Spanning trees, Lattice Green functions andCalabi-Yau equations

Tony Guttmann

MASCOS — University of Melbourne

Talk outline

Content

Spanning trees on a lattice

Lattice Green functions

Calculation of spanning tree constants

Digression: Connection between spanning trees and the Isingmodel and dimer coverings.

Calabi-Yau equations

Four-dimensional Green functions

Higher-dimensional Green functions

What is this about?

I want to highlight the connection between Lattice Green functions,spanning trees on a lattice, dimer coverings and the Ising model.The first two topics are connected at a fundamental level, whereasthe second two are related to the first two only in the case of twodimensional systems.

What is a spanning tree

A spanning tree on a graph G is a loop-free connected graphconnecting all sites of G .The number of spanning trees on a graph G we denote TG .For a large class of graphs the number of spanning trees growsexponentially with the number of sites of the lattice.For a regular lattice L of N sites, the number of spanning trees onL, denoted TL(N) grows like eλN for N large.The limit limN→∞

1N log TL(N) = λL exists, is greater than zero,

and depends on the lattice L.We call this limit λL the spanning tree constant.

How to calculate TG and λL

The standard graph-theoretical method is by construction of theLaplacian matrix.For a graph with n vertices this is an n × n matrix Q = D− A.Here A is the n × n adjacency matrix of the graph, and D is ann × n diagonal matrix whose entry di ,i = κi where κi is the degreeof site i .The row and column sums of Q vanish, hence one eigenvalue iszero. A standard result in graph theory is then:

TG = Any cofactor of Q =1

n

n−1∏i=1

δi ,

where {δi} is the set of n − 1 non-zero-eigenvalues.

How to calculate TG and λL II

Spanning trees are also related to a special value of the Tuttepolynomial.The Tutte polynomial for a graph G is

T (G , x , y) =∑

G ′⊆G

(x − 1)k(G ′)−k(G)(y − 1)c(G ′).

Here k(G ) is the number of connected components of graph G,c(G ) is the number of independent circuits in G , and the sum isover all spanning subgraphs G ′ of G .Note that k(G ) = 1 for connected graphs G .In the limit x → 1, y → 1 the only non-vanishing terms will bewhen k(G ′) = k(G ) = 1 and c(G ′) = 0, namely the spanningtrees.Hence

T (G , 1, 1) = TG .

How to calculate TG and λL III

If the underlying graph is a lattice, TL(N) can be expressed as thepartition function of a lattice model.

ZN(q, v) =∑G∈L

qnv e ,

the sum is over all graphs G in L with n components and e edges.Up to a multiplicative constant, this is just the partition functionfor a q-state Potts model when q ∈ N.Set v = qα, eliminate e by using the relation e = N + c − n, wherec is the number of circuits in G , then

ZN(q, qα) = qNα∑G∈L

qαc+n(1−α).

As q → 0, with α = 1, the leading terms correspond to tree graphs(c = 0). More precisely,

TL(N) = limq→0

qα(1−N)−1ZN(q, qα).

How to calculate TG and λL IV

Hence ZN(q, q) generates a forest of trees on the lattice, while for0 < α < 1 the leading terms are the spanning trees, correspondingto c = 0, and n = 1.More precisely,

TL(N) = limq→0

qα(1−N)−1ZN(q, qα).

This correspondence then allows one to calculate TL(N) as thepartition function of an ice-type model on a related lattice.In this correspondence, the partition function is evaluated as aPfaffian, and the spanning tree constant is given as ad-dimensional integral.

How to calculate TG and λL V

This integral may be written as

λL = log q +1

(2π)d

∫ π

−πdk1 · · ·

∫ π

−πdkd log(1− Λ(L)),

where q is now the co-ordination number of the lattice.Λ(L) is the structure function of the lattice.The co-ordination number is the number of nearest neighbours perlattice site. The structure function is the Fourier transform of thediscrete step probability function.There is a lot of literature on the problem of evaluating λL for avariety of lattices L. In two dimensions this can be done exactly,but not generally in higher dimension (HBCC excepted).

How to calculate TG and λL VI

In two dimensions the results for the square, triangular andhoneycomb lattice were first given by Wu in 1977. They are:

λsq =1

4π2

∫ π

−π

∫ π

−πdk1dk2 log[4− 2(cos k1 + cos k2)]

=4

π(1− 1

32+

1

52− 1

72+ . . .) =

4G

π= 1.1662436 . . .

