As part of a summer school associated to the Hausdorff Institute for Mathematics Program on Combinatorial Optimization, I will be giving some lectures on “Semi-definite extended formulations and sums of squares.”

I wanted to post here a draft of the lecture notes. These extend and complete the series of posts here on non-negative and psd rank and lifts of polytopes. They also incorporate many corrections, and have exercises of varying levels of difficulty. The bibliographic references are sparse at the moment because I am posting them from somewhere in the Adriatic (where wifi is also sparse).

We will now prove Talagrand’s majorizing measures theorem, showing that the generic chaining bound is tight for Gaussian processes. The proof here will be a bit more long-winded than the proof from Talagrand’s book, but also (I think), a bit more accessible as well. Most importantly, we will highlight the key idea with a simple combinatorial argument.

Theorem 1 Let be a Gaussian process, and let be a sequence of subsets such that and for . Then,

In order to make things slightly easier to work with, we look at an essentially equivalent way to state (1). Consider a Gaussian process and a sequence of increasing partitions of , where increasing means that is a refinement of for . Say that such a sequence is admissible if and for all . Also, for a partition and a point , we will use the notation for the unique set in which contains .

By choosing to be any set of points with one element in each piece of the partition , (1) yields,

We can now state our main theorem, which shows that this is essentially the only way to bound .

Theorem 2 There is a constant such that for any Gaussian process , there exists an admissible sequence which satisfies,

Theorem 3 For some constants and , the following holds. Suppose is a Gaussian process, and let be such that for . Then,

We will use only Theorem 3 and the fact that whenever to prove Theorem 2 (so, in fact, Theorem 2 holds with replaced by more general functionals satisfying an inequality like (4)).

The partitioning scheme

First, we will specify the partitioning scheme to form an admissible sequence , and then we will move on to its analysis. As discussed in earlier posts, we may assume that is finite. Every set will have a value associated with it, such that is always an upper bound on the radius of the set , i.e. there exists a point such that .

Initially, we set and . Now, we assume that we have constructed , and show how to form the partition . To do this, we will break every set into at most pieces. This will ensure that

Let be the constant from Theorem 3. Put , and let . We partition into pieces as follows. First, choose which maximizes the value

Then, set . We put .

A remark: The whole idea here is that we have chosen the “largest possible piece,” (in terms of -value), but we have done this with respect to the ball, while we cut out the ball. The reason for this will not become completely clear until the analysis, but we can offer a short explanation here. Looking at the lower bound (4), observe that the balls are disjoint under the assumptions, but we only get “credit” for the balls. When we apply this lower bound, it seems that we are throwing a lot of the space away. At some point, we will have to make sure that this thrown away part doesn’t have all the interesting stuff! The reason for our choice of vs. is essentially this: We want to guarantee that if we miss the interesting stuff at this level, then the previous level took care of it. To have this be the case, we will have to look forward (a level down), which (sort of) explains our choice of optimizing for the ball.

Now we continue in this fashion. Let be the remaining space after we have cut out pieces. For , choose to maximize the value

For , set , and put .

So far, we have been chopping the space into smaller pieces. If for some , we have finished our construction of . But maybe we have already chopped out pieces, and still some remains. In that case, we put , i.e. we throw everything else into . Since we cannot reduce our estimate on the radius, we also put .

We continue this process until is exhausted, i.e. eventually for some large enough, only contains singletons. This completes our description of the partitioning.

The tree

For the analysis, it will help to consider our partitioning process as having constructed a tree (in the most natural way). The root of the the tree is the set , and its children are the sets of , and so on. Let’s call this tree . It will help to draw and describe in a specific way. First, we will assign values to the edges of the tree. If and is a child of (i.e., and ), then the edge is given value:

If we define the value of a root-leaf path in as the sum of the edge lengths on that path, then for any ,

simply using .

Thus in order to prove Theorem 2, which states that for some ,

it will suffice to show that for some (other) constant , for any root-leaf path in , we have

Before doing this, we will fix a convention for drawing parts of . If a node has children , we will draw them from left to right. We will call an edge a right turn and every other edge will be referred to as a left turn. Note that some node may not have any right turn coming out of it (if the partitioning finished before the last step). Also, observe that along a left turn, the radius always drops by a factor of , while along a right turn, it remains the same.

