Super frustrated lattice models

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Super frustrated lattice models

Tristan Kuen

January 14, 2016

Bachelors project mathematics and physics

Supervisors: prof. dr. Kareljan Schoutens, dr. Raf Bocklandt

Korteweg-de Vries Instituut voor Wiskunde & Instituut voor Theoretische Fysica Amsterdam

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Abstract

A model in statistical physics is studied in a general setting and then applied to the five platonic solids, where their vertices are the lattice sites. The Hamiltonion is defined as the anti-commutor of two supercharges: a raising and a lowering operator, which put or take away fermions on the lattice. As a result of the supersymmetry, the zero energy states become elements of the (co)homology of the supercharges. By calculating the homologies of the independence complex of a given lattice, its ground states can be found. In this way the number of ground states is found for all the platonic solids.

Title: Super frustrated lattice models

Authors: Tristan Kuen, tristankuen@gmail.com, 10211594 Supervisors: prof. dr. Kareljan Schoutens, dr. Raf Bocklandt Date: January 14, 2016

Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam

Science Park 105-107, 1098 XG Amsterdam http://www.science.uva.nl/math

Institute of Theoretical Physics Amsterdam Institute of Physics, Universiteit van Amsterdam Science Park 904, 1098 XH Amsterdam

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1 Introduction

The scope of this thesis is to understand a theory of two-dimensional lattice models which arise in statistical physics, and to expand it to certain other lattices. Ground state degeneracy, or superfrustration, is commonly found in the studied model. Par-ticularly interesting it becomes on the triangular lattice on a torus, for there it grows exponentially with the size of the lattice. A generic description of the amount of ground state on the triangular lattice has, however, yet to be found.

The problem was presented to me by professor Kareljan Schoutens, who had first pub-lished it together with Fendley and De Boer in 2003 [2]. The problem of counting the ground states is a rather mathematical one. For the one-dimensional chain of general length, they could determine the complete energy spectrum, using the Bethe ansatz. For other lattices, for example the square lattice on the torus, they managed to determine the ground states [4]. The approach by Schoutens and his collegues, among whom Jakob Jonsson and Liza Huijse, was to use (co)homology.

In this thesis the same method is applied to a few examples of lattices with spherical topology. The mathmatical side of the thesis therefor consists mastering some theory about simplicial homology. Raf Bocklandt was glad to help me with that.

Chapter 2 is dedicated to describing the model and defining (co)homology. In chap-ter 3 the model is applied on a one-dimensional chain, already solved by Fendley and Schoutens. In chapter 4 and 5 it finally is applied on spherical lattices: the platonic solids and the bucky ball. For the latter, unfortunately, the solution is only obtained partially.

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2 Preambles

In this chapter several concepts are outlined and defined, which will be important throughout the rest of the thesis. In the first section the physical model is defined, and the fundamental features are discussed. Then two mathematical sections follow. In the first simplicial (co)homology is defined. The second one introduces a lemma that is used extensively to determine homology. These sections are presented at the start of the thesis, because they follow standard mathematical constructions, so with sufficiant knowledge they may be skipped in reading.

2.1 The model

Like Fendley, Schoutens and De Boer [2], we define a physical grid model by describing its Hamiltonian. Let G = (V, E) be an arbitrary simple graph. The graphs typically arising in this thesis will be planar (or wrapped around a sphere) and in some way symmetrical. We can interpret this graph physically as a lattice of atoms in a solid or molecule. On every atom there is place for one more fermion (an electron, in the simplest case). Spin does not play a role. In our model the fermions have the property that they cannot be situated at neighbouring sites, therefore called hard-core fermions. Every possible fermion configuration will be a quantum state of the whole lattice, and thus a basis vector in the state space H (Hilbert space).

First, we define the usual fermion operator c†_i, which adds a fermion on site i, together with its hermitian conjugate, ci, which deletes a fermion on i. These satisfy the following

usual fermionic relations: {ci, c †

j} = δij and {ci, cj} = {c †

i, cj†} = 0. From the second

relation follows that both operators are nilpotent, i.e. only one fermion is allowed on a lattice site. A projection operator P :=

Q

j next to i(1 − c †

jcj) is introduced, in order to

provide for the hard core property. Clearly, P = 1 only holds if the neigbouring sites

of i are empty in the state P works on.

Second, the supercharges Q and Q† are defined as Q := P ciP and its hermitian

conjugate Q† :=P c†_iP. The term supercharge comes from the supersymmetry that

they carry; Q† is the raising operator, adding a fermion to the lattice, and Q is the low-ering operator, deleting one. That is, F :=P

ic †

ici, which gives the fermion number of a

state, acts like F (Q |φi) = (f − 1)Q |φi and F (Q†|φi) = (f + 1)Q†_{|φi, for F |φi = f |φi.}

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Using above definitions, the Hamiltonian is defined as the anti-commuter of Q and Q†:

H := {Q, Q†} = QQ†+ Q†Q. Working out the Hamiltonian with given relations for ciand c

†

i, the following is obtained:

H =X i X j next to i Pc † icjP<j>+ X i P.

Here the fist term gives the energy for a fermion to hop to a neighbouring site (as long as that one is empty as well as all of its neighbours), the second term can be interpreted as a nearest-neighbour repulsion. It also follows that H is positive definite, so its eigen-values have E ≥ 0.

The way the Hamiltonian is defined provides a characteristical spectrum of energy eigenvalues, that makes studying this model interesting particularly interesting. First, note that both Q and Q†commute with H. If |φi is an eigenstate, i.e. H |φi = Eφ|φi =,

H, with fermion number f − 1. In the same way this holds for Q†|φi is an eigenstate with fermion number f + 1. The supersymmetry infers a very particular structure in the energy spectrum. All eigenstates with E ≥ 0 form doublets: they appear in pairs, having the same energy but differing by one in fermion number. All zero energy states, or groundstates, are singlets. Another way of saying: doublets are two dimensional rep-resentations of the supersymmetry algebra, singlets are one dimensional reprep-resentations. This feature provides a useful tool to count the number of ground states, namely the Witten index. The Witten index is generally defined as

W := tr[(−1)Fe−βH],

i.e. the partition function with an extra factor ±1, dependent on the number of fermions. In the basis of the energy eigenstate, the operator has as diagonal elements ±1 for the states with E = 0, and ±1 times an power of e for states with E > 0. But because the latter states appear as doublets in the spectrum, i.e. they have the same energy but differ 1 in the fermion number, they appear in the matrix once with and once without a minus. Consequently, they cancel out in the sum. So we may as well leave the e power out of the sum; now regrouping states with the same fermion number, we write

W =X

f

(−1)fdim(Hf),

where Hf is the subspace of the Hilbert space spanned by the f -particle states, its di-mension is just the number of linearly independent states with f particles. Because states with E > 0 cancel out, the only non-zero contribution to this sum comes from the ground states. If for a system all the ground states have the same f , |W | clearly equals the number of ground states. But it may happen that several ground states differ an

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odd number in f , then |W | becomes less than the actual number. Hence W is a lower boundary for the number of ground states.

