Tilings and Plane partitions
Zo¨e Schroot
July 15, 2015
Bachelor Project
Supervisors: prof. dr. Bernard Nienhuis and prof. dr. Eric Opdam
Korteweg-De Vries Institute for Mathematics Faculty of Science
Abstract
We start with looking at aperiodic square-triangle tilings with one tile added, as a model for quasicrystals, and investigate the integrability of the tiling models. Thereafter we look at a tiling of rhombi inside a hexagon with boundary conditions corresponding to young diagrams, and introduce a way of counting the total number of different tilings satisfying the boundary conditions by means of non-intersecting lattice paths. Since these tiling of rhombi coincide with plane partitions we continue with looking at partition functions for plane partitions with fixed limit shapes. To obtain these partition functions we use representation theory of extensions of gl∞onV
∞
2 and the vertex operators, which are closely related to partitions. The partition function is expressed in Schur functions because of the connection between the vertex operators and Schur functions.
Title: Tilings and Plane partitions
Author: Zo¨e Schroot, zoe.schroot@student.uva.nl, 10399437 Supervisors: prof. dr. Bernard Nienhuis and prof. dr. Eric Opdam Date: July 15, 2015
Korteweg-De Vries Institute for Mathematics University of Amsterdam
Science Park 904, 1098 XH Amsterdam http://www.science.uva.nl/math
Contents
1. Introduction 4
2. Quasicrystals 5
2.1. Two shields . . . 6
2.2. One shield . . . 9
3. Plane partitions with boundary 13 3.1. Puzzles and mosaics . . . 13
3.2. Rhombi . . . 14
3.3. Lindstr¨om-Gessel-Viennot . . . 16
3.4. Counting plane partitions . . . 18
3.4.1. Implementation . . . 23
4. Partition function for plane partitions 25 4.1. Skew Schur functions . . . 25
4.2. Representation theory . . . 31
4.2.1. Fermionic Fock space . . . 32
4.2.2. Clifford algebra . . . 32
4.2.3. Representation of gl∞ . . . 33
4.2.4. Representation of a∞ . . . 35
4.2.5. The loop algebra and the vertex operator . . . 37
4.3. Partition functions . . . 41
5. Discussion 48 A. Populaire samenvatting 50 B. Mathematica code 52 B.1. Plane partitions two boundaries . . . 52
B.2. Partition function . . . 53
1. Introduction
Already 4000 years before Christ the Sumerians decorated their houses and temples with mosaics in geometric patterns. Tiling of a surface with geometric shapes without gaps or overlap, are used in every known human society in one way or another. Besides their decorative or spiritual meaning, they are also very useful in mathematics and physics. A tiling can be seen as a model for a two dimensional solid, when we implement some
Figure 1.1.: Mosaics from the Stone-Cone Temple in ancient Sumerian city of Uruk IV. (BrokenSphere / Wikime-dia Commons)
physical structure.
Even though the concept of tilings is really old, there are still a lot unknown aspects. One of them is integrability, which is closely related to the possibility to calculate some physical properties exact. Not many tilings are known to be integrable. For this reason we will exam-ine the integrability of certain tilings. These particular tilings can be viewed as a model for quasicrystals, structures which really exist in nature.
Besides this also finite tilings have interest-ing properties. In particular the finite tilinterest-ing which is called mosaic in mathematics has some remarkable properties. If we fix the boundary conditions, and count all possible mosaics subject to these boundary conditions, we find a Littlewood-Richardson coefficient.
This coefficient is closely related to Schur functions, which play an important role in representation theory. Therefore we will look at a tiling of rhombi, inside a hexagon where we impose some comparable boundary conditions. Our goal is to find a, prefer-ably closed, formula for this counting problem, in the hope that this will give us a similar coefficient.
In the first chapter we will introduce quasicrystals and examine the integrability of two quasiperiodic tilings, which can be seen as a model for a quasicrystal. The second chapter is about the counting problem of rhombi we discussed before. Here we explain the connection between this problem and plane partitions. Moreover we introduce our approach to this counting problem, where we count the number of tilings with help of non-intersecting lattice paths. In the third chapter we continue on with plane partitions, especially their partition function. We will use representation theory to find the partition function for plane partitions with an asymptotic limit shape.
2. Quasicrystals
Until the sixties one thought that all tilings of the plane where periodic, but it turned out there are ways of tiling the plane which are not periodic. The most famous ex-ample is the Penrose tiling, a tiling consisting of just two different tiles which produce only non-periodic tilings of the plane, see figure 2.1. The tiling is named after the
Figure 2.1.: A part of a Penrose tiling mathematician Roger Penrose, who proposed
this tiling in 1976. Even more surprising was the discovery of such a non-periodic structure in nature. In 1982 Shechtman discovered the first quasicrystal, and he obtained a Nobel prise in chemistry for this. Up until then peo-ple universally accepted that the internal or-der of crystals was achieved through a peri-odic filling of space, even though this was never proven. They treated order and periodicity synonymously, but in a quasicrystal this is not the case, it is an order structure even though it is not filling the space periodically. [2]
The word quasicrystal is just an abbrevia-tion of quasiperiodic crystal, this is also how you should interpreted it. A quasicrystal is a packing of two or more distinct structural units, called unit cells, with long-range ori-entational order, minimal separation between
atomic sites and long-range quasiperiodic translational order. Long-range orientational order means that each edge of each unit cell is oriented in one of the directions of the set of ”star” axes. Quasiperiodic translational order means that the density function is of the packing is a quasiperiodic function. A function is called quasiperiodic if it can be expressed as a countable linear combination of periodic functions where the ratio of at least some of the periods is irrational. But only if integral linear combinations of a finite number of wave vectors span the all the wave vectors in the density function, a crystal is called quasiperiodic. In addition to these requirements one can impose the condition that there is no nontrivial translation which leaves the packing invariant, to make sure the quasicrystal is strictly non-periodic and hence unequal to a ordinary crystal. [1, 2]
To identify the atomic and molecular structure of a crystal physicists look at the diffraction pattern of a incident x-ray, electron or neutron beams. If the scattering in-side the crystal is weak, i.e. each wave is scattered just once, the diffraction pattern is closely related to the Fourier transformation of the density function. Thus because a
quasicrystal has a quasiperiodic density function the diffraction pattern of a quasicrys-tal consists of countable many Bragg peaks, which corresponds to delta functions. The diffraction pattern of classical crystals can posses only two, three, four, and six-fold rotational symmetries, but quasicrystals admit also different orders of rotational sym-metry. Shechtman observed a ten-fold symmetric diffraction pattern, during his routine investigations.
In two dimensions a quasicrystal corresponds to a quasiperiodic tiling. For example, the random square-triangle tiling is an example for a twelve-fold symmetric quasiperiodic tiling. In 1988 the researchers Chen, Li and Kuo found the first quasicristals with a
Figure 2.2.: A part of a random square-triangle tiling
structure corresponding to the square-triangle tiling, in the system V-Ni and V-Ni-Si. [3]
An important property of a tiling is integrability. In physics a system is called in-tegrable if it is possible to solve the system with the Bethe ansatz. With help of the Bethe ansatz it is possible to diagonalise the transfer matrix and as a consequence one can to compute the exact partition function, which gives access to a lot other physical quantities. The square-triangle tiling is known to be integrable. We will add some tiles to this tiling, such that the tiling remains quasiperiodic and investigate the integrability of this newly obtained tiling. We will do this by trying to construct bijections between the configurations of this tiling and an other, where the other tiling in integrable. If we find such a bijecetion this immediately shows our obtained tiling is integrable.
2.1. Two shields
We are going to look at the square-triangle tiling with one tile added, see figure 2.3 for all possible tiles. The last tile in figure 2.3 will be called a shield, and is added in two orientations. From the pure mathematical point of view it would be logical to add the shield in all four possible orientations in which they fit into the square-triangle tiling, such that the twelve-fold symmetry of the tiling is conserved. The shield in the other two orientations correspond to rotating our shields with 90 degree. But there is a reason for
Figure 2.3.: The possible tiles in the square-triangle tiling + shields.
this restriction. In section 3.1 we introduce mosaics and puzzles. There we tile a hexagon with three young diagrams inside corners with tiles from the square-triangle tiling, and the total number of possible tilings gives a coefficient from the the product formula for Schur functions. Now consider the same object, where we allow the two shields as in 2.3 and count all possible tilings. This number is conjectured to be the coefficient from the product formula of a generalisation of the Schur function, the Grothendieck polynomials (personal communication P. Zinn-Justin, February 5, 2015).
