1 A 4-Dimensional Graph Has at Least 9 Edges Math Colloquium 2015 December 2 Sonoma State University...

transcript

A 4-Dimensional Graph Has at Least 9 Edges

Math Colloquium2015 December 2

Sonoma State University

Roger House

What we're going to talk about

What is a graph?

What is the dimension of a graph?

What is the least number of edges a four-dimensional graph must have?

What's a graph?

Remember these graphs?

And these?

And graphs you can eat

None of these graphs are the kinds of graphs we're interested

in at the moment

A different kind of graph

We're going to deal with much different kinds of graphs, namely, mathematical graphs.

Definition of a graph

Definition: A graph G consists of a finite nonempty set V of vertices together with a set E of unordered pairs of distinct vertices of V. The pair e = {u,v} of vertices in E is called an edge of G.

There will be a quiz on the definition

But not on the one above

On the one below

A friendlier definition

A graph is dots connected by lines.

Dots and Lines by Richard J. Trudeau

Fancy terminology

The dots are called vertices. (Or points, or nodes, or ...)

The lines are called edges. (The graph above has 7 vertices and 9 edges.)

Two vertices connected by an edge are adjacent.

An edge is incident with its two vertices. The degree of a vertex is the number of

edges incident with the vertex.

Examples of graphs

Yet more graphs

Let's start at the beginning

What is the simplest, most basic graph?The empty set?No. The definition says the set of vertices must

be nonemptySo this is it:

What is the next simplest graph?

How many vertices?Seems like 2 must be the answerSo this is the next simplest graph:

What is the third simplest graph?

Three vertices?Are there any more graphs with 2 vertices? How about this one:

Are there any more graphs with 2 vertices?

Where do we go now?

Three vertices

More graphs with 3 vertices

Wait a minute!

Didn't we miss some?

These two graphs are isomorphic

iso-morphic = "same shape" = "equal form"

The Graph Isomorphism Problem

Is there an efficient algorithm for determining if two finite graphs are isomorphic?

None is knownBut there is no proof that one does not existNEWS BULLITEN 2015 Nov. 6: Laszlo Babai

has reported that he has a quasipolynomial time algorithm for the graph isomorphism problem

This is not an "efficient" algorithm, but it is a significant advance

Academic commercial

To learn more about isomorphism and the joys of abstract algebra:

Math 320 Modern Algebra I Math 420 Modern Algebra II

Graphs with V = 4

E = 0 E = 1

Graphs with V = 4

E = 2 E = 2

Graphs with V = 4

E = 3 E = 3

Graphs with V = 4

E = 4 E = 4

Graphs with V = 4

E = 5 E = 6

Kn - complete graph on n vertices

Kn has n(n-1)/2 edges

K4 has 6 edges K5 has 10 edges

Complete graphs K1 through K6

Find all graphs with 5 vertices

Each dotted line is an edge or not an edgeHow many different configurations of edges are there?

Find all graphs with 5 vertices

K5 has 5(5-1)/2 = 10 edges

So there are 10 dotted lines, each one either an edge or not an edge

So there are 210 = 1024 possible configurationsSo there are 1024 graphs with 5 vertices, right?No!Remember isomorphic graphs?There are 10 configurations with only one edgeThey're all isomorphic, so only 1 graph results

Are there enough graphs? V max E 2E #graphs 1 0 1 1 2 1 2 2 3 3 8 4 4 6 64 11 5 10 1,024 34 6 15 32,768 156 7 21 2,097,152 1,044 8 28 268,435,456 12,346 9 36 68,719,476,736 274,668 10 45 35,184,372,088,832 12,005,168

Consider these graphs ...

And these ...

What's the structure?

Complete bipartite graphs

If the set of all vertices can be partitioned into sets V and W such that every edge connects a vertex in V to a vertex in W, then the graph is called bipartite

If a bipartite graph has as many edges as possible, then it is called a complete bipartite graph

Notation: Km,n

, where m = |V| and n = |W|

How many edges in Km,n

| = mn

Not all bipartite graphs are complete

Cyclic graphs

A tree is a connected graph with no cycles

What if a single edge is added to a tree?

For a tree: |V| = |E| + 1

In general: |E| |V| - 1

What now?

