1. Introduction

The Klein-Plotkin-Rao (KPR) Theorem is a powerful statement about the geometry of planar graphs and their generalizations. Here, I’ll present a new, very simple proof of the theorem that was discovered in joint work with Cyrus Rashtchian. (This will appear in a preprint soon, together with some new results.) In the next post, I’ll give some applications in geometry and algorithms.

Recall that a graph is planar if it can be drawn in the plane without any edge crossings. Wagner’s theorem gives an intrinsic characterization of planar graphs in terms of excluded minors. Recall that a graph is a minor of a graph if can be obtained from by a sequence of (i) edge and vertex deletions and (ii) contraction of edges. A graph excludes as a minor if is not a minor of . Kuratowki’s theorem states that planar graphs are precisely those which exclude both and as minors, where we use and to denote the complete graph on vertices and the complete -by- bipartite graph, respectively. In this post, we are particularly concerned with -minor-free graphs, i.e. those which exclude as a minor for some .

I’ll first state and prove a simpler version of the KPR theorem. In the next post, I’ll discuss a stronger statement (in the language of random partitions) that follows directly from the proof. Then using these partitions, we will show that the observable diameter of every -minor-free graph is “large,” and use that fact to prove an upper bound on the uniform multi-commodity flow gap in such graphs.

2. Low-diameter graph partitioning

Consider a finite graph equipped with its shortest-path metric (much of what we say here extends to infinite graphs). For now, all the edges of will have length one, although we will generalize to arbitrary weighted graphs for some applications in the next post.

Given a subset , we write A weak but simpler version of the KPR Theorem can be stated as follows.

Theorem 1 Let be a graph that excludes as a minor. Then for every number , there exists a partition such that for every and at most an -fraction of edges of go between different sets in the partition.

The theorem was originally proved with a dependence of , but this was improved to by Fakcharoenphol and Talwar. Today I will prove the bound. The partitioning will be accomplished via an iterative operation which we will call chopping.

Consider any connected graph and a number . We will describe an operation which we call a -chop of . Let be any node of , which we will call the “root node” of the chop, and let (the “initial offset”).

Figure 1

The chopping operation is as follows: We partition , where , and the rest of the sets are the disjoint annuli,

for . See Figure 1 for an example of these cuts (the red and blue alternating regions) on a grid graph. Of course, since is finite, eventually the annuli are empty.

We define a -chop on a possibly disconnected graph as the partition arising from doing a -chop on each of its connected components. Finally, we define a -chop on a sequence of disjoint sets as the result of doing a -chop on each of the induced graphs for . Thus if we have an initial partition of , then a -chop of produces a refined partition of . See Figure 2 for the result of two iterated -chops applied to the grid graph. The yellow circles represent the root nodes in the second chop.

Figure 2

We can now state the main technical result needed to prove Theorem 1.

Lemma 2 If excludes as a minor, then for any , any sequence of iterated -chops on results in a partition such that for each .

Observe that refers to the diameter in , the shortest-path metric on . Also, note that we do not constrain the root node or the initial offset of the chops. Klein, Plotkin, and Rao prove this lemma with a dependence of on the diameter. FT use a more complicated approach.

To see how Lemma 2 implies Theorem 1, one proceeds as follows. First, let be large enough so that setting in Lemma 2 yields a partition into sets of diameter at most . After fixing the root node for a -chop of , one can consider the initial offsets . An edge can be cut (i.e. and end up in distinct sets of the partition) in at most one of these offsets. Thus there exists an offset that cuts only a -fraction of edges. Since we perform iterated -chops, there exists a choice of initial offsets that cuts at most -fraction of edges. That completes the reduction.

3. A sketch

Before moving onto the formal argument, I’ll present a simple sketch that contains the main ideas. The proof is by contradiction; if we perform a sequence of -chops and the diameter of any remaining piece fails to be , then we will construct a minor in .

First, we give an equivalent characterization of when a graph has a graph as a minor: There exist disjoint connected subsets , one corresponding to each vertex of . We call these supernodes. Furthermore, there should be an edge between supernodes whenever there is an edge between the corresponding vertices in .

Figure 3

Now, the proof is by induction. Note that the base case is trivial since a minor is a single vertex. By induction, we can assume that if a sequence of chops fails, then there must be a minor contained in some offending annulus. See Figure 3. If we could ensure that every supernode of the minor touched the upper boundary of the annulus as in Figure 3, we could easily construct a minor and be done, by simplying choosing the -st supernode to be a ball around .

