# CS294 Lecture 3: Cheeger Inequalities

In which we generalize the notion of normalized Laplacian to irregular graphs, we extend the basic spectral graph theory results from last lecture to irregular graphs, and we prove the easy direction of Cheeger’s inequalities.

1. Irregular Graphs

Let ${G=(V,E)}$ be an undirected graph, not necessarily regular. We will assume that every vertex has non-zero degree. We would like to define a normalized Laplacian matrix associated to ${G}$ so that the properties we proved last time are true: that the multiplicity of 0 as an eigenvalue is equal to the number of connected components of ${G}$, that the largest eigenvalue is at most 2, and that it is 2 if and only if (a connected component of) the graph is bipartite.

# CS294 Lecture 2: Basics of Spectral Graph Theory

In which we introduce the Laplacian matrix and we prove our first results in spectral graph theory.

1. The Basics of Spectral Graph Theory

Given an undirected graph ${G= (V,E)}$, the approach of spectral graph theory is to associate a symmetric real-valued matrix to ${G}$, and to related the eigenvalues of the matrix to combinatorial properties of ${G}$.

For the sake of this lecture, we will restrict ourselves to the case in which ${G}$ is a ${d}$-regular graph, and we will then see how to extend our results to apply to irregular graphs as well.

The most natural matrix to associate to ${G}$ is the adjacency matrix ${A}$ such that ${A_{i,j} = 1}$ if ${\{ i,j\} \in E}$ and ${A_{i,j} = 0}$ otherwise. In the second part of the course, in which we will study expander graphs, the adjacency matrix will indeed be the most convenient matrix to work with. For the sake of the algorithms that we will analyze in the first part of the course, however, a slight variation called the normalized Laplacian is more convenient.

# References on Laplacian eigenvalues and graph properties

After my lectures in the “boot camp” of the spectral graph theory program at the Simons Institute, I promised I would post some references, because I stated all results without attribution.

Here is a a first draft.

If you notice that work that you know of (for example, your work) is misrepresented or absent, please let me know and I will edit the document. (If possible, when you do so, do not compare me to Stalin and cc your message to half a dozen prominent people — true story.)

# The Power Method

This week, the topic of my online course on graph partitioning and expanders is the computation of approximate eigenvalues and eigenvectors with the power method.

If $M$ is a positive semidefinite matrix (a symmetric matrix all whose eigenvalues are nonnegative), then the power method is simply to pick a random vector $x\in \{ -1,+1 \}^n$, and compute $y:= M^k x$. If $k$ is of the order of $\frac 1 \epsilon \log \frac n \epsilon$, then one has a constant probability that

$\frac {y^T M y}{y^T y} \geq (1-\epsilon) \max_{x} \frac {x^T M x}{x^T x} = (1-\epsilon) \lambda_1$

where $\lambda_1$ is the largest eigenvalue of $M$. If we are interested in the Laplacian matrix $L = I - \frac 1d A$ of a $d$-regular graph, where $A$ is the adjacency matrix of the graph, this gives a way to compute an approximation of the largest eigenvalue, and a vector of approximately maximum Rayleigh quotient, which is useful to approximate Max Cut, but not to apply spectral partitioning algorithms. For those, we need a vector that approximates the eigenvector of the second smallest eigenvalue.

Equivalently, we want to approximate the second largest eigenvalue of the adjacency matrix $A$. The power method is easy to adjust to compute the second largest eigenvalue instead of the largest (if we know an eigenvector of the largest eigenvalue): after you pick the random vector, subtract the component of the vector that is parallel to the eigenvector of the largest eigenvalue. In the case of the adjacency matrix of a regular graph, subtract from every coordinate of the random vector the average of the coordinates.

The adjacency matrix is not positive semidefinite, but we can adjust it to be by adding a multiple of the identity matrix. For example we can work with $\frac 12 I + \frac 1{2d} A$. Then the power method reduces to the following procedure: pick randomly $x \sim \{ -1,1\}$, then subtract $\sum_i x_i/n$ from every entry of $x$, then repeat the following process $k = O\left( \frac 1 \epsilon \log \frac n \epsilon \right)$ times: for every entry $i$, assign $x_i := \frac 12 x_i + \frac 1 {2d} \sum_{j: (i,j) \in E} x_j$, that is, replace the value that the vector assigns to vertex $i$ with a convex combination of the current value and the current value of the neighbors. (Note that one iteration can be executed in time $O(|V|+|E|)$.