It is known that the square-triangle tiling is a integrable system, but would it still be an integrable system if we add the shield in two orientations? If it is possible to find a bijective map from this tiling to an integrable system, we can be sure our tiling is integrable. In this section we are going to look for such a map.
Figure 2.4.: The black lines represent all possible orientations of the edges of the tiles, these will be projected on the nearest red line.
At first, notice that our tiling does not fit on a lattice, therefore we are going to perform a projection in such a way that our tiling will fit on a triangular lattice. All edges of the tiles are oriented in one of the directions of the black lines in figure 2.4, between the black lines there is an an-gle of exactly thirty degrees. Every edge will be projected to the nearest red line, who represent the directions of the axis of a triangular lattice, and are located at an angle of 15 degrees from the nearest black line. Against loss of information we will label an edge with + if it has to be rotated by 15 degrees to align with the nearest red line, and a − if the edge has to be rotated by −15 degrees to align with the nearest red line. After this projection the tiles will look as in figure 2.5. -- - + + + -- -+ + + + -+ + -+ + - -+ + -+ -+ + -- + - +
Clearly all tiles can be built out of the building blocks in figure 2.6, and their 180 degree rotation. + + + -- - + -+ +
-Figure 2.6.: Building blocks of the projected tiles.
Since in each orientation there are six triangular building blocks it seems natural to map them on to a triangular lattice with tricoloured edges, because such a tiling can be built out of six similar triangles and there 180 degree rotation (figure 2.7).
Figure 2.7.: Building blocks of the tricoloured triangular tiling.
Let’s construct this map. For simplicity we will let the map conserve orientation of edges. Note that every building block has to be send to a triangle with tree differently coloured edges, therefore w.l.o.g. we can send he empty triangle to the following triangle:
This means the horizontal empty edge is mapped to the red horizontal edge etc. In this way all empty edges of building blocks will fit nicely together. From this follows where the empty edges of the other building blocks are mapped to. For example
+
-Since every building block has to be send to a triangle with tricoloured edges, there remain two options for the colouring of the dotted edges, red and green. Note that we already send the horizontal empty edge to a horizontal red edge, hence the horizontal edge labelled with + has to be send to the horizontal green edge. This is because we do not want the image of two building blocks being able to connected, if they originally could not. For the remaining edge, labelled with - there is only one option left, it has to be send to a red edge with the same orientation. Thus we get:
+
-Continuing this argumentation will give us the entire map, which can be summarised as follows:
+ + +
--
-It seems like we constructed a nice map form our tiling to the tricoloured triangular lattice, such that adjacent building blocks match if and only if the colour of the image of the respective edges coincide. But there is one ambiguity. For that we have to go back to the origin of the empty triangle. The empty triangle comes form the inside of the projection of a shield, but the surrounding building blocks are arranged in such a way that it is not possible to match two empty triangles. Therefore also in the image of the constructed map certain matching of triangles will be excluded.
Up until now we only now that the tricoloured triangle tiling is integrable if we allow all possible combinations of tiles. Consequently we can not say anything about the integrability of the tilings where we exclude certain combinations of tiles. For that we still do not know anything about the integrability of our tiling and we can not use the same solving methods for the triangle tiling to solve our tiling.
2.2. One shield
This time we will look at the square triangle tiling with the shield added in only one orientation, to prevent the problems we encountered with two shields. We will let the plus triangle predominate in the tiling. Then only a row of squares with their plus side oriented towards the plus triangles can separate two domains filled with plus triangles. We will call these rows of squares domain walls. The domain walls can meet in three different ways, using the remaining three unused tiles.
There is also an other tiling with similar behaviour, the triangular trimers, where one triangle predominates. This is a tiling of the triangular lattice such that each point of the lattice is part of exactly one triangle of the tiling.
0 5 1 0 1 2 3 4 5 0 0 5 1 4 2
Figure 2.9.: A tiling of triangular trimers where the trimers occupy all positions of one sub-lattice. There are six sub-lattices numbered by 1, . . . , 5 .
The triangular lattice has six sub-lattices, in figure 2.9 one sub-lattice is completely occupied. This particular configuration does not admit local changes, but it is possible to flip a whole line of trimers. These lines of flipped tiles will be part of one of the three odd sub-lattices. We will call these lines domain walls, because they separate areas covered with triangles in the sub-lattice which is numbered by 0. There are three possible in which the domain walls can meet, see figure 2.10. In two of these meetings a triangle from the sub-lattice 2 or 4 is needed. Tiles from the same sub-lattice got the same colour in the figures.
These tilings with triangular trimers are predominated by one triangle, in the 0 sub-lattice, and has domain walls which meet in three different ways. This is similar to the square-triangle tiling with one shield added. Furthermore the triangular trimer model is integrable, this is proven in [4]. Therefore we will try to find a bijecetion between the two mentioned tilings.
Figure 2.10.: The three possible ways domain walls can meet in a triangular trimer tiling with one predominating triangle. Here the gray lines represent the triangular lattice, and are no part of the tiling.
To do this we will first look at the domain walls. Because of the orientation of the domain walls the only possibility to send the domain walls to one another is the following.
(b)
(c)
Here the lines represent the domain walls, the red ones are the domain walls of the square-triangle tiling in the same order as we saw before. Note that the domain walls in the square-triangle tiling do not meet exactly in the middle, they are all chiral. On the other hand, one of the the domain walls in the triangular trimer tiling is achiral. This already indicates some complications for finding a bijection between the two tilings.
Suppose if we move the domain walls of the square-triangle tiling according to (c). Then the domain walls in the first possible way to meet, as in (a), will become more acentric, however they will still be chiral, exactly as desired in the image. But the domain walls in the second way of meeting, as in (b), will become central, and therefore achiral, which does not coincide with the desired image. Therefore this does not work. For a bijection between the configurations of the two tilings at least the map (c) should be satisfied, but this appears to be impossible. So this time we could not even construct a map between these two tilings, and consequently we can not say anything about the integrability of the square-triangle tiling with one shield added.
3. Plane partitions with boundary
Inspired on an object called a mosaic we will study plane partitions. This arises from the problem of counting the number possible tilings inside a given boundary. We will introduce all objects, the problem and thereafter present our approach this counting problem, where our main ingredient is lattice paths.
3.1. Puzzles and mosaics
Until now we have only considered unrestricted tilings, but also tilings with boundary conditions can have some interesting properties. In this section we are going to look at one specific restriction on the boundary of the square-triangle tiling.
Consider a hexagon AA0BB0CC0 with sides AA0 = BB0 = CC0 = n and AB = BC = CA = m with n, m ∈ Z and has an angle of 150◦ at A, B and C and an angle of 90◦at A0, B0 and C0. In each of the corners A, B and C we will nest a transformed young diagram as in figure 3.1. We will fill the remaining space with tiles from the square-triangle tiling. This construction is called a mosaic. [5]
Figure 3.1.: An example for a mosaic with young diagrams corresponding to λ = (3, 1), µ = (2, 1) and ν = (2) and all sides of length 3. From P. Zinn-Justin’s puzzle viewer http://www.lpthe.jussieu.fr/ pzinn/puzzles/.
We can perform the same projection as in the square triangle tiling on a mosaic, the projection is called a puzzle. There is an example in figure 3.2. The young diagrams will be projected to a string of pluses and minuses, which still define the boundary conditions. The length of the sides of the triangle is n + m.
Figure 3.2.: The puzzle corresponding to the mosaic in figure 3.1. From P. Zinn-Justin’s puzzle viewer http://www.lpthe.jussieu.fr/ pzinn/puzzles/.
A lot of people thought about the problem of counting the number of puzzles for fixed young diagrams. One of the result is that the number of puzzles with boundary λ, µ and ν is equal to cν
λµ, the Littlewood-Richardson coefficient.[7]
The Littlewood–Richardson coefficients are the coefficients such that sλsµ =
X
ν
cνλ,µsν,
with sλ the famous Schur function. But there are a lot of other places where these
coefficients appear.
3.2. Rhombi
Inspired on the remarkable result of counting puzzles or mosaics we want to look at a similar problem, but now with a tiling of rhombi. Ideally we could also find such an expression for this problem. We will look at a regular hexagon tiled with rhombi, here we are also able fix a boundary.
There is also an other interpretation of this problem. We can see this tiling as a corner filled with some cubes, see figure 3.3. The boxes are placed in such a way that the height is weakly decreasing when you walk away from the corner. This is as if all walls exert a gravitational force on the boxes, such that the boxes move in to the corner. Then the boundary condition corresponds to a restricting line on the walls of the corner, which the cubes may not pass.
