Analysis on Graphs. Alexander Grigoryan Lecture Notes. University of Bielefeld, WS 2009/10

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1 Analysis on Graphs Alexander Grigoryan Lecture Notes University of Bielefeld, WS 009/0

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3 Contents The Laplace operator on graphs 5. Thenotionofagraph Cayley graphs Randomwalks TheLaplaceoperator TheDirichletproblem... 8 Spectral properties of the Laplace operator 33. Green s formula EigenvaluesoftheLaplaceoperator Convergencetoequilibrium Moreabouttheeigenvalues Products of graphs Eigenvalues and mixing times in Z n m Geometric bounds for the eigenvalues Cheeger s inequality Estimating λ frombelowviadiameter Expansion rate Eigenvalues on infinite graphs DirichletLaplaceoperator Cheeger s inequality Isoperimetricinequalities SolvingtheDirichletproblembyiterations IsoperimetricinequalitiesonCayleygraphs The heat kernel and the type problem On-diagonalupperestimatesoftheheatkernel On-diagonallowerbounds The type problem Volumetestsforrecurrence Isoperimetrictestsfortransience...7 3

4 4 CONTENTS

5 Chapter The Laplace operator on graphs. The notion of a graph AgraphΓ is a couple (V,E) wherev is a set of vertices, that is, an arbitrary set, whose elements are called vertices, and E is a set of edges, thatis,e consists of some couples (x, y) wherex, y V. We write x» y (x is connected to y, orx is joint to y, orx is adjacent to y, orx is a neighbor of y )if(x, y) E. Graphs are normally represented graphically as a set of points on a plane, and if x» y then one connects the corresponding points on the plane by a line. There are two versions of the definition of edges:. The couples (x, y) are ordered, that is, (x, y) and(y, x) are considered as different (unless x = y). In this case, the graph is called directed or oriented.. The couples (x, y) are unordered, that is, (x, y) =(y, x). In this case, x» y is equivalent to y» x, and the graph is called undirected or unoriented. Unless otherwise specified, all graphs will be undirected. Theedge(x, y) will be normally denoted by xy, andx, y are called the endpoints of this edge. The edge xx with the same endpoints (should it exist) is called a loop. A graph Γ is called simple if it has no loops. AgraphΓ is called locally finite if each vertex has a finite number of edges. For each point x, define its degree deg (x) =#fy V : x» yg, that is, deg (x) is the number of the edges with endpoint x. A graph Γ is called finite if the number of vertices is finite. Of course, a finite graph is locally finite. We start with a simple observation. Lemma. (Double counting of edges) On any simple finite graph (V,E), the following identity holds: deg (x) =#E. x V 5 A.Grigoryan Lecture

6 6 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS Proof. Let n =#V and let us enumerate all vertices as,,..., n. Consider the adjacency matrix A =(a ij ) n i,j= of the graph Γ that is defined as follows: ½, i» j a ij = 0, i 6» j. This matrix is symmetric and the sum of the entries in the row i (and in the column i) isequaltodeg(i) sothat à n n n! n deg (i) = a ij = a ij. i= i= The entries a ij in the last summation are 0 and, and a ij =ifandonlyif(i, j) is an edge. In this case i 6= j and (j, i) is also an edge. Therefore, each edge ij contributes to this sum the value twice: as (i, j) andas(j, i). Hence, which finishes the proof. j= n a ij =#E, i,j= Example. Consider some examples of graphs.. A complete graph K n. The set of vertices is V = f,,..., ng, andtheedges are defined as follows: i» j for any two distinct i, j V. That is, any two distinct points in V are connected. Hence, the number of edges in K n is n i= i,j= deg (i) = n (n ).. A complete bipartite graph K n,m. The set of vertices is V = f,.., n, n +,..., n + mg, and the edges are defined as follows: i» j if and only if either i<nand j n or i n and j<n. That is, the set of vertices is split into two groups: S = f,..., ng and S = fn +,..., mg, and the vertices are connected if and only if they belong to the different groups. The number of edges in K n,m is equal to nm. 3. A lattice graph Z. The set of vertices V consists of all integers, and the integers x,y are connected if and only if jx yj =. 4. A lattice graph Z n. The set of vertices consists of all n-tuples (x,..., x n )where x i are integers, and (x,..., x n )» (y,..., y n ) if and only if n jx i y i j =. i= That is, x i is different from y i for exactly one value of the index i, and jx i y i j = for this value of i.

7 .. THE NOTION OF A GRAPH 7 Definition. A weighted graph is a couple (Γ,μ)whereΓ =(V,E) is a graph and μ xy is a non-negative function on V V such that. μ xy = μ yx ;. μ xy > 0 if and only if x» y. Alternatively, μ xy can be considered as a positive function on the set E of edges, that is extended to be 0 on non-edge pairs (x, y). The weight μ is called simple if μ xy =foralledgesx» y. Any weight μ xy gives rise to a function on vertices as follows: μ (x) = μ xy. (.) y,y x Then μ (x) is called the weight of a vertex x. For example, if the weight μ xy is simple then μ (x) = deg(x). The following lemma extends Lemma.. Lemma. On any simple finite weighted graph (Γ,μ), μ (x) = μ ξ. ξ E x V Proof. Rewrite (.) in the form μ (x) = y V μ xy wherethesummationisextendedtoally V. This does not change the sum in (.) because we add only non-edges (x, y) whereμ xy = 0. Therefore, we obtain μ (x) = μ xy = μ xy = μ = xy μ ξ. x V x V y V x,y V x,y:x y ξ E Definition. A finite sequence fx k g n k=0 of vertices on a graph is called a path if x k» x k+ for all k =0,,..., n. The number n of edges in the path is referred to as the length of the path. Definition. Agraph(V,E) is called connected if, for any two vertices x, y V, there is a path connecting x and y, thatis,apathfx k g n k=0 such that x 0 = x and x n = y. If (V,E) is connected then define the graph distance d (x, y) between any two distinct vertices x, y as follows: if x 6= y then d (x, y) is the minimal length of a path that connects x and y, andifx = y then d (x, y) =0. The connectedness here is needed to ensure that d (x, y) < for any two points. Lemma.3 On any connected graph, the graph distance is a metric, so that (V,d) isametricspace. Proof. We need to check the following axioms of a metric.

