Spanning trees, Lattice Green functions and Calabi-Yau equations

Transcription

1 Spanning trees, Lattice Green functions and Calabi-Yau equations Tony Guttmann MASCOS University of Melbourne

2 Talk outline Content Spanning trees on a lattice Lattice Green functions Calculation of spanning tree constants Digression: Connection between spanning trees and the Ising model and dimer coverings. Calabi-Yau equations Four-dimensional Green functions Higher-dimensional Green functions

3 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, whereas the second two are related to the first two only in the case of two dimensional systems.

4 What is a spanning tree A spanning tree on a graph G is a loop-free connected graph connecting all sites of G. The number of spanning trees on a graph G we denote T G. For a large class of graphs the number of spanning trees grows exponentially with the number of sites of the lattice. For a regular lattice L of N sites, the number of spanning trees on L, denoted T L (N) grows like e λn for N large. 1 The limit lim N N log T L(N) = λ L exists, is greater than zero, and depends on the lattice L. We call this limit λ L the spanning tree constant.

5 How to calculate T G and λ L The standard graph-theoretical method is by construction of the Laplacian 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 an n n diagonal matrix whose entry d i,i = κ i where κ i is the degree of site i. The row and column sums of Q vanish, hence one eigenvalue is zero. A standard result in graph theory is then: T G = Any cofactor of Q = 1 n n 1 i=1 where {δ i } is the set of n 1 non-zero-eigenvalues. δ i,

6 How to calculate T G and λ L II Spanning trees are also related to a special value of the Tutte polynomial. The Tutte polynomial for a graph G is T (G, x, y) = (x 1) k(g ) k(g) (y 1) c(g ). G 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 is over 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 be when k(g ) = k(g) = 1 and c(g ) = 0, namely the spanning trees. Hence T (G, 1, 1) = T G.

7 How to calculate T G and λ L III If the underlying graph is a lattice, T L (N) can be expressed as the partition function of a lattice model. Z N (q, v) = G L q n v 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 function for a q-state Potts model when q N. Set v = q α, eliminate e by using the relation e = N + c n, where c is the number of circuits in G, then Z N (q, q α ) = q Nα G L q αc+n(1 α). As q 0, with α = 1, the leading terms correspond to tree graphs (c = 0). More precisely, T L (N) = lim q 0 q α(1 N) 1 Z N (q, q α ).

8 How to calculate T G and λ L IV Hence Z N (q, q) generates a forest of trees on the lattice, while for 0 < α < 1 the leading terms are the spanning trees, corresponding to c = 0, and n = 1. More precisely, T L (N) = lim q 0 q α(1 N) 1 Z N (q, q α ). This correspondence then allows one to calculate T L (N) as the partition function of an ice-type model on a related lattice. In this correspondence, the partition function is evaluated as a Pfaffian, and the spanning tree constant is given as a d-dimensional integral.

9 How to calculate T G and λ L V This integral may be written as λ L = log q + 1 (2π) d π π dk 1 π π dk d 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 per lattice site. The structure function is the Fourier transform of the discrete step probability function. There is a lot of literature on the problem of evaluating λ L for a variety of lattices L. In two dimensions this can be done exactly, but not generally in higher dimension (HBCC excepted).

10 How to calculate T G and λ L VI In two dimensions the results for the square, triangular and honeycomb lattice were first given by Wu in They are: λ sq = 1 π π 4π 2 dk 1 dk 2 log[4 2(cos k 1 + cos k 2 )] π π = 4 π ( ) = 4G π = λ tri = 1 4π 2 π π π π dk 1,2 log[6 2(cos k 1 + cos k 2 + cos(k 1 + k 2 ))] = 3 3 π ( ) = λ honey = λ tri /2 G is Catalan s constant. A theorem linking λ L on lattice L to λ L on the dual lattice L gives the honeycomb result.

11 Generalising the integral λ L It will turn out to be useful to generalise the integral and form what I call a spanning tree generating function (STGF). This is defined as T L (z) = log q + 1 (2π) d π π Clearly, λ L = T L (1). It immediately follows that z dt L(z) dz π dk 1 dk d log[1/z Λ(L)]. (1) π = 1 π π 1 (2π) d dk 1 dk d π π 1 zλ(l) = P L(0, z). (2) The last integral may be recognised as the lattice Green function (LGF) of lattice L.