The problem is that if we started from a graph whose Laplacian matrix has a second smallest eigenvalue $\lambda_2$, the matrix $\frac 12 I + \frac 1{2d} A$ has second largest eigenvalue $1- \frac {\lambda_2}2$, and if the power method finds a vector of Rayleigh quotient at least $(1-\epsilon) \cdot \left( 1- \frac {\lambda_2}2 \right)$ for $\frac 12 I + \frac 1{2d} A$, then that vector has Rayleigh quotient about $\lambda_2 - 2\epsilon$ for $L$, and unless we choose $\epsilon$ of the same order as $\lambda_2$ we get nothing. This means that the number of iterations has to be about $1/\lambda_2$, which can be quite large.

The video below (taken from this week’s lecture) shows how slowly the power method progresses on a small cycle with 31 vertices. It goes faster on the hypercube, which has a much larger $\lambda_2$.

A better way to apply the power method to find small eigenvalues of the Laplacian is to apply the power method to the pseudoinverse $L^+$ of the Laplacian. If the Laplacian of a connected graph has eigenvalues $0 = \lambda_1 < \lambda_2 \leq \cdots \leq \lambda_n$, then the pseudoinverse $L^+$ has eigenvalues $0, \frac 1 {\lambda_2}, \cdots, \frac 1 {\lambda_n}$ with the same eigenvectors, so approximately finding the largest eigenvalue of $L^+$ is the same problem as approximately finding the second smallest eigenvalue of $L$.

Although we do not have fast algorithms to compute $L^+$, what we need to run the power method is, for a given $x$, to find the $y$ such that $L y = x$, that is, to solve the linear system $Ly = x$ in $y$ given $L$ and $x$.

For this problem, Spielman and Teng gave an algorithm nearly linear in the number of nonzero of $L$, and new algorithms have been developed more recently (and with some promise of being practical) by Koutis, Miller and Peng and by Kelner, Orecchia, Sidford and Zhu.

Coincidentally, just this week, Nisheeth Vishnoi has completed his monograph Lx=b on algorithms to solve such linear systems and their applications. It’s going to be great summer reading for those long days at the beach.

# Proof of the Cheeger inequality in manifolds

Having (non-rigorously) defined the Laplacian operator in manifolds in the previous post, we turn to the proof of the Cheeger inequality in manifolds, which we restate below.

Theorem 1 (Cheeger’s inequality) Let ${M}$ be an ${n}$-dimensional smooth, compact, Riemann manifold without boundary with metric ${g}$, let ${L:= - {\rm div} \nabla}$ be the Laplace-Beltrami operator on ${M}$, let ${0=\lambda_1 \leq \lambda_2 \leq \cdots }$ be the eigenvalues of ${L}$, and define the Cheeger constant of ${M}$ to be

$\displaystyle h(M):= \inf_{S\subseteq M : \ 0 < \mu(S) \leq \frac 12 \mu(M)} \ \frac{\mu_{n-1}(\partial(S))}{\mu(S)}$

where the ${\partial (S)}$ is the boundary of ${S}$, ${\mu}$ is the ${n}$-dimensional measure, and ${\mu_{n-1}}$ is ${(n-1)}$-th dimensional measure defined using ${g}$. Then

$\displaystyle h(M) \leq 2 \sqrt{\lambda_2} \ \ \ \ \ (1)$

We begin by recalling the proof of the analogous result in graphs, and then we will repeat the same steps in the context of manifolds.

Theorem 2 (Cheeger’s inequality in graphs) Let ${G=(V,E)}$ be a ${d}$-regular graph, ${A}$ be its adjacency matrix, ${L:= I - \frac 1d A}$ be its normalized Laplacian matrix, ${0 = \lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_{|V|}}$ be the eigenvalues of ${L}$, and define ${\mu(S):= d \cdot |S|}$ for every subset of vertices ${S\subseteq V}$. Define the conductance of ${G}$ as

$\displaystyle \phi(G) := \min_{S\subseteq V: \ 0 < \mu(S) \leq \frac 12 \mu(V) } \frac{| \partial S|}{\mu(S)}$

where ${\partial S}$ is the number of edges with one endpoint in ${S}$ and one endpoint in ${\bar S}$. Then

$\displaystyle \phi(G) \leq \sqrt{2 \lambda_2} \ \ \ \ \ (2)$

1. Proof the Cheeger inequality in graphs

We will use the variational characterization of the eigenvalues of the Laplacian ${L}$ of a graph ${G}$.

