymbol{P}}_2}  & \left\lbrace
 \log_2 \left|\boldsymbol{I}_{N_r} + \boldsymbol{G} \hat{\boldsymbol{P}}_2
 \boldsymbol{G}^H  \right| \right\rbrace \\
\text{s.t} & \text{Trace}\left( \hat{\boldsymbol{P}}_2 \right) = p_{\mathrm{max}},
\end{array}
\end{equation}
where $\boldsymbol{G} = \boldsymbol{Q}^{-\frac{1}{2}} \boldsymbol{H}_{2,2}
 \boldsymbol{V}_{2} \left( \boldsymbol{V}_2^H  \boldsymbol{V}_2 \right)^{-\frac{1}{2}}
 \in \mathbb{C}^{N_r \times N_t}$. The idea here is to solve the a
 priori non-trivial optimization problem defined by expression (\ref{EqOptimizationProblem2}) by
 introducing an equivalent channel matrix $\boldsymbol{G}$ to
 simplify the problem. Using $\boldsymbol{G}$ we can then apply a singular value decomposition to the new channel such that
$\boldsymbol{G} = \boldsymbol{E} \boldsymbol{\Delta}
\boldsymbol{Z}^H$, where $\boldsymbol{E} \in \mathbb{C}^{N_r \times
N_r}$ and $\boldsymbol{Z} \in \mathbb{C}^{N_t \times N_t}$ are unitary
matrices, and the matrix $\boldsymbol{\Delta} \in \left(\mathbb{R^+}\right)^{N_r
\times N_t}$ contains the singular values $\eta_1, \dots, \eta_{\min\left\lbrace N_r, N_t \right\rbrace}$ of
$\boldsymbol{G}$. Under these assumptions, the optimal solution
$\boldsymbol{P}_2^* = \boldsymbol{Z}^H \hat{\boldsymbol{P}}_2
\boldsymbol{Z}$ is
\begin{equation}\label{Eqwater-filling2}
\forall n \in \left\lbrace 1,\ldots,N_t\right\rbrace, \quad \boldsymbol{P}_2^*(n,n) = \left[\rho - \frac{1}{\eta_n^2}\right]^+,
\end{equation}
where $\rho$ is a Lagrangian multiplier that is determined to
satisfy $\displaystyle\sum_{j =1}^{N} \boldsymbol{P}_2^*(j,j) = p_{\mathrm{max}}$. Once $\boldsymbol{P}_2^*$ has been obtained, then the optimal power allocation matrix \cite{Cioffi2004} is
\begin{equation}\label{EqOptimalPower2}
\boldsymbol{P}_2 = \left(\boldsymbol{V}_2^H \boldsymbol{V}_2 \right)^{-\frac{1}{2}}
 \boldsymbol{Z} \hat{\boldsymbol{P}}_2^* \boldsymbol{Z}^H
  \left(\boldsymbol{V}_2^H \boldsymbol{V}_2  \right)^{-\frac{1}{2}}
\end{equation}
The constant $\alpha$ in (\ref{EqOrthogonalityCondition}) is tuned
such that the condition (\ref{EqPowerConstraints}) is met for $i =
2$.

\section{Numerical Results}
In this section we show numerical examples to illustrate
 the performance of our interference alignment strategy.

Considering the same number of antennas at the receiver and
 the transmitter, we analyze the number of unused singular
 values or free dimensions available for the secondary link
  as well as its achieved data rate. Recall that the primary
  link is interference-free, therefore its data rate corresponds
  to the single user case rate \cite{Cioffi2004}.

