CS294 Lecture 6: Higher-order Cheeger Inequality

In which we state an analog of Cheeger’s inequalities for the ${k}$ -th smallest Laplacian eigenvalue, and we discuss the connection between this result and the analysis of spectral partitioning algorithms

1. Cheeger-type Inequalities for ${\lambda_k}$

Let ${G=(V,E)}$ be an undirected ${d}$ -regular graph, ${A}$ its adjacency matrix, ${L = I - \frac 1d A}$ its normalized Laplacian matrix, and ${0 = \lambda_1 \leq \cdots \leq \lambda_n \leq 2}$ be the eigenvalues of ${L}$ , counted with multiplicities and listed in non-decreasing order.

In Handout 2, we proved that ${\lambda_k = 0}$ if and only if ${G}$ has at least ${k}$ connected components, that is, if and only if there are ${k}$ disjoint sets ${S_1,\ldots,S_k}$ such that ${\phi(S_i) = 0}$ for ${i=1,\ldots,k}$ . In this lecture and the next one we will prove a robust version of this fact.

First we introduce the notion of higher-order expansion. If ${S_1,\ldots,S_k}$ is a collection of disjoint sets, then their order- ${k}$ expansion is defined as

$\displaystyle \phi_k (S_1,\ldots,S_k) = \max_{i=1,\ldots,k} \ \ \phi(S_i)$

and the order- ${k}$ expansion of a graph ${G}$ is

$\displaystyle \phi_k(G )= \min_ {S_1,\ldots,S_k \ \rm disjoint } \ \ \phi(S_1,\ldots,S_k)$

If the edges of a graph represent a relation of similarity of affinity, a low-expansion collection of sets ${S_1,\ldots,S_k}$ represents an interesting notion of clustering, because the vertices in each set ${S_i}$ are more related to each other than to the rest of the graph. (Additional properties are desirable in a good clustering, and we will discuss this later.)

We will prove the following higher-order Cheeger inequalities:

$\displaystyle \frac { \lambda_k } 2 \leq \phi_k(G) \leq O(k^{3.5} ) \sqrt {\lambda_k}$

Stronger upper bounds are known, but the bound above is easier to prove from scratch. It is known that ${\phi_k (G) \leq O(k^2) \sqrt{\lambda_k}}$ and that ${\phi_k(G) \leq O_\epsilon ( \sqrt {\log k} ) \cdot \sqrt {\lambda_{(1+\epsilon) \cdot k} } }$ .

2. The Easy Direction

As usual, the direction ${\frac { \lambda_k } 2 \leq \phi_k(G) }$ is the easy one, and it comes from viewing ${\lambda_k}$ as a sort of continuous relaxation of the problem of minimizing order- ${k}$ expansion.

Recall that, in order to prove the easy direction of Cheeger’s inequality for ${\lambda_2}$ , we proved that if ${{\bf x}}$ and ${{\bf y}}$ are two orthogonal vectors, both of Rayleigh quotient at most ${\epsilon}$ , then the Rayleigh quotient of their sum is at most ${2\epsilon}$ . A similar argument could be made to show that the Rayleigh quotient of the sum of ${k}$ such vectors is at most ${k\epsilon}$ . Such results hold for every positive semidefinite matrix.

In the special case of the Laplacian of a graph, and of vectors that are not just orthogonal but actually disjointly supported, then we can lose only a factor of 2 instead of a factor of ${k}$ . (The support of a vector is the set of its non-zero coordinates; two vectors are disjointly supported if their supports are disjoint.)

Lemma 1 Let ${{\bf x}^{(1)}, \ldots, {\bf x}^{(k)}}$ be disjointly supported vectors. Then

$\displaystyle R_L \left( \sum_i{\bf x}^{(i)} \right) \leq 2 \cdot \max_{i=1,\ldots, k} R_L({\bf x}^{(i)} )$

Proof: We just have to prove that, for every edge ${\{ u, v \}}$ ,

$\displaystyle \left( \sum_i x^{(i)}_u - x^{(i)}_v \right) ^2 \leq 2 \sum_i ( x^{(i)}_u - x^{(i)}_v ) ^2$

The support disjointness implies that there is an index ${j}$ such that ${x^{(i)}_u = 0}$ for ${i\neq j}$ , and an index ${k}$ such that ${x^{(i)}_v =0}$ for ${i\neq k}$ . If ${j=k}$ , then