Thus we need to enforce this extra property of our minor. The (very simple) idea is contained in Figure 4.

Figure 4

After finding a minor that intersects the annuli, we extend the supernodes to touch the upper boundary of the annulus from the preceding chop (which is represented by the purple line in the picture). The point is that we can choose these paths to be contained above the red boundary (and thus disjoint from the supernodes), and also each of length at most since the width of the “purple” annulus is only . The same can be done for . (If we didn’t care that the paths have to be above the red boundary, we could choose them of length only .)

The only issue is that we need these new paths to be disjoint. Since the paths are always short (length at most ), we can enforce this by making sure that each supernode contains a representative and these representatives are pairwise far apart; then we grow the paths from the representatives. Initially, the representatives will be apart, and then as we go up the inductive chain, they will get closer by at most at every step. By choosing the initial separation large enough, they will remain disjoint. That’s the sketch; it should be possible to reproduce the proof from the sketch alone, but we now present a more formal proof.

4. The proof

We need a couple definitions. First, given a subset of vertices and a number , we say that a set is -dense in if every element of can reach an element of by a path of length that is contained completely in (the induced graph on ). Second, we say that an -minor is -represented by if every supernode of contains a representative from and these representatives are pairwise distance more than apart in the metric (the global shortest-path metric on ). We now state a lemma that we can prove by induction and implies Lemma 2.

Lemma 3 Let and be any numbers. Suppose is a graph and there is any sequence of iterated -chops on that leaves a component of diameter more than . Then for any set that is -dense in , one can find a -minor that is -represented by .

One can recover Lemma 2 by setting and .

Proof: We proceed by induction on . The case is trivial since a minor is simply a single vertex. Thus we may assume that .

The following figure will be a useful reference.

Figure 5

In general, we argue as follows. Let be any set satisfying the assumptions of the lemma. Assume there is a sequence of iterated -chops that leaves a component of diameter more than . Then there must be some annulus of the first -chop such that iterated -chops on leaves a component of diameter more than .

Suppose the first -chop has root node and initial offset . Let be the set of nodes at distance exactly , i.e. the upper boundary of . Observe that is -dense in by construction. Thus by induction, there is a -minor in that is -represented by . Let be the supernodes of this minor, and let be the representatives which are further than apart.

We now extend this to a -minor which is -represented by . First, we may assume that . Otherwise, all the points of lie in a ball of radius at most , and hence has diameter at most . In particular, we know that for every .

Now for each , we choose a point such that is connected to by a path of length at most in . This can be done by first going up a shortest-path from to of length to reach a point , and then choosing any point of within distance of (which can always be done since is -dense in ). We add this path to to get a new supernode . Observe that the sets are all connected and pairwise disjoint since the new paths are outside the annulus and the paths themselves are pairwise disjoint because they are each of length at most , but they emanate from representatives that are more than apart. In fact, this also shows that the representatives are further than apart, as required.

Finally, we construct a new supernode as follows. For each , let be the closest node to , and let be the closest node in to . For each , let be a shortest-path from to without its endpoint , and let be a shortest-path to , including . We now set . We claim that forms a -minor which is -represented by . First, it is clear that since (again, because is -dense in ). For the same reason, the path is disjoint from all the sets .

Thus the only possible obstruction to having a valid -minor is if some path intersects a set for . We now show that this cannot happen. We know that if intersects , then it must have already traveled distance at least away from . But contains a node adjacent to (by construction), which means it continues an additional distance of (the distance between and . This additional distance is also moving away from , implying that intersects , which is impossible. This completes the proof.

5. The dependence on

The best-known lower bound requires the conclusion of Theorem 1 to cut at least an -fraction of edges. (One can take to be an -vertex 3-regular expander graph, which obviously excludes as a minor. Now it is easy to see that for some constant , partitioning into pieces of diameter at most must cut at least an -fraction of edges.) In some special cases, e.g. graphs of genus (which exclude as a minor for some constant ), one can reduce the bound to (see this joint work with Sidiropoulos). This leads to the following open problem.

Open problem: Show that under the assumptions of Theorem 1, one can find a partition that cuts only an -fraction of edges.

A positive resolution would yield an optimal unifom multi-commodity flow/cut gap for -minor-free graphs. (See the next post for details.)