We now make two observations about computing the value up to a universal constant.

Observation (1): In computing the value of a root-leaf path , we only need to consider right turns.

To see this, suppose that we have a right turn followed by a consecutive sequence of left turns. If the value of the right turn is , then the value of the following sequence of left turns is, in total, at most

In other words, because the radius decreases by a factor of along every left turn, their values decrease geometrically, making the whole sum comparable to the preceding right turn. (Recall that , so indeed the sum is geometric.)

If the problem of possibly of having no right turn in the path bothers you, note that we could artificially add an initial right turn into the root with value . This is justified since always holds. A different way of saying this is that if the path really contained no right turn, then its value is , and we can easily prove (6).

Observation (2): In computing the value of a root-leaf path , we need only consider the last right turn in any consecutive sequence of right turns.

Consider a sequence of consecutive right turns, and the fact that the radius does not decrease. The values (taking away the factor) look like . In other words, they are geometrically increasing, and thus using only the last right turn in every sequence, we only lose a constant factor.

We will abbreviate last right turn to LRT, and write to denote the value of , just counting last right turns. By the two observations, to show (6) (and hence finish the proof), it suffices to show that, for every root-leaf path in ,

The analysis

Recall that our tree has values on the edges, defined in (5). We will also put some natural values on the nodes. For a node (which, recall, is just a subset ), we put . So the edges have values and the nodes have values. Thus given any subset of nodes and edges in , we can talk about the value of the subset, which will be the sum of the values of the objects it contains. We will prove (7) by a sequence of inequalities on subsets.

Fix a root-leaf path , for which we will prove (7). Let’s prove the fundamental inequality now. We will consider two consecutive LRTs along . (If there is only one LRT in , then we are done by the preceding remarks.) See the figure below. The dashed lines represent a (possibly empty) sequence of left turns and then right turns. The two LRTs are marked.

We will prove the following inequality, which is the heart of the proof. One should understand that the inequality is on the values of the subsets marked in red. The first subset contains two nodes, and the second contains two nodes and an edge.

With this inequality proved, the proof is complete. Let’s see why. We start with the first LRT. Since for any node in , we have the inequality:

This gets us started. Now we apply the inequality of Figure A repeatedly to each pair of consecutive LRTs in the path . What do we have when we’ve exhausted the path ? Well, precisely all the LRTs in are marked, yielding , as desired.

The LRT inequality

Now we are left to prove the inequality in Figure A. First, let’s label some of the nodes. Let , and suppose that . The purple values are not the radii of the corresponding nodes, but they are upper bounds on the radii, recalling that along every left turn, the radius decreases by a factor of . Since there are at least two left turns in the picture, we get a upper bound on the radius of .

Part of the inequality is easy: We have since . So we can transfer the red mark from to . We are thus left to prove that

This will allow us to transfer the red mark from to the LRT coming out of and to .

When was partitioned into pieces, this was by our greedy partitioning algorithm using centers . Since we cut out the ball around each center, we have for all . Applying the Sudakov inequality (Theorem 3), we have

where the last line follows from the greedy manner in which the ‘s were chosen.

But now we claim that

This follows from two facts. First, (since actually). Secondly, the radius of is at most ! But was chosen to maximize the value of over all balls of radius , so in particular its -value is at least that of the ball containing .

Combining (9) and the preceding inequality, we prove (8), and thus that the inequality of Figure A is valid. This completes the proof.

In order to prove that the chaining argument is tight, we will need some additional properties of Gaussian processes. For the chaining upper bound, we used a series of union bounds specified by a tree structure. As a first step in producing a good lower bound, we will look at a way in which the union bound is tight.

Theorem 1 (Sudakov inequality) For some constant , the following holds. Let be a Gaussian process such that for every distinct , we have . Then,

The claim is an elementary calculation for a sequence of i.i.d. random variables (i.e. ). We will reduce the general case to this one using Slepian’s comparison lemma.

Lemma 2 (Slepian’s Lemma) Let and be two Gaussian processes such that for all ,

Then .

