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 be an -dimensional smooth, compact, Riemann manifold without boundary with metric , let be the Laplace-Beltrami operator on , let be the eigenvalues of , and define the **Cheeger constant* of to be

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where the is the boundary of , is the -dimensional measure, and is -th dimensional measure defined using . Then

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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 be a -regular graph, be its adjacency matrix, be its normalized Laplacian matrix, be the eigenvalues of , and define for every subset of vertices . Define the conductance of as*

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* where is the number of edges with one endpoint in and one endpoint in . Then

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**1. Proof the Cheeger inequality in graphs **

We will use the *variational characterization* of the eigenvalues of the Laplacian of a graph .

and if is a minimizer in the above expression then

Following the definition of we see that

and so the minimum in (3) is 0, and it is achieved for . This means that

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

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 Rayleigh quotient and denote it by :

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

**Lemma 3 (Rounding of embeddings)** * For every non-negative vector there is a value such that
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**Lemma 4 (Embedding of into )** * For every non-negative vector , we have
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**Lemma 5 (From an eigenvector to a non-negative vector)** * For every such that there is a non-negative such that and such that*

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Now let us start from a function that optimizes (4), so that and , then apply Lemma 5 to find a function such that the volume of the vertices having positive coordinates in is at most and such that . Then consider the vector such that ; by Lemma 4, we have , and by Lemma 3 there is a threshold such that the set has conductance . Since is a subset of the vertices having positive coordinates in , we have , and so

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

*Proof:* of Lemma 3. For each threshold , define the set

The idea of the proof is that if we pick at random then the probability that an edge belongs to is proportional to and the probability that is proportional to , so that the expected number of edges in is proportional to the numerator of and the expected number of vertices in is proportional to the denominator of ; if is small, it is not possible for to always be large for every .

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

because we can write , where if and otherwise, and we see that only the values of between and make , so .

We also have

and if we denote by the threshold such that is smallest among all the , then

so that

*Proof:* of Lemma 3. Let us consider the numerator of ; it is:

(we used Cauchy-Swarz)

(we used the definition of and Cauchy-Swarz again)

And so

*Proof:* of Lemma 5. Let be the median of , and consider defined as . We have

because the numerators of and are the same (the additive term cancels). The denominators are such that

because and the vector are orthogonal, and so by Pythagoras’s theorem the length-squared of equals the length-squared of plus the length-squared of .

Let us define and so that . We use the following fact:

**Fact 6** * Let be disjointly supported non-negative vectors (“disjointly supported” means that they are non-zero on disjoint subsets of coordinates), then
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*Proof:* The numerator of is

and, using orthogonality and Pythagoras’s theorem, the denominator of is

The fact now follows from the inequality

The lemma now follows by observing that and are non-negative and disjointly supported, so

and that both and have at most non-zero coordinate.

**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 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 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 is symmetric, then for every we have

and the property 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

on functions , and more generally

for functions , where is a vector space with inner product , then we can say that an operator is *self-adjoint* if

for all (appropriately restricted) functions . If is compact, this property is true for the Laplacian, and, in particular, and are adjoints of each others, that is,

(The discrete analog would be that is the transpose of .)

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

and if is a minimizer of the above, then

(The minimization is quantified over all functions that are square-integrable, and the minimum is achieved because if 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 and are adjoint, we have

so

where the Rayleigh quotient

is always non-negative, and it is zero for constant , so we see that and

By analogy with the graph case, we define the “ Rayleigh quotient”

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

**Lemma 7 (Rounding of embeddings)** * For every non-negative function there is a value such that
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where the Cheeger constant of a subset of the manifold is

**Lemma 8 (Embedding of into )** * For every non-negative function , we have
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**Lemma 9 (From an eigenfunction to a non-negative function)** * For every function such that there is a non-negative such that and such that*

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Let us see the proof of these lemmas.

*Proof:* of Lemma 7. For each threshold , define the set

Let be a threshold for which is minimized

We will integrate the numerator and denominator of over all . The *coarea formula* for nonnegative functions is

and we also have

which combine to

so that

*Proof:* of Lemma 8. Let us consider the numerator of ; it is:

We can apply the chain rule, and see that

which implies

and, after applying Caucy-Swarz,

And so

*Proof:* of Lemma 9. Let be a median of , and consider defined as . We have

because the numerators of and are the same (the derivatives of functions that differ by a constant are identical) and the denominators are such that

where we used the fact the integral of is zero.

Let us define and so that . We use the following fact:

**Fact 10** * Let be disjointly supported non-negative functions (“disjointly supported” means that they are non-zero on disjoint subsets of inputs), then
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*Proof:* We begin with the following observation: if is a non-negative function, and , then , because has to be a local minimum.

Consider the expression occurring in the numerator of . We have

But

because for every at least one of or is zero, and so at least one of or is zero.

Using this fact, we have that the numerator of is equal to the sum of the numerators of and :

and the denominator of is also the sum of the denominators of and :

because for every . The fact now follows from the inequality

The lemma now follows by observing that and are non-negative and disjointly supported, so

and that both and have a support of volume at most .

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

The Rayleigh quotient is defined slightly differently in the continuous case. It would correspond to defining it as

in the discrete case.

If are disjointly supported and nonnegative, the sum of the numerators of the Rayleigh quotients and can be strictly smaller than the numerator of , while we always have equality in the continuous case. In the discrete case, the sum of the numerators of and can be up to twice the numerator of (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

corresponds to the step

In the continuous case, and are “infinitesimally close”, so we can approximate by .