In this section, we will provide a device to calculate the resistances and the critical exponents of the Besov spaces on K. We first recall some formal notions and techniques on electric network theory [DS, LP].
Let N = (V, c) denote the (electric) network with vertex setV (finite or count-ably infinite) and conductancec:V×V →[0,∞) (c(x, y) =c(y, x) for allx, y∈V).
The edge setE={(x, y)∈(V×V)\∆ :c(x, y)>0}. An edge (x, y)∈E is referred as aresistor(orconductor) with resistancerxy =r(x, y) =c(x, y)−1. The energy of f ∈`(V) on N is given by
EN[f] = 1 2
X
x,y∈V
c(x, y)(f(x)−f(y))2 (6.1) as in (2.4). Also we can define the effective resistance RN(A, B) between two nonempty subsetsA, B⊂V as in (4.2).
Definition 6.1. For two networks N1 = (V1, c1) and N2 = (V2, c2) with a set of common vertices U ⊂V1∩V2, #U ≥2, we say that N1 and N2 are equivalent on U if for anyf ∈`(U),
inf{EN1[g1] :g1 ∈`(V1), g1|U =f}= inf{EN2[g2] :g2 ∈`(V2), g2|U =f}. (6.2) It is easy to show that if N1 and N2 are equivalent on U, then they are also equivalent on any U0 ⊂U. As a result,RN1(A, B) =RN2(A, B) for any nonempty A, B⊂U.
The two most basic transformations to reduce networks to equivalent ones are theseries lawand theparallel lawof resistance. The third one is the ∆-Y transform (or star-triangle Law): let N1 be the triangle shaped network with V1 = {x, y, z}
as on the left of Figure 1, and let N2 be the starlike network on the right with V2 =V1∪ {p}; for the two network to be equivalent, the resistances are related by
∆-Y x
y z
x
y z
p
rxy rzx
ryz
Rx
Ry Rz
Figure 1: ∆-Y transform
Rx= rxyrzx
rxy +ryz+rzx
, Ry = rxyryz
rxy +ryz+rzx
, Rz= rzxryz
rxy+ryz+rzx
respectively. For some network N = {V, c}, #V > 3 with proper symmetry, we can add one vertex and transform it to an equivalent starlike network (see the examples in the sequel and [K] for more details); we regard such transformation as a generalized∆-Y transform.
More generally, we have from Proposition 4.1, that if V = V◦∪∂V, #∂V ≥2 then forf ∈`(∂V),
min{EN[g] :g∈`(V), g|∂V =f}= 1 2
X
x,y∈∂V,x6=y
c∗(x, y)(f(x)−f(y))2. (6.3)
Then the network N∗ = {∂V, c∗} is equivalent to N on ∂V. For proper ∂V, the graph of network N∗ may contain a complete subgraph Kn. In this case, we say that the transform N → N∗ is a local completion. For example, as in Figure 2, let
∂V ={x1, x2, . . . , x5}, then the graph ofN∗ is a complete graph K5.
K5 x1
x2
x3 x4
x5
x1
x2
x3 x4
x5
Figure 2: Local completion
Besides the above mentioned transformations, there are other basic tools in net-work reduction we will use: cutting and shorting, and the Rayleigh’s monotonicity law, namely,if some resistances of resistors in a network are increased (decreased), then the effective resistance between any two points in the graph can only increase (decrease).
Example 6.2. Cantor middle third set Let S1(ξ) = 13ξ and S2(ξ) = 13(ξ+ 2) on R. Then the self-similar set K is the Cantor middle-third set with ratio r = 13. It is totally disconnected and the Hausdorff dimension is α = log 2log 3. The critical exponents β1∗ =β2∗ =β3∗ =∞.
#
1 2
11 22
111 222
Figure 3: The limiting resistance for Cantor set
Indeed, for λ∈(0,12) (rα= 12), the resistance between 0 and 1 (see Figure 3) is
and Theorem 5.9 implies the result. 2
Example 6.3. Sierpinski gasketIt is the self-similar setKgenerated by the maps Si(ξ) = 12(ξ−ei−1)+ei−1 wheree0 = 0andei, i= 1, . . . , N−1are the standard basis (The critical exponent is known in [Jo].)
∆-Y
Figure 4: Cutting in Sierpinski gasket, N = 3
We only prove the case N = 3, the other case is the same (the reader is also advised to useN = 2 to get a clearer picture). By symmetry, it suffices to find the resistance R(λ)(1∞,2∞). We denoteR(λ)n =R(λ)n (1n,2n) for short.