There is a fairly elementary proof of Slepian’s Lemma (see, e.g. the Ledoux-Talagrand book), if one is satisfied with the weaker conclusion , which suffices for our purposes.

To see that Lemma 2 yields Theorem 1, take a family with for all and consider the associated variables where is a family of i.i.d. random variables. It is straightforward to verify that (1) holds, hence by the lemma, , and the result follows from the i.i.d. case.

The Sudakov inequality gives us “one level” of a lower bound; the following strengthening will allow us to use it recursively. If we have a Gaussian process and , we will use the notation

For and , we also use the notation

Here is the main theorem of this post; its statement is all we will require for our proof of the majorizing measures theorem:

Theorem 3 For some constants and , the following holds. Suppose is a Gaussian process, and let be such that for . Then,

The proof of the preceding theorem relies on the a strong concentration property for Gaussian processes. First, we recall the classical isoperimetric inequality for Gaussian space (see, for instance, (2.9) here).
We remind the reader that for a function ,

Theorem 4 Let , and let , where is the standard -dimensional Gaussian measure. Then,

Using this, we can prove the following remarkable fact.

Theorem 5 Let be a Gaussian process, then

A notable aspect of this statement is that only the maximum variance affects the concentration, not the number of random variables. We now prove Theorem 5 using Theorem 4.

Proof: We will prove it in the case , but of course our bound is independent of . The idea is that given a Gaussian process , we can write

for , where are standard i.i.d. normals, and the matrix is a matrix of real coefficients. In this case, if is a standard -dimensional Gaussian, then the vector is distributed as .

If we put , then Theorem 4 yields (3) as long as . It is easy to see that

But is just the maximum norm of any row of , and the norm of row is

Using this theorem, we are ready to prove Theorem 3. I will only give a sketch here, but filling in the details is not too difficult.

Assume that the conditions of Theorem 3 hold. Pick an arbitrary , and recall that we can write

so we could hope that for some , we simultaneously have , yielding

The problem, of course, is that the events we are discussing are not independent.

This is where Theorem 5 comes in. For any , all the variances of the variables are bounded by . This implies that we can choose a constant such that

So, in fact, we can expect that none of the random variables will deviate from its expected value by more than . Which means we can (morally) replace (4) by

When it is impossible to achieve a low-distortion embedding of some space into another space , we can consider more lenient kinds of mappings which are still suitable for many applications. For example, consider the unweighted -cycle . It is known that any embedding of into a tree metric incurs distortion . On the other hand, if we delete a uniformly random edge of , this leaves us with a random tree (actually a path) such that for any , we have

In other words, in expectation the distortion is only 2.

Let be a finite metric space, and let be a family of finite metric spaces. A stochastic embedding from into is a random pair where and is a non-contractive mapping, i.e. such that for all . The distortion of is defined by

Theorem 1 Every -point metric space admits a stochastic embedding into the family of tree metrics, with distortion .

We will need the random partitioning theorem we proved last time:

Theorem 2 For every , there is a -bounded random partition of which satisfies, for every and ,

From partitions to trees. Before proving the theorem, we discuss a general way of constructing a tree metric from a sequence of partitions. Assume (by perhaps scaling the metric first) that for all with . Let be partitions of , where is -bounded. We will assume that and is a partition of into singleton sets.

Now we inductively construct a tree metric as follows. The nodes of the tree will be of the form for and . The root is . In general, if the tree has a node of the form for , then will have children

The length of an edge of the form is . This specifies the entire tree . We also specify a map by . We leave the following claim to the reader.

Claim 1 For every ,

where is the largest index with .

Note, in particular, that is non-contracting because if for some , then since is -bounded, implying that .

Proof: Again, assume that for all . For , let be the -bounded random partition guaranteed by Theorem 2. Let be a partition into singletons, and let . Finally, let be the tree constructed above, and let be the corresponding (random) non-contractive mapping.

Now, fix and an integer such that . Using Claim 1 and (1), we have,

where in the penultimate line, we have evaluated three disjoint telescoping sums: For any numbers ,

Since every tree embeds isometrically into , this offers an alternate proof of Bourgain’s theorem when the target space is . Since we know that expander graphs require distortion into , this also shows that Theorem 1 is asymptotically tight.