8 8 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS. Positivity: 0 d (x, y) <, andd (x, y) =0ifandonlyifx = y.. Symmetry: d (x, y) =d (y, x). 3. The triangle inequality: d (x, y) d (x, z)+d (z,y). The first two properties are obvious for the graph distance. To prove the triangle inequality, choose a shortest path fx k g n k=0 connecting x and z, a shortest path fy k g m k=0 connecting y and z, sothat d (x, z) =n and d (z,y) =m. Then the sequence x,..., x n,z,y,..., y m is a path connecting x and y, that has the length n + m, which implies that d (x, y) n + m = d (x, z)+d (z,y). Lemma.4 If (V,E) is a connected locally finite graph then the set of vertices V is either finite or countable. Proof. Fix a reference point x V and consider the set B n : fy V : d (x, y) ng, that is a ball with respect to the distance d. Let us prove by induction in n that #B n <. Inductive step for n = 0 is trivial because B 0 = fxg. Inductive step: assuming that B n is finite, let us prove that B n+ is finite. It suffices to prove that B n+ n B n is finite. For any vertex y B n+ n B n, we have d (x, y) =n + so that there is a path fx k g n+ k=0 from x to y of length n +. Consider the vertex z = x n. Clearly, the path fx k g n k=0 connects x and z and has the length n, which implies that d (x, z) n and, hence, z B n. On the other hand, we have by construction z» y. Hence, we have shown that every vertex y B n+ n B n is connected to one of the vertices in B n. However, the number of vertices in B n is finite, and each of them has finitely many neighbors. Therefore, the total number of the neighbors of B n is finite, which implies # (B n+ n B n ) < and #B n+ <. Finally, observe that V = S n= B n because for any y V we have d (x, y) < so that y belongs to some B n. Then V is either finite or countable as a countable union of finite sets.

9 .. CAYLEY GRAPHS 9. Cayley graphs A.Grigoryan Lecture Here we discuss a large class of graphs that originate from groups. Recall that a group (G, )isasetg equipped with a binary operation that satisfies the following properties:. for all x, y G, x y is an element of G;. associative law: x (y z) =(x y) y; 3. there exists a neutral element e such that x e = e x = x for all x G; 4. there exists the inverse x element for any x G, such that x x = x x = e. If in addition the operation is commutative, that is, x y = y x then the group G is called abelian or commutative. In the case of abelian group, one uses the additive notation. Namely, the group operation is denoted + instead of, the neutral element is denoted by 0 instead of e, and the inverse element is denoted by x rather than x. Example.. Consider the set Z of all integers with the operation +. Then (Z, +) is an abelian group where the neutral element is the number 0 and the inverse of x isthenegativeofx.. Fix an integer q and consider the set Z q of all residues (Restklassen) modulo q, with the operation +. In other words, the elements of Z q are the equivalence classes of integers modulo q. Namely, one says that two integers x, y are congruent modulo q and writes x = y mod q if x y is divisible by q. This relation is an equivalence relation and gives rise to q equivalence classes, that are called the residues and are denoted by 0,,..., q as integers, so that the integer k belongs to the residue k. The addition in Z q is inherited from Z as follows: x + y = z in Z q, x + y = z mod q in Z. Then (Z q, +) is an abelian group, the neutral element is 0, and the inverse of x is q x (except for x =0). For example, consider Z = f0, g. Apart from trivial sums x +0=x, wehave the following rules in this group: + = 0 and =. IfZ 3 = f0,, g, wehave + =, +=0, += =, =.

10 0 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS Given two groups, say (A, +), (B,+), one can consider a direct product of them: the group (A B,+) that consists of pairs (a, b) wherea A and b B with the operation (a, b)+(a 0,b 0 )=(a + a 0,b+ b 0 ) The neutral element of A B is (0 A, 0 B ), and the inverse to (a, b) is( a, b). More generally, given n groups (A k, +) where k =,..., n, define their direct product (A A... A n, +) as the set of all sequences (a k ) n k= where a k A k, with the operation (a,..., a n )+(a 0,..., a 0 n)=(a + a 0,..., a n + a 0 n). The neutral element is (0 A,..., 0 An )andtheinverseis (a,..., a n )=( a,..., a n ). If the groups are abelian then their product is also abelian. Example.. The group Z n is defined as the direct product Z Z {z... Z } of n n copies of the group Z.. The group (Z Z 3, +) consists of pairs (a, b) wherea is a residue mod and b is a residue mod 3. For example, we have the following sum in this group: whence it follows that (, ) = (, ). (, ) + (, ) = (0, 0), What is the relation of the groups to graphs? Groups give rise to a class of graphs that are called Cayley graphs. Let (G, ) beagroupands be a subset of G with the property that if x S then x S and that e/ S. SuchasetS will be called symmetric. AgroupG and a symmetric set S ½ G determine a graph (V,E) asfollows:the set V of vertices coincides with G, andthesete ofedgesisdefined as follows: x» y, x y S. (.) Note that the relation x» y is symmetric in x, y, thatis,x» y implies y» x, because y x = x y S. Hence, the graph (V,E) is undirected. The fact that e / S implies that x 6» x because x x = e/ S. Hence, the graph (V,E) containsnoloops. Definition. The graph (V,E) defined as above is denoted by (G, S) and is called the Cayley graph of the group G with the edge generating set S. There may be many different Cayley graphs of the same group since they depend also on the choice of S. It follows from the construction that deg (x) =#S for any x V. In particular, if S is finite then the graph (V,E) is locally finite. In what

11 .. CAYLEY GRAPHS follows, we will consider only locally finite Cayley graphs and always assume that they are equipped with a simple weight. If the group operation is + then (.) becomes x» y, y x S, x y S. In this case, the symmetry of S means that 0 / S and if x S then also x S. Example.. G = Z with + and S = f, g. Then x» y if x y =or x y =, that is, if x and y are neighbors on the real axis:... ² ² ² ²... If S = f, g then x» y if jx yj =orjx yj = so that we obtain a different graph.. Let G = Z n with +. Let S consist of points (x,...,x n ) Z n such that exactly one of x i is equal to and the others are 0; that is ( ) n S = (x,...,x n ) Z m : jx i j =. For example, in the case n =wehave i= S = f(, 0), (, 0), (0, ), (0, )g. The connection x» y means that x y has exactly one coordinate, and all others are 0; equivalently, this means that n jx i y i j =. i= Hence, the Cayley graph of (Z n,s) is exactly the standard lattice graph Z n.inthe case n =, it looks as follows: j j j j j ² ² ² ² ² j j j j j ² ² ² ² ² j j j j j ² ² ² ² ² j j j j j ² ² ² ² ² j j j j j ² ² ² ² ² j j j j j Consider now another edge generating set S on Z with two more elements: S = f(, 0), (, 0), (0, ), (0, ), (, ), (, )g.

