# The majorizing measures theorem

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 ${\{X_t\}_{t \in T}}$ be a Gaussian process, and let ${T_0 \subseteq T_1 \subseteq \cdots \subseteq T}$ be a sequence of subsets such that ${|T_0|=1}$ and ${|T_n| \leq 2^{2^{n}}}$ for ${n \geq 1}$. Then,

$\displaystyle \mathop{\mathbb E}\sup_{t \in T} X_t \leq O(1) \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} \, d(t, T_n). \ \ \ \ \ (1)$

In order to make things slightly easier to work with, we look at an essentially equivalent way to state (1). Consider a Gaussian process ${\{X_t\}_{t \in T}}$ and a sequence of increasing partitions ${\{\mathcal A_n\}_{n \geq 0}}$ of ${T}$, where increasing means that ${\mathcal A_{n+1}}$ is a refinement of ${\mathcal A_n}$ for ${n \geq 0}$. Say that such a sequence ${\{\mathcal A_n\}}$ is admissible if ${\mathcal A_0 = \{T\}}$ and ${|\mathcal A_n| \leq 2^{2^n}}$ for all ${n \geq 1}$. Also, for a partition ${P}$ and a point ${t \in T}$, we will use the notation ${P(t)}$ for the unique set in ${P}$ which contains ${t}$.

By choosing ${T_n}$ to be any set of points with one element in each piece of the partition ${\mathcal A_n}$, (1) yields,

$\displaystyle \mathop{\mathbb E}\sup_{t \in T} X_t \leq O(1) \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} \,\mathrm{diam}(\mathcal A_n(t)). \ \ \ \ \ (2)$

We can now state our main theorem, which shows that this is essentially the only way to bound ${\mathop{\mathbb E} \sup_{t \in T} X_t}$.

Theorem 2 There is a constant ${L > 0}$ such that for any Gaussian process ${\{X_t\}_{t \in T}}$, there exists an admissible sequence ${\{\mathcal A_n\}}$ which satisfies,

$\displaystyle \mathop{\mathbb E} \sup_{t \in T} X_t \geq L \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} \,\mathrm{diam}(\mathcal A_n(t)). \ \ \ \ \ (3)$

Recall that for a subset ${A \subseteq T}$, we defined ${g(A) = \mathop{\mathbb E} \sup_{t \in A} X_t}$, and in the last post, we proved the following “Sudakov inequality.”

Theorem 3 For some constants ${\kappa > 0}$ and ${r \geq 4}$, the following holds. Suppose ${\{X_t\}_{t \in T}}$ is a Gaussian process, and let ${t_1, t_2, \ldots, t_m \in T}$ be such that ${d(t_i,t_j) \geq \alpha}$ for ${i \neq j}$. Then,

$\displaystyle g(T) \geq \kappa \alpha \sqrt{\log_2 m} + \min_{i=1,2,\ldots,m} g(B(t_i, \alpha/r)). \ \ \ \ \ (4)$

We will use only Theorem 3 and the fact that ${g(A) \leq g(B)}$ whenever ${A \subseteq B}$ to prove Theorem 2 (so, in fact, Theorem 2 holds with ${\mathop{\mathbb E} \sup_{t \in T} X_t}$ 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 ${\{\mathcal A_n\}}$, and then we will move on to its analysis. As discussed in earlier posts, we may assume that ${T}$ is finite. Every set ${C \in \mathcal A_n}$ will have a value ${\mathrm{rad}(C)}$ associated with it, such that ${\mathrm{rad}(C)}$ is always an upper bound on the radius of the set ${C}$, i.e. there exists a point ${x \in C}$ such that ${C \subseteq B(x, \mathrm{rad}(C))}$.

Initially, we set ${\mathcal A_0 = \{T\}}$ and ${\mathrm{rad}(T) = \mathrm{diam}(T)}$. Now, we assume that we have constructed ${\mathcal A_n}$, and show how to form the partition ${\mathcal A_{n+1}}$. To do this, we will break every set ${C \in \mathcal A_n}$ into at most ${2^{2^n}}$ pieces. This will ensure that

$\displaystyle |\mathcal A_{n+1}| \leq 2^{2^n} \cdot |\mathcal A_n| \leq 2^{2^n} \cdot 2^{2^n} = 2^{2^{n+1}}.$

Let ${r}$ be the constant from Theorem 3. Put ${m = 2^{2^n}}$, and let ${\Delta = \mathrm{rad}(C)}$. We partition ${C}$ into ${m}$ pieces as follows. First, choose ${t_1 \in C}$ which maximizes the value

$\displaystyle g(B(t_1, \Delta/r^2) \cap C).$

Then, set ${C_1 = B(t_1, \Delta/r) \cap C}$. We put ${\mathrm{rad}(C_1) = \Delta/r}$.

A remark: The whole idea here is that we have chosen the “largest possible piece,” (in terms of ${g}$-value), but we have done this with respect to the ${\Delta/r^2}$ ball, while we cut out the ${\Delta/r}$ 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 ${B(t_i, \alpha/3)}$ are disjoint under the assumptions, but we only get “credit” for the ${B(t_i, \alpha/r)}$ 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 ${\Delta/r}$ vs. ${\Delta/r^2}$ 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 ${\Delta/r^2}$ ball.

Now we continue in this fashion. Let ${D_{\ell} = C \setminus \bigcup_{i=1}^{\ell-1} C_i}$ be the remaining space after we have cut out ${\ell-1}$ pieces. For ${\ell \leq m}$, choose ${t_{\ell} \in C}$ to maximize the value

$\displaystyle g(B(t_{\ell}, \Delta/r^2) \cap D_{\ell}).$

For ${\ell < m}$, set ${C_{\ell} = B(t_{\ell}, \Delta/r) \cap D_{\ell}}$, and put ${\mathrm{rad}(C_{\ell}) = \Delta/r}$.

So far, we have been chopping the space into smaller pieces. If ${D_{\ell} = \emptyset}$ for some ${\ell \leq m}$, we have finished our construction of ${\mathcal A_{n+1}}$. But maybe we have already chopped out ${m-1}$ pieces, and still some remains. In that case, we put ${C_m = D_m}$, i.e. we throw everything else into ${C_m}$. Since we cannot reduce our estimate on the radius, we also put ${\mathrm{rad}(C_m) = \Delta}$.

