Asaf Nachmias' Lectures in PIMS Probability Summer School 2014 Random Walks on Random Fractals 3 Contents 1. June 3 3 1.1. Random walk interpretation of effective resistance 6 1.2. Network simplifications 7 2. June 5 7 2.1. Infinite network 13 2.2. Nash-William's inequality 14 3. June 6 14 3.1. Random walk on the infinite incipient cluster 16 4. June 9 17 5. June 10 21 6. June 12 26 7. June 13 32 7.1. Random walks on planar graphs 32 8. June 17 37 9. June 19 40 9.1. Distributional limits 42 10. June 20 42 11. June 23 46 12. June 24 49 13. June 26 50 14. June 27 53 1. June 3 Reference: Chapter 8 of "Probability on Trees: an introductory climb" by Yuval Peres. We start with a definition. Definition 1.1. A network is a finite connected graph endowed with non-negative edge weights fcege2E called conductances. We call the reciprocal re = 1=ce the edge resistances. We have two distinct vertices a and z. We will be considering the process of travelling from a to z. In our studies, harmonic functions would be analogous to voltages. Definition 1.2. A function h : V ! R is harmonic at a vertex x 2 V if X Cxy h(x) = h(y) π y:y∼x x P where πx = y:y∼x Cxy and y ∼ x ) y is a neighbour of x. Definition 1.3. A voltage is a function V on the vertices that is harmonic on G − fa; zg. The point here is that the voltage function is completely determined by V (a) and V (z). Lemma 1.4. Let h : V ! R be a voltage such that h(a) = h(z) = 0. Then h(x) = 0 for all x 2 V . Proof. Let it not be so, and let the maximum M > 0 be attained at some point x 2 G − fa; zg. Then if x1; x2; :::; xm be the neighbouring vertices, we then must have m X Cx;x h(x) = i h(x ) π i i=1 x 4 and so either there exists some i for which h(xi) > h(x) = M or for each i, we have h(xi) = M. The former cannot happen since M is the maximum, and so the latter is the only possibility. But proceeding in this way, we find that every vertex must have voltage M, which contradicts the boundary values. Existence: Why must such a function exist? Say jV j = n, then we simply have to solve n − 2 linear equations for n − 2 variables (since the boundary values are already given to us). Alternatively consider the network random walk Xk with transition probability Cxy px;y = : πx Note that this random walk is reversible, i.e. πxpx;y = πypy;x for all x; y 2 V: Put h(x) = Px(Xn visits z before hitting a): We claim that this is a harmonic function. This is because, say if x1; x2; :::; xm are the neighbours of x then in the very first step after starting from x, the walker will visit some xi; 1 ≤ i ≤ m. Then we should have m X Px(Xn visits z before hitting a) = Px(X1 = xi;Xn visits z before hitting a) i=1 m X = Px(X1 = xi)P (Xn visits z before hitting ajX1 = xi) [Markov property] i=1 m X = px;xi h(xi) i=1 m X Cx;x = i h(x ) π i i=1 x which proves that it is indeed harmonic. Note that h(z) = 1 and h(a) = 0. In general if you want a voltage with boundary values g(a) and g(z) then put g(x) = g(a) + h(x)[g(z) − g(a)]: Now we shall define current flow which is defined only on directed graphs. Definition 1.5. A flow from a to z is a function θ : E ! R satisfying: (i) θ(x; y) = −θ(y; x); (ii) for all v 2 V − fa; zg, we have X θ(v; w) = 0 w:w∼v and this is the Kirchoff's law which states that flow in must equal flow out at any vertex. Given a voltage h, the current flow associated with it is I(x; y) = [h(y) − h(x)]Cx;y: This has the following properties: (i) antisymmetric; (ii) Kirchoff's law holds by harmonicity. To see the latter, note that X X X X I(x; y) = Cx;y[h(y) − h(x)] = h(y)Cx;y − h(x) Cx;y = h(x)πx − h(x)πx = 0 y:y∼x y∼x y∼x y∼x 5 using harmonicity of h. A current flow satisfies the cycle law if for any cycle