12 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS The corresponding graph (Z,S) is shown here: j Á j Á j Á j Á j Á ² ² ² ² ² Á j Á j Á j Á j Á j Á ² ² ² ² ² Á j Á j Á j Á j Á j Á ² ² ² ² ² Á j Á j Á j Á j Á j Á ² ² ² ² ² Á j Á j Á j Á j Á j Á ² ² ² ² ² Á j Á j Á j Á j Á j 3. Let G = Z = f0, g. The only possibility for S is S = fg (note that = ). The graph (Z,S)coincideswithK and is shown here: ² ² 4. Let G = Z q where q>, and S = f g. That is, each residue k =0,,...,q has two neighbors: k andk +. For example, 0 has the neighbors and q. The graph (Z q,s) is called the q-cycle and is denoted by C q. Here are the graphs C 3 and C 4 : ² ² ² C 3 = Á j, C 4 = j j ² ² ² ² 5. Consider Z q with the symmetric set S = Z q nf0g. That is, every two distinct elements x, y Z q are connected by an edge. Hence, the resulting Cayley graph is the complete graph K q. 6. Let G = Z n := Z Z... Z {z }, that consists of n-tuples (x,..., x n )of n residues mod, that is, each x i is 0 or. Let S consist of all elements (x,..., x n ). such that exactly one x i is equal to and all others are 0. Then the graph (Z n,s)is called the n-dimensional binary cube and is denoted by f0, g n, analogously to the geometric n-dimensional cube [0, ] n. Clearly, f0, g = K and f0, g = C 4. The graph f0, g 3 is shown here in two ways: f0, g 3 = ² ² Á j Á j ² j ² j j ² j ² j Á j Á ² ² = ² ² j  Á j j ² ² j j j j j j ² ² j j Á  j ² ² 7. Let G = Z q Z. Then G consists of pairs (x, y) wherex Z q and y Z. Then G can be split into two disjoint subsets G 0 = Z q f0g = f(x, 0) : x Z q g G = Z q fg = f(x, ) : x Z q g,

13 .3. RANDOM WALKS 3 each having q elements. Set S = G.Then (x, a)» (y, b), a b =, a 6= b. In other words, (x, a)» (y, b) if and only if the points (x, a) and(y, b) belongto different subsets G 0,G.Hence,thegraph(Z q Z,S) coincides with the complete bipartite graph K q,q with the partition G 0,G. Definition. Agraph(V,E) is called D-regular, if all vertices x V have the same degree D (that is, each vertex has D edges). A graph is called regular if it is D-regular for some D. Of course, there are plenty of examples of non-regular graphs. Clearly, all Cayley graphs are regular. All regular graphs thatwehavediscussedabove,werecayley graphs. However, there regular graphs that are not Cayley graphs (cf. Exercise 8)..3 Random walks Consider a classical problem from Probability theory. Let fx k g k= be a sequence of independent random variables, taking values and with probabilities / each; that is, P (x k =)=P (x k = ) = /. Consider the sum n = x x n and ask, what is a likely behavior of n for large n? Historically, this type of problem came from the game theory (and the gambling practice): at each integer value of time, a player either wins with probability / or loses with the same probability. The games at different times are independent. Then n represents the gain at time n if n > 0, and the loss at time n is n < 0. Of course, the mean value of n, that is, the expectation, is 0 because E (x k )= ( ) = 0 and n E ( n )= E (x k )=0. k= The games with this property are called fair games or martingales. However, the deviation of n fromtheaveragevalue0canbestillsignificant in any particular game. We will adopt a geometric point of view on n as follows. Note that n Z and n is defined inductively as follows: 0 =0and n+ n is equal to or with equal probabilities /. Hence, we can consider n as a position on Z of a walker that jumps at any time n from its current position to a neighboring integer, either right or left, with equal probabilities /, and independently of the previous movements. Such a random process is called a random walk. Note that this random walk is related to the graph structure of Z: namely, a walker moves at each step along the edges of this graph. Hence, n can be regarded as a random walk on the graph Z. Similarly, one can define a random walk on Z N : at any time n =0,,,..., let n be the position of a walker in Z N. It starts at time 0 at the origin, and at time

14 4 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS n + it moves with equal probability / (N) by one of the vectors e,..., e N, where e,..., e n is the canonical basis in R n.thatis, 0 =0and P ( n+ n = e k )= N. We always assume that the random walk in question has the Markov property: the choiceofthemoveatanytimen is independent of the previous movement. The following picture of the trace of a random walk on Z was copied from Wikipedia: A.Grigoryan Lecture Figure.: More generally, one can define a random walk on any locally finite graph (V,E). Namely, imagine a walker that at any time n =0,,... has a random position n at one of the vertices of V that is defined as follows: 0 = x 0 is a given vertex, and n+ is obtained from n by moving with equal probabilities along one of the edges of n,thatis, P ( n+ = y j n = x) = ½ / deg (x), y» x, 0, y 6» x. (.3) The random walk f n g defined in this way, is called a simple random walk on (V,E). The adjective simple refers to the fact that the walker moves to the neighboring vertices with equal probabilities. A simple random walk is a particular case of a Markov chain. Givenafinite or countable set V (that is called a state space), a Markov kernel on V is any function P (x, y) :V V! [0, +) with the property that P (x, y) =8x V. (.4) y V

