# tcs math – some mathematics of theoretical computer science

## October 13, 2008

### Lecture 5: Uniformizing graphs, multi-flows, and eigenvalues

In the previous lecture, we gave an upper bound on the second eigenvalue of the Laplacian of (bounded degree) planar graphs in order to analyze a simple spectral partitioning algorithm.  A natural question is whether these bounds extend to more general families of graphs.  Well-known generalizations of planar graphs are those which can be embedded on a surface of fixed genus, and, more generally, families of graphs that arise by forbidding minors.  In fact, Spielman and Teng conjectured that for any graph excluding $K_h$ as a minor, one should have $\lambda_2 \lesssim \mathrm{poly}(h) d_{\max}/n$.   Of course planar graphs have genus 0, and by Wagner’s theorem, are precisely the graphs which exclude $K_5$ and $K_{3,3}$ as minors.  In this lecture, we will follow an intrinsic approach of Biswal, myself, and Rao which, in particular, is able to resolve the conjecture of Spielman and Teng.  First, we see why even pushing the conformal approach to bounded genus graphs is difficult.

### Bounded genus graphs

For graphs of bounded genus, there is hope to use an approach based on conformal mappings.  In 1980, Yang and Yau proved that

$\displaystyle \lambda_2(M) \lesssim \frac{g+1}{\mathrm{vol}(M)}$

for any compact Riemannian surface $M$ of genus $g$.  (Note that for the Laplace-Beltrami operator, one usually writes $\lambda_1$ as the first non-zero eigenvalue, rather than $\lambda_2$.)  In analog with Hersch’s proof of the genus 0 case, they use Riemann-Roch to obtain a degree-$(g+1)$ conformal mapping to the Riemann sphere, then try to pull back a second eigenfunction.  A factor of the degree is lost in the Rayleigh quotient (hence the $g+1$ factor in the preceding bound), and Hersch’s Möbius trick is still required.

An analogous proof for graphs $G$ of bounded genus would proceed by constructing a circle packing of $G$ on the sphere $S^2$, but instead of the circles having disjoint interiors, we would be assured that every point of $S^2$ is contained in at most $g$ circles.  Unfortunately, such a result is impossible (this has to do with the handling of branch points in the discrete setting).  Kelner has to take a different approach in his proof that $\lambda_2(G) \leq d_{\max}^{O(1)} (g+1)/n$ for graphs $G$ of genus at most $g$.

He starts with a circle packing of $G$ on a compact surface $\mathbb S_0$ of genus $g$ (whose existence follows from results of Beardon and Stephenon and He and Schramm).  Then Kelner randomly subdivides $G$ repeatedly, and these subdivisions give progressively better approximations to some sequence of surfaces $\{\mathbb S_i\}$.  Once the approximation is of high enough quality, one applies Riemann-Roch to $\mathbb S_k$, and infers something about a subdivision of $G$.  The final element is to track how the second eigenvalue of $G$ changes (in expectation) under random subdivision.

Needless to say, this approach is already quite delicate, and for graphs that can’t be equipped with some kind of conformal structure, we seem to have reached a dead end.  In this lecture, we’ll see how to use intrinsic deformations of the geometry of $G$ in order to bound its eigenvalues.  Eventually, this will reduce to the study of certain kinds of multi-commodity flows.

### Metrics on graphs

Let $G=(V,E)$ be an arbitrary n-vertex graph with maximum degree $d_{\max}$.  Recall that we can write

$\displaystyle \lambda_2 = \min_{f \neq 0 : \sum_{x \in V} f(x)=0} \frac{\sum_{xy \in E} |f(x)-f(y)|^2}{\sum_{x \in V} f(x)^2}.$

where $f : V \to \mathbb R$.  (Also recall that we can replace $\mathbb R$ by any Hilbert space, and the same formula holds.)  The first step is to prepare this equality for “non-linearization” by getting rid of the linear condition $\sum_{x \in V} f(x)=0$ and the sum $\sum_{x \in V} f(x)^2$.  (This is a popular sort of passage in the non-linear geometry of Banach spaces, which also plays a rather important role in applications to the theoretical CS.)  The goal is to get only terms that look like $|f(x)-f(y)|$.  Fortunately, there is a well-known way to do this:

$\displaystyle \lambda_2 = 2 n \cdot \min_{f : V \to \mathbb R} \frac{\sum_{xy \in E} |f(x)-f(y)|^2}{\sum_{x,y \in V} |f(x)-f(y)|^2},$

which follows easily from the equality $\sum_{x,y \in V} |f(x)-f(y)|^2 = 2n \sum_{x \in V} f(x)^2$ when $\sum_{x \in V} f(x)=0$.