We continue this process until ${T}$ is exhausted, i.e. eventually for some ${n}$ large enough, ${\mathcal A_n}$ 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 ${T}$, and its children are the sets of ${\mathcal A_1}$, and so on. Let’s call this tree ${\mathcal W}$. It will help to draw and describe ${\mathcal W}$ in a specific way. First, we will assign values to the edges of the tree. If ${C \in \mathcal A_n}$ and ${C_j}$ is a child of ${C}$ (i.e., ${C_j \in \mathcal A_{n+1}}$ and ${C_j \subseteq C}$), then the edge ${(C,C_j)}$ is given value:

$\displaystyle \kappa \cdot \frac{\mathrm{rad}(C)}{r} \cdot 2^{n/2}, \ \ \ \ \ (5)$

where ${\kappa}$ and ${r}$ are the constants from Theorem 3.

If we define the value of a root-leaf path in ${\mathcal W}$ as the sum of the edge lengths on that path, then for any ${t \in T}$,

$\displaystyle \sum_{n \geq 0} 2^{n/2}\,\mathrm{diam}(\mathcal A_n(t)) \leq 2 \frac{r}{\kappa} \left(\textrm{value of the path from the root to } t\right),$

simply using ${\mathrm{diam}(\mathcal A_n(t)) \leq 2\, \mathrm{rad}(\mathcal A_n(t))}$.

Thus in order to prove Theorem 2, which states that for some ${L > 0}$,

$\displaystyle g(T) \geq L \sup_{t \in T} \sum_{n \geq 0} 2^{n/2}\,\mathrm{diam}(\mathcal A_n(t)),$

it will suffice to show that for some (other) constant ${L > 0}$, for any root-leaf path ${P}$ in ${\mathcal W}$, we have

$\displaystyle g(T) \geq L \cdot \mathrm{value}(P). \ \ \ \ \ (6)$

Before doing this, we will fix a convention for drawing parts of ${\mathcal W}$. If a node ${C \in \mathcal A_n}$ has children ${C_1, C_2, \ldots, C_m}$, we will draw them from left to right. We will call an edge ${(C,C_m)}$ a right turn and every other edge will be referred to as a left turn. Note that some node ${C}$ 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 ${r}$, while along a right turn, it remains the same.

We now make two observations about computing the value ${\mathrm{value}(P)}$ up to a universal constant.

Observation (1): In computing the value of a root-leaf path ${P}$, 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 ${\frac{\kappa}{r} \Delta 2^{n/2}}$, then the value of the following sequence of left turns is, in total, at most

$\displaystyle \frac{\kappa}{r} \sum_{j=1}^{\infty} 2^{(n+j)/2} \frac{\Delta}{r^j} \leq O(1) \frac{\kappa}{r} \Delta 2^{n/2}.$

In other words, because the radius decreases by a factor of ${r}$ along every left turn, their values decrease geometrically, making the whole sum comparable to the preceding right turn. (Recall that ${r \geq 4}$, so indeed the sum is geometric.)

If the problem of possibly of having no right turn in the path ${P}$ bothers you, note that we could artificially add an initial right turn into the root with value ${\mathrm{diam}(T)}$. This is justified since ${g(T) \geq \frac12 \mathrm{diam}(T)}$ always holds. A different way of saying this is that if the path really contained no right turn, then its value is ${O(\mathrm{diam}(T))}$, and we can easily prove (6).

Observation (2): In computing the value of a root-leaf path ${P}$, 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 ${\kappa/r}$ factor) look like ${\Delta 2^{n/2}, \Delta 2^{(n+1)/2}, \Delta 2^{(n+2)/2}, \ldots}$. 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 ${\mathrm{value}_{\mathrm{LRT}}(P)}$ to denote the value of ${P}$, 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 ${P}$ in ${\mathcal W}$,

$\displaystyle 2 \cdot g(T) \geq \mathrm{value}_{\mathrm{LRT}}(P). \ \ \ \ \ (7)$

The analysis

Recall that our tree ${\mathcal W}$ has values on the edges, defined in (5). We will also put some natural values on the nodes. For a node ${C}$ (which, recall, is just a subset ${C \subseteq T}$), we put ${\mathrm{value}(C) = g(C)}$. So the edges have values and the nodes have values. Thus given any subset of nodes and edges in ${\mathcal W}$, 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 ${P}$, for which we will prove (7). Let’s prove the fundamental inequality now. We will consider two consecutive LRTs along ${P}$. (If there is only one LRT in ${P}$, 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 $g(T) \geq g(C)$ for any node $C$ in $\mathcal W$, 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 ${P}$. What do we have when we’ve exhausted the path $P$? Well, precisely all the LRTs in ${P}$ are marked, yielding ${2 \cdot g(T) \geq \mathrm{value}_{\mathrm{LRT}}(P)}$, 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 ${\Delta = \mathrm{rad}(C)}$, and suppose that ${C \in \mathcal A_n}$. 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 ${r}$. Since there are at least two left turns in the picture, we get a ${\Delta/r^2}$ upper bound on the radius of ${J}$.

Part of the inequality is easy: We have ${g(A) \geq g(B)}$ since ${B \subseteq A}$. So we can transfer the red mark from ${A}$ to ${B}$. We are thus left to prove that

$\displaystyle g(C) \geq \frac{\kappa}{r} \Delta 2^{n/2} + g(J). \ \ \ \ \ (8)$

This will allow us to transfer the red mark from ${C}$ to the LRT coming out of ${C}$ and to ${J}$.

