Two pages of the book, stuck together

April 12, 2018 ~ James ~ Leave a comment

Notes for tomorrow’s lecture on Approximation by ultrametrics are now up. (Yes, notes from the preceding two lectures still coming.)

(The combination of power and simplicity in this lecture is pretty incredible.)

Metrical task systems on a weighted star

April 9, 2018 ~ James ~ Leave a comment

Notes for the third lecture Metrical task systems on a weighted star are up.

Navigating a convex body online

April 6, 2018 ~ James ~ Leave a comment

Notes for the second lecture of our course Competitive analysis via convex optimization are up: Navigating a convex body online.

Competitive analysis, a new config

April 1, 2018 ~ James ~ Leave a comment

After not generating any new posts for a long time, I am happy to have finally switched platforms. The new site is hosted on gitpages at tcsmath.github.io. While the domain name tcsmath.org should point there, I am trying to work around well-known https certificate issues that cause Chrome to have a nervous breakdown.

The motivation for switching platforms is a course I’ve just started co-teaching with Seb Bubeck on Competitive analysis via convex optimization. The notes for the first lecture are live: Regret minimization and competitive analysis.

Follmer’s drift, Ito’s lemma, and the log-Sobolev inequality

November 21, 2015November 23, 2015 ~ James ~ Leave a comment

1. Construction of Föllmer’s drift

In a previous post, we saw how an entropy-optimal drift process could be used to prove the Brascamp-Lieb inequalities. Our main tool was a result of Föllmer that we now recall and justify. Afterward, we will use it to prove the Gaussian log-Sobolev inequality.

Consider ${f : \mathbb R^n \rightarrow \mathbb R_+}$ with ${\int f \,d\gamma_n = 1}$ , where ${\gamma_n}$ is the standard Gaussian measure on ${\mathbb R^n}$ . Let ${\{B_t\}}$ denote an ${n}$ -dimensional Brownian motion with ${B_0=0}$ . We consider all processes of the form

$\displaystyle W_t = B_t + \int_0^t v_s\,ds\,, \ \ \ \ \ (1)$

where ${\{v_s\}}$ is a progressively measurable drift and such that ${W_1}$ has law ${f\,d\gamma_n}$ .

Theorem 1 (Föllmer) It holds that

$\displaystyle D(f d\gamma_n \,\|\, d\gamma_n) = \min D(W_{[0,1]} \,\|\, B_{[0,1]}) = \min \frac12 \int_0^1 \mathop{\mathbb E}\,\|v_t\|^2\,dt\,,$

where the minima are over all processes of the form (1).

Proof: In the preceding post (Lemma 2), we have already seen that for any drift of the form (1), it holds that

$\displaystyle D(f d\gamma_n \,\|\,d\gamma_n) \leq \frac12 \int_0^1 \mathop{\mathbb E}\,\|v_t\|^2\,dt = D(W_{[0,1]} \,\|\, B_{[0,1]})\,,$

thus we need only exhibit a drift ${\{v_t\}}$ achieving equality.

We define

$\displaystyle v_t = \nabla \log P_{1-t} f(W_t) = \frac{\nabla P_{1-t} f(W_t)}{P_{1-t} f(W_t)}\,,$

where ${\{P_t\}}$ is the Brownian semigroup defined by

$\displaystyle P_t f(x) = \mathop{\mathbb E}[f(x + B_t)]\,.$

As we saw in the previous post (Lemma 2), the chain rule yields

$\displaystyle D(W_{[0,1]} \,\|\, B_{[0,1]}) = \frac12 \int_0^1 \mathop{\mathbb E}\,\|v_t\|^2\,dt\,. \ \ \ \ \ (2)$

We are left to show that ${W_1}$ has law ${f \,d\gamma_n}$ and ${D(W_{[0,1]} \,\|\, B_{[0,1]}) = D(f d\gamma_n \,\|\,d\gamma_n)}$ .

We will prove the first fact using Girsanov’s theorem to argue about the change of measure between ${\{W_t\}}$ and ${\{B_t\}}$ . As in the previous post, we will argue somewhat informally using the heuristic that the law of ${dB_t}$ is a Gaussian random variable in ${\mathbb R^n}$ with covariance ${dt \cdot I}$ . Itô’s formula states that this heuristic is justified (see our use of the formula below).

The following lemma says that, given any sample path ${\{W_s : s \in [0,t]\}}$ of our process up to time ${s}$ , the probability that Brownian motion (without drift) would have “done the same thing” is ${\frac{1}{M_t}}$ .

Remark 1 I chose to present various steps in the next proof at varying levels of formality. The arguments have the same structure as corresponding formal proofs, but I thought (perhaps naïvely) that this would be instructive.

Lemma 2 Let ${\mu_t}$ denote the law of ${\{W_s : s \in [0,t]\}}$ . If we define

$\displaystyle M_t = \exp\left(-\int_0^t \langle v_s,dB_s\rangle - \frac12 \int_0^t \|v_s\|^2\,ds\right)\,,$

then under the measure ${\nu_t}$ given by

$\displaystyle d\nu_t = M_t \,d\mu_t\,,$

the process ${\{W_s : s \in [0,t]\}}$ has the same law as ${\{B_s : s \in [0,t]\}}$ .

