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Functional van Lint-Seidel Relative and Gerzon Universal Bounds

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12 August 2024

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12 August 2024

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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.
Keywords: 
Subject: Computer Science and Mathematics  -   Analysis

MSC:  52A40; 05C50

1. Introduction

Let d N and γ [ 0 , 1 ] . Recall that a collection { τ j } j = 1 n of unit vectors in R d is said to be γ -equiangular lines [1,2] if
| τ j , τ k | = γ , 1 j , k n , j k .
A fundamental problem associated with equiangular lines is the following.
Problem 1.1.
Given d N and γ [ 0 , 1 ] , what is the upper bound on n such that there exists a collection { τ j } j = 1 n of γ-equiangular lines in R d ?
Two answers to Problem (1.1) 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.
[2,3] (van Lint-Seidel Relative Bound) Let { τ j } j = 1 n be γ-equiangular lines in R d . Then
n ( 1 d γ 2 ) d ( 1 γ 2 ) .
In particular, if
γ < 1 d ,
then
n d ( 1 γ 2 ) 1 d γ 2 .
Theorem 1.3.
[4] (Gerzon Universal Bound) Let { τ j } j = 1 n be γ-equiangular lines in R d . Then
n d ( d + 1 ) 2 .
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 1.2 and 1.3.

2. Functional Equiangular Lines

In the paper, R denotes a subring of R and M is a rank d module over R . By M * we mean the module of all homomorphisms from M to R . Module of all homomorphisms from M to M is denoted by Mor ( M ) .
Definition 2.1.
Let { τ j } j = 1 n be a collection in a module M of rank d over R and { f j } j = 1 n be a collection in M * . Let γ 0 . The pair ( { f j } j = 1 n , { τ j } j = 1 n ) is said to befunctional γ -equiangularif following conditions hold.
(i)
f j ( τ j ) = 1 for all 1 j n .
(ii)
| f j ( τ k ) f k ( τ j ) | = γ 2 for all 1 j , k n , j k .
(iii)
The operator
S f , τ : M x j = 1 n f j ( x ) τ j M
is similar (through invertible operator) to a diagonal operator.
Theorem 2.2.
(Functional van Lint-Seidel Relative Bound) If ( { f j } j = 1 n , { τ j } j = 1 n ) is functional γ-equiangular for rank d module M , then
n ( 1 d γ 2 ) d ( 1 γ 2 ) .
In particular, if
γ < 1 d ,
then
n d ( 1 γ 2 ) 1 d γ 2 .
Proof. 
Define
S f , τ : M x S f , τ x j = 1 n f j ( x ) τ j M .
Since S f , τ is diagonalizable,
n 2 = j = 1 n f j ( τ j ) 2 = ( Tra ( S f , τ ) ) 2 = k = 1 d λ k 2 d k = 1 d λ k 2 = d Tra ( S f , τ 2 ) = d j = 1 n k = 1 n f j ( τ k ) f k ( τ j ) = d j = 1 n f j ( τ j ) 2 + d j , k = 1 , j k n f j ( τ k ) f k ( τ j ) = d n 2 + d j , k = 1 , j k n f j ( τ k ) f k ( τ j ) d n 2 + d j , k = 1 , j k n | f j ( τ k ) f k ( τ j ) | = d n 2 + d ( n 2 n ) γ .
Therefore
n d n + d ( n 1 ) γ .
We are unable to derive Gerzon bound for functional equiangular lines. However, we derive following Gerzon bound for functionals.
Theorem 2.3.
(Functional Gerzon Universal Bound) Let { τ j } j = 1 n be a collection in a module M of rank d over R and { f j } j = 1 n be a collection in M * . Assume the following.
(i)
f j ( τ j ) = 1 for all 1 j n .
(ii)
There is a γ 1 such that f j ( τ k ) f k ( τ j ) = γ 2 for all 1 j , k n , j k .
Then
n d 2 .
Proof. 
For 1 j n , define
τ j f j : M x ( τ j f j ) f j ( x ) τ j M
We show that the collection { τ j f j } j = 1 n in Mor ( M ) is linearly independent over R . Let c 1 , , c n R be such that
j = 1 n c j ( τ j f j ) = 0 .
Let 1 k n be fixed. Then previous equation gives
j = 1 n c j ( τ j f j ) ( t a u k f k ) = 0 .
By taking trace we get
0 = j = 1 n c j Tra ( ( τ j f j ) ( τ k f k ) ) = j = 1 n c j f j ( τ k ) f k ( τ j ) = j = 1 , j k n c j f j ( τ k ) f k ( τ j ) + c k f k ( τ k ) 2 = j = 1 , j k n c j γ 2 + c k = γ 2 j = 1 n c j c k + c k = γ 2 j = 1 n c j + ( 1 γ 2 ) c k .
Therefore
c k = γ 2 γ 2 1 j = 1 n c j = : c , 1 k n .
Now
0 = Tra j = 1 n c j ( τ j f j ) = Tra j = 1 n c ( τ j f j ) = j = 1 n c Tra ( τ j f j ) = j = 1 n c f j ( τ j ) = c n .
Hence c = 0 . Therefore { τ j f j } j = 1 n is linearly independent. Since the rank of Mor ( M ) is d 2 we must have n d 2 . □

References

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  3. van Lint, J.H.; Seidel, J.J. Equilateral point sets in elliptic geometry. Indag. Math. 1966, 28, 335–348. [Google Scholar] [CrossRef]
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