Now all of the properties of the model only depend on the choice of the lattice. It turns out that for bigger systems it is not only difficult to determine the whole spectrum, but also the ground states and even the Witten index, for the number of possible systems grows exponentially with the size of the lattice.

2.2 Simplicial homology and cohomology

In this section simplicial homology and cohomology are defined, following Munkres [7]. A few notational matters to start with. A n-simplex is the convex hull of n + 1 vertices vi, and will be denoted as Vn+1_i vi, but more often as [v1, ..., vn+1]. An m-face of an

n-simplex σ is any m-simplex, whose vertices are also vertices of σ (0-faces are usually referred to as vertices, 1-simplices as edges and (n − 1)-simplices as facets). A simplicial complex K is a collection of simplices, such that any face of a simplex σ ∈ K is also in K, and the intersection of any two simplices σ1 and σ2 is either empty or a face of both

σ1 and σ2.

Let the ith chain group Ci(K, Z) be defined for a simplicial complex K, as the free

abelian group over Z ofi-simplices in K. We have to orientation into account though: as an i-simplex can be represented in multiple ways as a sequence of vertices [v1, ..., vi+1],

therefore we define the relations:

[v1, ..., vi+1] = [s(v1), ..., s(vi+1)] if s is an even permutation,

[v1, ..., vi+1] = −[s(v1), ..., s(vi+1)] if s is odd.

Now let Ki be the set of all i-simplices σj in K (indexed with j), then Ki forms a basis

for Ci(K, Z) and every element in Ci(K, Z) can be written as a formal sum of σj in Ki:

c = X

σj∈Ki

cjσ,

where the coefficients ci are elements from Z (but in general can be elements from any

field). Elements from Ci(K, Z) are called i-chains.

The ith boundary operator ∂i is defined as follows:

∂i : Ci(K, Z) → Ci−1(K, Z) (2.1) X σj∈Ki cjσ 7→ X σ∈Ki cj∂i(σj); (2.2)

where ∂i is defined on the basis elements of Ci(K, Z) as

∂i(σj) = ∂i([v1, ..., vi]) = i

X

k=1

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where ˆvk means that the vertex vk is omitted in the sequence. The most basic result of

this definition is that ∂i−1◦ ∂i = 0.

Now the chain complex is the sequence of chain groups Ci together with the boundary

operators ∂i, as shown in figure 2.1. The assumption that the first chain group is zero,

comes from the fact that we only treat finite simplicial complexes, so the simplices have a maximum size; correspondingly, also the last chain group is zero, as the boundary of a vertix is empty. Sometimes with chain complex is meant L

iCi(K, Z), without further

notice. In this way we can talk about ∂, as a homomorphism on the chain complex itself (thus an endomorphism); clearly for any i ∂i is just a homomorphism from one subspace

of the complex to another. The same holds for homology and cohomology, introduced below. 0 ∂ ... Ci(K, Z) ∂i+1 Ci−1(K, Z) ∂i ∂i−1 _... 0 ∂

Figure 2.1: The chain complex of K

The cycles Zi(K) in Ci(K, Z) are the elements for which ∂i gives zero, so Zi(K) :=

ker(∂i). The boundaries Bi(K) are the elements of Ci(K, Z) which are in the image of

∂i+1. Then the ith homology group Hi of K is defined as

Hi(K, Z) := Zi(K)/Bi(K).

Closely related, and often the same (as groups), is the construction of cohomology. Let the cochain complex be the vector space of linear functions from the chain complex to the field Z, i.e. the dual of the chain complex, indexed with superscript. An n-cochain σn _{thus works on an n-chain c}

n, with σn(cn) ∈ Z. A natural basis for the nth

cochain group is the set of n-cochains which are 1 on one n-chain and 0 on everything else. The coboundary operater then is defined through how ∂†(σn_{) works on (n +}

1)-chains: ∂†(σn_)(c

n+1) = σn(∂n+1cn+1). So the coboundary makes (n + 1)-cochains from

n-cochains, whereas the boundary operator makes (n − 1)-chains from n-chains.

0 ∂ ... † Ci_{(K, Z)} ∂i−1 Ci+1_{(K, Z)} ∂i ... ∂i+1 0 ∂†

Figure 2.2: The cochain complex of K

The resulting reversed sequence, shown in figure 2.2, is the cochain complex. Cocy-cles and coboundaries are defined in a totally consequent way, as the kernels respec-tively the images of the coboundary operator. The ith cohomology group then becomes Hi_{(K, Z) := Z}i/Bi. In the context of simplicial cohomology it makes sense to neglect the fact that cochains are functions, and just talk about sums of simplices.

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2.3 The tic-tac-toe lemma

Calculating the (co)homology directly is often quite complex. In this section the tic-tac-toe lemma is introduced, a tool to calculate the (co)homology in two steps. The lemma arises in the theory of spectral sequences. This proof is not presented, but can be found in Bott and Tu [1]. For concistence, the notation and examples in this section will always refer to cohomology, not homology, as this is mostly used further on in the thesis. First, the notion of a doubly graded sequence is needed. Let the vertices from the simplicial complex be partitioned into two different sets S1 and S2.This then naturally

‘divides’ the boundary operator in two: ∂† = ∂1+ ∂2, where ∂1 only works on S1 and ∂2

only on S2 (note that that the superscript here does not refer to the group it acts on,

as in the previous section). In this way a doubly graded chain complex, usually denoted with K, is obtained, where ∂1 works horizontally and ∂2 vertically, see 2.3.

K0,0 K1,0 K2,0 K3,0 ... K0,1 K1,1 K2,1 K3,1 ... K0,2 K1,2 K2,2 K3,2 ... ... ... ... ... ... ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 _∂1 _∂1 _∂1 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2

Figure 2.3: A doubly graded cochain complex.

If this is compared to the simple chain complex from the previous section, one sees that Ci =

L

j+k=iKj,k, i.e. the sum along a diagonal in figure 2.3. Now H2, the

co-homology with respect to ∂2, can be calculated, by treating each column of the doubly

graded cochain complex as a a cochain in itself. This results in a lattice of cohomology groups, see figure 2.4.

Now it is possible to calculate H12 _{:= H}1_(H2_{), i.e. the cohomology with respect to ∂}1

on the lattice H2_{, now treating every horizontal line as a cochain complex. Note that}

the cochains, which are cocycles with respect to H2, are not directly related to the sim-plicial complex anymore; first calculating H1 _{and then H}1 _{may also result in something}

different. The tic-tac-toe lemma describes under what condition it makes sense to use this approach.