In Fig. \ref{FigSingularValues} we show the number of unused
singular values in the primary link as a function of the number of
antennas and $SNR = \frac{p_{\mathrm{max}}}{\sigma^2}$. Note that in
low SNR regime, the transmitter attempts to concentrate all its
power in the best singular values leaving all the others unused. On
the contrary, in high SNR regime the primary transmitter tends to
spread its power among all its available singular values. Thus, in
the first case the opportunistic link has plenty of free dimensions,
while in the second one, it is effectively limited. This power
allocation behavior has been also reported in \cite{Valenzuela2002}
and \cite{Valenzuela2002a}. Similarly, it is observed that
increasing the number of antennas leads on average, to a linear
scaling of the unused singular values. In Fig.
\ref{FigDataRateOptimal}, we show the achieved data rate of
secondary link when optimal power allocation is implemented
($R_{\mathrm{2,optimal}}$) as a function of the number of antennas
and the SNR. Therein, it is shown that at very low and very high
SNRs the data rate approaches zero bits/sec. In low SNR regime this
effect is natural since detection is difficult due to the noise.
However, at high SNRs it is due to the fact that the primary
transmitter does not leave any unused singular value. Nonetheless,
at intermediate SNRs, significant data rates are achieved by the
secondary link. Note that increasing the number of antennas always
leads to higher data rates for the secondary link. In Fig.
\ref{FigDiffR12Optimal}, we plot the data rate of the primary $R_{\mathrm{1}}$ and secondary link
for the case of uniform $R_{\mathrm{2,uniform}}$ and optimal $R_{\mathrm{2,optimal}}$ power allocation. Note that in high
SNR regime, the $R_{\mathrm{2,uniform}}$ and $R_{\mathrm{2,optimal}}$
performs similarly. It is due to the fact that few or even none of
the singular values are left unused by the primary link, therefore the uniform power allocation does not differ
from the optimal in average. In contrast, at intermediate SNRs the
difference in performance is more significant for a large number of antennas.
\begin{figure}
\begin{center}
\includegraphics[width=\linewidth]{Figures/FreeDimensions.eps}
\caption{\label{FigSingularValues} Average number of unused
singular values in the primary link as a function of the total number of antennas and the $SNR = \frac{p_{\mathrm{max}}}{\sigma^2}$. The SNR and the number of antennas $N_r = N_t$ are assumed the same for the primary and secondary links}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=\linewidth]{Figures/SecondaryRateOptimal.eps}
\caption{\label{FigDataRateOptimal} Average data rate of the secondary link when optimal power allocation is implemented as a function of the number of antennas $N_r = N_t$ and $SNR = \frac{p_{\mathrm{max}}}{\sigma^2}$. The SNR and the number of antennas are assumed the same for the primary and secondary link.}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=\linewidth]{Figures/DataRates.eps}
\caption{\label{FigDiffR12Optimal} Average data rate achieved by the primary $R_{1}$ and secondary link for uniform $R_{\mathrm{2,uniform}}$ and optimal $R_{\mathrm{2,optimal}}$ power allocation as a function of their $SNR = \frac{p_{\mathrm{max}}}{\sigma^2}$. The dashed lines correspond to $N_r = N_t = 20$ antennas. The solid lines correspond to $N_r = N_t = 3$ antennas.}
\end{center}
\end{figure}

\section{Conclusions}\label{SecConclusions}

We provided a novel interference alignment scheme which allows an opportunistic
point-to-point MIMO link to co-exist with a similar pre-existing primary link  on the same
fully utilized band without generating any additional interference. The proposed scheme
exploits the fact that the licensed transmitter, while performing  water-filling power
allocation on its MIMO channel (in order to maximize its single user rate)  will leave
some singular values unused when constrained by power limitations.  Hence, no
transmission takes place along the corresponding spatial directions. We proposed a linear
pre-coder for the opportunistic radio, which perfectly aligns the transmitted signal with
such unused dimensions. We also provided a power allocation scheme which maximizes the
data rate of the opportunistic link. Numerical results show that significant data rates
are achieved by the secondary link even for a reduced number of antennas. Further studies
will extend the novel  approach to multi-user multi-carrier systems and  the case of
incomplete CSI.
\bibliographystyle{IEEEtran}
\bibliography{spectrumsharing}

\end{document}


TITLE: If $\epsilon_1$ and $\epsilon_2$ are 2 elementary families of sets in $X_1$ and $X_2$ respectively then the cross product is an elementary family too.
QUESTION [1 upvotes]: I am trying to show that if $\epsilon_1$ and $\epsilon_2$ are 2 elementary families of sets in $X_1$ and $X_2$ respectively then $$\epsilon_1 \times \epsilon_2=\{E_1 \times E_2 | E_1 \in \epsilon_1, E_2 \in \epsilon_2\}$$ is an elementary family of sets on $X_1 \times X_2$.

