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\fancyhead[LO]{\textbf{Finite Field Grothendieck Inequality, Finite Field Johnson-Lindenstrauss Flattening and Finite Field Bourgain-Tzafriri Restricted Invertibility Problems}}
\fancyhead[RE]{\textbf{K. MAHESH KRISHNA}}


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{\bf{Finite Field Grothendieck Inequality, Finite Field Johnson-Lindenstrauss Flattening and Finite Field Bourgain-Tzafriri Restricted Invertibility Problems}}\\
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\textbf{K. Mahesh Krishna}\\
School of Mathematics and Natural Sciences\\
Chanakya University Global Campus\\
NH-648, Haraluru Village\\
Devanahalli Taluk, 	Bengaluru  North District\\
Karnataka State 562 110 India\\
Email: kmaheshak@gmail.com\\

Date: \today
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\textbf{Abstract}: We ask for finite field versions of following three: (1) Grothendieck Inequality, (2) Johnson-Lindenstrauss Flattening Lemma, (3) Bourgain-Tzafriri Restricted Invertibility Theorem.

\textbf{Keywords}:  Grothendieck Inequality, Johnson-Lindenstrauss Lemma, Bourgain-Tzafriri Restricted Invertibility, Finite Field.

\textbf{Mathematics Subject Classification (2020)}: 12E20, 46B20, 90C27.\\

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\section{Finite Field Grothendieck Inequality Problem}
The Grothendieck inequality was derived in the following form by Lindenstrauss and Pelczynski in 1968 \cite{LINDENSTRAUSS}. 
  \begin{theorem}\cite{BLEI, RIETZ, GROTHENDIECK, FRIEDLANDLIMZHANG, ALBIACKALTON, DIESTELFOURIESWART, PISIERGROTHENDIECK, LINDENSTRAUSS}\label{G}
  	 (\textbf{Grothendieck Inequality} or \textbf{The Fundamental Theorem in the Metric Theory of Tensor Products}) Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. There is a universal constant $K_G^{\mathbb{K}}>0$ satisfying the following: For every  Hilbert space $\mathcal{H}$ over $\mathbb{K}$ and for all $m, n \in \mathbb{N}$, if a scalar matrix $[a_{j,k}]_{1\leq j \leq m, 1\leq k \leq n}$ satisfies
  	\begin{align*}
  		\left|\sum_{j=1}^m \sum_{k=1}^na_{j,k}s_jt_k\right|\leq \left(\max_{1\leq j \leq m}|s_j|\right) \left(\max_{1\leq k \leq n}|t_k|\right),\quad  \forall s_j , t_k \in \mathbb{K}, 
  	\end{align*}
  then 
  \begin{align*}
  \left| \sum_{j=1}^m \sum_{k=1}^na_{j,k}\langle u_j, v_k\rangle\right|	\leq K^\mathbb{K}_G\left(\max_{1\leq j \leq m}\|u_j\|\right)  \left(\max_{1\leq k \leq n}\|v_k\|\right), \quad  \forall u_j , v_k \in \mathcal{H}. 
  \end{align*}
Further, 
\begin{align*}
	1.67\leq K^\mathbb{R}_G\leq 1.79, \quad 1.33\leq K^\mathbb{C}_G\leq 1.41.
\end{align*}
  	 \end{theorem}
  We formulate  problem which is  finite field analogue of Theorem \ref{G}. 	Let $p$ be a prime,  $\mathbb{Z}_p\coloneqq \{0, 1, \dots, p-1\}$ be the standard field of integers modulo $p$. Given $n \in \mathbb{Z}_p$, we define 
   \begin{align*}