$\displaystyle \left( \sum_i x^{(i)}_u - x^{(i)}_v \right) ^2 = ( x^{(j)}_u - x^{(j)}_v )^2 = \sum_i ( x^{(i)}_u - x^{(i)}_v ) ^2$

and, if ${j\neq k}$ , then

$\displaystyle \left( \sum_i x^{(i)}_u - x^{(i)}_v \right) ^2 = (x^{(j)}_u - x^{(k)}_v )^2$

$\displaystyle \leq 2 (x^{(j)}_u)^2 + 2 ( x^{(k)}_v )^2 = 2 \sum_i ( x^{(i)}_u - x^{(i)}_v ) ^2$

and now, using also the fact that disjoint support implies orthogonality, we have

$\displaystyle R_L \left( \sum_i{\bf x}^{(i)} \right) = \frac {\sum_{\{ u,v \}} \left( \sum_i x^{(i)}_u - x^{(i)}_v \right) ^2 } { \left \| \sum_i{\bf x}^{(i)} \right\| ^2 }$

$\displaystyle \leq 2 \frac { \sum_i \sum_{ \{ u,v \} \in E} ( x^{(i)}_u - x^{(i)}_v ) ^2 } { \sum_i || {\bf x}^{(i)} ||^2}$

$\displaystyle \leq 2 \max_{i=1,\ldots,k} \ \ R_L( {\bf x}^{(i)} )$

$\Box$

To finish the proof of the easy direction, let ${S_1,\ldots,S_k}$ be sets such that ${\phi(S_i) \leq \phi(G)}$ for every ${i}$ . Consider the ${k}$ -dimensional space ${X}$ of linear combinations of the indicator vectors ${{\bf 1}_{S_i}}$ of such sets. The indicator vectors have Rayleigh quotient at most ${\phi(G)}$ and are disjointly supported, so all their linear combinations have Rayleigh quotient at most ${2\phi(G)}$ . We have found a ${k}$ -dimensional space of vectors all of Rayleigh quotient ${\leq 2 \phi(G)}$ , which proves ${\lambda_k \leq 2 \phi(G)}$ .

3. The Difficult Direction: Main Lemma

We will prove the following result

Lemma 2 (Main) Let ${{\bf x}^{(1)},\ldots, {\bf x}^{(k)}}$ be orthonormal vectors. Then we can find disjointly supported non-negative vectors ${{\bf y}^{(1)},\ldots, {\bf y}^{(k)}}$ such that for every ${i=1,\ldots,k}$

$\displaystyle R_L( {\bf y}^{(i)} ) \leq O(k^7) \cdot \max_{j=1,\ldots,k} R_L({\bf x}^{(j)} )$

By applying the Main Lemma to the eigenvectors of ${\lambda_1,\ldots,\lambda_k}$ , we get disjointly supported vectors ${{\bf y}^{(1)},\ldots, {\bf y}^{(k)}}$ all of Rayleigh quotient at most ${O(k^7) \cdot \lambda_k}$ . In a past lecture, we proved that for every non-negative vector ${{\bf y}}$ there is a subset ${S}$ of its support such that ${\phi(S) \leq \sqrt { 2 R_L({\bf y})}}$ , and applying this fact to the vectors ${{\bf y}^{(1)},\ldots, {\bf y}^{(k)}}$ we find ${k}$ disjoint sets all of expansion at most ${O(k^{3.5} ) \cdot \sqrt{\lambda_k}}$ , proving

$\displaystyle \phi_k (G) \leq O(k^{3.5}) \cdot \sqrt { \lambda_k }$

It is possible, with a more involved proof, to improve the ${O(k^7)}$ factor in the conclusion of the Main Lemma to ${O(k^6)}$ , implying that ${ \phi_k (G) \leq O(k^{3}) \cdot \sqrt { \lambda_k }}$ . A different approach, which we will not discuss, is used to show that, given ${k}$ orthonormal vectors, one can find ${k}$ disjoint sets ${S_1,\ldots,S_k}$ such that, for all ${i}$ ,

$\displaystyle \phi(S_i) \leq O(k^2) \cdot \sqrt{ \max_{j=1,\ldots,k} R_L({\bf x}^{(j)} ) }$

implying ${ \phi_k (G) \leq O(k^{2}) \cdot \sqrt { \lambda_k }}$ , which is the best known bound.

