cardysformula

8/2/2019 cardysformula

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Reading Group

Probability in Graphs

Dialid Santiago

February 9, 2012


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Notation

Let

be the triangular lattice with density p = 12

of open or black vertices.


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Notation

Let



and

be the hexagonal lattice.


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Proof of

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Lemma 2

Conclusion

Notation

Let



and


Let D = be an open simply connected domain in 2 and assume that D isa Jordan curve.


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Conclusion

Notation

Let



and



Let T be the unit equilateral triangle.


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Lemma 2

Conclusion

Notation

Let



and



Let T be the unit equilateral triangle.We are interested on the site percolation on

D.


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Conclusion

Notation

Let



and



Let T be the unit equilateral triangle.We are interested on the site percolation on

D.In particular ifd= 1

nand D = Twe have

n =1

n T.


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Summary

We are interested on the event:

Figure: The event {ac bx D}


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Summary



More precisely in

[ac bx D] as 0 (1)


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Summary



More precisely in

[ac bx D] as 0 (1)Cardys formula tells us the value of the limit.


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Lemma 1

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Proof of

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Lemma 2

Conclusion

Summary

As we saw last time there exists a conformal map from D to T the unit

equilateral triangle with vertices

A = 0, A = 1 A2 = ei3

Figure: The conformal map is a bijection from D to the interior ofT, and extends uniquelyto the boudaries.


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Lemma 2

Conclusion

Summary

As we saw last time there exists a conformal map from D to T the unit

equilateral triangle with vertices

A = 0, A = 1 A2 = ei3

Moreover such can be extended to the boundary D in such a way that itbecomes a homeomorphism.

Figure: The conformal map is a bijection from D to the interior ofT, and extends uniquelyto the boudaries.


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Cardys Formula

Cardys Formula

[ac bx D] |AX| 0. (2)


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Summary

Translating through

Figure: Applying the map we get the event {A1A2 AX T}


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Sumary


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Lemma 1

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Proof of

Lemma 1

Lemma 2

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Sumary


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Proof of

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Sumary

We defined the events

Enj (z) with j = 1, ,

2.


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Sumary



2.

and the functions

Hnj (z) = (E

nj (z)) j = 1, ,

2.

S


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Sumary



2.

and the functions

Hnj (z) = (E

nj (z)) j = 1, ,

2.

It was proved that these functions

are uniformly Holder on V( n).

S


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Sumary



2.

and the functions

Hnj (z) = (E

nj (z)) j = 1, ,

2.

It was proved that these functions

are uniformly Holder on V( n).Figure: An illustration of the event E

n

2 .

S


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Lemma 1

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Lemma 1

Lemma 2

Conclusion

Sumary

The proof was completed under the assumption of the analyticity of the

functions

G1 = H1 + H + H2

G2 = H1 + H + 2H2

A i lt


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Proof of

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A previous result

Lemma 1

We have that

[En1(z1)\En1(z)] = [En(z2)\En(z)]= [En2 (z3)\En2 (z)]

Exploration Process


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Proof of

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Exploration Process

Definition

Exploration Process


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Exploration Process

Definition

Suppose that all vertices just outside the arc A1A2 of n are black.

Exploration Process


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Exploration Process

Definition


all vertices just outside the arc AA2 are white.

Exploration Process


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Exploration Process

Definition



The exploration path is defined to be the unique path n on the edges of thedual graph, beginning immediately above A2 and descending to A1A2 such

that.

Exploration Process


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Exploration Process

Definition



The exploration path is defined to be the unique path n on the edges of thedual graph, beginning immediately above A2 and descending to A1A2 such

that.

As we traverse n from top to bottom the vertex immediately to our left(respectively, right), looking along the path from A2 , is white (respectively,

black).

Exploration Process


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Exploration Process

Exploration Process


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Exploration Process

Proof of Lemma 1


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Proof of

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Conclusion

Event (1)-(3)

Notice that the event En1(z1) \En1(z) occurs iff there exist disjoint paths l1, l2 and l3of n such that

Proof of Lemma 1


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Proof of

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Event (1)-(3)

Notice that the event En1(z1) \En1(z) occurs iff there exist disjoint paths l1, l2 and l3of

n such that1 l1 is white and joins s1 to AA2

Proof of Lemma 1


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Event (1)-(3)



2 l2 is black and joins s2 to A1A2

Proof of Lemma 1


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Event (1)-(3)



2 l2 is black and joins s2 to A1A2

3 l3 is black and joins s3 to A1A

Proof of Lemma 1


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Proof of Lemma 1


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Proof of

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Proof of Lemma 1


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Remarks

Proof of Lemma 1


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Remarks

Now just notice that on this event the exploration path n passes through zand arrives at z along the edge < z3,z >

Proof of Lemma 1


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Lemma 1

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Proof of

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Remarks

Now just notice that on this event the exploration path n passes through zand arrives at z along the edge < z3,z >

Furthermore, up to the time at which it hits z it lies in the region of n between

l1 and l2.