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H2 0,0 H1,02 H2,02 H3,02 ... H2 0,1 H1,12 H2,12 H3,12 ... H2 0,2 H1,22 H2,22 H3,22 ... ... ... ... ... ... ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂1 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2

Figure 2.4: The cohomology with respect to ∂2 of a doubly graded cochain complex.

Lemma 1. H12 equals (i.e. is isomorphic to) H, only if all but one line in H12 are zero. So if there exists a k, such that if H_j,i12= 0 for all i 6= k, then H_j,k12 = Hj.

It often proves not so difficult to find a lattice partition such that the resulting H12

satisfies the condition of the lemma, for example, if S2 is an independent set (in other

words, there are no edges between vertices in S2), H2 will become zero on every but the

first line. However, it may still be difficult to calculate H1 _{from H}2_.

(Co)chain complexes with their (co)homologies, as well as doubly graded variants, are examples of spectral sequences. A spectral sequence is a collection of three sequences:

(Ei)i≥0 a sequence of objects called sheets,

(di : Ei → Ei)i≥0 a sequence of endomorphisms called boundary maps,

(φi : Ei → H(Ei−1))i≥1 a sequence of isomorphisms.

A regular (co)chain complex C and its (co)homology H form a spectral sequence in the sense that E0 = C and Ei = H for i > 0, di = ∂ for all i, and φi = Id, the identity, for

all i. This works because ∂ working on H is, by definition of H, the zero map, so taking the homology again doesn’t change anything anymore.

The doubly graded construction gives a slightly more interesting spectral sequence. Here E0 = C, E1 = H2 and Ei = H12 for i > 1; d0 = ∂2, d1 = ∂1 and for i > 1 di can be

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3 A one-dimensional chain

In this chapter the model, defined in §2.1, is applied to a (closed) chain. This is not the lattice of particular interest, but several constructions and intuitions follow from having treated the one-dimensional case. First, for an explicit example the Hamiltonian and its eigenstates will be calculated. In the second section the connection with homology is made clear. In the last section the homology for a general chain of length n is determined.

3.1 The hexagon

Let G be the closed chain of six sites (i.e. a hexagon), see figure 3.1.

v1 v5 v6 v2 v3 v4

Figure 3.1: The hexagon lattice

To specify the states of the state space H, Dirac notation is used. Let |0i denote the empty state, |1i the state with a fermion on site v1, |3, 6i the state with a fermion on

sites v3 and v6 and so on. In order to consequently apply the fermionic property that

|a, bi = − |b, ai, basis vectors are fixed. Underneath the chosen basis vectors are listed, f denoting the number of fermions and fd the dimension of the subspace of H with f

fermions. The Witten index can already be calculated to be W = 1 − 6 + 9 − 2 = 2, so at least two ground states must be found.

f df State

0 1 |0i

1 6 |1i , |2i , |3i , |4i , |5i , |6i

2 9 |1, 3i , |2, 4i , |3, 5i , |4, 6i , |5, 1i , |6, 2i , |1, 4i , |2, 5i , |3, 6i 3 2 |1, 3, 5i , |2, 4, 6i

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In order to find the energy eigen states, the Hamiltonion will be written in matrix form. This is done by first letting H work on each of the basis states - due to symmetry one of the states per f have to be worked out, and only two for f = 2.

H |0i = (QQ†+ Q†Q) |0i = Q(|1i + |2i + |3i + |4i + |5i + |6i) + 0 = 6 |0i ,

as no fermion can be deleted from an empty lattice, so Q |0i = 0. The same is done for a one-particle state.

H |1i =(QQ†+ Q†Q) |1i = Q(|3, 1i + |4, 1i + |5, 1i) + Q†|0i =

|1i − |3i + |1i − |4i + |1i − |5i + |1i + |2i + |3i + |4i + |5i + |6i = 4 |1i + |2i + |6i . For the two-particle state |1, 3i is considered, on which a third fermion can be added; and |1, 4i, which is already maximal:

H |1, 3i =(QQ†+ Q†Q) |1, 3i = Q |5, 1, 3i + Q†(|3i − |1i) = |1, 3i − |5, 3i + |5, 1i + |5, 3i + |6, 3i + |1, 3i − |3, 1i − |4, 1i − |5, 1i = 3 |1, 3i + |1, 4i − |3, 6i , and H |1, 4i =(QQ†+ Q†Q) |1, 4i = 0 + Q†(|4i − |1i) = |6, 4i + |1, 4i + |2, 4i − |3, 1i − |4, 1i − |5, 1i = 2 |1, 4i + |1, 3i − |4, 2i + |6, 4i − |5, 1i .

Finally, the three-particle state, represented by |1, 3, 5i:

H |1, 3, 5i = (QQ†+ Q†Q) |1, 3, 5i = 0 + Q†(|3, 5i − |1, 5i + |1, 3i) = 3 |1, 3, 5i Below, H is expressed as a matrix relative to the standard basis, where |ii = ei+1,

|i, i + 2i = ei+7, |i, i + 3i = ei+13 and |i, i + 2, i + 4i = ei+16.

H =                                 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 4 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 4 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 4 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 4 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 1 0 −1 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 −1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 −1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 −1 −1 0 0 0 0 0 0 0 0 0 1 1 0 −1 −1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 −1 −1 0 2 0 0 0 0 0 0 0 0 0 0 −1 0 1 1 0 −1 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3                                

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The eigensystem is obtained by simply diagonalising above matrix. Note that this procedure can only be followed for very small systems, as the size of the matrix blows up with the number of lattice sites. The eigenvalues are presented in figure 3.2, see also Huijse and Schoutens (2008) [4]. Overlapping circles have the same E, the dotted lines show which states forms doublets. As expected, there are two ground states, both with two particles. 1 2 3 f 0 1 2 3 4 5 6 E

Figure 3.2: Energy spectrum of the hexagon chain The two ground states are:

|αi =2 |1, 3i + |2, 4i − |3, 5i − 2 |4, 6i − |5, 1i + |6, 2i − 3 |1, 4i + 3 |3, 6i |βi = |1, 3i + 2 |2, 4i + |3, 5i − |4, 6i − 2 |5, 1i − |6, 2i − 3 |1, 4i − 3 |2, 5i .

3.2 Ground states as harmonic elements

In this section the constructions of homology and cohomology are linked to ground states of the Hamiltionian in the supersymmetric model. The example of the hexagon, described in the previous section, is calculated again using the new technique.

For a graph G = (V, E) the independence complex Σ(G) be defined as the simplicial complex which contains all the independent sets of G as simplices. An independent set I in G is a set of vertices v ∈ V , such that E of G(I) is empty, i.e. there are no edges between points in I. In this way, if n = #(I),V

v∈Iv is an n-simplex in Σ(G).

