On isometric sequences of colored spaces Mitsugu Hirasaka (jont work - - PowerPoint PPT Presentation

On isometric sequences of colored spaces

Mitsugu Hirasaka (jont work with Masashi Shinohara)

Pusan National University

Combinatorics Seminar at Shanghai Jiao Tong University December 24, 2017.

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Introduction

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Introduction

I like to use a white board.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances).

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle. That’s all.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle. That’s all. Though I am conservative, let me add one new relation, say β to the existing distance, say α.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle. That’s all. Though I am conservative, let me add one new relation, say β to the existing distance, say α. The distances from the fourth points to the former three points are βββ, ββα or βαα.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle. That’s all. Though I am conservative, let me add one new relation, say β to the existing distance, say α. The distances from the fourth points to the former three points are βββ, ββα or βαα. We can draw the following pictures.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle. That’s all. Though I am conservative, let me add one new relation, say β to the existing distance, say α. The distances from the fourth points to the former three points are βββ, ββα or βαα. We can draw the following pictures. Moreover, we can draw other kinds of four points without regular triangles.

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Introduction

I like to use a white board. Because I feel something created from the vacant white space. Let me put one point in this Euclidean plane. Let me make one of his friends. Since I am conservative, I want to avoid to make new relations (distances). We can put another point forming a regular triangle. That’s all. Though I am conservative, let me add one new relation, say β to the existing distance, say α. The distances from the fourth points to the former three points are βββ, ββα or βαα. We can draw the following pictures. Moreover, we can draw other kinds of four points without regular triangles. Can we add the fifth point to these pictures in which only α and β appear?

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Distances and Triangles

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Distances and Triangles

Do you remember LLL, LAL and ALA?

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Distances and Triangles

Do you remember LLL, LAL and ALA? These are properties for two triangles to be congruent.

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Distances and Triangles

Do you remember LLL, LAL and ALA? These are properties for two triangles to be congruent. How many congruence classes of triangles are there?

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Distances and Triangles

Do you remember LLL, LAL and ALA? These are properties for two triangles to be congruent. How many congruence classes of triangles are there? For X ⊆ Rn we set A(X) := {d(x, y) | x, y ∈ X, x = y}, and

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Distances and Triangles

Do you remember LLL, LAL and ALA? These are properties for two triangles to be congruent. How many congruence classes of triangles are there? For X ⊆ Rn we set A(X) := {d(x, y) | x, y ∈ X, x = y}, and we denote by T(X) the set of congruence classes of triangles in X.

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Distances and Triangles

Do you remember LLL, LAL and ALA? These are properties for two triangles to be congruent. How many congruence classes of triangles are there? For X ⊆ Rn we set A(X) := {d(x, y) | x, y ∈ X, x = y}, and we denote by T(X) the set of congruence classes of triangles in X. We observe that |A(X)| ≤ |T(X)| if |X| ≥ 5.

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Distances and Triangles

Do you remember LLL, LAL and ALA? These are properties for two triangles to be congruent. How many congruence classes of triangles are there? For X ⊆ Rn we set A(X) := {d(x, y) | x, y ∈ X, x = y}, and we denote by T(X) the set of congruence classes of triangles in X. We observe that |A(X)| ≤ |T(X)| if |X| ≥ 5. Here we aim to prove it and show what happens if the equality holds.

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Examples

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Examples

(i) It is easy to find X ⊆ Rn with |A(X)| = |X|

2

• if n is enough large,

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Examples

(i) It is easy to find X ⊆ Rn with |A(X)| = |X|

2

• if n is enough large,

so that |T(X)| = |X|

3

• .

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Examples

(i) It is easy to find X ⊆ Rn with |A(X)| = |X|

2

• if n is enough large,

so that |T(X)| = |X|

3

• .

(ii) If X ⊆ R3 forms an octahedron, then |A(X)| = 2 and |T(X)| = 2.

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Examples

(i) It is easy to find X ⊆ Rn with |A(X)| = |X|

2

• if n is enough large,

so that |T(X)| = |X|

3

• .