When ${C}$ was partitioned into ${m = 2^{2^n}}$ pieces, this was by our greedy partitioning algorithm using centers ${t_1, t_2, \ldots, t_m}$. Since we cut out the ${\Delta/r}$ ball around each center, we have ${d(t_i, t_j) \geq \Delta/r}$ for all ${i \neq j}$. Applying the Sudakov inequality (Theorem 3), we have

$\displaystyle \begin{array}{rl} g(C) &\displaystyle \geq \kappa \frac{\Delta}{r} \sqrt{\log_2 m} + \min_{i=1,\ldots,m} g(B(t_i, \Delta/r^2)) \\ \\ &\displaystyle = \frac{\kappa}{r} \Delta 2^{n/2} + \min_{i=1,\ldots,m} g(B(t_i, \Delta/r^2)) \\ \\ &\displaystyle \geq \frac{\kappa}{r} \Delta 2^{n/2} + \min_{i=1,\ldots,m} g(B(t_i, \Delta/r^2) \cap D_i) \\ \\ &\displaystyle = \frac{\kappa}{r} \Delta 2^{n/2} + g(B(t_m, \Delta/r^2) \cap D_m), \\ \end{array}$

where the last line follows from the greedy manner in which the ${t_i}$‘s were chosen.

But now we claim that

$\displaystyle g(B(t_m, \Delta/r^2) \cap D_m) \geq g(J). \ \ \ \ \ (9)$

This follows from two facts. First, ${J \subseteq D_m}$ (since ${D_m=C_m}$ actually). Secondly, the radius of ${J}$ is at most ${\Delta/r^2}$! But ${t_m}$ was chosen to maximize the value of ${g(B(t_m, \Delta/r^2) \cap D_m)}$ over all balls of radius ${\Delta/r^2}$, so in particular its ${g}$-value is at least that of the ${\Delta/r^2}$ ball containing ${J}$.

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

# Majorizing measures: Gaussian tools

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 $C > 0$, the following holds. Let ${\{X_t\}_{t \in T}}$ be a Gaussian process such that for every distinct ${s,t \in T}$, we have ${d(s,t) \geq \alpha}$. Then,

$\displaystyle \mathop{\mathbb E} \sup_{t \in T} X_t \geq C \alpha \sqrt{\log |T|}.$

The claim is an elementary calculation for a sequence of i.i.d. ${N(0,1)}$ random variables ${g_1, g_2, \ldots, g_n}$ (i.e. ${\mathop{\mathbb E} \sup_i g_i \geq C\sqrt{\log n}}$). We will reduce the general case to this one using Slepian’s comparison lemma.

Lemma 2 (Slepian’s Lemma) Let ${\{X_t\}_{t \in T}}$ and ${\{Y_t\}_{t \in T}}$ be two Gaussian processes such that for all ${s,t \in T}$,

$\displaystyle \mathop{\mathbb E} \,|X_s - X_t|^2 \geq \mathop{\mathbb E} \,|Y_s - Y_t|^2. \ \ \ \ \ (1)$

Then ${\mathop{\mathbb E} \sup_{t \in T} X_t \geq \mathop{\mathbb E} \sup_{t \in T} Y_t}$.

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 ${2\, \mathop{\mathbb E}\,|X_s-X_t|^2 \geq \mathop{\mathbb E}\,|Y_s-Y_t|^2}$, which suffices for our purposes.

To see that Lemma 2 yields Theorem 1, take a family ${\{X_t\}_{t \in T}}$ with ${d(s,t) \geq \alpha}$ for all ${s \neq t \in T}$ and consider the associated variables ${Y_t = \frac{\alpha}{\sqrt{2}} g_t}$ where ${\{g_t\}_{t \in T}}$ is a family of i.i.d. ${N(0,1)}$ random variables. It is straightforward to verify that (1) holds, hence by the lemma, ${\mathop{\mathbb E} \sup_{t \in T} X_t \geq \frac{\alpha}{\sqrt{2}} \mathop{\mathbb E} \sup_{t \in T} g_t}$, 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 ${\{X_t\}_{t \in T}}$ and ${A \subseteq T}$, we will use the notation

$\displaystyle g(A) = \mathop{\mathbb E} \sup_{t \in A} X_t.$

For ${t \in T}$ and ${R \geq 0}$, we also use the notation

$\displaystyle B(t,R) = \{ s \in T : d(s,t) \leq R \}.$

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 $C > 0$ and ${r > 1}$, the following holds. Suppose ${\{X_t\}_{t \in T}}$ is a Gaussian process, and let ${t_1, t_2, \ldots, t_m \in T}$ be such that ${d(t_i,t_j) \geq \alpha}$ for ${i \neq j}$. Then,

$\displaystyle g(T) \geq C \alpha \sqrt{\log m} + \min_{i=1,2,\ldots,m} g(B(t_i, \alpha/r)).$

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 ${F : \mathbb R^n \rightarrow \mathbb R}$,

$\displaystyle \|F\|_{\mathrm{Lip}} = \sup_{x \neq y \in \mathbb R^n} \frac{|F(x)-F(y)|}{\|x-y\|}.$

Theorem 4 Let ${F : \mathbb R^n \rightarrow \mathbb R}$, and let ${\mu = \int F \,d\gamma_n}$, where ${\gamma_n}$ is the standard ${n}$-dimensional Gaussian measure. Then,

$\displaystyle \gamma_n\left(x \in \mathbb R^n : |F(x)-\mu| > \lambda\right) \leq 2\,\exp\left(\frac{-\lambda^2}{2 \|F\|_{\mathrm{Lip}}}\right). \ \ \ \ \ (2)$

Using this, we can prove the following remarkable fact.

Theorem 5 Let ${\{X_t\}_{t \in T}}$ be a Gaussian process, then

$\displaystyle \mathbb P\left(\left|\sup_{t \in T} X_t - \mathop{\mathbb E} \sup_{t \in T} X_t\right| > \lambda\right) \leq 2\,\exp\left(\frac{-\lambda^2}{2 \sup_{t \in T} \mathop{\mathbb E}(X_t^2)}\right). \ \ \ \ \ (3)$

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 ${|T|=n}$, but of course our bound is independent of ${n}$. The idea is that given a Gaussian process ${\{X_1, X_2, \ldots, X_n\}}$, we can write

$\displaystyle X_i = a_{i1} \,g_1 + a_{i2}\, g_2 + \cdots + a_{in}\, g_n,$

for ${i=1,2,\ldots, n}$, where ${\{g_i\}_{i=1}^n}$ are standard i.i.d. normals, and the matrix ${A=(a_{i,j})}$ is a matrix of real coefficients. In this case, if ${g = (g_1, g_2, \ldots, g_n)}$ is a standard ${n}$-dimensional Gaussian, then the vector ${Ag}$ is distributed as ${(X_1, X_2, \ldots, X_n)}$.

