Euclidean Jordan algebras and some new inequalities over the

parameters of a strongly regular graph

LUIS VIEIRA

Faculty of Engineering of University of Porto

Department of Civil Engineering

Street D Roberto Frias, 4200 46 Porto

PORTUGAL

Abstract: Let’s consider a primitive strongly regular graph Gand it’s adjacency matrix A. Next we consider the

Euclidean subalgebra Aof the Euclidean Jordan algebra of real symmetric matrices of order n, with the Jordan

product and with the inner product of two matrices as being the usual trace of two matrices. Finally, we make a

spectral analysis of an Hadamard series of an element of Ato establish some new conditions over the spectrum

and the parameters of the primitive strongly regular graph G.

Key–Words: Typing manuscripts, L

Received: August 11, 2021. Revised: July 13, 2022. Accepted: August 14, 2022. Published: September 20, 2022.

1 Introduction

For a precise description of Euclidean Jordan algebras

one must cite the monograph book, Analysis on Sym-

metric cones, of Jacques Faraut and Adam Kor´

anyi,

see [1].

The Euclidean jordan algebras become a good

theoretical environment to develop may applications

in many branches of research of mathematics, see for

instance [2–12] but our main goal is recurring to this

theory to develop some properties over the spectrum

of some discrete structures like the strongly regular

graphs and the association schemes, see for instance

[13–18].

This paper is organized as follows. In the section

2 we present some notes about Euclidean Jordan alge-

bras, namely the more relevant notions about ﬁnite di-

mensional real Euclidean Jordan algebras. In the fol-

lowing section we present some notes about strongly

regular graphs necessary for a clear exposition of this

paper. Finally, in the last section we present two new

inequalities over the parameters and the spectrum of

a primitive strongly regular graph in the environment

of Euclidean Jordan algebras. On one new inequality

we establish a new relation between the parameters

and one eigenvalue of a strongly regular graph, see

inequality (29), and in the other new inequality we es-

tablished a relation between only the parameters of a

regular graph, see the inequality (30).

2 Some Notes on Euclidean Jordan

Algebras

In this section we present the more relevant deﬁnitions

and results of the theory of Euclidean Jordan algebras

relevant for this paper.

For good monographs about Jordan algebras we

must cite the Book, “A taste of Jordan Algebras” writ-

ten by Kevin McCrimmon, see [19], and “Statistical

Applications of Jordan Algebras” written by James.

D. Malley, see [20].

A real ﬁnite dimensional Jordan algebra Ais an

algebra with an operation of multiplication of vectors

?such that for any of its elements xand ywe have:

x?y =y ? x,

x2??(x?y) = x ? (x2?? y),

where x2?=x ? x. And for any natural number kthe

powers of order k, are deﬁned in the following way:

x0?=e, x1?=x,

xk? =x?x(k−1)?, k ≥2.

An element eof a real ﬁnite dimensional Eu-

clidean Jordan algebra Ais an unit element of Aif

e? x =x ? e=xfor any element xin A.

Example 1 Let’s consider the ﬁnite dimensional al-

gebra Aover Rof real symmetric matrices of order n

with the usual operations of addiction of matrices and

of multiplication of a matrix by a real number. Then,

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considering the operation ?, instead of the usual op-

eration of multiplication of matrices, deﬁned for any

xand yin Aby x?y =xy+yx

2,then Ais a Jordan

algebra. Indeed, let xand ybe elements of A, then

we have the following calculations:

x?y =xy +yx

2=yx +xy

2=y ? x.

Firstly, we must say that for any element xof Awe

have x2?=x2where x2represent the usual square

of a symmetric matrix of order n. Indeed, x2?=

xx+xx

2=x2+x2

2=x2.

Next, we will show that x2??(x?y) = x?(x2??y).

Since, we have:

x2??(x?y) = x2?(x?y)+(x?y)x2?

=x2xy+yx

2+xy+yx

2x2

=x2(xy +yx)+(xy +yx)x2

=x2xy +x2yx +xyx2+yxx2

=x3y+x2yx +xyx2+yx3

and since

x ? (x2?? y) = x(x2?? y)+(x2?? y)x

xx2y+yx2

2+x2y+yx2

2x

=x(x2y+yx2)+(x2y+yx2)x

=x3y+xyx2+x2yx +yx3

=x3y+x2yx +xyx2+yx3

So, we have proved that x2??(x?y) = x?(x2??x).

for any xand yof A.And, therefore we conclude that

Ais a Jordan real. We will denote sometimes this Eu-

clidean Jordan algebra Aby the notation Sym(n, R).