(ii) If X ⊆ R3 forms an octahedron, then |A(X)| = 2 and |T(X)| = 2. (iii) If X ⊆ R2 forms a regular hexagon, then |A(X)| = 3 and |T(X)| = 3.

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Examples

(i) It is easy to find X ⊆ Rn with |A(X)| = |X|

2

• if n is enough large,

so that |T(X)| = |X|

3

• .

(ii) If X ⊆ R3 forms an octahedron, then |A(X)| = 2 and |T(X)| = 2. (iii) If X ⊆ R2 forms a regular hexagon, then |A(X)| = 3 and |T(X)| = 3. (iv) If X ⊆ Rn is a regular simplex, then |A(X)| = |T(X)| = 1.

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Generalize X ⊆ Rn to colored spaces

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that r(U) = r(f (U)) for each U ∈ Y

2

• .

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that r(U) = r(f (U)) for each U ∈ Y

2

• .

We shall write Y ≃r Z if Y is isometric to Z.

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that r(U) = r(f (U)) for each U ∈ Y

2

• .

We shall write Y ≃r Z if Y is isometric to Z. We set Ak(r) = {[Y ] | Y ∈ X

k

• } where [Y ] = {Z | Y ≃r Z}.

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that r(U) = r(f (U)) for each U ∈ Y

2

• .

We shall write Y ≃r Z if Y is isometric to Z. We set Ak(r) = {[Y ] | Y ∈ X

k

• } where [Y ] = {Z | Y ≃r Z}.

The sequence (a1(r), a2(r), · · · , a|X|(r)) is called the isometric sequence

• f (X, r) where X is a finite set and ak(r) = |Ak(r)|.

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that r(U) = r(f (U)) for each U ∈ Y

2

• .

We shall write Y ≃r Z if Y is isometric to Z. We set Ak(r) = {[Y ] | Y ∈ X

k

• } where [Y ] = {Z | Y ≃r Z}.

The sequence (a1(r), a2(r), · · · , a|X|(r)) is called the isometric sequence

• f (X, r) where X is a finite set and ak(r) = |Ak(r)|.

If X ⊆ Rn is finite, then (X, r) is a colored space

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Generalize X ⊆ Rn to colored spaces

For a set X and positive integer k we denote by X

k

• the set of k-subsets of X.

A pair (X, r) is called a colored space if r is a function whose domain is X

2

• .

For Y , Z ⊆ X we say that Y is isometric to Z if there exists a bijection f : Y → Z such that r(U) = r(f (U)) for each U ∈ Y

2

• .

We shall write Y ≃r Z if Y is isometric to Z. We set Ak(r) = {[Y ] | Y ∈ X

k

• } where [Y ] = {Z | Y ≃r Z}.

The sequence (a1(r), a2(r), · · · , a|X|(r)) is called the isometric sequence

• f (X, r) where X is a finite set and ak(r) = |Ak(r)|.

If X ⊆ Rn is finite, then (X, r) is a colored space where r({x, y}) := d(x, y) for {x, y} ∈ X

2

• , and A(X) = A2(r) and

T(X) = A3(r).

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a2(r) ≤ a3(r) if |X| ≥ 5

Definition

For colored spaces (X, r) and (X, r1) we say that r1 is a fusion of r if {r−1(c) | c ∈ Im(r)} is a refinement of {r−1

1 (c) | c ∈ Im(r1)}

For each (X, r) with 1 < a2(r) < |X|

2

• we can find a fusion r1 of r such

that a2(r) − a2(r1) = 1 and a3(r) − a3(r1) ≥ 1, or a2(r) − a2(r1) = 2 and a3(r) − a3(r1) ≥ 2. Repeating this argument we obtain that, if a2(r) > a3(r), then 2 = a2(r) > a3(r) = 1, which implies |X| ≤ 5, a contradiction.

