~-
THE RAMANUJAN JOURNAL, 10,205-214,2005 @ 2005 Springer Science + Business Media, Inc. Manufactured in the Netherlands.
On the Entropy of Spanning Trees on a Large Triangular Lattice M.L. GLASSER
[email protected]
Department of' Physics, Clarkson University, Potsdam, New York 13699
F.Y.WU Department {!f'Physics,Northeastern University,Boston, Massachusetts 02115 Received October 24, 2003; Accepted December 11, 2003
Abstract. The double integral representing the entropy Sui of spanning trees on a large triangular lattice is evaluated using two different methods, one algebraic and one graphical. Both methods lead to the same result 2Jr Stri
= (4n2)-1 1 0 =
Key words:
(3J3/n)(l
2Jr
de -
5-2
10
dcpln[6 - 2cose - 2coscp - 2cos(e + cp)]
+ r2
-
11-2 + 13-2
-..
.).
triangular lattice, spanning tree, dilogarithm
2000 Mathematics
Subject Classification:
Primary-05AI6,
33B30, 82B20
1. Introduction It is weIJ-known that spanning trees on an n- vertex graph G can be enumerated by computing the eigenvalues of the Laplacian matrix Q associated with G as 1 n-l NST(G)
= - TI Ai,
(1)
n i=l
= 1,2,..., n - I, are the n - I nonzero eigenvalues of Q [1]. In physics one often deals with lattices. For large lattices the number of spanning trees N ST grows exponentialJy in n. This permits one to define the entropy of spanning trees for a class of lattices G as the limiting value where Ai, i
.
So
I
= n--+oo hm -In n
NST(G).
(2)
The resulting finite number So is a quantity of physical interest. For regular lattices in d dimensions the eigenvalues Ai can be computed by standard means [2] leading to expressions of So in terms of d-dimensional definite integrals. The
...J
206
GLASSER AND WU
resulting integrals, which are independent of boundary conditions, have been given for a number of regular lattices [3], but very few of the definite integrals have been evaluated in closed forms. To be sure, the entropy for the square lattice has been known and computed [4]. It is found to be {2IT Ssg
=
(4n2)-1
10
(IT
de 10
d4>ln(4 - 2cose
~Ch(~) = ~ 1- ~ + ~ - ~ + ~ -.. n [ 32 52 72 92
- 2cos4»
=
. J
4 =-G n = 1.l662436l6...
(3)
where
.
00
CI2(e) =
L sm(ne) n=l
(4)
n2
is the Clausen's function and G is the Catalan constant. The algebra reducing the double integral in (3) into the series given in the third line is straightforward and can be found in [5,6]. The only other published closed-form numerical result on the entropy of spanning trees is that of the triangular lattice [7] for which one has (X SIr;
= (4n2)-1 10 = ~Ch ~
d4>In[6 - 2 cos e - 2 cos 4>- 2 cos(e + 4»]
()
n
3
= 3~ n
=
(IT
de 10
[
1-~+~-~+~-"' 52
72
1.615329736097....
112
132
J (5)
However, no details leading from the double integral in (5) to the ficnalseries expression have been published. The expression in the second line is new. The purpose of this note is to provide some details of the steps leading to the second and third lines in (5) which, as we shall see, are nontrivial. We evaluate the integral in (5) using two different methods: a direct evaluation and a graphical approach. We first evaluate the integral algebraically, and then in an alternate approach we use graphical considerations to convert the spanning tree problem to a Potts model which is in turn related to an F model solved by Baxter [8]. In either case, the same series expression in (5) is deduced.'
ON THE ENTROPY OF SPANNING TREES ON A LARGE TRIANGULAR
LATTICE
207
2. Algebraic approach In this section we evaluate the integral (5) directly. Using the integration formula 2Jr
I
-
2n
dcpIn [2A + 2B cas cp+ 2C sin cp]= In[A+
10
J A2 -
B2 -
C21,
one of the two integrations in (5) can be carried out, giving Jr
I
Stri
= -n
de In[3 - cas e + J(7 - cas e)(l - cas e)].
10
(6)
Next, we use the identity 3
-
case + )(7
cose)(1
-
case)
-
= ~(v 3 +)3 + v2)(2v+)3 + V2)
(7)
where v = sinCe/2). Then (6) becomes
Stri
(8)
= In (~) + II + /z,
where Jr/2
2 IIIl
= -n
de In(m sine + J3 + sin2e),
10
m = 1,2
(9)
after a change of variable e /2 --f e. To evaluate II, we consider the more general integral 2
=
f(a) such that
f(../3)= II. 2 -
no
-n
Jr/2
10
In(sine+Ja2+sin2e)
Using
1
Jr
2
Jr
de Incase = -
no
1
de Incose = -In 2,
we have Jr
2
f(O)
=
f'ea)
=
de In(2 sine)de n 0 2a Jr/2 -
=
n 0 (sine+)a2+sin2e)Ja2+sin2e 2 - tan-I a.
