Proof of Proposition 5.9

The structure of this subsection is similar to the previous two subsections. It may help the reader to refresh the core ideas of this subsection, described on page 77.

Now, let us consider our domain

A2 ={∃ 0lasso, ∂∗C_Λmax 0 ←→1∗ ∂∗C_Λmax 1}

={∃ 0lasso, ∂∗C_Λmax 0 ←→1∗ ∂∗C_Λmax 1, C_Λmax 0 > C_Λmax 1}, (5.22) where the second identity follows from the fact that the maximal0circuit is weakly

0connected toΛc _{and, therefore, is larger than every 1circuit.}

As before, we state some fundamental relations between σ and m(σ). Remark 5.19 Let σ ∈A2_{. Then the following properties hold:}

σ−1(1)∪C_extmin 0_Cmax 1

Λ (σ)(σ)⊃m(σ) −1

(1) (5.23)

σ−1(0) =m(σ)−1(0)∪C_extmin 0_Cmax 1

Λ (σ)(σ) (5.24) i_Cfill_(σ)∩Cmin 0

extCmax 1

Λ (σ)(σ)∩C

max 0_{(σ)6=∅ (5.25)}

Proof: First of all, recall that

m(σ) =₁σ−1_(1)∪i_Cfill_(σ)

and that for our case

i_Cfill_{(σ) = max}i C

C_extmin 0_Cmax 1

Λ (σ)(σ)∪σ

−1_(1).

(5.26) These definitions immediately imply the first two Properties (5.23) and (5.24).

Recall that the maximal0circuit[σ]is larger than the maximal1circuit[σ], since it is weakly 0connected[σ] to Λc_{. Because of this, the minimal 0circuit[σ] in the}

exterior of the maximal 1circuit[σ], C_extmin 0_Cmax 1

Λ (σ)(σ), exists and is contained in the

annulus

C_Λmax 1(σ), C_Λmax 0(σ) :=extC_Λmax 1(σ)∩intC_Λmax 0(σ)∪C_Λmax 0(σ).

Consequently, each node of the maximal0circuit[σ]that is∗weakly1∗connected[σ]

to the maximal 1circuit[σ] also belongs to the minimal 0circuit[σ] in the exterior of the maximal 1circuit[σ]. Because of

such a node exists and, therefore,

C_extmin 0_Cmax 1

Λ (σ)(σ)∩C max 0

(σ)6=∅

follows. Moreover, at least one of these nodes in

C_extmin 0_Cmax 1

Λ (σ)(σ)∩C

max 0_(σ)

has to be weakly 0connected[σ] toΛc _in

i

C_extmin 0_Cmax 1

Λ (σ)(σ), ∂ ∗

Λh :=extC_extmin 0_Cmax 1

Λ (σ)(σ)∩Λ.

This is the case because a 0lasso[σ] exists and, therefore, we could follow a0path from ∂(Λc) to the maximal 0circuit[σ] and then through the maximal 0circuit[σ]

toCmin 0 extCmax 1

Λ (σ)

(σ); the first node x inCmin 0 extCmax 1

Λ (σ)

(σ)reached this way satisfies the required feature.

If a node of Cmin 0

extC_Λmax 1(σ)(σ) is∗weakly 0∗connected toΛ

c _{in the annulus}

i

C_extmin 0_Cmax 1

Λ (σ)(σ), ∂ ∗

Λh ,

then it also belongs to iCfill (σ), since it cannot be “circumvented“ by a1path of

i_Cfill_(σ)Def.

= maxiC

C_extmin 0_Cmax 1

Λ (σ)(σ)∪σ

−1_(1).

Consequently, the node x also belongs to the circuitiCfill(σ). Summing up, the node x belongs to

i_Cfill_(σ)∩Cmin 0 extCmax 1

Λ (σ)(σ)∩C

max 0_(σ),

which implies the last Property (5.25).

Once again, we need further (more involved) relations between a configuration

σ ∈A2 and the corresponding configurationm(σ). More precisely, we have to find a circuit i_Cempty_{(m(σ)) that changes m(σ)to σ if it is emptied. Our candidate is}

i_Cempty_{(m(σ)) := min}i_CCmax 1

Λ (m(σ))∪m(σ) −1

(0)∩extC_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ))

= miniCiC_Λmax 1(m(σ))∪m(σ)−1(0)∩extC_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ))

,

(5.27) where the identity follows fromi_Cmax 1

Λ (m(σ))≤CΛmax 1(m(σ))andiCΛmax 1(m(σ))⊂

Lemma 5.20 Let σ ∈ A2. Then the maximal induced 1circuit[m(σ)] equals the circuitiCfill_{(σ)and the minimal induced 0circuit[σ] in the exterior of the maximal}