My book, Folland, defines an elementary family to be a collection
  $\epsilon$ of subsets of $X$ such that:
1) The empty set is in $\epsilon$
2) If $E,F \in \epsilon$, then $E \cap F \in \epsilon$
3) If $E \in \epsilon$, then $E^c$ is a finite disjoint union of
  members of $\epsilon$

For this problem I am trying to show the following:
1) The empty set is in $\epsilon_1 \times \epsilon_2$ is clear since it is in both $\epsilon_1$ and $\epsilon_2$.
I am not sure about 2) and 3)...
2) I want to show that if $E,F \in \epsilon_1 \times \epsilon_2$, then $E \cap F \in \epsilon_1 \times \epsilon_2$
We know that there exists $E_i,E_m \in \epsilon_1$ and $E_j,E_n \in \epsilon_2$ so that $E=E_i \times E_j$ and $F=E_m \times E_n$. Then, $E \cap F=(E_i \times E_j) \cap (E_m \times E_n)$ where both terms in parentheses are in $\epsilon_1 \times \epsilon_2$ by definition. So we have that $E \cap F \in \epsilon_1 \times \epsilon_2$. I am wondering if this is correct...
3) I want to show that If $E \in \epsilon_1 \times \epsilon_2$, then $E^c$ is a finite disjoint union of members of $\epsilon_1 \times \epsilon_2$.
Again, we know that there exists $E_i \in \epsilon_1$ and $E_j \in \epsilon_2$ so that $E=E_i \times E_j$. I then said that $E^c=(E_i \times E_j)^c = (E_i^c \times X_1) \cup (E_j^c \times X_2)$ but I am stuck because $(E_i^c \times X_1)$ and $(E_j^c \times X_2)$ are not disjoint as far as I understand...
Thank you for any help you can give me.

REPLY [1 votes]: For 3, let $E\in\varepsilon_1$ and $E'\in\varepsilon_2$. Then
$X_1=E_0\cup E_1\cup\cdots\cup E_m$ where this is a disjoint union of
elements of $\varepsilon_1$ and $E_0=E$. Likewise
$X_2=E_0'\cup E_1'\cup\cdots\cup E_n'$ where this is a disjoint union of
elements of $\varepsilon_2$ and $E_0'=E'$.
Then $(E\times E')^c$ is the disjoint union of the $E_i\times E_j'$
for $(i,j)\ne(0,0)$.


TITLE: Curved Space and Great Circles in General Relativity
QUESTION [1 upvotes]: I have been studying general relativity by B. Schutz. It says locally straight lines on a sphere does not remain parallel, as great circles do intersect on a sphere.
Below is a picture where two cirlces, $C^1$ (great circle) and $C^2$ (not a great circle) do remain parallel if we extend it to the entire sphere, $S^2$. So why the author claimed that locally straight lines usually do not remain parallel on curved space.

REPLY [6 votes]: Your curve $C_2$ is not a straight line.
There is an easy way to see that $C_2$ isn't a straight line without having to resort to differential geometry. A straight line is the shortest distance between two points$^1$, so consider two points on $C_2$ on opposite sides of the pole:

I've drawn $C_2$ as the red curve on a perspective diagram and also in the side on view. We'll take the sphere to have unit radius for convenience.
$C_2$ is a circle of radius $r = \sin\theta$ so the distance between opposite points measured along the circle is just half the circumference:
$$ s_1 = \tfrac{1}{2}2\pi r = \pi\sin\theta $$
Compare this with the distance between the points if we follow the line through the pole shown on the right diagram by the dashed line above the sphere. This is a segment of a great circle so the length of the segment is:
$$ s_2 = 2\theta $$
And for $0 \le \theta \le \pi/2$ the distance $s_2$ is always smaller than $s_1$. That means the curve $C_2$ cannot be a straight line because it isn't the shortest distance between the points. Rather than go though the algebra to show that $\pi\sin\theta \ge 2\theta$ I simply drew a graph to show this:

Note that the two distances are equal only for the trivial case where $C_2$ has zero radius and $\theta = \pi/2$ i.e. when $C_2$ is a great circle.

$^1$ not strictly true. A straight line extremises the path length, but let's overlook this to keep the argument simple.

REPLY [1 votes]: It's easy to get confused about locally straight lines on a sphere, especially when you do it by eye as you have.
Instead, a method should be adopted. Now, the straight line on a plane is simply the shortest distance between two points. We can draw this easily enough by eye, but another method is to stretch a string between the two points and pull it taut.
If we now do this on the sphere you'll find the great circles and these are different from the circles that you have drawn on the sphere. You should also see quite quickly that any two will always intersect. So the geometry is not Euclidean.