Proof of Lemma 1


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Figure: The exploration path n passes through z and arrives at z along the edge < z3,z > .

Proof of Lemma 1


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Under last considerations

Proof of Lemma 1


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Conclusion


The states of vertices of n lying below l1 l2 are independent Bernoulli variables.

Proof of Lemma 1


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Lemma 1

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Proof of

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Lemma 2

Conclusion


The states of vertices of n lying below l1 l2 are independent Bernoulli variables.Thus the conditional probability if a black path l3 satisfying the condition (3) is thesame as that of a white path.

Proof of Lemma 1


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Conclusion


The states of vertices of n lying below l1 l2 are independent Bernoulli variables.Thus the conditional probability if a black path l3 satisfying the condition (3) is thesame as that of a white path.We make the measure preserving change, and then we interchange the colourswhite and black to conclude that the event

En1(z1)\En1(z)

Proof of Lemma 1


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En1(z1)\En1(z)

has the same probability as the event that there exists disjoint paths l1, l2 and l3of

n such that

Proof of Lemma 1


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En1(z1)\En1(z)


n such that

1 l1 is black (before white) and joins s1 to AA2

Proof of Lemma 1


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Lemma 2

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Thus the conditional probability if a black path l3 satisfying the condition (3) is thesame as that of a white path.We make the measure preserving change, and then we interchange the colourswhite and black to conclude that the event

En1(z1)\En1(z)


n such that

1 l1 is black (before white) and joins s1 to AA22 l2 is white (before black) and joins s2 to A1A2

Proof of Lemma 1


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En1(z1)\En1(z)


n such that

1 l1 is black (before white) and joins s1 to AA22 l2 is white (before black) and joins s2 to A1A23 l3 is black and joins s3 to A1A

Proof of Lemma 1


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En1(z1)\En1(z)


n such that


That is precisely the eventE

n(z2)\En(z),

Proof of Lemma 1


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Lemma 1

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Proof of

Lemma 1

Lemma 2

Conclusion


The states of vertices of

n lying below l1 l2 are independent Bernoulli variables.Thus the conditional probability if a black path l3 satisfying the condition (3) is thesame as that of a white path.We make the measure preserving change, and then we interchange the colourswhite and black to conclude that the event

En1(z1)\En1(z)


n such that


That is precisely the eventE

n(z2)\En(z),

and the lemma is proved.

Moreras Theorem


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Lemma 2

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Morera

Iff : R is continuous on the open region R, and

fdz = 0, (3)

for all closed curves in R, then f is analytic.

Lemma 2


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Lemma 2

Let be an equilateral triangle contained in the interior ofTwith sides parallel tothose ofT. Then

Hn1 (z)dz =

Hn(z)

dz + O(n)

=

Hn2

(z)

2dz + O(n),

where is given in Lemma 5.48.

Proof of Lemma 2


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Conclusion

Let be an equilateral triangle contained in the interior ofTwith sides parallel

to those ofT.

Proof of Lemma 2


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Lemma 1

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Lemma 2

Conclusion

Let be an equilateral triangle contained in the interior ofTwith sides parallel

to those ofT.Let n be the subgraph of

n lying with .

Proof of Lemma 2


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Figure: An illustration.

Proof of Lemma 2


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Define the set Dn asDn = { downward pointing faces of n},

Proof of Lemma 2


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and let be a vector such that:

Proof of Lemma 2


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Ifz is the centre of a facet ofDn, then z + is the centre of a neighboring face.

Proof of Lemma 2


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Ifz is the centre of a facet ofDn, then z + is the centre of a neighboring face.That is, {i, i, i

2

}/n3

Proof of Lemma 2


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2

}/n3Consider

hnj (z, ) = (E

n2 (z + )/E

n2 (z))

R di

Proof of Lemma 2


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2

}/n3Consider

hnj (z, ) = (E

n2 (z + )/E

n2 (z))

At this point we can invoke Lemma 1.