15 .3. RANDOM WALKS 5 If V is countable then the summation here is understood as a series. AnyMarkovkerneldefines a Markov chain f n g n=0 as a sequence of random variables with values in V such that the following identity holds P ( n+ = y j n = x) =P (x, y), (.5) and that the behavior of the process at any time n onwards is independent of the past. The latter requirement is called the Markov property and it will be considered in details below. Observe that the rule (.3) defining the random walk on a graph (V,E) canbe also written in the form (.5) where P (x, y) = ½ deg(x) y» x, 0, y 6» x. (.6) The condition (.4) is obviously satisfied because P (x, y) = P (x, y) = deg (x) =. y x y x y V Hence, the random walk on a graph is a particular case of a Markov chain, with a specific Markov kernel (.6). Let us discuss the Markov property. The exact meaning of it is given by the following identity: P ( = x, = x,..., n+ = x n+ j 0 = x) = P ( = x, = x,..., n = x n j 0 = x) P ( n+ = x n+ j n = x n )(.7) that postulates the independence of the jump from x n to x n+ from the previous path. Using (.5) and (.7), one obtains by induction that P ( = x, = x,..., n = x n j 0 = x) =P (x, x ) P (x,x )...P (x n,x n ). (.8) Obviously, (.8) implies back (.7). In fact, (.8) can be used to actually construct the Markov chain. Indeed, it is not obvious that there exists a sequence of random variables satisfying (.5) and (.7). Proposition.5 The Markov chain exists for any Markov kernel. Proof. This is the only statement in this course that requires a substantial use of the foundations of Probability Theory. Indeed, it is about construction of a probability space (Ω, P) anddefining a sequence f n g n=0 of random variables satisfying the required conditions. The set Ω will be taken to be the set V of all sequences fx k g k= of points of V, that represent the final outcome of the process. In order to construct a probability measure P on Ω, wefirst construct a probability measure P (n) of the set of finite sequences fx k g n k=. Note that the set of sequences of n points of V is nothing other than the product V n = V {z... V }. Hence, our n strategy is as follows: first construct a probability measure P () on V,thenP (n) on

16 6 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS V n, and then extend it to a measure P on V. In fact, we will construct a family of probability measures P x indexed by a point x V,sothatP x is associated with a Markov chain starting from the point 0 = x. Fix a point x V and observe that P (x, ) determines a probability measure P () x on V as follows: for any subset A ½ V,set P () x (A) = y A P (x, y). Clearly, P x is σ-additive, that is, P () x à G! A k = k= P x (A k ), k= and P () x (V )=by(.4). Next, define a probability measure P (n) x on the product V n = V {z... V } n follows. Firstly, define the measure of any point (x,..., x n ) V n by P (n) x (x,..., x n )=P (x, x ) P (x,x )...P (x n,x n ), (.9) andthenextendittoallsetsa½v n by P (n) x (A) = (x,...,x n ) A P (n) x (x,..., x n ). Let us verify that it is indeed a probability measure, that is, P x (n) (V n )=. The inductive basis was proved above, let us make the inductive step from n to n + = = = x,...,x n+ V x,...,x n+ V x,...,x n V x n+ V x,...,x n V = =, P (n) x (x,..., x n+ ) P (x, x ) P (x,x )...P (x n,x n ) P (x n,x n+ ) P (x, x ) P (x,x )...P (x n,x n ) P (x n,x n+ ) P (x, x ) P (x,x )...P (x n,x n ) x n+ V P (x n,x n+ ) wherewehaveuse(.4)andtheinductivehypothesis. n o The sequence of measures P (n) x constructed above is consistence in the n= following sense. Fix two positive integers n<m.then every point (x,..., x n ) V n Note that measure P (n) x is not aproductmeasureofn copies of P () x have been P (x; x ) P (x; x ) :::P (x; x n ) : as since the latter would

17 .3. RANDOM WALKS 7 can be regarded as a subset of V m that consists of all sequence where the first n terms are exactly x,.., x n, and the rest terms are arbitrary. Then we have P (n) x (x,..., x n )=P (m) x (x,..., x n ), (.0) that is proved as follows: P x (m) (x,..., x n ) = P (m) x (x,..., x n,x n+,..., x m ) x n+,...,x m V = P (x, x ) P (x,x )...P (x n,x n ) P (x n,x n+ )...P (x m,x m ) x n+,...,x m V = P (x, x ) P (x,x )...P (x n,x n ) P (x n,x n+ )...P (x m,x m ) x n+,...,x m V = P (n) x (x,..., x n ) P x (n m) n V n m = P (n) x (x,..., x n ). Inthesameway,anysubsetA ½ V n admits the cylindrical extension A 0 to a subset of V m as follows: a sequence (x,..., x m ) belongs to A 0 if (x,..., x n ) A. It follows from (.0) that P (n) x (A) =P x (m) (A 0 ). (.) This is a Kolmogorov consistency condition, that allows to extend a sequence of measures P (n) x on V n toameasureonv. Consider first cylindrical subsets of V, that is, sets of the form where A is a subset of V n,andset A 0 = ffx k g k= :(x,..., x n ) Ag (.) P x (A 0 )=P (n) x (A). (.3) Due to the consistency condition (.), this definition does not depend on the choice of n. Kolmogorov s extension theorem says that the functional P x defined in this way on cylindrical subsets of V, extends uniquely to a probability measure on the minimal σ-algebra containing all cylindrical sets. Now we define the probability space Ω as V endowed with the family fp x g of probability measures. The random variable n is a function on Ω with values in V that is defined by n (fx k g k= )=x n. Then (.9) can be rewritten in the form P x ( = x,..., n = x n )=P (x, x ) P (x,x )...P (x n,x n ). (.4) The identity (.4) together with (.4) are the only properties of Markov chains that we need here and in what follows. Let us use (.4) to prove that the sequence f n g is indeed a Markov chain with the Markov kernel P (x, y). We need to verify (.5) and (.8). The latter is obviously equivalent to (.4). To prove the former, write P x ( n = y) = P x ( = x,..., n = x n, n = y) = x,...,x n V x,...,x n V P (x, x ) P (x,x )...P (x n,y)

18 8 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS and P x ( n = y, n+ = z) = = P x ( = x,..., n = x n, n = y, n+ = z) x,...,x n V x,...,x n V P (x, x ) P (x,x )...P (x n,y) P (y, z) whence P x ( n+ = z j n = y) = P x ( n = y, n+ = z) P x ( n = y) = P (y, z), which is equivalent to (.5). Given a Markov chain f n g with a Markov kernel P (x, y), note that by (.4) P (x, y) =P x ( = y) so that P (x, ) is the distribution of.denotebyp n (x, ) the distribution of n, that is, P n (x, y) =P x ( n = y). The function P n (x, y) is called the transition function or the transition probability of the Markov chain. Indeed, it fully describes what happens to random walk at time n. Forafixed n, thefunctionp n (x, y) is also called the n-step transition function. It is easy to deduce a recurrence relation between P n and P n+. Proposition.6 For any Markov chain, we have P n+ (x, y) = z V P n (x, z) P (z, y). (.5) Proof. By (.4), we have P n (x, z) = P x ( n = z) = P x ( = x,..., n = x n, n = z) = x,...,x n V x,...,x n V P (x, x ) P (x,x )...P (x n,z). Applying the same argument to P n+ (x, y), we obtain P n+ (x, y) = P (x, x ) P (x,x )...P (x n,x n ) P (x n,y) x,...,x n V 0 P (x, x ) P (x,x )...P (x n,z) A P (z, y) z V x,...,x n V = P n (x, z) P (z,y), z V which finishes the proof.

