A bit more on expanders…

So the lecture notes I posted a year ago giving a “simpler proof” of expansion for the Margulis-Gabber-Galil graphs turns out to be very similar to arguments of M. Berger (1991) and Y. Shalom (1999) for proving that has property (T). See Section 4.2 of the book of Bekka, de la Harpe, and Valette.

Anyone interested in expander graphs should take a look at Amir Yehudayoff’s recent article in the SIGACT newsletter: Proving expansion in three steps. Amir uses a ping-pong lemma argument (which is essentially the same as the “combinatorial lemma” in the preceding post) to exemplify the “opening” of the three part strategy of Bourgain and Gamburd to proving expansion in Cayley graphs.

Gabber-Galil analysis of Margulis’ expanders

I’m currently teaching a course on spectral graph theory, and I decided to lecture on Margulis’ expander construction and the analysis of its spectral gap by Gabber and Galil. I had never realized how intuitive the analysis could be; the lectures notes I had seen didn’t quite reflect this. In particular, we won’t do any unsightly calculations. Everything I’m about to say is probably well-known, but I thought I would write it down anyway.

The idea is to first start with an initial “expanding object,” and then try to construct a family of graphs out of it. First, consider the infinite graph with vertex set . The edges are given by two maps and , where and . So the edges are and . Clearly is not adjacent to anything. Except for the origin, this graph is an expander in the following sense.

Lemma 1 For any subset , we have

where denotes the edges leaving .

The following simple proof is inspired by a paper of Linial and London.

Proof: First consider a subset that does not intersect the coordinate axes. Let represent the four quadrants of , and let . Consider that and . The latter fact follows because if then . Similarly, and and , while all the respective pairs and and are disjoint.

Combining this with the fact that and are bijections on immediately shows that is at least

Dealing with the coordinate axes is not too hard. Suppose now that is arbitrary. Let . The preceding argument shows . We will show that . Averaging these two inequalities with weights and completes the proof.

Let , and . Then we have the bounds:

The first equation follows since each element of is connected to exactly two elements of , and each element of is connected to exactly two elements of . For instance, is connected to and , while is connected to and .

The second follows because every point of is connected to two unique points of , e.g. is connected to and . The final inequality follows from the fact that from our first argument (since does not touch the axes), and because has no edges to . Summing these three inequalities yields .

Of course, is not a finite graph, so for a parameter , we define the graph with vertex set , where . There are four types of edges in : A vertex is connected to the vertices

This yields a graph of degree at most 8.

Theorem 2 There is a constant such that for every ,

where is the second-smallest eigenvalue of the Laplacian on .

Of course, the graphs now have torsion, and thus our expansion result for is not, a priori, very useful. The non-trivial idea of the proof is that the torsion doesn’t matter, making Lemma 1 applicable. This is not difficult to show using some Fourier analysis. It turns out to be better though, if we first move to the continuous torus.

Recall that,

Let be the 2-dimensional torus equipped with the Lebesgue measure, and consider the complex Hilbert space

equipped with the inner product .

We might also define a related value,

Note that this is just defined as a number; the eigenvalue notation is merely suggestive, but we have not introduced an operator on the torus.

Claim 1 There is some such that for any , we have \

Proof: We simply sketch a proof, which is rather intuitive. Suppose we are given some map such that . Define its continuous extension as follows: Under the natural embedding of into , every point sits inside a grid square with four corners . Writing a local convex combination , we define

Now it is elementary to verify that . It is also easy to verify that for some . (Even if , we still get a contribution from to on this square because we are taking a weighted average.)

Finally, for any , there is a path of length in connecting each of the corners of ‘s square to the corners of ‘s square. A similar fact holds for . In fact, this is the only place where we need to use the fact that edges of the form and are present in . Thus any contribution to can be charged against a term in , and similarly for .

Now our goal is to show that . We will use the Fourier transform to “remove the torsion.” The point is that and , being shift operators, will act rather nicely on the Fourier basis.