12 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. The existence of known exact results for some lattice Green functions can now be integrated and provide new, simpler, representations for the spanning tree generating function, as well as new integral identities. Similarly, if the LGF P L (0, z) is known, it can be integrated to give the STGF and spanning tree constant. More precisely, we have P(0, z) T L (z) = log q dz. (3) z

13 The connection between STGFs and LGFs. II Note We have restored the constant of integration, log q, lost in the differentiation. When the indefinite integral can be evaluated, and the STGF found, the spanning tree constant can be found by evaluating the STGF at z = 1. In most cases, particularly for lattices of dimension greater than 2, the integral cannot be performed. In that case we can still get an expression for the spanning tree constant. 1 λ L = log q 0 P(0, z) 1 dz. z

14 Lattice Green functions For a regular lattice, what is the probability that a walker starting at the origin will be at position l after n steps? The PGF is known as the Lattice Green Function (LGF). It is P( l; z) = 1 π π (2π) d π π exp( i l. k)d d k 1 zλ(. k) So that [z n ]P( l; z) =the probability that a walker starting at the origin will be at l after n steps. Again, Λ( k) is the structure function of the lattice walk, and is given by the discrete Fourier transform of the individual step probabilities. For example, for the d-dimensional hypercubic lattice, it is Λ( k) = 1 d (cos k 1 + cos k 2 + cos k d ).

15 One dimension. The linear chain P lc ( 0; z) = 1 π 2π π dk 1 z cos k = 1 = 1 z 2 k 0 ( 2k k ) (z 2 ) 2k The corresponding spanning tree generating function is T lc (z) = log 2+ 1 π ( ) 1 log(1/z cos k)dk = arctanh 2π π 1 z 2 λ lc = T lc (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 the number of sites!

16 Two dimensions The probability of return to the origin is 1 1/P( 0; 1). P( 0; 1) = 1 π π (2π) 2 π π dk 1 dk 2 1 Λ( k) Since P( 0; 1) diverges for two-dimensional lattices, this leads to the well-known result that the probability of return in two dimensions is certain.

17 2d lattices Now we systematically exploit the LGF STGF connection. For example, for the square lattice P( 0; z) = 2 π K(z), so and so T sq (z) = log 4 2 K(z) dz π z = log 4 log z z2 8 4 F 3 (1, 1, 3 2, 3 2 ; 2, 2, 2; z2 ), λ sq = 4G π = log π K(z) 1 z dz = log F 3 (1, 1, 3 2, 3 2 ; 2, 2, 2; 1) = 3F 2 ( 1 2, 1 2, 1 2 ; 1, 3 2 ; 1)

18 Triangular lattice Similarly, for the triangular lattice, T tri (z) = log 6 2 π where k = 4z 2 (3 z) z(3 z)(1+z), 6 3K(k) πz(3 z) (3 z)(1 + z) dz and so [ 1 6 ] 3K(k) λ tri = log 6 0 πz 2 (3 z) (3 z)(1 + z) 1 dz z = 3 3 π ( ) = 1 ( ) π Ψ (5/6) 3 2π =

19 LGF coefficients give 2d identities It is instructive to consider the coefficients in the expansion of the LGFs. These give the number of returns to the origin after a given number of steps. We have: P( 0; z) = a n ( z q )n n 0 where q is the co-ordination number of the lattice. Thus q = 3, 4, 6 for the honeycomb, square, triangular lattices. For the honeycomb lattice, For the square lattice, a 2n = n j=0 a 2n = ( n j ) 2 ( 2j j ( ) 2n 2. n ).

20 Triangular lattice For the triangular lattice, n ( ) n a n = ( 3) k j b j, j j=0 where b j = a 2j (honeycomb). 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 by expressing both sides in terms of hypergeometric functions.) ( ) π Ψ (5/6) = log 3 2π m>0 1 2m ( 1 3 ) 2m m j=0 ( 2j j )( ) m 2 (hc) j

21 Triangular lattice λ tri = log 6 n>0 1 n = 1 3 ( π Ψ (5/6) 2π ( ) 1 n n 6 j=0 ) Since λ tri /2 = λ honey, it follows that = n>0 log n>0 1 n 1 n ( 1 9 ( 3) n j ( n j ) j k=0 = ) n n j=0 ( 2j j ( ) 1 n n ( n ( 3) n j 6 j j=0 )( ) n 2 j ) j k=0 ( ) j 2 ( ) 2k k k ( ) j 2 ( ) 2k. k k

22 Three dimensions In d = 3 for z = 1, P( 0; 1) gives the famous Watson integrals. Encountered by van Peype, a student of Kramers, who solved the b.c.c. case, but Watson did the s.c and f.c.c cases. The structure functions are: Λ( k) = 1 3 (cos k 1 + cos k 2 + cos k 3 ). sc Λ( k) = (cos k 1 cos k 2 cos k 3 ). bcc Λ( k) = 1 3 (cos k 1 cos k 2 + cos k 2 cos k 3 + cos k 1 cos k 3 ). fcc.