Thus if we want to bound $\lambda_2 = O(1/n)$, we need to find an $f : V \to \mathbb R$ for which the latter ratio (without the $2n$) is $O(1/n^2)$.  Now, for someone who works a lot with linear programming relaxations, it’s very natural to consider a “relaxation”

$\displaystyle \gamma(G) = \min_{d} \frac{\sum_{xy \in E} d(x,y)^2}{\sum_{x,y \in V} d(x,y)^2},$

where the minimization is over all pseudo-metrics d, i.e. symmetric non-negative functions $d : V \times V \to \mathbb R$ which satisfy the triangle inequality, but might have $d(x,y)=0$ even for $x \neq y$.  Certainly $\gamma(G) \leq \lambda_2/2n$, but Bourgain’s embedding theorem (which states that every n-point metric space embeds into a Hilbert space with distortion at most $O(\log n)$) also assures us that $\lambda_2(G) \leq O(n \log^2 n) \gamma(G)$.  Since we are trying to show that $\gamma(G) = O(1/n^2)$, this $O(\log^2 n)$ term is morally negligible.  One can see the paper for a more advanced embedding argument that doesn’t lose this factor, but for now we concentrate on proving that $\gamma(G) = O(1/n^2)$.  The embedding theorems allow us to concentrate on finding an intrinsic metric on the graph with small “Rayleigh quotient,” without having to worry about an eventual geometric representation.

As a brief preview… we are going to find a good metric $d$ by taking a certain kind of all-pairs multi-commodity flow at optimality, and weighting the edges by their congestion in the optimal flow.  Thus as the flow spreads out on the graph, it has the effect of “uniformizing” its geometry.

### Discrete Riemannian metrics, convexification, and duality

Let’s now assume that $G$ is planar.  We want to show that $\gamma(G) = O(d_{\max}/n^2)$.  First, let’s restrict ourselves to vertex weighted metrics on $G$.  Given any non-negative weight function $\omega : V \to \mathbb R$, we can define the length of a path $P$ in $G$ by summing the weights of vertices along it:  $\mathsf{len}_{\omega}(P) = \sum_{x \in P} \omega(x)$.  Then we can define a vertex-weighted shortest-path pseudo-metric on $G$ in the natural way

$\displaystyle \mathsf{dist}_{\omega}(x,y) = \min \left\{ \mathsf{len}_{\omega}(P) : P \in \mathcal P_{xy}\right\},$

where $\mathcal P_{uv}$ is the set of all u-v paths in $G$.  We also have the nice relationship

$\displaystyle \sum_{xy \in E} \mathsf{dist}_{\omega}(x,y)^2 \leq 2 d_{\max} \sum_{x \in V} \omega(x)^2,\qquad(1)$

since $\mathsf{dist}_{\omega}(x,y) \leq [\omega(x)+\omega(y)]^2$.

So if we define

$\displaystyle \Lambda_0(\omega) = \frac{\displaystyle \sum_{x \in V} \omega(x)^2}{\displaystyle \sum_{x,y \in V} \mathsf{dist}_{\omega}(x,y)^2}$

then by (1), we have $\gamma(G) \leq 2 d_{\max} \min_{\omega} \Lambda_0(\omega)$.

Examples. Let’s try to exhibit weights $\omega$ for two well-known examples:  the grid, and the complete binary tree.

For the $\sqrt{n} \times \sqrt{n}$ grid, we can simply take $\omega(x)=1$ for all $x \in V$.  Clearly $\sum_{x \in V} \omega(x)^2 = n$.  On the other hand, a random pair of points in the grid is $\Omega(\sqrt{n})$ apart, hence $\sum_{x,y \in V} \mathsf{dist}_{\omega}(x,y)^2 \approx n^2 \cdot (\sqrt{n})^2 = n^3$.  It follows that $\Lambda_0(\omega) = O(1/n^2)$, as desired.

For the complete binary tree with root $r$, we can simply put $\omega(r)=1$ and $\omega(x)=0$ for $x \neq r$.  (Astute readers will guess the geometrically decreasing weights are actually the optimal choice.)  In this case, $\sum_{x \in V} \omega(x)^2 = 1$, while all the pairs $x,y$ on opposite sides of the root have $\mathsf{dist}_{\omega}(x,y)=1$.  It again follows that $\Lambda_0(\omega) = O(1/n^2)$.  Our goal is to provide such a weight $\omega$ for any planar graph.

A natural way to study the optimal weight $\omega$ is via duality; unfortunately, $\min_{\omega} \Lambda_0(\omega)$ cannot be written as a convex program.  Thus we will pass to a “convexified” objetive:

$\displaystyle \Lambda(\omega) = \frac{\displaystyle \sqrt{\sum_{x \in V} \omega(x)^2}}{\displaystyle \sum_{x,y \in V} \mathsf{dist}_{\omega}(x,y)}$

The key here is that we have replaced the denominator by a linear form, which allows us to write $\min_{\omega} \Lambda(\omega)$ as a convex program.  To relate it to $\Lambda_0$, note that by Cauchy-Schwarz, for every $\omega,$

$\Lambda_0(\omega) \leq n^2 \Lambda(\omega)^2,$

so our task reduces to finding an $\omega$ for which $\Lambda(\omega)=O(1/n^2)$.  To see why we are not going lose much in the Cauchy-Schwarz bound, note that in the two “extremal” examples above, $\mathsf{dist}_{\omega}(x,y)$ was concentrated near its maximum value (the tight case for Cauchy-Schwarz).