19 .3. RANDOM WALKS 9 Corollary.7 For any fixed n, P n (x, y) is also a Markov kernel on V. Proof. We need only to verify that P n (x, y) =. y V For n = this is given, the inductive step from n to n + follows from (.5): P n+ (x, y) = P n (x, z) P (z, y) y V y V z V = P n (x, z) P (z, y) z V y V = P n (x, z) =. z V Corollary.8 We have, for all positive integers n, k, P n+k (x, y) = z V P n (x, z) P k (z,y). (.6) Proof. Induction in k. Thecasek = is covered by (.5). The inductive step from k to k +: P n+(k+) (x, y) = P n+k (x, w) P (w, y) w V = P n (x, z) P k (z, w) P (w, y) w V z V = P n (x, z) P k (z, w) P (w, y) z V w V = P n (x, z) P k+ (z,y). z V NowweimposeonerestrictiononaMarkovchain. Definition. AMarkovkernelP (x, y) is called reversible if there exists a positive function μ (x) on the state space V, such that P (x, y) μ (x) =P (y, x) μ (y). (.7) A Markov chain is called reversible if its Markov kernel is reversible. It follows by induction from (.7) and (.5) that P n (x, y) is also a reversible Markov kernel. The condition (.7) means that the function μ xy := P (x, y) μ (x)

20 0 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS is symmetric in x, y. For example, this is the case when P (x, y) is symmetric in x, y; then we just take μ (x) =. However, the reversibility condition can be satisfied for non-symmetric Markov kernel as well. For example, in the case of a simple random walk on a graph (V,E), we have by (.6) ½, x» y μ xy = P (x, y)deg(x) = 0, x 6» y. which is symmetric. Hence, a simple random walk is a reversible Markov chain. Any reversible Markov chain on V gives rise to a graph structure on V as follows. Define the set E of edges on V by the condition x» y, μ xy > 0. Then μ xy canbeconsideredasaweighton(v,e) (cf. Section.). Note that the function μ (x) can be recovered from μ xy by the identity μ xy = P (x, y) μ (x) =μ (x), y,y x y V which matches (.). Let Γ =(V,E) be a graph. Recall that a non-negative function μ xy on V V is called a weight, if μ xy = μ yx and μ xy > 0 if and only if x» y. Acouple(V,μ) (or (Γ,μ)) is called a weighted graph. Note that the information about the edges is contained in the weight μ so that the set E of edges is omitted in the notation (V,μ). As we have seen above, any reversible Markov kernel on V determines a weighted graph (V,μ). Conversely, a weighted graph (V,μ) determines a reversible Markov kernel on V provided the set V is finite or countable and 0 < y V μ xy < for all x V. (.8) For example, the positivity condition in (.8) holds if the graph (V,E) that is determined by the weight μ xy, has no isolated vertices, that is, the vertices without neighbors, and the finiteness condition holds if the graph (V,E) is locally finite, so that the summation in (.8) has finitely many positive terms. The full condition (.8) is satisfied for locally finite graphs without isolated vertices. If (.8) holds, then the weight on vertices μ (x) = y V μ xy is finite and positive for all x, and we can set P (x, y) = μ xy μ (x) (.9) A.Grigoryan Lecture so that the reversibility condition (.7) is obviously satisfied. In this context, a reversible Markov chain is also referred to as a random walk on a weighted graph.

21 .3. RANDOM WALKS From now on, we stay in the following setting: we have a weighted graph (V,μ) satisfying (.8), the associated reversible Markov kernel P (x, y), and the corresponding random walk (= Markov chain) f n g. Fix a point x 0 V and consider the functions v n (x) =P x0 ( n = x) =P n (x 0,x) and u n (x) =P x ( n = x 0 )=P n (x, x 0 ) The function v n (x) is the distribution of n at time n. By Corollary.7, we have P x V v n (x) =. Function u n (x) is somewhat more convenient to be dealt with. The function v n and u n are related follows: v n (x) μ (x 0 )=P n (x 0,x) μ (x 0 )=P n (x, x 0 ) μ (x) =u n (x) μ (x), where we have used the reversibility of P n. Hence, we have the identity v n (x) = u n (x) μ (x). (.0) μ (x 0 ) Extend u n and v n to n = 0 by setting u 0 = v 0 = {x0 },where A denotes the indicator function of a set A ½ V, that is, the function that has value at any point of A and value 0 outside A. Corollary.9 For any reversible Markov chain, we have, for all x V and n = 0,,,..., v n+ (x) = μ (y) v n (y) μ xy (.) y (a forward equation) and u n+ (x) = μ (x) u n (y) μ xy (.) y (a backward equation). Proof. If n =0then(.)becomes P (x 0,x)= μ (x 0 ) μ xx 0 which is a defining identity for P.Forn, we obtain, using (.9) and (.5), y μ (y) v n (y) μ yx = P n (x 0,y) P (y, x) y = P n+ (x 0,x)=v n+ (x), which proves (.). Substituting (.0) into (.), we obtain (.).