We recall that if and we define by , then forms an orthonormal Hilbert basis for . In particular, every can be written uniquely as

where . The Fourier transform is defined as a map , where . In particular, is a linear isometry.

Define and . Then for any , we have

In particular, for any , we have and . The final thing to note is that . So now if we simply apply the Fourier transform (a linear isometry) to the expression in (1), we get a reformulation that is precisely

Here we have simply replaced by in (1), and then written .

But now recall our initial infinite graph on . If we denote by the Laplacian on , then we can rewrite this as,

In other words, it is precisely the first Dirichlet eigenvalue on , subject to the boundary condition .

But now the discrete Cheeger inequality tells us that

where is the minimal expansion of any set not containing the origin. Thus we have indeed unwrapped the torsion and returned to our initial question. Lemma 1 shows that , yielding the desired lower bound on .

The pseudorandom subspace problem

In this post, I will talk about the existence and construction of subspaces of which are “almost Euclidean.” In the next few, I’ll discuss the relationship between such subspaces, compressed sensing, and error-correcting codes over the reals.

A good starting point is Dvoretzky’s theorem:

For every and , the following holds. Let be the Euclidean norm on , and let be an arbitrary norm. Then there exists a subspace with , and a number so that for every ,

.

Here, is a constant that depends only on .

The theorem as stated is due to Vitali Milman, who gave the (optimal) dependence of on the dimension of , and proved that the above fact actually holds with high probability when is chosen uniformly at random from all subspaces of this dimension. In other words, for , a random -dimensional subspace of an arbitrary -dimensional normed space is almost Euclidean. Milman’s proof was one of the starting points for the “local theory” of Banach spaces, which views Banach spaces through the lens of sequences of finite-dimensional subspaces whose dimension goes to infinity. It’s a very TCS-like philosophy; see the book of Milman and Schechtman.

One can associate to any norm its unit ball , which will be a centrally symmetric convex body. Thus we can restate Milman’s proof of Dvoretzky’s theorem geometrically: If is an arbitrary convex body, and is a random -dimensional hyperplane through the origin, then with high probability, will almost be a Euclidean ball. This is quite surprising if one starts, for instance, with a polytope like the unit cube or the cross polytope. The intersection is again a polytope, but which is approximated closely by a smooth ball. Here’s a pictorial representation lifted from a talk I once gave:

(Note: Strictly speaking, the intersection will only be an ellipsoid, and one might have to pass to a subsection to actually get a Euclidean sphere.)

Euclidean subspaces of and Error-Correcting Codes.

It is not too difficult to see that the dependence is tight when we consider equipped with the norm (see e.g. Claim 3.10 in these notes), but rather remarkably, results of Kasin and Figiel, Lindenstrauss, and Milman show that if we consider the norm, we can find Euclidean subspaces of proportional dimension, i.e. .

For any , Cauchy-Schwarz gives us

To this end, for a subspace , let’s define

which measures how well-spread the -mass is among the coordinates of vectors . For instance, if , then every non-zero has at least non-zero coordinates (because ). This has an obvious analogy with the setting of linear error-correcting codes, where one would like the same property from the kernel of the check matrix. Over the next few posts, I’ll show how such subspaces actually give rise to error-correcting codes over that always have efficient decoding algorithms.

The Kasin and Figiel-Lindenstrauss-Milman results actually describe two different regimes. Kasin shows that for every , there exists a subspace with , and . The FLM result shows that for every , one can get and . In coding terminology, the former approach maximizes the rate of the code, while the latter maximizes the minimum distance. In this post, we will be more interested in Kasin’s “nearly full dimensional” setting, since this has the most relevance to sensing and coding. Work of Indyk (see also this) shows that the “nearly isometric” setting is useful for applications in high-dimensional nearest-neighbor search.