Problematically, the Khot-Vishnoi gap is rather inefficient: To achieve the optimal gap for Unique Games with alphabet size , one needs an instance of size . As far as I know, there is no obstacle to achieving a gap instance where the number of variables is only .

The Walsh-Hadamard code

The Khot-Vishnoi construction is based on the Hadamard code.
(See Section 5.2 here for a complete description.) If we use to denote the Hilbert space of real-valued functions , then the Walsh-Hadamard basis of is the set of functions of the form

where .

Of course, for two such sets , we have the orthogonality relations,

In their construction, the variables are essentially all functions of the form , of which there are , while there are only basis elements which act as the alphabet for the underlying Unique Games instance. This is what leads to the exponential relationship between the number of variables and the label size.

A PSD lifting question

In an effort to improve this dependence, one could start with a much larger set of nearly orthogonal vectors, and then somehow lift them to a higher-dimensional space where they would become orthogonal. In order for the value of the SDP not to blow up, it would be necessary that this map has some kind of Lipschitz property. We are thus led to the following (possibly naïve) question.

Let be the smallest number such that the following holds. (Here, denotes the -dimensional unit sphere and denotes the unit-sphere of .)

There exists a map such that and whenever satisfy , we have .

(Recall that .)

One can show that

by randomly partitioning so that all vectors satisfying end up in different sets of the partition, and then mapping all the points in a set to a different orthogonal vector.

My question is simply: Is a better dependence on possible? Can one rule out that could be independent of ? Note that any solution which randomly maps points to orthogonal vectors must incur such a blowup (this is essentially rounding the SDP to an integral solution).

]]> http://tcsmath.wordpress.com/2011/02/23/psd-lifting-and-unique-games-integrality-gaps/feed/ 2 James http://tcsmath.wordpress.com/2011/01/07/reading-for-a-rainy-day/ http://tcsmath.wordpress.com/2011/01/07/reading-for-a-rainy-day/#comments Fri, 07 Jan 2011 11:16:11 +0000 James Lee http://tcsmath.wordpress.com/?p=1331 ]]> Today it’s raining in both Seattle and Paris. Here are a few things I think are worth reading while the weather clears up.

• Stop collecting coupons. Recently, Batson, Spielman, and Srivastava gave a beautiful sparsification result for graphs: Every graph has a linear-sized spectral sparsifier. Here is one statement of their result (taken from Srivastava’s thesis):
  

  Theorem [BSS]: For every and , the following holds. For every , there are non-negative numbers of which at most are non-zero, and such that for all ,

  

  Assaf Naor has given a very nice account of some recent geometric applications of this idea. These involves breaking the “coupon collecting” barrier from a number of results which were previously based on random sampling. There is still an outstanding open problem left open here: Embedding -dimensional subspaces of into .

• Lecture notes that are lecture notes. Some people write lecture notes like they are preliminary book drafts. These lecture notes by Ryan O’Donnell for an undergraduate course on “Probability and Computing” are like transcribed lectures. They’re conversational, with philosophical asides–a great example of the style.

• Metaphors in systolic geometry. Larry Guth and Nets Katz recently made awesome progress on the Erdos distance problem in the plane. On a different note, here is a great informal survey of Guth on Gromov’s systolic inequality (which itself answers the question—when would an isometric embedding into ever be employed for the usefulness of the ambient space?!)

• An important epsilon. Finally, there is the recent result of Gharan, Saberi, and Singh giving a approximation for the Traveling Salesman Problem in unweighted graphs. I haven’t gotten a chance to digest the paper yet. Here’s hoping someone else will write an overview of the key ideas.

Notes from the program and related courses and talks will be posted at the Metric 2011 blog.

1. Cover times and the Gaussian free field

Consider a finite, connected graph and the simple random walk on (which, at every step, moves from a vertex to a uniformly random neighbor). If we let denote the first (random) time at which every vertex of has been visited, and we use to denote expectation over the random walk started at , then the cover time of is defined by

On the other hand, consider the following Gaussian process , called the (discrete) Gaussian free field (GFF) on . Such a process is specified uniquely by its covariance structure. Fix some , and put . Then the rest of the structure is specified by the relation

for all , where the effective resistance between and when is thought of as an electrical network. Equivalently, the density of is proportional to

where the sum is over edges of .
One of our main theorems can be stated as follows.

Theorem 1 (Ding-L-Peres) For every graph , one has

where denotes the Gaussian free field on .

GFF on the 2D lattice; courtesy of Scott Sheffield.