1circuit[σ] equals i_Cempty_{(m(σ)), in short}

i_Cmax 1 Λ (m(σ)) = i_Cfill_{(σ) (5.28)} i_Cempty_{(m(σ)) =}i_Cmin 0 extCmax 1 Λ (σ)(σ). (5.29)

Proof: On the one hand, the inequality iC_Λmax 1(m(σ))≥i_Cfill_{(σ) follows from}

σ−1(1)∪i_Cfill_{(σ) =m(σ)}−1

(1),

see (5.4). On the other hand, we also know that

i_Cfill_(σ)Def._{= max}i_CCmin 0 extCmax 1 Λ (σ)(σ)∪σ −1 (1) (5.23) ≥i_Cmax 1 Λ (m(σ))

holds. Both inequalities together imply (5.28), which immediately leads to

C_extmin 0_Cmax 1

Λ (σ)(σ)∩

i_Cfill_(σ)⊂i_Cmax 1_(m(σ))⊂Cmax 1_{(m(σ)). (5.30)}

We further know that

C_extmin 0_Cmax 1

Λ (σ)(σ)≤ i

C_extmin 0_Cmax 1

Λ (σ)(σ)≤ i

Cfill(σ). (5.31) holds. Indeed, the first inequality is obvious and the second one follows from the definition of i_Cfill_{(σ) asmax}i_CCmin 0

extCmax 1 Λ (σ) (σ)∪σ−1(1). A further fact is i_Cfill_(σ)∩Cmin 0 extCmax 1 Λ (σ)(σ) (5.25) 6 = ∅. (5.32)

These two Relations (5.31) and (5.32) prove that all1∗circuits in inti_Cfill_{(σ) have}

to be contained in intCmin 0 extCmax 1

Λ (σ)

(σ), which gives us the following inclusion

C_extmin 0_Cmax 1

Λ (σ)(σ)⊂extC max 1∗ inti_Cfill_(σ)(σ)

(5.3)

= extC_intmax 1∗i_Cfill_(σ)(m(σ))

(5.28)

= extC_intmax 1∗i_Cmax 1

Λ (m(σ))(m(σ)) = extC max 1∗ intCmax 1

Λ (m(σ))(m(σ)).

Therefore,

C_extmin 0_Cmax 1

Λ (σ)(σ)∩(

i_Cfill_(σ))c_⊂m(σ)−1

(0)∩extC_intmax 1∗_Cmax 1

since the inclusion

C_extmin 0_Cmax 1

Λ (σ)(σ)∩

i_Cfill_(σ)c

⊂m(σ)−1(0)

follows from C_extmin 0_Cmax 1

Λ (σ)(σ)⊂σ

−1_{(0) and (5.3).}

Next, by definition of i_Cfill_{(σ) andm(σ), we know that} i_Cfill_(σ)⊂m(σ)−1_(1)∩

extCmax 1

Λ (σ), which immediately implies

C_Λmax 1(σ)≤C_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ)). (5.34)

So, on the one side,

i_Cempty_{(m(σ)) = min}i_CCmax 1

Λ (m(σ))∪m(σ)

−1_(0)∩

extC_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ))

(5.30),(5.33)

≤miniCC_extmin 0_Cmax 1

Λ (σ)(σ)∩

i_Cfill_(σ)∪Cmin 0 extCmax 1

Λ (σ)(σ)∩(

i_Cfill_(σ))c

=iC_extmin 0_Cmax 1 Λ (σ)(σ)

holds. On the other side, it is the case that

i_Cempty_(m(σ))(5.27)_{= min}i_Ci_Cmax 1

Λ (m(σ))∪m(σ)

−1_(0)∩

extC_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ))

(5.28)

= miniCiCfill(σ)∪m(σ)−1(0)∩extC_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ))

(5.31)

≥miniCC_extmin 0_Cmax 1

Λ (σ)(σ)∪m(σ) −1

(0)∩extC_intmax 1∗_Cmax 1

Λ (m(σ))(m(σ))

(5.34)

≥miniCC_extmin 0_Cmax 1

Λ (σ)(σ)∪m(σ) −1 (0)∩extC_Λmax 1(σ) m(σ)−1_(0)⊂σ−1₍₀₎ ≥miniC

C_extmin 0_Cmax 1

Λ (σ)(σ)∪σ −1

(0)∩extC_Λmax 1(σ)

=iC_extmin 0_Cmax 1

Λ (σ)(σ).

Taking both inequalities together yields (5.29). Now, we have gathered enough to finally analyse the image of m|A2 .

Lemma 5.21 Let σ∈A2. Then

m(σ)∈ {∃ 1∗lasso, C_Λmax 1 ←→1∗ C_intmax 1_Cmax 1

Λ }.