Proof: We argue by analogy with the discrete proof. First, let us define the infinitesimal “transition kernel” of Brownian motion using our heuristic that ${dB_t}$ has covariance ${dt \cdot I}$ :

$\displaystyle p(x,y) = \frac{e^{-\|x-y\|^2/2dt}}{(2\pi dt)^{n/2}}\,.$

We can also compute the (time-inhomogeneous) transition kernel ${q_t}$ of ${\{W_t\}}$ :

$\displaystyle q_t(x,y) = \frac{e^{-\|v_t dt + x - y\|^2/2dt}}{(2\pi dt)^{n/2}} = p(x,y) e^{-\frac12 \|v_t\|^2 dt} e^{-\langle v_t, x-y\rangle}\,.$

Here we are using that ${dW_t = dB_t + v_t\,dt}$ and ${v_t}$ is deterministic conditioned on the past, thus the law of ${dW_t}$ is a normal with mean ${v_t\,dt}$ and covariance ${dt \cdot I}$ .

To avoid confusion of derivatives, let’s use ${\alpha_t}$ for the density of ${\mu_t}$ and ${\beta_t}$ for the density of Brownian motion (recall that these are densities on paths). Now let us relate the density ${\alpha_{t+dt}}$ to the density ${\alpha_{t}}$ . We use here the notations ${\{\hat W_t, \hat v_t, \hat B_t\}}$ to denote a (non-random) sample path of ${\{W_t\}}$ :

$\displaystyle \begin{array}{lll} \alpha_{t+dt}(\hat W_{[0,t+dt]}) &= \alpha_t(\hat W_{[0,t]}) q_t(\hat W_t, \hat W_{t+dt}) \\ &= \alpha_t(\hat W_{[0,t]}) p(\hat W_t, \hat W_{t+dt}) e^{-\frac12 \|\hat v_t\|^2\,dt-\langle \hat v_t,\hat W_t-\hat W_{t+dt}\rangle} \\ &= \alpha_t(\hat W_{[0,t]}) p(\hat W_t, \hat W_{t+dt}) e^{-\frac12 \|\hat v_t\|^2\,dt+\langle \hat v_t,d \hat W_t\rangle} \\ &= \alpha_t(\hat W_{[0,t]}) p(\hat W_t, \hat W_{t+dt}) e^{\frac12 \|\hat v_t\|^2\,dt+\langle \hat v_t, d \hat B_t\rangle}\,, \end{array}$

where the last line uses ${d\hat W_t = d\hat B_t + \hat v_t\,dt}$ .

Now by “heuristic” induction, we can assume ${\alpha_t(\hat W_{[0,t]})=\frac{1}{M_t} \beta_t(\hat W_{[0,t]})}$ , yielding

$\displaystyle \begin{array}{lll} \alpha_{t+dt}(\hat W_{[0,t+dt]}) &= \frac{1}{M_t} \beta_t(\hat W_{[0,t]}) p(\hat W_t, \hat W_{t+dt}) e^{\frac12 \|\hat v_t\|^2\,dt+\langle \hat v_t, d \hat B_t\rangle} \\ &= \frac{1}{M_{t+dt}} \beta_t(\hat W_{[0,t]}) p(\hat W_t, \hat W_{t+dt}) \\ &= \frac{1}{M_{t+dt}} \beta_{t+dt}(\hat W_{[0,t+dt]})\,. \end{array}$

In the last line, we used the fact that ${p}$ is the infinitesimal transition kernel for Brownian motion. $\Box$

Now we will show that

$\displaystyle P_{1-t} f(W_t) = \exp\left(\frac12 \int_0^t \|v_s\|^2\,ds + \int_0^t \langle v_s, dB_s\rangle\right) = \frac{1}{M_t}\,. \ \ \ \ \ (3)$

From Lemma 2, it will follow that ${W_t}$ has the law ${(P_{1-t} f)\cdot d\nu_t}$ where ${d\nu_t}$ is the law of ${B_t}$ . In particular, ${W_1}$ has the law ${f\,d\nu_1 = f\,d\gamma_n}$ which was our first goal.

Given our preceding less formal arguments, let us use a proper stochastic calculus argument to establish (3). To do that we need a way to calculate

$\displaystyle d \log P_{1-t} f(W_t) \quad \textrm{``}= \log P_{1-t-dt} f(W_{t+dt}) - \log P_{1-t} f(W_t)\textrm{''} \ \ \ \ \ (4)$

Notice that this involves both time and space derivatives.

