Random Walks and Electric Networks Rich Schwartz March 9, 2020 Most of the things in these notes can be found in the book Random Walks and Electric Networks, by Peter Doyle and Laurie Snell. I highly recommend this excellent book. These notes cover some of the main points in the book, but they do not always do things as they are in the book. 1 Harmonic Functions Let G be a graph. The function f : G ! R (defined on the vertex set of G) is called harmonic at a vertex v if k 1 X f(v) = f(wi): (1) k i=1 Here k is the degree of v and w1; :::; wk are the vertices incident to v. Notice the similarity between this equation and Equation 12. Suppose that the disjoint vertex sets A and B have been fixed. Suppose that f is some function on A [ B. We call the function F : G ! R a harmonic extension of f if F = f on A[B and F is harmonic on G−A−B. We will show that each f has a unique harmonic extension. Lemma 1.1 If F1 and F2 are two harmonic extensions of f then F1 = F2. Proof: Let g = F1 − F2. Note that g is a harmonic extension of the 0 function on A [ B. There must be some vertex v where g achieves its max- imum. But g(v) is the average of the values of g at the vertices incident to v. This is only possible if g also takes its max at all the vertices incident to v. Continuing outward from v, we see that g must take its max everywhere. But then this common value must be 0. ♠ 1 Lemma 1.2 Every function f on A [ B has a harmonic extension. Proof: Let V be the vector space of functions defined on G − A − B. Let ∆f : V ! V be the map 1 ∆ g(v) = v − X g(w ): (2) f k i As usual k is the degree of v and w1; :::; wk are the vertices incident to v. The map ∆f is an affine transformation from V into V . (It is matrix multi- plication followed by translation which depends on the function f.) Suppose that ∆f (F1) = ∆f (F2). We extend g to that g = 0 on A [ B. Consider g = F1 − F2. Then, at each vertex v 2 G − A − B, the value g(v) is the average of its neighbors. But then the same argument as above shows that g cannot have a nonzero maximum or a nonzero minimum. Hence g is identically 0. (This step uses the fact that A [ B is nonempty, so that g = 0 at some vertex.) We have shown that ∆f is one to one. But then ∆ must be onto. In particular, the 0 function on G − A − B is in the image of ∆f . That is, there exists F 2 V such that ∆f (F ) is the 0 function. By definition F is a harmonic extension of F . ♠ Electric Flow: We keep the same notation as above. We imagine G as an electric network made of wires which all have the same resistance. We think of A as a set of sources and B as a set of sinks. Imagine using some kind of battery to keep all the vertices of A at voltage 0 and all the vertices of B at voltage 1. Electric current will then flow through the network. Let V be the harmonic extension of the function that is equal to come constant Va on A and equal to 0 on B. The current ivw that flows from vertex v to vertex w if V (v) − V (w). This definition obeys the usual laws one learns about in physics. First, δV = iR; where R is the resistance and δV denotes the change in voltage. Here we are setting R = 1 for every edge. Second, the flow of current into any vertex in G − A − B is the same as the total flow out. Below we will always make the choice of Va so that the current flowing out of a equals 1. In this ccase, we call Va the voltage function. 2 2 Effective Resistance Let V be the voltage function. Just to do things a bit more formally, we define X ia = (V (a) − V (x)): (3) x$a Again, Va is chosen so that ia = 1. Lemma 2.1 ia + ib = 0. Proof: This is a consequence of Kirkhoff's law. Here's the proof. Define ixy = V (x) − V (y) when x and y are connected by an edge and otherwise ixy = 0. Setting X ix = ixy; (4) y we have ix = 0 unless x = a or x = b. This is Kirkhoff's law, and it follows from the fact that V is harmonic on G − a − b. But then X X X ia + ib = ix = ixy = 0 (5) x x y The last sum is 0 because ixy = −iyx and we are summing over all pairs. ♠ Say that a nice flow is a function j : G × G ! R such that 1. jxy = 0 if x and y do not share an edge. 2. jxy = −jyx. 3. If x 2 G − a − b then jx = 0. That is, the amount of current flowing P into x is the same as the amount flowing out. Here jx = jxy. If, additionally, ja = 1, we call ja a unit flow. Example: Notice that if we take j = i then the axioms are satisfied. But, there are plenty of examples of unit flows other than the electric flow. Here is an example. Suppose that G0 is a graph obtained by adding some edges to G. Then the current i defined relative to (G; a; b) gives a unit flow relative to (G0; a; b) but this unit flow on G0 might not be the current defined relative to (G0; a; b). The extra wires might change the current. Here is a key equation. 3 Lemma 2.2 Suppose that j is a nice flow and φ : G ! R is any function. Then 1 X ja(φ(a) − φ(b)) = jxy(φ(x) − φ(y)): 2 xy Proof: Both the left hand side and the right hand side are linear functions of φ. So, to prove this equality it suffices to prove it on a basis in the vector space of such functions. Suppose that φ(a) = 1 and φ(x) = 0 otherwise. Then the two sides are equal just by definition. Suppose that φ(x) = 1 for some x 2 G − a − b. Then the left side is obviously 0, and the right side is given by k X ixyk ; i=1 and this vanishes by Kirkhoff's law. Here y1; :::; yk are the vertices incident to x. Finally, consider the case when φ is identically 1. Then both sides vanish. We've established the identity on a basis, so we're done. ♠ Now we come to the main point of the section. Define the energy dissi- pation 1 X 2 E(G; a; b) = ixy: (6) 2 xy 2 2 The factor of 1=2 is added because ixy = iyx and both terms appear in the sum. Theorem 2.3 R(G; a; b) = E(G; a; b). Proof: Applying Lemma 2.2 to the case j = i and φ = V (the voltage function) we have 1 X 1 X E(G; a; b) = ixyixy = ixy(V (x) − V (y)) = 2 xy 2 xy ia(Va − Vb) = Va − Vb = R(G; a; b): This completes the proof. ♠ 4 3 A Variational Principle Let (G; a; b) and V and ixy be as above. Here we explain a variational princi- ple that characterizes the current iix as the one which minimizes the energy dissipation. Define 1 X 2 E(j) = jxy: (7) 2 xy All we are assuming about j is that it is a nice flow. We call j a unit flow if ja = 1. We call j a null flow if ja = 0. We already know that R(G; a; b) = E(i). Here is the main result: Theorem 3.1 E(i) ≤ E(j). In other words, R(G; a; b) is the min of all E(j) taken over all unit flows j. Proof: The nice flows form a vector space. Let V be this vector space. There is a canonical inner product on V , namely 1 X hj; ki = jxykxy (8) 2 xy With this formalism, we have E(j) = hj; ji. Suppose that k is a null flow. Then X 1 X hk; ii = kxyixy = kxy(V (x) − V (y)) =∗ ka(V (a) − V (b)) = 0: (9) x;y 2 x;y The starred equality is Lemma 2.2. Now let j be any unit flow. We can write j = i + k where k 2 V is a null flow. But then E(j) = hi + k; i + ki = E(i) + E(k) ≥ E(i): Here have used the fact that hi; ki = 0. ♠ This result has a geometric interpretation. The set of null flows is a hyperplane through the origin in V and and set of unit flows is a translation of this hyperplane by the electric flow. By essentially the Pythagorean theorem, the square distance from a unit flow to the origin is minimized by i. 5 4 Rayleigh's Theorem Now we can prove the main technical result. Theorem 4.1 Let G0 be a graph obtained from G by adding some edges. Then R(G0; a; b) ≤ R(G; a; b).