A real ﬁnite dimensional Euclidean Jordan algebra is

a real ﬁnite dimensional Jordan algebra equipped with

the multiplication of vectors ?, and provided with an

inner product •|• such that for any three of it’s ele-

ments x,y, and zthe equality (1) is veriﬁed.

(x?y)|z=y|(x?z).(1)

Example 2 Let’s consider the Jordan algebra A=

Sym(n, R),equipped with the vector operation ?such

that x?y =xy+yx

2for any xand yof A,and provided

with the inner product •|• such that x|y=trace(x?y)

for any elements xand yof A.Then Ais an Euclidean

Jordan algebra, before showing that we will prove that

trace(x?y) = trace(xy)for any two of it’s elements

xand y.

Indeed, we have

trace(x?y) = trace(xy +yx

2trace(xy +yx)

2(trace(xy) + trace(yx))

2(trace(xy) + trace(xy))

2(2trace(xy))

=trace(xy).

Now, we consider a natural number k, and x, y and z

elements of A.The powers of order kof the element

x,xk? are deﬁned in following way.

x0?=e,

x1?=x,

xk? =x?x(k−1)?, k ≥2.

Next, we will show that (x?y)|z=y|(x?z).So, we

have the following calculations.

(x?y)|z=trace (x?y)z+z(x?y)

2

=trace((x?y)z)

=trace xy +yx

2z

=trace (xy)z+ (yx)z

2

=trace x(yz)

2+trace (yx)z

2

=trace (yz)x

2+trace y(xz)

2

=trace y(zx)

2+trace (xz)y

2

=trace y(zx)

2+trace y(xz)

2

=trace y(xz)

2+trace y(zx)

2

=trace yxz +zx

2

=trace yxz+zx

2+xz+zx

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=trace y ? xz +zx

2

=trace (y ? (x?z))

=y|(x?z)

The unit of this Euclidean Euclidean Jordan algebra

is the identity matrix eof order n. Indeed, we have:

e? x =ex+xe

2=x+x

2=2x

2=x=x ? e.

Let Abe a n dimensional real Euclidean Jordan

algebra with the vector product ?, the inner product

•|• and with the unit e.Then Ais a power associative

algebra, this is for any of it’s element xthe algebra

spanned by xand eis associative.

The rank of an element ain Ais the least natural

number ksuch that {e, a1?, . . . , ak?}is a linearly de-

pendent set and we write rank(a) = k. Since for any

a∈ A we have rank(x)≤n, then we deﬁne the rank

of Aas being the natural number r=rank(A) =

max{rank(a) : a∈ A}.An element aof Ais reg-

ular if rank(a) = r, Let xbe a regular element of

Aand r=rank(x).Then, there exist real scalars

β1(x), β2(x), . . . , βr−1(x)and βr(x)such that

xr? −β1(x)xr−1?+· · · + (−1)rβr(x)x0?= 0,(2)

where 0is the null vector of A.Taking into account

(2) we conclude that the polynomial

p(x, λ) = λr−β1(x)λr−1+· · · + (−1)rβr(x)(3)

is the minimal polynomial of x. When xis not regular

the minimal polynomial of xhas a degree less than

r. The roots of the minimal polynomial of xare the

eigenvalues of x.

An element x∈ A is an idempotent if x2?=x.

Two idempotent aand bare orthogonal if a ? b = 0.

The set {g1, g2, . . . , gl}is a complete system of or-

thogonal idempotent if g2?

i=gi, for i= 1, . . . , l,gi?

gj= 0, if i6=jand 1≤i, j ≤l, and Pl

i=1 gi=e.

An idempotent is primitive if is a nonzero idempotent

of Aand cannot be written as a sum of two nonzero or-

thogonal idempotent. We say that {g1, g2, . . . , gk}is a

Jordan frame if {g1, g2, . . . , gk}is a complete system

of orthogonal idempotent such that each idempotent

is primitive.

