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Domino tilingsMathematical tools

A few ideas of the proof

The arctic circle

Vincent Beffara

ISSMYS August 28, 2012

Vincent Beffara The arctic circle


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DefinitionsRandom tilingsOther kinds of tilings

What is adomino?



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What is adomino?

A domino is a rectangle of dimensions 2 1, oriented eithervertically or horizontally.



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What is a dominotiling?


D i ili D fi i i


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Aplanar domainis a bounded region in the plane formed of

adjacent 1 1 squares. Usually it will have no hole.


D i tili D fi iti


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Aplanar domainis a bounded region in the plane formed of

adjacent 1 1 squares. Usually it will have no hole.

Adomino tilingof a domain is a way to cover it with dominoswithout overlaps.


Domino tilings Definitions


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What is arandomdomino tiling?

For a given domain, there are only finitely many ways (possibly

none at all) to cover is with dominos. Assuming that there arecoverings, we can pick one at random, giving the same probabilityto all of them. This is what we will always mean by a randomdomino tiling.




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How does a random domino tiling look like?




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How does a random domino tiling look like?




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How to generate a random domino tiling?

There are efficient ways, but they are tricky to describe. Here is asimple one:




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gMathematical tools

A few ideas of the proofRandom tilingsOther kinds of tilings



1 Start from your favorite tiling;




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gMathematical tools





2 Look for a pair of adjacent, parallel dominos forming a 2 2square; pick such a pair at random;




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gMathematical tools






3 Flipthe pair to the other possible configuration within the2 2 square;




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Mathematical toolsA few ideas of the proof

Random tilingsOther kinds of tilings






4 Go back to step2.


Domino tilingsM h i l l

DefinitionsR d ili


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4 Go back to step2.

5 Except, stop after a long enough time.


Domino tilingsM th ti l t ls

DefinitionsR d tili s


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Tilings of the Aztec diamond

TheAztec diamondof size n is this domain:



DefinitionsRandom tilings


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Tilings of the Aztec diamond

TheAztec diamondof size n is this domain:





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video



http://circle.mp4/http://circle.mp4/


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Thats a bit strange. What is happening?





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A few ideas of the proofg

Other kinds of tilings


We are selecting a tiling at random, and we see a deterministicshapeappearing.





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A few ideas of the proof Other kinds of tilings


We are selecting a tiling at random, and we see a deterministicshapeappearing. What that means is, most domino tilings of theAztec diamond are frozen outside the disk.

We need to understand a few things now:





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We need to understand a few things now:Why do the dominos align in the corners?



A f id f h f

DefinitionsRandom tilingsO h ki d f ili


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Why do the dominos align in the corners?

What is the shape, is it really a circle?



A f id f th f

DefinitionsRandom tilingsOth ki d f tili


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How come in the first square there is no frozen region?






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How come in the fake Aztec diamond everything is frozen?






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How come in the fake Aztec diamond everything is frozen?

The right answer to the last question holds the key to all theothers.






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One can tile a hexagon with lozenges






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One can tile a hexagon with lozenges






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p g

One can even tile a more complicated shape






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p g

One can even tile a more complicated shape






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To sum up:

In many cases we see a surprising behavior: a deterministic shapeoccurs inside a random object. We would like to understand it:


Domino tilingsMathematical toolsA few ideas of the proof



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To sum up:


Why is there a deterministic shape appearing?





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To sum up:


Why is there a deterministic shape appearing?What is the shape? How to determine it?





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To sum up:



What kind of mathematical tools can we use?





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To sum up:



What kind of mathematical tools can we use?

The main idea is that, since we select one out of many tilings, themain step will involvecounting tilingshaving a certain behavior.



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Counting possible tilingsThe height functionStatement of the arctic circle theorem


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Easy test: Parity of the number of squares

We will need to count all possible tilings of a given shape if wewant to understand why a circle appears. The first step is to

determine whether a domain has a tiling at all ...

Very easy remark: a domino hastwosquares, so a domain with anoddnumber of squares cannot be covered by dominos.





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Easy test: Parity of the number of squares (too easy)

We will need to count all possible tilings of a given shape if wewant to understand why a circle appears. The first step is to

determine whether a domain has a tiling at all ...

Very easy remark: a domino hastwosquares, so a domain with anoddnumber of squares cannot be covered by dominos.





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Easy test 2: Bipartition

Here is a nice trick: draw the domain on a (infinite) chess board.It has white and black squares, and every domino has one of each.This gives us a better criterion:





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If a domain has a different number of white and black squares, itcannot be covered by dominos.




