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\fancyhead[LO]{\textbf{Functional van Lint-Seidel Relative and Gerzon Universal Bounds}}
\fancyhead[RE]{\textbf{K. MAHESH KRISHNA}}


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{\bf{Functional van Lint-Seidel Relative and Gerzon Universal Bounds}}\\
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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 introduce the notion of  functional equiangular lines in   finite rank modules over subrings of $\mathbb{R}$. We show that  van Lint-Seidel relative  and Gerzon universal bounds hold for this most general lines.

\textbf{Keywords}:  Equiangular lines.

\textbf{Mathematics Subject Classification (2020)}: 52A40, 05C50.\\

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\section{Introduction}
Let $d \in \mathbb{N}$ and $\gamma \in  [0,1]$. Recall that a collection $\{\tau_j\}_{j=1}^n$  of unit vectors  in $\mathbb{R}^d$	is said to be \textbf{$\gamma$-equiangular  lines} \cite{HAANTJES, LEMMENSSEIDEL} if 
	\begin{align*}
			|\langle \tau_j, \tau_k\rangle|=\gamma, \quad \forall 1\leq j, k \leq n, j \neq k.
	\end{align*}	
A fundamental problem associated with equiangular lines is  the following.
\begin{problem}\label{1} 
	Given $d \in \mathbb{N}$ and $\gamma \in  [0,1]$, what is the upper bound on $n$ such that there exists a collection $\{\tau_j\}_{j=1}^n$  of  $\gamma$-equiangular lines in $\mathbb{R}^d$?
	\end{problem}
Two  answers to Problem (\ref{1}) which are driving forces in the study of equiangular lines is the following relative bound  of van Lint and Seidel \cite{VANLINTSEIDEL, LEMMENSSEIDEL} and universal bound of Gerzon \cite{WALDRONBOOK}.
\begin{theorem} \cite{VANLINTSEIDEL, LEMMENSSEIDEL} (\textbf{van Lint-Seidel Relative Bound}) \label{VSTHEOREM}
Let $\{\tau_j\}_{j=1}^n$  	be  $\gamma$-equiangular  lines in $\mathbb{R}^d$. 
Then 
\begin{align*}
n(1-d\gamma^2) \leq d(1-\gamma^2).
\end{align*}
In particular, if 
\begin{align*}
	\gamma <\frac{1}{\sqrt{d}},
\end{align*}
then 
\begin{align*}
	n\leq \frac{d(1-\gamma^2)}{1-d\gamma^2}.
\end{align*}
\end{theorem}
\begin{theorem} \cite{WALDRONBOOK} \label{GERZONBOUND}
(\textbf{Gerzon Universal Bound}) 
Let $\{\tau_j\}_{j=1}^n$  	be  $\gamma$-equiangular  lines in $\mathbb{R}^d$. 
Then 
\begin{align*}
n \leq \frac{d(d+1)}{2}.
\end{align*}	
\end{theorem}
The notion of functional  equiangular lines is hinted in \cite{KRISHNA}. In this paper, we define it in most general form   and derive functional forms of Theorems \ref{VSTHEOREM} and \ref{GERZONBOUND}.




