Conductance, the Normalized Laplacian, and Cheeger s Inequality

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Spectral Graph Theory Lecture 6 Conductance, the Normalized Laplacian, and Cheeger s Inequality Daniel A. Spielman September 17, 2012 6.1 About these notes These notes are not necessarily an accurate representation of what happened in class. The notes written before class say what I think I should say. The notes written after class way what I wish I said. I think that these notes are mostly correct. 6.2 Overview As the title suggests, in this lecture I will introduce conductance, a measure of the quality of a cut, and the normalized Laplacian matrix of a graph. I will then prove Cheeger s inequality, which relates the second-smallest eigenvalue of the normalized Laplacian to the conductance of a graph. Cheeger [Che70] first proved his famous inequality for manifolds. Many discrete versions of Cheeger s inequality were proved in the late 80 s [SJ89, LS88, AM85, Alo86, Dod84, Var85]. Some of these consider the walk matrix (which we will see in a week or two) instead of the normalized Laplacian, and some consider the isoperimetic ratio instead of conductance. The proof that I present today follows an approach developed by Luca Trevisan [Tre11]. 6.3 Conductance Back in Lecture 2, we related to isoperimetric ratio of a subset of the vertices to the second eigenvalue of the Laplacian. We proved that for every S V where s = S / V and θ(s) λ 2 (1 s), θ(s) def = (S). S Re-arranging terms slightly, this can be stated as V (S) S V S λ 2. 6-1

Lecture 6: September 17, 2012 6-2 Cheeger s inequality provides a relation in the other direction. However, the relation is tighter and cleaner when we look at two slightly different quantities: the conductance of the set and the second eigenvalue of the normalized Laplacian. The formula for conductance has a different denominator that depends upon the sum of the degrees of the vertices in S. I will write d(s) for the sum of the degrees of the vertices in S. Thus, d(v ) is twice the number of edges in the graph. We define the conductance of S to be φ(s) def = (S) min(d(s), d(v S)). Note that many similar, although sometimes slightly different, definitions appear in the literature. We define the conductance of a graph G to be φ G def = min S V φ(s). The conductance of a graph is more useful in many applications than the isoperimetric number. I usually find that conductance is the more useful quantity when you are concerned about edges, and that isoperimetric ratio is most useful when you are concerned about vertices. Conductance is particularly useful when studying random walks in graphs. 6.4 The Normalized Laplacian It seems natural to try to relate the conductance to the following generalized Rayleigh quotient: If we make the change of variables y T Ly y T Dy. (6.1) D 1/2 y = x, then this ratio becomes x T D 1/2 LD 1/2 x x T. x That is an ordinary Rayleigh quotient, which we understand a little better. The matrix in the middle is called the normalized Laplacian (see [Chu97]). We reserve the letter N for this matrix: N def = D 1/2 LD 1/2. This matrix often proves more useful when examining graphs in which nodes have different degrees. We will let 0 = ν 1 ν 2 ν n denote the eigenvalues of N. My first real goal in this lecture will be to prove an analog of the relation that we already know between the isoperimetric ratio and the second eigenvalue of the ordinary Laplacian: ν 2 /2 φ G. (6.2)

Lecture 6: September 17, 2012 6-3 We will then prove Cheeger s inequality, which is much more interesting. φ G 2ν 2, The eigenvector of eigenvalue 0 of N is d 1/2, by which I mean the vector whose entry for vertex u is the square root of the degree of u. Observe that D 1/2 LD 1/2 d 1/2 = D 1/2 L1 = D 1/2 0 = 0. The eigenvector of ν 2 is given by x T N x arg min x d 1/2 x T x. Transfering back into the variable y, and observing that we find We will now use this formula to prove (6.2). Lemma 6.4.1. For every S V, x T d 1/2 = y T D 1/2 d 1/2 = y T d, ν 2 = min y d y T Ly y T Dy. θ(s) ν 2 /2. Proof. As in Lecture 2, we would like to again use χ S as a test vector. But, it is not orthogonal to d. To fix this, we subtrat a constant. Set where You should now check that y T d = 0: We already know that y = χ S σ1, σ = d(s)/d(v ). y T d = χ T S d σ1 T d = d(s) (d(s)/d(v ))d(v ) = 0. y T Ly = (S). It remains to compute y T Dy. If you remember the previous computation like this, you would guess that it is d(s)(1 σ) = d(s)d(v S)/d(V ), and you would be right: y T Dy = u S d(u)(1 σ) 2 + u S d(u)σ 2 = d(s)(1 σ) 2 + d(v S)σ 2 = d(s) 2d(S)σ + d(v )σ 2 = d(s) d(s)σ = d(s)d(v S)/d(V ).