If we put ${F(x)=\max \{ (Ax)_i : i=1,\ldots,n\}}$, then Theorem 4 yields (3) as long as ${\|F\|_{\mathrm{Lip}} \leq \max_i \sqrt{\mathop{\mathbb E}(X_i^2)}}$. It is easy to see that

$\displaystyle \|F\|_{\mathrm{Lip}} = \|A\|_{2 \rightarrow \infty} = \sup_{\|x\|_2 = 1} \|A x\|_{\infty}.$

But ${\|A\|_{2 \rightarrow \infty}}$ is just the maximum ${\ell_2}$ norm of any row of ${A}$, and the ${\ell_2}$ norm of row ${i}$ is

$\displaystyle \sqrt{\sum_{j=1}^n a_{ij}^2} = \sqrt{\mathop{\mathbb E}(X_i^2)}.$

$\Box$

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 ${t_0 \in T}$, and recall that we can write

$\displaystyle g(T) = \mathop{\mathbb E} \sup_{t \in T} X_t = \mathop{\mathbb E} \sup_{t \in T} (X_t - X_{t_0})$

since our gaussians are centered.

Now, by Theorem 1,

$\displaystyle \mathop{\mathbb E} \max_{i=1,\ldots,m} \left(X_{t_i} - X_{t_0}\right) \geq C \alpha \sqrt{\log m}.$

Suppose that ${t_1}$ achieves this. By definition,

$\displaystyle g(B(t_1, \alpha/r)) = \mathop{\mathbb E} \sup_{t \in B(t_1, \alpha/r)} \left(X_t - X_{t_1}\right),$

so we could hope that for some ${t \in B(t_1,\alpha/r)}$, we simultaneously have ${X_t - X_{t_1} \geq g(B(t_1,\alpha/r))}$, yielding

$\displaystyle X_t - X_{t_0} = (X_t - X_{t_1}) + (X_{t_1} - X_{t_0}) \geq C\alpha \sqrt{\log m} + g(B(t_1, \alpha/r)). \ \ \ \ \ (4)$

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

This is where Theorem 5 comes in. For any ${i}$, all the variances of the variables ${\{X_t - X_{t_i} : t \in B(t_i,\alpha/r)\}}$ are bounded by ${d(t,t_i)^2 \leq (\alpha/r)^2}$. This implies that we can choose a constant ${c_0 > 0}$ such that

$\displaystyle \mathbb P\left(\left|\sup_{t \in B(t_i,\alpha/r)} X_t - g(B(t_i, \alpha/r))\right| > c_0 (\alpha/r) \sqrt{\log m}\right) \leq m^{-2}.$

So, in fact, we can expect that none of the ${m}$ random variables ${\sup_{t \in B(t_i,\alpha/r)} X_t}$ will deviate from its expected value by more than ${c_0 (\alpha/r) \sqrt{\log m}}$. Which means we can (morally) replace (4) by

$\displaystyle \begin{array}{rl} X_t - X_{t_0} &= (X_t - X_{t_1}) + (X_{t_1} - X_{t_0}) \\ &\geq C\alpha \sqrt{\log m} + g(B(t_1, \alpha/r)) - c_0 (\alpha/r) \sqrt{\log m}. \end{array}$

But now by choosing ${r = 2 C c_0}$, the error term is absorbed.

# Random tree embeddings

Random embeddings

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

$\displaystyle d_{C'_n}(x,y) \geq d_{C_n}(x,y) \geq \frac12\, \mathop{\mathbb E}[d_{C'_n}(x,y)].$

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

Let ${(X,d)}$ be a finite metric space, and let ${\mathcal F}$ be a family of finite metric spaces. A stochastic embedding from ${X}$ into ${\mathcal F}$ is a random pair ${(F,Y)}$ where ${Y \in \mathcal F}$ and ${F : X \rightarrow Y}$ is a non-contractive mapping, i.e. such that ${d_Y(F(x),F(y)) \geq d_X(x,y)}$ for all ${x,y \in X}$. The distortion of ${F}$ is defined by

$\displaystyle \max_{x,y \in X} \frac{\mathop{\mathbb E} \left[d_Y(F(x),F(y))\right]}{d_X(x,y)}\,.$

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 ${n}$-point metric space ${(X,d)}$ admits a stochastic embedding into the family of tree metrics, with distortion ${O(\log n)}$.

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

Theorem 2 For every ${\Delta > 0}$, there is a ${\Delta}$-bounded random partition ${\mathcal P}$ of ${X}$ which satisfies, for every ${x \in X}$ and ${0 < r \leq \Delta/8}$,

$\displaystyle \mathop{\mathbb P}\left(B(x,r) \nsubseteq \mathcal P(x)\right) \leq \frac{8r}{\Delta} H\left(|B(x,\Delta/8)|,|B(x,\Delta)|\right). \ \ \ \ \ (1)$

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 ${1 < d(x,y) \leq 2^M}$ for all ${x,y \in X}$ with ${x \neq y}$. Let ${P_0, P_1, \ldots, P_M}$ be partitions of ${X}$, where ${P_k}$ is ${2^k}$-bounded. We will assume that ${P_M = \{X\}}$ and ${P_0}$ is a partition of ${X}$ into singleton sets.

Now we inductively construct a tree metric ${T = T(P_0, P_1, \ldots, P_M)}$ as follows. The nodes of the tree will be of the form ${(k,S)}$ for ${k \in \{0,1,\ldots,M\}}$ and ${S \subseteq X}$. The root is ${(M,X)}$. In general, if the tree has a node of the form ${(j,S)}$ for ${j > 0}$, then ${(j,S)}$ will have children

$\displaystyle \{ (j-1, S \cap T) : T \in P_{j-1} \textrm{ and } S \cap T \neq \emptyset\}.$

The length of an edge of the form ${\{(j,S), (j-1,S')\}}$ is ${2^{j}}$. This specifies the entire tree ${T}$. We also specify a map ${F : X \rightarrow T}$ by ${F(x) = (0,\{x\})}$. We leave the following claim to the reader.

Claim 1 For every ${x,y \in X}$,

$\displaystyle d_{T}(F(x),F(y)) = 2 \sum_{k=1}^{j+1} 2^k = 2^{j+3}-4.$

where ${j \in \{0,1,\ldots,M\}}$ is the largest index with ${P_j(x) \neq P_j(y)}$.