Now each each element in Σ(G) represents exactly a basisvector in H of the model -an n-simplex represents a state with n + 1 fermions. Then the ith chain group of Σ(G)

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Ci(Σ(G), C) equals Hi+1, the subspace of the Hilbert space spanned by the states with

i + 1 particles. Note that the orientation of the simplices works the same as for states. Furthermore, the chain group is takes over C instead of C, for which it was defined in §2.1. This is needed to make it a complete Hilbert space. The construction of the homology is completely analogous.

The supercharge Q makes from a n-particle state an n − 1-particle state, so Q : Ci(Σ(G), C) → Ci−1(Σ(G), C), and Q of a one-particle state is a zero-particle one, so

Q : C0(Σ(G), C) → C. Thus the Hilbert subspaces form an augmented chain complex,

on which Q acts as a boundary operator. Note that Q handles orientation in the same way ∂i does.

Similarly, Q†is a coboundary operator on the augmented dual chain complex - the same chain complex, but with Ci_{(Σ(G), C) = hom(C}i(Σ(G), C), C) instead of Ci(Σ(G), C).

The elements of Ci_{(Σ(G), C) are of course linear operators. Thus instead of elements}

of the Hilbert space, they are elements of its dual, which could be notated with bra’s in Dirac notation. However, a pairing vi = vi is implied, that considers both spaces the

same.

Here vi _{are of course the basisvectors of C}0_{(Σ(G), C), which are the vertices of Σ(G).}

The basisvectors of C1_{(Σ(G), C) then are v}i∧ vj _{and so on (although an isomorphism is}

implied, sub and superscript to denote homology vs. cohomology will still be used when relevant). From now on, for sake of neatness, reference to C will be omitted in context of homology, i.e. the ith (co)chain group will be notated as Ci(Σ(G)) and Ci(Σ(G))

respectively.

It is clear that the ground states with i − 1 particles, must be the harmonic elements of the ith homology group of Σ(G), as they must vanish under both Q and Q†, and at the same time cannot be in the image of Q and Q†. This means that if the homology and cohomology of a Σ(G) can be determined, its groundstates will be easily recognised. In this way the ground states of the hexagon chain will now be determined. Write_{7 for} G. v1 v4 v2 v6 v3 v5 v1 v4 v2 v6 v3 v5 v1 v3 v5 v4 v6 v2

Figure 3.3: The hexagon, its independence complex and the orientated independence complex

Herefrom the (augmented) chain and cochain complexes are derived, as presented in figures 3.4 and 3.5. The resulting reduced homology and cohomology are presented in

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figures 3.6 and 3.7. Further on in the thesis, the specifications ‘augmented’ and ‘reduced’ will be implied but omitted in the description.

0 C2(Σ(7), C) = h[v₁, v3, v5] [v2, v4, v6]i C1(Σ(7), C) Q = h[v1, v3], [v3, v5] [v5, v3], [v2, v4] [v4, v6], [v6, v2] [v1, v4], [v2, v5] [v3, v6]i C0(Σ(7), C) Q = h[v₁], [v2], [v3] [v4], [v5], [v6]i C Q 0,

Figure 3.4: The augmented chain complex of the hexagon.

0 _C C0(Σ(7), C) Q† C1(Σ(7), C) Q† C2(Σ(7), C) Q† 0, Figure 3.5: The augemented cochain complex of the hexagon.

As the first (co)homology group is the only non-vanishing one, that one will be elab-orated. The cycles in C1(Σ(7) that are not boundaries are easy to identify from figure

3.3(c), namely the elements [v1, v3] + [v3, v6] − [v4, v6] − [v1, v4], [v3, v5] − [v2, v5] − [v6, v2] −

[v3, v6] and [v5, v1] + [v1, v4] − [v2, v4] + [v2, v5], together with all possible linear

combina-tions. These three are not linearly independent however, because the first plus the last differ a boundary from [v1, v3, v5] and [v2, v4, v6]. Thus two of them span ˜H1.

˜ H2(Σ(7)) = 0 ˜ H1(Σ(7)) Q ∼= C2 ˜ H0(Σ(7)) Q = 0 0 Q

Figure 3.6: The reduced homology of the hexagon.

The first cohomology group is a little bit less transparant. The co-cycles are the 1-simplices that do not appear in any 2-simplex, [v1, v4], [v2, v5] and [v3, v6], and for each

triangle vertex a tuple consisting of the in and outgoing edge, e.g. for v1 [v5, v1] − [v1, v3].

Per triangle, we have that one of these tuples is linearly dependent of the other two. The co-boundaries are given of the form (for v1) [v5, v1] − [v1, v3] − [v1, v4]. If these are

quotiented out, the following equivalence relations are obtained: [v5, v1] − [v1, v3] = [v1, v4] = −[v2, v6] + [v4, v6],

[v1, v3] − [v3, v5] = [v3, v6] = −[v4, v6] + [v6, v2] and

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0 Q H˜0_(Σ(₇₎₎ † = 0 ˜ H1_(Σ(₇₎₎ Q† ∼= C2 ˜ H2_(Σ(_7)). Q† = 0 Figure 3.7: The reduced cohomology of the hexagon.

but because the third is a linear combination of the first two, only two equivalence classes remain, so ˜H1_(Σ(_{7)) ∼}₌_C2_.

Two “square” cycles will generate the homology, but to let them be elements of the cohomology some boundaries of the triangles must be added. The folllowing cycles are elements of the homology:

|αi = [v2, v4] + [v4, v6] + [v6, v2] − [v1, v3] − [v3, v5] − [v5, v1] + 3([v3, v6] − [v4, v6] − [v1, v4] + [v1, v3]) |βi = [v1, v3] + [v3, v5] + [v5, v1] − [v2, v4] − [v4, v6] − [v6, v2] − 3([v5, v1] + [v2, v5] + [v1, v4] − [v2, v4]),

and if rewritten in the following way:

|αi = [v1, v3] − [v3, v5] − [v3, v6] + [v1, v3] − [v5, v1] + [v1, v4] − [v4, v6] + [v6, v2] − [v3, v6] − [v4, v6] + [v2, v4] + [v1, v4] + 5([v3, v6] − [v1, v4]) |βi = [v2, v3] − [v5, v1] + [v1, v4] + [v1, v3] − [v5, v1] + [v2, v5] + [v2, v4] − [v4, v6] + [v1, v4] + [v2, v4] − [v6, v2] + [v2, v5] − 5([v1, v4] + [v2, v5]),

it is clear that they are also elements of the cohomology. So they are harmonic elements. One can check that they are the same as the states described in the previous section.

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3.3 The N -chain

In this section the cohomology of the closed N -chain is dealt with, as in [2], using the tic-tac-toe lemma described in section §2.3. Hereby only the fermion number of the ground states can be found, to discover the specific eigenstates for the general N -chain one can use the Bethe-ansatz, which will also give all the other energy eigenstates. Notation will be as described, with Q1 acting on S1, Q2 acting on S2 and Q1+ Q2 = Q†.