$\displaystyle \lambda_1 = \min_{f \in {\mathbb R}^v} \frac {f^T Lf}{f^T f} \ \ \ \ \ (3)$

and if ${f_1}$ is a minimizer in the above expression then

$\displaystyle \lambda_2 = \min_{f \in {\mathbb R}^V: \ f \perp f_1} \frac {f^T Lf}{f^T f}$

Following the definition of ${L}$ we see that

$\displaystyle f^T L f = \frac 1d \sum_{(u,v)\in E} |f_u - f_v |^2$

and so the minimum in (3) is 0, and it is achieved for ${f = (1,\cdots,1)}$. This means that

$\displaystyle \lambda_2 = \min_{f \in {\mathbb R}^V: \ \sum_v f_v = 0} \frac {\sum_{(u,v) \in E} |f_u-f_v|^2}{d \sum_v f_v^2} \ \ \ \ \ (4)$

The expression in the right-hand-side of (4) is an important one, and it is called the Rayleigh quotient of ${f}$, which we will denote by ${R(f)}$:

$\displaystyle R(f):= \frac {\sum_{(u,v) \in E} |f_u-f_v|^2}{d \sum_v f_v^2}$

It is also useful to consider the variant of the Rayleigh quotient where there are no squares; this does not have a standard name, so let us call it the ${\ell_1}$ Rayleigh quotient and denote it by ${R_1}$:

$\displaystyle R_1(g):= \frac {\sum_{(u,v) \in E} |g_u-g_v|}{d \sum_v |g_v|}$

The proof of the graph Cheeger inequality now continues with the proof of the following three facts.

Lemma 3 (Rounding of ${\ell_1}$ embeddings) For every non-negative vector ${g\in {\mathbb R}^V_{\geq 0}}$ there is a value ${t\geq 0}$ such that

$\displaystyle \phi(\{ v: g(v) > t \}) \leq R_1(g)$

Lemma 4 (Embedding of ${\ell_2^2}$ into ${\ell_1}$) For every non-negative vector ${f\in {\mathbb R}^V_{\geq 0}}$, we have

$\displaystyle R_1(f^2) \leq \sqrt{2 R(f)}$

Lemma 5 (From an eigenvector to a non-negative vector) For every ${f\in {\mathbb R}^V}$ such that ${\sum_v f_v =0}$ there is a non-negative ${f'\in {\mathbb R}^V_{\geq 0}}$ such that ${\mu(\{ v: f'(v) >0 \}) \leq \frac 12 \mu(V)}$ and such that

$\displaystyle R(f') \leq R(f)$

Now let us start from a function ${f}$ that optimizes (4), so that ${\sum_v f_v = 0}$ and ${R(f) = \lambda_2}$, then apply Lemma 5 to find a function ${f'}$ such that the volume of the vertices having positive coordinates in ${f'}$ is at most ${\frac 12 \mu(V)}$ and such that ${R(f') \leq R(f) = \lambda_2}$. Then consider the vector ${g\in {\mathbb R}^V}$ such that ${g_v := f'^2_v}$; by Lemma 4, we have ${R_1(g) \leq \sqrt{2 R(f')} \leq \sqrt{2\lambda_2}}$, and by Lemma 3 there is a threshold ${t}$ such that the set ${S:= \{ v: g(v) > t \}}$ has conductance ${h(S) \leq \sqrt {2\lambda_2}}$. Since ${S}$ is a subset of the vertices having positive coordinates in ${g}$, we have ${\mu(S) \leq \frac 12 \mu(V)}$, and so

$\displaystyle \phi(G) \leq \sqrt{2 \lambda_2 }$

which is the Cheeger inequality for graphs. It remains to prove the three lemmas.