22 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS In particular, for a simple random walk we have μ xy =forx» y and μ (x) = deg (x) so that we obtain the following identities: v n+ (x) = y x deg (y) v n (y). u n+ (x) = deg (x) u n (y). y x The last identity means that u n+ (x) is the mean-value of u n (y) taken at the points y» x. Note that in the case of a regular graph, when deg (x) const, we have u n v n by (.0). Example. Let us compute the function u n (x) on the lattice graph Z. Since Z is regular, u n = v n so that u n (x) the distribution of n at time n provided 0 =0.We evaluate inductively u n using the initial condition u 0 = {0} and the recurrence relation: u n+ (x) = (u (x +)+u (x )). (.3) The computation of u n (x) forn =,, 3, 4andjxj 4areshownhere(allempty fields are 0): x Z n =0 n = n = 4 4 n = n = n = One can observe (and prove) that u n (x)! 0asn!. Example. Consider also computation of u n (x) onthegraphc 3 =(Z 3, f g). The formula (.3) is still true provided that one understands x as a residue mod 3. We

23 .3. RANDOM WALKS 3 have then the following computations for n =,..., 6: x Z 3 0 n =0 0 0 n = 0 n = 4 n =3 3 8 n =4 5 6 n =5 3 n = Here one can observe that the function u n (x) converges to a constant function /3, and later we will prove this. Hence, for large n, the probability that n visits a given point is nearly /3, which should be expected. The following table contains a similar computation of u n on C 5 =(Z 5, f g) x Z n = n = n = n = n = n = Here u n (x) approaches to but the convergence is much more slower than in the 5 case of C 3. Consider one more example: a complete graph K 5.Inthiscase,functionu n (x) satisfies the identity u n+ (x) = u n (y). 4 y6=x The computation shows the following values of u n (x): x K n = n = n = n = n =

24 4 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS This time the convergence to the constant /5 occurs much faster, than in the previous example, although C 5 and K 5 has the same number of vertices. The extra edges in K 5 allows a quicker mixing than in the case of C 5. As we will see, for finite graphs it is typically the case that the transition function u n (x) converges to a constant as n!.forthefunctionv n this means that v n (x) = u n (x) μ (x) μ 0 (x)! cμ (x) as n! for some constant c. The constant c is determined by the requirement that cμ (x) is a probability measure on V, that is, from the identity c x V μ (x) =. Hence, cμ (x) is asymptotically the distribution of n as n!.. The function cμ (x)on V is called the stationary measure or the equilibrium measure of the Markov chain. One of the problems for finite graphs that will be discussed in this course, is therateofconvergenceofv n (x) to the equilibrium measure. The point is that n can be considered for large n as a random variable with the distribution function cμ (x) so that we obtain a natural generator of a random variable with a prescribed law. However, in order to be able to use this, one should know for which n the distribution of n is close enough to the equilibrium measure. The value of n, for whichthisisthecase,iscalledthemixing time. For infinite graphs the transition functions u n (x) andv n (x) typically converge to 0 as n!, and an interesting question is to determine the rate of convergence to 0. For example, we will show that, for a simple random walk in Z N, v n (x) ' n N/ as n!. The distribution function v n (x) is very sensitive to the geometry of the underlying graph. Another interesting question that arises on infinite graphs, is to distinguish the following two alternatives in the behavior of a random walk n on a graph:. n returns infinitely often to a given point x 0 with probability,. n visits x 0 finitely many times and then never comes back, also with probability. In the first case, the random walk is called recurrent, and in the second case - transient. By a theorem of Polya, a simple random walk in Z N is recurrent if and only if N..4 The Laplace operator Let f (x) be a function on R. Recallthat f 0 (x) = lim h 0 f (x + h) f (x) h

25 .4. THE LAPLACE OPERATOR 5 so that f 0 (x) ¼ f (x + h) f (x) h ¼ f (x) f (x h) h for small h. The operators f(x+h) f(x) and f(x) f(x h) are called the difference operators and can be considered as numerical approximations of the derivative. What h h would be the approximation of the second derivative? f 00 (x) ¼ f 0 (x + h) f 0 (x) ¼ h = f(x+h) f(x) h f (x + h) f (x)+f (x h) h = h µ f (x + h)+f (x h) f (x). f(x) f(x h) h h Hence, f 00 is determined by the average value of f at neighboring points x + h and x h, minusf (x). For functions f (x, y) onr, one can develop similarly numerical approximations for second order partial derivatives f x and f y,andthenforthelaplace operator f = f x + f y. More generally, for a function f of n variables x,..., x n, the Laplace operator is defined by n f f =. (.4) x k= k This operator in the three-dimensional space was discovered by Pierre-Simon Laplace in while investigating the movement of the planets in the solar system, using the gravitational law of Newton. It turns out that the Laplace operator occurs in most of the equations of mathematical physics: ² wave propagation, ² heat propagation, ² diffusion processes, ² electromagnetic phenomena (Maxwell s equations), ² quantum mechanics (the Schrödinger equation). The two-dimensional Laplace operator (.4) admits the following approximation: f (x, y) ¼ 4 µ f (x + h, y)+f (x h, y)+f (x, y + h)+f (x, y h) f (x) h 4 that is, f (x, y) is determined by the average value of f at neighboring points (x + h, y), (x h, y), (x, y + h), (x, y h) minus the value at (x, y). This observation motivates us to define a discrete version of the Laplace operator on any graph as follows. Definition. Let (V,E) be a locally finite graph without isolated points (so that 0 < deg (x) < for all x V ). For any function f : V! R, define the function

26 6 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS A.Grigoryan Lecture f by f (x) = deg (x) f (y) f (x). The operator on functions on V is called the Laplace operator of (V,E). In words, f (x) isthedifference between the arithmetic mean of f (y) atall vertices y» x and f (x). Note that the set R of values of f can be replaced by any vector space over R, for example, by C. For example, on the lattice graph Z we have while on Z f (x) = y x f (x +)+f (x ) f (x), f (x +,y)+f (x,y)+f (x, y +)+f (x, y ) f (x, y) = f (x). 4 The notion of the Laplace operator can be extended to weighted graphs as follows. Definition. Let (V,μ) be a locally finite weighted graph without isolated points. For any function f : V! R, define the function μ f by μ f (x) = μ (x) f (y) μ xy f (x). (.5) y The operator μ acting on functions on V, is called the weighted Laplace operator of (V,μ). Note that the summation in (.5) can be restricted to y» x because otherwise μ xy =0. Hence, μ f (x) isthedifference between the weighted average of f (y) at the vertices y» x and f (x). The Laplace operator is a particular case of the weighted Laplace operator when the weight μ is simple, that is, when μ xy =for all x» y. Denote by F the set of all real-valued functions on V. Then F is obviously a linear space with respect to addition of functions and multiplication by a constant. Then μ canberegardedasanoperatorinf, thatis, μ : F!F.Notethat μ is a linear operator on F, thatis, μ (λf + g) =λ μ f + μ g for all functions f,g F and λ R, which obvious from (.5). Another useful property to mention: μ const = 0 (a similar property holds for the differential Laplace operator). Indeed, if f (x) c then f (y) μ μ (x) xy = c μ μ (x) xy = c whence the claim follows. y y