Example 3 Let’s consider the Euclidean Jordan al-

gebra A=Sym(n, R)with the Jordan product ?such

that x ? y =xy+yx

2,∀x, y ∈ A and the inner product

•|• such that for any xand yelements of Awe have

x|y=trace(x?y).Let’s consider iand jbe natu-

ral numbers such that 1≤i, j ≤n, the matrices Eij

of Asuch that the only non null entry of Eij is the

entry ij and it’s value is 1.Then, the set of matrices

B1={E11, E22,· · · , Enn}is a Jordan frame of A

and the set of matrices B2={E11 +E22,Pn

i=3 Eii}

is a complete system of orthogonal idempotent of A.

Theorem 1 ( [1], p. 43). Let Vbe a real Eu-

clidean Jordan algebra. Then for xin Vthere exist

unique real numbers λ1, λ2, . . . , λk,all distinct, and

a unique complete system of orthogonal idempotent

{g1, g2, . . . , gk}such that

x=λ1g1+λ2g2+· · · +λkgk.(4)

The numbers λj’s of (4) are the eigenvalues of xand

the decomposition (4) is the ﬁrst spectral decomposi-

tion of x.

Theorem 2 ( [1], p. 44). Let Vbe a real Euclidean

Jordan algebra with rank(V) = r. Then for each x

in Vthere exists a Jordan frame {g1, g2,· · · , gr}and

real numbers λ1,· · · , λr−1and λrsuch that

x=λ1g1+λ2g2+· · · +λrgr.(5)

The decomposition (5) is called the second spectral

decomposition of x.

3 Some results about strongly regu-

lar graphs

Along this paper we consider only non empty, simple

and non complete graphs. By simple graphs we mean

graphs without loops and parallel edges. Strongly reg-

ular graphs were ﬁrstly introduced by R. C. Bose in

the paper [21].

One says that Gis the complement of the graph

Gif it has the same set o vertices as Gand if any of its

two distinct vertices are adjacent vertices in Gif and

only if are non adjacent vertices in G.

A non null and non complete graph G, whose

order is greater or equal than 3is called a strongly

regular graph with parameters (n, k;λ, µ)if Gis

k−regular graph such that any pair of adjacent ver-

tices have λcommon neighbor vertices and any pair

of non adjacent vertices have µcommon neighbor ver-

tices.

If Gis a (n, k;λ, µ)strongly regular graph then

the complement graph of G, G is a (n, n −k−1; n−

2k+µ−2, n −2k+λ)strongly regular graph.

Let’s consider a graph G. We call a set of edges

and vertices a walk of vertices in Gto every sequence

v0e1v1e2. . . el−1vl−1elvlsuch that v1, v2, . . . , vl−1

and vlare vertices and e1, e2...,el−1, elare edges of

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Gand each edge eihas extreme vertices vi−1and ver-

tice vifor i= 1,· · · , l. The walk is closed if v0=vl

and is open otherwise. One says that a walk in Gis a

path if all the vertices vis are distinct with the excep-

tion of the initial vertex v0and the ﬁnal vertex vl.

One says that a path is a closed path or a cycle if

the initial vertex and ﬁnal vertex of the path are the

same.

A graph Gis connected if for any pair of distinct

vertices exists a path that joins them. A (n, k;λ, µ)

strongly regular graph Gis primitive if and only if G

and Gare connected. Otherwise one says that Gis

disconnected.

A primitive strongly regular graph (n, k;λ, µ)is a

non primitive strongly regular graph if and only if µ=

kor µ= 0.In the following text we only consider

primitive strongly regular graphs.

From now we only consider primitive strongly

regular graphs.

Let Gbe a (n, k;λ, µ)strongly regular graph.

The adjacency matrix of G,A= [aij ], is a binary ma-

trix of order nsuch that aij = 1, if the vertex iis adja-

cent to jand 0otherwise. The adjacency matrix of G

satisﬁes the equation A2=kIn+λA+µ(Jn−A−In),

where Jnis the all ones matrix of order n. It is

well known (see, for instance, [22]) that the eigen-

values of Aare k,θand τ, where θand τare given

by θ= (λ−µ+p(λ−µ)2+ 4(k−µ))/2and

τ= (λ−µ−p(λ−µ)2+ 4(k−µ))/2,(see [22]).