   	|n|\coloneqq \text{ unique } a \text{ such that } 0\leq a \leq p-1 \text{ and }n \equiv a (\text{mod }p).
   \end{align*}
Since $p$ is a prime, we have the following:
\begin{enumerate}[\upshape (i)]
	\item If $n \in \mathbb{Z}_p$ is such that $|n|=0$, then $n=0$.
	\item $|mn|\leq |m||n|$ for all $m, n \in \mathbb{Z}_p$.
	\item $|m+n|\leq |m|+|n|$ for all $m, n \in \mathbb{Z}_p$.
\end{enumerate}
   For  $d \in \mathbb{N}$, let $\mathbb{Z}_p^d$ be the standard vector space over  $\mathbb{Z}_p$. We define  the inner product 
   \begin{align*}
   	\langle (a_j)_{j=1}^d,(b_j)_{j=1}^d\rangle \coloneqq \sum_{j=1}^da_jb_j \in \mathbb{Z}_p,  \quad \forall (a_j)_{j=1}^d,(b_j)_{j=1}^d \in \mathbb{Z}_p^d
   \end{align*}
   and the norm 
   \begin{align*}
   	\|(c_j)_{j=1}^d\|^2\coloneqq \sum_{j=1}^d|c_j|^2 \in \{0, 1, \dots, d(p-1)^2\}, \quad \forall (c_j)_{j=1}^d\in 	\mathbb{Z}_p^d.
   \end{align*}
Note that 
\begin{align*}
	\sum_{j=1}^d|c_j^2|\leq \sum_{j=1}^d|c_j|^2=\|(c_j)_{j=1}^d\|^2 , \quad \forall (c_j)_{j=1}^d\in 	\mathbb{Z}_p^d.
\end{align*}
   Inner product and norm will have following properties:
   	\begin{enumerate}[\upshape (i)]
   	\item If $x \in \mathbb{Z}_p^d$ is such that $\langle x,y \rangle =0$ for all $y \in\mathbb{Z}_p^d$, then $x=0$.
   	\item $\langle nx, y \rangle =n\langle x, y \rangle$ for all $ n \in \mathbb{Z}_p$, for all $x,y \in \mathbb{Z}_p^d$.
   	\item $\langle x, y \rangle =\langle y, x \rangle$ for all $x,y \in \mathbb{Z}_p^d$.
   	\item $\langle x+y, z \rangle =\langle x,  z \rangle+\langle y, z\rangle$ for all $x,y,z \in \mathbb{Z}_p^d$.
   	\item $|\langle x, y \rangle |\leq \|x\|\|y\|$ for all $x,y \in \mathbb{Z}_p^d$.
   	 	\item If $x \in \mathbb{Z}_p^d$ is such that $\|x\| =0$, then $x=0$.
   	 	\item $\|x\|^2 \in \mathbb{N}\cup \{0\}$ for all $x\in \mathbb{Z}_p^d$.
   	\item $\|n x\|\leq |n|\|x\|$ for all $ n \in \mathbb{Z}_p$, for all $x \in \mathbb{Z}_p^d$.
   	\item $\|x+y\|\leq \|x\|+\|y\|$ for all $x,y \in \mathbb{Z}_p^d$.
   \end{enumerate}

Based on Theorem \ref{G}, we formulate following problem.
   \begin{problem}
 	\textbf{(Finite Field  Grothendieck Inequality Problem)} 	Let $p$ be a prime. Whether there is a  universal constant $K_p>0$ (which may depend upon $p$) satisfying the following property: for every $d \in \mathbb{N}$, for all $m,n \in \mathbb{N}$,	if $[a_{j,k}]_{1\leq j \leq m, 1\leq k \leq n}\in \mathbb{M}_{m\times n}(\mathbb{Z}_p)$ satisfies 
 		\begin{align*}
 		\left|\sum_{j=1}^m \sum_{k=1}^na_{j,k}s_jt_k\right|\leq \left(\max_{1\leq j \leq m}|s_j|\right) \left(\max_{1\leq k \leq n}|t_k|\right),\quad  \forall s_j , t_k \in \mathbb{Z}_p, 
 	\end{align*}
 	then 
 	\begin{align*}
 		\left| \sum_{j=1}^m \sum_{k=1}^na_{j,k}\langle \mathbf{u}_j, \mathbf{v}_k\rangle\right| 	\leq K_p\left(\max_{1\leq j \leq m}\|\mathbf{u}_j\|\right)  \left(\max_{1\leq k \leq n}\|\mathbf{v}_k\|\right), \quad  \forall \mathbf{u}_j , \mathbf{v}_k \in \mathbb{Z}_p^d.
 	\end{align*}
 \end{problem}
 




