PROPERTIES OF SOME ALGEBRAICALLY DEFINED DIGRAPHS Aleksandr Kodess, - - PowerPoint PPT Presentation

PROPERTIES OF SOME ALGEBRAICALLY DEFINED DIGRAPHS

Aleksandr Kodess, Felix Lazebnik

Department of Mathematical Sciences University of Delaware

Modern Trends of Algebraic Graph Theory

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is a digraph?

v1 v2 v3 v4 v5

Definition A digraph is a pair D = (V, A)

• f:

a set V, whose elements are called vertices or nodes a set A of ordered pairs of vertices, called arcs, directed edges, or arrows

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is an algebraic digraph D(q; f)?

Let Fq be a finite field with q elements; f : F2

q → Fq be a bivariate polynomial.

Definition An algebraic digraph, denoted D(q; f), is a digraph whose vertex set is F2

q

arc set consits of ordered pairs

• (x1, x2), (y1, y2)
• with the

relation x2 + y2 = f(x1, y1),

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is an algebraic digraph D(q; f)?

Let Fq be a finite field with q elements; f : F2

q → Fq be a bivariate polynomial.

Definition An algebraic digraph, denoted D(q; f), is a digraph whose vertex set is F2

q

arc set consits of ordered pairs

• (x1, x2), (y1, y2)
• with the

relation x2 + y2 = f(x1, y1),

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is an algebraic digraph D(q; f)?

Let Fq be a finite field with q elements; f : F2

q → Fq be a bivariate polynomial.

Definition An algebraic digraph, denoted D(q; f), is a digraph whose vertex set is F2

q

arc set consits of ordered pairs

• (x1, x2), (y1, y2)
• with the

relation x2 + y2 = f(x1, y1),

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is an algebraic digraph D(q; f)?

Let Fq be a finite field with q elements; f : F2

q → Fq be a bivariate polynomial.

Definition An algebraic digraph, denoted D(q; f), is a digraph whose vertex set is F2

q

arc set consits of ordered pairs

• (x1, x2), (y1, y2)
• with the

relation x2 + y2 = f(x1, y1),

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is an algebraic digraph D(q; f)?

Let Fq be a finite field with q elements; f : F2

q → Fq be a bivariate polynomial.

Definition An algebraic digraph, denoted D(q; f), is a digraph whose vertex set is F2

q

arc set consits of ordered pairs

• (x1, x2), (y1, y2)
• with the

relation x2 + y2 = f(x1, y1),

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Example of D(q; f)

Example of D(q; f) V(D) = F2

q

f : F2

q → Fq

There is an arc from vertex (x1, x2) to vertex (y1, y2) if and

• nly if

x2 + y2 = x2 + xy + y2 + 1 = f1(x1, y1) x2 + y2 = x2 + xy + y2 = f2(x1, y1) x2 + y2 = xy = f3(x1, y1) Easy to argue that if q is odd, then D(q; f1) ∼ = D(q; f2) ∼ = D(q; f3).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Example of D(q; f)

Example of D(q; f) V(D) = F2

q

f : F2

q → Fq

There is an arc from vertex (x1, x2) to vertex (y1, y2) if and

• nly if

x2 + y2 = x2 + xy + y2 + 1 = f1(x1, y1) x2 + y2 = x2 + xy + y2 = f2(x1, y1) x2 + y2 = xy = f3(x1, y1) Easy to argue that if q is odd, then D(q; f1) ∼ = D(q; f2) ∼ = D(q; f3).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Example of D(q; f)

Example of D(q; f) V(D) = F2

q

f : F2

q → Fq

There is an arc from vertex (x1, x2) to vertex (y1, y2) if and

• nly if

x2 + y2 = x2 + xy + y2 + 1 = f1(x1, y1) x2 + y2 = x2 + xy + y2 = f2(x1, y1) x2 + y2 = xy = f3(x1, y1) Easy to argue that if q is odd, then D(q; f1) ∼ = D(q; f2) ∼ = D(q; f3).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Example of D(q; f)