TITLE: Does the existence of a nondegenerate paring of finite-dimensional vector spaces $V$ and $W$ imply that $V \simeq W^*$?
QUESTION [2 upvotes]: In Bott & Tu, Differential Forms in Algebraic Topology, at the beginning of the discussion of Poincaré duality (page 44 in my edition), I find the following definition and claim:

A pairing between two finite-dimensional vector spaces $\langle\ ,\ \rangle \colon V \otimes W \to \mathbb{R}$ is said to be nondegenerate if $\langle v, w \rangle = 0$ for all $w$ implies $v = 0$; equivalently, the map $v \mapsto \langle v,\ \rangle$ should define an isomorphism $V \xrightarrow{\sim} W^*$.

I can see immediately how a pairing implies a map $V \to W^*$, and it's obvious that if $V$ is nondegenerate such a map must be injective. But why is it surjective? In particular, why does the following counterexample not work?
Let $V = \mathbb{R}$ and $W = \mathbb{W}^2$. The function $(v_1) \times (w_1, w_2) \mapsto v_1 w_1$ is a nondegenerate pairing, but the vector spaces $V$ and $W^*$ have different dimension. What am I missing here?

REPLY [4 votes]: You're right: according to this asymmetric definition, there is no reason for the map to be surjective. The standard definition for a non-degenerate pairing asks for both the left and the right radical to vanish. This appears to have been fixed in my version of the book:

Note that according to this standard definition, your pairing is degenerate because $\langle a, (0,b)\rangle = 0$ for all $a$ even though $(0,b) \neq 0$.


TITLE: Calculate the percentile correspond and the percentile of the sampling distribution
QUESTION [1 upvotes]: The SAT placement test for admission to American universities had an average of μ = 1026 and a standard deviation σ = 209. The distribution of marks is approximately normal.
a) Ramon had a grade of 1100. If the notes have a normal distribution, to what percentile does Ramon's note correspond?
b) Now consider the average X¯ of the scores of 70 randomly selected students. If X¯ = 1100, at what percentile of the sampling distribution of X¯ does this result correspond?
c) Which of your two calculations, a) or b), is less precise since SAT notes do not follow exactly a normal law?
Here is my approach:
a) I calculated z-score = 0.354 then I found that the proportion correspond is 0.6368. So what is the percentile correspond? 63.68 or 64?
b) I obtained z-score =2.96 so the proportion is 0.9985. Then what is the percentile?

REPLY [0 votes]: If you are less familiar with R, you can also use Excel of a TI-84 calculator.
(a) Use the normalCDF function which takes the parameters normalCDF(lower bound, upper bound, mean, standard deviation)
In your case, your lower bound is technically $- \infty$, but entering a large, arbitrary negative number will do. Your upper bound is the grade of 1100. This would look like the following:
normalCDF(-1000000, 1100, 1026, 209) which yields 0.6383555986 (so approximately the 64th percentile... you could keep it as the 63.68th percentile if you really wanted to be exact.)
(b) Perform the same calculation as in (a), but change your standard deviation to $\frac{209}{\sqrt{70}}$.
(c) For Part C, how many students are there in part (a). Assuming Normality, possibly part (b) would be better since you know you are getting the average of 70 scores. If part (a) is bringing in a larger sample size $n$, then it would be more normal.
Typically, the larger the $n$, the more Normal.


TITLE: Solutions based on augmented matrix form
QUESTION [0 upvotes]: Find the set of values of a for which the system of equations below has a unique solution.
$$x-2y-2z=-7$$
$$2x+(a-9)y-10z=-11$$
$$3x-6y+2az=-29$$
I rewrote the equations in the augmented matrix form as:
$$ \left[
\begin{array}{ccc|c}
  1&-2&-2&-7\\
  2&(a-9)&-10&-11\\
  3&-6&2a&-29
\end{array}
\right] $$
And then used row reduction:
$$ \left[
\begin{array}{ccc|c}
  1&-2&-2&-7\\
  0&(a-5)&-6&3\\
  0&0&(2a+6)&-8
\end{array}
\right] $$
After this step however, I do not know how to come to the conclusion as to which values of a will give a unique solution.
I know $a\neq-3$ because then the last equation in the reduced form would be inconsistent, since $2(-3)+6\neq-8$.
Is my concept correct here?
The other answer is $a\neq5$ but I do not understand why this is so. Can somebody help me out please?
Note: This question is taken from a CIE A Level Past Paper