One deﬁnes the eigenvalues of Gas being the eigen-

values of A. And, we also know that the multiplicities

fθand fτof the eigenvalues θof τare given respec-

tively by the relations (6) and (7).

fθ=1

2n−1 + 2k+ (n−1)(λ−µ)

τ−θ,(6)

fτ=1

2n−1−2k+ (n−1)(λ−µ)

τ−θ.(7)

Since fθand fτare integer positive numbers, then the

conditions present on (8) and on (9) are known as in-

tegrability conditions

fθ∈N,(8)

fτ∈N.(9)

In the context of strongly regular graphs one

of the problems to analyse is to know if given

the real numbers n, k, λ and µif there exists a

(n, k;λ, µ)strongly regular graph. The more refer-

enced admissibility conditions for the existence of a

(n, k;λ, µ)strongly regular graph are the inequalities

(10),(11),(12),(12),(13), and (14).

k(k−1−a)=(n−k−1)µ, (10)

(τ+ 1)(k+τ+ 2θτ)≤(k+τ)(θ+ 1)2,(11)

(θ+ 1)(k+θ+ 2θτ)≤(k+θ)(τ+ 1)2,(12)

n≤1

2fθ(fθ+ 3),(13)

n≤1

2fτ(fτ+ 3),(14)

The inequalities (11) and (12) are known as the

Krein conditions of the strongly regular graph G, and

the inequalities (13) and (14) are known as the abso-

lute bounds. In the next section we establish some

new inequalities over the spectrum of a strongly regu-

lar graph and it’s parameters, over certain conditions,

but relating only the parameters of a strongly regular

graph or only one eigenvalue of the strongly regular

graph and it’s parameters.

4 Some new inequalities over the

parameters of a strongly regular

graph

Let’s Gbe a primitive (n, k;λ, µ)strongly regular

such that 0< µ < k −1, k < n

2, λ > µ, and a posi-

tive real number such that λk +|τ|3+ > (k−µ) +

(λ−µ)λ+µk and such that λK+|τ|3+ > (λ−µ)µ+

µk, A it’s adjacency matrix and ﬁnally let’s consider

the 3-dimension Euclidean subalgebra Aof rank three

of the Euclidean Jordan algebra Sym(n, R)spanned

by Inand the natural powers of A. Next, let’s consider

the unique Jordan frame B={G1, G2, G3}where we

have: G1=1

nIn+1

nA+1

n(Jn−A−In) = Jn

G2=|τ|n+τ−k

n(θ−τ)In+n+τ−k

n(θ−τ)A+τ−k

n(θ−τ)(Jn−A−In,

G3=θn+k−θ

n(θ−τ)In+−n+k−θ

n(θ−τ)A+k−θ

n(θ−τ)(Jn−A−In).

Now, we know that A2=kIn+λA+µ(Jn−A−In)

where Jnis the matrix where each of it’s entries is the

real number 1.And so after some algebraic manipula-

tion we conclude that (15) is veriﬁed.

A2= (k−µ)In+ (λ−µ)A+µJn(15)

and therefore we conclude (16)

A3= (k−µ)A+ (λ−µ)A2+µkJn).(16)

And, noting that

A2=kIn+λA +µ(Jn−A−In

we deduce the equality A3= (k−µ)A+(λ−µ)(kIn+

λA +µ(Jn−A−In)) +µkJn

Hence, we can write the inequality (17)

A3+|τ|3In= (λk +|τ|3)In+

+ ((k−µ)+(λ−µ)λ+µk)A+

+ ((λ−µ)µ+µk)(Jn−A−In).(17)

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Now, since λk+|τ|3+ > (k−µ)+(λ−µ)λ+µk

and λK +|τ|3+ > (λ−µ)µ+µk then, let’s consider

the Hadammard series S=P+∞

k=0 A3+|τ|3In

λk+|τ|3+k◦.

Considering the notation

α1=λk +|τ|3,

α2= (k−µ)+(λ−µ)λ+µk,

α3= (λ−µ)µ+µk,

α0=λk +|τ|3+

we can write S=1

1−α1

α0

In+1

1−α2

α0

A+

1−α3

α0

(Jn−A−In).Next, let’s consider the element

G3◦Sof A.So we conclude that (18) is veriﬁed.