In addition to the current sequence on Talagrand’s majorizing measures theory, I’m also going to be putting up a series of lecture notes about embeddings of finite metric spaces. The first few will concern random partitions of metric spaces, and their many applications.

Random partitions

Often when one is confronted with the problem of analyzing some problem on a metric space , a natural way to proceed is by divide and conquer: Break the space into small pieces, do something (perhaps recursively) on each piece, and then glue these local solutions together to achieve a global result. This is certainly an important theme in areas like differential geometry, harmonic analysis, and computational geometry.

In the metric setting, “small” often means pieces of small diameter. Of course we could just break the space into singletons , but this ignores the second important aspect in a “divide and conquer” approach: what goes on at the boundary. If we are going to combine our local solutions together effectively, we want the “boundary” between distinct pieces of the partition to be small. Formalizing this idea in various ways leads to a number of interesting ideas which I’ll explore in a sequence of posts.

Example: The unit square

Consider the unit square in the plane, equipped with the metric. If we break the square into pieces of side length , then every piece has diameter at most , and a good fraction of the space is far from the boundary of the partition. In fact, it is easy to see that e.g. 60% of the measure is at least distance from the boundary (the red dotted line).

But there is a problem with using this as a motivating example. First, observe that we had both a metric structure (the distance) and a measure (in this case, the uniform measure on ). In many potential applications, there is not necessarily a natural measure to work with (e.g. for finite metric spaces, where counting the number of points is often a poor way of measuring the “size” of various pieces).

To escape this apparent conundrum, note that the uniform measure is really just a distraction: The same result holds for any measure on . This is a simple application of the probabilistic method: If we uniformly shift the -grid at random, then for any point , we have

Thus, by averaging, for any measure on there exists a partition where 60% of the -measure is -far from the boundary. This leads us to the idea that, in many situations, instead of talking about measures, it is better to talk about distributions over partitions. In this case, we want the “boundary” to be small on “average.”

Lipschitz random partitions

We will work primarily with finite metric spaces to avoid having to deal with continuous probability spaces; much of the theory carries over to general metric spaces without significant effort (but the development becomes less clean, as this requires certain measurability assumptions in the theorem statements).

Let be a finite metric space. If is a partition of , and , we will write for the unique set in containing . We say that is -bounded if for all . We will also say that a random partition is -bounded if it is supported only on -bounded partitions of .

Let’s now look at one way to formalize the idea that the “boundary” of a random partition is small.

A random partition of is -Lipschitz if, for every ,

Intuitively, the boundary is small if nearby points tend to end up in the same piece of the partition. There is a tradeoff between a random partition being -bounded and -Lipschitz. As increases, we expect that we can make , and hence the “boundary effect,” smaller and smaller. The following theorem can be derived from work of Leighton and Rao, or Linial and Saks. The form stated below comes from work of Bartal.

Theorem 1
If is an -point metric space and , then for every , there is an -Lipschitz, -bounded random partition of .

We will prove a more general fact that will be essential later. For positive numbers , define (note that ). The next theorem and proof are from Calinescu, Karloff, and Rabani.

Theorem 2
For every , there is a -bounded random partition of which satisfies, for every and ,

Observe that Theorem 1 follows from Theorem 2 by setting , and noting that implies . We also use for .

Proof:
Suppose that . Let be chosen uniformly at random, and let be a uniformly random bijection . We will think of as giving a random ordering of . We define our random partition as follows.

For , define

It is straightforward that is a partition of , with perhaps many of the sets being empty. Furthermore, by construction, is -bounded. Thus we are left to verify (1).

To this end, fix and , and enumerate the points of so that . Let . We will say that a point sees if , and we will say that cuts if .
(In the picture below, and see , while does not. Only cuts .)

Observe that for any ,

Using this language, we can now reveal our analysis strategy: Let be the minimal element (according to the ordering ) which sees . Then can only occur if also cuts . The point is that the fate of is not decided until some point sees . Then, if does not cut , we have , hence .