\section{Functional Equiangular Lines}
In  the paper, $\mathcal{R}$ denotes a subring of  $\mathbb{R}$ and $\mathcal{M}$ is a rank  $d$ module over $\mathcal{R}$. By $\mathcal{M}^*$ we mean the module of all homomorphisms from $\mathcal{M}$ to $\mathcal{R}$. Module of all homomorphisms from $\mathcal{M}$ to $\mathcal{M}$ is denoted by $\operatorname{Mor}(\mathcal{M})$.
\begin{definition}\label{EF}
	Let $\{\tau_j\}_{j=1}^n$ be a collection in a module  $\mathcal{M}$ of rank  $d$  over  $\mathcal{R}$  and 	$\{f_j\}_{j=1}^n$ be a collection in  $\mathcal{M}^*.$ Let $\gamma \geq 0$. The pair  $ (\{f_j \}_{j=1}^n, \{\tau_j \}_{j=1}^n) $   is said to be 	\textbf{functional $\gamma$-equiangular} if following conditions hold. 
	\begin{enumerate}[\upshape(i)]
		\item $f_j(\tau_j)=1$ for all $1\leq j \leq n$.
		\item $	|f_j(\tau_k)f_k(\tau_j)|=\gamma^2$ for all $  1\leq j,k \leq n, j\neq k.$
		\item The operator 
	\begin{align*}
		S_{f,\tau} : \mathcal{M}\ni x \mapsto \sum_{j=1}^nf_j(x) \tau_j \in \mathcal{M}
	\end{align*}
	is similar (through invertible operator) to a diagonal operator.   
	\end{enumerate}
\end{definition}
\begin{theorem} (\textbf{Functional van Lint-Seidel Relative Bound}) 
	If  $ (\{f_j \}_{j=1}^n, \{\tau_j \}_{j=1}^n) $   is 	functional $\gamma$-equiangular for rank  $d$ module $\mathcal{M}$, then 
	\begin{align*}
		n(1-d\gamma^2) \leq d(1-\gamma^2).
	\end{align*}
	In particular, if 
	\begin{align*}
		\gamma <\frac{1}{\sqrt{d}},
	\end{align*}
	then 
	\begin{align*}
		n\leq \frac{d(1-\gamma^2)}{1-d\gamma^2}.
	\end{align*}
\end{theorem}
\begin{proof}
	Define \begin{align*}
		S_{f, \tau}:\mathcal{M}\ni x \mapsto S_{f, \tau}x\coloneqq \sum_{j=1}^n
		f_j(x)\tau_j \in\mathcal{M}.
	\end{align*}
Let $\lambda_1, \dots, \lambda_d$ be eigenvalues of $S_{f, \tau}$. Since $S_{f, \tau}$ is diagonalizable,  
\begin{align*}
n^2&=\left(\sum_{j=1}^n
	f_j(\tau_j)\right)^2=(\operatorname{Tra}(S_{f,\tau}))^2=\left(\sum_{k=1}^d
	\lambda_k\right)^2\\
	&\leq d \sum_{k=1}^d
	\lambda_k^2 =d\operatorname{Tra}(S^2_{f,\tau})=d\sum_{j=1}^n\sum_{k=1}^nf_j(\tau_k)f_k(\tau_j)\\
	&=d\sum_{j=1}^nf_j(\tau_j)^2+d\sum_{j,k=1, j\neq k}^nf_j(\tau_k)f_k(\tau_j)=dn+d\sum_{j,k=1, j\neq k}^nf_j(\tau_k)f_k(\tau_j)\\
	&\leq dn+d\sum_{j,k=1, j\neq k}^n|f_j(\tau_k)f_k(\tau_j)|= dn+ d(n^2-n)\gamma^2.
\end{align*}	
Therefore 
\begin{align*}
	n \leq d+d(n-1)\gamma^2 \implies 	n(1-d\gamma^2) \leq d(1-\gamma^2).
\end{align*}
\end{proof}
We are unable to derive Gerzon universal bound for functional equiangular lines. However, we derive following  Gerzon bound for functionals.
\begin{theorem}	(\textbf{Functional Gerzon Universal Bound}) 
	Let $\{\tau_j\}_{j=1}^n$ be a collection in a module  $\mathcal{M}$ of rank  $d$  over  $\mathcal{R}$  and 	$\{f_j\}_{j=1}^n$ be a collection in  $\mathcal{M}^*.$ Assume the following.
		\begin{enumerate}[\upshape(i)]
		\item $f_j(\tau_j)=1$ for all $1\leq j \leq n$.
		\item There is a $\gamma\neq 1$ such that $	f_j(\tau_k)f_k(\tau_j)=\gamma^2$ for all $  1\leq j,k \leq n, j\neq k.$
		\end{enumerate}
	Then 
	\begin{align*}
		n \leq d^2.	
	\end{align*}
\end{theorem}
\begin{proof}
	For $1\leq j \leq n$, define 
	\begin{align*}
		\tau_j \otimes f_j: \mathcal{M} \ni x \mapsto(\tau_j \otimes f_j)(x)\coloneqq f_j(x)\tau_j \in \mathcal{M}
	\end{align*}
	We show that the collection $\{\tau_j \otimes f_j\}_{j=1}^n$ in $\operatorname{Mor}(\mathcal{M})$  is linearly independent  over  $\mathcal{R}$. Let $c_1, \dots,  c_n \in \mathcal{R}$ be such that 
	\begin{align*}
		\sum_{j=1}^{n}c_j(\tau_j \otimes f_j)=0.
	\end{align*}
	Let $1\leq k \leq n$ be fixed. Then previous equation gives 
	\begin{align*}
		\sum_{j=1}^{n}c_j(\tau_j \otimes f_j)(\tau_k \otimes f_k)	=0.
	\end{align*}
	By taking trace we get
	\begin{align*}
		0&=\sum_{j=1}^{n}c_j\operatorname{Tra}((\tau_j \otimes f_j)(\tau_k \otimes f_k))=\sum_{j=1}^{n}c_jf_j(\tau_k)f_k(\tau_j)\\
		&=\sum_{j=1, j \neq k}^{n}c_jf_j(\tau_k)f_k(\tau_j)+c_kf_k(\tau_k)^2=\sum_{j=1, j \neq k}^{n}c_j\gamma^2+c_k\\
		&=\gamma^2\left(\sum_{j=1}^{n}c_j-c_k\right)+c_k=\gamma^2\left(\sum_{j=1}^{n}c_j\right)+(1-\gamma^2)c_k.
	\end{align*}
	Therefore 
	\begin{align*}
		c_k=\frac{\gamma^2}{\gamma^2-1}\sum_{j=1}^{n}c_j=:c, \quad \forall 1\le k \leq n.	
	\end{align*}
	Now 
	\begin{align*}
		0&=\operatorname{Tra}\left(\sum_{j=1}^{n}c_j(\tau_j \otimes f_j)\right)=\operatorname{Tra}\left(\sum_{j=1}^{n}c(\tau_j \otimes f_j)\right)\\
		&=\sum_{j=1}^{n}c\operatorname{Tra}(\tau_j \otimes f_j)=\sum_{j=1}^{n}cf_j(\tau_j)=cn.
\end{align*}
Hence $c=0$. Therefore $\{\tau_j \otimes f_j\}_{j=1}^n$ is linearly independent.  Since the rank of $\operatorname{Mor}(\mathcal{M})$ is $d^2$ we must have $n\leq d^2$.
\end{proof}















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