Note, in particular, that ${F : X \rightarrow T}$ is non-contracting because if ${2^j < d(x,y) \leq 2^{j+1}}$ for some ${j \geq 0}$, then ${P_j(x) \neq P_j(y)}$ since ${P_j}$ is ${2^{j}}$-bounded, implying that ${d_{T}(F(x),F(y)) \geq 2^{j+3}-4 \geq 2^{j+1}}$.

Now we are ready to prove Theorem 1.

Proof: Again, assume that ${1 < d(x,y) \leq 2^M}$ for all ${x,y \in X}$. For ${0 < k < M}$, let ${\mathcal P_k}$ be the ${2^k}$-bounded random partition guaranteed by Theorem 2. Let ${\mathcal P_0}$ be a partition into singletons, and let ${\mathcal P_M = \{X\}}$. Finally, let ${\mathcal T=\mathcal T(\mathcal P_0, \ldots, \mathcal P_M)}$ be the tree constructed above, and let ${F : X \rightarrow \mathcal T}$ be the corresponding (random) non-contractive mapping.

Now, fix ${x,y \in X}$ and an integer ${h}$ such that ${2^h < d(x,y) \leq 2^{h+1}}$. Using Claim 1 and (1), we have,

$\displaystyle \mathop{\mathbb E}\left[d_{\mathcal T}(F(x),F(y))\right] \leq \sum_{j=0}^M \mathop{\mathbb P}[\mathcal P_j(x) \neq \mathcal P_j(y)] \cdot 2^{j+3}$

$\displaystyle \leq O(d(x,y)) + \sum_{j=h+3}^M \frac{8\,d(x,y)}{2^j} 2^{j+3} H\left(|B(x, 2^{j-3})|,|B(x,2^{j})|\right)$

$\displaystyle \leq O(d(x,y)) + \sum_{j=h+3}^M \frac{8\,d(x,y)}{2^j} 2^{j+3} H\left(|B(x, 2^{j-3})|,|B(x,2^{j})|\right)$

$\displaystyle \leq O(d(x,y)) + O(1) \,d(x,y) \sum_{j=h+3}^M H\left(|B(x, 2^{j-3})|,|B(x,2^{j})|\right)$

$\displaystyle \leq O(H(1,n))\,d(x,y)$

$\displaystyle \leq O(\log n)\,d(x,y),$

where in the penultimate line, we have evaluated three disjoint telescoping sums: For any numbers ${1 \leq a_1 \leq a_2 \leq \cdots \leq a_k}$,

$\displaystyle H(a_1,a_2)+H(a_2,a_3)+\cdots+H(a_{k-1},a_k)=H(a_1,a_k).$

$\Box$

Since every tree embeds isometrically into ${\ell_1}$, this offers an alternate proof of Bourgain’s theorem when the target space is ${\ell_1}$. Since we know that expander graphs require ${\Omega(\log n)}$ distortion into ${{\ell}_1}$, this also shows that Theorem 1 is asymptotically tight.

# Random partitions of metric spaces

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 ${(X,d)}$, 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 ${X = \{x_1\} \cup \{x_2\} \cup \cdots}$, 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 ${[0,1]^2}$ in the plane, equipped with the ${\ell_1}$ metric. If we break the square into pieces of side length ${\delta/2}$, then every piece has diameter at most ${\delta}$, 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 ${\delta/20}$ from the boundary (the red dotted line).

2

But there is a problem with using this as a motivating example. First, observe that we had both a metric structure (the ${\ell_1}$ distance) and a measure (in this case, the uniform measure on ${[0,1]^2}$). 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 ${[0,1]^2}$. This is a simple application of the probabilistic method: If we uniformly shift the ${\delta/2}$-grid at random, then for any point ${x \in [0,1]^2}$, we have

$\displaystyle {\mathbb P}(x \textrm{ is } \delta/20\textrm{-far from the boundary}) \geq 0.6.$

Thus, by averaging, for any measure ${\mu}$ on ${[0,1]^2}$ there exists a partition where 60% of the ${\mu}$-measure is ${\delta/20}$-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 ${(X,d)}$ be a finite metric space. If ${P}$ is a partition of ${X}$, and ${x \in X}$, we will write ${P(x)}$ for the unique set in ${P}$ containing ${x}$. We say that ${P}$ is ${\Delta}$-bounded if ${\mathrm{diam}(S) \leq \Delta}$ for all ${S \in P}$. We will also say that a random partition ${\mathcal P}$ is ${{\Delta}}$-bounded if it is supported only on ${\Delta}$-bounded partitions of ${X}$.

Let’s now look at one way to formalize the idea that the “boundary” of a random partition ${\mathcal P}$ is small.

A random partition ${\mathcal P}$ of ${X}$ is ${{L}}$-Lipschitz if, for every ${x,y \in X}$,

$\displaystyle \mathbb P\left(\mathcal P(x) \neq \mathcal P(y)\right) \leq L \cdot d(x,y).$

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 ${\mathcal P}$ being ${\Delta}$-bounded and ${{L}}$-Lipschitz. As ${\Delta}$ increases, we expect that we can make ${{L}}$, 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 ${(X,d)}$ is an ${n}$-point metric space and ${n \geq 2}$, then for every ${\Delta > 0}$, there is an ${\frac{O(\log n)}{\Delta}}$-Lipschitz, ${\Delta}$-bounded random partition ${\mathcal P}$ of ${X}$.

We will prove a more general fact that will be essential later. For positive numbers ${a,b}$, define ${H(a,b) = \sum_{i=a+1}^b \frac{1}{i}}$ (note that ${H(a,a)=0}$). The next theorem and proof are from Calinescu, Karloff, and Rabani.

Theorem 2

For every ${\Delta > 0}$, there is a ${\Delta}$-bounded random partition ${\mathcal P}$ of ${X}$ which satisfies, for every ${x \in X}$ and ${0 < r \leq \Delta/8}$,

Observe that Theorem 1 follows from Theorem 2 by setting ${r = d(x,y)}$, and noting that ${\mathcal P(x) \neq \mathcal P(y)}$ implies ${B(x,r) \nsubseteq \mathcal P(x)}$. We also use ${H(1,n) = O(\log n)}$ for ${n \geq 2}$.