λtri =1

4π2

∫ π

−π

∫ π

−πdk1,2 log[6− 2(cos k1 + cos k2 + cos(k1 + k2))]

=3√

3

π(1− 1

52+

1

72− 1

112+

1

132+ . . .) = 1.6153297 . . .

λhoney = λtri/2

G is Catalan’s constant. A theorem linking λL on lattice L to λL∗on the dual lattice L∗ gives the honeycomb result.

Generalising the integral λL

It will turn out to be useful to generalise the integral and formwhat I call a spanning tree generating function (STGF).This is defined as

TL(z) = log q +1

(2π)d

∫ π

−πdk1 · · ·

∫ π

−πdkd log[1/z − Λ(L)]. (1)

Clearly, λL = TL(1). It immediately follows that

− zdTL(z)

dz=

1

(2π)d

∫ π

−πdk1 · · ·

∫ π

−πdkd

1

1− zΛ(L)= PL(0, z).

(2)The last integral may be recognised as the lattice Green function(LGF) of lattice L.

The connection between spanning trees and LGFs.

This connection is not new, though has not been widely used.Here we exploit this connection in a systematic manner. Theexistence of known exact results for some lattice Green functionscan now be integrated and provide new, simpler, representationsfor the spanning tree generating function, as well as new integralidentities.Similarly, if the LGF PL(0, z) is known, it can be integrated to givethe STGF and spanning tree constant.More precisely, we have

TL(z) = log q −∫

P(0, z)

zdz . (3)

The connection between STGFs and LGFs. II

Note

I We have restored the constant of integration, log q, lost in thedifferentiation.

I When the indefinite integral can be evaluated, and the STGFfound, the spanning tree constant can be found by evaluatingthe STGF at z = 1.

I In most cases, particularly for lattices of dimension greaterthan 2, the integral cannot be performed. In that case we canstill get an expression for the spanning tree constant.

I

λL = log q −∫ 1

0

P(0, z)− 1

zdz .

Lattice Green functions

I For a regular lattice, what is the probability that a walkerstarting at the origin will be at position ~l after n steps?

I The PGF is known as the Lattice Green Function (LGF).

I It is

P(~l ; z) =1

(2π)d

∫ π

−π· · ·

∫ π

−π

exp(−i~l .~k)dd~k

1− zΛ(~k).

I So that [zn]P(~l ; z) =the probability that a walker starting atthe origin will be at ~l after n steps.

I Again, Λ(~k) is the structure function of the lattice walk, andis given by the discrete Fourier transform of the individualstep probabilities.

I For example, for the d-dimensional hypercubic lattice, it isΛ(~k) = 1

d (cos k1 + cos k2 + · · · cos kd).

One dimension. The linear chain

I

Plc(~0; z) =1

2π

∫ π

−π

dk

1− z cos k=

1√1− z2

=∑k≥0

(2k

k

) (z

2

)2k

The corresponding spanning tree generating function is

I

Tlc(z) = log 2+1

2π

∫ π

−πlog(1/z−cos k)dk = arctanh

(1√

1− z2

)I λlc = Tlc(1) is undefined, which is to be expected, as the

number of spanning trees is exactly 1 for the linear chain,which is not an exponentially increasing function of thenumber of sites!

Two dimensions

I The probability of return to the origin is 1− 1/P(~0; 1).

I

P(~0; 1) =1

(2π)2

∫ π

−π

∫ π

−π

dk1 dk2

1− Λ(~k)

I Since P(~0; 1) diverges for two-dimensional lattices, this leadsto the well-known result that the probability of return in twodimensions is certain.