E B


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If a domain has a different number of white and black squares, itcannot be covered by dominos.

We will keep this trick from now on: the domain is assumed to be

drawn on a chess board. This explains the difference in colorsbetween the dominos in the previous picture:




Th ll 4 f d i


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There are really 4 types of dominos




R hl h ili h ?


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Roughly, how many tilings are there anyway?

Take the easy case of a 2n 2n square S. We can give two bounds

on the number of tilings N(S) it admits:




R hl h tili th ?


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Each little square is in a domino, and there are only 4 kinds ofdominos; giving the kind of all these dominos specifies thetiling.




R hl h tili th ?


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Each little square is in a domino, and there are only 4 kinds ofdominos; giving the kind of all these dominos specifies thetiling. (It says nothing about overlaps, so we get a number

which is much bigger than the real one!)




Roughly how many tilings are there anyway?


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which is much bigger than the real one!) Hence:

N(S) 44n2.






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N(S) 44n2.

One can look for tilings obtained by gluing 2 2 squares

together, n2

of them. Each small square can be covered intwo ways.







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N(S) 44n2.


together, n2

of them. Each small square can be covered intwo ways. (Of course not every tiling is of that kind, so thenumber we get is much too small!)







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N(S) 44n2.


together, n2

of them. Each small square can be covered intwo ways. (Of course not every tiling is of that kind, so thenumber we get is much too small!) Hence:

N(S) 2n2.







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2n2 N(S) 44n2 .







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2n2

N(S) 44n2

.

Hence:

N(S) cn2.

Thats only partly a joke, it isthe right answer.







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2n2

N(S) 44n2

.

Hence:

N(S) cn2.

Thats only partly a joke, it isthe right answer. In fact, c is knownexplicitly:

c= exp

4

k=0

(1)k

(2k+ 1)2

5.354 . . .





The height function


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The height function

A domino tiling is a complicated mathematical object. We wouldlike to encode it into a simpler one...

The height function is defined by a simple local rule: around ablack square, turning counterclockwise, it increases by 1 alongevery edge,except when crossing a domino(in which case itdecreases by 3).

Of course one needs to check that this is well-defined (that there issuch a function).





The height function along the boundary


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One important remark: the height function on the boundary of agiven domaindoes not depend on the tiling. On our 3 favoritedomains, different things occur:







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On the square, the boundary value is almost flat;







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g g y



On the Aztec diamond, it looks like a jigsaw;







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g g y



On the Aztec diamond, it looks like a jigsaw;

On the fake Aztec diamond, it looks like a wedge.

Because a height function cannot oscillate much, if the boundarycondition is very wild, then it has little choice inside the domain.


Domino tilings


Counting possible tilings

The height functionStatement of the arctic circle theorem

Statement of the main theorem


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Take a sequence of domains of the same shape, but larger andlarger;


Domino tilings






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Take a sequence of domains of the same shape, but larger andlarger; or equivalently, draw a finer and finer grid on a fixeddomain.


Domino tilings






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Take a sequence of domains of the same shape, but larger andlarger; or equivalently, draw a finer and finer grid on a fixeddomain. Then look at the height function generated by a randomtiling of each one.


Domino tilings






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Take a sequence of domains of the same shape, but larger andlarger; or equivalently, draw a finer and finer grid on a fixeddomain. Then look at the height function generated by a randomtiling of each one. (And normalize it correctly.)


Domino tilings






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Theorem

The height function approaches a deterministic, well characterizedshape


Domino tilings






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Theorem

The height function approaches a deterministic, well characterizedshape (which depends on the domain shape)


Domino tilings






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Theorem

The height function approaches a deterministic, well characterizedshape (which depends on the domain shape) (and on the details of the boundary).


Domino tilings






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Theorem

The height function approaches a deterministic, well characterizedshape (which depends on the domain shape) (and on the details of the boundary).

Where is the arctic circle?


Domino tilings




Localizing the arctic circle


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Here is what the theorem says in our favorite domains:


Domino tilings






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On the square, the height function is asymptotically constant


Domino tilings






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On the square, the height function is asymptotically constantso the random tiling is uniform;


Domino tilings






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On the fake arctic circle, the height function is asymptoticallyaffine


Domino tilings






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On the fake arctic circle, the height function is asymptoticallyaffine (we know that the tiling is completely aligned already);


Domino tilings






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On the arctic circle, the shape of the height function becomesinteresting:


Domino tilings






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it is slanted in the corners;


Domino tilings






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and it is flat in the middle


Domino tilings






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and it is flat in the middleso the random tiling is interesting as well.