Example of D(q; f) V(D) = F2

q

f : F2

q → Fq

There is an arc from vertex (x1, x2) to vertex (y1, y2) if and

• nly if

x2 + y2 = x2 + xy + y2 + 1 = f1(x1, y1) x2 + y2 = x2 + xy + y2 = f2(x1, y1) x2 + y2 = xy = f3(x1, y1) Easy to argue that if q is odd, then D(q; f1) ∼ = D(q; f2) ∼ = D(q; f3).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Simple Observation

Simple isomorphisms Let q be an odd prime power, and f ∈ Fq[x, y]. Let f1(x, y) = f(x, y) − f(0, 0), and f ∗(x, y) = f1(x, y) − f(x, 0) − f(0, y). The following statements hold: D(q; f) ∼ = D(q; f1). If, in addition, f is a symmetric polynomial, then D(q; f) ∼ = D(q; f1) ∼ = D(q; f ∗).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Simple Observation

Simple isomorphisms Let q be an odd prime power, and f ∈ Fq[x, y]. Let f1(x, y) = f(x, y) − f(0, 0), and f ∗(x, y) = f1(x, y) − f(x, 0) − f(0, y). The following statements hold: D(q; f) ∼ = D(q; f1). If, in addition, f is a symmetric polynomial, then D(q; f) ∼ = D(q; f1) ∼ = D(q; f ∗).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Simple Observation

Simple isomorphisms Let q be an odd prime power, and f ∈ Fq[x, y]. Let f1(x, y) = f(x, y) − f(0, 0), and f ∗(x, y) = f1(x, y) − f(x, 0) − f(0, y). The following statements hold: D(q; f) ∼ = D(q; f1). If, in addition, f is a symmetric polynomial, then D(q; f) ∼ = D(q; f1) ∼ = D(q; f ∗).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

What is a monomial algebraic digraph?

Definition A monomial algebraic digraph, denoted D(q; m, n), is an algebraic digraph in which vertex set V is F2

q

there is an arc from (x1, x2) to (y1, y2) if and only if x2 + y2 = xm

1 yn 1 .

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

D(3; 1, 2)

0,1 0,2 1,2 2,2 1,1 2,1 2,0 1,0 0,0

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Motivation

Work of: Lazebnik, Woldar (2001) Lazebnik, Ustimenko (1993, 1995, 1996) Viglione (2001) Dmytrenko, Lazebnik, Viglione (2005) Bipartite undirected graph BΓn V(BΓn) = Pn ∪ Ln, both Pn and Ln are copies of Fn

q

point (p) = (p1, . . . , pn) is adjacent to line [l] = (l1, . . . , ln) if l2 + p2 = f2(p1, l1) l3 + p3 = f3(p1, l1, p2, l2) . . . ln + pn = fn(p1, l1, p2, l2, . . . , pn−1, ln−1).

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Properties of BΓn

BΓn admits neighbor-complete coloring, i.e. every color is uniquely represented among the neighbors of each vertex covering properties of BΓn. For instance, BΓn covers BΓk for n > k. embedded spectra properties. For instance, spec(BΓk) ⊆ spec(BΓn) for k < n. edge-decompostiion properties. We have BΓn decomposing Kqn,qn.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Properties of BΓn

BΓn admits neighbor-complete coloring, i.e. every color is uniquely represented among the neighbors of each vertex covering properties of BΓn. For instance, BΓn covers BΓk for n > k. embedded spectra properties. For instance, spec(BΓk) ⊆ spec(BΓn) for k < n. edge-decompostiion properties. We have BΓn decomposing Kqn,qn.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Properties of BΓn