Proof: of Lemma 3. For each threshold ${t}$, define the set

$\displaystyle S_t := \{ v: g_v > t \}$

The idea of the proof is that if we pick ${t}$ at random then the probability that an edge belongs to ${\partial S_t}$ is proportional to ${|g_u - g_v|}$ and the probability that ${v\in S_t}$ is proportional to ${|g_v|}$, so that the expected number of edges in ${\partial S_t}$ is proportional to the numerator of ${R_1(g)}$ and the expected number of vertices in ${S_t}$ is proportional to the denominator of ${R_1(g)}$; if ${R_1(g)}$ is small, it is not possible for ${\partial S_t/\mu(S_t)}$ to always be large for every ${t}$.

To avoid having to normalize the range of ${t}$ to be between ${0}$ and ${1}$, instead of taking averages over a random choice of ${t}$, we will consider the integral over all values of ${t}$. We have

$\displaystyle \int_0^\infty |\partial (S_t) | {\rm d} t = \sum_{u,v} |g_u - g_v |$

because we can write ${|\partial (S_t)| = \sum_{(u,v) \in E} I_{u,v} (t)}$, where ${I_{u,v} (t) = 1}$ if ${(u,v) \in \partial S_t}$ and ${I_{u,v}(t) = 0}$ otherwise, and we see that only the values of ${t}$ between ${g_u}$ and ${g_v}$ make ${I_{u,v}(t) =1}$, so ${\int_{0}^\infty I_{u,v} (t) {\rm d} t = |g_u - g_v|}$.

We also have

$\displaystyle \int_0^\infty d |S_t| {\rm d} t = d \sum_v |g_v|$

and if we denote by ${t^*}$ the threshold such that ${\phi(S_{t^*})}$ is smallest among all the ${\phi(S)}$, then

$\displaystyle \sum_{u,v} |g_u - g_v | = \int_0^\infty |\partial (S_t) | {\rm d} t$

$\displaystyle \geq \int_0^\infty h(S_{t^*}) d|S_t| {\rm d} t$

$\displaystyle = h(S_{t^*}) d \sum_v |g_v|$

so that

$\displaystyle h(S_{t^*}) \leq \frac { \sum_{u,v} |g_u - g_v | }{d \sum_v |g_v| } = R_1(g)$

$\Box$

Proof: of Lemma 3. Let us consider the numerator of ${R_1(f^2)}$; it is:

$\displaystyle \sum_{(u,v)\in E} |f^2_u - f^2_v|$

$\displaystyle = \sum_{(u,v)\in E} |f_u - f_v| \cdot (f_u + f_v)$

$\displaystyle \leq \sqrt{\sum_{(u,v)\in E} |f_u - f_v|^2} \sqrt{ \sum_{(u,v)\in E}(f_u + f_v)^2 }$

(we used Cauchy-Swarz)

$\displaystyle \leq \sqrt{R(f) \cdot d\sum_v f_v^2} \sqrt{ \sum_{(u,v)\in E} 2f_u^2 + 2f_v^2 }$

(we used the definition of ${R_2}$ and Cauchy-Swarz again)

$\displaystyle = \sqrt{R(f) \cdot d\sum_v |f_v|^2} \sqrt{ 2 d\sum_{v} f_v^2 }$

$\displaystyle = \sqrt{2 R(f) } \cdot d \sum_v f_v^2$

And so

$\displaystyle R_1(f^2) = \frac {\sum_{(u,v)\in E} |f^2_u - f^2_v|}{d \sum_v f_v^2} \leq \sqrt{2 R_2(f) }$

$\Box$

Proof: of Lemma 5. Let ${m}$ be the median of ${f}$, and consider ${\bar f}$ defined as ${\bar f_v := f_v - m}$. We have

$\displaystyle R(\bar f) \leq R(f)$

because the numerators of ${R(\bar f)}$ and ${R(f)}$ are the same (the additive term ${-m}$ cancels). The denominators are such that

$\displaystyle \sum_v \bar f_v^2 = || \bar f||^2 = || f||^2 + || - m \cdot {\bf 1} ||^2 \geq || f||^2 = \sum_v f_v^2$

because ${f}$ and the vector ${{\bf 1} = (1,\ldots,1)}$ are orthogonal, and so by Pythagoras’s theorem the length-squared of ${\bar f = f - m {\bf 1}}$ equals the length-squared of ${f}$ plus the length-squared of ${-m {\bf 1}}$.