2d lattices

I Now we systematically exploit the LGF–STGF connection. Forexample, for the square lattice P(~0; z) = 2

πK (z), so

Tsq(z) = log 4− 2

π

∫K (z)

zdz

= log 4− log z − z2

84F3(1, 1,

3

2,3

2; 2, 2, 2; z2),

I and so

λsq =4G

π= log 4−

∫ 1

0

2πK (z)− 1

zdz = log 4

− 1

84F3(1, 1,

3

2,3

2; 2, 2, 2; 1) = 3F2(

1

2,1

2,1

2; 1,

3

2; 1)

Triangular lattice

I Similarly, for the triangular lattice,

Ttri (z) = log 6− 2

π

∫6√

3K (k)

πz(3− z)√

(3− z)(1 + z)dz

where k = 4z2

(3−z)√

z(3−z)(1+z),

I and so

λtri = log 6−∫ 1

0

[6√

3K (k)

πz2(3− z)√

(3− z)(1 + z)− 1

z

]dz

=3√

3

π(1− 1

52+

1

72− 1

112+ . . .) =

1√3

(π − Ψ′(5/6)

2π

)= 1.6153297 . . .

LGF coefficients give 2d identities

I It is instructive to consider the coefficients in the expansion ofthe LGFs. These give the number of returns to the originafter a given number of steps.

I We have:P(~0; z) =

∑n≥0

an(z

q)n

I where q is the co-ordination number of the lattice. Thusq = 3, 4, 6 for the honeycomb, square, triangular lattices.

I For the honeycomb lattice,

a2n =n∑

j=0

(n

j

)2(2j

j

).

I For the square lattice,

a2n =

(2n

n

)2

.

Triangular lattice

I For the triangular lattice,

an =n∑

j=0

(n

j

)(−3)k−jbj ,

where bj = a2j(honeycomb).I From the connection between the LGFs, and the expressions

for their coefficients, we have the following identites:∑m>0

(2m

m

)2 1

m

(1

4

)2m

= log 16− 8G

π(square)

(this is quite straightforward and follows immediately byexpressing both sides in terms of hypergeometric functions.)

I

1

2√

3

(π − Ψ′(5/6)

2π

)= log 3−

∑m>0

1

2m

(1

3

)2m m∑j=0

(2j

j

)(m

j

)2

(hc)

Triangular lattice

I

λtri = log 6−∑n>0

1

n

(1

6

)n n∑j=0

(−3)n−j

(n

j

) j∑k=0

(j

k

)2(2k

k

)

=1√3

(π − Ψ′(5/6)

2π

)= 1.615329736 . . .

I Since λtri/2 = λhoney , it follows that

log2

3+

∑n>0

1

n

(1

9

)n n∑j=0

(2j

j

)(n

j

)2

=∑n>0

1

n

(1

6

)n n∑j=0

(−3)n−j

(n

j

) j∑k=0

(j

k

)2(2k

k

).

Three dimensions

I In d = 3 for z = 1, P(~0; 1) gives the famous Watson integrals.

I Encountered by van Peype, a student of Kramers, who solvedthe b.c.c. case, but Watson did the s.c and f.c.c cases.

I The structure functions are:

Λ(~k) =1

3(cos k1 + cos k2 + cos k3). sc

Λ(~k) = (cos k1 cos k2 cos k3). bcc

Λ(~k) =1

3(cos k1 cos k2 + cos k2 cos k3 + cos k1 cos k3). fcc.

Watson integrals

I

P(~0; 1) =1

(2π)3

∫ π

−π

∫ π

−π

∫ π

−π

dk1 dk2 dk3

1− Λ(~k)

I

P(~0; 1)sc =1

32π3(√

3− 1)[Γ(1/24)Γ(11/24)]2 ≈ 1.516386

I

P(~0; 1)bcc =1

214/3π4[Γ(1/4)]4 ≈ 1.3932039

I

P(~0; 1)fcc =9

4π3[Γ(1/3)]6 ≈ 1.344661

Simple cubic lattice generating function

I

P(~0; z) =1

(π)3

∫ π

0

∫ π

0

∫ π

0

dk1 dk2 dk3

1− z3(cos k1 + cos k2 + cos k3)

I Joyce (1998) showed that this could be expressed as

P(~0; z) =1− 9ξ4

(1− ξ)3(1 + 3ξ)

[2

πK (k1)

]2

;

I where

k21 =

16ξ3

(1− ξ)3(1 + 3ξ);

ξ = (1 +√

1− z2)−1/2(1−√

1− z2/9)1/2

Body-centred cubic lattice generating function

I

P(~0; z) =1

(π)3

∫ π

0

∫ π

0

∫ π

0

dk1 dk2 dk3

1− z(cos k1 cos k2 cos k3).