    \section{Finite Field Johnson-Lindenstrauss Flattening Problem}
   Johnson and  Lindenstrauss showed that points in high dimensional Euclidean space can be mapped to low dimensional Euclidean space with small distortion to distances between every pair of points,  using Lipschitz maps
    \cite{JOHNSONLINDENSTRAUSS}.
\begin{theorem}\cite{MATOUSEKBOOK, JOHNSONLINDENSTRAUSS}\label{JL}
	\textbf{(Johnson-Lindenstrauss Flattening Lemma)} Let $M, N\in \mathbb{N}$ and  $\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_M\in \mathbb{R}^N$. For every $0<\varepsilon<1$, there exists a Lipschitz map $f: \mathbb{R}^N\to \mathbb{R}^m$  and a real number $r>0$ such that 
		\begin{align*}
			r(1-\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|\leq \|f(\mathbf{x}_j)-f(\mathbf{x}_k)\|\leq r(1+\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|, \quad \forall 1 \leq j, k\leq M, 
		\end{align*}
		where 
		\begin{align*}
			m=O\left(\frac{\log M}{\varepsilon^2}\right).
	\end{align*}
\end{theorem}
Over the period, some improvements of Theorem \ref{JL} have been  obtained. We recall them. 
\begin{theorem}\cite{FRANKLMAEHARA, JOHNSONLINDENSTRAUSS} \textbf{(Johnson-Lindenstrauss  Flattening  Lemma: Frankl-Maehara form)}
		Let $0<\varepsilon<\frac{1}{2}$   and $ M\in \mathbb{N}$. Define 
		\begin{align*}
			m (\varepsilon, M)\coloneqq \left\lceil 9\frac{1}{\varepsilon^2-\frac{2\varepsilon^3}{3}}\log M \right \rceil +1.
		\end{align*} 
		If  $M> m (\varepsilon, M)$, then for every  $\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_M\in \mathbb{R}^M$,  there exists 	a map $f: \{\mathbf{x}_j\}_{j=1}^M\to \mathbb{R}^m$   such that 
		\begin{align*}
			(1-\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|^2\leq \|f(\mathbf{x}_j)-f(\mathbf{x}_k)\|^2\leq (1+\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|^2, \quad \forall 1 \leq j, k\leq M.
	\end{align*}
\end{theorem}
\begin{theorem}\cite{DASGUPTAGUPTA, JOHNSONLINDENSTRAUSS}
	\textbf{(Johnson-Lindenstrauss Flattening Lemma: Dasgupta-Gupta form)}
		Let $M, N\in \mathbb{N}$ and  $\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_M\in \mathbb{R}^N$. Let $0<\varepsilon<1$. Choose any natural number $m$ such that 
		\begin{align*}
			m>4\frac{1}{\frac{\varepsilon^2}{2}-\frac{\varepsilon^3}{3}}\log M. 
		\end{align*}
		Then there exists a  map $f: \mathbb{R}^N\to \mathbb{R}^m$   such that 
		\begin{align*}
			(1-\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|^2\leq \|f(\mathbf{x}_j)-f(\mathbf{x}_k)\|^2\leq (1+\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|^2, \quad \forall 1 \leq j, k\leq M.
		\end{align*}
		The map $f$ can be determined in randomized polynomial time.
\end{theorem}
\begin{theorem} \cite{JOHNSONLINDENSTRAUSS, FOUCARTRAUHUT} \textbf{(Johnson-Lindenstrauss Flattening Lemma: matrix form)} \label{MF}
		There is a universal constant $C>0$ satisfying the following. Let $0<\varepsilon<1$, $M,N \in \mathbb{N}$ and $\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_M\in \mathbb{R}^N$. For each natural number 
		\begin{align*}
			m>\frac{C}{\varepsilon^2}\log M,
		\end{align*}
		there exists a matrix $M\in \mathbb{M}_{m\times N}(\mathbb{R})$ such that 
		\begin{align*}
			(1-\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|^2\leq \|M(\mathbf{x}_j-\mathbf{x}_k)\|^2\leq (1+\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|^2, \quad \forall 1 \leq j, k\leq M.
	\end{align*}
\end{theorem}
Based on Theorem \ref{MF}, we formulate following problem.
\begin{problem}\textbf{(Finite Field Johnson-Lindenstrauss Flattening Problem)}
	Let $p$ be a prime.  What is the best function $\phi: (0,1)\times \mathbb{N} \to (0,\infty)$ satisfying the following: 	There is a universal constant $C_p>0$ (which may depend upon $p$) satisfying the following. Let $0<\varepsilon<1$, $M,N \in \mathbb{N}$ and $\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_M\in \mathbb{Z}_p^N$. For each natural number 
	\begin{align*}
		m>C_p\phi (\varepsilon, M),
	\end{align*}
	there exists a matrix $M\in \mathbb{M}_{m\times N}(\mathbb{Z}_p)$ such that 
	\begin{align*}
		(1-\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\| \leq  \|M(\mathbf{x}_j-\mathbf{x}_k)\| \leq (1+\varepsilon)\|\mathbf{x}_j-\mathbf{x}_k\|, \quad \forall 1 \leq j, k\leq M.
\end{align*}
\end{problem}