Domino tilings





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Domino tilings


Counting tilings of a given slope


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We saw before the approximate number of tilings in a box of size n

(with a straight boundary, which makes the height functionroughly constant along the boundary).


Domino tilings




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Now, tune the boundary to make the height function have a slope:along the boundary,

h(x, y) ax+by.


Domino tilings



W b f h i b f ili i b f i


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h(x, y) ax+by.

It is still the case that the number of tilings inside is exponential inthe surface of the square, like before.


Domino tilings



W b f h i b f ili i b f i


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h(x, y) ax+by.

It is still the case that the number of tilings inside is exponential inthe surface of the square, like before. This can be shown using

sub-additivity.

Vincent Beffara The arctic circle Domino tilings



W b f th i t b f tili i b f i


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h(x, y) ax+by.

It is still the case that the number of tilings inside is exponential inthe surface of the square, like before. This can be shown using

sub-additivity. But the constant is different!

N c(a, b)(2n)2.



The value of c(a, b)


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Thats where things get complicated...





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Thats where things get complicated... but there is an exact

formula for it. Actually this is a very interesting story starting here.





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Thats where things get complicated... but there is an exact

formula for it. Actually this is a very interesting story starting here.

For completeness:

c(a, b) = expL(p1) +L(p2) +L(p3) +L(p4)

where

L(x) = x

0 log2sin tdt is the Lobachevsky function;

2(p1 p2) =a;

2(p4 p3) =b;p1+p2+p3+p4 = 1;

sin(p1) sin(p2) = sin(p3)sin(p4).



Counting tilings given an approximate height function

( )


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Consider a (smooth) shape h for the height function to follow.




C id ( h) h h f h h i h f i f ll


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Locally the shape looks flat, so we know how many possibilitiesthere are for the local tiling: something like

c(a, b)(n)2

where a and bdepend on the local slope.




C id ( h) h h f h h i h f i f ll


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c(a, b)(n)2

where a and bdepend on the local slope. Namely, a=xh,b=yh.




C id ( th) h h f th h i ht f ti t f ll


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c(a, b)(n)2

where a and bdepend on the local slope. Namely, a=xh,b=yh.

To put things in a formula: The number of tilings giving rise to aheight function close to h is approximately

N(h) exp

log(c(xh, yh)) dxdy

C(h)n

2.



The most likely height function


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N(h) C(h)n2

.





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N(h) C(h)n2

.

So, the larger N(h) is, the more likely it is to observe a heightfunction that is close to h.





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N(h) C(h)n2

.

So, the larger N(h) is, the more likely it is to observe a heightfunction that is close to h. But remember, the boundary value ofhis fixed, and h cannot be too wild...





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N(h) C(h)n2

.


There is a unique h0 maximizing N(h) under the constraints that hhas to satisfy. h0 is themost likely (asymptotic) height function.





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N(h) C(h)n2

.


There is a unique h0 maximizing N(h) under the constraints that hhas to satisfy. h0 is themost likely (asymptotic) height function.

TheoremThe height function of a uniform random domino tiling, in anydomain, is very likely to be close toh0.



Large deviations

What is the probability that the height function H of a random


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at s t e p obab ty t at t e e g t u ct o H o a a do

tiling is close to some fixed function h?



Large deviations

What is the probability that the height function Hof a random


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p y g


P[H h] {tilings s.t. H h}

{tilings}

C(h)n2

C(h)n2

Vi t B ff Th ti i l


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Large deviations

What is the probability that the height function Hof a random


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p y g


P[H h] {tilings s.t. H h}

{tilings}

C(h)n2

C(h)n2

But: ifc1


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P[H h] C(h)C(h0)

n

2

.

Vi t B ff Th ti i l Domino tilingsMathematical tools


Large deviations


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P[H h] C(h)C(h0)

n

2

.

There are two cases now, giving the proof of the theorem:




Large deviations


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P[H h] C(h)C(h0)

n

2

.


Ifh=h0, then P[H h] 1;




Large deviations


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P[H h] C(h)C(h0)

n

2

.


Ifh=h0, then P[H h] 1;Ifh =h0, then it isexponentially small, like

P[H h] ec(h)n2

where c= log(C(h)/C(h0))> 0.




Large deviations


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P[H h] C(h)C(h0)

n

2

.


Ifh=h0, then P[H h] 1;Ifh =h0, then it isexponentially small, like

P[H h] ec(h)n2

where c= log(C(h)/C(h0))> 0.

This is known as alarge deviation theorem.


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