Note that, in all the known arguments, the bounds still hold if one replaces ${\lambda_k}$ by the (possibly smaller) quantity

$\displaystyle \inf_{ {\bf x}^{(1)}, \ldots, {\bf x}^{(k)}\ \rm orthonormal} \ \ \ \max_{i=1,\ldots,k} R_L({\bf x}^{(i)} ) \ \ \ \ \ (1)$

There are graphs, however, in which

$\displaystyle \phi_k(G) \geq \Omega ( \sqrt k ) \cdot \sqrt { \inf_{ {\bf x}^{(k)}, \ldots, {\bf x}^{(k)}\ \rm orthonormal} \ \ \ \max_{i=1,\ldots,k} R_L({\bf x}^{(i)} )}$

so, if a bound of the form ${\phi_k(G) \leq (\log k)^{O(1)} \cdot \sqrt{\lambda_k}}$ is true, then, in order to prove it, we need to develop new techniques that distinguish between ${\lambda_k}$ and the quantity (1).

4. The Spectral Embedding

Given orthonormal vectors ${{\bf x}^{(1)},\ldots, {\bf x}^{(k)}}$ as in the premise of the Main Lemma, we define the mapping ${F: V \rightarrow {\mathbb R}^k}$

$\displaystyle F(v) := ( x^{(1)}_v , \ldots, x^{(k)}_v ) \ \ \ \ \ (2)$

If ${{\bf x}^{(1)},\ldots, {\bf x}^{(k)}}$ are the eigenvectors of the ${k}$ smallest Laplacian eigenvalues of ${L}$ , then ${F(\cdot)}$ is called the spectral embedding of ${G}$ into ${{\mathbb R}^k}$ . Spectral clustering algorithms compute such an embedding, and then find clusters of nodes by clustering the points ${\{ F(v) : v\in V \}}$ using geometric clustering algorithms, such as ${k}$ -means, according either to Euclidian distance, or to the normalized distance function

$\displaystyle dist(u,v) := \left \| \ \frac {F(u)} { ||F(u) ||} - \frac {F(v)} { ||F(v) ||} \right \| \ \ \ \ \ (3)$

Our construction of disjointly supported vectors with small Rayleigh quotient will proceed similarly, by working only with the points ${\{ v : F(v) \}}$ and forgetting the edge structure of the graph, and by making use of the above distance function.

To develop some intuition about the spectral mapping, we introduce a notion of Laplacian Rayleigh quotient for a mapping ${f: V\rightarrow {\mathbb R}^k}$ , defined, by formally replacing absolute values with norms, as

$\displaystyle R_L(f) := \frac { \sum_{ \{ u,v \}} || f(u) - f(v) ||^2}{d \sum_v || f(v)||^2}$

For a mapping ${F: V \rightarrow {\mathbb R}^k}$ defined in terms of ${k}$ orthonormal vectors ${{\bf x}^{(i)}}$ as in (2), we have

$\displaystyle R_L(F) = \frac { \sum_{ \{ u,v \}} \sum_i \ (x^{(i)}_u - x^{(i)}_v ) ^2}{d \sum_v \sum_i (x_v^{(i)} )^2 }$

$\displaystyle = \frac { \sum_i \sum_{ \{ u,v \}} \ (x^{(i)}_u - x^{(i)}_v ) ^2}{d k }$

$\displaystyle = \frac 1k \sum_i \frac { \sum_{ \{ u,v \}} \ (x^{(i)}_u - x^{(i)}_v ) ^2 } {d}$

$\displaystyle = \frac 1k \sum_i R_L( {\bf x}^{(i)} )$

$\displaystyle \leq \max_{i=1,\ldots, k} \ \ R_L( {\bf x}^{(i)} )$

In particular, if ${{\bf x}^{(i)}}$ are the eigenvectors of the ${k}$ smallest Laplacian eigenvalues, then ${R_L(F) \leq \lambda_k}$ .

Let us use this setup to prove again that if ${\lambda_k=0}$ then ${G}$ has at least ${k}$ connected components. If ${\lambda_k=0}$ , and we construct ${F(\cdot)}$ using the eigenvectors of the smallest Laplacian eigenvalues, then ${R_L(F) = 0}$ , which means that ${F(u) = F(v)}$ for every edge ${\{ u,v \}}$ , and so ${F(u)=F(v)}$ for every ${u}$ and ${v}$ which are in the same connected component. Equivalently, if ${F(u) \neq F(v)}$ , then ${u}$ and ${v}$ are in different connected component. For every point in the range ${\{ F(v): v\in V \}}$ in the range of ${F(\cdot)}$ , let us consider its pre-image, and let ${S_1,\ldots,S_t}$ be the sets constructed in this way. Clearly, every set has expansion zero.