Starting with a closed N -chain with N = 3n (for integer n), we divide the chain into two sublattices. We chose S2 to be the sublattice consisting of every third site, and S1

the remaining 2n sites. If we would write down the doubly graded co-chain complex of the independence complex, in the first row we would get all the states with zero particles on S1, and zero until n particles on S1, because S1 contains n two-chains, on which can

be only one particle.

... ...

Figure 3.8: The closed chain of N = 3n sites

Note that if there is a particle on a site of S2, the two neighbouring in S1 must be

empty. This state will not be in any cohomology group of Q2, for it is the coboundary of a state with no particle on that site. So the elements of the cohomology of Q2 _must

have an empty S2_{, and every site in S}

2 must have an occupied neighbour, to prevent it

to be a coboundary. This implies that the condition of the tic-tac-toe lemma, that only one row of H12 _{can contain non-zero groups, must be satisfied. There are only two such}

states:

... ...

Figure 3.9: The two cohomology cycles of Q2

So all cohomology groups vanish, exept for H2

n,0, which is generated by two (n −

1)-simplices. Now computing H_n,012 is trivial: because both H_n−1,02 and H_n+1,02 are zero, every element of H2

n,0 must be a cycle with respect to Q1 and cannot be a coboundary.

Thus H12

n,0 equals Hn,02 , and is two dimensional, which implies that the closed N -chain

has two zero energy states with n particles.

For the N -chain with N = 3n − 1 particles, we partition the lattice in the same way, but now on S1 there are only n − 1 two-chains and 1 one-chain. The property that every

site in S2 must have an occupied neighbour, yields here as well. There are n states with

n particles that have that effect, because for every site in S2 there is such a state in which

both neighbours are occupied. There are n − 1 such states with n − 1 particles, for every two-chain in S1 there is a desired state which leaves it empty, but the one-chain has to

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be always occupied. This leaves us with an n-dimensional H_n,02 and an n − 1-dimensional H2

n−1,0, all the other groups are zero. Because Q1 on an element in Hn−1,02 gives one in

H2

n,0, Hn,012 must be one-dimensional, so there is only one ground state and it contains n

particles.

... ...

Figure 3.10: The closed chain with N = 3n + 1 sites, an n-particle state In the same fashion, for the N -chain with N = 3n + 1, H2

n,0will be n − 1-dimensional,

and H_n+1,02 will be n-dimensional, which yields a H_n+1,012 that is one-dimensional, so there is also only one ground state, with n + 1 particles.

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4 The platonic solids

In this chapter the zero energy states on the platonic solids are determined, as an exercise of studying the model in two dimensions on a spherical topology. They are arranged in order of the number of vertices. In each paragraph first the Witten index is stated, then the number of ground states follows as a proposition, and finally for all, but the dodecahedron, a basis for the harmonic elements is given. The numbers of ground states is determined using cohomology. The chain complexes of which the cohomologies and homologies are calculated, are always based on the independence complexes of the given lattice. The field used is Z; no complex numbers are needed, but one can without alterations replace Z by C, to be more precise concerning the physics. We note on forehand that the zero’th homology is always empty - there are no (co)cycles - so it won’t be discussed in every section. To be consistent we notate the harmonic elements in simplex-notation, instead of ket-notation.

4.1 The tetrahedron

In figure 4.1 the tetrahedron with its independence complex is presented. The only possible states are either empty or have one particle, so |W | = |1 − 4| = 3.

Proposition 1. The tetrahedron has three one-particle ground states.

Proof. As H2 _{is zero, every element in H}1 _{is a cocycle. The coboundaries are generated}

by [v1] + [v2] + [v3] + [v4], so H1 must be three-dimensional.

It is easy to see that a basis for the ground states is {[v1] − [v2], [v1] − [v3], [v1] − [v4]}.

v1 v2 v3 v4 v1 v2 v3 v4

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4.2 The octahedron

On the octahedron states with up to two particles are allowed, |W | = |1 − 6 + 3| = 2. Proposition 2. The octahedron has two two-particle ground states.

Proof. In C1 _{the cocycles are generated by [v}

1] + [v6], [v2] + [v4] and [v3] + [v5]. The

coboundaries are generated by [v1] + [v2] + [v3] + [v4] + [v5] + [v6], so H1 is two-dimensional

(setting [v1] + [v2] + [v3] + [v4] + [v5] + [v6] = 0 makes one cocycle linearly dependent

of the other two). In C2, every element is a cocycle, but also a coboundary, so H2 is

zero.

The difference of any two vertices makes a cycle in C1 (e.g. Q(v1− v2) = 0, so a basis

for the harmonic elements is {[v1] + [v6] − [v2] − [v4], [v1] + [v6] − [v3] − [v5]}.

v1 v2 v6 v4 v3 v5 v1 v6 v2 v4 v3 v5

Figure 4.2: The octahedron and its independence complex

4.3 The cube

On the cube states with up to four particles are allowed. We have W = 1−8+16−8+2 = 3. Although the homology may be calculated directly, the tic-tac-toe lemma will give a quicker way.

Proposition 3. The cube has three two-particle ground states.

Proof. Let S2 be {v2, v8}, which is a set of isolated points, and S1be {v1, v3, v4, v5, v6, v7}.

As we have seen in the proof of the 1-dimensional chain, both v2 and v8 must have

an occupied neighbour to remain in the cohomology of Q2_{, for the isolated site has a}

vanishing homology. This means that both from {v1, v3, v6} and {v4, v5, v7} one site

must be occupied. The only configurations which allow this are [v4, v6], [v3, v5] and

[v1, v7]. So H0,22 = Z3, where H0,22 is the cohomology group of Q2 with zero particles

on S2 and two on S1. All the other cohomology groups are zero, so automatically

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v3 v2 v4 v1 v8 v5 v7 v6 v1 v3 v6 v8 v7 v5 v4 v2

Figure 4.3: The cube and its independence complex

To determine the ground states themselves, we examine H2 and H2 closer. The

cocycles Z in C2 _{are edges [v}

1, v7], [v2, v8], [v3, v5] and [v4, v6], together with the sums

of three edges of a tetrahedron that come together in one vertix, e.g. [v1, v8] + [v3, v8] +

[v6, v8]. For each tetrahedron there are four of such sums, but only three are linearly

independent.

The coboundaries B in C2 are the sums of all edges that come together in one point, e.g.

[v1, v8] + [v2, v8] + [v3, v8] + [v6, v8]. In Z/B the coboundaries are zero, therefore three

cocycles remain:

[v3, v1] + [v6, v1] + [v8, v1] =[v1, v7] = −[v2, v7] − [v4, v7] − [v5, v7]

[v1, v3] + [v6, v3] + [v8, v3] =[v3, v5] = −[v2, v5] − [v4, v5] − [v7, v5]

−[v1, v6] − [v3, v6] − [v8, v6] =[v4, v6] = [v2, v4] + [v5, v4] + [v7, v4],

because [v2, v8] is linearly dependent of the other three cocycles.