23 Watson integrals P( 0; 1) = 1 π π π dk 1 dk 2 dk 3 (2π) 3 π π π 1 Λ( k) P( 0; 1) sc = 1 32π 3 ( 3 1)[Γ(1/24)Γ(11/24)] P( 0; 1) bcc = /3 π 4 [Γ(1/4)] P( 0; 1) fcc = 9 4π 3 [Γ(1/3)]

24 Simple cubic lattice generating function P( 0; z) = 1 π π π dk 1 dk 2 dk 3 (π) z 3 (cos k 1 + cos k 2 + cos k 3 ) Joyce (1998) showed that this could be expressed as P( 0; z) = 1 9ξ 4 [ ] 2 2 (1 ξ) 3 (1 + 3ξ) π K(k 1) ; where k 2 1 = 16ξ 3 (1 ξ) 3 (1 + 3ξ) ; ξ = (1 + 1 z 2 ) 1/2 (1 1 z 2 /9) 1/2

25 Body-centred cubic lattice generating function P( 0; z) = 1 π π π dk 1 dk 2 dk 3 (π) z(cos k 1 cos k 2 cos k 3 ). Maradudin et al. (1960) showed that this could be expressed as [ ] 2 2 P( 0; z) = π K(k 2) where k 2 2 = z 2.

26 Face-centred cubic lattice generating function P( 0; z) = 1 π π π dk 1 dk 2 dk 3 (π) z 3 (c 1c 2 + c 1 c 3 + c 2 c 3 ) where c i = cos k i. Joyce (1998) showed that this could be expressed as P( 0; z) = (1 + 3ξ 2 ) 2 [ ] 2 2 (1 ξ) 3 (1 + 3ξ) π K(k 3) ; where k 2 3 = 16ξ 3 (1 ξ) 3 (1 + 3ξ) ; ξ = (1 + 1 z) 1 ( z).

27 Connection between spanning trees and the Ising model and dimer coverings There is a close connection between the STC λ and the FE of the Ising model at T c. Unfortunately, this seems to be true only for planar lattices The Onsager soln. for the FE of the sq-lattice Ising model is F (v) = ( ) 2 log 1 v π π 8π 2 log[(1 + v 2 ) 2 π π 2v(1 v 2 )(cos x + cos y)]dx.dy where v = tanh( J k B T ). At the critical temperature, v = v c = 2 1, this simplifies to F (v c ) = 1 2 log(2) + 1 8π 2 = (λ sq + log 2)/2 π π π π log[4 2(cos x + cos y)]dx.dy

28 If b n is the no. of distinct dimer coverings of an n n sq. lattice (n even), then, (Kasteleyn and Fisher and Temperley), n/2 b n = 2 n/2 n/2 j=1 k=1 ( cos 2 jπ n kπ cos2 n + 1 In the infinite lattice limit, 1 lim n n log b n = 1 π π 16π 2 log[4+2(cos x+cos y)]dx.dy = G π π π, where G is Catalan s constant. ). V. sim. to the expn.for λ sq, but for a (critical) sign change. However Temperley (1974) pointed out that if one considers dimers on a (2n 1) (2n 1) site lattice, with one boundary site removed, the relevant sign changes, and the integral agrees with that for the spanning tree constant.

29 We can derive further identities by considering the corresponding results for 3d lattices. For the diamond lattice, a 2n = n j=0 ( n j ) 2 ( 2j j For the simple cubic lattice, ( ) n 2n a 2n = n j=0 )( ) 2n 2j. n j ( n j For the body-centred cubic lattice, ( ) 2n 3 a 2n =, n j=0 ) 2 ( 2j j while for the face-centred cubic lattice, (Bailey et al.) n ( ) n a n = ( 4) n j b j, where b j = a 2j (diam). j ).