I Maradudin et al. (1960) showed that this could be expressedas

P(~0; z) =

[2

πK (k2)

]2

I where

k22 =

1

2− 1

2

√1− z2.

Face-centred cubic lattice generating function

I

P(~0; z) =1

(π)3

∫ π

0

∫ π

0

∫ π

0

dk1 dk2 dk3

1− z3(c1c2 + c1c3 + c2c3)

where ci = cos ki .

I Joyce (1998) showed that this could be expressed as

P(~0; z) =(1 + 3ξ2)2

(1− ξ)3(1 + 3ξ)

[2

πK (k3)

]2

;

I where

k23 =

16ξ3

(1− ξ)3(1 + 3ξ);

ξ = (1 +√

1− z)−1(−1 +√

1 + 3z).

Connection between spanning trees and the Ising modeland dimer coverings

I There is a close connection between the STC λ and the FE ofthe Ising model at Tc .

I Unfortunately, this seems to be true only for planar lattices

I The Onsager soln. for the FE of the sq-lattice Ising model is

F (v) = log

(2

1− v2

)+

1

8π2

∫ π

−π

∫ π

−πlog[(1 + v2)2

− 2v(1− v2)(cos x + cos y)]dx .dy

where v = tanh( JkBT ). At the critical temperature,

v = vc =√

2− 1, this simplifies to

F (vc) =1

2log(2) +

1

8π2

∫ π

−π

∫ π

−πlog[4− 2(cos x + cos y)]dx .dy

= (λsq + log 2)/2

I If bn is the no. of distinct dimer coverings of an n × n sq.lattice (n even), then, (Kasteleyn and Fisher and Temperley),

bn = 2n/2

n/2∏j=1

n/2∏k=1

(cos2 jπ

n + 1+ cos2 kπ

n + 1

).

I In the infinite lattice limit,

limn→∞

1

nlog bn =

1

16π2

∫ π

−π

∫ π

−πlog[4+2(cos x+cos y)]dx .dy =

G

π,

where G is Catalan’s constant.

I V. sim. to the expn.for λsq, but for a (critical) sign change.

I However Temperley (1974) pointed out that if one considersdimers on a (2n − 1)× (2n − 1) site lattice, with oneboundary site removed, the relevant sign changes, and theintegral agrees with that for the spanning tree constant.

I We can derive further identities by considering thecorresponding results for 3d lattices.

I For the diamond lattice,

a2n =n∑

j=0

(n

j

)2(2j

j

)(2n − 2j

n − j

).

I For the simple cubic lattice,

a2n =

(2n

n

) n∑j=0

(n

j

)2(2j

j

).

I For the body-centred cubic lattice,

a2n =

(2n

n

)3

,

I while for the face-centred cubic lattice, (Bailey et al.)

an =n∑

j=0

(n

j

)(−4)n−jbj , where bj = a2j(diam).

I So we can now express the spanning tree constants in termsof sums of binomial coefficients, such as

λd = log 4−∑n≥1

(1

4

)2n 1

2n

n∑j=0

(n

j

)2(2j

j

)(2n − 2j

n − j

)I Since λd = λfcc/2, this gives rise to the identity

∑n>0

1

n

(1

12

)n n∑j=0

(n

j

)(−4)n−j

j∑k=0

(j

k

)2(2k

k

)(2j − 2k

j − k

)

= log3

4+

∑n≥1

(1

4

)2n 1

n

n∑j=0

(n

j

)2(2j

j

)(2n − 2j

n − j

)

HBCC lattice

I For the d-dimensional hbcc lattice, the stgf is

Td(z) =1

πd

∫ π

0dθ1 · · ·

∫ π

0dθd log

(1

z− cos(θ1) · · · cos(θd)

)=

1

πd

∫ π

0dθ1 · · ·

∫ π

0dθd log (1− z cos(θ1) · · · cos(θd))− log(z)

= d log(2)− log(z)− 1

2

∞∑l=1

z l

l

((2l)!

22l(l!)2

)d

So the growth constant λc is

λhbccc (d) = d log(2)− 1

2

∞∑l=1

1

l

(2l

l

)d ( z

4d

)l

= d log(2)− 1

2d+1 d+2Fd+1(1, 1,3

2, · · · ,

3

2; 2, 2, · · · , 2; 1),

a result first given by Chang and Shrock.