This concludes a crash introduction to graph theory

We move on to our next main question:

The dimension of a graph

In 1965, the illustrious trio P. Erdős, F. Harary, and W.T. Tutte published a paper entitled

On the dimenion of a graph

The first thing they did in this paper was to define the dimension of a graph

The dimension of a graph

The dimension of a graph G, denoted dim(G), is the minimum n such that G has a unit-distance representation in n, i.e., every edge is of length 1. The vertices of G are mapped to distinct points of n, but edges may cross

Intuitively: Given a graph, construct a model where every edge is of length 1, and then figure out whether it can live in 1-space, 2-space, 3-space, 4-space, ..., n-space

What is the dimension of K1,1

dim(K1,1

dim(K1,2

dim(K1,3

dim(K2,2

dim(K2,3

in three dimensions

A, B to origin: √3/2

circle: radius = /2

dim(K2,3

What about Kn,m

for n,m ≥ 3?

There are a lot of points on that circle

dim(K2,m

) = 3 for m 3

What is dim(K3,3

What's a lower bound on dim(K3,3

Since K2,3

is a subgraph of K3,3

dim(K2,3

) dim(K3,3

So dim(K3,3

) ≥ 3

A lower bound for dim(G)

Thm: If H is a subgraph of G, dim(H) dim(G)Proof: Say dim(G) = nConsider an embedding of G in n Remove vertices and edges of G so that only H

is leftH is embedded in n so dim(H) n

dim(K3,3

We know dim(K3,3

Might dim(K3,3

) = 3?

Consider three spheres SA, S

B and S

c of radius 1

centered at A, B, and C, respectively

SA and S

B intersect in a circle on which vertices 1,

2, and 3 lie

This circle must also lie on SC

This can only happen if SC is one of S

A or S

Which means C = A or C = B ⇒

So dim(K3,3

In fact, dim(K3,3

) 4To show this, we consider two circles of radius

1/2 in 4:

C1: x2 + y2 = ½

C2: z2 + w2 = ½

Pick any point P = (x, y, 0, 0) on circle C1

Pick any point Q = (0, 0, z, w) on circle C2

The distance between P and Q is the square root of

(x-0)2 + (y-0)2 + (0-z)2 + (0-w)2 = x2 + y2 + z2 + w2 = (x2 + y2) + (z2 + w2) = ½ + ½ = 1

So every point P on circle C1 is at a distance 1

from every point Q on circle C2

Pick any three distinct points on C1 and call them

A, B, and C

Pick any three distinct points on C2 and call them

1, 2, and 3Insert an edge of length 1 from each letter to

each digit

We have an embedding of K3,3

What is the dimension of Km,n

What about dim(Km,n

) for m,n ≥ 3?

There are a lot of points on those two circles

dim(Km,n

) = 4 for m,n ≥ 3

What is the dimension of Kn?

K3 can be represented as an equilateral triangle:

dim(K3) = 2

K4 can be represented as a regular tetrahedron

dim(K4) = 3

K5 can be represented as ...? dim(K

5) = ?

An exercise for the perspicacious student: Show that dim(K

n) = n - 1

Basic results about dimension

dim(Kn) = n - 1

dim(Kn - e) = n - 2

dim(K1,1

) = dim(K1,2

) = 1, dim(K1,m

) = 2 for m3

dim(K2,2

) = 2, dim(K2,m

) = 3 for m 3

dim(Km,n

) = 4 for m, n 3

dim(Cn) = 2 for C

n a cyclic graph of order n 3

dim(tree) 2if H is a subgraph of G then dim(H) dim(G)

Questions

Now we have answered our first two questions:

What is a graph?

Only one question is left:

What is the least number of edges a four-dimensional graph must have?

Exactly what is the question?

In 2009 The Mathematical Coloring Book by Alexander Soifer was published

In this book a question posed by Paul Erdős in 1991 appears

What is the smallest number of edges in a graph G if dim(G) = 4?

In the rest of this talk we will answer this question

Where to start?

We have seen two graphs of dimension 4

K5: V=5, E=10 K

3,3: V=6, E=9

Minimum number of vertices?

Since dim(Kn) = n-1, we know dim(K

5) = 4

Since dim(Kn - e) = n-2, we know dim(K

5 - e) = 3,

so every proper subgraph of K5 has

dimension at most 3So a four dimensional graph with a minimum

number of edges must have more than 5 vertices

Therefore we need to look at graphs with 6 or more vertices

Number of edges?

