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Tilings of irregular hexagons
Large Conformal maps from probability to physis
May 23-28, 2010
Monte Verita, Asona, Tiino GGI
Bernard Nienhuis
University of Amsterdam
Outline
The model
Why do physiists look at this model
Why do mathematiian look at this model
Bethe Ansatz
Yield transition
Conformal behavior
1
The model: hexagonal net
2
The hexagons an grow at the expense of their neighbors,
with the onstraint that all sides are non-negative.
3
The model
Why do physiists look at this model
Why do mathematiian look at this model
Bethe Ansatz
Yield transition
Conformal behavior
4
An monolayer of atoms adsorbed on a rystal surfae,
suh that the latties are inommensurable.
5
An monolayer of atoms adsorbed on a rystal surfae,
suh that the latties are inommensurable.
6
An monolayer of atoms adsorbed on a rystal surfae,
suh that the latties are inommensurable.
7
An monolayer of atoms adsorbed on a rystal surfae,
suh that the latties are inommensurable.
8
An monolayer of atoms adsorbed on a rystal surfae,
suh that the latties are inommensurable.
Hexagonal network of domain walls (Villain 1980)
9
This random paking
of disks is equivalent
to a tiling of squares
and triangles.
(given the appropriate
size ratio, and maximal paking density)
10
When there are far more triangles than squares, the tiling behaves
like a hexagonal net. (Oxborrow, Henley 1993)
11
In the trimer model, analogous to the dimer model, every site of
the triangular lattie part of one triangular trimer. (Verberkmoes,
BN 1999)
Equivalently the hexagonal lattie is tiled with 'triominos'.
12
When the trimers
predominantly point
the same way, the
trimers with minority
orientation are sorted
out along a hexagonal
net.
The dierent olors
represent dierent
sublatties.
(The Y vertex omes
in two hiralities)
13
Fores in disk pakings
Imagine the disks to
be perfetly hard.
Eah disk is held in
plae by the ontat
fores from its six
neighbors.
For eah disk two
balane equations,
and three fores to
solve them.
(Ostoji,
BN
2005)[2m℄
14
F1
F1
F2
F3
F6
F5
F4
F2
F6
F3
F5
F4
F2
F1
F6
F3
F5
F4
Represent the fores by linesegments perpendiular to the diretion
of the fore with length proportional to the strength.
Then these linesegments form a losed hexagon.(Maxwell)
Plaing these in
one gure gives a
hexagonal net
15
The model
Why do physiists look at this model
Why do mathematiian look at this model
Bethe Ansatz
Yield transition
Conformal behavior
16
Sums of hermitian matries
Let be a weakly dereasing sequene of real numbers, and H
be set of hermitian matries with eigenvalues .
For what , and does
M +M +M =0
(1)
have solutions in H , H and H , respetively?
In other words: if we know (only) the spetrum of two hermitian
matries, what does that tell us of the spetrum of their sum?
Horn (1962) onjetured that a set of inequalities (together with
the obvious ondition on the trae) are suÆient onditions for
(1) to have a solution.
In 2000 Knutson and Tao proved this to be orret, and formulated
it in terms of the hexagonal net:
17
Knutson & Tao: The
matries with spetra , and an add to zero if the
spetra an be onneted
with a hexagonal net.
Even stronger:
with uniform measure
on the matries and on
the hexagonal net, the
partition funtion of the
hexagonal net measures
the relative probability
that is the spetrum
of M + M .
18
For integer sequenes (or equivalent Young tableaux):
Let V be a representation
of U(n). Then the number
of times that V ours in
V V is given by the number
of (integer) hexagonal nets
that onnet , and .
These Littlewood-Rihardson
oeÆients are also naturally
represented as tilings with
squares and triangles.
(Zinn-Justin, 2009)
19
The model
Why do physiists look at this model
Why do mathematiian look at this model
Bethe Ansatz
Yield transition
Conformal behavior
20
Some alulational tools
Transfer matrix: add one row of disks
Use the Bethe Ansatz to diagonalize the transfer matrix:
The positions of the diagonal lines proeed with plane waves as
weight funtion, between ollisions (i.e. events that two lines oinide). On ollision the wave numbers may be interhanged.
This suggests that the eigenstates of the transfer matrix an be
written as a linear ombination of produts
of
plane
waves:
0
1
X
X
(x1; x2; : : : ; y1; y2; : : :) = A exp ix p ( ) + iy q ( )A
N
;
;
j =1
j j
j j
21
This turns out to \work" (i.e. solve the eigenvalue equation).
This implies that multiple ollisions an be written as a sequene
of binary ollisions.
The eigenvalue equation T = results in the following
onsisteny onditions (B.A.eqns) on the wave numbers:
Y
Y
u
(u v e ) = u
(u v e ) = v v
=1
=1
where u = exp(ip ) and v = exp( iq ), and e is the weight of
one unit vertial line.
In these equations we want to take two (independent) limits:
(i) The disrete hexagonal to the ontinuum, and
(ii) the number N of hexagons in a row to 1.
The seond limit turns the (logarithm of the) produts into integrals.
N
j
k
k
j
k
F
k
j
N
k
k
j
j
j
F
22
In the limit we get integral equations:
Z dy
g (y )
f (z ) = Fu +
2 i (z + y )
Z dz f (z)
g (y ) = Fv +
2i (y z)
with the restrition that
Z g(y) dy Z f (z) dz
=
=
1
2 i
2 i
The entropy per added hexagon is then given by
Z dy Z dz
log(
y z )
S=
2 i
2 i
The funtion f is analyti exept in a ut ! ,
and g is analyti exept in a ut + ! + .