Alternatively, we could see this tiling as a rectangle composed of small cubes where some cubes are removed. But a cube may only be removed if there is no direction in which the cube still has both neighbours on the opposite sides. This could be interpreted as a frozen crystal where some of atoms are removed, because one of the corners starts to melt.
For now we will use the interpretation of filling a corner with cubes, since this is the visualisation of a well known object, called a plane partition.
Definition 3.1. A plane partition is a two-dimensional array of non-negative integers (πi,j)i,j∈N with finite support (i.e. finitely many nonzero entries) such that
πi,j ≥ πi,j+1 and πi,j ≥ πi+1,j for all i, j ∈ N. [6]
A plane partition can be visualised by a stack of πi,j boxes above the point (i, j) on
the plane.
Our goal is to count the number of plane partitions, with a boundary corresponding to the young diagrams of three partitions λ, µ and ν. In figure 3.3 we see that the light blue area in the lowest corner looks like a shifted version of the young diagram of the partition (3, 2, 1). That is exactly what is meant by a boundary corresponding to the partition λ = (3, 2, 1).
First we will rewrite this problem in therms of paths, because there is a lot known about the problem of counting paths. We can do that by adding a line to two of the three tiles, as in figure 3.4.
Figure 3.4.: The rhombi with a line added.
Then the tiling becomes as in figure 3.5. There we see that a plane partition corre-sponds to a set of non-intersecting paths starting from the edge on the lower right and walking there way up to the upper left edge.
Now we only need to count all possible tuples of non-intersecting paths inside a bound-ary given by three partitions.
Figure 3.5.: Paths in a plane partition as result of the new tiles.
3.3. Lindstr¨
om-Gessel-Viennot
There is a really useful lemma for counting tuples of non-intersecting paths, now known as the Lindstr¨om-Gessel-Viennot lemma. This lemma was already stated and proved in 1973 by Lindstr¨om, but its full potential became known in 1985 when Gessel an Viennot published a paper full of interesting applications of this lemma. [9]
Before we state the lemma we need to introduce some new concepts and notation. We are going to look at a locally finite directed acyclic graph D, which means that each vertex of D has finite degree, and the directed edges do not form a cycle. To each edge e of D we will assign a weight w(e), and define for a path P the weight of a path w(P ) as the product of the weight of its edges. For two vertices a and b let p(a, b) denote sum of the weights of all possible paths from a to b, i.e. p(a, b) =P
P :a→bω(P ). Notice that
if we assign a weight of 1 to every edge p(a, b) counts the number of paths from a to b, this will be important for our purpose.
Consider two k-tuples of vertices of D, the sources u = (u1, u2, . . . , uk) and the sinks
v = (v1, v2, . . . , vk). We will look at k-tuples of paths P = (P1, P2, . . . , Pn), where Pi
is a path between the source ui and the sink vi, these k-tuples are called k-paths. The
weight of a k-path is the product of the weights of the paths it consists of.
Definition 3.2. A k-path P = (P1, P2, . . . , Pn) is called non-intersecting if the paths
Pi and Pj do not have a vertex in common whenever i 6= j.
Now denote P (u, v) for the sum of all weighted k-paths from u to v, and N (u, v) for the sum of the weights of all non-intersecting k-paths from u to v. Furthermore we will use the notation σ(v) for the k-tuple of vertices vσ(1), vσ(2), ..., vσ(k) for the permutation
At last we will define a matrix M =
p(u1, v1) p(u1, v2) · · · p(u1, vk)
p(u2, v1) p(u2, v2) · · · p(u2, vk)
..
. ... . .. ...
p(uk, v1) p(uk, v2) · · · p(uk, vk)
.
After the introduction of all of the notation we can finally state and proof the lemma of Lindstr¨om Gessel and Viennot.
Lemma 3.3. Lindstr¨om-Gessel-Viennot
For the k-tuples of vertices u = (u1, u2, . . . , uk) and v = (v1, v2, . . . , vk) of a locally
finite directed acyclic graph D, we have det(M ) = X
σ∈Sk
sign(σ)N (u, σ(v)). [8] (3.1)
Proof. First of all we can write the determinant as follows; det M = X σ∈Sn sign(σ) · p u1, vσ(1) p u2, vσ(2) · · · p un, vσ(n) = X σ∈Sk sign(σ)P (u, σ(v)).
Then the statement becomes X
σ∈Sk
sign(σ)P (u, σ(v)) = X
σ∈Sk
sign(σ)N (u, σ(v)).
Thus we need to prove that all paths which are not non-intersecting vanish in the sum. We will do that by defining an bijective involution, such that the non-intersecting paths are fixed points, but the others aren’t. This involution should also preserve the weight of a k-path, but changes the end points such that the permutation corresponding to the new end points has the opposite sign compared to the permutation corresponding to the original endpoints.
Let’s define such a map f : [
σ∈Sk
P (u, σ(v)) → [
σ∈Sk
P (u, σ(v)).
For a k-path P = (P1, P2, . . . , Pn), let i be smallest index such that Pi has an intersection.
Let x be the first point of intersection along Pi and then let j be the least index bigger
than i such that x lies on Pj. Then we will define f (P ) to be the same as P but with the
tails of the two paths Pi and Pj switched. Thus now the new Pi consist of the first part
of the old Pi until the point x, thereafter it will follow the path which was originally Pj,
see figure 3.6. If the k-path is non-intersecting it will be mapped to itself.
Clearly this map preserves the weight of a k-path, is bijective and an involution. The only fixed points of the involution are the non-intersecting paths. For the others we have: if P ∈ P (u, σ(v)) and f (P ) ∈ P (u, τ (v)), then sign(σ) = −sign(τ ). Therefore P
Figure 3.6.: A not non-intersecting k-path with its image after applying the involution. will be cancelled by f (P ) in the sumP
σ∈Sksign(σ)P (u, σ(v)), and only the fixed point of the involution remain. Consequently
X
σ∈Sk
sign(σ)P (u, σ(v)) = X
σ∈Sk
sign(σ)N (u, σ(v)).
We will use this lemma to count the number of non-intersecting paths which will represent plane partitions as described in the previous chapter. For this purpose we will give all edges a weight of 1. It is possible to count the number of non-intersecting paths if our pair of k-vertices (u, v) is nonpermutable, this means there are no non-intersecting k-paths from u to σ(v) unless σ is the identity permutation. Then the lemma of Lindstr¨om-Gessel-Viennot 3.1 will give us:
N (u, v) = X
σ∈Sk
sign(σ)P (u, σ(v)) = det(M ). (3.2)
This is exactly what we are going to use to count the number of non-intersecting paths. Most of the time this result will be meant, when we talk about the lemma of Lindstr¨ om-Gessel-Viennot.
The similar ideas used for the lemma of Lindstr¨om-Gessel-Viennot appeared already earlier, in the Slater determinant, which gives a formula for the wave function of a multi-fermionic system in quantum mechanics. The Slater determinant may be seen as an ”ancestor” of the Lindstr¨om-Gessel-Viennot determinant. One could think of the lemma of Lindstr¨om-Gessel-Viennot as an instance of the fermion-boson correspondence. In the boson case all paths are allowed, and therefore the total number of allowed k-paths is the permanent of the matrix M . The fermions need to satisfy the Pauli exclusion principle, consequently the paths should be non-intersecting, and the total number of k-paths is the determinant of M , analogous to the Slater determinant.
3.4. Counting plane partitions
To count the number of plane partitions with a given boundary we first need to rewrite our problem in therms of non-intersecting lattice paths. We will write this prob-lem in therms of lattice paths, because Z2 with the edges (x, y) → (x, y + 1) and
(x, y) → (x − 1, y) is exactly a locally finite directed acyclic graph. The we can use the lemma of Lindstr¨om-Gessel-Viennot to find the number of non-intersecting paths, which correspond to plane partitions.
Suppose we want to count the number of plane partitions with a boundary corre-sponding to the partitions λ = (3, 2), µ = (3, 2, 1) and ν = (2, 1, 1), as in figure 3.7. Here λ = (λ1, λ2, . . .) is placed in the lowest corner and is oriented such that λ1is aligned
with the lower left edge of the hexagon, µ is placed in the upper left corner such that µ1 is aligned with the left vertical edge, and ν is placed in the upper right corner with
ν1 alongside of the right vertical edge.
Figure 3.7.: The frame of a plane partition with a boundary corresponding to λ = (3, 2), µ = (3, 2, 1) and ν = (2, 1, 1).
We will deform this such that it will fit nicely on a square lattice, precisely as in figure 3.8.