1
l
na
= 0,
de
(10)
208
GLASSER AND WU
The reduction of j'(a) is effected by noting that (-lsin2 8 + a2+sin 8)-1 sin8)/a2. Hence,l(a) = (2/n)Ti2(a) and 2.
= -Ti2(-v3) n
II
= (Jsin28
r..
+ a2-
(11)
where a tan-I t -dt 0 t a3 a5 =a--+---+'" 32 52
=
Ti2(a)
(12)
1
a7 72
is the inverse tangent integral function described in [9]. To evaluate 12, we introduce the following identity
2 sin e + -l 3 + sin28 =
~
1 + u(8) 1 + u(8)'
cos 8
a ::::8 ::::n /2,
(13)
where u(8)
=
2sin8 -l3 + sin28'
Substituting (13) into h in (9) and making use of (10), we obtain
h
1
=-
2
7r/2
1
In3 - In2 + -
1 = -ln3
.J3
-ln2+
-
2
In
10
n
2n
1+ U(8) de [ 1 - u(e) ] 1
I
10
I+U In -
( )
(1 - u2/4).JI=U2
1- u
after an obvious change of integration variable in the last step. Next we express the factor in the integrand as
1
I
I
= 2[ I -
1 - u2/4
u/2
1
=
1+ u/2 ]
and set x
= J(1
- u)/(1 + u).
After a little reduction we obtain 1
2.J3
2
n
12 = - In 3 - In 2 - -
1 In x
2.J3 dx - 0 3x2 + 1 n
1
1 In x
10
-dx. x2 + 3
du
(14)
ON THE ENTROPY OF SPANNING TREES ON A LARGE TRIANGULAR
LATTICE
209
(, From integration by parts one has 1
10
In x X2 + a2
I a
d x---
( )1
1 -tan dx
-I
0 x
x a
() (15)
= (~)Ti2(~). Thisyields
= ~ In 3 -
h
In 2 + ~ [ Ti2 (~)
+ Ti2
(~) ].
(16)
To express II and 12 in terms of the Clausen's series (4), we have from Eqs. (2.6), (4.31) and (4.] 8) of [9] the identities
n
Tiz(y-I) = Tiz(y)- -Iny, 2
y>0
(17)
I Tiz(tan8) = 8 In(tan8) + -2 [C\z(28) + C\z(n - 28)] , 2n = ~C\z ~ . c\z
(3 )
Setting y
= ,j3
8>0
(3 )
3
(18) (19)
in (17)ande = n /3 in (18), we obtain after making use of (19) Tiz(~)
= ~C\z(~) + ~ In3
(~,j3 ) = 6 (~3 ) + ~ln3. 12
Ti2
~C\z
(20)
Thus, combining (11) and (16) with (8), we obtain
Sir;= ~C\z n
(~ )
(21)
3
as given in the second line in (5). To express SIr; in the form of the series given in (5), we note that
(~ )
CI2
.3
--S,j3 2
-
(22)
where OO
S
=
~
I
I
I
I
[ (6/11+ 1)2 + (6/11+ 2)2 - (6/11+ 4)2 - (6/11+ 5)2]
= SI +
I
-S2 22
GLASSER AND WU
210 with OO
51
1
= m=O L [ (6m + OO
52
=
~
1
1)2 - (6m + 5)2] '
1
1
[ (3m + 1)2 - (3m + 2)2] .
Separating terms in 52 with odd and even denominators, we have 1 52 It follows that 52
= (4/5)51
= 5]
- 2252.
hence 5 = (6/5)5], and from (21) and (22) we obtain
3v'3 5tri=-517T = 3v'3 1-~+~-~+~-"' 7T [ 52 72
, 112
132
]
This is the result given in the third line in (5).
3. Graphical approach In this section we evaluate the integral (5) with the help of a graphical mapping. The medial graph (or the surrounding lattice) ,[' of a triangular lattice ,[ is the kagome lattice shown in Fig. 1, where we have shaded its triangular faces for convenience. We start from a well-known equivalence between the q-state Potts model on ,[ and an ice-rule vertex model on,[' [10-12], Denote the respective partition functions by Zpottsand Zkag-ice'Then the equivalence reads Zpotts(q, fqx)
= qn/2Zkag-ice(1, l,x,x,
~
Figure J. a point.
t-1 +xt2, t +xt-2)
w
(23)
The shrinking of a kagome lattice into a triangular lattice. Each shaded triangular face is shrunk into
ON THE ENTROPY OF SPANNING TREES ON A LARGE TRIANGULAR
LATTICE
211
>;:~:t1XX x
x Figure 2.
The six ice-rule configurations
t-I + xt2
t + xt-2
of the kagome lattice and the associated vertex weights.
where x
=
(eK
-
I)/~, (24)
t3 + t-3 = ~,
K being the nearest-neighbor interaction of the Potts model, and the arguments of Zkag-ice are weights2 of the ice-rule model as depicted in Fig. 2. On the other hand, it is also known that by taking an appropriate q -+ a limit the Potts partition function generates spanning trees [7,13]. Write the spanning tree generating function in a slightly more general form by attaching a weight x to each line segment of the
tree. Since the number ofIines in a spanning tree on an n-vertex graph is fixed at n - 1,this merely adds an overall factor xn-l. Then, the equivalence reads [7, 13] (25)
xn-l NST = q->O \im q-Cn+2)/2Zpotts(q, ~x).