Proof:
Suppose that ${|X|=n}$. Let ${\alpha \in [\frac14, \frac12]}$ be chosen uniformly at random, and let ${\pi}$ be a uniformly random bijection ${\pi : \{1,2,\ldots,n\} \rightarrow X}$. We will think of ${\pi}$ as giving a random ordering of ${X}$. We define our random partition ${\mathcal P}$ as follows.

For ${i \geq 1}$, define

$\displaystyle C_i = B(\pi(i), \alpha \Delta) \setminus \bigcup_{j=1}^{i-1} C_j.$

It is straightforward that ${\mathcal P = C_1 \cup C_2 \cup \cdots \cup C_n}$ is a partition of $X$, with perhaps many of the sets ${C_i}$ being empty. Furthermore, by construction, ${\mathcal P}$ is ${\Delta}$-bounded. Thus we are left to verify (1).

To this end, fix ${x \in X}$ and ${r > 0}$, and enumerate the points of ${X = \{y_1, y_2, \ldots, y_n\}}$ so that ${d(x,y_1) \leq d(x,y_2) \leq \cdots \leq d(x,y_n)}$. Let ${B = B(x,r)}$. We will say that a point ${y_i}$ sees ${B}$ if ${d(y_i, x) \leq \alpha \Delta + r}$, and we will say that ${y_i}$ cuts ${B}$ if ${\alpha \Delta - r\leq d(y_i, x) \leq \alpha \Delta + r}$.
(In the picture below, $y_1$ and $y_2$ see $B$, while $y_3$ does not. Only $y_2$ cuts $B$.)

Observe that for any ${y \in X}$,

Using this language, we can now reveal our analysis strategy: Let ${y_{\min}}$ be the minimal element (according to the ordering ${\pi}$) which sees ${B}$. Then ${B(x,r) \nsubseteq \mathcal P(x)}$ can only occur if ${y_{\min}}$ also cuts ${B}$. The point is that the fate of ${B(x,r)}$ is not decided until some point sees ${B}$. Then, if ${y_{\min}}$ does not cut ${B}$, we have ${B(x,r) \subseteq C_{\pi^{-1}(y_{\min})}}$, hence ${B(x,r) \subseteq \mathcal P(x)}$.

Thus we can write,

To analyze the latter sum, first note that if ${y \notin B(x,\Delta/2+r)}$, then ${y}$ can never see ${B}$ since ${\alpha \Delta \leq \Delta/2}$ always. On the other hand, if ${y \in B(x,\Delta/4-r)}$ then ${y}$ always sees ${B}$, but can never cut ${B}$ since ${\alpha \Delta \geq \Delta/4}$ always.
Recalling that ${r \leq \Delta/8}$ by assumption, and setting ${a=|B(x,\Delta/8)|}$ and let ${b=|B(x,\Delta)|}$, we can use (3) to write

Now we come to the heart of the analysis: For any ${i \geq 1}$,

The idea is that if ${y_i}$ cuts ${B}$, then ${\alpha \Delta \geq d(x,y_i)-r}$. Thus if any ${y_j}$ for ${j < i}$ comes before ${y_i}$ in the ${\pi}$-ordering, then ${y_j}$ also sees ${B}$ since ${d(x,y_j) \leq d(x,y_i)}$, hence ${y_i \neq y_{\min}}$. But the probability that ${y_i}$ is the ${\pi}$-minimal element of ${\{y_1, \ldots, y_i\}}$ is precisely ${\frac{1}{i}}$, proving (5).

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

$\displaystyle \mathbb P\left(B(x,r) \nsubseteq \mathcal P(x)\right) \leq \sum_{i=a+1}^b \mathbb P\left(y_i \textrm{ cuts } B\right) \cdot \mathbb P\left(y_i = y_{\min}\,|\, y_i \textrm{ cuts } B\right) \leq \frac{8r}{\Delta} \sum_{i=a+1}^b \frac{1}{i}\,.$

$\Box$

Tightness for expanders. Before ending this section, let us mention that Theorem 1 is optimal up to a constant factor. Let ${\{G_k\}}$ be a family of degree-3 expander graphs, equipped with their shortest-path metric ${d_k}$. Assume that ${|E(S, \bar S)| \geq c|S|}$ for some ${c > 0}$ and all ${|S| \leq \frac12 |G_k|}$. Let ${\Delta_k = \frac{\log |G_k|}{10}}$, and suppose that ${G_k}$ admits a ${\Delta_k}$-bounded ${{L}}$-Lipschitz random partition. By averaging, this means there is a fixed ${\Delta_k}$-bounded partition ${P}$ of ${G_k}$ which cuts at most an ${{L}}$-fraction of edges. But every ${S \in P}$ has ${\mathrm{diam}(S) \leq \Delta_k}$, which implies ${|S| < n/2}$. Hence the partition must cut an ${\Omega(1)}$ fraction of edges, implying ${L = \Omega(1)}$.

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.

# The generic chaining

In the last post, we considered a Gaussian process ${\{X_t\}_{t \in T}}$ and were trying to find upper bounds on the quantity ${\mathop{\mathbb E}\sup_{t \in T} X_t}$. 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 ${T}$. First, recall that we fixed ${t_0 \in T}$, and we have used the observation that ${\mathop{\mathbb E}\sup_{t \in T} X_t = \mathop{\mathbb E}\sup_{t \in T} (X_t-X_{t_0})}$.

Now, assume that ${T}$ is finite. Write ${T_0 = \{t_0\}}$, and consider a sequence of subsets ${\{T_n\}}$ such that ${T_0 \subseteq T_1 \subseteq T_2 \subseteq \cdots \subseteq T}$. We will assume that for some large enough ${m}$, we have ${T_n = T}$ for ${n \geq m}$. For every ${n \geq 0}$, let ${\pi_n : T \rightarrow T_n}$ denote a “closest point map” which sends $t \in T$ to the closest point in $T_n$.