Here is the assertion that and are within universal constant factors. In particular, we use this theorem to give a deterministic -approximation to the cover time, answering a question of Aldous and Fill, and to resolve the Winkler-Zuckerman blanket time conjectures.
This follows some partial progress on these questions by Kahn, Kim, Lovasz, and Vu (1999).

2. Derandomizing the cover time

The cover time is one of the few basic graph parameters which can easily be computed by a randomized polynomial-time algorithm, but for which we don’t know of a deterministic counterpart. More precisely, for every , we can compute a -approximation to the cover time by simulating the random walk enough times and taking the median estimate, but even given the above results, the best we can do in deterministic polynomial-time is an -approximation.

We now describe one conjectural path to a better derandomization. Let denote the maximal hitting time in , where is the expected number of steps needed to hit from a random walk started at . We prove the following more precise estimate.

Theorem 2 There is a constant such that for every graph ,

This prompts the following conjecture, which describes a potentially deeper connection between cover times and the GFF.

Conjecture 1 For a sequence of graphs with ,

where denotes the GFF on .

Here, we use to denote . The conjecture holds in some interesting cases, including the complete graph, the discrete torus, and complete -ary trees. (See the full paper for details; except for the complete graph, these exact estimates are entire papers in themselves.)

Since the proof of Theorem 1 makes heavy use of the Fernique-Talagrand majorizing measures theory for estimating , and this theory is bound to lose a large multiplicative constant, it seems that new techniques will be needed in order to resolve the conjecture. In particular, using the isomorphism theory discussed below, it seems that understanding the structure of the near-maxima, i.e. those points which achieve , will be an essential part of any study.

The second part of such a derandomization strategy is the ability to compute a deterministic -approximation to .

Question 1 For every , is there a deterministic polynomial-time algorithm that, given a finite set of points , computes a -approximation to the value

where is a standard -dimensional Gaussian? Is this possible if we know that has the covariance structure of a Gaussian free field?

3. Understanding the Dynkin Isomorphism Theory

Besides majorizing measures, another major tool used in our work is the theory of isomorphisms between Markov processes and Gaussian processes. We now switch to considering the continuous-time random walk on a graph . This makes the same transitions as the simple discrete-time walk, but now spends an exponential (with mean one) amount of time at every vertex. We define the local time at at time by

when we have run the random walk for time .

Work of Ray and Knight in the 1960′s characterized the local times of Brownian motion, and then in 1980, Dynkin described a general connection between the local times of Markov processes and associated Gaussian processes. The version we use is due to Eisenbaum, Kaspi, Marcus, Rosen, and Shi (2000).

Theorem 3 Let and fix some , which is the origin of the random walk. Let be given, and define the (random) time . If is the GFF with , then

Note that on the left hand side, the local times and the GFF are independent. This remarkable theorem (and many others like it) are proved in the book of Marcus and Rosen. In the introduction, the authors describe the “wonderful, mysterious isomorphism theorems.” They continue,

Another confession we must make is that we do not really understand the actual relationship between local times… and their associated Gaussian processes. If one asks us, as is often the case during lectures, to give an intuitive description… we must answer that we cannot. We leave this extremely interesting question to you.

So I will now pass the question along. The proof of the isomorphism theorems proceeds by taking Laplace transforms and then doing some involved combinatorics. It’s analogous to the situation in enumerative combinatorics where we have a generating function proof of some equality, but not a bijective proof where you can really get your hands on what’s happening.

What is the simplest isomorphism-esque statement which has no intuitive proof? The following lemma is used in the proof of the original Ray-Knight theorem on Brownian motion (see Lemma 6.32 in the Morters-Peres book). It can be proved in a few lines using Laplace transforms, yet one suspects there should be an explicit coupling.

Lemma 4 For any , the following holds. Suppose that is a standard normal, are i.i.d. standard exponentials, and is Poisson with parameter . If all these random variables are independent, then

Amazing! The last open question is to explain this equality of distributions in a satisfactory manner, as a first step to understanding what’s really going on in the isomorphism theory.

]]> http://tcsmath.wordpress.com/2010/12/09/open-question-cover-times-and-the-gaussian-free-field/feed/ 3 James GFF http://tcsmath.wordpress.com/2010/07/18/the-majorizing-measures-theorem/ http://tcsmath.wordpress.com/2010/07/18/the-majorizing-measures-theorem/#comments Sun, 18 Jul 2010 11:05:00 +0000 James Lee http://tcsmath.wordpress.com/?p=1257 ]]> 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.