Proof: By (5.4), the circuit iCfill_{(σ) is a 1circuit[m(σ)], which, by (5.25), inter-}

sects the maximal 0circuit[σ]. The intersection is ∗weakly 1∗connected[σ] to Λc_.

Since also σ−1(1)⊂m(σ)−1(1), we can conclude that

So, it only remains to show that

m(A2)∈ {! C_Λmax 1←→1∗ C_intmax 1_Cmax 1

Λ }. (5.35)

SinceCmin 0 extCmax 1

Λ (σ)

(σ) is the minimal0circuit[σ] outside of the maximal1circuit[σ], every node of Cmin 0

extCmax 1 Λ (σ)

(σ) is ∗weakly 1∗connected[σ, m(σ)] to the maximal

1circuit[σ], i.e., for all x∈Cmin 0 extCmax 1

Λ (σ)

(σ)

∂∗x←→1∗ C_Λmax 1(σ) (5.36)

with respect to σ and, therefore, also with respect to m(σ) = ₁σ−1_(1)∪i_Cfill_(σ). A

further fact is

i

Cfill(σ)∩C_extmin 0_Cmax 1 Λ (σ)(σ)

(5.25)

6

= ∅. (5.37)

A consequence of (5.36) and (5.37) is

∂∗iCfill(σ)←→1∗ C_Λmax 1(σ) (5.38) with respect to σ and m(σ). Therefore, by (5.28), the statement (5.35) will be proved once we have shown that the right side of (5.38) equalsCmax 1

intC_Λmax 1(m(σ))(m(σ)).

Summing up, we know everything but the last identity in the following:

C_Λmax 1(m(σ))←→1∗ i_Cmax 1 Λ (m(σ)) (5.28) = ∂∗iCfill(σ) (5.38) 1∗ ←→ 1∗

←→C_Λmax 1(σ)=! C_intmax 1_Cmax 1

Λ (m(σ))(m(σ)),

where the first 1∗connection is obvious. The latter identity will be proven in the remainder.

"≤": By the definition of i_Cfill_{(σ), we know that C}max 1

Λ (σ) ⊂ intiCfill(σ) (5.28)

=

intiCmax 1

Λ (m(σ)).A consequence of this, together withm(σ) =1σ−1_(1)∪i_Cfill_(σ),

isCmax 1

Λ (σ)≤Cintmax 1Cmax 1 Λ (m(σ))

(m(σ)).

"≥": It is the case that

C_intmax 1_Cmax 1

Λ (m(σ))(m(σ)) =C max 1 inti_Cmax 1

Λ (m(σ))(m(σ)) (5.28)

=C_intmax 1i_Cfill_(σ)(m(σ))

(5.3)

=C_intmax 1i_Cfill_(σ)(σ)

which concludes the proof. Last, we have to show the invertibility of m|A2. Our candidate for the inverse

map is

m−1 :m(A2)→! A2; m(σ)7→1−₁m(σ)−1_(0)∪i_Cempty_(m(σ))

!

=σ ,

which is illustrated in Figure 5.2.





0

Figure 5.2: In this figure the white squares represent nodes with 0spin[σ, m(σ)]. The gray squares stand for nodes with0spin[σ]and1spin[m(σ)]. The black squares are nodes with 1spin[σ, m(σ)]. The circuit iCfill_{(σ) is indicated by a red curve.}

The circuit i_Cempty_{(m(σ))is indicated by a blue curve. The maximal1circuit[σ]is}

indicated by a green curve.

Lemma 5.22 Let σ ∈ A2. Then 1−₁m(σ)−1_(0)∪i_Cempty_(m(σ)) = σ , i.e., the map

mΛ|A2 is injective.

Proof: It is sufficient to show that

i_Cfill_(σ)∩σ−1

(0)=! iCempty(m(σ))∩m(σ)−1(1)

"⊂" This direction is a consequence of i_Cfill_(σ)(5.28)₌ i_Cmax 1 Λ (m(σ))⊂m(σ) −1 (1) and i_Cfill_(σ)∩σ−1 (0) (5.2)⊂ i_Cmin 0 extCmax 1 Λ (σ)(σ) (5.29) = iCempty(m(σ)).

"⊃" This implication follows from

i_Cempty_(m(σ))(5.29)₌ i_Cmin 0 extCmax 1 Λ (σ)(σ)⊂σ −1 (0) and i_Cempty_{(m(σ))∩m(σ)}−1 (1) (5.27)⊂ i_Cmax 1 Λ (m(σ)) (5.28) = iCfill(σ),

which concludes the proof.

5.2

The Connection to the Widom-Rowlinson Model

In document Carstens, Sebastian Maurice (2012): Percolation analysis of the two-dimensional Widom-Rowlinson lattice model. Dissertation, LMU München: Fakultät für Mathematik, Informatik und Statistik (Page 100-107)