To estimate the upper bound, we delete the edges (ϑ, i), (ijk, jik), fori6=j ∈Σ, k= 0,1, . . . , n−2 in the subgraph ofXn(see Figure 4). Then we get a new subgraph consisting of 3 copies of Xn−1 with 3 horizontal edges (ijn−1, jin−1), i 6= j ∈ Σ at level n connecting them; we label these copies by 1,2,3 such that the copy i contains the vertex in. Then apply the the ∆-Y transform to the three vertices in Ai := {ijn−1 : j ∈ Σ} at the n-th level of each copy to get a starlike tree with center in∗, i ∈Σ respectively. As the resistance between any pair of vertices in Ai equals 3λRn−1, it follows that the resistance between in∗ and a vertex in Ai in the corresponding starlike tree is 3λ2 Rn−1. Moreover, between any pair in∗, j∗n, i 6= j, there is a 3-step path [in∗, ijn−1, jin−1, j∗n]. Replacing these paths with resistors, we get a triangle with vertices{in∗ :i∈Σ}and each side has resistance 3λRn−1+ (3λ)n.
By applying the monotonicity law and the series law, R(λ)n ≤R(1n,1n∗) +R(1n∗,2n∗) +R(2n∗,2n)
= 3λ
2 R(λ)n−1+2
3(3λR(λ)n−1+ (3λ)n) +3λ 2 R(λ)n−1
= 5λR(λ)n−1+ 2·3n−1λn.
Hence R(λ)(1∞,2∞) = limn→∞R(λ)n = 0 for λ ∈ (0,15). By Proposition 5.3 and Theorem 5.4, we have β1∗ ≤β3∗ ≤ log 5log 2.
To obtain the lower bound of the critical exponent, we need another technique.
We reassign the conductance on the n-th level of the subgraph Xn (n ≥ 1): for µ >0, letec(x,x−) = (3λ)−|x| forx∈Xn, and let
ec(x,y) =
( (3λ)−|x|, if|x|< n,
µ−1(3λ)−n, if|x|=n, forx∼hy∈Xn.
Denote the resistance between 1nand 2nwith respect to the aboveecbyRn(λ,µ). Then apply the generalized ∆-Y transforms to each triangle (x,x1,x2,x3) forx∈ Jn−1, and then replace each pair{x,x0}by a singlex(see Figure 5 forN = 2 for a clearer illustration; Figure 6 for N = 3 corresponds to the dotted box in Figure 5).
We have
R(λ,µ)n ≥ 2µ
µ+ 3(3λ)n+R(λ,φ(µ))n−1 , (6.4)
whereφis given by the parallel resistance formula φ(µ)−1 =
3λ
2µ µ+ 3+µ
−1
+ 1. (6.5)
∆-Y
Figure 5: µ-parameter and shorting forN = 2
∆-Y Shorting
Figure 6: µ-parameter and shorting forN = 3
The equation φ(µ) = µ has a solution µ∈ (0,1) if and only if λ > 15. With such fixed point µ, by (6.4), we have R(λ)(1∞,2∞) ≥limn→∞R(λ,µ)n ≥ R(λ,µ)1 > 0. By Theorem 5.7, we have log 5log 2 ≤β1∗ ≤β3∗, and completes the proof. 2 In the next example, we adjust the above method slightly for the new situation with two different effective resistances of (i∞, j∞).
Example 6.4. Pentagasket The pentagasket K is the attractor K of the five similitudes Si(ξ) = 3− To determine the critical exponent, we need to calculate the resistances
R(λ)(1∞,2∞) and R(λ)(1∞,3∞).
∆-Y Figure 7, and using the same technique as before, we have
An≤R(1n,1n∗) +R(1n∗,3n∗) +R(3n∗,3n)
To obtain the lower bound of the critical exponents, we reassign the conductance on the bottom of the subgraph Xn (n ≥ 1) with two parameters µ1 and µ2 : for x∈ Jn−2, and then replace each complete subgraph K4 by a starlike network with greater energy (Figure 8). By a direct calculation, the conductancec∗inK4 is given by
c∗(x1,x11) = µ1+ 4
µ1+ 2, c∗(x1,x13) =c∗(x1,x14) = µ1+ 3 µ1+ 2,
K4 Shorting
and the resistances in the star are given by ρ1 = µ1(µ1+ 2)
µ21+ 5µ1+ 5, ρ2 = µ1(µ1+ 2)2 (µ1+ 1)(µ21+ 5µ1+ 5). By the monotonicity law and series law,
A(µn1,µ2)≥2ρ2(5λ)n+A(φn−11(µ1,µ2),φ2(µ1,µ2)), (6.6) (same inequality holds if we replace A by B) where φ1 and φ2 are given by the parallel resistance formulas
More computational issues on the critical exponent of nested fractals can be found in [K]. Finally, we give an example thatβ1∗6=β3∗.