Thus we can write,

To analyze the latter sum, first note that if , then can never see since always. On the other hand, if then always sees , but can never cut since always.
Recalling that by assumption, and setting and let , we can use (3) to write

Now we come to the heart of the analysis: For any ,

The idea is that if cuts , then . Thus if any for comes before in the -ordering, then also sees since , hence . But the probability that is the -minimal element of is precisely , proving (5).

To finish the proof, we combine (2), (4), and (5), yielding

Tightness for expanders. Before ending this section, let us mention that Theorem 1 is optimal up to a constant factor. Let be a family of degree-3 expander graphs, equipped with their shortest-path metric . Assume that for some and all . Let , and suppose that admits a -bounded -Lipschitz random partition. By averaging, this means there is a fixed -bounded partition of which cuts at most an -fraction of edges. But every has , which implies . Hence the partition must cut an fraction of edges, implying .

In the next post, we’ll see how these random partitions allow us to reduce may questions on finite metric spaces to questions on trees.

In the last post, we considered a Gaussian process and were trying to find upper bounds on the quantity . We saw that one could hope to improve over the union bound by clustering the points and then taking mini union bounds in each cluster.

Hierarchical clustering

To specify a clustering, we’ll take a sequence of progressively finer approximations to our set . First, recall that we fixed , and we have used the observation that .

Now, assume that is finite. Write , and consider a sequence of subsets such that . We will assume that for some large enough , we have for . For every , let denote a “closest point map” which sends to the closest point in .

The main point is that we can now write, for any ,

This decomposition is where the term “chaining” arises, and now the idea is to bound the probability that is large in terms of the segments in the chain.

What should look like?

One question that arises is how we should think about choosing the approximations . We are trading off two measures of quality: The denser is in the set (or, more precisely, in the set ) the smaller the variances of the segments will be. On the other hand, the larger is, the more segments we’ll have to take a union bound over.

So far, we haven’t used any property of our random variables except for the fact that they are centered. To make a more informed decision about how to choose the sets , let’s recall the classical Gaussian concentration bound.

Lemma 1 For every and ,

This should look familiar: is a mean-zero Gaussian with variance .

Now, a first instinct might be to choose the sets to be progressively denser in . In this case, a natural choice would be to insist on something like being a -net in . If one continues down this path in the right way, a similar theory would develop. We’re going to take a different route and consider the other side of the tradeoff.

Instead of insisting that has a certain level of accuracy, we’ll insist that is at most a certain size. Should we require or , or use some other function? To figure out the right bound, we look at (2). Suppose that are i.i.d. random variables. In that case, applying (2) and a union bound, we see that to achieve

we need to select . If we look instead at points instead of points, the bound grows to . Thus we can generally square the number of points before the union bound has to pay a constant factor increase. This suggests that the right scaling is something like . So we’ll require that for all .

The generic chaining

This leads us to the generic chaining bound, due to Fernique (though the formulation we state here is from Talagrand).

Theorem 2 Let be a Gaussian process, and let be a sequence of subsets such that and for . Then,

Proof: As before, let denote the closest point map and let . Using (2), for any , , and , we have

Now, the number of pairs can be bounded by , so we have

Theorem 1.2 gives us a fairly natural way to upper bound the expected supremum using a hierarchical clustering of . Rather amazingly, as we’ll see in the next post, this upper bound is tight. Talagrand’s majorizing measure theorem states that if we take the best choice of in Theorem 1.2, then the upper bound in (3) is within a constant factor of .

It’s most intuitive to start with the geometric viewpoint, in which case an -simplex is defined to be the convex hull of affinely independent points in .Â These points are called the vertices of the simplex.Â Here are examples for

A simplicial complexis then a collection of simplices glued together along lower-dimensional simplices.Â More formally, if is a (geometric) simplex, then a face of is a subset formed by taking the convex hull of a subset of the vertices of .

Finally, a (geometric) simplicial complex is a collection of simplices such that

If and is a face of , then , and

If and , then is a face of both and .

Property (1) gives us downward closure, and property (2) specifies how simplices can be glued together (only along faces).Â For instance, the first picture depicts a simplicial complex.Â The second does not.