How many sets do we have? We claim that the range of ${F(\cdot)}$ must contain at least ${k}$ distinct points, and so ${t\geq k}$ and ${G}$ has at least ${k}$ connected component. To prove the claim, consider the matrix ${X}$ whose rows are the vectors ${{\bf x}^{(i)}}$ ; since the rows of ${X}$ are linearly independent, ${X}$ has full rank ${k}$ ; but if the range of ${F(\cdot)}$ contained ${\leq k-1}$ distinct points, then ${X}$ would have ${\leq k-1}$ distinct columns, and so its rank would be ${\leq k-1}$ .

Our proof of the higher-order Cheeger inequality will be somewhat analogous to the previous argument: we will use the fact that, if the Rayleigh quotient of ${F(\cdot)}$ is small, then the endpoints of edges ${\{ u,v \}}$ are typically close, in the sense that the distance defined in (3) between ${u}$ and ${v}$ will typically be small; we will also use the fact that, because the ${{\bf x}^{(i)}}$ are orthonormal, ${F(\cdot)}$ tends to “spread out” vertices across ${{\mathbb R}^k}$ , so that we can find ${k}$ regions of ${{\mathbb R}^k}$ each containing a large (in a certain weighted sense) number of vertices, and such that the regions are well-separated according to the distance (3), implying that there are few edges crossing from one region to the other, so that the vertices in each region are a non-expanding set. (This is an imprecise description of the argument, but it conveys the basic intuition.)

5. Overview of the Proof of the Main Lemma

We will break up the proof of the Main Lemma into the following two Lemmas.

Lemma 3 (Well-Separated Sets) Given a function ${F: V \rightarrow {\mathbb R}^k}$ defined in terms of ${k}$ orthonormal vectors as in (2), we can find ${k}$ disjoint subsets of vertices ${A_1,\ldots, A_k}$ such that

For every ${i=1,\ldots,k}$ , ${\sum_{v\in A_i} ||F(v)||^2 \geq \frac 14}$
For every ${u}$ and ${v}$ belonging to different sets, ${dist(u,v) \geq \Omega(k^{-3})}$

Lemma 4 (Localization) Given a function ${F: V \rightarrow {\mathbb R}^k}$ defined in terms of ${k}$ orthonormal vectors as in (2), and ${t}$ sets ${A_1,\ldots,A_t}$ such that, for every ${i=1,\ldots, t}$ , ${\sum_{v \in A_i} || F(v)||^2 \geq \frac 14}$ and,for every ${u,v}$ in different sets ${dist(u,v) \geq \delta}$ , we can construct ${t}$ disjointly supported vectors ${{\bf y}^{(1)},\ldots, {\bf y}^{(t)}}$ such that for every ${i=1,\ldots, t}$ , we have

$\displaystyle R_L( {\bf y}^{(t)} ) \leq O(k \cdot \delta^{-2}) \cdot R_L(F)$

5 thoughts on “CS294 Lecture 6: Higher-order Cheeger Inequality”

No-Stupid-Questions on February 10, 2016 at 3:06 pm said:

For the second equation, I believe that the k-way edge expansion (of a graph) should be defined as the __minimum__ edge expansion of any choice of k disjoint subsets.
luca on February 10, 2016 at 3:25 pm said:

Thanks, I corrected it
Arnold Filtser on February 18, 2016 at 2:22 pm said:

Thank you for the great note.
Two small corrections (w.r.t pdf file):
eq. (1) from pg. 4. should be x^(1),…,x^(k) and not x^(k),…,x^(k).
The second equation in pg. 5. While the conclusion is true, the first equality seems to me to be simply wrong.
luca on February 18, 2016 at 2:39 pm said:

Thanks. That equality is correct, but I guess it was not obvious, so I added some intermediate steps
Yu Zhao on February 18, 2016 at 5:56 pm said:

Typo in last line of Lemma 4, R_L(y^(i)) not R_L(y^(t))

in theory

"Marge, I agree with you – in theory. In theory, communism works. In theory." — Homer Simpson

CS294 Lecture 6: Higher-order Cheeger Inequality

5 thoughts on “CS294 Lecture 6: Higher-order Cheeger Inequality”

Leave a Reply Cancel reply

Share this:

Like this:

Related

5 thoughts on “CS294 Lecture 6: Higher-order Cheeger Inequality”

Leave a Reply Cancel reply