Let us define ${f^+_v := \min\{ 0, \bar f_v\}}$ and ${f^-_v := \min \{ 0, -\bar f_v\}}$ so that ${\bar f = f^+ - f^-}$. We use the following fact:

Fact 6 Let ${a,b \in {\mathbb R}^V_{\geq 0}}$ be disjointly supported non-negative vectors (“disjointly supported” means that they are non-zero on disjoint subsets of coordinates), then

$\displaystyle \min\{ R(a) , R(b) \} \leq R(a-b)$

Proof: The numerator of ${R(a-b)}$ is

$\displaystyle \sum_{(u,v)\in E} |a_v - b_v + b_u - a_v|^2 \geq \sum_{(u,v)\in E} |a_v - a_u|^2 + |b_v - b_u|^2$

and, using orthogonality and Pythagoras’s theorem, the denominator of ${R(a-b)}$ is

$\displaystyle d || a- b||^2 = d||a ||^2 + d|| b||^2$

The fact now follows from the inequality

$\displaystyle \min \left \{ \frac {n_1}{d_1} , \frac{n_2}{d_2} \right \} \leq \frac{n_1+n_2}{d_1+d_2}$

$\Box$

The lemma now follows by observing that ${f^+}$ and ${f^-}$ are non-negative and disjointly supported, so

$\displaystyle \min\{ R(f^+) , R(f^-) \} \leq R(\bar f) \leq R(f)$

and that both ${f^+}$ and ${f^-}$ have at most ${n/2}$ non-zero coordinate. $\Box$

2. Proof of the Cheeger inequality in manifolds

We will now translate the proof of the graph Cheeger inequality to the setting of manifolds.

As you may remember, we started off by saying that ${L}$ is symmetric and so all its eigenvalues are real and they are given by the variational characterization. Now we are already in trouble because the operator ${L}$ on manifolds cannot be thought of as a matrix, so what does it mean for it to be symmetric? The consequence of symmetry that is exploited in the analysis of the spectrum of symmetric matrices is the fact that if ${A}$ is symmetric, then for every ${x,y}$ we have

$\displaystyle \langle x,Ay \rangle = x^T Ay = (A^Tx)^Ty = (Ax)^Ty = \langle Ax,y \rangle$

and the property ${\langle x, Ay \rangle = \langle Ax , y \rangle}$ makes no references to coordinates, and it is well defined even for linear operators over infinite-dimensional spaces, provided that there is a notion of inner product. If we the define the inner product

$\displaystyle \langle f,g \rangle := \int_M f(x)\cdot g(x) \ {\rm d} \mu(x)$

on functions ${f: M \rightarrow {\mathbb R}}$, and more generally

$\displaystyle \langle f,g \rangle := \int_M \langle f(x), g(x)\rangle_X \ {\rm d} \mu(x)$

for functions ${f: M \rightarrow X}$, where ${X}$ is a vector space with inner product ${\langle \cdot,\cdot,\rangle_X}$, then we can say that an operator ${A}$ is self-adjoint if

$\displaystyle \langle f, Ag \rangle = \langle Af ,g \rangle$

for all (appropriately restricted) functions ${f,g}$. If ${M}$ is compact, this property is true for the Laplacian, and, in particular, ${-{\rm div}}$ and ${\nabla}$ are adjoints of each others, that is,

$\displaystyle \langle \nabla f, g \rangle = \langle f, - {\rm div} g \rangle$

(The discrete analog would be that ${C^T}$ is the transpose of ${C}$.)

Self-adjointness (and appropriate conditions on ${M}$) imply a version of the spectral theorem and of the variational characterization. In particular, all eigenvalues of ${L}$ are real, and if there is a minimum one then it is

$\displaystyle \lambda_1 = \min_{f: M \rightarrow {\mathbb R}} \frac {\langle f, Lf \rangle}{\langle f,f\rangle}$

and if ${f_1}$ is a minimizer of the above, then

$\displaystyle \lambda_2 = \min_{f: M \rightarrow {\mathbb R},\ \langle f, f_1 \rangle = 0} \frac {\langle f, Lf \rangle}{\langle f,f\rangle}$

(The minimization is quantified over all functions that are square-integrable, and the minimum is achieved because if ${M}$ is compact then the space of such functions is also compact and the cost function that we are minimizing is continuous. In this post, whenever we talk about “all functions,” it should be understood that we are restricting to whatever space of functions makes sense in the context.)