  \section{Finite Field Bourgain-Tzafriri Restricted Invertibility Problem}
   Let $\mathbb{R}^d$ be the standard $d$-dimensional Euclidean  space with canonical orthonormal basis $\{\mathbf{e}_j\}_{j=1}^d$.  In 1987, Bourgain and Tzafriri  showed that every matrix can be inverted restrictively with control over the norm   \cite{BOURGAINTZAFRIRI}.
  \begin{theorem}\cite{BOURGAINTZAFRIRI, VERSHYNIN2}\label{BOURGAINTZARIRITHEOREM}  \textbf{(Bourgain-Tzafriri Restricted Invertibility Theorem)}
  		There are universal constants $A>0$, $c>0$ satisfying the following property. If  $d \in \mathbb{N}$, and  $T:\mathbb{R}^d \to \mathbb{R}^d$ is a linear operator   with  $\|T\mathbf{e}_j\|=1$, $\forall 1\leq j\leq d$, then there exists a subset $\sigma \subseteq \{1, \dots, d\}$ of cardinality 
  		\begin{align*}
  			\operatorname{Card}(\sigma)\geq \frac{cd}{\|T\|^2}
  		\end{align*}
  		such that 
  		\begin{align*}
  			\left\|T\left(\sum_{j \in \sigma}a_j\mathbf{e}_j\right)\right\|^2\geq A \sum_{j \in \sigma}|a_j|^2, \quad \forall a_j \in  \mathbb{R}, \forall j \in \sigma.
  	\end{align*}
  \end{theorem}  
It is clear that we can state Theorem \ref{BOURGAINTZARIRITHEOREM} also in following way. 
\begin{theorem}\cite{BOURGAINTZAFRIRI, VERSHYNIN2} \textbf{(Bourgain-Tzafriri Restricted Invertibility Theorem)}
	There are universal constants $A>0$, $c>0$ satisfying the following property. If  $d \in \mathbb{N}$, and  $T:\mathbb{R}^d \to \mathbb{R}^d$ is a linear operator   with  
	\begin{align*}
		\|T\mathbf{e}_1\|^2=\cdots =\|T\mathbf{e}_d\|^2=:r \quad \text{(say)}, 
	\end{align*}
 then there exists a subset $\sigma \subseteq \{1, \dots, d\}$ of cardinality 
	\begin{align*}
		\operatorname{Card}(\sigma)\geq \frac{rcd}{\|T\|^2}
	\end{align*}
	such that 
	\begin{align*}
		\left\|T\left(\sum_{j \in \sigma}a_j\mathbf{e}_j\right)\right\|^2\geq rA \sum_{j \in \sigma}|a_j|^2, \quad \forall a_j \in  \mathbb{R}, \forall j \in \sigma.
	\end{align*}
\end{theorem}  
  Spielman and Srivastava gave a simple proof of Theorem \ref{BOURGAINTZARIRITHEOREM} by proving a generalization of  Theorem \ref{BOURGAINTZARIRITHEOREM}  \cite{SPIELMANSRIVASTAVA} due to Vershynin \cite{VERSHYNIN}.
