In the book “Elementary number theory, group theory and Ramanujan graphs“, Sarnak et. al. gave an elementary construction of expander graphs. We decided to go through the construction in the small seminar and I am recently assigned to give a talk about the girth estimate of such graphs.

Given graph (finite and undirected) , we will denote the set of vertices by and the set of edges . The graph is assumed to be equipped with the standard metric where each edge has length .

The **Cheeger constant** (or isoperimetric constant of a graph, see this pervious post) is defined to be:

Here the notation denote the set of edges connecting an element in to an element outside of .

Note that this is indeed like our usual isoperimetric inequalities since it’s the smallest possible ratio between size of the boundary and size of the set it encloses. In other words, this calculates the most efficient way of using small boundary to enclose areas as large as possible.

It’s of interest to find graphs with large Cheeger constant (since small Cheeger constant is easy to make: take two large graphs and connect them with a single edge).

It’s also intuitive that as the number of edges going out from each vertice become large, the Cheeger constant will become large. Hence it make sense to restrict ourselves to graphs where there are exactly edges shearing each vertex, those are called **-regular graphs**.

If you play around a little bit, you will find that it’s not easy to build large k-regular graphs with Cheeger constant larger than a fixed number, say, .

**Definition: ** A sequence of k-regular graphs where is said to be an **expander family** if there exists constant where for all .

By random methods due to Erdos, we can prove that expander families exist. However an explicit construction is much harder.

**Definition:** The **girth** of is the smallest non-trivial cycle contained in . (Doesn’t this smell like systole? :-P)

In the case of trees, since it does not contain any non-trivial cycle, define the girth to be infinity.

The book constructs for us, given pair of primes where is large (but fixed) and , a graph -regular graph with

where .

Note that the bound is strictly positive and independent of . Giving us for each , as q runs through primes larger than is a -regular expander family.

In fact, this constructs for us an infinite family of expander families: a -regular one for each prime and the uniform bound on Cheeger constant gets larger as becomes larger.

One of the crucial step in proving this is to bound the girth of the graph , i.e. they showed that and if the quadratic reciprocity then . This is what I am going to do in this post.

Let be the set of quaternions with coefficient, i.e.

Fix odd prime , let

where the norm on is the usual .

consists of points with only odd first coordinate or points lying on spheres of radius and having only even first coordinate. One can easily check is closed under multiplication.

Define equivalence relation on by

if there exists s.t. .

Let , let be the quotient map.

Since we know , carries an induced multiplication with unit.

In elementary number theory, we know that the equation has exactly integer solutions. Hence the sphere of radius in contain points.

In each -tuple exactly one is of a different parity from the rest, depending on whether or . Restricting to solutions where the first coordinate is non-negative, having different parity from the rest (in case the first coordinate is , we pick one of the two solutions to be canonical), this way we obtain exactly solutions.

Let be this set of points on the sphere. Note that the s represent the solutions where the first coordinate is exactly .

Check that generates .

We have and . By definition and is injective on . Let .

Consider the Cayley graph , this is a -regular graph. Since generares , is connected.

**Claim:** is a tree.

Suppose not, let a non-trivial cycle. . Since is a Cayley graph, we may assume .

Hence , for some .

Since for all , the word cannot contain either or , hence cannot be further reduced.

in means for some we have

.

Since every word in with norm must have a unique factorization where is a reduces word of length in and .

Contradiction. Establishes the claim.

Now we reduce the group mod :

One can check that .

Let , is a central subgroup.

For , . Which means we have well defined homomorphism

.

Let , if we have is injective on and hence $latex .

Now we are ready to define our expanding family:

.

Since generates , generates . Hence is -regular and connected.

**Theorem 1:**

Let be a cycle in , there is such that for .

Let , . Note that from the above arguement we know is a reduced word, hence . In particular, this implies cannot all be .

Also, since is reduced, .

By Lemma, since , hence divide , we conclude

We deduce hence for all cycle. i.e. .

**Theorem 2:** If does not divide and is not a square mod (i.e. ), then .

For any cycle of length as above, we have , i.e. .

Since , we have is even. Let .

Note that , we also have

Hence .

Since , .

If we will have . Then .

But we know that , one of must be divisible by , hence .

Conclude , , hence . Contradiction.

September 19, 2011 at 4:54 am

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