BΓn admits neighbor-complete coloring, i.e. every color is uniquely represented among the neighbors of each vertex covering properties of BΓn. For instance, BΓn covers BΓk for n > k. embedded spectra properties. For instance, spec(BΓk) ⊆ spec(BΓn) for k < n. edge-decompostiion properties. We have BΓn decomposing Kqn,qn.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Properties of BΓn

BΓn admits neighbor-complete coloring, i.e. every color is uniquely represented among the neighbors of each vertex covering properties of BΓn. For instance, BΓn covers BΓk for n > k. embedded spectra properties. For instance, spec(BΓk) ⊆ spec(BΓn) for k < n. edge-decompostiion properties. We have BΓn decomposing Kqn,qn.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

An application of BΓn with a certain specialization

Let Cn denote the cycle of length n ≥ 3 ex(v, {C3, C4, . . . , C2k}) denote the greatest number of edges in a graph or order v which contains no subgraphs isomorphic to any C3, . . . , C2k. Theorem (Lazebnik, Ustimenko, Woldar 1995) ex(v, {C3, C4, . . . , C2k}) ≥ ckv1+

2 3k−3+ǫ ,

where ck is a positive function if k, and ǫ = 0 if k = 5 is odd, and ǫ = 1 if k is even. This lower bounds comes from BΓn with a certain choice of defining functions fi, 2 ≤ i ≤ n.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

An application of BΓn with a certain specialization

Let Cn denote the cycle of length n ≥ 3 ex(v, {C3, C4, . . . , C2k}) denote the greatest number of edges in a graph or order v which contains no subgraphs isomorphic to any C3, . . . , C2k. Theorem (Lazebnik, Ustimenko, Woldar 1995) ex(v, {C3, C4, . . . , C2k}) ≥ ckv1+

2 3k−3+ǫ ,

where ck is a positive function if k, and ǫ = 0 if k = 5 is odd, and ǫ = 1 if k is even. This lower bounds comes from BΓn with a certain choice of defining functions fi, 2 ≤ i ≤ n.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some Upper Bounds on ex(v, {C3, C4, . . . , C2k})

Erd˝

• s, Bondy–Simonovits, 1974

ex(v, {C3, C4, . . . , C2k}) ≤ 20kv1+(1/k) , for v sufficiently large. Verstraëte, 2000 ex(v, {C3, C4, . . . , C2k}) ≤ 8(k − 1)v1+(1/k), for v sufficiently large. Pikhurko, 2012 ex(v, {C3, C4, . . . , C2k}) ≤ (k − 1)v1+(1/k) + O(v) . Bukh, Jiang, 2014 ex(v, {C3, C4, . . . , C2k}) ≤ 80

• k log k · v1+(1/k) + O(v) .

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some Upper Bounds on ex(v, {C3, C4, . . . , C2k})

Erd˝

• s, Bondy–Simonovits, 1974

ex(v, {C3, C4, . . . , C2k}) ≤ 20kv1+(1/k) , for v sufficiently large. Verstraëte, 2000 ex(v, {C3, C4, . . . , C2k}) ≤ 8(k − 1)v1+(1/k), for v sufficiently large. Pikhurko, 2012 ex(v, {C3, C4, . . . , C2k}) ≤ (k − 1)v1+(1/k) + O(v) . Bukh, Jiang, 2014 ex(v, {C3, C4, . . . , C2k}) ≤ 80

• k log k · v1+(1/k) + O(v) .

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some Upper Bounds on ex(v, {C3, C4, . . . , C2k})

Erd˝

• s, Bondy–Simonovits, 1974

ex(v, {C3, C4, . . . , C2k}) ≤ 20kv1+(1/k) , for v sufficiently large. Verstraëte, 2000 ex(v, {C3, C4, . . . , C2k}) ≤ 8(k − 1)v1+(1/k), for v sufficiently large. Pikhurko, 2012 ex(v, {C3, C4, . . . , C2k}) ≤ (k − 1)v1+(1/k) + O(v) . Bukh, Jiang, 2014 ex(v, {C3, C4, . . . , C2k}) ≤ 80

• k log k · v1+(1/k) + O(v) .