Elements of the corresponding homology group are easier to identify: in figure 4.3 we see six two-cycles that are not boundaries, namely all trapzoids between the two tetrahedrons. Clearly they are not linearly independent (at most three can be). A set of ground states thus reads:

3[v1, v7] + [v5, v3] + [v2, v8] + [v4, v6] + [v3, v1] + [v8, v1] + [v6, v1] + [v7, v2] + [v7, v4] + [v7, v5], 3[v4, v6] + [v1, v7] + [v3, v5] + [v8, v2] + [v2, v4] + [v5, v4] + [v7, v4] + [v6, v3] + [v6, v8] + [v6, v1] and 3[v3, v5] + [v2, v8] + [v4, v6] + [v7, v1] + [v1, v3] + [v6, v3] + [v8, v3] + [v5, v2] + [v5, v4] + [v5, v7].

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4.4 The icosahedron

In figure 4.4 the icosahedral lattice G = (V, E) is represented with coordinates, it con-sists of two outer points v1 and v12, and two pentagons (later they will be referred to

as rings). The independence complex is obtained by switching the two pentagons and permuting its vertices, and adding six edges to adversively situated vertices (the curved lines in the figure). Then Σ(G) contains 20 2-simplices, 36 1-simplices and 12 0-simplices (vertices), this gives us W = 1 − 12 + 36 − 20 = 5.

We can understand this number also from the Euler-characteristic from the great icosahedron. The great icosahedron consists of all triangles that are not in the icosa-hedron, and which edges are also not in the icosahedron. Thus the difference with the independence complex of the icosahdron, as shown below, is that it does not contain the six edges between two opposite points, and has its triangles intersect. The Euler-characteristic is defined (for any grpah) as the number of vertices minus the number of edges plus the number of faces. Evidently, this number is the same for the great icosahe-dron as for the icosaheicosahe-dron, namely 2. To obtain the Witten-index, we have to account for the empty state, as well as for the states on opposite vertices, which gives 1−2+6 = 5. The cohomology will be calculated using tic-tac-toe, with this time all sheets ex-plicitely written out.

v1 v2 v8 v12 v10 v5 v6 v7 v11 v3 v9 v4 v1 v7 v5 v12 v4 v8 v10 v9 v3 v2 v6 v11

Figure 4.4: The icosahedral lattice and its independence complex

Proposition 4. The icosahedron has six two-particle and one three-particle ground states.

Proof. The vertex set is partitioned into two subsets, S2 = {v1, v12} and S1 = V \ S2.

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every column the cohomology of Q2 is considered.

The first column first column totally vanishes: in K0,0 there are no coycles; in K0,1

only [v1] + [v12] is a cocycle, but also the boundary of the empty state; in K0,2 every

cocycle is a coboundary. K0,0 = Z K1,0 = Z10 K2,0 = Z25 K3,0 = Z10 0 K0,1 = Z2 K1,1 = Z10 K2,1 = Z10 0 K0,2 = Z 0 0 0 Q1 Q1 Q1 Q1 Q1 Q1 Q1 Q1 Q1 Q2 Q2 Q2 Q2 Q2 Q2 Q2 Q2

Figure 4.5: The doubly graded chain complex for Σ(G)

In the second column, K1,0 is generated by the states with one particle on the two

rings. In K1,1, a particle in the upper (i.e. in the independence complex) ring can only

form a two-state with v1, and one in the lower ring with v12. So Q2 maps every basis

element of K1,0 to a basis element in K1,1, so the kernel must be trivial and this leaves

us with H2

1,0 zero. At the same time this implies that every basis element in K1,1 is a

boundary, so H_1,12 must be zero as well.

In the third column, K2,0 is generated by all the 1-simplices between vertices on the

two rings, there are 25 different ones. All the 1-simplices that lie either on the upper or the lower ring itself (for instance [v7, v9], [v2, v4]), are send by Q2 to the triangles from

v1 to the upper ring and from v12 to the lower ring - these form a basis of K2,1. ‘Askew’

and ‘curved’ elements (i.e. one particle on each ring, see the independence complex in figure 4.4) in K2,0 are sent to zero and thus remain in the cohomology group, whereas

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Finally in the fourth column, K3,0 is generated by the ten triangles between the upper

and lower ring. These are all sent to zero, as there are no four-particle states. This results in H_3,02 = K3,0. H2 0,0 = 0 H2 1,0 = 0 H2 2,0 = Z15 H2 0,3 = Z10 0 H2 0,1 = 0 H2 1,1 = 0 H2 2,1 = 0 0 H2 0,2 = 0 Q1 Q1 Q1 Q1 Q1 Q1 Q1 Q2 Q2 Q2 Q2 Q2

Figure 4.6: H2 _{for the icosahedron with S}

2 = {v1, v12} and S1 = {v2, ..., v11}

Only on the first row non-zero cohomology groups are left, so the tic-tactoe lemma can be applied. Now H12 _{:= H}1_(H2_{) is calculated. Because H}2

1,0 is zero, there are no

coboundaries in H_2,02 relative to Q1. The cocycles are generated by all the ‘curved’ ele-ments, together with the 1-cochain which is the sum of all the ‘askew’ elements in H2

2,0,

so the dimension is 6. Every element in H2

3,0 is a cycle, so to find H3,012 we just have to quotient out the

coboundary relations. Every elementary 1-chain in H2

2,0 which consists of a state with

a particle in each ring, is sent to the sum of the two triangles that contain that edge. We can choose orientation conveniently such that it will give a minus sign for one, e.g. Q1_([v

2, v9]) = [v5, v2, v9] − [v2, v11, v9], which gives rise to the relation [v5, v2, v9] =

[v2, v11, v9] in H3,012. As this yields for all adjacent triangles in H3,02 , this leaves only one

equivalence class in H12 3,0.

So there are six two-particle ground states and one three-particle ground state. With some trial and error and symmetry considerations, the ground states are determined to

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0 Q 0 1 H12 2,0 Q1 = Z6 H12 3,0 Q1 = Z 0 Q1

Figure 4.7: H12 for the icosahedron with S2 = {v1, v12} and S1 = {v2, ..., v11}

be as follows. The two-particle sates are of the form:

|φ2i = −5 |1, 12i − |2, 10i − |3, 11i − |4, 7i − |5, 8i − |6, 9i

+ |1, 7i + |1, 8i + |1, 9i + |1, 10i + |1, 11i + |2, 12i + |3, 12i + |4, 12i + |5, 12i + |6, 12i ;

i.e. for each pair of opposite particles a, b, the state is 5[a, b] plus all the other opposite pairs, plus all vertices [i, a] and [i, b], for all possible i. The three-particle state is a sum of all triangles:

|φ3i = |1, 7, 9i + |1, 9, 11i + |1, 11, 8i + |1, 8, 10i + |1, 10, 7i

+ |7, 5, 9i + |9, 2, 11i + |11, 4, 8i + |8, 6, 10i + |10, 3, 7i + |5, 2, 9i + |2, 4, 11i + |4, 6, 8i + |6, 3, 10i + |3, 5, 7i + |2, 5, 12i + |4, 2, 12i + |6, 4, 12i + |3, 6, 12i + |5, 3, 12i .