The main point is that we can now write, for any ${t \in T}$,

$\displaystyle X_t - X_{t_0} = \sum_{n \geq 1} X_{\pi_n(t)} - X_{\pi_{n-1}(t)}. \ \ \ \ \ (1)$

This decomposition is where the term “chaining” arises, and now the idea is to bound the probability that ${X_t - X_{t_0}}$ is large in terms of the segments in the chain.

What should ${T_n}$ look like?

One question that arises is how we should think about choosing the approximations ${T_n}$. We are trading off two measures of quality: The denser ${T_n}$ is in the set ${T}$ (or, more precisely, in the set ${T_{n-1}}$) the smaller the variances of the segments ${X_{\pi_n(t)}-X_{\pi_{n-1}(t)}}$ will be. On the other hand, the larger ${T_n}$ 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 ${\{T_n\}}$, let’s recall the classical Gaussian concentration bound.

Lemma 1 For every ${s,t \in T}$ and ${\lambda > 0}$,

$\displaystyle \mathop{\mathbb P}(X_s - X_t > \lambda) \leq \exp\left(-\frac{\lambda^2}{2\, d(s,t)^2}\right). \ \ \ \ \ (2)$

This should look familiar: ${X_s-X_t}$ is a mean-zero Gaussian with variance ${d(s,t)^2}$.

Now, a first instinct might be to choose the sets ${T_n}$ to be progressively denser in ${T}$. In this case, a natural choice would be to insist on something like ${T_n}$ being a ${2^{-n}}$-net in ${T}$. 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 ${T_n}$ has a certain level of accuracy, we’ll insist that ${T_n}$ is at most a certain size. Should we require ${|T_n| \leq n}$ or ${|T_n| \leq 2^n}$, or use some other function? To figure out the right bound, we look at (2). Suppose that ${g_1, g_2, \ldots, g_m}$ are i.i.d. ${N(0,1)}$ random variables. In that case, applying (2) and a union bound, we see that to achieve

$\displaystyle \mathop{\mathbb P}(\exists i : g_i > B) \leq m \mathop{\mathbb P}(g_1 > B) < 1,$

we need to select ${B \asymp \sqrt{\log m}}$. If we look instead at ${m^2}$ points instead of ${m}$ points, the bound grows to ${\sqrt{2 \log m}}$. 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 ${|T_{n+1}| = |T_n|^2}$. So we’ll require that ${|T_n| \leq 2^{2^n}}$ for all ${n \geq 1}$.

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 ${\{X_t\}_{t \in T}}$ be a Gaussian process, and let ${T_0 \subseteq T_1 \subseteq \cdots \subseteq T}$ be a sequence of subsets such that ${|T_0|=1}$ and ${|T_n| \leq 2^{2^{n}}}$ for ${n \geq 1}$. Then,

$\displaystyle \mathop{\mathbb E}\sup_{t \in T} X_t \leq O(1) \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} d(t, T_n). \ \ \ \ \ (3)$

Proof: As before, let ${\pi_n : T \rightarrow T_n}$ denote the closest point map and let ${T_0 = \{t_0\}}$. Using (2), for any ${n \geq 1}$, ${t \in T}$, and ${u > 0}$, we have

$\displaystyle \mathop{\mathbb P}\left(|X_{\pi_n(t)} - X_{\pi_{n-1}(t)}| > u 2^{n/2} d(\pi_n(t),\pi_{n-1}(t))\right) \leq \exp\left(-\frac{u^2}{2} 2^n\right).$

Now, the number of pairs ${(\pi_n(t),\pi_{n-1}(t))}$ can be bounded by ${|T_n| \cdot |T_{n-1}| \leq 2^{2^{n+1}}}$, so we have

$\displaystyle \mathop{\mathbb P}\left(\exists t : |X_{\pi_n(t)} - X_{\pi_{n-1}(t)}| > u 2^{n/2} d(\pi_n(t),\pi_{n-1}(t))\right) \leq 2^{2^{n+1}} \exp\left(-\frac{u^2}{2} 2^n\right). \ \ \ \ \ (4)$

If we define the event

$\displaystyle \Omega_u = \left\{ \forall n \geq 1, t \in T : |X_{\pi_n(t)} - X_{\pi_{n-1}(t)}| \leq u 2^{n/2} d(\pi_n(t),\pi_{n-1}(t))\right\},$

then summing (4) yields,

$\displaystyle \mathop{\mathbb P}(\overline{\Omega_u}) \leq \sum_{n \geq 1} 2^{2^{n+1}} \exp\left(-\frac{u^2}{2} 2^n\right) \leq O(1)\, e^{-u^2} \ \ \ \ \ (5)$

for ${u \geq 4}$, since we get geometrically decreasing summands.

Write

$\displaystyle S = \sup_{t \in T} \sum_{n \geq 1} 2^{n/2} d(\pi_n(t), \pi_{n-1}(t)).$

Note that if ${\Omega_u}$ occurs, then ${\sup_{t \in T} (X_t - X_{t_0}) \leq uS}$. Thus (5) implies that for $u \geq 4$,

$\displaystyle \mathop{\mathbb P}(\sup_{t \in T} X_t - X_{t_0} > uS) \leq O(1) \, e^{-u^2},$

which implies that

$\displaystyle \mathop{\mathbb E} \sup_{t \in T} X_t \leq O(S) \leq O(1) \sup_{t \in T} \sum_{n \geq 1} 2^{n/2} d(\pi_n(t), \pi_{n-1}(t)). \ \ \ \ \ (6)$

Finally, by the triangle inequality,

$\displaystyle d(\pi_n(t), \pi_{n-1}(t)) \leq d(t, T_n) + d(t, T_{n-1}) \leq 2\,d(t,T_{n-1}).$

Plugging this into (6) recovers (3). $\Box$

Theorem 1.2 gives us a fairly natural way to upper bound the expected supremum using a hierarchical clustering of ${T}$. 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 ${\{T_n\}}$ in Theorem 1.2, then the upper bound in (3) is within a constant factor of ${\mathop{\mathbb E} \sup_{t \in T} X_t}$.

# Lecture 8a. A primer on simplicial complexes and collapsibility

Before we can apply more advanced fixed point theorems to the Evasiveness Conjecture, we need a little background on simplicial complexes, and everything starts with simplices.