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Exploring the case where a2(r) = a3(r) = 2

Set A2(r) = {α, β}. Suppose |X| ≥ 6. Then the graph (X, r−1(α)) or (X, r−1(β)) contains a triangle since the Ramsey number R(3, 3) equals 6. Thus, we may assume ααα ∈ A3(r). If ααβ / ∈ A3(r), then (X, r−1(α)) is a disjoint union of cliques. This implies that αββ ∈ A3(r), and hence, A3(r) = {ααα, αββ}. If ααβ ∈ A3(r), then A3(r) = {ααα, ααβ}. In the former case (X, r−1(β)) is a complete bipartite graph. In the latter case (X, r−1(β)) is a matching.

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Examples of colored spaces with a2(r) = a3(r)

(i) A3(r) = {ααα, αββ, αβγi | i = 1, 2, . . . , a2(r) − 2}. (ii) A3(r) = {ααα, ααβi | i = 1, 2, . . . , a2(r) − 1}. (iii) A3(r) = {ααα, αββ, αγγ, βγδ}.

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Main results

Theorem (H, Shinohara)

Every finite colored space (X, r) with 4 ≤ a2(r) = a3(r) and 9 ≤ |X| is isomorphic to one of the above examples.

Remark

(1) We have already classified the case where a2(r) = a3(r) ≤ 3; (2) Several sporadic examples exist such that |X| ≤ 8 and a2(r) = a3(r); (3) The proof is done by induction on a2(r); (4) We spent much pages to prove the first step, i.e., a2(r) = 4.

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Questions and Known results

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Questions and Known results

(1) When can ak(r) = 1 appear?

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1.

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear?

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2.

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2. (3) Is an isometric sequence unimodal?

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2. (3) Is an isometric sequence unimodal? Open

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2. (3) Is an isometric sequence unimodal? Open (4) Can we improve the inequality |X| ≤ n+s

s

• for X ⊆ Rn where

s = |A(X)| by using |T(X)|?

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2. (3) Is an isometric sequence unimodal? Open (4) Can we improve the inequality |X| ≤ n+s

s

• for X ⊆ Rn where

s = |A(X)| by using |T(X)|? I have no idea to approach.

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2. (3) Is an isometric sequence unimodal? Open (4) Can we improve the inequality |X| ≤ n+s

s

• for X ⊆ Rn where

s = |A(X)| by using |T(X)|? I have no idea to approach. (5) Find colored spaces (X, r) such that, for all U, V ∈ X

3

• ,

U ≃r V if and only if Uσ = V for some σ ∈ Aut(X, r).

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Questions and Known results

(1) When can ak(r) = 1 appear? If ak(r) = 1 with 1 < k < |X| − 1, then a2(r) = 1. (2) When can a2(r) = 2 appear? If ak(r) = 2 with 4 ≤ k ≤

1+√ 1+4|X| 2

, then a2(r) = 2. (3) Is an isometric sequence unimodal? Open (4) Can we improve the inequality |X| ≤ n+s

s

• for X ⊆ Rn where

s = |A(X)| by using |T(X)|? I have no idea to approach. (5) Find colored spaces (X, r) such that, for all U, V ∈ X

3

• ,

U ≃r V if and only if Uσ = V for some σ ∈ Aut(X, r). I know only examples.

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References

• E. Bannai, Et. Bannai, D. Stanton,

An upper bound for the cardinality of an s-distance subset in real Euclidean space II, Combinatorica 3 (1983), no. 2, 147-152. S.J. Einhorn and I.J. Schoenberg, On euclidean sets having only two distances between points. I. II. Nederl. Akad.

• Wetensch. Proc. Ser. A 69 Indag. Math. 28 (1966), 479–488, 489–504.
• D. G. Larman, C. A. Rogers, J. J. Seidel,

On two-distance sets in Euclidean space, Bull. London Math. Soc. 9 (1977), no. 3, 261-267.

• P. Lisonek,

New maximal two-distance sets. (English summary) J. Combin. Theory Ser. A 77 (1997), no. 2, 318-338.

• M. Hirasaka, M. Shinohara,

Characterization of finite metric spaces by their isometric sequences submitted to Electric J. Comb.

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Thank you for your attention.

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