It is easy to see that all triangles become zero onder Q†; and because every side appears twice in Q(|φ3i), with opposite signs, Q(|φ3i) = 0.

4.5 The dodecahedron

The dodecahedron consists of 20 vertices, 30 edges and 12 pentagons. Calculating all the possible states for a dodecahedral lattice, is already a tedious job to do by hand. The calculation done by a computer, the Witten index reads W = 1 − 20 + 160 − 660 + 1510 − 1912 + 1240 − 320 + 5 = 4. Luckily, it is possible to directly determine the homol-ogy without having to determine the chain complex first. Therefore the independence complex is not needed either.

As shown in figure 4.8, the dodecahedron allows states with up to eight particles -these correspond with the corner vertices of the five possible inscribed cubes. It is easy to see that eight is the limit, for every pentagon onlly allows two particles, and every vertex on the dodecahedron is part of three pentagons, so we have 2 × 12/3 = 8.

Proposition 5. The dodecahedron has four six-particle states.

Proof. For S2 _{the eight points of an inscribed cube, as in 4.8, are chosen, specific}

coor-dinates are not needed. The motivation is as follows: S2 must be an as large as possible independent set, because this reduces the amount of points and thus states on S1_{, which}

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Figure 4.8: The dodecahedron, and with an inscribed cube

simplifies the homology. Because all vertices in S2 _{are independent, the only states that}

do not vanish in the homology of Q2 are the ones for which every vertex in S2 has a neighbour, for the homology of an isolated state is empty (see section 3.3). There are only four such states, represented in 4.10. Here the vertices of S2 _{are represented by}

squares, the edges of the cube are drawn in red, because they are not edges in the do-decahedron, but in the chain complex of its indipendence complex.

0 H2 6,0 Q1 = Z4 0 Q1

Figure 4.9: H2 for the dodecahedron with S2 an 8-particle state

This results in a homology of Q2 as in figure 4.9. Applying the tic-tac-toe lemma, the full homology H0,6 = H0,612 = H0,62 = Z4.

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5 The truncated icosahedron

In this chapter the truncated icosahedron, or bucky call, is discussed. Other than the platonic solids, which served more as a theoretical exercise, the buky ball lattice appears in nature as the C60 molecule. It is, however, not clear, that or how, the model will give

an accurate description of this molecule. An attempt to calculate the cohomology using tic-tac-toe has turned out to be not so fruitful, as calculating H1 _{is rather non-trivial.}

In this chapter the procedure followed so far will be presented, with a conjecture for the final answer.

Figure 5.1: The truncated icosahedral lattice

5.1 The cohomology with tic-tac-toe

First some general notions about the lattice. The truncated icosahedron contains 60 vertices, 90 edges and 32 facets, of which 20 hexagons and 12 pentagons. The lattice allows states with up to 24 particles: as every vertex lies on a pentagon, and only two

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particles can be placed on every pentagon, 24 is indeed a maximum. Unfortunately, for states with only a few particles, there are already trillions of possibilities. Thus the Witten index turns out to be hard to determine, both algebraically as numerically.

Now a partition of the lattice vertices in S1 and S2 will be chosen, in such a fashion

that S2 is a set of independent points. As used several times throughout this thesis, the

(co)homology on an isolated site vanishes. This automatically makes the cohomology H2 _{zero everywhere, but on the first line (in the doubly graded cohomology lattice).}

Then it is immediately clear that the conditions of lemma 1 are satisfied. Our choice is presented in figure 5.2, where S2 is marked by squares and S1 by circles. From the

figure it is also clear that S1 consists of 18 pairs.

Figure 5.2: The truncated icosahedral lattice partitioned into S1 and S2.

The states that remain in the cohomology of Q2_{, are the configurations in which every}

square has an occupied neighbour. Every circle in S1 neighbours two squares in S2.

As there are 24 squares, at least 12 circles must be occupied to obtain a state in H2_.

Indeed, such a configuration is found in figure 5.3. In the figure the eight hexagons with three squares are shaded gray. Between each two oppositely situated shaded hexagons, the lattice partition is the same. Therefore the given configuration can be placed in four different ways, yielding four cocycles in H2

12,0. Note that the configuration is not

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a1 a2 b1 b2 c1 c2 f2 f1 e2 e1 d2 d1

Figure 5.3: State with 12 particles in H2

Thus so far, H2 _{is in the following shape:}

0 H2 12,0 Q1 = Z4 H2 13,0 Q1 H2 14,0 Q1 H2 15,0 Q1 H2 16,0 Q1 H2 17,0 Q1 H2 18,0 Q1 0 Q1

Figure 5.4: H2 for the dodecahedron with S2 an 8-particle state

where H2

12,0 is merely an assumption (it has not been explicitely prooved that there

are no other 12-particle states). Here the problem becomes clear: classifying all the states in H2 _{in order to calculate H}12_{. Until now, it is only clear that ground states}

have fermion number 12 up to 18. In the next section, H2

18,0 will be considered a bit

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5.2 The cohomology with 18 particles

In this section a conjecture about the shape of H12

18,0 is presented and motivated. First,

note that every element in H_18,02 is a cocycle, because H_19,02 is zero. The coboundaries in H2

18,0 are generated by the images of the elements in H17,02 .

For an element in H_17,02 , as for every element in H2, every square must have an occu-pied neighbour. An example of en element in H2

17,0is the state in figure 5.3, on which 5

additional fermions are placed on pairs b until f . On which site of the pair it is placed, bears no importance, as every obtained state will be in H_17,02 . Call this state σ. Then Q1(σ) = a1 ∧ σ + a2 ∧ σ. Now let τ be any element of H17,02 . It is clear that H18,02 with

the relations a1∧ τ = −a2∧ τ for all τ in H17,02 equals H18,012 .

So far it has not yet been made clear what the elements of H_17,02 and H_18,02 exactly are, and thus, how many there are, apart from the condition that every square must have an occupied neighbour. One may characterise all the states in H_18,02 by starting with the state in figure 5.3, and adding six more particles on sites a1 to f1.

A flip of one pair (i.e. switch the occupied and unoccupied site of a pair) is allowed, only if the resulting configuration is also an element of H_18,02 ; this is equivalent with the condition that the configuration without a particle on that pair, is an element of H2

17,0.