G3◦S=θn +k−θ

n(θ−τ)

1−α1

α0

In+

+−n+k−θ

n(θ−τ)

1−α2

α0

A+

+k−θ

n(θ−τ)

1−α3

α0

(Jn−A−In)(18)

Now, we consider the spectral decomposition q3◦S=

q31G1+q32G2+q33G3.We deduce that

q31 =θn +k−θ

n(θ−τ)

1−α1

α0

+−n+k−θ

n(θ−τ)

1−α2

α0

+k−θ

n(θ−τ)

1−α3

α0

(n−k−1).(19)

Since θn+k−θ

n(θ−τ)+−n+k−θ

n(θ−τ)k+k−θ

n(θ−τ)(n−k−1) = 0,

from (19) we conclude that:

q31 =θn +k−θ

n(θ−τ) 1

1−α1

α0

−1

1−α3

α0!+

+−n+k−θ

n(θ−τ) 1

1−α2

α0

−1

1−α3

α0!k

Now, from a spectral analysis of S3◦Swe conclude

that q3i≥0,for i= 1,· · · ,3,and therefore since

q31 ≥0so we can write inequality (20).

θn +k−θ

n−k+θ 1

1−α1

α0

−1

1−α3

α0!≥

≥ 1

1−α2

α0

−1

1−α3

α0!k. (20)

From an algebraic manipulation of (20) we deduce the

inequality (21).

θn +k−θ

n−k+θ1

α0−α1

−1

α0−α3≥

≥1

α0−α2

−1

α0−α3k. (21)

So, from (21) we deduce the inequality (22).

θn +k−θ

n−k+θα1−α3

(α0−α1)(α0−α3)

≥α2−α3

(α0−α2)(α0−α3)k.

(22)

By, rewriting the inequality (22) we obtain the in-

equality (23).

θn +k−θ

n−k+θα1−α3

α0−α1≥

≥α2−α3

α0−α2k. (23)

After, some calculations from (23), and noting that

α0−α1=and α0−α2= (λ−µ)(k−λ)−(k−

µ) + |τ|3+, considering α4= (λ−µ)(k−λ)−

(k−µ) + |τ|3, we deduce (24).

θn +k−θ

n−k+θ(α1−α3)≥

α4+(α2−α3)k. (24)

Hence, we conclude that

θn +k−θ

n−k+θ(α1−α3)≥(α2−α3)k. (25)

But, since α2−α3= (k−µ)+(λ−µ)2and α1−α3=

(λ−µ)(k−µ) + |τ|3then from (25) we deduce (26).

θn +k−θ

n−k+θ((λ−µ)(k−µ) + |τ|3)

≥((k−µ)+(λ−µ)2)k. (26)

Next, we suppose that k < n

2,then in this case we

conclude that θn +k−θ≤2θ+1

2and 1

n−k+θ≤2

nand

therefore from (26) we conclude that the inequality

(27) is veriﬁed.

(2θ+ 1)((λ−µ)(k−µ) + |τ|3)

≥((k−µ)+(λ−µ)2)k. (27)

Next, since |τ|<k−µ

λ−µwe obtain (28).

(2θ+ 1)((λ−µ)4(k−µ)+(k−µ)3)

≥((k−µ)(λ−µ)3+ (λ−µ)5)k. (28)

Then, we have establish the Theorem 3.

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Theorem 3 Let Gbe a primitive (n, k;λ, µ)strongly

regular graph such that 0< µ < k −1, λ > µ, k < n

then we have the inequality (29).

(2θ+ 1)((λ−µ)4(k−µ)+(k−µ)3)

≥(k−µ)(λ−µ)3+ (λ−µ)5.(29)

Making a similar spectral analysis of the element

G3◦Sand analyzing the eigenvalue q33 of G3◦S

we deduce the inequality (30) presented on Theorem

Theorem 4 Let Gbe a primitive (n, k;λ, µ)strongly

regular graph such that 0< µ < k −1, λ > µ, k < n

then we have the inequality (30).

(((λ−µ)4(k−µ)+(k−µ)3))

≥1

3((k−µ)(λ−µ)3+ (λ−µ)5).(30)

5 Conclusion

The research of this paper allow us to establish some

new inequalities over the parameters of a primitive

strongly regular graph and it’s spectrum, but estab-

lishing relations over only the parameters of a primi-

tive strongly regular graph or over the parameters of a

primitive strongly regular and one of it’s eigenvalues.

In future research we will establish relations but relax-

ing the conditions over the parameters of a primitive

strongly regular graph. To achieve that we will use

spectral analysis of the Hadamard power series of the

power of order n of the adjacency matrix of a primitive

strongly regular graph with an asymptotic algebraic

approach or with others spectral analysis methods.

Acknowledgements: Luis Vieira was partially sup-

ported by CMUP (UID/MAT/00144/2019), which is

funded by FCT with national (MCTES) and European

structural funds through the programs FEDER, under

the partnership agreement PT2020. .

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