HSC latticeI For the d-dimensional simple cubic lattice, the STGF is

Td(z) =1

πd

∫ π

0..

∫ π

0dθ1,..,d log

(1

z− 1

dcos(θ1) + .. cos(θd)

)=

1

πd

∫ π

0..

∫ π

0dθ1..d log

(1− z

d[cos(θ1) + .. cos(θd)]

)− log(z)

= log(2d)− log(z)− 1

2

∞∑l=1

z l

la(d)l

a(2)l =

(2l

l

) l∑j=0

(l

j

)2

=

(2l

l

)2

a(3)l =

(2l

l

) l∑j=0

(l

j

)2(2j

j

)=

(2l

l

)3F2(

1

2,−l ,−l ; 1, 1; 4)

(2l

l

) l∑j=0

(l

j

)2(2j

j

)(2l − 2j

l − j

)=

(2l

l

)2

4F3(1

2,−l ,−l ,−l ; 1, 1,

1

2−l ; 1)

What are Calabi-Yau ODEs?

I A class of ODE that are pivotal in string theory. Here weconsider only 4th order ODEs, (corresponding to the case ofCalabi-Yau threefolds).

I Consider ODE’s of the form

y (s) + as−1(z)y (s−1) + · · ·+ a1(z)y ′ + a0(z)y(z) = 0,

where {ai} are meromorphic fns. of z .

Maximal unipotent monodromy MUM

I Then the roots of the indicial equation

λ(λ− 1) · · · (λ− s + 1) + as−1(0)λ(λ− 1) · · · (λ− s + 2)+

+ · · ·+ a1(0)λ + a0(0) = 0

determine the exponents of the ODE at the origin.

I The DE is said to be of Maximal Unipotent Monodromy(MUM) if all the exponents at 0 are zero.

I Consider a 4th order ODE which is MUM:

y (4) + a3(z)y (3) + a2(z)y ′′ + a1(z)y ′ + a0(z)y = 0.

It has four solutions, y0, y1, y2, y3.

I The C-Y condition is a1 = 12a2a3 − 1

8a33 + a′2 − 3

4a3a′3 − 1

2a′′3 .

Maximal unipotent monodromy MUM-II

I Being MUM means that:

y0 = 1 +∑n≥1

anzn;

y1 = y0 log z +∑n≥1

bnzn;

y2 =1

2y0 log2 z +

∑n≥1

bnzn

log z +∑n≥1

cnzn;

y3 =1

6y0 log3 z +

1

2

∑n≥1

bnzn

log2 z +

∑n≥1

cnzn

log z

+∑n≥1

dnzn;

Yukawa coupling and Instanton numbers

I Define q = exp(y1/y0) =∑

n≥1 tnzn, (the inverse functionz = z(q) =

∑unq

n is the mirror map in C-Y language).

I Then the Yukawa coupling K (q) is given by

K (q) =

(q

d

dq

)2 (y2

y0

)= 1 +

∞∑k=1

nkqk

1− qk.

I nk are called instanton numbers, and two C-Y equations areconsidered equivalent if they have the same instantonnumbers.

I Usually (exception later), Nk = N0nk/k3 are integers, whereN0 is a small integer—usually 1, 2 or 3.

I We don’t know the combinatorial significance of thecoefficients of K (q) or of the modified instanton numbers Nk .

Almkvist and Zudilin list

I Almkvist and Zudilin have catalogued a large number of 4th

order ODEs by ordering them in terms of increasing degree, k.

I We use the operator θ = z ddz , and write the 4th order ODE as

Df (z) = 0, where D = θ4 + zP1(θ) + z2P2(θ) . . . + zkPk(θ)where Pl , l = 1 . . . k are polynomials of degree 4 in θ.

I Thus ODEs of 1st degree take the form [θ4 + zP1(θ)]f (z) = 0.

I Almkvist et al found exactly 14 such ODEs. All can be solved.Their solutions have coefficients expressed as finite sums ofproducts of binomial coefficients.

Hyper body-centred cubic lattice

I

P(~0; z) =1

(π)4

∫ π

0· · ·

∫ π

0

dk1 dk2 dk3 dk4

1− z(cos k1 cos k2 cos k3 cos k4).

