arc index of spatial graphs - waseda university · 2016. 8. 7. · arc index of spatial graphs...

Arc index of spatial graphs

Sungjong No

jointwork with Minjung Lee and Seungsang Oh

International Workshop on Spatial Graphs 2016

August 5, 2016

Arc index of spatial graphs Sungjong No

Introduction Examples Proof of Main Theorem

Contents

1 Introduction

2 Examples

3 Proof of Main Theorem



Definition

• An arc presentation of a spatial graph G is an ambientisotopic image of G contained in the union of finitely manyhalf planes, called pages, with a common boundary line,called binding axis in such a way that each half planecontains a properly embedded single arc.

• An arc index α(G) of a graph G is a minimum of thenumber of arcs for arc presentations of a graph G.

=



Definition

• To describe our results, decompose the spatial graph G intocut components by splitting and cutting along a maximalset of disjoint 2-spheres, each of which is either disjointfrom G or meeting G only at a vertex, while eachcomponent of their complement intersects with G.



Definition

• A graph is called a bouquet if it is consist of one vertex andloops.

• If a cut-component is itself a bouquet spatial graph, we callit a bouquet cut-component .

bouquet bouquet cut-components



Theorems for arc indices of knots

Bae-Park (2000)

Let K be an any nontrivial knot. Then α(K) ≤ c(K) + 2.Moreover if K is a non-alternating prime knot, thenα(K) ≤ c(K) + 1.

Jin-Park (2010)

Let K be an any non-alternating prime knot. Thenα(K) ≤ c(K).



Theorem (Lee, No, Oh)

Let G be any spatial graph with e edges and b bouquetcut-components. Then

α(G) ≤ c(G) + e+ b.

Furthermore, this is the lowest possible upper bound.



• trivial θn-curve

=

Naturally α(θn) = n (c(θn) = 0, e = n, b = 0)



• 31 knot(regard it as a graph consist of one vertex and oneloop)

=

α(31) = 5 (c(31) = 3, e = 1, b = 1)



• 51 θ-curve

=

• α(51) ≤ 8 (c(51) = 5, e = 3, b = 0)



Wheel diagram and spoke

3

2

1

4

5

{1,4}

{2,5}

{1,4}

{3,5}

{2,4}

{1,3}

3

2

1

4

5{1,4} spoke

A wheel diagram is consist of α(G) spokes. To each spoke,assign the pair of numbers {a, b} when the arc corresponding tothe spoke connects the points numbered a and b on the binding.



• Any spatial graph can be converted to a spoke diagram byspoking algorithm.

=

3

33

3

33

4 4 3

33

4 45

5 {3,5}3

3

4 45

{3,5}

3

3

44

{2,5}

2 3

3

4 4

{3,5}

{2,5} {2,6}

6

34

6{3,5}

{2,5} {2,6}

{3,7}

{4,7}

{3,5}

{2,5} {2,6}

{3,7}

{4,7}{1,4}

{1,3}

{1,6}

= 3

2

1

4

7

6

5



• From now, we consider the diagram is of a cut component.

• Select a pivot vertex v0 of the diagram and we willconstruct a new diagram in three different ways accordingto the following three types;(Type 1) The vertex v comes from a crossing of Dm.(Type 2) The vertex v comes from a vertex of Dm which isnot v0.(Type 3) The pulling edge e is a loop, i.e., v = v0.



(Type 1)

{1,4}

0v

3

1

i

j

ve e `

2e `

1e `0v

i

j

e `2e `

1e `

kk

0v

i e `2e `

1e `

k

j k0v

i e `2e `

k

{j,k}

gnikopsgnidils

D `mD

m+1D



(Type 2)

gnikopsgnidils 0v

e

2e

1e

4e

3e

kk

j k

ik k j `

{1,4}

0v

3

1

i

j

ve

2e

1e

4e 3e j `

0v

i

j

e

2e

1e

4e

3e

kkkkk

j `

0v

2e

3e

kk

{j,k}

{i,k}`{j ,k}



(Type 3)

gnikopsgnidils

{1,4}

0v

3

1

i

j

e

0v

i

jk

k

0v j k

i k

{j,k}

{i,k}

0v



• By spoking algorithm, the sum of spokes, regions andvertices is unchanged for Type 1 and 2 cases, while greaterthan by 1 for Type 3 case.

• If the diagram is a bouquet cut component, then Type 3case is necessary while the spoking algorithm is processing.

• We can adjust the spoking algorithm without increasingthe number of cut components.



Example

3

33

3

33

4 4 {3,5}3

3

4 45

{3,5}

3

3

44

{2,5}

2 3

3

4 4

{3,5}

{2,5} {2,6}

6

34

6{3,5}

{2,5} {2,6}

{3,7}

{4,7}

{3,5}

{2,5} {2,6}

{3,7}

{4,7}{1,4}

{1,3}

{1,6}

regions : 8vertices :2spoke : 0







For every step, r(the number of regions)+s(the number ofspokes)+v(the number of vertices) is not changed.



In the original diagram, c(G) + e+ 2 = r + v (s = 0). And inthe final diagram, r + v + s = s+ 2. So, c(G) + e = s. But finals is the number of arcs of this diagram. So, α(G) ≤ c(G) + e.



Example of Type 3


regions : 2 vertices :1spoke : 7




G1 G2 GGlue or combine cut-components at the intersecting vertex likethis figure. Then we get the inequality α(G) ≤ c(G) + e+ b.The proof is completed for the original graph G.



Thank you


arc index of spatial graphs - waseda university · 2016. 8. 7. · arc index of spatial graphs...

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