REPLY [0 votes]: Yes your concept is correct. After row reduction your system becomes
$$x-2y-2z=-7$$
$$(a-5)y-6z=3$$
$$(2a+6)z=-8$$
If $2a+6=0$ or $a=-3$, then the last equation $\textbf{is not}$ satisfied for any $z,$ but we are looking for a unique solution. Thus $a\neq -3$ which gives $z=-\frac{4}{a+3}.$
Now have $z$ and substituting into the second equation and re-arranging gives
$$(a-5)y=3+6z=\frac{3(a-5)}{a+3}$$
Now if $a=5$ then the equation $\textbf{is}$ satisfied for any $y$ (since we will have $0=0$). However we want a unique solution. Hence $a\neq 5$, and we obtain that $$y=\frac{3}{a+3}$$
and finally substituting into the first equation gives
$$x=-\frac{7a+23}{a+3}$$
Thus for $a\neq -3$ and $a\neq 5$ we obtain the unique solution
$$(x,y,z)=\left(-\frac{7a+23}{a+3},\frac{3}{a+3},-\frac{4}{a+3}\right)$$

REPLY [0 votes]: When you write $2(-3)+6 \neq -8$ I think you are making a hidden mistake.
Your last equation is in fact $(2a+6) z = 8$, and the question we are asking here is if there is a value for $a$ when there can't exist a solution, and we can easily find such a case when $a = -3$.
If we simply ask the question looking at the last 2 rows; Is there an $a$ such that no value for $y$ and $z$ can satisfy the equations $(a-5)y -6z = -11$ and $(2a+6)z = 8$?
For a nice simple linear system like this, the only way for that to happen is if the right hand sides are linearly dependent.
It not also becomes clear that $a = 5$ will turn our system of equations to
$-6z = -11$ and $16z = 8$, which clearly has no solution.


TITLE: Let $ R $ be a p.i.d. and $ A\in M_n(R) $. If $ \det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.
QUESTION [5 upvotes]: Let $ R $ be a p.i.d. and $ A\in M_n(R) $. If $ \det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.

I know that we can express $ A $ as products of elementary matrices when we require $ R $ to be Euclidean. Since we can get rid of matrices  whose left uppermost elements are like $$ \begin{pmatrix}x & s\\ y & t\end{pmatrix} .$$
(Note: Furthermore, I know that if we require $ R $ to be a field, and $ \det(A)=1 $, we can even express $ A $ as products of transvection matrices. Transvection matrices generate SLn(R))

Now back to this question, it seems to me that it suffices to prove that invertible matrices of type: $$ \begin{pmatrix}x & s\\ y & t\end{pmatrix} $$ where $ ax+by=d $, $ \gcd(a, b)=d $ and $ s=bd^{-1}, t=-ad^{-1} $(Here the inverse is obtained by cancellation in $ R $) can be expressed into products of elementary matrices. Since $ \det(A)=1 $, then $ d=1 $ and we have: $ \gcd(a, b)=1, s=b, t=-a $. Well how to move on?

EDIT: I have changed my title and statement in a less misleading way since I find out that we can't express $ A $ in such a way under the given assumption generally.

REPLY [2 votes]: As pointed out by @darijgrinberg in the comment, we have already had a counterexample in the cited article. (p23):
Consider the ring of integers in $ \mathbb Q(\sqrt{-19}) $: $ I=\mathbb Z\left[\frac{1+\sqrt{-19}}{2}\right] $ which is also a p.i.d. (Ref: An example of a PID which is not a Euclidean), then according to Theorem (6.1) in the article, we know that $ I $ is not a $ GE_2 $-ring.

Definition of $ \mathbf{GE_n} $: Every invertible $ n\times n $ matrix is a product of elementary $
n\times n $ matrices.

Furthermore, on the top of page23, the author gives an explicit matrix in $ GL_2(I) $ but not in $ GE_2(I) $:
$$ \begin{pmatrix} 3-\theta& 2+\theta\\ -3-2\theta& 5-2\theta \end{pmatrix} $$ where $ \theta^2-\theta+5=0 $. By direct computation, we know that $$ \det\begin{pmatrix} 3-\theta& 2+\theta\\ -3-2\theta& 5-2\theta \end{pmatrix}=1 .$$ So generally we cannot express a matrix with determinant $ 1 $ in $ M_n(R) $ as products of elementary matrices.