The red points in figure 3.7 and 3.8 represent the starting points and the end points of the paths. The sources are the end points of the light blue young diagram and named u1 up to uk from left to right, and the sinks are the end points of the dark blue young
diagram and named v1 up to vk from left to right. Because partitions are decreasing,
both x and y values of the sources and sinks in figure 3.8 are strictly increasing, precisely because of that the pair of k-vertices (u, v) is nonpermutable. Consequently by (3.2) we can count the number of non-intersecting paths from the sources and sinks by calculating the determinant of M . Now we only need to count the number of paths from each starting point to the end points.
If there is no boundary in the upper right corner, it is easy to count the number of paths from an begin point to an end point, because we can just use binomial coefficients. See appendix of some examples. But if this is not the case have to count the number of paths from the lower right to the upper left in an square where a young diagram is excluded, see figure 3.9.
Figure 3.9.: An example for a configuration where we need to count the number of paths from the lower right to the upper left.
Here we may not enter the purple young diagram, thus all paths should stay inside the blue boundary. This blue boundary also exactly shaped like a young diagram;
Notice that we could also count the number of paths inside this partition by counting all paths to most left purple dot and adding all possible paths from the bottom right to the right purple dot. Then we get
were each diagram represents the number of paths inside this diagram. This way we can continue, the next step would be:
= +
This way we can count the number of paths inside a young diagram recursively. In therms of partitions the previous diagrams state:
#(3, 2, 1) = #(3, 2) + #(2, 1) and
#(3, 2) = #(3, 1) + #(1),
where #λ is the numbers of paths inside the young diagram corresponding to λ. In general we get
#(λ1, λ2, . . . , λk) = #(λ1, λ2, . . . , λk− 1) + #(λ1− λk, . . . , λk−1− λk, 0). (3.3)
Here we can continue until k = 1, because then we know #(λ1) = λ11+1 = λ1+ 1.
Of course it is even more convenient to stop the recursion as soon as the young diagram is in the shape of a rectangle, because then we can use binomial coefficients to find the number of paths inside the young diagram. Then it would be even more efficient to use recursion from the other end of the young diagram as well such that the number of paths inside the young diagram will be expressed in therms of the biggest rectangle and some other therms. For example:
= +
= + +
thus the number of paths inside λ = (3, 2, 1) is equal to 42+ 31+ 31+ 21 = 6+3+3+2 = 14, here we applied one more step of the recursion to find the number of paths inside the last young diagram of the figure.
We could also use a more general method for counting the number of paths inside a young diagram, as in figure 3.9. First we will assign a different weight to each edge inside a rectangle of length n and height m, an edge (i, h) → (i − 1, h) gets weight xn−i−1,h and
an edge (i, h) → (i, h + 1) gets weight yn−i,h+1, see figure 3.10.
x1,0 x2,0 xn,0 x1,1 x1,m xn,m y0,1 y0,2 y0,m y1,1 yn,1 yn,m
Figure 3.10.: Weights of the edges.
Now we want to find p((n, 0), (0, m)), the sum of weighted paths from (n, 0) to (0, m) inside the rectangle. We can do this recursively. Instead of counting all weighed paths to (0, m) we could count the paths to (1, m) and to (0, m − 1). We can do this because if we go to (1, m) we only need to take one more step to the left to get to (0, m), all other paths have to go through (0, m − 1) and subsequently a step up to (0, m). This corresponds to the recursion
p((n, 0), (0, m)) = xn,m· p((n, 0), (1, m)) + yn,m· p((n, 0), (0, m − 1)).
We can continue this procedure, which leads to the recursion formula
p((n, 0), (a, b)) = xn−a,b· p((n, 0), (a + 1, b)) + yn−a,b· p((n, 0), (a, b − 1)),
with end conditions p((n, 0), (n, b)) =Qb
i=1y0,iand p((n, 0), (a, 0)) =
Qn−a
i=1 xi,0. The end
conditions correspond to the weight of the unique straight path from (n, 0) to (n, b) and from (n, 0) to (a, 0).
Now that we have p((n, 0), (0, m)) we can substitute 0 for all edges in the interior of the young diagram ν and 1 for all other edges to get the number of paths inside the allowed area of the rectangle. This way all paths in the forbidden area will vanish, and all others remain with weight 1, thus p((n, 0), (0, m)) gives us the desired number of paths.
Similarly we can count the number of paths inside a rectangle with some edges ex-cluded, by substituting 0 for the weights of all excluded edges and 1 for the remaining edges. But because of the long recursion and huge number of variables this is not very efficient.
3.4.1. Implementation
Initially we will define a function to count the paths inside a plane partition as in figure 3.9. We will do this according to the recursion 3.3 described in the previous section. f[λ_]:=
f[λ] =f[DeleteCases[ReplacePart[λ, -1 →λ[[-1]] - 1], 0]] + f[DeleteCases[λ-λ[[-1]], 0]] f[{n}] =n+1
f[{}] = 1
This function precisely executes the recursion
#(λ1, λ2, . . . , λk) = #(λ1, λ2, . . . , λk− 1) + #(λ1− λk, . . . , λk−1− λk, 0).
with begin conditions f ({n}) = n + 1 and f ({}) = 1. We need the second condition because we delete all zeros inside the remaining young diagrams. Moreover it remembers all previous calculated results.
For example
In[4]:= f[{3, 2, 1}] Out[4]= 14
this function gives the same result for the number of paths inside (3, 2, 1) as we calculated before.
Now we will calculate the number of paths from a to b with a partition ν in the upper right corner at position (n, n). For this you need to determine how the relevant rectangle looks like, and which edges are excluded because of the young diagram ν, subsequently apply f to the result. This exactly how pad(a, b, ν, n) is defined.
pad[a_,b_,ν_, n_]:= Module[
{v=Select[If[n+1 -a[[1]] ≤ Length[ν], ν[[n+1 -a[[1]] ;; -1]] - (n-b[[2]]), {}],
# ≥ 0 &]}, Piecewise[{{0, (a[[1]] -b[[1]]) < 0 || (b[[2]] -a[[2]]) < 0}}, f[Sort[DeleteCases[ConstantArray[(b[[2]] -a[[2]]), (a[[1]] -b[[1]])]
-Join[v, ConstantArray[0, (a[[1]] -b[[1]]) - Length[v]]], 0], Greater] ]]]
At last we can define a function to count the number of plane partitions with bound-aries λ, µ and ν, where the edges of the hexagon have length n.
Paden[λ_,μ_,ν_,n_]:= Module[{p=Join[λ, ConstantArray[0,n-Length[λ]]],
l=Join[μ, ConstantArray[0,n-Length[μ]]],v=Join[ν, ConstantArray[0,n-Length[ν]]]}, Module[
{P=Table[pad[{n+h-p[[h+1]], h}, {i, n+i-l[[i+1]]},v, 2n-1], {i, 0, n-1}, {h, 0,n-1}]}, {MatrixForm[P], Det[P]}]]
diagrams corresponding to λ and respectively µ, as in figure 3.8. With help of the func-tion pad we can find the number of paths from the starting points to the end points where we may not enter ν. Now formula 3.2 will give us the number of non-intersecting n-paths from the starting points to the end points with a boundary corresponding to ν and therefore the number of plane partitions with boundaries λ, µ and ν.
Notice that we do not lose generality with the assumption of a frame in the shape of a regular hexagon, since by the choice of the boundaries we can get the number of a tilings/ plane partitions inside every possible hexagon.
In figure 3.11 are some examples for the number of paths inside plane partitions with boundary λ, µ and ν. In[7]:= Grid[Table[Apply[Paden,i], {i, {{{3, 2, 1}, {3, 2}, {2, 1}, 3}, {{2, 2, 1}, {2, 2, 1}, {1, 1}, 3}, {{2, 2, 1, 1}, {2, 2, 1}, {3, 1}, 4}, {{3, 2, 2, 1, 1}, {4, 2, 2, 1, 1}, {3, 2, 2}, 5}, {{6, 3, 2, 1}, {4, 2, 2, 2, 2}, {5, 4, 1, 1, 1}, 6}}}], Dividers → All] Out[7]= 1 0 0 0 2 1 0 1 9 17 2 1 0 1 2 1 0 1 6 16 6 4 1 0 4 6 5 1 1 5 20 15 0 1 21 34 4173 3 1 0 0 0 5 20 15 7 1 1 15 20 21 7 0 7 21 70 56 0 1 7 55 65 180 936 1 5 1 0 0 0 0 35 56 36 10 1 0 21 70 84 45 10 0 7 56 126 120 44 0 1 28 126 209 110 0 0 10 156 636 502 115 270 668
Figure 3.11.: Some examples for the number of plane partitions with boundary λ, µ and ν and the corresponding determinant M .