Itô’s lemma. Suppose we have a continuously differentiable function ${F : \mathbb R \times [0,1] \rightarrow \mathbb R}$ that we write as ${F(x,t)}$ where ${x}$ is a space variable and ${t}$ is a time variable. We can expand ${d F}$ via its Taylor series:

$\displaystyle d F = \partial_t F \,dt + \partial_x F\,dx + \frac12 \partial_x^2 F\,dx^2 + \frac12 \partial_x \partial_t F\,dx\,dt + \cdots\,.$

Normally we could eliminate the terms ${dx^2, dx\, dt}$ , etc. since they are lower order as ${dx,dt \rightarrow 0}$ . But recall that for Brownian motion we have the heuristic ${\mathop{\mathbb E}[dB_t^2]=dt}$ . Thus we cannot eliminate the second-order space derivative if we plan to plug in ${x=B_t}$ (or ${x=W_t}$ , a process driven by Brownian motion). Itô’s lemma says that this consideration alone gives us the correct result:

$\displaystyle d F(W_t,t) = \partial_t F(W_t,t)\,dt + \partial_x F(W_t,t)\,dW_t + \frac12 \partial_x^2 F(W_t,t)\,dt\,.$

This generalizes in a straightforward way to the higher dimensional setting ${F : \mathbb R^n \times [0,1] \rightarrow \mathbb R}$ .

With Itô’s lemma in hand, let us continue to calculate the derivative

$\displaystyle \begin{array}{lll} d P_{1-t} f(W_t) &= - \Delta P_{1-t} f(W_t)\,dt + \langle \nabla P_{1-t} f(W_t), dW_t\rangle + \Delta P_{1-t} f(W_t) \,dt \\ &= \langle \nabla P_{1-t} f(W_t), dW_t\rangle \\ &= P_{1-t} f(W_t) \,\langle v_t, dW_t\rangle\,. \end{array}$

For the time derivative (the first term), we have employed the heat equation

$\displaystyle \partial_t P_{1-t} f = - \Delta P_{1-t} f\,,$

where ${\Delta = \frac12 \sum_{i=1}^n \partial_{x_i}^2}$ is the Laplacian on ${\mathbb R^n}$ .

Note that the heat equation was already contained in our “infinitesimal density” ${p}$ in the proof of Lemma 2, or in the representation ${P_t = e^{t \Delta}}$ , and Itô’s lemma was also contained in our heuristic that ${dB_t}$ has covariance ${dt \cdot I}$ .

Using Itô’s formula again yields

$\displaystyle d \log P_{1-t} f(W_t) = \langle v_t, dW_t\rangle - \frac12 \|v_t\|^2\,dt = \frac12 \|v_t\|^2\,dt + \langle v_t,dB_t\rangle\,.$

giving our desired conclusion (3).

Our final task is to establish optimality: ${D\left(W_{[0,1]} \,\|\, B_{[0,1]}\right) = D(W_1\,\|\,B_1)}$ . We apply the formula (3):

$\displaystyle D(W_1\,\|\,B_1) = \mathop{\mathbb E}[\log f(W_1)] = \mathop{\mathbb E}\left[\frac12 \int_0^1 \|v_t\|^2\,dt\right],$

where we used ${\mathop{\mathbb E}[\langle v_t,dB_t\rangle]=0}$ . Combined with (2), this completes the proof of the theorem. $\Box$

2. The Gaussian log-Sobolev inequality

Consider again a measurable ${f : \mathbb R^n \rightarrow \mathbb R_+}$ with ${\int f\,d\gamma_n=1}$ . Let us define ${\mathrm{Ent}_{\gamma_n}(f) = D(f\,d\gamma_n \,\|\,d\gamma_n)}$ . Then the classical log-Sobolev inequality in Gaussian space asserts that

$\displaystyle \mathrm{Ent}_{\gamma_n}(f) \leq \frac12 \int \frac{\|\nabla f\|^2}{f}\,d\gamma_n\,. \ \ \ \ \ (5)$

First, we discuss the correct way to interpret this. Define the Ornstein-Uhlenbeck semi-group ${\{U_t\}}$ by its action

$\displaystyle U_t f(x) = \mathop{\mathbb E}[f(e^{-t} x + \sqrt{1-e^{-2t}} B_1)]\,.$

This is the natural stationary diffusion process on Gaussian space. For every measurable ${f}$ , we have

$\displaystyle U_t f \rightarrow \int f d\gamma_n \quad \textrm{ as } t \to \infty$

or equivalently

$\displaystyle \mathrm{Ent}_{\gamma_n}(U_t f) \rightarrow 0 \quad \textrm{ as } t \to \infty$

The log-Sobolev inequality yields quantitative convergence in the relative entropy distance as follows: Define the Fisher information

$\displaystyle I(f) = \int \frac{\|\nabla f\|^2}{f} \,d\gamma_n\,.$

One can check that

$\displaystyle \frac{d}{dt} \mathrm{Ent}_{\gamma_n} (U_t f)\Big|_{t=0} = - I(f)\,,$

thus the Fisher information describes the instantaneous decay of the relative entropy of ${f}$ under diffusion.

So we can rewrite the log-Sobolev inequality as:

$\displaystyle - \frac{d}{dt} \mathrm{Ent}_{\gamma_n}(U_t f)\Big|_{t=0} \geq 2 \mathrm{Ent}_{\gamma_n}(f)\,.$

This expresses the intuitive fact that when the relative entropy is large, its rate of decay toward equilibrium is faster.