First, let’s recall the bound we proved earlier.

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,

Recall that for a subset , we defined , and in the last post, we proved the following “Sudakov inequality.”

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:

where and are the constants from Theorem 3.

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.

Figure A.

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.

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

since our gaussians are centered.

Now, by Theorem 1,

Suppose that achieves this. By definition,

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

But now by choosing , the error term is absorbed.

The capacity of quantum channels

A mixed quantum state on is represented by a density matrix , which is Hermitian, positive semi-definite matrix with . The von Neumann entropy of is given by

If is the space of matrices with complex entries, then a quantum channel is a mapping

for some , which sends PSD matrices to PSD matrices and preserves the trace. Finally, the minimal output entropy of a quantum channel is

where ranges over all mixed states on . Since the entropy is a concave function, this minimum is achieved at a pure state.

Work of Shor showed that the following conjecture is equivalent to a number of long-standing problems in quantum information theory.

Additivity conjecture for minimal output von Neumann entropy: Is is true that, for every two quantum channels and , we have:

Observe that we always have above, by considering the product state where and are minimizers for and , respectively. The conjecture is about whether we can get a smaller output entropy by entangling the the inputs to .

Hastings’ counterexample and Dvoretzky’s theorem

Hastings proved that the conjecture is false: By appropriately choosing randomly constructed quantum channels, it is possible to achieve

Remarkably, it has recently been shown by Aubrun, Szarek, and Werner that this counterexample follows from an appropriate version of Dvoretzky’s theorem on almost-Euclidean subspaces of normed spaces (for a refresher, see this earlier post on pseudorandom subspaces).

Let be the space of matrices with complex entries. We will consider two norms on matrices : The Hilbert-Schmidt norm , and the Schatten -norm , where . This latter quantity is precisely the norm of the singular values of . Dvoretzky’s theorem tells us (qualitatively) that an appropriate random subspace of the normed space will be nearly Euclidean.

More precisely, for every , a uniformly random subspace of dimension will, with high probability, satisfy: For every ,

i.e. it will be -Euclidean. This strong quantitative version follows along the lines of enhancements to Milman’s proof of Dvoretzky’s theorem.

Oscillations and chaining

The proof of the preceding quantitative bound is based on a version of Dvoretzky’s theorem for Lipschitz functions, which I will state here in the real case for simplicity. We will use for the -dimensional unit sphere, and then for a function and a subset , we write

where is the average value of on .

Theorem 1 If is a 1-Lipschitz function, then for every , if is a random subspace of dimension , then with high probability,

A standard way to prove this lemma would be to use the fact that every -Lipschitz function is highly concentrated on (Levy’s lemma) and then to take a union bound over an -net on whose size is bounded by . Unfortunately, this leads to an extra term on the right-hand side of (1) which, remarkably, would lead to a version of Dvoretzky’s theorem that is too weak to imply Hastings’ counterexample to the additivity conjecture.

Gordon originally proved Theorem 1 using comparison results for Gaussian processes. (In fact, we will see a main technical step called Slepian’s Lemma in our next post on majorizing measures.) Alternately, Schechtman showed that one can use concentration of measure and a more sophisticated chaining argument, of the type discussed in our previous post.

Later, Arora extended this to work in for every fixed . What properties of are really needed to get such an algorithm?

Certainly a key property is that the volume of a ball of radius in only grows like . This ensures that one can choose an -net of size at most in a ball of radius , which is essential for using dynamic programming. In my opinion, this leads to the most fascinating problem left open in this area:

Is bounded volume growth the only property needed to get a PTAS?

This would imply that the use of Euclidean geometry in Arora’s algorithm is non-essential. We can state the question formally as follows. Let be a metric space, and let be the smallest number such that for every ,every ball of radius in can be covered by balls of radius . Is there a PTAS for TSP in ? (In other words, the running time should be bounded by a polynomial in whose degree depends only on .)

The problem is open, though there are some partial results by Talwar.

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

We will now argue that every finite metric space admits a surprisingly good stochastic embedding into tree metrics. The next result is from Fakcharoenphol, Rao, and Talwar, following work by Bartal and Alon, Karp, Peleg, and West.

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 .

Now we are ready to prove Theorem 1.

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.

Using Claim 1 and (1), we have for every ,

where in the penultimate line, we have evaluated a telescoping sum, i.e. 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.