The search for explicit constructions

Both approaches show that the bounds hold for a random subspace of the proper dimension (where “random” can mean different things;we’ll see one example below). In light of this, many authors have asked whether there are explicit constructions of such subspaces exist, see e.g. Johnson and Schechtman (Section 2.2), Milman (Problem 8), and Szarek (Section 4). As I’ll discuss in future posts, this would also yield explicit constructions of codes over the reals, and of compressed sensing matrices.

Depending on the circles you run in, “explicit” can mean different things. Let’s fix a TCS-style definition: An explicit construction means a deterministic algorithm which, give as input, outputs a description of a subspace (e.g. in the form of a set of basis vectors), and runs in time . Fortunately, the constructions I’ll discuss later will satisfy just about everyone’s sense of explicitness.

In light of the difficulty of obtaining explicit constructions, people have started looking at weaker results and partial derandomizations. One starting point is Kasin’s method of proof: He shows, for instance, that if one chooses a uniformly random sign matrix (i.e. one whose entries are chosen independently and uniformly from ), then with high probability, one has . Of course this bears a strong resemblance to random parity check codes. Clearly we also get .

In work of Guruswami, myself, and Razborov, we show that there exist explicit sign matrices for which (recall that the trivial bound is ). Our approach is inspired by the construction and analysis of expander codes. Preceding our work, Artstein-Avidan and Milman pursued a different direction, by asking whether one can reduce the dependence on the number of random bits from the trivial bound (and still achieve ). Using random walks on expander graphs, they showed that one only needs random bits to construct such a subspace. Lovett and Sodin later reduced this dependency to . (Indyk’s approach based on Nisan’s pseudorandom generator can also be used to get .) In upcoming work with Guruswami and Wigderson, we show that the dependence can be reduced to for any .

Kernels of random sign matrices

It makes sense to end this discussion with an analysis of the random case. We will prove that if is a random sign matrix (assume for simplicity that is even), then with high probability . It might help first to recall why a uniformly random matrix with entries is almost surely the check matrix of a good linear code.

Let be the finite field of order 2. We would like to show that, with high probability, there does not exist a non-zero vector with hamming weight , and which satisfies (this equation is considered over ). The proof is simple: Let be the rows of . If , then Therefore, for any non-zero , we have

(**)

On the other hand, for small enough, there are fewer than non-zero vectors of hamming weight at most , so a union bound finishes the argument.

The proof that proceeds along similar lines, except that we can no longer take a naive union bound since there are now infinitely many “bad” vectors. Of course the solution is to take a sufficiently fine discretization of the set of bad vectors. This proceeds in three steps:

1. If is far from , then any with small is also far from .
2. Any fixed vector with is far from with very high probability. This is the analog of (**) in the coding setting.
3. There exists a small set of vectors that well-approximates the set of bad vectors.

Combining (1), (2), and (3) we will conclude that every bad vector is far from the kernel of (and, in particular, not contained in the kernel).

To verify (1), we need to show that almost surely the operator norm is small, because if we define

,

then

In order to proceed, we first need a tail bound on random Bernoulli sums, which follows immediately from Hoeffding’s inequality.

Lemma 1: If and is chosen uniformly at random, then

In particular, for and , we have . We conclude that if we take and , then

(4)

observing that , and applying Lemma 1. Now we can prove that is usually not too big.

Theorem 1: If is a random sign matrix, then

.

Proof:

For convenience, let , and let and be -nets on the unit spheres of and , respectively. In other words, for every with , there should exist a point for which , and similarly for . It is well-known that one can choose such nets with and (see, e.g. the book of Milman and Schechtman mentioned earlier).

So applying (4) and a union bound, we have

So let’s assume that no such and exist. Let and be arbitrary vectors with , and write and , where and , where (by choosing successive approximations). Then we have,

.

We conclude by nothing that .

This concludes the verification of (1). In the next post, we’ll see how (2) and (3) are proved.