Based on Theorem \ref{BOURGAINTZARIRITHEOREM}, we formulate finite field  problem. Given a linear operator $T:\mathbb{Z}_p^d \to \mathbb{Z}_p^d$, we define 
\begin{align*}
	\|T\|\coloneqq \max \left\{\frac{\|T\mathbf{x}\|}{\|\mathbf{x}\|}: \mathbf{x}\in \mathbb{Z}_p^d, \mathbf{x} \neq 0\right\}.
\end{align*}
  \begin{problem}
  	\textbf{(Finite Field Bourgain-Tzafriri Restricted Invertibility Problem)}
  	Let $p$ be a prime.  	Whether there are universal constants $A_p>0$, $c_p>0$ (which may depend upon $p$) satisfying the following property: If  $d \in \mathbb{N}$, and  $T:\mathbb{Z}_p^d \to \mathbb{Z}_p^d$ is a linear operator   with  $\|T\mathbf{e}_j\|=1$, $\forall 1\leq j\leq d$, then there exists a subset $\sigma \subseteq \{1, \dots, d\}$ of cardinality 
  		\begin{align*}
  			\operatorname{Card}(\sigma)\geq \frac{c_pd}{\|T\|^2}
  		\end{align*}
  		such that 
  		\begin{align*}
  			\left\|T\left(\sum_{j \in \sigma}a_j\mathbf{e}_j\right)\right\|^2\geq A_p \sum_{j \in \sigma}|a_j|^2, \quad \forall a_j \in  \mathbb{Z}_p, \forall j \in \sigma.
  	\end{align*}
  \end{problem}
    \begin{problem}
  	\textbf{(Finite Field Bourgain-Tzafriri Restricted Invertibility Problem)}
  	Let $p$ be a prime.  	Whether there are universal constants $A_p>0$, $c_p>0$ (which may depend upon $p$) satisfying the following property: If  $d \in \mathbb{N}$, and  $T:\mathbb{Z}_p^d \to \mathbb{Z}_p^d$ is a linear operator   with   
  	\begin{align*}
  	\|T\mathbf{e}_1\|^2=\cdots =\|T\mathbf{e}_d\|^2=:r \quad \text{(say)}, 
  \end{align*}
  	then there exists a subset $\sigma \subseteq \{1, \dots, d\}$ of cardinality 
  	\begin{align*}
  		\operatorname{Card}(\sigma)\geq \frac{rc_pd}{\|T\|^2}
  	\end{align*}
  	such that 
  	\begin{align*}
  		\left\|T\left(\sum_{j \in \sigma}a_j\mathbf{e}_j\right)\right\|^2\geq rA_p \sum_{j \in \sigma}|a_j|^2, \quad \forall a_j \in  \mathbb{Z}_p, \forall j \in \sigma.
  	\end{align*}
  \end{problem}
  
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