4. Partition function for plane
partitions
In this chapter we are going to look at an other approach regarding the problem of counting plane partitions with a given boundary. We will construct the partition function for plane partitions as in [10]. Here they use some more sophisticated mathematics than we used before, therefore we have to go through some new concepts first.
4.1. Skew Schur functions
Symmetric polynomials are often used in algebra, as we saw in Galois theory. In this sec-tion we will highlight one specific family of symmetric polynomials, the Schur funcsec-tions, and its generalisation the skew Schur functions. Schur functions are especially important in representation theory because of their relation with the irreducible representations of Sn, GLn and young tableaux. Because of there diversity there are many definitions for
these polynomials.
We will introduce the polynomials as in [11], starting with the combinatorial approach. For a more extensive explanation about Schur functions you should look at the book The symmetric group, Representations, combinatorial algorithms, and symmetric functions, reference [11], and for some notes on skew Schur functions, reference [12]
Let λ be a partition. Recall that the Young diagram of shape λ is a collection of boxes, where the rows are left-justified and the i-th row has length λi. For example the
diagram corresponding to λ = (4, 2, 1) is
.
By reflecting the Young diagram along its main diagonal we obtain a new Young diagram, corresponding to the transpose partition λt. For our previous example we get
, thus λt = (3, 2, 1, 1).
For partitions λ and µ we use the notation µ ⊂ λ if λ1 ≥ µ1, λ2 ≥ µ2, etc. The two
For partitions λ and µ such that µ ⊂ λ we can define the skew Young diagram of shape λ/µ as the Young diagram of shape λ without the boxes corresponding to the diagram of µ.
A (skew) Young tableau is obtained by filling the boxes of the (skew) Young diagram with positive integers. A tableau is called semi-standard if the entries of the rows are weakly increasing and the columns are strictly increasing. An example for a semi-standard skew Young tableau with λ = (3, 2, 1, 1) and µ = (1, 1) is
3 3 4 1
2 .
For a skew Young tableau T of shape λ/µ let T(i,j)denote the filling of the box positioned
at the ith row and and jth column, and define
xT = Y
1≤i≤length(λ), µi<j≤λi
xT(i,j),
where length((λ1, . . . , λl)) = l.
Definition 4.1. The skew Schur function sλ/µ corresponding to the partition λ/µ is
defined as sλ/µ(x) = X SSYT T , shape(T )=λ/µ xT.
The Schur function sλ is defined as sλ/∅
Clearly all identities for skew Schur functions automatically hold for ordinary Schur functions. An important remark is that the skew Schur functions are symmetric, for a prove look at proposition 4.4.2 of [11].
Some important identities for calculating skew Schur functions are the Jacobi-Trudi identities. The prove of these identities relies on the lemma of Lindstr¨om-Gessel-Viennot, that is why it is suitable to go through this prove. First we need to define some basic symmetric polynomials
Definition 4.2. Let n ∈ Z>0. The nth elementary symmetric polynomial is
en(x) =
X
1≤i1<i2<...<in
xi1xi2· · · xin
and the nth complete homogeneous symmetric polynomial is hn(x) =
X
1≤i1≤i2≤...≤ln
xi1xi2· · · xin.
Theorem 4.3. (Jacobi-Trudi Determinants)
Let λ = (λ1, . . . , λn) and µ = (µ1, . . . , µn) be partitions such that µ ⊂ λ, where we
allow µ to contain zeros. Then we have
sλ/µ = det(hλi−µj+j−i) n i,j=1 = hλ1−µ1 hλ1−µ2+1 . . . hλ1−µn+n−1 hλ2−µ1−1 hλ2−µ2 . . . hλ2−µn+n−2 .. . ... . .. ... hλn−µ1−n+1 hλn−µ2−n+2 . . . hλn−µn , and sλt/µt = det(eλ i−µj+j−i) n i,j=1 = eλ1−µ1 eλ1−µ2+1 . . . eλ1−µn+n−1 eλ2−µ1−1 eλ2−µ2 . . . eλ2−µn+n−2 .. . ... . .. ... eλn−µ1−n+1 eλn−µ2−n+2 . . . eλn−µn .
Proof. The crucial insight for this prove is that one can view both tableaux and determi-nants as lattice paths. Consider paths P on the plane Z2with the edges (x, y) → (x, y+1)
and (x, y) → (x + 1, y). Denote every edge of P with si such that
P = (s1, s2, s3, . . .),
then each si corresponds to a step northward of a step eastward on the plane. In the
following figure there is shown such a path.
s1 s2 s3 s4 s5 s6 s7 s8
Now define two different labellings for a path P . The e-labelling Le assigns to each
eastward oriented edge si the label
Le(si) = i.
The h-labelling Lh which assigns to each eastward edge si the label
1
4 5
7
(a) The e-labelling.
1
3 3
4
(b) The h-labelling. Figure 4.1.: The two labellings.
where ∆y is the difference in height between the starting point of P and si. The labellings
applied to our example path gives the next pair of diagrams.
For our purposes it is convenient to extend Z2 by addition of some points at infinity. For each x ∈ Z add a point (x, ∞) on the vertical line through (x, 0), such that a path can only reach (x, ∞) by ending with an infinite number of northward steps along this line.
Let P be a path with a finite number of eastward steps. Then define two weightings ωe and ωh of P corresponding to the e-labelling and the h-labelling, respectively, by
ωe(P ) = Y eastward si∈p xLe(si) and ωh(P ) = Y eastward si∈p xLh(si).
Notice that for P from (a, b) to (a + n, ∞) the weight ωe(P ) is always square free and
corresponds exactly to one monomial of en(x), likewise ωh(P ) can be any monomial and
corresponds to exactly one therm of hn(x). Therefore
pe((a, b), (a + n, ∞)) = X P :(a,b)→(a+n,∞) ωe(P ) = en(x) and ph((a, b), (a + n, ∞)) = X P :(a,b)→(a+n,∞) ωh(P ) = hn(x).
Let λ = (λ1, . . . , λn) be given, and consider the sources u = (u1, . . . , un), w =
(w1, . . . , wn) and sinks v = (v1, . . . , vn), where
uj = (−j + µj, 0), wj = (−j + µj, j)
and
Both pairs of k-vertices (u, v) and (w, v) are nonpermutable, this is because the x-values of u, w and v are strictly decreasing and the y x-values are weakly increasing. By the choice of vertices hλi−µj−i+j =
P
P :uj→viωh(P ) and eλi−µj−i+j = P
P :wj→viωe(P ). Now apply the lemma of Lindstr¨om Gessel and Viennot and get
det(Mh) = det(hλi−µj+j−i)
n
i,j=1 = Nh(u, v)
and
det(Me) = det(eλi−µj+j−i)
n
i,j=1= Ne(w, v),
where Nhis the sum of h-weighted non-intersecting k-paths and Nethe sum of e-weighted
non-intersecting k-paths.
Now we only need to show sλ = Nh(u, v) and sλt = Ne(w, v). For this we will construct a bijection between the labelled non-intersecting k-paths and semi-standard Young tableaux.
First we will look at the case with the h-labelling. Consider the h-labelled k-path P = (P 1, . . . , Pn) running from u to v, now construct the ith row of the tableau by
reading of the labelling of Pi. This results in a tableau of shape λ.
For example the following non-intersecting k-path, corresponding to λ = (4, 2, 1) and µ = ∅; u1 u2 u3 v1 v2 v3 4 2 3 1 2 2 4
gives the tableau
1 2 2 4 2 3
4 .
By definition the rows are weakly increasing and because of the non-intersecting condi-tion and the choice of initial points the rows are strictly increasing.
Conversely, given a tableau we can construct a k-path by letting Pibe the path starting
row of the tableau minus one. This k-path is non-intersecting because of the column strictness of the tableau.
Obviously the weight of the k-path and the corresponding tableau are the same. For this reason and the bijection between k-paths and tableaux we get
sλ = det(hλi−µj+j−i)
n i,j=1.
For the e-labelling there is a similar argument. This time we construct a column of a tableau by reading of the labelling of Pi. The resulting tableau has shape λt/µt,
the columns are strictly increasing by definition and because of the starting points and the non-intersecting property the rows are weakly increasing. We can also construct a inverse in a similar manner as we did for the h-labelling.
To illustrate the process an example for λ = (4, 2, 1) and µ = ∅ ;
u1 u2 u3 v1 v2 v3 4 2 4 1 3 4 7
gives the tableau
1 2 4 3 4 4
7 .
Because both the k-path and corresponding tableau have the same weight we get sλt/µt = det(eλ
i−µj+j−i)
n i,j=1.