Martingale property of the optimal drift. Now for the proof of (5). Let ${dW_t = dB_t + v_t\,dt}$ be the entropy-optimal process with ${W_1 \sim f \,d\gamma_n}$ . We need one more fact about ${\{v_t\}}$ : The optimal drift is a martingale, i.e. ${\mathop{\mathbb E}[v_t \mid v_s] = v_s}$ for ${s < t}$ .

Let’s give two arguments to support this.

Argument one: Brownian bridges. First, note that by the chain rule for relative entropy, we have:

$\displaystyle D(W_{[0,1]} \,\|\, B_{[0,1]}) = D(W_1 \,\|\, B_1) + \int D(W_{[0,1]} \,\|\, B_{[0,1]} \mid W_1=B_1=x) f(x) d\gamma_n(x)\,.$

But from optimality, we know that the latter expectation is zero. Therefore ${f \,d\gamma_n}$ -almost surely, we have

$\displaystyle D(W_{[0,1]} \,\| B_{[0,1]} \mid W_1=B_1=x) = 0\,.$

This implies that if we condition on the endpoint ${x}$ , then ${W_{[0,1]}}$ is a Brownian bridge (i.e., a Brownian motion conditioned to start at ${0}$ and end at ${x}$ ).

This implies that ${\mathop{\mathbb E}[v_t \mid v_s, W_1=x] = v_s}$ , as one can check that a Brownian bridge ${\{\hat B_t\}}$ with endpoint ${x}$ is described by the drift process ${d\hat B_t = dB_t + \frac{x-\hat B_t}{1-t}\,dt}$ , and

$\displaystyle \mathop{\mathbb E}\left[\frac{x-\hat B_t}{1-t} \,\Big|\, B_{[0,s]}\right] = \frac{x-\hat B_s}{1-s}\,.$

That seemed complicated. There is a simpler way to see this: Given ${\hat B_s}$ and any bridge ${\gamma}$ from ${\hat B_s}$ to ${x}$ , every “permutation” of the infinitesimal steps in ${\gamma}$ has the same law (by commutativity, they all land at ${x}$ ). Thus the marginal law of ${dB_t + v_t\,dt}$ at every point ${t \geq s}$ should be the same. In particular,

$\displaystyle \mathop{\mathbb E}[v_t\,dt \mid v_s] = \mathop{\mathbb E}[dB_t + v_t\,dt \mid v_s] = \mathop{\mathbb E}[dB_s + v_s \,ds \mid v_s] = v_s\,ds\,.$

Argument two: Change of measure. There is a more succinct (though perhaps more opaque) way to see that ${\{v_t\}}$ is a martingale. Note that the process ${\nabla P_{1-t} f(B_t) = P_{1-t} \nabla f(B_t)}$ is a Doob martingale. But we have ${v_t = \frac{\nabla P_{1-t} f(W_t)}{P_{1-t} f(W_t)}}$ and we also know that ${\frac{1}{P_{1-t} f(W_t)} = \frac{1}{M_t}}$ is precisely the change of measure that makes ${\{W_t\}}$ into Brownian motion.

Proof of the log-Sobolev inequality. In any case, now we are ready for the proof of (5). It also comes straight from Lehec’s paper. Since ${\{v_t\}}$ is a martingale, we have ${\mathop{\mathbb E}\,\|v_t\|^2 \leq \mathop{\mathbb E}\,\|v_1\|^2}$ . So by Theorem 1:

$\displaystyle \mathrm{Ent}_{\gamma_n}(f) = \frac12 \int_0^1 \mathop{\mathbb E}\,\|v_t\|^2\,dt \leq \frac12 \mathop{\mathbb E}\,\|v_1\|^2 = \frac12 \mathop{\mathbb E}\, \frac{\|\nabla f(W_1)\|^2}{f(W_1)^2} = \frac12 \mathop{\mathbb E}\, \frac{\|\nabla f(B_1)\|^2}{f(B_1)}\,.$

The latter quantity is $\frac12 I(f)$ . In the last equality, we used the fact that ${\frac{1}{f(W_1)}}$ is precisely the change of measure that turns ${\{W_t\}}$ into Brownian motion.

A geometric proof of Okamura-Seymour

November 18, 2015November 19, 2015 ~ James ~ 4 Comments

After using it in the last post on non-positively curved surfaces, I thought it might be nice to give a simple proof of the Okamura-Seymour theorem in the dual setting. This argument arose out of conversations in 2007 with Amit Chakrabarti and former UW undergrad Justin Vincent. I was later informed by Yuri Rabinovich that he and Ilan Newman discovered a similar proof.

Note that establishing a node-capacitated version of the Okamura-Seymour theorem was an open question of Chekuri and Kawarabayashi. Resolving it positively is somewhat more difficult.

Theorem 1 Given a weighted planar graph ${G = (V,E)}$ and a face ${F}$ of ${G}$ , there is a non-expansive embedding ${\varphi:V\rightarrow L_1}$ such that ${\varphi|_F}$ is an isometry.