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some Upper Bounds on ex(v, {C3, C4, . . . , C2k})

Erd˝

• s, Bondy–Simonovits, 1974

ex(v, {C3, C4, . . . , C2k}) ≤ 20kv1+(1/k) , for v sufficiently large. Verstraëte, 2000 ex(v, {C3, C4, . . . , C2k}) ≤ 8(k − 1)v1+(1/k), for v sufficiently large. Pikhurko, 2012 ex(v, {C3, C4, . . . , C2k}) ≤ (k − 1)v1+(1/k) + O(v) . Bukh, Jiang, 2014 ex(v, {C3, C4, . . . , C2k}) ≤ 80

• k log k · v1+(1/k) + O(v) .

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Questions studied

Strong Connectivity of D(q; f) Classification into Isomorphism Classes

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Questions studied

Strong Connectivity of D(q; f) Classification into Isomorphism Classes

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Strong Connectivity

Definition A directed graph is called strongly connected if it contains a directed path from u to v

and a directed path from v to

u for every pair of distinct vertices u, v.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

D(3; 1, 2) revisited

0,1 0,2 1,2 2,2 1,1 2,1 2,0 1,0 0,0

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

(Strong) Connectivity of Algebraic Digraphs

Definition (Linked alternating sums) For any function f : F2

q → Fq, any positive integer k ≥ 2 and any

k-tuple (x1, . . . , xk) ∈ Fk

q we call the sum

f(x1, x2) − f(x2, x3) + · · · + (−1)k−1f(xk−1, xk) a linked alternating sum of f of length k − 1. Let LSk(f) be the set of all possible linked alternating sums of f of length k − 1, and let LS(f) :=

• k≥2

LSk(f). We call LS(f) the set of linked alternating sums of f.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some additional notation

Im(f) — image of f : F2

q → Fq

Im(f) := {(f(x, y), : x, y ∈ Fq}. Affine subspaces H0 and H1 For any vectors α1, . . . , αd ∈ Fq, let H0 and H1 be defined as H0 = H0(α1, . . . , αd) = d

• i=1

siαi : all si in Fp,

d

• i=1

si = 0

• ,

H1 = H1(α1, . . . , αd) = d

• i=1

siαi : all si in Fp,

d

• i=1

si = 1

• .

Then H0 is a subspace of Fq, and H1 is an affine subspace of Fq.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Property of the set LS(f)

Trivially, LS(f) ⊆ Im(f) Either LS(f) = Im(f), span of image of f in Fq, OR LS(f) = H0 ∪ H1, union of two affine planes in Fq This leads to the following

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Property of the set LS(f)

Trivially, LS(f) ⊆ Im(f) Either LS(f) = Im(f), span of image of f in Fq, OR LS(f) = H0 ∪ H1, union of two affine planes in Fq This leads to the following

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Property of the set LS(f)

Trivially, LS(f) ⊆ Im(f) Either LS(f) = Im(f), span of image of f in Fq, OR LS(f) = H0 ∪ H1, union of two affine planes in Fq This leads to the following

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Strong Connectivity of D(p; f)

Theorem Let p be an odd prime, and let f : Fp × Fp → Fp. The following statements hold. (i) If f is not a symmetric function, i.e. there exist x1, x2 ∈ Fp for which f(x1, x2) = f(x2, x1), then D(p; f) is strong. (ii) If f is a symmetric function, let f ∗(x, y) = f(x, y) − f(x, 0) − f(0, y) − f(0, 0). If f ∗ is a nonzero polynomial, then D(p; f) is strong. If f ∗ is the zero polynomial, then D(p; f) has (p + 1)/2 strong