Now that we have a convex objective, we can find its dual.  If you are used to dealing with programs where the triangle inequality occurs in the primal, you already know that some kind of flow is going to rear its head in the dual.

Multi-flows and $\ell_2$-congestion.

Let $\mathcal P = \bigcup_{u,v \in V} \mathcal P_{uv}$ be the set of all paths in $G$.  A flow in G will simply be a non-negative mapping $F : \mathcal P \to \mathbb R$ which assigns values to paths.  We will define the flow from u to v as the quantity $F[u,v] = \sum_{p \in \mathcal P_{uv}} F(p)$.  A complete flow in G is a flow F such that $F[u,v] = 1$ for all $u,v \in V$.  Finally, we define the congestion of the flow at a vertex v by $C_F(v) = \sum_{p : v \in p} F(p)$, i.e. the amount of flow going through $v$.  The $\ell_2$-congestion is now the quantity

$\displaystyle \mathsf{con}_2(F) = \sqrt{\sum_{v \in V} C_F(v)^2}$

Working out the Lagrangian multipliers and using strong duality (Slater’s condition) yields the following.

Theorem (Duality): For any graph $G$, we have

$\displaystyle \min_{\omega} \Lambda(\omega) = (\min_F \mathsf{con_2(F)})^{-1}$

where the minimums are over all vertex weights $\omega$ and over all complete flows $F$.

Thus in order to show the existence of a weight $\omega : V \to \mathbb R$ with $\Lambda(\omega) = O(1/n^2)$, we need to show that for every complete flow F in G, $\mathsf{con}_2(F) \gtrsim n^2$.

### Congestion, drawings, and crossings

First, let’s revisit our examples.  In the $\sqrt{n} \times \sqrt{n}$ grid, a complete flow has total “length” about $n^2 \sqrt{n}$ since about $n^2$ pairs need to send flow down a path of length at least $\sqrt{n}$.  To minimize the $\ell_2$-congestion, we would spread this out evenly, putting $\frac{n^2 \sqrt{n}}{n} = n\sqrt{n}$ congestion at every node, yelding $\mathsf{con}_2(F) \gtrsim n^2$ for any complete flow $F$.  In the complete binary tree, the argument is even easier since clearly $\Omega(n^2)$ flow has to go through the root.

Randomized rounding.

First, let’s round to an integral flow, i.e. such that the u-v flow is sent along a single u-v path for every pair of vertices.  Let $F^*$ be the random integral flow that arises in the following way:  For every pair of vertices u and v, let choose a path $P \in \mathcal P_{uv}$ with probability $F(P)$.  It is easy to check that

$\mathbb E[\mathsf{con}_2(F^*)] \leq \mathsf{con}_2(F) + n^{3/2},$

Notice that $n^{3/2}$ is the amount of $\ell_2$-congestion incurred by any complete flow congesting against itself (every vertex sends out flow n, so even that incurs $n^{3/2}$ congestion).

Thus we can assume that our flow is integral.  But now since our graph $G$ is planar, it has a drawing in the plane.  Now, I claim that every complete integral flow $F$ induces a drawing of the complete graph $K_n$ on n vertices in the plane, via the drawing of $G$.  Furthermore, edges of $K_n$ cross only at vertices of $G$.

In the picture, the graph $G$ is represented by black edges, while some of the edges of $K_n$ are drawn in colors.

How many edge crossings occur in the drawing of $K_n$?  Well, we have no idea how many crossings there are “inside” a vertex of $G$, but if $k$ edges go in, we can make sure there are at most ${k \choose 2} = O(k^2)$ crossings (in the worst case, every edge crosses every other).  But this immediately implies that we have a drawing of $K_n$ in the plane with at most $\sum_{v \in V} C_F(v)^2$ crossings.  Now, by the well-known crossing number inequality (found independently by Leighton, and also Ajtai, Chvatal, Newborn, and Szemeredi), the number of such crossings is always at least $\Omega(n^4)$, implying that $\mathsf{con}_2(F) \gtrsim n^2$ for any complete flow in $G$.  This completes our alternate argument that $\lambda_2 \leq O(d_{\max}/n)$ for planar graphs.

In fact, it is trivial to prove the crossing inequality for the complete graph:  Every copy of $K_5$ must induce one crossing, now double count (it is easy to check that every crossing involving less than four points can be removed, so every crossing involves four points, hence we overcount by a factor of n, representing the one free vertex).  This kind of argument extends immediately to bounded genus graphs via similar crossing arguments.  For more advanced arguments, including the excluded-minor case conjectured by Spielman and Teng, one has to use not only complete graphs, but dense graphs, in which case the full power of the crossing number inequality is needed.  The beautiful probabilistic proof of Chazelle, Sharir, and Welzl (see the exposition on Terry Tao’s blog) can be appropriately generalized to excluded-minor graphs, and even to families of higher-dimensional graphs.