Functional van Lint-Seidel Relative and Gerzon Universal Bounds

1. Introduction

Let $d\in \mathbb{N}$ and $\gamma \in [0,1]$. Recall that a collection ${\left\{{\tau }_j\right\}}_{j=1}^n$ of unit vectors in ${\mathbb{R}}^d$ is said to be $\gamma$-equiangular lines [1,2] if

$$\begin{array}{c}\hfill |\langle {\tau }_j,{\tau }_k\rangle |=\gamma ,\forall 1\le j,k\le n,j\ne k.\end{array}$$

A fundamental problem associated with equiangular lines is the following.

Problem 1.

Given $d\in \mathbb{N}$ and $\gamma \in [0,1]$, what is the upper bound on n such that there exists a collection ${\left\{{\tau }_j\right\}}_{j=1}^n$ of γ-equiangular lines in ${\mathbb{R}}^d$?

Two answers to Problem (1) which are driving forces in the study of equiangular lines is the following relative bound of van Lint and Seidel [2,3] and universal bound of Gerzon [4].

Theorem 1.

[2,3] (van Lint-Seidel Relative Bound) Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be γ-equiangular lines in ${\mathbb{R}}^d$. Then

$$\begin{array}{c}\hfill n(1-d{\gamma }^2)\le d(1-{\gamma }^2).\end{array}$$

In particular, if

$$\begin{array}{c}\hfill \gamma <{\displaystyle \frac{1}{\sqrt{d}}},\end{array}$$

then

$$\begin{array}{c}\hfill n\le {\displaystyle \frac{d(1-{\gamma }^2)}{1-d{\gamma }^2}}.\end{array}$$

Theorem 2.

[4] (Gerzon Universal Bound) Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be γ-equiangular lines in ${\mathbb{R}}^d$. Then

$$\begin{array}{c}\hfill n\le {\displaystyle \frac{d(d+1)}{2}}.\end{array}$$

The notion of functional equiangular lines is hinted in [5]. In this paper, we define it in most general form and derive functional forms of Theorems 1 and 2.

2. 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 $Mor\left(\mathcal{M}\right)$.

Definition 1.

Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be a collection in a module $\mathcal{M}$ of rank d over $\mathcal{R}$ and ${\left\{f_j\right\}}_{j=1}^n$ be a collection in ${\mathcal{M}}^{*}.$ Let $\gamma \ge 0$. The pair $({\left\{f_j\right\}}_{j=1}^n,{\left\{{\tau }_j\right\}}_{j=1}^n)$ is said to be functional $\gamma$-equiangular if following conditions hold.

(i)

$f_j\left({\tau }_j\right)=1$ for all $1\le j\le n$.

(ii)

$|f_j\left({\tau }_k\right)f_k\left({\tau }_j\right)|={\gamma }^2$ for all $1\le j,k\le n,j\ne k.$

(iii)

The operator

$$\begin{array}{c}\hfill S_{f,\tau }:\mathcal{M}\ni x\mapsto \sum _{j=1}^n{f}_j\left(x\right){\tau }_j\in \mathcal{M}\end{array}$$

is similar (through invertible operator) to a diagonal operator.

Theorem 3.  (Functional van Lint-Seidel Relative Bound) If $({\left\{f_j\right\}}_{j=1}^n,{\left\{{\tau }_j\right\}}_{j=1}^n)$ is functional γ-equiangular for rank d module $\mathcal{M}$, then

$$\begin{array}{c}\hfill n(1-d{\gamma }^2)\le d(1-{\gamma }^2).\end{array}$$

In particular, if

$$\begin{array}{c}\hfill \gamma <{\displaystyle \frac{1}{\sqrt{d}}},\end{array}$$

then

$$\begin{array}{c}\hfill n\le {\displaystyle \frac{d(1-{\gamma }^2)}{1-d{\gamma }^2}}.\end{array}$$

Proof. 

Define

$$\begin{array}{c}\hfill S_{f,\tau }:\mathcal{M}\ni x\mapsto S_{f,\tau }x\sum _{j=1}^n{f}_j\left(x\right){\tau }_j\in \mathcal{M}.\end{array}$$