From the integral equations it follows that for on the ut in f :
f ( + ") f ( ") = g ( + )
23
In two limits we an make progress with these equations:
(i) In the limit that the uts in f and g are very far apart.
(ii) In the speial ase that the end points of the uts in f (z) and
in g(z + ) oinide, the monodromy struture simplies.
1.5
1.5
+ =2
=2
+ =2 = =2
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
+ =2
-1
-0.5
0
0.5
=
2
1
1.5
-1.5
-1.5
+ =2 = -1
-0.5
0
=
0.5
2
1
24
1.5
The oinidene of the endpoints of the uts (one omplex equation) requires the symmetry ondition Fu = Fv: the asending lines
and desending lines are equal on average, and a speial tuning of
the parameter , whih ontrols Fw the length of the vertial lines.
This ase was solved as follows (Kalugin (1994)):
Passing through both uts 6 times leaves both f and g invariant
This immediately suggests an unfoliation of the Riemann sheets,
to nd new variables in whih the solution is analyti.
On an assume the o-inidene to happen, leaving the speial
value of where this happens, and the position of the endpoint
as unknowns. The known properties of the funtions leaves
enough equations to solve for these unknowns.
The result for the partition sum per hexagon is
e = 6F exp( 3=2)
with F = Fu = Fv = Fw the mean side length in all three diretions.
S
25
The model
Why do physiists look at this model
Why do mathematiian look at this model
Bethe Ansatz
Yield transition
Conformal behavior
26
Questions:
Is the distribution sale invariant
... onformally invariant
partition funtion
orrelation funtions
nature of the yield transition:
27
In the ontinuous ase only the ratios of Fu, Fv and Fw matter.
There are two free parameters.
In the enter we have an exat
expression for the partition funtion.
We know that it vanishes at the
boundary of the triangle, (where one
of the F 's vanishes).
+ 2dyi (z +g(y) y)
Z dz f (z)
g (y ) = Fv +
2i (y z)
( )=
f z
Fu
Z
When 1 the denominator of the integrand is always big, and
an be expanded in powers of 1=. In this limit Fw is very small.
28
This is a tedious proess, as the funti ons f and g as well as
the integration boundaries and have to be alulated order by
order. But it leads to a onsistent series
For ompat notation I introdue: a = Fu 1 1 and b = Fv 1 1.
In that notation:
1(a2 + b2) 1 (a3 + b3) + 1 (a4 + b4) + 1a2b2+
S = log( )
6
2
20
6
5 (a5+b5)+ 2 a2b2(a+b)+ 9 a3b3 1 (a6+b6) 1 a2b2(a2+b2)
12
3
4 2
42
6
7 (a7 + b7) 17 a2b2(a3 + b3) 23 a3b3(a + b) + 20
12
30
so far up to 10 orders. Conversion from S(Fu; Fv; ) to S(Fu; Fv; Fw)
is still needed, but does not seem to improve the esthetis.
What is surprising is that the partition sum an be expanded like
this near an instability.
29
The model
Why do physiists look at this model
Why do mathematiian look at this model
Bethe Ansatz
Yield transition
Conformal behavior
30
Sale- and onformal invariane
In a onformally invariant system one nds
2
= exp(SN + vF 6N vF N + o(1=N ))
Here S is the bulk free energy and gives the dominant size
dependene, and vF (Fermi veloity) is a geometri fator relating
the sales in the 'time-like' and 'spae-like' diretion.
The fat that the gaps are inversely proportional to the (linear)
system size learly indiates sale invariane: these orretions
play the role of inverse orrelation lengths. If a orrelation length
is proportional to the linear system size there is apparently no
intrinsi sale.
A typial harater of onformally invariant systems is that there
are many integer valued or , the signature of the
Virasoro algebra.
What do we nd in the hexagonal net?
j
j
j
j
j
k
31
In the isotropi point the spetrum has all these harateristis,
and a entral harge = 2.
In an anisotropi hexagonal net, one
may need to nd v empirially, (rather
than by a priori geometri arguments).
To avoid this ompliation we look at
the square-triangle tiling in the square
phase.
This is equivalent to a phase with staks of elongated hexagons,
but in the square triangle tiling the symmetry is more obvious.
F
32
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
The eetive entral harge
as funtion of the triangle
density. In the right half the
triangles dominate and form
an isotropi hexagonal net.
In the left half the squares
dominate and form a utuating square net of domain
walls.
In the spetrum we nd many instanes with j j = 1, but
with omplex values of .
These ndings an be explained as follows: The diagonally running
domain walls form two families of fermions. The B.A. equations
show that they interat only with the members of the other family.
The pitures suggest that the spae-time of eah speies of fermions
is elongated in the diretion of the orresponding domain wall.
j
k
33
The harateristi spae-like sale is the inverse density.
The harateristi time-like sale is the time between ollisions.
34
Beause the spae-time strething is dierent for the two families
of fermions, it an not be orreted with a vF or a tilt in the
transfer matrix.
But the theory gives explains preisely the relation between the
value of the eetive entral harge, and the imaginary part of
.
Also it predits that the eetive entral harge is larger than 2,
when the transfer matrix is oriented axially w.r.t. the domain walls.
j
35
many hexagonal nets are seen in physis
they are a useful tool in mathematis
Satisfy Bethe Ansatz, but not always 'solvable'
Yield transition aessible analytially
Strange onformal behavior
36