27 .4. THE LAPLACE OPERATOR 7 Recall that the corresponding reversible Markov kernel is given by so that we can write P (x, y) = μ xy μ (x) μ f (x) = y P (x, y) f (y) f (x). Consider the Markov kernel also as an operator on functions as follows: Pf (x) = y P (x, y) f (y). This operator P is called the Markov operator. Hence, the Laplace operator μ and the Markov operator P are related by a simple identity where id is the identical operator in F. μ = P id, Example. Let us approximate f 00 (x) onr using di erent values h and h for the steps of x: f 00 (x) ¼ f 0 (x + h ) f 0 (x) ¼ µ f (x + h ) f (x) f (x) f (x h ) h h h h = µ f (x + h )+ µ f (x h ) f (x) h h h = h µ h + h " h + h + h h µ f (x + h )+ f (x h ) h h # f (x) : Hence, we obtain the weighted average of f (x + h )andf (x h )withtheweights h and h, respectively. ThisaveragecanberealizedasaweightedLaplaceoperatorasfollows. Considera sequence of reals fx k g kz that is de ned by the rules ½ xk + h x 0 =0; x k+ = ; x k + h ; k is even k is odd. For example, x = h, x = h + h, x = h, x = h h,etc. SetV = fx k g kv de ne the edge set E on V by x k» x k+.nowde netheweight¹ xy on edges by and Then we have and, for any function f on V, ¹ f (x k ) = = ½ =h ; k is even, ¹ xk x k+ = =h ; k is odd. ¹ (x k )=¹ xk x k+ + ¹ xk x k ==h +=h ³f (x k+ ) ¹ ¹ (x k ) xk xk+ + f (x k ) ¹ xk xk ½ h f (x k + h )+ h f (x k h ) ; k is even, =h +=h h f (x k + h )+ h f (x k h ) ; k is odd.

28 8 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS.5 The Dirichlet problem Broadly speaking, the Dirichlet problem is a boundary value problem of the following type: find a function u in a domain Ω assuming that u is known in Ω and u is known at the boundary Ω. For example, if Ω is an interval (0, ) then this problem becomesasfollows:find a function u (x) on[0, ] such that ½ u 00 (x) =f (x) for all x (0, ) u (0) = a and u () = b where the function f and the reals a, b are given. This problem can be solved by repeated integrations, provided f is continuous. A similar problem for n-dimensional Laplace operator = P n k= is stated as follows: given a bounded open domain x k Ω ½ R n, find a function u in the closure Ω that satisfies the conditions ½ u (x) =f (x) for all x Ω, (.6) u (x) =g (x) for all x Ω, where f and g are given functions. Under certain natural hypotheses, this problem can be solved, and a solution is unique. One of the sources of the Dirichlet problem is Electrical Engineering. If u (x) is the potential of an electrostatic field in Ω ½ R 3 then u satisfies in Ω the equation u = f where f (x) is the density of a charge inside Ω, whilethevaluesofu at the boundary are determined by the exterior conditions. For example, if the surface Ω is a metal then it is equipotential so that u (x) =conston Ω. Another source of the Dirichlet problem is Thermodynamics. If u (x) is a stationary temperature at a point x in a domain Ω then u satisfies the equation u = f where f (x) is the heat source at the point x. Again the values of u at Ω are determined by the exterior conditions. Let us consider an analogous problem on a graph that, in particular, arises from a dissertation of the problem (.6) for numerical purposes. Theorem.0 Let (V,μ) be a connected locally finite weighted graph (V,μ), and let Ω be a subset of V. Consider the following Dirichlet problem: ½ μ u (x) =f (x) for all x Ω, u (x) =g (x) for all x Ω c, (.7) where u : V! R is an unknown function while the functions f : Ω! R and g : Ω c! R are given. If Ω is finite and Ω c is non-empty then, for all functions f,g as above, the Dirichlet problem (.7) has a unique solution. Note that, by the second condition in (.7), the function u is already defined outside Ω, so the problem is to construct an extension of u to Ω that would satisfy the equation μ u = f in Ω. Define the vertex boundary of Ω as follows: Ω = fy Ω c : y» x for some x Ωg.

29 .5. THE DIRICHLET PROBLEM 9 Observe that the Laplace equation μ u (x) =f (x) forx Ω involves the values u (y) at neighboring vertices y of x, and any neighboring point y belongs to either Ω or to Ω. Hence, the equation μ u (x) =f (x) uses the prescribed values of u only at the boundary Ω, which means that the second condition in (.7) can be restricted to Ω as follows: u (x) =g (x) for all x Ω. This condition (as well as the second condition in (.7) is called the boundary condition. If Ω c is empty then the statement of Theorem.0 is not true. For example, in this case any constant function u satisfies the same equation μ u =0sothatthere is no uniqueness. The existence also fails in this case, see Exercises. The proof of Theorem.0 is based on the following maximum principle. A function u : V! R is called subharmonic in Ω if μ u (x) 0 for all x Ω, and superharmonic in Ω if μ u (x) 0 for all x Ω. A function u is called harmonic in Ω if it is both subharmonic and superharmonic, that is, if it satisfies the Laplace equation μ u = 0. For example, the constant function is harmonic on all sets. Lemma. (A maximum/minimum principle) Let (V,μ) be a connected locally finite weighted graph and let Ω be a non-empty finite subset of V such that Ω c is non-empty. Then, for any function u : V! R, that is subharmonic in Ω, we have max u sup u, Ω Ω c and for any function u : V! R, that is superharmonic in Ω, we have min u inf u. Ω Ω c Proof. It suffices to prove the first claim. If sup Ω c u =+then there is nothing to prove. If sup Ω c u<then, by replacing u by u + const, we can assume that sup Ω c u =0. Set M =max u Ω and show that M 0, which will settle the claim. Assume from the contrary that M>0and consider the set S := fx V : u (x) =Mg. (.8) Clearly, S ½ Ω and S is non-empty. Claim. If x S then all neighbors of x also belong to S. Indeed, we have μ u (x) 0 which can be rewritten in the form u (x) P (x, y) u (y). Since u (y) M for all y V,wehave P (x, y) u (y) M P (x, y) =M. y x y x y x