Example 6.5. Cantor set×interval Let Σ ={1,2,3,4,5,6}and let p1 = 0, p2 = (0,13), p3 = (0,23), p4 = (23,0), p5 = (23,13), p6 = (23,23) in R2. For i∈Σ, let Si(ξ) =
1
3ξ +pi on R2. Then the self-similar set K is the product of a Cantor middle-third set and a unit interval (see the associated augmented tree in Figure 9), and α = dimHK = log 2log 3 + 1 = log 6log 3. The λ-NRW has conductance (6λ)−n on the n-th level (rα = 16). The critical exponents are
β1∗ = 2 at λ= 1
9; β2∗ =β3∗ =∞
#
Figure 9: The graph for Cantor set×inteval
First we show that R(λ)(1∞,4∞) > 0 for any λ > 0. For n ≥ 1, consider a
Figure 10: Shorting in Cantor set×interval
Next we consider the effective resistance R(λ)(1∞,3∞) by using a similar short-ing device as in previous examples. Denote R(λ)n = R(λ)n (1n,3n) for short. As in Example 6.4, we reassign the conductance on the bottom of the subgraphXnby an additional factorµ−1, and by the same method applied to triangles (x,x1,x3) (also to (x,x4,x6), see Figure 10), we have
R(λ,µ)n ≥ 2µ
µ+ 1(6λ)n+R(λ,φ(µ))n−1 , (6.8)
whereφis given by
φ(µ)−1= 2
6λ 2µ
µ+ 1+µ −1
+ 1. (6.9)
The equationφ(µ) =µhas a solution µ∈(0,1) if and only if λ > 19.
With such fixed point µ, by (6.8), we have R(λ)(1∞,3∞) ≥ limn→∞R(λ,µ)n ≥ R(λ,µ)1 > 0. On the other hand, we show that if R(λ)(1∞,3∞) > 0, then λ ≥ 19. Without loss of generality, we assume that 0< λ < 1/6. For n≥1, let fn be the energy minimizer (harmonic function) onXnwith boundary conditions fn(1n) = 1 and fn(3n) = 0. Then Rn(1n,3n) = EXn[fn]−1. By Corollary 5.2 (iv) ⇒ (iii), let C1:= supn≥1EXn[fn] = (infn≥1Rn(1n,3n))−1 <∞. Pick a positive integer n1 such thatP∞
n=n1+1(6λ)n< 36C1
1. Then for n≥n1,
|fn(1n)−fn(1n1)|2 ≤ EXn[fn]RXn(1n,1n1)≤C1
n
X
k=n1+1
(6λ)k ≤ 1
36, (6.10) which implies fn(1n1) ≥ 56. Analogously we have fn(3n1) ≤ 16. Let m = n−n1. With a similar argument as in (6.10), for z∈ {1,4}m,
|fn(1n1z)−fn(1n1)|2 ≤ EXn[fn]RXn(1n1z,1n1)≤ 1 36,
which impliesfn(1n1z)≥ 23. Analogously we havefn(3n1w)≤ 13 for allw∈ {3,6}m. Now, for z = i1i2· · ·im ∈ {1,4}m, denote the word j1j2· · ·jm ∈ {3,6}m with jk = ik+ 2 for all k by z0. Note that for each z ∈ {1,4}m, there is a horizontal path with length 3n−1 from 1n1z to 3n1(z0). The resistance on such path is given by RJn(1n1z,3n1(z0)) = (3n−1)(6λ)n. Counting the energy on these 2m disjoint horizontal paths, we get
C1≥ EXn[fn]≥ X
z∈{1,4}m
[fn(1n1z)−fn(3n1(z0))]2
RJn(1n1z,3n1(z0)) ≥ 2n−n1 9(3n−1)(6λ)n
for arbitrary n≥ n1. Hence λ≥ 19 and the claim follows. By Proposition 5.3, we
have β1∗= 2. 2
Remark. To investigate the situation that β1∗ < β3∗, it is natural to study the products of self-similar sets. But in general, ifK1 andK2 are connected self-similar sets, then the critical exponent of the productK1×K2 satisfies
β1∗ ≤max{dimHK1,dimHK2}+ 1≤dimHK1+ dimHK2 =α.
Although the criteria in the last section cannot be applied directly, it still has a similar link between the effective resistance of EX and the energy on the product (see [K] for more details). For example, in the product [0,1]×SG, the effective resistancesR(λ)(i∞, j∞) have two critical exponentsλ∗1 = 14 and λ∗3= 15 for various i, j, while 2 =β1∗ < log 5log 2 =β3∗ < α= log 6log 2. With a similar technique as in Example 6.5, it follows thatβ1∗ = 2 if one of Ki is a unit interval. To generalize the results above, we may leave a conjecture as
β1∗(K1×K2) = min{β1∗(K1), β1∗(K2)}, and β3∗(K1×K2) = max{β3∗(K1), β3∗(K2)}.