30 So we can now express the spanning tree constants in terms of sums of binomial coefficients, such as λ d = log 4 n 1 ( ) 1 2n 1 4 2n n j=0 ( n j ) 2 ( 2j j Since λ d = λ fcc /2, this gives rise to the identity n>0 ( 1 1 n 12 = log n 1 ) n n j=0 ( ) 1 2n 1 4 n ( ) n ( 4) n j j n j=0 ( n j j k=0 ) 2 ( 2j j ( j k )( ) 2n 2j n j ) 2 ( 2k k )( ) 2n 2j n j )( ) 2j 2k j k

31 HBCC lattice For the d-dimensional hbcc lattice, the stgf is T d (z) = 1 π d π = 1 π π d dθ π = d log(2) log(z) 1 2 π dθ 1 So the growth constant λ c is 0 0 ( ) 1 dθ d log z cos(θ 1) cos(θ d ) dθ d log (1 z cos(θ 1 ) cos(θ d )) log(z) l=1 λ hbcc c (d) = d log(2) 1 2 z l l l=1 ( ) (2l)! d 2 2l (l!) 2 1 l ( ) 2l d ( z ) l l 4 d = d log(2) 1 2 d+1 d+2 F d+1 (1, 1, 3 2,, 3 ; 2, 2,, 2; 1), 2 a result first given by Chang and Shrock.

32 HSC lattice For the d-dimensional simple cubic lattice, the STGF is T d (z) = 1 π π ( 1 π d.. dθ 1,..,d log 0 0 z 1 ) d cos(θ 1) +.. cos(θ d ) = 1 π π ( π d.. dθ 1..d log 1 z ) 0 0 d [cos(θ 1) +.. cos(θ d )] log(z) = log(2d) log(z) 1 z l 2 l a(d) l ( 2l l a (3) l = ) l j=0 ( 2l l ( l j a (2) l = j=0 ( 2l l l=1 ) l j=0 ) l ( ) l 2 ( ) 2j = j j ) 2 ( 2j j )( ) 2l 2j = l j ( ) l 2 = j ( ) 2l 2 l ( ) 2l 3F 2 ( 1, l, l; 1, 1; 4) l 2 ( ) 2l 2 4F 3 ( 1 l 2, l, l, l; 1, 1, 1 2 l;

33 What are Calabi-Yau ODEs? A class of ODE that are pivotal in string theory. Here we consider only 4 th order ODEs, (corresponding to the case of Calabi-Yau threefolds). Consider ODE s of the form y (s) + a s 1 (z)y (s 1) + + a 1 (z)y + a 0 (z)y(z) = 0, where {a i } are meromorphic fns. of z.

34 Maximal unipotent monodromy MUM Then the roots of the indicial equation λ(λ 1) (λ s + 1) + ã s 1 (0)λ(λ 1) (λ s + 2)+ + + ã 1 (0)λ + ã 0 (0) = 0 determine the exponents of the ODE at the origin. The DE is said to be of Maximal Unipotent Monodromy (MUM) if all the exponents at 0 are zero. Consider a 4 th order ODE which is MUM: y (4) + a 3 (z)y (3) + a 2 (z)y + a 1 (z)y + a 0 (z)y = 0. It has four solutions, y 0, y 1, y 2, y 3. The C-Y condition is a 1 = 1 2 a 2a a3 3 + a a 3a a 3.

35 Maximal unipotent monodromy MUM-II Being MUM means that: y 0 = 1 + n 1 a n z n ; y 1 = y 0 log z + n 1 b n z n ; y 2 = 1 2 y 0 log 2 z + b n z n log z + c n z n ; n 1 n 1 y 3 = 1 6 y 0 log 3 z + 1 b n z n log 2 z + c n z n log z 2 n 1 n 1 + n 1 d n z n ;

36 Yukawa coupling and Instanton numbers Define q = exp(y 1 /y 0 ) = n 1 tn z n, (the inverse function z = z(q) = u n q n is the mirror map in C-Y language). Then the Yukawa coupling K(q) is given by ( K(q) = q d ) 2 ( ) y2 n k q k = 1 + dq 1 q k. y 0 k=1 n k are called instanton numbers, and two C-Y equations are considered equivalent if they have the same instanton numbers. Usually (exception later), N k = N 0 n k /k 3 are integers, where N 0 is a small integer usually 1, 2 or 3. We don t know the combinatorial significance of the coefficients of K(q) or of the modified instanton numbers N k.