There is also a remarkable relation between semi-standard skew Young tableaux and sequences of interlacing partitions. Let us call sequence of partitions λ(0), λ(1), . . . , λ(n) interlacing if λ(0) ≺ λ(1) ≺ . . . ≺ λ(n).
Let T be a semi-standard Young tableau of shape λ/µ, and denote its the maximal number by n. Define λ(i) as the partition corresponding to the diagram of µ where we add all boxes of T with a filling less or equal to i. Clearly λ(0) = µ and λ(n) = λ, and because columns of T are strictly increasing and the rows increase weakly, the sequence of partitions λ(0), λ(1), . . . , λ(n) is interlacing. For example 1 1 3 2 2 2 3 3 corresponds to the sequence of diagrams
λ(0) = , λ(1) = , λ(2) = , λ(3) = ,
which gives the interlacing sequence (2, 1) ≺ (4, 1) ≺ (4, 3, 1) ≺ (5, 3, 2, 1).
This correspondence gives a bijection between semi-standard tableaux of shape λ/µ and sequences of interlacing partitions µ = λ(0) ≺ λ(1) ≺ . . . ≺ λ(n) = λ.
Consequently sλ/µ(x1, . . . , xn, 0, 0, . . . ) = X µ≺λ(1)≺...λ(n−1)≺λ x|λ(1)|−|µ|1 x|λ(2)|−|λ(1)|2 · · · x|λ(n−1)|−|λ|n , where |λ| =P
i≥0λi. This gives
sλ/µ(x1, 0, 0, . . . ) = ( x|λ|−|µ|1 if µ ≺ λ 0 if µ 6≺ λ, and therefore sλ/µ(x1, . . . , xn, 0, 0, . . . ) = X µ≺λ(1)≺...λ(n−1)≺λ sλ(1)/µ(x1)sλ(2)/λ(1)(x2) . . . sλ/λ(n−1)(xn). (4.1) We will use this peculiar relation later on.
4.2. Representation theory
To derive the partition function, probably better known as the generating function in mathematics, the authors of [10] use a lot of representation theory. This is not explained
in the article but really worthwhile to go through. Similar formalisms are often used in mathematical physics, and illustrate the wide applications of representation theory. We will look at representations of infinite dimensional lie algebras and their extensions, the Clifford algebra and vertex operators. Therefor we use [14], [15], and the appendix of [13].
4.2.1. Fermionic Fock space
First we will define the vector space we will use for our representations. Let V be a complex inner product space with orthonormal basisei : i ∈ Z + 12 . Define F = V
∞ 2 V as the vector space spanned by the vectors
eI = ei1 ∧ ei2 ∧ ei3 ∧ ei4∧ . . . ,
where I = {ik : k ∈ Z} with ik ∈ Z +12 such that ik < ik−1 for all k and in = in−1− 1 for
n 1. We will equip F with the inner product (., .) such that the collection of vectors {eI} forms an orthonormal basis. This inner product arises naturally from the inner
product on V . The space F is often called the fermionic Fock space. For k ∈ Z + 12 define the wedging and contracting operators ψk and ψ
∗
k on F by
ψk(ei1 ∧ ei2 ∧ ei3 ∧ ei4 ∧ . . .) = ek∧ ei1 ∧ ei2 ∧ ei3 ∧ ei4 ∧ . . .
and
ψk∗(ei1∧ei2∧ei3∧ei4∧. . .) = (
0 if k 6= in for all n ∈ Z
(−1)s−1ei1 ∧ . . . ∧ eis−1 ∧ eis+1∧ . . . if k = is
.
These operators satisfy the following identities: ψiψj + ψjψi = 0, ψi∗ψ ∗ j + ψ ∗ jψ ∗ i = 0 and ψiψ∗j + ψ ∗ jψi = δij,
for all i, j ∈ Z +12. Furthermore ψk and ψ∗kare adjoint operators, i.e (ψkv, w) = (v, ψ∗kw)
for all v, w ∈ F .
The infinite Clifford algebra Cl(V ⊕ V∗) is generated by elements with precisely the defining relations as above. Therefore F is a representation of the Clifford algebra. This representation is clearly irreducible because with a combination of the operators ψk and
ψk∗ we can construct every element of F out of an arbitrary basis vector eI.
4.2.2. Clifford algebra
In case you do not know the Clifford algebra or need a reminder there will be some more explanation here. We will focus on the infinite dimensional Clifford algebra, since that is the most relevant to us, but it is not hard to deduce the finite case from there.
Definition 4.4. For a vector space V over a field K we define the tensor algebra T (V ) as the space T (V ) = ∞ M k=0 V⊗k = K ⊕ V ⊕ (V ⊗ V ) ⊕ (V ⊗ V ⊗ V ) ⊕ · · · ,
where multiplication is the tensor product.
The Clifford algebra is a quotient of the tensor algebra.
Definition 4.5. Let V be a vector space over a field K with a symmetric bilinear from h., .i. Define the Clifford algebra as
T (V )/J,
with J the two-sided ideal generated by the elements of the form x ⊕ y + y ⊕ x − 2hx, yi for all x, y ∈ V .
Note that is we choose the bilinear form to be zero the Clifford algebra will become the exterior algebra V∗
V .
Now consider the inner product space V over C with orthonormal basis {ei : i ∈ Z}.
Then the dual space V∗ has a dual basis {e∗i : i ∈ Z}. On V ⊕ V∗ we can define a sym-metric bilinear form h., .i as the linear extension of the form generated by the relations
hei, eji = 0 = he∗i, e ∗ ji and he ∗ i, eji = hej, e∗ii = 1 2δij for all i, j ∈ Z. The corresponding Clifford algebra has the defining relations
eiej + ejei = 0, e∗ie ∗ j + e ∗ je ∗ i = 0 and eie∗j + e ∗ jei = δij,
this follows directly from substituting the basis elements of V and V∗ in the relation for the generators of the ideal.
Now it is clear that (F, ϕ), where ϕ : Cl(V ⊕ V∗) → EndC(F ) is the homomorphism defined by ek7→ ψk and e∗k 7→ ψk∗ is a representation of Cl(V ⊕ V∗).
4.2.3. Representation of gl
∞Now we will look at a representation of the Lie algebra gl∞on the fermionic Fock space,
where we use the same operators as for Cl(V ). This is important because we will extend this representation to a bigger Lie algebra, which is needed to find the partition function for plane partitions.
The Lie algebra gl∞ is defined by:
gl∞=
n
a = (aij)i,j∈Z+1
2 : only finitely many aij are nonzero o
where the Lie bracket is the ordinary matrix commutator. We will look at indices in Z + 12 because we did this for V as well and this will be convenient later on. This Lie
algebra arises from the Lie group GL∞, which is defined as follows:
GL∞ =
n
A = (aij)i,j∈Z+1
2 : A is invertible and only finitely many aij− δij are nonzero o
, with group operation being matrix multiplication.
We will denote Eij for the matrix with 1 on the entry (i, j) and 0 elsewhere. The
collection of matrices {Eij}Z+1
2 forms a basis of gl∞. We can express the commutation relations of gl∞ in therms of the commutation relations of only Eij. We have
[Eij, Emn] = δjmEin− δniEmj.
Now we will look at a representation of gl∞ on F . For that cause we will first look
at GL∞. Notice that GL∞ acts naturally on V , as a consequence GL∞ also acts on the
wedge product. Therefore it is natural to represent GL∞ on F by
π(A)(ei1 ∧ ei2 ∧ ei3. . .) = Aei1 ∧ Aei2 ∧ Aei3 ∧ . . . ,
which is the natural action of GL∞. Because gl∞ is the Lie algebra of GL∞, gl∞ acts
diagonal on F . This leads to the representation ρ of gl∞ on F defined as
ρ(a)(ei1 ∧ ei2 ∧ ei3. . .) = aei1 ∧ ei2 ∧ ei3. . . + ei1 ∧ aei2 ∧ ei3. . . + . . . . Then we see ρ(Eij)(ei1 ∧ ei2 ∧ ei3 ∧ . . .) = ( 0 if j 6= in for all n ∈ Z ei1 ∧ . . . ∧ eis−1 ∧ ei∧ eis+1 ∧ . . . if j = is = ψiψj∗(ei1 ∧ ei2 ∧ ei3 ∧ . . .).