Lecture 6: September 17, 2012 6-4 So, ν 2 y T Ly y T Dy = (S) d(v ) d(s)d(v S). As the larger of d(s) and d(v S) is at least half of d(v ), we find (S) ν 2 2 min(d(s), d(v S)). 6.5 Cheeger s Inequality Cheeger s inequality proves that if we have a vector y, orthogonoal to d, for which the generalized Rayleigh quotient (6.1) is small, then one can obtain a set of small conductance from y. We obtain such a set by carefully choosing a real number t, and setting S = {u : y(u) t}. Theorem 6.5.1. Let y be a vector orthogonal to d. Then, there is a number t for which the set S = {u : y(u) < t} satisfies φ(s) 2 y T Ly y T Dy. Before proving the theorem, I wish to make one small point about the denominator in the expression above. It is essentially minimized when y T d = 0, at least with regards to shifts. Lemma 6.5.2. Let v s = y + s1. Then, the minimum of v T s Dv T s v T s d = 0. is achieved at the s for which Proof. The derivative with respect to s is 2d T v s, and the minimum is achieved when this derivative is zero. Proof of Theorem 6.5.1. Define and assume, without loss of generality, that ρ = y T Ly y T Dy, y(1) y(2) y(n). We begin with some normalization. Let j be the least number for which j d(u) d(v )/2. u=1

Lecture 6: September 17, 2012 6-5 We would prefer a vector that is centered at j. So, set z = y y(j)1. This vector z satisfies z (j) = 0, and, by Lemma 6.5.2, Also assume that z has been scaled so that z T Lz z T Dz ρ. z (1) 2 + z (n) 2 = 1. Recall that φ(s) = (S) min(d(s), d(v S)). We will define a distribution on S for which we can prove that This implies 1 that there is some S for which which means φ(s) 2ρ. E [ (S) ] 2ρ E [min(d(s), d(v S))]. (S) 2ρ min(d(s), d(v S)), The distribution on S will be induced from a distribution on a random number t, which we will specify in a moment. We will set S = {u : z (u) < t}. Trevisan had the remarkable idea of choosing t between z (1) and z (n) with probability density 2 t. That is, the probability that t lies in the interval [a, b] is b t=a To see that the total probability is 1, observe that z (n) as z (1) z (j) z (n) and z (j) = 0. t=z (1) 2 t = sgn(b)b 2 sgn(a)a 2. 2 t = z (n) 2 + z (1) 2 = 1, We now derive a formula for the denominator of φ. Begin by observing that E [d(s)] = u Pr [u S] d(u) = u Pr [z (u) t] d(u). The result of our centering of z at j is that t < 0 = d(s) = min(d(s), d(v S)), t 0 = d(v S) = min(d(s), d(v S)). 1 If this is not immediately clear, note that it is equivalent to assert that E [ 2ρ min(d(s), d(v S)) (S) ] 0, which means that there must be some S for which the expression is non-negative. and