37 Almkvist and Zudilin list Almkvist and Zudilin have catalogued a large number of 4 th order ODEs by ordering them in terms of increasing degree, k. We use the operator θ = z d dz, and write the 4th order ODE as Df (z) = 0, where D = θ 4 + zp 1 (θ) + z 2 P 2 (θ)... + z k P k (θ) where P l, l = 1... k are polynomials of degree 4 in θ. Thus ODEs of 1st degree take the form [θ 4 + zp 1 (θ)]f (z) = 0. Almkvist et al found exactly 14 such ODEs. All can be solved. Their solutions have coefficients expressed as finite sums of products of binomial coefficients.

38 Hyper body-centred cubic lattice P( 0; z) = 1 π π dk 1 dk 2 dk 3 dk 4 (π) z(cos k 1 cos k 2 cos k 3 cos k 4 ). P( 0; z) = 1 π 4 = n=0 n=0 z n ( π 0 ) 4 cos n kdk ( 1 2 ) n( 1 2 ) n( 1 2 ) n( 1 2 ) n z 2n (1) n (1) n (1) n n! = 4 F 3 ( 1 2, 1 2, 1 2, 1 2 ; 1, 1, 1; z2 ) = n=0 ( ) 2n 4 ( z ) 2n n 16

39 Hyper body-centred cubic lattice-cont. This admits to no further simplification, or special values at z = 1, though the series is rapidly convergent, so we can evaluate the integral at z = 1, giving P( 0; 1) It is the C-Y ODE #3 on the list of Almkvist et al. With θ = z d dz, the LGF satisfies DP( 0; 16z) = 0, where D = θ 4 256z(θ )4 Because of the simple structure, we can also consider higher dimensions. For example, with d = 5, P( 0; z) = 5 F 4 ( 1 2, 1 2, 1 2, 1 2, 1 2 ; 1, 1, 1, 1; z2 ) = n=0 ( ) 2n 5 ( z ) 2n n 32 which satisfies a 5 th order Fuchsian ODE.

40 Hyper cubic lattice: Glasser and Guttmann 1994 P( 0; z) = 1 π π dk 1 dk 2 dk 3 dk 4 (π) z 4 (c 1 + c 2 + c 3 + c 4 ), where c i = cos k i. Use the identity 1 λ = 0 exp( λt)dt, P( 0; z) = = since I 0 (z) = 1 π a n = 1 (π) 4 0 π 0 e t I 4 0 π 0 ez cos θ dθ. 0 π e t 0 ( zt ) = 4 n=0 4 e (ztc j /4) dtd k j=1 a n z n, [ ( 1 2 ) ] 3 n 4F 3 ( n n! 2, 1 n 2, n, 1 2 ; 1 2 n, 1 n, 1; 1) 2

41 Hyper cubic lattice With θ = z d dz, the LGF satisfies DP( 0; z) = 0, where D = θ 4 4z(2θ + 1) 2 (5θ 2 + 5θ + 2) z 2 (θ + 1) 2 (2θ + 1)(2θ + 3) This is equation 16 on the list of Almkvist et al., who give a n = = ( ) 2n n ( 2n n j+k+l+m=n ) n k=0 ( n k ) 2 ( 2k k ( n! ) 2 j!k!l!m! )( ) 2n 2k. n k Almkvist (2007) gives a further 15 distinct expressions for a n, all involving single or double sums of products of binomial coefficients.

42 d-dimensional generalisations From the work of Glasser and Montaldi and Guttmann and Prellberg, we can write for the d-dimensional hyper-cubic LGF [(2dz) 2n ]P d ( 0; z) = ( ) 2n n k 1 +k k d =n ( n! k 1!k 2!... k d! The 5 dimensional LGFs satisfies a 5th degree ODE, which can be pulled back to a degree 4 ODE, that is also C-Y. This is also the case for the 5d bcc. All the ODEs are in A & Z s list. ) 2