Define F(m) as the linear span of semi-infinite monomials of the form e(m)I = ei1∧ . . . ∧ ein−1 ∧ em−n+1/2∧ em−n−1/2∧ . . . ,
i.e. there is a n ∈ Z such that ik = m−k +1/2 for all k ≥ n. Then F can be decomposed
as the direct sum F =L
m∈ZF(m). Also note that ψiψj∗ maps F(m) to itself. Moreover
e(m)I = ψi1ψ
∗
m−1/2· · · ψin−1ψ
∗
m−n+3/2em−1/2∧ em−3/2∧ em−5/2∧ . . . ,
therefore every F(m) is a irreducible representation of gl
∞. The vector
v0(m)= em−1/2∧ em−3/2∧ em−5/2∧ . . .
is often called the vacuum vector, or the highest weight vector. This vector is often used as a reference, and the other vectors are viewed as a variation of this vector. A more physical interpretation would be to see the vacuum vector as a ground state and the others as excitations. Vectors in F(m) can be parametrised by partitions as follows, for a partition λ define
v(m)λ = eλ1+m−1/2∧ eλ2+m−3/2∧ eλ3+m−5/2∧ . . .
These vectors clearly span F(m). We will use this notation from now on, and it will
prove useful in the application to plane partitions. Here we use the convention that every partition ends on an infinite array of zero’s, i.e. λ = (λ1, . . . , λl, 0, 0, . . .). We will
4.2.4. Representation of a
∞The representation of gl∞ is useful, but it would be favourable to have a representation
of a bit bigger matrix lie algebra. We will try to extend this representation to the finite band matrices, which naturally gives rise to the concept of central extensions.
First we define the bigger lie algebra a∞ of finite band matrices as
a∞= n A = (aij)i,j∈Z+1 2 : aij = 0 for |i − j| 0 o ,
where the lie bracket is still the matrix commutator. Note that the product of two matrices of a∞ is well defined and again a matrix in a∞. Also the lie algebra gl∞ is a
subalgebra of a∞.
The matrices of a∞ are finite linear combinations of the matrices
dk =
X
i∈Z+12
ciEi,i+k with ci ∈ C.
If we try our representation on dk we get for k 6= 0 that ρ(dk)v (m)
λ is a finite linear
combination of semi-infinite monomials in F(m), that is fine. But for k = 0 we get
ρ(d0)v (m) 0 = X i≥m−1/2 civ (m) 0 ,
which does not converge in general. To solve this problem we modify our representation ρ and call the new representation r which is given by
r(Eij) = ( ψiψj∗ if i 6= j or i = j > 0 ψ∗jψi if j = i < 0 . Now r(d0)v (m) 0 = ( Pm i=1ci−1/2v (m) 0 if m > 0 Pm+1 i=0 ci−1/2v (m) 0 if m < 0 ,
these are finite sums, and hence the anomaly is solved. But mind that this modification does alter the commutation relations, which result in
[r(Eij), r(Ekl)] = r([Eij, Ekl]) + α(Eij, Ekl)I,
where α is a two-cocycle defined by
α(Eij, Eji) = −α(Eji, Eij) = 1 if i < 0 and j > 0
α(Eij, Ekl) = 0 otherwise.
Because of additional scalar in the commutator r is not a representation of a∞, but by
extending r linear it can be made into a representation of the central extension of a∞,
Definition 4.6. A central extension of a Lie algebra g is a short exact sequence of Lie algebra homomorphisms
0 → z→i bg → g → 0,π such that i(z) is contained in the centre ofbg.
As vector spaces the short exact sequence splits as bg= g0⊕ z. Suppose this happens via the section σ : g → g0, then
β : g × g → z, (x, y) 7→ [σ(x), σ(y)] − σ([x, y]) is a two-cocycle, i.e β is bilinear, alternating and satisfies
β(x, [y, z]) + β(y, [z, x]) + β(z, [x, y]) = 0.
Conversely, given a two-cocycle β we can construct a central extension bgβ = g ⊕ z, as
vector spaces, with Lie bracket [g + z, g0+ z0] = [g, g0] + β(g, g0).
In our case we can construct a central extension of a∞ with help of α. We call this
Lie algebra a∞, as vector spaces we have a∞ = a∞⊕ CK with K in the centre of a∞,
and Lie bracket
[A, B] = [A, B]a∞ + α(A, B)K for all A, B ∈ a∞⊂ a∞.
Extending r to a∞by letting r(K) = I, gives us a representation of level 1 of a∞, which
satisfies all desired relations.
We can write α(Eij, Emn) in the form Tr(Eij[D, Emn]a∞) with D the matrix
D = X
i∈Z<0+12
ξ + X
i∈Z≥0+12
(ξ + 1),
for some ξ ∈ C. This is quite easy to see. If we start with D =P
a,bdabEab we get Eij[D, Emn] = Eij X a,b dab(EabEmn− EmnEab) =X a,b dab(δajδbmEin− δjmδnaEnb) = djmEin− X b dnbδjmEib.
Thus in case i = n and j 6= m we have
0 = α(Eij, Emn) = Tr(djmEii) = djm,
thus D should be diagonal. Now the expression inside the trace reduces to djjδjmEin−
dnnδjmEin, which only nonzero if i = n and j = m. If we also impose the condition
Thus D should be as mentioned above. By definition α(Eij, Emn) = Tr(Eij[D, Emn]a∞) and the trace in this expression is well defined for all matrices A and B. This is because
Tr(A[D, B]a∞) = X i,j∈Z+1 2, i<0, j>0 (bjiaij − bijaji),
this is a finite sum, and therefore well defined.
Maybe you are wondering why we would go through all this just to write a simple expression in a different form. Well that is just because this way of writing is used more often and now it is immediately clear how α works on general matrices.
4.2.5. The loop algebra and the vertex operator
The algebra a∞ is a big algebra in which we can embed a lot of other algebras. We will
look at the embedding of the loop algebra of gl1in a∞, and the resulting embedding of the
central extension of the loop algebra in a∞. Thereafter we will define the vertex operators
over the Fock space, which arise from the representation of the central extension of the loop algebra.
We will start with the definition of a loop algebra, which is a special case of the Kac-Moody algebras. Here we use chapter four of [16].
Definition 4.7. The loop algebra of a Lie algebra g, which we denote with Lg, is the space C[t, t−1] ⊗ g with Lie bracket
[tn⊗ x, tm⊗ y] = tn+m⊗ [x, y]
It is usual to use the short hand notation xn = tn⊗ x for all x ∈ g and n ∈ Z, then
the Lie bracket becomes
[xn, yn] = [x, y]n+m
Now fix an invariant symmetric bilinear form (., .) : g × g → C on g. Invariant means ([a, b], c) = (a, [b, c]).
Let c be a central element of Lg. Then β(xn, yn) = nδn,−m(x, y)c is a two-cocycle with
values in the one dimensional Lie algebra Cc. This cocycle leads to the central extension c
Lg = Lg ⊕ Cc with Lie bracket
[xn, ym] = [x, y]n+m+ nδn,−m(x, y)c
We want to look at the loop algebra of gl1. This algebra Lgl1 is commutative, since
gl1 is, and is generated by the elements Jn= tn⊗ 1 for n ∈ Z. We will fix the invariant
bilinear form on gl1 as (x, y) = xy, this clearly satisfies all desired relations, since gl1
is commutative. This leads to the central extension of Lgl1, which is dLgl1 = Lgl1 ⊕ Cc
with bracket
We can embed dLgl1 in a∞ by the map f : dLgl1 → a∞ which sends Jn7→ X i∈Z+1 2 Ei,i+n and c 7→ K.
It is easy to verify that f preserves the algebra structure of dLgl1.
Define for k 6= 0 the operator αn = r(f (Jn)), explicitly we have
αk =
X
i∈Z+1 2
ψiψi+k∗ if k ∈ Z \ {0},
in physical applications these operators are called the free bosons. The operators ψk and
ψk∗are called the free fermions, and the process of introducing the αk is the bosonization.
The operators αk have the following commutation relations;
[αn, αm] = nδn,−m [αn, ψk] = ψk−n [αn, ψk∗] = −ψ ∗ k+n.
Now define the vertex operators as the formal power series Γ+(x) = exp( X n∈Z>0 xn n α−n) and Γ−(x) = exp( X n∈Z>0 xn n αn). Note that Γ−(x)v0(m) = v (m) 0 ,
for all n > 0 and m ∈ Z, because in this case αnv(m)0 = 0. Moreover, since ψk and ψ∗k
are adjoint operators, we get (αnv, w) = (v, α−nw) for all v, w ∈ F and therefore
(Γ−(x)v, w) = (v, Γ+(x)w) = (Γ+(x)w, v).