From the property that ${-{\rm div}}$ and ${\nabla}$ are adjoint, we have

$\displaystyle \langle f, Lf \rangle = \langle f, -{\rm div} \nabla f \rangle = \langle \nabla f,\nabla f \rangle$

so

$\displaystyle \lambda_1 = \min_{f:M \rightarrow {\mathbb R}} \frac{ \int_M \langle \nabla f(x),\nabla f(x) \rangle \ {\rm d} \mu(x) } {\int M f^2(x) \ {\rm d} \mu (x)}$

where the Rayleigh quotient

$\displaystyle R(f):= \frac{ \int_M \langle \nabla f(x),\nabla f(x) \rangle \ {\rm d} \mu(x) } {\int M f^2(x) \ {\rm d} \mu (x)} = \frac {\int_M ||\nabla f(x)||^2 \ {\rm d} \mu (x)}{\int_M f^2(x) \ {\rm d} \mu (x)}$

is always non-negative, and it is zero for constant ${f\equiv 1}$, so we see that ${\lambda_1=0}$ and

$\displaystyle \lambda_2 = \min_{f:M \rightarrow {\mathbb R} : \int_M f=0 \ {\rm d} \mu} R(f)$

By analogy with the graph case, we define the “${\ell_1}$ Rayleigh quotient”

$\displaystyle R_1(g) := \frac {\int_M ||\nabla g(x)|| \ {\rm d} \mu (x)}{\int_M |g(x)| \ {\rm d} \mu (x)}$

And we can prove the analogs of the lemmas that we proved for graphs.

Lemma 7 (Rounding of ${\ell_1}$ embeddings) For every non-negative function ${g: V \rightarrow {\mathbb R}_{\geq 0}}$ there is a value ${t\geq 0}$ such that

$\displaystyle h(\{ x: g(x) > t \}) \leq R_1(g)$

where the Cheeger constant ${h(S)}$ of a subset ${S\subseteq M}$ of the manifold is

$\displaystyle h(S) := \frac{\mu_{n-1} (\partial S)}{\mu(S)}$

Lemma 8 (Embedding of ${\ell_2^2}$ into ${\ell_1}$) For every non-negative function ${f; m \rightarrow {\mathbb R}_{\geq 0}}$, we have

$\displaystyle R_1(f^2) \leq \sqrt{2 R(f)}$

Lemma 9 (From an eigenfunction to a non-negative function) For every function ${f: M \rightarrow R}$ such that ${\int_M f \ {\rm d} \mu =0}$ there is a non-negative ${f': M \rightarrow {\mathbb R}_{\geq 0}}$ such that ${\mu(\{ x: f'(x) >0 \}) \leq \frac 12 \mu(V)}$ and such that

$\displaystyle R(f') \leq R_2(f)$

Let us see the proof of these lemmas.

Proof: of Lemma 7. For each threshold ${t}$, define the set

$\displaystyle S_t := \{ x: g(x) > t \}$

Let ${t*}$ be a threshold for which ${h(S_{t*})}$ is minimized

We will integrate the numerator and denominator of ${h(S_t)}$ over all ${t}$. The coarea formula for nonnegative functions is

$\displaystyle \int_M || \nabla g || {\rm d} \mu = \int_{0}^\infty \mu_{n-1} ( \partial \{ x: g(x) > t \}) {\rm d} t$

and we also have

$\displaystyle \int_M |g| {\rm d} \mu = \int_{0}^\infty \mu ( \{ x: g(x) > t \}) {\rm d} t$

which combine to

$\displaystyle \int_M || \nabla g || {\rm d} t = \int_0^\infty \mu_{n-1} ( \partial S_t) {\rm d} t$

$\displaystyle = \int_0^\infty h(S_t) \mu(S_t) {\rm d} t$

$\displaystyle \geq h(S_{t^*}) \int_0^\infty \mu(S_t) {\rm d} t$

$\displaystyle = h(S_{t^*}) \int_M g {\rm d} t$

so that

$\displaystyle h(S_{t^*}) \leq \frac{\int_M || \nabla g || {\rm d} t }{ \int_M g {\rm d} t} = R_1(g)$