I

P(~0; z) =1

π4

∞∑n=0

zn

(∫ π

0cosn kdk

)4

I

=∞∑

n=0

(12)n(

12)n(

12)n(

12)n

(1)n(1)n(1)nn!z2n

I

= 4F3(1

2,1

2,1

2,1

2; 1, 1, 1; z2) =

∞∑n=0

(2n

n

)4 ( z

16

)2n

Hyper body-centred cubic lattice-cont.

I This admits to no further simplification, or special values atz = 1, though the series is rapidly convergent, so we canevaluate the integral at z = 1, giving P(~0; 1) ≈ 1.118636.

I It is the C-Y ODE #3 on the list of Almkvist et al.

I With θ = z ddz , the LGF satisfies DP(~0; 16z) = 0, where

D = θ4 − 256z(θ +1

2)4

I Because of the simple structure, we can also consider higherdimensions. For example, with d = 5,

P(~0; z) = 5F4(1

2,1

2,1

2,1

2,1

2; 1, 1, 1, 1; z2)

=∞∑

n=0

(2n

n

)5 ( z

32

)2n

which satisfies a 5th order Fuchsian ODE.

Hyper cubic lattice: Glasser and Guttmann 1994

I

P(~0; z) =1

(π)4

∫ π

0· · ·

∫ π

0

dk1 dk2 dk3 dk4

1− z4(c1 + c2 + c3 + c4)

,

where ci = cos ki .I Use the identity 1

λ =∫∞0 exp(−λt)dt,

I

P(~0; z) =1

(π)4

∫ ∞

0

∫ π

0· · ·

∫ π

0e−t

4∏j=1

e(ztcj/4)dtd~k

=

∫ ∞

0e−t I 4

0

(zt

4

)=

∞∑n=0

anzn,

since I0(z) = 1π

∫ π0 ez cos θdθ.

I

an =

[(12)n

n!

]3

4F3(−n

2, 1− n

2,−n,

1

2;1

2− n,

1

2− n, 1; 1)

Hyper cubic lattice

I With θ = z ddz , the LGF satisfies DP(~0; z) = 0, where

D = θ4 − 4z(2θ + 1)2(5θ2 + 5θ + 2) +

+ 28z2(θ + 1)2(2θ + 1)(2θ + 3)

I This is equation 16 on the list of Almkvist et al., who give

an =

(2n

n

) ∑j+k+l+m=n

(n!

j!k!l!m!

)2

=

(2n

n

) n∑k=0

(n

k

)2(2k

k

)(2n − 2k

n − k

).

I Almkvist (2007) gives a further 15 distinct expressions for an,all involving single or double sums of products of binomialcoefficients.

d -dimensional generalisations

I From the work of Glasser and Montaldi and Guttmann andPrellberg, we can write for the d-dimensional hyper-cubic LGF

[(2dz)2n]Pd(~0; z) =

(2n

n

) ∑k1+k2+...kd=n

(n!

k1!k2! . . . kd !

)2

I The 5 dimensional LGFs satisfies a 5th degree ODE, which canbe “pulled back” to a degree 4 ODE, that is also C-Y. This isalso the case for the 5d bcc. All the ODEs are in A & Z’s list.

Hyper face-centred cubic lattice

I

P(~0; z) =1

(π)4

∫ π

0· · ·

∫ π

0

dk1 dk2 dk3 dk4

1− z6λ

,

where λ = (c1c2 + c1c3 + c1c4 + c2c3 + c2c4 + c3c4) andci = cos ki .

I Set a = 1− z6(c2c3 + c2c4 + c3c4); b = z

6(c2 + c3 + c4).I Then the integrand is [a− b cos k1]

−1.I Use

1

π

∫ π

0

dθ

a− b cos θ=

1√a2 − b2

=1√

(a + b)(a− b)

to eliminate k1.I Next write (a + b)(a− b) = e(c − cos k2)(d − cos k2), where

c , d , e are independent of k2, and use∫ π

0

dθ√(c − cos θ)(d − cos θ)

=2K (k)√

(c − 1)(d + 1)

to eliminate k2, where k2 = 2(c−d)(c−1)(d+1) .