Lecture 6: September 17, 2012 6-6 That is, for u < j, u is in the smaller set if t < 0; and, for u j, u is in the smaller set if t 0. So, E [min(d(s), d(v S))] = Pr [z (u) < t and t < 0] d(u) + Pr [z (u) > t and t 0] d(u) u<j u j = z (u) 2 d(u) + z (u) 2 d(u) u<j u j = u z (u) 2 d(u) = z T Dz. We now turn to the numerator. An edge (u, v) with z (u) z (v) is on the boundary of S if z (u) t < z (v). We previously computed the probability that this happens: { z (u) sgn(z (v))z (v) 2 sgn(z (u))z (u) 2 2 z (v) 2 when sgn(u) = sgn(v), = z (u) 2 + z (v) 2 when sgn(u) sgn(v). We now show that both of these terms are upper bounded by z (u) z (v) ( z (u) + z (v) ). Regardless of the signs, z (u) 2 z (v) 2 = (z (u) z (v))(z (u) + z (v)) z (u) z (v) ( z (u) + z (v) ). When sgn(u) = sgn(v), So, z (u) 2 + z (v) 2 (z (u) z (v)) 2 = z (u) z (v) ( z (u) + z (v) ). E [ (S) ] = Pr [(u, v) (S)] z (u) z (v) ( z (u) + z (v) ). (6.3) Applying the Cauchy-Schwartz inequality, we find that this is at most (z (u) z (v)) 2 ( z (u) + z (v) ) 2. (6.4) We have defined ρ so that the term under the left-hand square root is at most ρ u z (u) 2 d(u). To bound the right-hand square root, we observe ( z (u) + z (v) ) 2 2 z (u) 2 + z (v) 2 = 2 u z (u) 2 d(u).

Lecture 6: September 17, 2012 6-7 Putting all these inequalities together yields E [ (S) ] ρ u z (u) 2 d(u) 2 u z (u) 2 d(u) = 2ρ u z (u) 2 d(u) = 2ρ E [min (d(s), d(v S))]. I wish to point out two important features of this proof: 1. This proof does not require y to be an eigenvector it obtains a cut from any vector y that is orthogonal to d. 2. This proof goes through almost without change for weighted graphs. The main difference is that for weighted graphs we measure the sum of the weights of edges on the boundary instead of their number. The main difference in the proof is that lines (6.3) and (6.4) become E [w( (S))] = Pr [(u, v) (S)] w u,v z (u) z (v) ( z (u) + z (v) )w u,v z (u) z (v) ( z (u) + z (v) )w u,v w u,v (z (u) z (v)) 2 w u,v ( z (u) + z (v) ) 2, and we observe that w u,v ( z (u) + z (v) ) 2 2 u z (u) 2 d(u). The only drawback that I see to the approach that we took in this proof is that the application of Cauchy-Schwartz is a little mysterious. Shang-Hua Teng and I came up with a proof that avoids this by introducing one inequality for each edge. If you want to see that proof, look at my notes from 2009. References [Alo86] N. Alon. Eigenvalues and expanders. Combinatorica, 6(2):83 96, 1986. [AM85] Noga Alon and V. D. Milman. λ 1, isoperimetric inequalities for graphs, and superconcentrators. J. Comb. Theory, Ser. B, 38(1):73 88, 1985.

Lecture 6: September 17, 2012 6-8 [Che70] J. Cheeger. A lower bound for smallest eigenvalue of the Laplacian. In Problems in Analysis, pages 195 199, Princeton University Press, 1970. [Chu97] F. R. K. Chung. Spectral Graph Theory. American Mathematical Society, 1997. [Dod84] Jozef Dodziuk. Difference equations, isoperimetric inequality and transience of certain random walks. Transactions of the American Mathematical Society, 284(2):787 794, 1984. [LS88] [SJ89] Gregory F. Lawler and Alan D. Sokal. Bounds on the l 2 spectrum for Markov chains and Markov processes: A generalization of Cheeger s inequality. Transactions of the American Mathematical Society, 309(2):557 580, 1988. Alistair Sinclair and Mark Jerrum. Approximate counting, uniform generation and rapidly mixing Markov chains. Information and Computation, 82(1):93 133, July 1989. [Tre11] Luca Trevisan. Lecture 4 from cs359g: Graph partitioning and expanders, stanford university, January 2011. available at http://theory.stanford.edu/ trevisan/cs359g/lecture04.pdf. [Var85] N. Th. Varopoulos. Isoperimetric inequalities and Markov chains. Journal of Functional Analysis, 63(2):215 239, 1985.