$\Box$

Proof: of Lemma 8. Let us consider the numerator of ${R_1(f^2)}$; it is:

$\displaystyle \int_M || \nabla f^2|| {\rm d} \mu$

We can apply the chain rule, and see that

$\displaystyle \nabla f^2(x) = 2f(x) \cdot \nabla f(x)$

which implies

$\displaystyle \int_M || \nabla f^2 || {\rm d} \mu$

$\displaystyle = \int_M || 2f(x) \nabla f(x)|| {\rm d} \mu(x)$

$\displaystyle = \int_M 2|f(x)| \cdot ||\nabla f(x)|| {\rm d} \mu (x)$

and, after applying Caucy-Swarz,

$\displaystyle \leq \sqrt{\int_M 4 f^2(x) {\rm d} \mu(x)} \cdot \sqrt{\int_M ||\nabla f(x)||^2 {\rm d} \mu(x)}$

$\displaystyle = 2 \cdot \left( \int_M f^2 {\rm d} \mu \right) \cdot \sqrt{ R(f)}$

And so

$\displaystyle R_1(f^2) \leq 2 \sqrt{R_2(f) }$

$\Box$

Proof: of Lemma 9. Let ${m}$ be a median of ${f}$, and consider ${\bar f}$ defined as ${\bar f(x) := f(x) - m}$. We have

$\displaystyle R(\bar f) \leq R(f)$

because the numerators of ${R(\bar f)}$ and ${R(f)}$ are the same (the derivatives of functions that differ by a constant are identical) and the denominators are such that

$\displaystyle \int_M \bar f^2{\rm d} \mu$

$\displaystyle = \int_M f^2 - 2mf + m^2 {\rm d} \mu$

$\displaystyle = \int_M f^2 + m^2 {\rm d} \mu$

$\displaystyle \geq \int_M f^2 {\rm d} \mu$

where we used the fact the integral of ${f}$ is zero.

Let us define ${f^+(x) := \min\{ 0, \bar f(x)\}}$ and ${f^-_v := \min \{ 0, -\bar f(x)\}}$ so that ${\bar f(x) = f^+(x) - f^-(x)}$. We use the following fact:

Fact 10 Let ${a,b : M \rightarrow {\mathbb R}_{\geq 0}}$ be disjointly supported non-negative functions (“disjointly supported” means that they are non-zero on disjoint subsets of inputs), then

$\displaystyle \min\{ R(a) , R(b) \} \leq R(a-b)$

Proof: We begin with the following observation: if ${a}$ is a non-negative function, and ${a(x)=0}$, then ${\nabla a(x) = \{ \bf 0 \}}$, because ${x}$ has to be a local minimum.

Consider the expression ${||\nabla (a-b)||^2}$ occurring in the numerator of ${R(a-b)}$. We have

$\displaystyle ||\nabla (a-b)||^2$

$\displaystyle = || \nabla a - \nabla b ||^2$

$\displaystyle = || \nabla a||^2 + || \nabla b||^2 - 2 \langle \nabla a,\nabla b \rangle$

But

$\displaystyle \langle \nabla a,\nabla b \rangle = 0$

because for every ${x}$ at least one of ${a(x)}$ or ${b(x)}$ is zero, and so at least one of ${\nabla a(x)}$ or ${\nabla b(x)}$ is zero.

Using this fact, we have that the numerator of ${R(a-b)}$ is equal to the sum of the numerators of ${R(a)}$ and ${R(b)}$:

$\displaystyle \int_M ||\nabla a-b||^2 {\rm d} \mu = \int_M ||\nabla a-b||^2 {\rm d} \mu + \int_M ||\nabla a-b||^2 {\rm d}$

and the denominator of ${R(a-b)}$ is also the sum of the denominators of ${R(a)}$ and ${R(b)}$:

$\displaystyle =\int_M (a-b)^2 {\rm d} \mu$

$\displaystyle =\int_M a^2 {\rm d} \mu + \int_M b^2 {\rm d} \mu - 2 \int_M ab {\rm d} \mu$