Let ${\lambda }_1,\dots ,{\lambda }_d$ be eigenvalues of $S_{f,\tau }$. Since $S_{f,\tau }$ is diagonalizable,

$$\begin{array}{cc}\hfill n^2& ={\left(\sum _{j=1}^n{f}_j\left({\tau }_j\right)\right)}^2={(Tra\left(S_{f,\tau }\right))}^2={\left(\sum _{k=1}^d{\lambda }_k\right)}^2\hfill \\ \hfill & \le d\sum _{k=1}^d{\lambda }_k^2=dTra\left(S_{f,\tau }^2\right)=d\sum _{j=1}^n\sum _{k=1}^n{f}_j\left({\tau }_k\right)f_k\left({\tau }_j\right)\hfill \\ \hfill & =d\sum _{j=1}^n{f}_j{\left({\tau }_j\right)}^2+d\sum _{j,k=1,j\ne k}^n{f}_j\left({\tau }_k\right)f_k\left({\tau }_j\right)=dn+d\sum _{j,k=1,j\ne k}^n{f}_j\left({\tau }_k\right)f_k\left({\tau }_j\right)\hfill \\ \hfill & \le dn+d\sum _{j,k=1,j\ne k}^n\left|f_j\left({\tau }_k\right)f_k\left({\tau }_j\right)\right|=dn+d(n^2-n){\gamma }^2.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill n\le d+d(n-1){\gamma }^2\Rightarrow n(1-d{\gamma }^2)\le d(1-{\gamma }^2).\end{array}$$

□

We are unable to derive Gerzon universal bound for functional equiangular lines. However, we derive following Gerzon bound for functionals.

Theorem 4.  (Functional Gerzon Universal Bound) Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be a collection in a module $\mathcal{M}$ of rank d over $\mathcal{R}$ and ${\left\{f_j\right\}}_{j=1}^n$ be a collection in ${\mathcal{M}}^{*}.$ Assume the following.

(i)

$f_j\left({\tau }_j\right)=1$ for all $1\le j\le n$.

(ii)

There is a $\gamma \ne 1$ such that $f_j\left({\tau }_k\right)f_k\left({\tau }_j\right)={\gamma }^2$ for all $1\le j,k\le n,j\ne k.$

Then

$$\begin{array}{c}\hfill n\le d^2.\end{array}$$

Proof. 

For $1\le j\le n$, define

$$\begin{array}{c}\hfill {\tau }_j\otimes f_j:\mathcal{M}\ni x\mapsto ({\tau }_j\otimes f_j)\left(x\right)f_j\left(x\right){\tau }_j\in \mathcal{M}\end{array}$$

We show that the collection ${\{{\tau }_j\otimes f_j\}}_{j=1}^n$ in $Mor\left(\mathcal{M}\right)$ is linearly independent over $\mathcal{R}$. Let $c_1,\dots ,c_n\in \mathcal{R}$ be such that

$$\begin{array}{c}\hfill \sum _{j=1}^n{c}_j({\tau }_j\otimes f_j)=0.\end{array}$$

Let $1\le k\le n$ be fixed. Then previous equation gives

$$\begin{array}{c}\hfill \sum _{j=1}^n{c}_j({\tau }_j\otimes f_j)({\tau }_k\otimes f_k)=0.\end{array}$$

By taking trace we get

$$\begin{array}{cc}\hfill 0& =\sum _{j=1}^n{c}_{j}Tra\left(({\tau }_j\otimes f_j)({\tau }_k\otimes f_k)\right)=\sum _{j=1}^n{c}_j{f}_j\left({\tau }_k\right)f_k\left({\tau }_j\right)\hfill \\ \hfill & =\sum _{j=1,j\ne k}^n{c}_j{f}_j\left({\tau }_k\right)f_k\left({\tau }_j\right)+c_k{f}_k{\left({\tau }_k\right)}^2=\sum _{j=1,j\ne k}^n{c}_j{\gamma }^2+c_k\hfill \\ \hfill & ={\gamma }^2\left(\sum _{j=1}^n{c}_j-c_k\right)+c_k={\gamma }^2\left(\sum _{j=1}^n{c}_j\right)+(1-{\gamma }^2)c_k.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill c_k={\displaystyle \frac{{\gamma }^2}{{\gamma }^2-1}}\sum _{j=1}^n{c}_j=:c,\forall 1\le k\le n.\end{array}$$

Now

$$\begin{array}{cc}\hfill 0& =Tra\left(\sum _{j=1}^n{c}_j({\tau }_j\otimes f_j)\right)=Tra\left(\sum _{j=1}^{n}c({\tau }_j\otimes f_j)\right)\hfill \\ \hfill & =\sum _{j=1}^{n}cTra({\tau }_j\otimes f_j)=\sum _{j=1}^{n}c{f}_j\left({\tau }_j\right)=cn.\hfill \end{array}$$

Hence $c=0$. Therefore ${\{{\tau }_j\otimes f_j\}}_{j=1}^n$ is linearly independent. Since the rank of $Mor\left(\mathcal{M}\right)$ is $d^2$ we must have $n\le d^2$. □