A flip makes from one state in H_18,02 another state in H_18,02 , but at the same time infers the coboundary condition that both states are equal, apart from a minus sign.

The conjecture now follows from notion that every state in H2

18,0 can be obtained by

starting with a random initial state, and then flip a path towards it. Every flip may allow other pairs to flip. Because all elements are linked by flips, all elements will be the same in H12

18,0.

Conjecture 1. The cohomology with 18 particles on the bucky ball is one-dimensional: H12

18,0= Z.

In figure 5.5 a state is presented that looks quite different from the one obtained from figure 5.3: the six pairs in the middle are oriented in exactly the opposite way. However, there is a path of flips from the one to the other. Unfortunately, the harder part of the proof is to show that there are no states that cannot be reached by flipping. We leave this for further research.

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a1 a2 b1 b2 c1 c2 f2 f1 e2 e1 d2 d1

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6 Discussion

The model studied in this thesis is still a source for much research. First of all, for the triangular and hexagonic lattice on the torus (and one can think of more examples) the homology is not yet understood. Secondly, the examples calculated and studied in this thesis touch on a new subject, namely lattices on a sphere.

However, from the calculated cases no generalisations of lattice size can be inferred. Some ways to generalise the studied spheric lattices include: inscribe a triangle on every triangle of the tetrahedron, octahedron or icosahedron, such that on every old triangle four new ones are created. This can be done indefinitely. Similarly, each pentagon of the dodecahedron can be cut into five triangles; in each of those again a triangle can be inscribed. On the cube one may draw a cubic or triangular grid of any size on every face, such that the vertices connect on the edges of the cube. All the resulting polyhedra, however, lack the property of the initial bodies, that every vertex has the same degree. One may guess that for enough extra vertices, this effect may wear off. For example, if one considers the icosahedron, every vertex added in the way described will have degree six, whereas only the initial twelve vertices will have degree five. But it could well be that those ‘disturbances’ play a key role in the resulting homology.

Aditionally, also the tools of calculating the homologies may be extended. There is a lot of theory about the calculation of homology, using specific sequences, of which the tic-tac-toe lemma is just one outcome. As was the case for the truncated icosahedron, tic-tac-toe prooved not sufficient enough to tackle the homology in a simple way.

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7 Populaire samenvatting

In de natuur zien we een enorme rijkdom aan verschillende stoffen, metalen, moleculaire stoffen, zouten. Elke van deze stoffen is opgebouwd uit dezelfde kleine bouwstenen: protonen, neutronen en elektronen. De kwantummechanica beschrijft het gedrag van hele kleine systemen, van enkele protonen of atomen. De statistische fysica probeert vat te krijgen op grote verzamelingen van deze kwantummechanische deeltjes, zo groot dat we ze waar kunnen nemen als de ‘normale’ stoffen die we kennen. Door naar de eigenschappen op atomair niveau te kijken, probeert men eigenschappen van de stof op normaal niveau af te leiden. Helaas zijn gebrek aan rekenkracht en de complexiteit van de formules altijd beperkende factor. Men moet veel concessies doen, factoren verwaar-lozen, zaken simplificeren, om tot een behapbaar systeem te komen, een model, waar dingen aan berekend kunnen worden. En dan moet men altijd weer rekening houden met de gedane concessies bij het duiden van het berekende.

Een andere manier om in het vakgebied te werk te gaan, is om juist bij een klein systeem te beginnen. Men definieert een model op atomair niveau, dat wil zeggen, op een wiskundige manier wordt beschreven hoe de deeltjes zich ten opzichte van elkaar gedragen. Dit gebeurt over het algemeen met behulp van de hamiltoniaan, de energie-operator. De hamiltoniaan vertelt hoe de energie ‘werkt’ in het systeem, en aan de hand van de energie kunnen de overige factoren worden berekend.

Precies zo ben ik te werk gegaan in dit bachelorproject. Mijn begeleider Kareljan Schoutens kwam met een model dat gedefinieerd was aan de hand van een hamilto-niaan. Dit model kun je vervolgens op verschillende roosters loslaten. Iets concreter, gegeven een willekeurig rooster van atoomkernen in een stof, beschrijft het model hoe de elektronen zich gedragen op dat rooster. Bij elke atoomkern kan een elektron zitten, onder de voorwaarde dat nooit twee naburige atoomkernen een elektron hebben. De elektronen stoten elkaar, zogezegd, dusdanig af, dat de aanwezigheid bij een atoomkern ervoor zorgt dat bij naburige kernen geen andere elektronen kunnen komen. Verder kun-nen elektrokun-nen op de kern zitten, of springen ze naar een naburige, ertussenin zweven kan niet.

Vervolgens kijken we naar welke energiewaarden bepaalde configuraties van elektronen op het rooster hebben. Kwantummechanica heeft ons al doen kennismaken met de notie van een discreet energiespectrum: een systeem kan in principe niet elke willekeurige energie hebben, maar slechts een waarde uit een discrete verzameling. Ons model heeft nog een extra eigenschap, het is namelijk supersymmetrisch. Dit betekent dat alle toe-standen met energie groter dan nul in paren voorkomen: twee toetoe-standen met dezelfde

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energie, waarvan de ene een elektron minder heeft dan de ander. Deze noemen we dou-blets. Alleen toestanden met nul energie komen als singlet voor (en ook alleen maar als singlet). Wat het model verder interessant maakt, is dat er op grote roosters heel veel toestanden zijn met nul energie. Dit is wat we superfrustratie noemen, een verschijnsel dat niet veel voorkomt (in modellen). Daarnaast is nog niet waargenomen bij bekende materialen.

Omdat de uitgangspunt een geconstrueerd model op micro-niveau was, met ook nog vrij specifieke eigenschappen, is niet meteen duidelijk of wat dit model beschrijft ook als zodanig in de natuur voorkomt. Er is nog genoeg onderzoek dat hieraan verricht kan worden. De roosters die in dit project zijn bestudeerd, zijn ook misschien niet allemaal relevant, in de zin van dat het bestaande systemen zijn. Wel zijn het oefeningen die de theorie in het algemeen vooruit kunnen helpen.

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Physical Review Letters 90, 120402, 2003.

[3] Huijse L and Schoutens K, Superfrustration of charge degrees of freedom, The Euro-pean Physical Journal B 64, 543-550, 2008.

[4] Huijse L and Schoutens K, Supersymmetry, lattice fermions, independence complexes and cohomology theory Adv. Theor. Math. Phys. 14, 643-694, 2010.

[5] Huijse L, Mehta D, Moran N, Schoutens K and Vala J, Supersymmetric lattice fermions on the triangular lattice: superfrustration and criticality, New Journal of Physics 14, 073002 (40pp), 2012.

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Super frustrated lattice models