For basis vectors vλ(m), vµ(m) ∈ F(m) we get (Γ−(x)v(m)λ , vµ(m)) = (Γ+(x)v (m)
µ , v(m)λ ), this is
really useful for calculating the action of Γ±, because if we know the action of one of
them on a basis vector, this relation immediately gives us how the other one acts. To determine the action of Γ± we will first define generating series for ψk and ψ∗k , which is
also called the formal Fourier transform of ψk and ψ∗k, or fermionic fields
ψ(z) = X k∈Z+12 zkψk and ψ(z)∗ = X k∈Z+12 z−kψ∗k.
Theorem 4.8. The following commutation relations hold: • Γ+(x)Γ−(y) = (1 − xy)Γ−(y)Γ+(x)
• Γ+(x)ψ(z) = (1 − z−1x)−1ψ(z)Γ+(x)
• Γ+(x)ψ(z)∗ = (1 − z−1x)ψ(z)∗Γ+(x)
• Γ−(x)ψ(z)∗ = (1 − zx)ψ(z)∗Γ−(x)
Proof. For the first identity we will use the Baker-Campbell-Hausdorff formula in case z = [X, Y ] commutes with both X and Y . This gives us
eXeY = eX+Y +z2 and eYeX = eX+Y − z 2, thus eXeY = eX+Y +z2e−(X+Y − z 2)eYeX = ezeYeX,
since [X + Y + z2, X + Y − z2] = 0 and the Baker-Campbell-Hausdorff formula in this case. So if we take X =P n∈Z>0 xn nα−n and Y = P n∈Z>0 xn nαn we have [X, Y ] = X n,m∈Z>0 xnym nm [α−n, αm] = X n,m∈Z>0 xnym nm (−nδ−n,−m) = X n∈Z>0 −(xy) m n = ln(1 − xy). Because this obviously commutes with both X and Y we get
Γ+(x)Γ−(y) = eXeY = (1 − xy)eYeX = (1 − xy)Γ−(y)Γ+(x).
For the other identities we use a different approach. Assume [X, A] = aA for some constant a and we want to calculate the commutation relations for ex and A. Now
define A(t) = etXAe−tX, because then we get d
dtA(t) = e
tX
[X, A]e−tX = aetXAe−tX = aA(t).
The solution to this differential equation with begin condition A(0) = A is clearly A(t) = eatA. Thus for t = 1 we get
eXA = eaeX.
We use this to compute the last four commutation relations. Let us calculate one example. Take X =P n∈Z>0 xn nα−n and A = ψ(z), then [X, A] = X n>0,k xnzk n [α−n, ψk] = X n>0,k xnzk n ψk+n =X n>0 xn nznψ(z) = − ln(1 − x z)ψ(z), hence Γ+(x)ψ(z) = (1 − z−1x)−1ψ(z)Γ+(x).
Now we want to find the action of Γ±. Here we need the fact that for a partition
λ = (λ1, . . . , λn) the corresponding vector can be written as
v(m)λ = ψλ1+m−1/2ψλ2+m−3/2· · · ψλn+m+(2n−1)/2ν
(m−n)
0 .
Therefore it is convenient to look at the action of Γ± on ψk. We can deduce this from
the commutation relations we proved earlier. We have Γ−(x)ψ(z) = (1 − zx)−1ψ(z)Γ−(x) = ∞ X n=0 znxn X k∈Z+12 zkψkΓ−(x) = X k∈Z+12 zk ∞ X n=0 xnψk−nΓ−(x)
Comparing coefficients of zk gives Γ−(x)ψk =P ∞ n=0x nψ k−nΓ−(x). Now we get Γ−(x)vλ(m) = ∞ X l1,...,ln=0 xl1+···+lnψ λ1−l1+m−1/2ψλ2−l2+m−3/2· · · ψλn−ln+m−(2n−1)/2v (m−n) 0 =X λ µ x|λ|−|µ|vµ(m),
because all other therms are zero or cancel. To see this assume that we have λk− lk <
λk+1, and further that λk − lk 6= λk+1 − lk+1− 1 otherwise the wedge product will be
zero. Now a therm in the sum corresponding to this situation cancels to the therm where λk− lk0 = λk+1− lk+1− 1 and λk+1− lk+10 = λk− lk+ 1 and the rest stays the same, thus in
the new therm we let the kth element in the wedge product go to the place where in the other term the (k+1)th element was and the (k+1)th element to the place of the previous kth element. This is possible because λk−lk< λk+1. These two therms cancel since they
have the opposite sign and the same power of x. Therefore only the therms corresponding to a interlacing partition remain. Because (Γ−(x)vλ(m), vµ(m)) = (Γ+(x)v
(m) µ , vλ(m)) and Γ−(x)v (m) λ = X λ µ x|λ|−|µ|vµ(m), we immediately get Γ+(x)v (m) λ = X λ≺µ x|µ|−|λ|vµ(m). Furthermore because of formula (4.1) we get
(Γ−(x1) · · · Γ−(xn)v (m) λ , v
(m)
µ ) = sλ/µ(x1, . . . , xn, 0, . . .).
There is one more operator we need to define, the grading operator. This operator L(m)0 is supposed to measure the size of the partition λ when it works on vλ(m). We define
this operator as L(m)0 = X k∈Z>m−12 kψkψk∗+ X k∈Z≤m−12 (kψkψ∗k− kI) = X k∈Z>m−12 kψkψk∗+ X k∈Z≤m−12 kψk∗ψk,
then it works exactly as desired. This operator operator satisfies the relations: L(m+1)0 ψkv (m) λ = ψk(k − (m + 1/2) + L (m) 0 )v (m) λ L(m−1)0 ψk∗vλ(m) = ψ∗k(−k + m − 1/2 + L(m)0 )vλ(m) Γ+(x)qL (m) 0 = qL (m) 0 Γ +(xq−1) Γ−(x)qL (m) 0 = qL (m) 0 Γ−(xq).
The last identities we get by looking at the action on basis vectors. For example we have q−L(m)0 Γ−(x)qL (m) 0 v(m) λ = q |λ|X λ µ x|λ|−|µ|q−|µ|vµ(m) = Γ−(qx)v(m)λ ,
this holds for each λ and therefore for all vectors of F(m).
Now we have all the knowledge to calculate some interesting result about plane par-titions. The vertex operators will help us to calculate the partitions function for plane partitions, even with boundary conditions.
4.3. Partition functions
To calculate partition functions for plane partitions it is important to understand the connection between the representation theory we did before and plane partitions. For this we will look at a plane partition on a slightly different way. We will cut the plane partition in diagonal slices, as in 4.2.
Then a plane partition corresponds to a sequence of partitions {λ(t)} with t ∈ Z such that λ(t) ≺ λ(t + 1) for t < 0 and λ(t) λ(t + 1) for t > 0. This interlacing property reminds us of the vertex operators Γ±, and indicates their usefulness.
We will look at the grand canonical ensemble for plane partitions weighted with q#boxes. Let us start with the simplest case, the partition function for unlimited plane partitions. Then the partition function is
Z = X
plane partitions π
q|π|,
where |π| is the sum of its elements, thus the number of boxes of π. With help of our vertex operators Γ± we can write this as
Z = ( ∞ Y t=0 qL(m)0 Γ −(1) ! qL(m)0 −∞ Y t=−1 Γ+(1)qL (m) 0 ! v0(m), v0(m)).
The product of Γ+(1)’s followed by Γ−(1)’s generate all possible sequences of interlacing
partitions, precisely corresponding to a plane partitions, while the operator qL(m)0 keeps track of the total volume of each plane partition.
In physics Γ±are called the transfer matrices and one usually uses the bra-ket notation.
Then the previous expression looks like * 0 ∞ Y t=0 qL0Γ −(1) ! qL0 −∞ Y t−10 Γ+(1)qL0 ! 0 + .
Because we used the mathematical notation this whole time we will continue using this notation.
The next step in the computation is to commute the operators qL(m)0 to the outside,
where we splits the middle one in half, thereafter we commute the Γ+ to the left and
the Γ− to the right. This results in
Z = ( Y n>0 Γ−(qn− 1 2) ! Y n≥0 Γ+(qn+ 1 2) ! v0(m), v0(m)) = Y n≥0, m>0 (1 − qn+m)−1( Y n≥0 Γ+(qn+ 1 2) ! Y n>0 Γ−(qn− 1 2) ! v0(m), v0(m)) =Y n>0 (1 − qn)−n, because Γ−(x)v0(m)= v (m) 0 and (Γ−(x)v(m)λ , v(m)µ ) = (Γ+(x)v (m)
µ , vλ(m)) the last inner
prod-uct becomes (v(m)0 , v(m)0 ) = 1. This is a famous result, known as MacMahon’s function, because Percy MacMahon was the first who calculated this partition function for plane partitions, better known as the generating function for plane partitions. Notice that in our case the generating function and partition function are exactly the same thing.