### Simplices

It’s most intuitive to start with the geometric viewpoint, in which case an $n$-simplex is defined to be the convex hull of $n+1$ affinely independent points in $\mathbb R^n$.  These points are called the vertices of the simplex.  Here are examples for $n=0,1,2,3.$

A simplicial complex is then a collection of simplices glued together along lower-dimensional simplices.  More formally, if $S \subseteq \mathbb R^n$ is a (geometric) simplex, then a face of $S$ is a subset $F \subseteq S$ formed by taking the convex hull of a subset of the vertices of $S$.

Finally, a (geometric) simplicial complex is a collection $\mathcal K$ of simplices such that

1. If $S \in \mathcal K$ and $F$ is a face of $S$, then $F \in \mathcal K$, and
2. If $S,S' \in \mathcal K$ and $S \cap S' \neq \emptyset$, then $S \cap S'$ is a face of both $S$ and $S'$.

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

# Lecture 7: The evasiveness conjecture

Continuing our look at some toplogical methods, today we’ll see the evasiveness conjecture in decision tree complexity.  In the next lecture, we’ll see how we can sometimes analyze the complexity using fixed point theorems, and their generalizations (like the Hopf index formula), following the work of Kahn, Saks, and Sturtevant.  These two lectures are co-blogged with Elisa Celis, with a lot of input from Lovasz’s lecture notes.

### Decision tree complexity and evasiveness

Consider a boolean function $f : \{0,1\}^n \to \{0,1\}$ on n bits.  We define the decision tree complexity of f as follows.  Given an unknown input $x \in \{0,1\}^n$, you are allowed to ask about the values of various bits of x, e.g. $x_{17}, x_{34}, x_3, \ldots$.  Your goal is to compute $f(x)$ using as few questions as possible, and your questions can be adaptive, depending on answers to previous questions.  The complexity of such a strategy is the maximum number of questions asked for any $x \in \{0,1\}^n$.  The decision tree complexity, written $D(f)$, is the minimum complexity of any strategy that computes f.  (There are many other interesting models of decision complexity, see e.g. this survey.)

Clearly $D(f) \leq n$, because we can trivially query all the bits of x, and then output f(x).  A function f is called evasive if this upper bound is met, i.e. $D(f)=n$.  As an example, consider the parity function $\mathsf{PARITY}(x) = x_1 \oplus x_2 \oplus \cdots \oplus x_n$, where $\oplus$ is addition modulo 2.  Clearly $\mathsf{PARITY}$ is evasive because after $n-1$ bits of x are asked about, the setting of the final bit determines the value of f.

For a more general example, consider any $f : \{0,1\}^n \to \{0,1\}$ such that $\#\{ x : f(x)=1 \}$ is odd.  In this case, for every $i =1,2,\ldots, n$, exactly one of $f|_{x_i=0}$ or $f|_{x_i=1}$ has the same property that the number of inputs resulting in a 1 is odd.  (These two functions are the natural restriction of f to functions on n-1 bits, which results from fixing the value of the ith bit.)  Thus an adversary could keep answering questions “$x_i$?” so that the restricted function retains this property.  Since the number of inputs yielding a 1 is always odd, the restricted function always takes both possible values, implying that f is evasive–the advesary ensures that the value cannot be determined until all n possible questions are asked.

For an example of a non-evasive property, think of a point $x \in \{0,1\}^{N \choose 2}$ as speciying a directed graph on $N$ vertices, where there is exactly one directed edges connecting every pair of vertices, and x specifies the direction of this edge (this is called a tournament).  Thinking of the vertices as players, a directed edge from u to v means that u defeats v.  Now $f(x)=1$ if the digraph specified by x has one vertex that defeats everyone else.  What is $D(f)$?

Well, first we can conduct a single elimination tournament, where vertex 1 plays vertex 2, and the winner players vertex 3, and the winner of that players vertex 4, etc.  At the end, there is only one remaining vertex $i$ that remains undefeated.  Now asking $N-2$ more questions, we can determine whether $i$ indeeds defeats everyone else.  The total number of questions was $(N-1)+(N-2) = 2N-3$, hence $D(f) \leq 2N-3 \ll {N \choose 2}$, implying that f is not evasive.

### Montone graph properties and the evasiveness conjecture

Let $m = {N \choose 2}$.  In general, we can encode an arbitrary undirected N-vertex graph as an element $G \in \{0,1\}^{m}$.  A function $f: \{0,1\}^m \to \{0,1\}$ is called a graph property if relabeling the vertices of $G$ doesn’t affect the value of $f(G)$.  The function f is monotone if the value of the function can never change from 1 to 0 when flipping one of the input bits from 0 to 1.  In the setting of graph properties, this corresponds to those which are maintained under addition of edges to the graph, e.g. $f(G) =$ “is G connected?” or $f(G)=$ “does G have a k-clique?”

Evasiveness Conjecture (Aanderaa-Karp-Rosenberg): Every non-trivial monotone graph property is evasive.

Here, non-trivial means that $f(\bar 0) \neq f(\bar 1)$, where $\bar 0$ and $\bar 1$ denote the all-zeros and all-ones strings, respectively.

For example, consider the example $f(G) =$ “is G connected?”  The adversary is simple:  When asked about a possible edge $\{i,j\}$, she answers NO unless this answer would imply that the graph is disconnected.  In other words, she answers NO unless she has answered NO already for all edges across a cut $(S,\bar S)$ except for $\{i,j\}$, in which case she has to answer YES.

Now, suppose there is a strategy which figures out the connectivity of $G$ without asking a question about some edge $\{i,j\}$.  In this case, the conclusion must be that $f(G) = \textrm{ YES}$ because the adversary always maintains that by answering everything in the future YES, she could force the graph to be connected.  In this case, the edges answered YES have to form a spanning tree of G (otherwise by answering all unasked questions NO, the graph would become disconnected).  Consider a path P from i to j in this YES spanning tree.  Let $\{u,v\}$ be the edge of P which was asked about last.  Clearly the adversary answered YES for $\{u,v\}$, but this contradicts the advesary’s strategy.  Since $\{i,j\}$ has not been asked yet, the adversary is safe to answer NO for $\{u,v\}$, and still later by answering YES on $\{i,j\}$, she could force the graph to be connected.  Thus no such strategy exists, and connectivity is evasive.