Let be a -regular graph, and let

be the eigenvalues of the adjacency matrix of counted with multiplicities and sorted in descending order.

How good can the spectral expansion of be?

**1. Simple Bounds **

The simplest bound comes from a trace method. We have

by using one definition of the trace and

using the other definition and observing that counts the paths that go from to in two steps, of which there are at least : follow an edge to a neighbor of , then follow the same edge back. (There could be more if has multiple edges or self-loops.)

So we have

and so

The condition is necessary to get lower bounds of ; in the clique, for example, we have and .

A trace argument does not give us a lower bound on , and in fact it is possible to have and , for example in the bipartite complete graph.

If the diameter of is at least 4, it is easy to see that . Let be two vertices at distance 4. Define a vector as follows: , if is a neighbor of , and if is a neighbor of . Note that there cannot be any edge between a neighbor of and a neighbor of . Then we see that , that (because there are edges, each counted twice, that give a contribution of to ) and that is orthogonal to .

**2. Nilli’s Proof of the Alon-Boppana Theorem **

Nilli’s proof of the Alon-Boppana theorem gives

where is the diameter of . This means that if one has a family of (constant) degree- graphs, and every graph in the family satisfies , then one must have . This is why families of *Ramanujan* graphs, in which , are special, and so hard to construct, or even to prove existence of.

Friedman proves a stronger bound, in which the error term goes down with the square of the diameter. Friedman’s proof is the one presented in the Hoory-Linial-Wigderson survey. I like Nilli’s proof, even if it is a bit messier than Friedman’s, because it starts off with something that *clearly* is going to work, but the first two or three ways you try to establish the bound don’t work (believe me, I tried, because I didn’t see why some steps in the proof had to be that way), but eventually you find the right way to break up the estimate and it works.

So here is Nilli’s proof.

We are going to use essentially the same vector that we used to analyze the spectrum of the infinite tree, although the analysis will be a bit different.

Let be two vertices in at distance , and call . Let be a neighbor of . We say that the distance of a vertex from is the smallest of the shortest path distance from to and from to .

We construct a vector as follows:

- if
- if is at distance from

Note that this is more or less the same vector we constructed in the case of the tree. The reason for talking about the distance from two vertices instead of one, is that we want to say that every vertex is adjacent to at most vertices whose value of is strictly smaller; in the case of a tree, the root is exceptional because it has neighbors whose value of is strictly smaller. There will be a step in the proof in which this choice really makes a difference.

We claim

It turns out that, in order to prove (1) it is easier to reason about the Laplacian matrix than about the adjacency matrix. Define to be the (non-normalized) Laplacian of . We have the following nice expression for the quadratic form of

For every vertex let us call (for **s**maller) the set of neighbors of such that . We always have . Let us call the set of vertices at distance exactly from .

Now we do the calculations:

Finally,

because decreases (or at least does not increase) for increasing .

Putting everything together we have

and so

Now we are finally almost done: define a vector with the same construction we did for , but using the set as the reference for the distance, where is a neighbor of . We then have

It is clearly not possible for a vertex at distance from to also be at distance from , otherwise we would have a path of length from to , so the vectors and are non-zero on disjoint subsets of coordinate, and hence are orthogonal.

But we can say more: we also have , because there cannot be an edge such that both and , because otherwise we would have a path of length from to .

This means that if we take any linear combination of and we have

so we have found a two-dimensional set of vectors whose Rayleigh quotient is at most the above expression, and so

What did just happen? The basic intuition is that, as in the infinite tree, we set weights to go down by every time we get away from the “root,” and we would like to argue that, for every node , we have

by reasoning that one of the neighbors must be closer to the root, and hence have value , while the other neighbors are all at least . This bound fails at the “leaves” of the construction, which is fine because they account for a small portion of , but it also fails at the root, which is not adjacent to any larger vertex. In the case of the infinite tree this is still ok, because the root also accounts for only a small portion of ; in general graphs, however, the “root” vertex might account for a very large fraction of .

Indeed, the root contributes 1 to , and each set contributes . If the size of grows much more slowly than , then the contribution of the root to is too large and we have a problem. In this case, however, for many levels , there have to be many vertices in that have fewer than edges going forward to , and in that case, for many vertices we will have that will be much more than .

Although it seems hopeless to balance this argument and charge the weight of the edge in just the right way to and to , the calculation with the Laplacian manages to do that automatically.

**3. A More Sophisticated “Trace” Method **

The Hoory-Linial-Wigderson survey also gives a very clean and conceptual proof that

via a trace argument. I am not sure who was the first to come up with this idea.

Let us pick two vertices at distance , and call , and . Fix . Then and are orthogonal vectors, because the coordinates on which is nonzero correspond to vertices at distance from and the coordinates on which is nonzero correspond to vertices at distance from , and these conditions cannot be simultaneously satisfied. This means that .

Since is orthogonal to , we have

but we also have

Now, is the number of walks of length in that start at and get back to . In every -regular graph, and for every start vertex, the number of such walks is at least the corresponding number in the infinite -ary tree. This is clear if is a Cayley graph; it takes a few minutes (at least I had to think about it for a bit) to see that it holds in every graph.

Ok, then, what is the number of closed walks of length in the infinite -ary tree? This is a long story, but it is a very well-studied question and there is a bound (not the tightest know, but sufficient for our purposes) giving

so,

I put “trace” in quote in the title of this section, but one can turn the argument into an honest-to-God application of the trace method, although there is no gain in doing so. Pick an even number . We can say that

and

which gives

for , which is implied by the bound based on diameter that we proved above.

Nice writeup. By the way, A. Nilli is a pseudonym for Noga Alon. (One of Noga’s daughters is named Nilli.)

— Ravi Boppana

In the simple bound section in last para, it seems that x^T A x can be >= 4 (instead of =4) since there is also contribution from edges between two neighbors or a, or betweem two neighbors of b?

Yes, it’s at least , I corrected it.

Ravi, I went on for 1385 words without revealing Noga’s secret identity and you give it away like this!

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