By rational approximation and subdivision of edges, we may assume that ${G}$ is unweighted. The following proof is constructive, and provides an explicit sequence of cuts on ${G}$ whose characteristic functions form the coordinates of the embedding ${\varphi}$ . Each such cut is obtained by applying the following lemma. Note that for a subset ${S}$ of vertices, we use the notation ${G/\partial S}$ , for the graph obtained from ${G}$ by contracting the edges across ${S}$

Lemma 2 (Face-preserving cut lemma) Let ${G = (V,E)}$ be a ${2}$ -connected bipartite planar graph and ${F}$ be the boundary of a face of ${G}$ . There exists a cut ${S\subseteq V}$ such that

$\displaystyle \forall\, u,v\in F,~ d_G(u,v) - d_{G/\partial S}(u,v) ~=~ |\mathbf{1}_S(u) - \mathbf{1}_S(v)| \, .$

Proof: Fix a plane embedding of ${G}$ that makes ${F}$ the boundary of the outer face. Since ${G}$ is ${2}$ -connected, ${F}$ is a cycle, and since ${G}$ is bipartite and has no parallel edges, ${\mathrm{len}(F)\geq 4}$ .

Consider an arbitrary pair ${(a,b)}$ of distinct vertices on ${F}$ . There is a unique path from ${a}$ to ${b}$ that runs along ${F}$ counterclockwise; call this path ${\lambda_{ab}}$ . Consider a path ${\gamma\in\mathcal P_{ab}}$ and a vertex ${w\in V}$ . We say that ${w}$ lies below ${\gamma}$ if, in the plane embedding of ${G}$ , ${w}$ lies in the closed subset of the plane bounded by ${\gamma}$ and ${\lambda_{ab}}$ .

(Note that the direction of ${\gamma}$ is significant in the definition of “lying below,” i.e., belowness with respect to a path in ${\mathcal P_{ab}}$ is not the same as belowness with respect to the reverse of the same path in ${\mathcal P_{ba}}$ .)

We say that ${w}$ lies strictly below ${\gamma}$ if ${w}$ lies below ${\gamma}$ and ${w\notin\gamma}$ . We use this notion of “lying below” to define a partial order on the paths in ${\mathcal P_{ab}}$ : for ${\gamma,\gamma'\in\mathcal P_{ab}}$ we say that ${\gamma'}$ is lower than ${\gamma}$ if every vertex in ${\gamma'}$ lies below ${\gamma}$ .

We now fix the pair ${(a,b)}$ and a path ${\pi\in\mathcal P_{ab}}$ so that the following properties hold:

${\mathrm{len}(\pi) = d_G(a,b) < \mathrm{len}(\lambda_{ab})}$ .
If ${a',b'\in\lambda_{ab}}$ are distinct vertices with ${a'}$ preceding ${b'}$ and ${d_G(a',b') < \mathrm{len}(\lambda_{a'b'})}$ , then ${a' = a}$ and ${b' = b}$ .
If ${\pi'\in\mathcal P_{ab}}$ is lower than ${\pi}$ and ${\mathrm{len}(\pi') = \mathrm{len}(\pi)}$ , then ${\pi' = \pi}$ .

Note that a suitable pair ${(a,b)}$ exists because ${\mathrm{len}(F)\ge4}$ . Finally, we define the cut ${S\subseteq V}$ as follows: ${S = \{v\in V:\, v}$ does not lie strictly below ${\pi\}}$ .

Claim 1 The cut ${S}$ satisfies the conditions of the lemma.

For the rest of this section, we fix the pair ${(a,b)}$ , the path ${\pi}$ and the cut ${S}$ as defined in the above proof.

Lemma 1 Let ${x}$ and ${y}$ be distinct vertices on ${\pi}$ and ${\beta\in\mathcal P_{xy}}$ be such that ${\emptyset\ne\mathrm{int}(\beta)\subseteq\bar S}$ . Then ${\mathrm{len}(\beta) - 2 \ge d_G(x,y)}$ .

Proof: If ${x = y}$ the lemma holds trivially, so assume ${x \ne y}$ . Also assume without loss of generality that ${x}$ precedes ${y}$ in the path ${\pi}$ . The conditions on ${\beta}$ imply that all vertices in ${\mathrm{int}(\beta)}$ lie strictly below ${\pi}$ . Therefore, the path ${\pi' := \pi[a:x]\circ\beta\circ\pi[y:b]}$ is lower than ${\pi}$ and distinct from ${\pi}$

By property (3), we have ${\mathrm{len}(\pi') > \mathrm{len}(\pi)}$ , which implies ${\mathrm{len}(\beta) > \mathrm{len}(\pi[x:y])}$ . Since ${G}$ is bipartite, the cycle formed by ${\beta}$ and ${\pi[x:y]}$ must have even length; therefore, ${\mathrm{len}(\beta) \ge \mathrm{len}(\pi[x:y]) + 2 = d_G(x,y) + 2}$ . $\Box$

Lemma 2 Let ${x}$ and ${y}$ be distinct vertices on ${\pi}$ and ${\gamma\in\mathcal P_{xy}}$ . Then ${\mathrm{len}(\gamma) - |\gamma\cap\partial S| \ge d_G(x,y)}$ .