fe1; e2; :::; emg we have m X rei I(ei) = 0: i=1 To see this, let the cycle be given by the vertices v1; v2; :::; vm where vm+1 = v1 and ei = vi ! vi+1. Then note that m m X X 1 r I(e ) = [h(v ) − h(v )]C ei i C i+1 i vi;vi+1 i=1 i=1 ei m X = [h(vi+1) − h(vi)] i=1 =0 From now on, all voltages have h(z) = 0. Definition 1.6. The strength of a flow θ is X jjθjj = − θ(a; x) x:x∼a (there has been a little confusion about the sign of the summation on the right hand side, but it's minor). Lemma 1.7. If θ is a flow from a to z satisfying the cycle law, and I is the current flow of a voltage h with h(z) = 0, and jjθjj = jjIjj, then θ = I. Proof. The function J = θ − I : E ! R is a flow satisfying the cycle law. This is because both θ and I satisfy the cycle law, and the cycle law expression is linear. Define n X g(x) = J(ei)rei i=1 where e1; e2; :::; em is a path from a to x. Clearly g is well-defined due to the cycle law being satisfied. Now we have to check that it is harmonic. For this we use Kirchoff's law. Let w1; w2; :::; wk be neighbours of v 2 V − fag, then if we consider the path P from a to v, then a path from a to wi would be P [ (v; wi) for all 1 ≤ i ≤ k. Hence we have X g(wi) = J(e)re + J(v; wi)r(v;wi); e2P and hence we get k k X Cv;wi X Cv;wi X g(wi) = [ J(e)re + J(v; wi)r(v;wi)] πv πv i=1 i=1 e2P k k X X Cv;wi X Cv;wi = J(e)re[ ] + J(v; wi)r(v;wi) πv πv e2P i=1 i=1 k k X 1 X X = J(e)re + J(v; wi) [as Cv;wi = πv and Cv;wi :r(v;wi) = 1] πv e2P i=1 i=1 X 1 X = J(e)re + J(v; wi) πv e2P w:w∼v =g(v) + 0 [ by Kirchoff's law which holds for J] 6 and thus g harmonic. Flow out of a is 0, since we have defined g along paths starting from a. Thus clearly g is harmonic at a too. This is because, if x1; x2; :::; xm are neighbours of a, then g(xi) = J(ei)rei where ei = (a; xi): Then m m X Ce X Ce i g(x ) = i J(e )r π i π i ei i=1 a i=1 a m 1 X = J(e ) π i a i=1 m m 1 X X = [ θ(e ) − I(e )] π i i a i=1 i=1 1 = [jjIjj − jjθjj] πa =0 since jjθjj = jjIjj. Hence it is harmonic on entire V (G), and hence it must be constant everywhere, since g(a) = 0 [note that the conductances are all nonnegative, hence if g is harmonic at a too, then the value of g on all its neighbours must be 0, and this keeps on progressing]. This proves that J = 0. Lemma 1.8. Gievn a network, the ratio Va − Vz jjIjj is independent of the choice of the voltage function V , where I is the current flow induced by V . Proof. We have already assumed that Vz = 0. And we know that the voltage values on the intermediate vertices are completely determined (using harmonicity) by the boundary values Va and Vz (or rather, in this case, just by Va, so that if instead of Va we consider c:Va, then every value on the intermediate vertices will get multiplied by c, and the ratio would remain unchanged. Hence proved. Definition 1.9. The effective resistance between a and z is given by V − V R (a $ z) = a z : eff jjIjj Here h(a) ≥ 0. 1.1. Random walk interpretation of effective resistance. Firstly, note that for any x 2 V , Va − Vx qx = Px( hit z before a) = : Va − Vz The reasoning behind this is as follows: When we consider starting from a, the probability of this event is clearly 0, and when we start from z, it is obviously 1. Now let us start at some x 2 V − fa; zg. Then clearly, in the very first step of the walk starting from x, the walker will go to one of the neighbours of x, hence we can write X Px( hit z before a) = Px(X1 = y; hit z before a) y:x∼y X = P [X1 = yjX0 = x]P [ hit z before ajX1 = y] y:x∼y X Cxy = P [ hit z before a] π y y:x∼y x 7 X Cxy = q π y y:x∼y x which means X Cxy q = q x π y y:x∼y x for all x 2 V − fa; zg, and hence q is harmonic on V − fa; zg, with boundary values qa = 0 and qz = 1.