Hn1 (z + )

H

n1 (z) = h

n1(z, )

hn1(z + ,

)

= hn(z, ) hn(z + ,).

R di

Proof of Lemma 2


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Conclusion

NowH

n(z + ) Hn(z) = hn(z, ) hn(z + ,),

and so there is a cancellation in

In =

zDn

[Hn1 (z + ) Hn1 (z)] zDn

[Hn(z + ) Hn(z)]

of all terms except those of the form

hn(z

,)for certain z lying in faces of n that abut

n. But there are just O(n) such z andtherefore the Holder condition ofHnj (z) implies that

|In| O(n1) (4)

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Proof of Lemma 2


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Conclusion

Consider the sum

Jn =

1

n(Ini + I

ni +

2In

i2 ),

where Inj is an abbreviation for the I jn

3

.

Reading

Proof of Lemma 2


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Proof of

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Lemma 2

Conclusion

Consider the sum

Jn =

1

n(Ini + I

ni +

2In

i2 ),


3

.

The terms of the form Hnj (z) do not contribute to Jn, since each is multiplied by

(1 + + 2)

n= 0.

Reading

Proof of Lemma 2


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Conclusion

Consider the sum

Jn =

1

n(Ini + I

ni +

2In

i2 ),


3

.


(1 + + 2)

n= 0.

The remaining terms of the form Hnj (z + ),Hnj (z + ) mostly disappear also,

as we said before.

Reading

Proof of Lemma 2


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Lemma 2

Conclusion

Consider the sum

Jn =

1

n(Ini + I

ni +

2In

i2 ),


3

.


(1 + + 2)

n= 0.

The remaining terms of the form Hnj (z + ),Hnj (z + ) mostly disappear also,

as we said before.

At the end we are left only with terms Hnj (z

) for certain z

at the centre ofupwards-pointing faces of n abutting

n.

Reading

Proof of Lemma 2


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Conclusion

Ifz is at the bottom, but not the corner, of n, then its contribution will be

1n

[(+ 2)Hn1 (z) (1 + )Hn(z)] = 1n

[Hn1 (z) Hn(z)/].

Reading

Proof of Lemma 2


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Conclusion


1n

[(+ 2)Hn1 (z) (1 + )Hn(z)] = 1n

[Hn1 (z) Hn(z)/].

When z is at the right edge ofn, we obtain

1

n[(+ 2)Hn1 (z

) (1 + )Hn(z)] = n

[Hn1 (z) Hn(z)/].

Reading

Proof of Lemma 2


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Proof of

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Lemma 2

Conclusion


1n

[(+ 2)Hn1 (z) (1 + )Hn(z)] = 1n

[Hn1 (z) Hn(z)/].


1

n[(+ 2)Hn1 (z

) (1 + )Hn(z)] = n

[Hn1 (z) Hn(z)/].

Finally when z is at the left edge of n, we obtain

1

n[(+ 2)Hn1 (z

) (1 + )Hn(z)] = 2

n[Hn1 (z

) Hn(z)/].

Reading

G

Proof of Lemma 2


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Lemma 2

Conclusion


1n

[(+ 2)Hn1 (z) (1 + )Hn(z)] = 1n

[Hn1 (z) Hn(z)/].


1

n[(+ 2)Hn1 (z

) (1 + )Hn(z)] = n

[Hn1 (z) Hn(z)/].

Finally when z is at the left edge of n, we obtain

1

n[(+ 2)Hn1 (z

) (1 + )Hn(z)] = 2

n[Hn1 (z

) Hn(z)/].

Therefore

n

[Hn1 (z) Hn(z)/]dz = Jn + O(n) = O(n), (5)

by (4), where the first O(n) term covers the fact that the z in this equations isa continuous rather than a discrete variable.

Reading

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Proof of Lemma 2


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Lemma 1

Lemma 2

Conclusion

Hence

[Hn1 (z) Hn(z)/]dz = O(n),

as we wanted.

Reading

Group

Proof of Lemma 2


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Lemma 2

Conclusion

Hence

[Hn1 (z) Hn(z)/]dz = O(n),

as we wanted.

By a similar argument we can prove

[Hn1 (z) Hn2 (z)/2]dz = O(n).

Reading

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Some Final Remarks


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Lemma 1

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Proof of

Lemma 1

Lemma 2

Conclusion

This concludes the proof of Cardys formula when the domain D is an

equilateral triangle.


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Lemma 1

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Proof of

Lemma 1

Lemma 2

Conclusion

Thank you!

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