Proof: If ${\gamma\cap\partial S = \emptyset}$ there is nothing to prove. If not, we can write

$\displaystyle \gamma ~=~ \alpha_1 \circ \beta_1 \circ \alpha_2 \circ \beta_2 \circ \cdots \circ \alpha_k \circ \beta_k \circ \alpha_{k+1}$

for some ${k\ge1}$ where, for all ${i}$ , ${\alpha_i\subseteq S}$ and ${\emptyset\ne\mathrm{int}(\beta_i)\subseteq \bar S}$ . Let ${x_i}$ and ${y_i}$ denote the endpoints of ${\beta_i}$ with ${x_i}$ preceding ${y_i}$ in the path ${\gamma}$ . Further, define ${y_0 := x}$ and ${x_{k+1} := y}$ .

By Lemma 1, we have ${d_G(x_i,y_i) \le \mathrm{len}(\beta_i) - 2}$ for ${1\le i\le k}$ . Since ${\pi}$ is a shortest path, we have ${d_G(y_{i-1},x_i) \le \mathrm{len}(\alpha_i)}$ for ${1\le i\le k+1}$ . Therefore

$\displaystyle d_G(x,y) ~\le~ \sum_{i=1}^k d_G(x_i,y_i) + \sum_{i=1}^{k+1} d_G(y_{i-1},x_i) ~\le~ \sum_{i=1}^k (\mathrm{len}(\beta_i) - 2) + \sum_{i=1}^{k+1} \mathrm{len}(\alpha_i)\,.$

The latter quantity is precisely ${\mathrm{len}(\gamma) - 2k}$ which completes the proof since ${|\gamma\cap\partial S| = 2k}$ . $\Box$

Proof of Claim 1: Let ${u,v\in F}$ be arbitrary and distinct. It is clear that ${d_{G/\partial S}(u,v) \le d_G(u,v) - |\mathbf{1}_S(u) - \mathbf{1}_S(v)|}$ , so it suffices to prove the opposite inequality. We begin by observing that

$\displaystyle d_{G/\partial S}(u,v) ~=~ \min\left\{ \mathrm{len}(\gamma) - |\gamma\cap\partial S|:\, \gamma\in\mathcal P_{uv}\right\} \, .$

Let ${\gamma}$ be a path in ${\mathcal P_{uv}}$ that achieves the minimum in the above expression. First, suppose ${\gamma\cap\partial S = \emptyset}$ . Then we must have ${\mathbf{1}_S(u) = \mathbf{1}_S(v)}$ . Now, ${d_{G/\partial S}(u,v) = \mathrm{len}(\gamma) \ge d_G(u,v)}$ , which implies ${d_{G/\partial S}(u,v) = d_G(u,v)}$ and we are done.

Next, suppose ${\gamma\cap\partial S \ne \emptyset}$ . Then, there exists at least one vertex in ${\gamma}$ that lies on ${\pi}$ . Let ${x}$ be the first such vertex and ${y}$ the last (according to the ordering in ${\gamma}$ ) and assume that ${x}$ precedes ${y}$ in the path ${\pi}$ . Let ${\hat\gamma := \gamma[x:y]}$ . Note that ${\hat\gamma}$ may be trivial, because we may have ${x = y}$ . Now, ${\gamma\cap\partial S = (\gamma[u:x]\cap\partial S) \cup (\hat\gamma\cap\partial S) \cup (\gamma[y:v]\cap\partial S)}$ , whence

$\displaystyle |\gamma\cap\partial S| ~=~ (1 - \mathbf{1}_S(u)) + |\hat\gamma\cap\partial S| + (1 - \mathbf{1}_S(v)) \, . \ \ \ \ \ (1)$

This gives

$\begin{array}{rcl} d_{G/\partial S}(u,v) & = & \mathrm{len}(\gamma) - (1 - \mathbf{1}_S(u)) - |\hat\gamma\cap\partial S| - (1 - \mathbf{1}_S(v)) \\ &\geq & \mathrm{len}(\gamma[u:x]) + \mathrm{len}(\hat\gamma) + \mathrm{len}(\gamma[y:v]) + (\mathbf{1}_S(u) + \mathbf{1}_S(v) - 2) - |\hat\gamma\cap\partial S| \nonumber \\ &\geq & \mathrm{len}(\gamma[u:x]) + d_G(x,y) + \mathrm{len}(\gamma[y:v]) + (\mathbf{1}_S(u) + \mathbf{1}_S(v) - 2) \, \ \ \ \ \ (**) \\ &\geq & d_G(u,v) + (\mathbf{1}_S(u) + \mathbf{1}_S(v) - 2) \end{array}$

where the first line follows from Eq. (1) and the definition of ${\gamma}$ and the third line is obtained by applying Lemma 2 to the path ${\hat\gamma}$ . If at least one of ${u,v}$ lies in ${S}$ , then ${\mathbf{1}_S(u) + \mathbf{1}_S(v) - 2 = -|\mathbf{1}_S(u) - \mathbf{1}_S(v)|}$ and we are done.