• components. The component containing the vertex (0, 0) is

isomorphic to the complete looped digraph − → K p. All other strong components are isomorphic to the complete bipartite digraph − → K p,p with no loops.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Problem What are necessary and sufficient conditions for (m1, n1) and (m2, n2) in order for the graphs D(q; m1, n1) and D(q; m2, n2) to be isomorphic? Idea to Solve it — find a strong easily computable digraph invariant number of cycles number of ’triangles’ with various orientations of arcs number of subdigraphs on 4 vertices number of ’looped’ paths

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Problem What are necessary and sufficient conditions for (m1, n1) and (m2, n2) in order for the graphs D(q; m1, n1) and D(q; m2, n2) to be isomorphic? Idea to Solve it — find a strong easily computable digraph invariant number of cycles number of ’triangles’ with various orientations of arcs number of subdigraphs on 4 vertices number of ’looped’ paths

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Problem What are necessary and sufficient conditions for (m1, n1) and (m2, n2) in order for the graphs D(q; m1, n1) and D(q; m2, n2) to be isomorphic? Idea to Solve it — find a strong easily computable digraph invariant number of cycles number of ’triangles’ with various orientations of arcs number of subdigraphs on 4 vertices number of ’looped’ paths

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Problem What are necessary and sufficient conditions for (m1, n1) and (m2, n2) in order for the graphs D(q; m1, n1) and D(q; m2, n2) to be isomorphic? Idea to Solve it — find a strong easily computable digraph invariant number of cycles number of ’triangles’ with various orientations of arcs number of subdigraphs on 4 vertices number of ’looped’ paths

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Problem What are necessary and sufficient conditions for (m1, n1) and (m2, n2) in order for the graphs D(q; m1, n1) and D(q; m2, n2) to be isomorphic? Idea to Solve it — find a strong easily computable digraph invariant number of cycles number of ’triangles’ with various orientations of arcs number of subdigraphs on 4 vertices number of ’looped’ paths

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Problem What are necessary and sufficient conditions for (m1, n1) and (m2, n2) in order for the graphs D(q; m1, n1) and D(q; m2, n2) to be isomorphic? Idea to Solve it — find a strong easily computable digraph invariant number of cycles number of ’triangles’ with various orientations of arcs number of subdigraphs on 4 vertices number of ’looped’ paths

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Dmytrenko, Lazebnik,Viglione, 2005 — undirected bipartite version This worked for undirected bipartite monomial graphs. They found the following simple graph invariants: number of C4 for all sufficiently large q’s number of Ks,t for all q’s. as multisets {m1, n1} = {m2, n2}, where x = gcd(q − 1, x) Directed version We have examples of non-isomorphic graphs that have equal numbers of 3, 4, 5, 6-cycles and a large collection of other subgraphs. Example: Digraphs D(169; 1, 5) and D(169; 1, 101) have equal number of directed cycles Ck, for 3 ≤ k ≤ 13, but they are not isomorphic.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Dmytrenko, Lazebnik,Viglione, 2005 — undirected bipartite version This worked for undirected bipartite monomial graphs. They found the following simple graph invariants: number of C4 for all sufficiently large q’s number of Ks,t for all q’s. as multisets {m1, n1} = {m2, n2}, where x = gcd(q − 1, x) Directed version We have examples of non-isomorphic graphs that have equal numbers of 3, 4, 5, 6-cycles and a large collection of other subgraphs. Example: Digraphs D(169; 1, 5) and D(169; 1, 101) have equal number of directed cycles Ck, for 3 ≤ k ≤ 13, but they are not isomorphic.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Isomorphism Classification Problem

Dmytrenko, Lazebnik,Viglione, 2005 — undirected bipartite version This worked for undirected bipartite monomial graphs. They found the following simple graph invariants: number of C4 for all sufficiently large q’s number of Ks,t for all q’s. as multisets {m1, n1} = {m2, n2}, where x = gcd(q − 1, x) Directed version We have examples of non-isomorphic graphs that have equal numbers of 3, 4, 5, 6-cycles and a large collection of other subgraphs. Example: Digraphs D(169; 1, 5) and D(169; 1, 101) have equal number of directed cycles Ck, for 3 ≤ k ≤ 13, but they are not isomorphic.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Open Conjecture