Hyper face-centred cubic lattice

I We are left with a two-dimensional integral, which wasexpanded as a power series in z and integrated term-by-termin Maple. We got to 40 terms in a few hours, then searchedfor an ODE.

I With θ = z ddz , the LGF satisfies DP(~0; z) = 0, where

D = θ4 + z(39θ4 − 30θ3 − 19θ2 − 4θ)

+ 2z2(16θ4 − 1070θ3 − 1057θ2 − 676θ − 192)

− 36z3(171θ3 + 566θ2 + 600θ + 316)(3θ + 2)

− 2533z4(+384θ4 + 1542θ3 + 2635θ2 + 2173θ + 702)

− 2633z5(1393θ3 + 5571θ2 + 8378θ + 4584)(1 + θ)

− 21035z6(31θ2 + 105θ + 98)(1 + θ)(θ + 2)

− 21237z7(θ + 1)(θ + 2)2(θ + 3).

Hyper face-centred cubic lattice

I This is a 4th order, degree 7 Calabi-Yau ODE with regularsingular points at 0, 1/24, -1/4, -1/8, -1/12, -1/18, and ∞.

I It is new, and one of only 3 known C-Y ODEs of degree 7(Almkvist, private communication).

I We do not yet have a nice expression for the series coefficientsin terms of binomial coefficients. Only

an =∑

i+j+k+l+m=n

(2i

i

)(2j

j

)(2k

k

)(l + m

m

)(2(n − i − j − k)

n − i − j − k

)(

n

2(n − i − j − k)

)(2(n − i − j − k)

l + m

)(2(i + j + k)− n

n − 2i − l −m

)(

4i + 2j + 2k + l + m − 2n

2i + j + m − n

)The mirror map for the differential equation gives Yukawacoupling K (q) whose instanton numbers, Nk , are3;−4; 64;−253; 4292;−25608; 442008;−3202512; 56565002;−457852636 . . .

Hyper diamond lattice

The structure function is:I Λ = 2 cos(k2 − k3) + 2 cos(k2 − k4) + 2 cos(k3 − k4) +

4 cos k1(cos k1 + cos k2 + cos k3 + cos k4) + 3.I Then I (and D Broadhurst) find that P(~0; z) is given by the

square of the 5-d multinomial coeffs:∑a2nz

n =∑

i+j+k+l+m=n

(n!

i !j!k!l!m!

)2

(z/5)n (n even).

I I gave the ODE for this in a 1993 paper with ThomasPrellberg, and it is simply related to the gen. fn. for 5dstaircase polygons! It isθ4 − z(35θ4 + 70θ3 + 63θ2 + 28θ + 5) + z2(θ + 1)2(259θ2 +518θ + 285)− 225z3(θ + 1)2(θ + 2)2,and number 34 on the Almkvist et al. list.

I Higher dimensional generalisations are obvious. They havebeen studied by Broadhurst who finds that nk/k2 are integers(rather than Ndnk/k3). Tested for D < 10, k < 100.

Other LGFs

An “obvious” structure function in 4d is

Λ = c1c2c3 + c1c2c4 + c2c3c4 + c1c3c4

It is satisfied by an 8th order ODE of degree 16, which is not MUM(D. Broadhurst)D. Broadhurst has also calculated the LGF for the 5d hyper-FCC.The structure function is

Λ = c1(c2 + c3 + c4 + c5) + c2(c3 + c4 + c5) + c3(c4 + c5) + c4c5

Then in a heroic calculation, Broadhurst showed that theunderlying ODE is of order 6 and degree 13, and is not MUM.

Conclusion

Conclusion

I We have made the connection between LGFs and STGFs Thisenables one to come up with a number of results (old andnew) in a systematic manner, including a number of integralidentities and binomial sum identities. We have given ODEsfor the 4-dimensional hyper-cubic, hyper-bcc hyper-fcc andhyper-diamond lattices, and found expressions for thecoefficients in the series expansion for all the lattices.

I All satisfy Calabi-Yau 4th order ODEs.

I We do not as yet have them in a unified form, as we do for 3dLGFs.

I We also have solutions for 5d LGFs for the hyper cubic, hyperbcc, hyper diamond and hyper fcc (Broadhurst) lattices.

I THE END -Thank you