Therefore, suppose ${u,v\in\bar S}$ . Let ${\beta = \lambda_{ab}}$ . For a path ${\alpha}$ and vertices ${w,z}$ on ${\alpha}$ , let us use ${|wz|_\alpha}$ as shorthand for ${\mathrm{len}(\alpha[w:z])}$ . By property (1), we have ${|ab|_\beta > |ab|_\pi}$ and since ${G}$ is bipartite, this means ${|ab|_\beta - |ab|_\pi \ge 2}$ . By property (2), we have ${d_G(u,b) = |ub|_\beta}$ and ${d_G(a,v) = |av|_\beta}$ . Using these facts, we now derive

$\displaystyle \begin{array}{rcl} \mathrm{len}(\gamma[u:x]) + d_G(x,y) + \mathrm{len}(\gamma[y:v]) & = & |ux|_\gamma + |xy|_\pi + |yv|_\gamma \\ & = & |ux|_\gamma + |xb|_\pi + |ay|_\pi + |yv|_\gamma - |ab|_\pi \\ &\ge& d_G(u,b) + d_G(a,v) - |ab|_\pi \\ & = & |ub|_\beta + |av|_\beta - |ab|_\pi \\ & = & |uv|_\beta + |ab|_\beta - |ab|_\pi \\ &\ge& |uv|_\beta + 2 \, . \end{array}$

Using this in (**) above and noting that ${\mathbf{1}_S(u) = \mathbf{1}_S(v) = 0}$ , we get ${d_{G/\partial S}(u,v) \ge |uv|_\beta + 2 - 2 = d_G(u,v)}$ . This completes the proof. $\Box$

Proof of Theorem 1: Assume that ${G}$ is ${2}$ -connected. We may also assume that ${G}$ is bipartite. To see why, note that subdividing every edge of ${G}$ by introducing one new vertex per edge leaves the metric ${d_G}$ essentially unchanged except for a scaling factor of ${2}$ .

We shall now prove the stronger statement that for every face ${f}$ of ${G}$ there exists a sequence of cuts ${S_1,S_2,\ldots,S_k}$ of ${G}$ such that for all ${u,v}$ on ${f}$ , we have ${d_G(u,v) = \sum_{i=1}^k |\mathbf{1}_{S_i}(u) - \mathbf{1}_{S_i}(v)|}$ and that for ${i\ne j}$ , ${\partial S_i\cap\partial S_j = \emptyset}$ . We prove this by induction on ${|V|}$ .

The result is trivial in the degenerate case when ${G}$ is a single edge. For any larger ${G}$ and any cut ${S\subseteq V}$ , the graph ${G/\partial S}$ is either a single edge or is ${2}$ -connected. Furthermore, contracting a cut preserves the parities of the lengths of all closed walks; therefore ${G/\partial S}$ is also bipartite.

Apply the face-preserving cut lemma (Lemma 1) to obtain a cut ${S_1\subseteq V}$ . By the above observations, we can apply the induction hypothesis to ${G/\partial S_1}$ to obtain cuts ${\hat S_2,\ldots,\hat S_k}$ of ${G/\partial S_1}$ corresponding to the image of ${f}$ in ${G/\partial S_1}$ . Each cut ${\hat S_i}$ induces a cut ${S_i}$ of ${G}$ . Clearly ${\partial S_1\cap\partial S_i = \emptyset}$ for any ${i > 1}$ . Finally, for any ${u,v\in f}$ , we have

$d_G(u,v) = d_{G/\partial S_1}(u,v) + |\mathbf{1}_{S_1}(u) - \mathbf{1}_{S_1}(v)| = \sum_{i=2}^k |\mathbf{1}_{S_i}(u) - \mathbf{1}_{S_i}(v)| + |\mathbf{1}_{S_1}(u) - \mathbf{1}_{S_1}(v)|$

where the first equality follows from the property of ${S_1}$ and the second follows from the induction hypothesis. This proves the theorem. $\Box$

Embeddings and extendable geodesics

November 18, 2015November 18, 2015 ~ James ~ 1 Comment

Isometric embedding of non-positively curved surfaces into ${L^1}$

As I have mentioned before, one of my favorite questions is whether the shortest-path metric on a planar graph embeds into ${L^1}$ with ${O(1)}$ distortion. This is equivalent to such graphs having an ${O(1)}$ -approximate multi-flow/min-cut theorem. We know that the distortion has to be at least 2. By a simple discretization and compactness argument, this is equivalent to the question of whether every simply-connected surface admits a bi-Lipschitz embedding into ${L^1}$ .

In a paper of Tasos Sidiropoulos, it is proved that every simply-connected surface of non-positive curvature admits a bi-Lipschitz embedding into ${L^1}$ . A followup work of Chalopin, Chepoi, and Naves shows that actually such a surface admits an isometric emedding into ${L^1}$ . In this post, we present a simple proof of this result that was observed in conversations with Tasos a few years ago—it follows rather quickly from the most classical theorem in this setting, the Okamura-Seymour theorem.