43 Hyper face-centred cubic lattice P( 0; z) = 1 π π dk 1 dk 2 dk 3 dk 4 (π) z 6 λ, where λ = (c 1 c 2 + c 1 c 3 + c 1 c 4 + c 2 c 3 + c 2 c 4 + c 3 c 4 ) and c i = cos k i. Set a = 1 z 6 (c 2c 3 + c 2 c 4 + c 3 c 4 ); b = z 6 (c 2 + c 3 + c 4 ). Then the integrand is [a b cos k 1 ] 1. Use 1 π dθ π 0 a b cos θ = 1 a 2 b = 1 2 (a + b)(a b) to eliminate k 1. Next write (a + b)(a b) = e(c cos k 2 )(d cos k 2 ), where c, d, e are independent of k 2, and use π 0 dθ (c cos θ)(d cos θ) = to eliminate k 2, where k 2 = 2(c d) (c 1)(d+1). 2K(k) (c 1)(d + 1)

44 Hyper face-centred cubic lattice We are left with a two-dimensional integral, which was expanded as a power series in z and integrated term-by-term in Maple. We got to 40 terms in a few hours, then searched for an ODE. With θ = z d dz, the LGF satisfies DP( 0; z) = 0, where D = θ 4 + z(39θ 4 30θ 3 19θ 2 4θ) + 2z 2 (16θ θ θ 2 676θ 192) 36z 3 (171θ θ θ + 316)(3θ + 2) z 4 (+384θ θ θ θ + 702) z 5 (1393θ θ θ )(1 + θ) z 6 (31θ θ + 98)(1 + θ)(θ + 2) z 7 (θ + 1)(θ + 2) 2 (θ + 3).

45 Hyper face-centred cubic lattice This is a 4th order, degree 7 Calabi-Yau ODE with regular singular points at 0, 1/24, -1/4, -1/8, -1/12, -1/18, and. It is new, and one of only 3 known C-Y ODEs of degree 7 (Almkvist, private communication). We do not yet have a nice expression for the series coefficients in terms of binomial coefficients. Only ( )( )( )( )( ) 2i 2j 2k l + m 2(n i j k) a n = i j k m n i j k i+j+k+l+m=n ( )( )( ) n 2(n i j k) 2(i + j + k) n 2(n i j k) l + m n 2i l m ( ) 4i + 2j + 2k + l + m 2n 2i + j + m n The mirror map for the differential equation gives Yukawa coupling K(q) whose instanton numbers, N k, are 3; 4; 64; 253; 4292; 25608; ; ; ; 45785

46 Hyper diamond lattice The structure function is: Λ = 2 cos(k 2 k 3 ) + 2 cos(k 2 k 4 ) + 2 cos(k 3 k 4 ) + 4 cos k 1 (cos k 1 + cos k 2 + cos k 3 + cos k 4 ) + 3. Then I (and D Broadhurst) find that P( 0; z) is given by the square of the 5-d multinomial coeffs: a2n z n = i+j+k+l+m=n ( n! i!j!k!l!m! ) 2 (z/5) n (n even). I gave the ODE for this in a 1993 paper with Thomas Prellberg, and it is simply related to the gen. fn. for 5d staircase polygons! It is θ 4 z(35θ θ θ θ + 5) + z 2 (θ + 1) 2 (259θ θ + 285) 225z 3 (θ + 1) 2 (θ + 2) 2, and number 34 on the Almkvist et al. list. Higher dimensional generalisations are obvious. They have been studied by Broadhurst who finds that n k /k 2 are integers (rather than N d n k /k 3 ). Tested for D < 10, k < 100.

47 Other LGFs An obvious structure function in 4d is Λ = c 1 c 2 c 3 + c 1 c 2 c 4 + c 2 c 3 c 4 + c 1 c 3 c 4 It is satisfied by an 8 th 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 = c 1 (c 2 + c 3 + c 4 + c 5 ) + c 2 (c 3 + c 4 + c 5 ) + c 3 (c 4 + c 5 ) + c 4 c 5 Then in a heroic calculation, Broadhurst showed that the underlying ODE is of order 6 and degree 13, and is not MUM.

48 Conclusion

49 Conclusion We have made the connection between LGFs and STGFs This enables one to come up with a number of results (old and new) in a systematic manner, including a number of integral identities and binomial sum identities. We have given ODEs for the 4-dimensional hyper-cubic, hyper-bcc hyper-fcc and hyper-diamond lattices, and found expressions for the coefficients in the series expansion for all the lattices. All satisfy Calabi-Yau 4th order ODEs. We do not as yet have them in a unified form, as we do for 3d LGFs. We also have solutions for 5d LGFs for the hyper cubic, hyper bcc, hyper diamond and hyper fcc (Broadhurst) lattices. THE END -Thank you