Remember that a connected graph must have at least one more vertex than it has edges, i.e., |E| |V| - 1

So, if |V| = 6, it must be that |E| 6 - 1 = 5 Therefore we need to look at graphs with 5 or

more edgesWe already have a four-dimensional graph with 9

edges (K3,3

), so we need not consider graphs with more than 9 edges

Maximum number of vertices?

Using |E| |V| - 1 again with |E| = 9, we have 9 |V| - 1, so |V| 10

Therefore we need to look at graphs with 10 or fewer vertices

To sum up: Find a four-dimensional graph with6 |V| 105 |E| 9

Fill in the blanks

# edge#vert

9 8 7 6 5

|V| = |E| + 1 is easy: A tree

# edge#vert

9 8 7 6 5

6 tree

7 tree

8 tree

9 tree

10 tree

dim(tree)

We return to the question: What if a single edge is added to a tree?

Add one edge, get one cycle

|V| = |E| is easy: A cycle

# edge#vert

9 8 7 6 5

6 cycle tree

7 cycle tree

8 cycle tree

9 cycle tree

10 tree

Add two edges and get what?

Two cycles with no common edge

Two cycles with edges in common

Reduce to essentials

Make all edges have unit length

3-routes

When two edges are added to a tree so that the result is two cycles sharing at least one common edge, the resulting graph is called a 3-route

A 3-route consists of three paths which have nothing in common except their end vertices

So there is a left path, a middle path, and a right path from one end vertex to the other end vertex

Another example of a 3-route

Another exercise for the perspicacious student: Show that dim(3-route) = 2 with one exception.

|V| = |E| - 1 is easy: A 3-route

# edge#vert

9 8 7 6 5

6 3-route cycle tree

9 cycle tree

10 tree

We're getting closer and closer

Now only these three cases are left:6 vertices, 8 edges6 vertices, 9 edges7 vertices, 9 edges

There are 156 graphs with 6 vertices and 1044 graphs with 7 vertices

Way too manyTo the rescue: An Atlas of Graphs by R.C. Read

and R.J. Wilson

6 vertices, 8 edges

Narrowing down

The Atlas lets us pare down to fewer graphs: 6 vertices, 8 edges: 24 graphs 6 vertices, 9 edges: 21 graphs 7 vertices, 9 edges: 131 graphs Total: 176 graphsIt's still too manyBut let's take a look at that Atlas page again

6 vertices, 8 edges

Cut Vertex

Narrowing down yet more

These graphs are not of interest: A graph with vertices of degree 0 or 1 A graph which is not connected A graph which has a cut vertexYet another exercise for the perspicacious

student: If G is a four-dimensional graph with a minimum number of edges, then G cannot contain a cut vertex

Terminology: A graph with no cut vertex is said to have vertex connectivity 2

All the blanks are filled in

# edge#vert

9 8 7 6 5

6 14 9 3-route cycle tree

7 20 3-route cycle tree

9 cycle tree

10 tree

Narrowed down enough?

So we are now down to this many graphs: 6 vertices, 8 edges: 9 graphs 6 vertices, 9 edges: 14 graphs 7 vertices, 9 edges: 20 graphs Total: 43 graphsLacking a brilliant flash of insight, we look at all of

the 43 graphs and see if we can embed them in 2-, 3-, or 4-dimensions

G147 (V=6, E=8)

G580 (V=7, E=9)

G146 (V=6, E=8)

G171 (V=6, E=9)

Only 39 more to go

Fortunately, we are running out of time, so you won't have to look at the details of the 39 remaining embeddings

Here is the breakdown by dimension: 2-dimensional: 27 graphs 3-dimensional: 15 graphs 4-dimensional: 1 graph Total: 43 graphsHere are all 43 embeddings:

2-dimensional - part 1

There's only one left

Of the 43 candidate graphs, 42 are 2- or 3-dimensional, leaving just one graph, which answers the original question

A 4-dimensional graph must have at least 9 edges, and there is only one 4-dimensional graph with 9 edges:

I had a lovely unit-distance drawing of K3,3 in 4-

dimensions, but I lost it. This will have to do:

4-dimensional graph with 9 edges

That's all, folks

For all the details see: rogerfhouse.com Thank you

1 A 4-Dimensional Graph Has at Least 9 Edges Math Colloquium 2015 December 2 Sonoma State University...

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