Suppose that ${(X,d)}$ is a geodesic metric space (i.e. the distance between any pair of points ${x,y \in X}$ is realized by a geodesic whose length is ${d(x,y)}$ ). One says that ${(X,d)}$ has non-positive curvature (in the sense of Busemann) if for any pair of geodesics ${\gamma : [a,b] \rightarrow X}$ and ${\gamma' : [a',b'] \rightarrow X}$ , the map ${D_{\gamma,\gamma'} : [0,1] \rightarrow \mathbb R}$ given by

$\displaystyle D_{\gamma,\gamma'}(t) = d\left(\vphantom{\bigoplus} \gamma((1-t)a + tb),\gamma'((1-t)a' + tb')\right)$

is convex.

Theorem 1 Suppose that ${\mathcal S}$ is homeomorphic to ${\mathbb R^2}$ and is endowed with a geodesic metric ${d}$ such that ${(\mathcal S,d)}$ has non-positive curvature. Then ${(\mathcal S,d)}$ embeds isometrically in ${L^1}$ .

We will access the non-positive curvature property through the following fact. We refer to the book Metric spaces of non-positive curvature.

Lemma 2 Every geodesic ${\gamma}$ in ${\mathcal S}$ can be extended to a bi-infinite geodesic line.

Proof of Theorem 1: By a standard compactness argument, it suffices to construct an isometric embedding for a finite subset ${X \subseteq \mathcal S}$ . Let ${C}$ denote the convex hull of ${X}$ . (A set ${C}$ is convex if for every ${x,y \in C}$ we have ${\gamma \subseteq C}$ for every geodesic ${\gamma}$ connecting ${x}$ to ${y}$ .)

It is an exercise to show that the boundary of ${C}$ is composed of a finite number of geodesics ${\mathcal F = \left\{[x_1, x_2], [x_2, x_3], \cdots, [x_{N-1},x_N], [x_N, x_1]\right\}}$ between points ${x_1, x_2, \ldots, x_N \in C}$ . For every pair ${x,y \in X}$ , let ${\stackrel{\leftrightarrow}{xy}}$ denote a geodesic line containing ${x}$ and ${y}$ which exists by Lemma 2. Consider the collection of sets ${\mathcal L = \mathcal F \cup \{ \stackrel{\leftrightarrow}{xy} : x,y \in X\}}$ , and let ${I}$ denote the set of intersection points between geodesics in ${\mathcal L}$ . Since ${\mathcal L}$ is a collection of geodesics, and all geodesics intersect at most once (an easy consequence of non-positive curvature), the set ${I}$ is finite.

Consider finally the set ${Z = \{x_1, \ldots, x_N\} \cup X \cup I}$ . The geodesics in ${\mathcal L}$ naturally endow ${Z}$ with the structure of a planar graph ${G=(Z,E)}$ , where two vertices ${x,y \in Z}$ are adjacent if they lie on a subset of some geodesic in ${\mathcal L}$ and the portion ${[x,y]}$ between ${x}$ and ${y}$ does not contain any other points of ${Z}$ . Note that ${F = Z \cap \partial C}$ is the outer face of ${G}$ in the natural drawing, where ${\partial C}$ is the boundary of ${C}$ (the union of the geodesics in ${\mathcal F}$ ).

We can put a path metric on this graph by defining the length of an edge ${\{x,y\}}$ as ${d(x,y)}$ . Let ${d_G}$ denote the induced shortest-path metric on the resulting graph. By construction, we have the following two properties.

If ${u,v \in X}$ or ${u,v \in F\, \cap \stackrel{\leftrightarrow}{xy}}$ for some ${x,y \in X}$ , then ${d_G(u,v)=d(u,v)}$ .
For every ${x,y \in X}$ , there is a shortest path ${P_{xy}}$ between two vertices in ${F}$ such that ${x,y \in P_{xy}}$ .

Both properties follow from our construction using the lines ${\{ \stackrel{\leftrightarrow}{xy} : x,y \in X\}}$ .

Now let us state the geometric (dual) version of the Okamura-Seymour theorem.

Theorem 3 (Okamura-Seymour dual version) For every planar graph ${G=(V,E)}$ and face ${F \subseteq V}$ , there is a ${1}$ -Lispchitz mapping ${\varphi : V \rightarrow L^1}$ such that ${\varphi|_F}$ is an isometry.

Let us apply this theorem to our graph ${G}$ and face ${F}$ . Consider ${x,y \in X}$ and ${\{u,v\} = F\, \cap \stackrel{\leftrightarrow}{xy}}$ . By property (1) above, we have ${d_G(u,v)=d(u,v)}$ . Since ${u,v \in F}$ , from Theorem 3, we have ${\|\varphi(u)-\varphi(v)\|=d_G(u,v)}$ . But property (2) above says that ${x}$ and ${y}$ lie on a ${u}$ – ${v}$ shortest-path ${P_{xy}}$ in ${G}$ . Since ${\varphi}$ is ${1}$ -Lipschitz, we conclude that it maps the whole path ${P_{xy}}$ isometrically, thus ${\|f(x)-f(y)\|_1=d_G(x,y)=d(x,y)}$ , showing that ${\varphi|_{X}}$ is an isometry, and completing the proof. $\Box$