$\displaystyle =\int_M a^2 {\rm d} \mu + \int_M b^2 {\rm d} \mu$

because ${a(x)b(x)=0}$ for every ${x}$. The fact now follows from the inequality

$\displaystyle \min \left \{ \frac {n_1}{d_1} , \frac{n_2}{d_2} \right \} \leq \frac{n_1+n_2}{d_1+d_2}$

$\Box$

The lemma now follows by observing that ${f^+}$ and ${f^-}$ are non-negative and disjointly supported, so

$\displaystyle \min\{ R(f^+) , R(f^-) \} \leq R(\bar f) \leq R(f)$

and that both ${f^+}$ and ${f^-}$ have a support of volume at most ${\frac 12 \mu(M)}$. $\Box$

If anybody is still reading, it is worth observing a couple of differences between the discrete proof and the continuous proof.

The ${\ell_1}$ Rayleigh quotient is defined slightly differently in the continuous case. It would correspond to defining it as

$\displaystyle \frac {\sum_{u\in V} \sqrt{\sum_{v: (u,v)\in E} (f_u-f_v)^2 } }{d \sum_v |f_v|}$

in the discrete case.

If ${a,b \in {\mathbb R}^V}$ are disjointly supported and nonnegative, the sum of the numerators of the Rayleigh quotients ${R(a)}$ and ${R(-b)}$ can be strictly smaller than the numerator of ${R(a-b)}$, while we always have equality in the continuous case. In the discrete case, the sum of the numerators of ${R(a)}$ and ${R(b)}$ can be up to twice the numerator of ${R(a+b)}$ (this fact is useful, but it did not come up in this proof), while again we have exact equality in the continuous case.

The chain rule calculation

$\displaystyle \nabla f^2 = 2f \nabla f$

corresponds to the step

$\displaystyle f_v^2 - f_u^2 = (f_v + f_u) \cdot (f_v - f_u)$

In the continuous case, ${f_v}$ and ${f_u}$ are “infinitesimally close”, so we can approximate ${f_v + f_u}$ by ${2f_v}$.

# The Cheeger inequality in manifolds

Readers of in theory have heard about Cheeger’s inequality a lot. It is a relation between the edge expansion (or, in graphs that are not regular, the conductance) of a graph and the second smallest eigenvalue of its Laplacian (a normalized version of the adjacency matrix). The inequality gives a worst-case analysis of the “sweep” algorithm for finding sparse cuts, it shows a necessary and sufficient for a graph to be an expander, and it relates the mixing time of a graph to its conductance.

Readers who have heard this story before will recall that a version of this result for vertex expansion was first proved by Alon and Milman, and the result for edge expansion appeared in a paper of Dodzuik, all from the mid-1980s. The result, however, is not called Cheeger’s inequality just because of Stigler’s rule: Cheeger proved in the 1970s a very related result on manifolds, of which the result on graphs is the discrete analog.

So, what is the actual Cheeger’s inequality?

Theorem 1 (Cheeger’s inequality) Let ${M}$ be an ${n}$-dimensional smooth, compact, Riemann manifold without boundary with metric ${g}$, let ${L:= - {\rm div} \nabla}$ be the Laplace-Beltrami operator on ${M}$, let ${0=\lambda_1 \leq \lambda_2 \leq \cdots }$ be the eigenvalues of ${L}$, and define the Cheeger constant of ${M}$ to be

$\displaystyle h(M):= \inf_{S\subseteq M : \ 0 < \mu(S) \leq \frac 12 \mu(M)} \ \frac{\mu_{n-1}(\partial(S))}{\mu(S)}$

where the ${\partial (S)}$ is the boundary of ${S}$, ${\mu}$ is the ${n}$-dimensional measure, and ${\mu_{n-1}}$ is ${(n-1)}$-th dimensional measure defined using ${g}$. Then

$\displaystyle h(M) \leq 2 \sqrt{\lambda_2} \ \ \ \ \ (1)$

The purpose of this post is to describe to the reader who knows nothing about differential geometry and who does not remember much multivariate calculus (that is, the reader who is in the position I was in a few weeks ago) what the above statement means, to describe the proof, and to see that it is in fact the same proof as the proof of the statement about graphs.

In this post we will define the terms appearing in the above theorem, and see their relation with analogous notions in graphs. In the next post we will see the proof.