30 30 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS Since u (x) = M, all inequalities in the above two lines must be equalities, whence it follows that u (y) =M for all y» x. This implies that all such y belong to S. Claim. Let S be a non-empty set of vertices of a connected graph (V,E) such that x S implies that all neighbors of x belong to S. ThenS = V. Indeed, let x S and y be any other vertex. Then there is a path fx k g n k=0 between x and y, thatis, x = x 0» x» x»...» x n = y. Since x 0 S and x» x 0,weobtainx S. Sincex» x,weobtainx S. By induction, we conclude that all x k S, whence y S. It follows from the two claims that the set (.8) must coincide with V,which is not possible since u (x) 0inΩ c. This contradiction shows that M 0. Proof of Theorem.0. Let us first prove the uniqueness. If we have two solutions u and u of (.7) then the difference u = u u satisfies the conditions ½ μ u (x) =0 forallx Ω, u (x) =0 forallx Ω c. We need to prove that u 0. Since u is both subharmonic and superharmonic in Ω, Lemma.yields 0=inf Ω c u min Ω u max u sup u =0, Ω Ω c whence u 0. Let us now prove the existence of a solution to (.7) for all f,g. For any x Ω, rewrite the equation μ u (x) =f (x) intheform P (x, y) u (y) u (x) =f (x) P (x, y) g (y), (.9) y x,y Ω y x,y Ω c where we have moved to the right hand side the terms with y Ω c andusedthat u (y) =g (y). Denote by F Ω the set of all real-valued functions u on Ω and observe that the left hand side of (.9) can be regarded as an operator in this space; denote it by Lu, thatis, Lu (x) = P (x, y) u (y) u (x), y x,y Ω for all x Ω. Rewrite the equation (.9) in the form Lu = h where h is the right hand side of (.9), which is a given function on Ω. NotethatF Ω is a linear space. Since the family {x} ªx Ω of indicator functions form obviously a basis in F Ω, we obtain that dim F Ω =#Ω <. Hence, the operator L : F Ω!F Ω is a linear operator in a finitely dimensional space, and the first part of the proof shows that Lu = 0 implies u = 0 (indeed, just set f =0andg = 0 in (.9), that is, the operator L is injective. By By Linear Algebra, any injective operator acting in the spaces of equal dimensions, must be bijective (alternatively, one can say that the injectivity of L implies that det L 6= 0 whence it follows that L is invertible and, hence, bijective). Hence, for any h F Ω, there is a solution u = L h F Ω,which finishes the proof.

31 .5. THE DIRICHLET PROBLEM 3 How to calculate numerically the solution of the Dirichlet problem? Denote N = #Ω and observe that solving the Dirichlet problem amounts to solving a linear system Lu = h where L is an N N matrix. If N is very large then the usual elimination method (not to say about inversion of matrices) requires too many operations. A more economical Jacobi s method uses an approximating sequence fu n g that is constructed as follows. Using that μ = P id, rewrite the equation μ u = f in the form u = Pu f and consider a sequence of functions fu n g given by the recurrence relation ½ Pun f in Ω u n+ = g in Ω c. The initial function u 0 can be chosen arbitrarily to satisfy the boundary condition; for example, take u 0 =0inΩ and u 0 = g in Ω c. In the case f = 0, we obtain the same recurrence relation u n+ = Pu n as for the distribution of the random walk, although now we have in addition some boundary values. Let us estimate the amount of computation for this method. Assuming that deg (x) is uniformly bounded, computation of Pu n (x) f (x) for all x Ω requires ' N operations, and this should be multiplied by the number of iterations. As we will see later (see Section 4.4), if Ω is a subset of Z m of a cubic shape then the number of iterations should be ' N /m. Hence, the Jacobi method requires ' N +/m operations. For comparison, the row reduction requires ' N 3 operations.ifm = then the Jacobi method requires also ' N 3 operations, but for higher dimensions m the Jacobi method is more economical. Example. Let us look at a numerical example in the lattice graph Z for the set = f; ; :::; 9g, for the boundary value problem ½ u (x) =0 in u (0) = 0; u(0) = : The exact solution is a linear function u (x) =x=0. Using the explicit expression for, write the approximating sequence in the form u n+ (x) = u n (x +)+u n (x ) ; x f; ; :::; 9g while u n (0) = 0 and u n (0) = for all n. Set u 0 (x) =0forx f; ; :::; 9g. The computations yield the following: x Z n = n = n = n = n = ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: n =50 0:00 0:084 0:7 0:6 0:35 0:45 0:55 0:68 0:77 0:88 :00... ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: n =8 0:00 0:097 0:9 0:9 0:39 0:49 0:59 0:69 0:79 0:897 :00 so that u 8 is rather close to the exact solution. Here N = 9 and, indeed, one needs ' N iterations to approach to the solution. For the row reduction method, one needs ' N of row operation, and each row operation requires ' N of elementary operations. Hence, one needs ' N 3 of elementary operation.

32 3 CHAPTER. THE LAPLACE OPERATOR ON GRAPHS

33 Chapter Spectral properties of the Laplace operator Let (V,μ) be a locally finite weighted graph without isolated points and μ be the weighted Laplace operator on (V,μ). A.Grigoryan Lecture Green s formula Letusconsiderthedifference operator r xy x, y V and maps F to R as follows: that is defined for any two vertices r xy f = f (y) f (x). The relation between the Laplace operator μ and the difference operator is given by μ f (x) = (r xy f) μ μ (x) xy = P (x, y)(r xy f) y y Indeed, the right hand side here is equal to (f (y) f (x)) P (x, y) = f (y) P (x, y) f (x) P (x, y) y y y = f (y) P (x, y) f (x) = μ f (x). y The following theorem is one of the main tools when working with the Laplace operator. For any subset Ω of V, denote by Ω c the complement of Ω, that is, Ω c = V n Ω. Theorem. (Green s formula) Let (V,μ) be a locally finite weighted graph without isolated points, and let Ω be a non-empty finite subset of V. Then, for any two functions f,g on V, μ f(x)g(x)μ(x) = (r xy f)(r xy g) μ xy + (r xy f) g(x)μ xy (.) x Ω x,y Ω x Ω y Ω c 33