Conjecture Let q be a prime power, 1 ≤ m, n, m′, n′ ≤ q − 1 integers. Then D(q; m1, n1) ∼ = D(q; m2, n2) if and only if there exists k coprime with q − 1 such that m2 ≡ km1 mod q − 1, n2 ≡ kn1 mod q − 1. Comments easy to prove sufficiency cannot prove necessity checked with sage for orders of up to 1000

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Open Conjecture

Conjecture Let q be a prime power, 1 ≤ m, n, m′, n′ ≤ q − 1 integers. Then D(q; m1, n1) ∼ = D(q; m2, n2) if and only if there exists k coprime with q − 1 such that m2 ≡ km1 mod q − 1, n2 ≡ kn1 mod q − 1. Comments easy to prove sufficiency cannot prove necessity checked with sage for orders of up to 1000

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Open Conjecture

Conjecture Let q be a prime power, 1 ≤ m, n, m′, n′ ≤ q − 1 integers. Then D(q; m1, n1) ∼ = D(q; m2, n2) if and only if there exists k coprime with q − 1 such that m2 ≡ km1 mod q − 1, n2 ≡ kn1 mod q − 1. Comments easy to prove sufficiency cannot prove necessity checked with sage for orders of up to 1000

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Open Conjecture

Conjecture Let q be a prime power, 1 ≤ m, n, m′, n′ ≤ q − 1 integers. Then D(q; m1, n1) ∼ = D(q; m2, n2) if and only if there exists k coprime with q − 1 such that m2 ≡ km1 mod q − 1, n2 ≡ kn1 mod q − 1. Comments easy to prove sufficiency cannot prove necessity checked with sage for orders of up to 1000

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some comments on Conjecture

Using reduction to bipartites graphs D(q; m1, n1) ∼ = D(q; m2, n2) implies {m1, n1} = {m2, n2} If, say, m1 = m2 and n1 = n2, then m2 ≡ k1m1 mod q − 1, n2 ≡ k2n1 mod q − 1. We cannot conclude k1 = k2. However, we have no examples of isomorphic digraphs with k1 = k2.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some comments on Conjecture

Using reduction to bipartites graphs D(q; m1, n1) ∼ = D(q; m2, n2) implies {m1, n1} = {m2, n2} If, say, m1 = m2 and n1 = n2, then m2 ≡ k1m1 mod q − 1, n2 ≡ k2n1 mod q − 1. We cannot conclude k1 = k2. However, we have no examples of isomorphic digraphs with k1 = k2.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Some comments on Conjecture

Using reduction to bipartites graphs D(q; m1, n1) ∼ = D(q; m2, n2) implies {m1, n1} = {m2, n2} If, say, m1 = m2 and n1 = n2, then m2 ≡ k1m1 mod q − 1, n2 ≡ k2n1 mod q − 1. We cannot conclude k1 = k2. However, we have no examples of isomorphic digraphs with k1 = k2.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs

Conjecture proved under restrictions for p > 2

Theorem Let q = p > 2 be prime and 1 ≤ m1, n1, m2, n2 ≤ p − 1 be

• integers. Suppose that D1 = D(p; m1, n1) ∼

= D2 = D(p; m2, n2). Assume moreover that there exists an isomorphism φ: V(D1) → V(D2) of the form φ: (x, y) → (f(x, y), g(x, y)), in which f depends on x only. If m1 = n1 or m2 = n2, then there exists an integer k, coprime with p − 1, such that m2 ≡ km1 mod p − 1, n2 ≡ kn1 mod p − 1.

Aleksandr Kodess, Felix Lazebnik Algebraic Digraphs