The equivalences (23) and (25) hold quite generally for finite lattices, provided that appropriate vertex weights are defined for the boundary [11]. We next combine (23) and (25) and take the thermodynamic limit n -+ 00. Assuming the two \imits on nand q commute, this leads to
Stri= -lnx
+ Zkag-ice(1,1,x, x, t-l + xt2, t + xt-2),
t = eilf/6
(26)
where Zkao-ice= lim n-Iln ~
n->oo
Zkao-ice. 0
We note in passing that the vertex weights in (26) now satisfy the free-fermion condition [14]
1 . 1+ x . x
= (t-I + xt2)
. (t
+ xt-2)
so Zkag-icecan be evaluated by standard means such as using Pfaffians [7]. This verifies the integral expression of Stri given in (5). Alternately, using a mapping introduced by Lin [15] (see also [16]), the kagome ice-rule model can be mapped into an ice-rule model on the triangular lattice. The trick is to 'shrink' each shaded face of the kagome lattice into a point as shown in Fig. 1. This results in a 2a-vertex ice-rule model on the triangular lattice which has 3 arrows in and 3 arrows out at every vertex.
212
GLASSER AND WU
",
'2
"3
* =A * =A+A
* *
£4 Figure 3.
=A+
=A+
The four types of ice-rule configurations
of the triangular
lattice and the associated shrinking
processes.
There are 4 different vertex types in the 20-vertex model: (i) six vertices in which the 3 incoming arrows are adjacent, (ii) two vertices in which the incoming and outgoing arrows alternate around the vertex, (iii) six vertices in which 2 incoming arrows are adjacent while the third in-arrow points in a direction opposite to one of the 2 adjacent incoming arrows, and (iv) six vertices in which 2 incoming arrows are adjacent while the third in-arrow points opposite to the other adjacent incoming arrow. These 4 configurations are shown in the left-side column in Fig. 3 to which we assign respective energies Ei, i = I, 2, 3, 4. The other 16 ice-rule configurations are obtained by rotations. The shrinking processes from which these ice-rule configurations are deduced are also shown in Fig. 3. Write the 4 weights as a, b, C, cl, respectively. We read off fi'om Fig. 3 to obtain
= e-EI/kT b = e-E2/kT C = e-EJ/kT cl = e-E4/kT a
= I . I .x
= (t + xt-2)3 + x3 = x(t + xt-2)2 + X2(t-l + xt2) = X2(t + xt-2) + X(t-I + xt~)2.
(27)
Thus, we have established the equivalence Zkag-ice(l, I,x,t-I
+xt2,t+xt-2)=Ztri-ice(a,b,c,c\
where Ztri-iceis defined similar to Zkag-ice'
(28)
Baxter [8] has considered the evaluation of Ztri-cel11 the subspace of c = fl. Now the right-hand side of (26) is independent of the value of x so we have the freedom to choose a
LATTICE
213
value of x for which Zice-trican be evaluated. It is readily verified that by taking x and t = ein/6 we have
= 1/,)3
ON THE ENTROPY OF SPANNING TREES ON A LARGE TRIANGULAR
b = 3a,
a = 1/,)3,
c=c'=2a
which is a case solved by Baxter. In this case Baxter obtained 00
dx
1 + e-x
1
Ztri-ice(a,3a, 2a, 2a) = Ina + P -00 x (eX- 1+ e-X)(l + r2x + e-4x) 1 3,)3 + ..!.._~+~ , = Ina + ---;;- [1 - 52 72 112 132 ] where P denotes the principal value. Combining (26) and (28) with (29) and a
(29)
= 1/,)3,
we obtain St n
=
3,)3 n
[
1-~+~-~+~-"' 52
72
,
112
132
]
This is the result given in the third line in (5) as first reported in [7]. Acknowledgment Work has been supported in part by NSF Grants DMR-0l21146 (MLG) and DMR-9980440 (FYW). We thank W. T. Lu for assistance in preparing the graphs. Notes I. The numerical values of StJ'jreported in [7] and [8] contain typos in the last two digits. 2. In comparing vertex weights with those in [12], it should be noted that for the triangular lattice which we are considering, roles of shaded and un shaded faces are reversed from those indicated in [12].
References 1. N.L. Biggs, Algebraic Graph Theory, Cambridge University Press (second edition), Cambridge, 1993. 2. W.J. Tzeng and F.Y. Wu, "Spanning trees on hypercubic lattices and nonorientahle surfaces," Lett. Appl. Math.
I3 (2000),19-25.
-
3. R. Shrock and F.Y. Wu, "Spanning trees on graphs and lattices in d dimensions," J. PhiS. A: Math. Gen. 33 (2000),3881-3902. 4. H.NV Temperley, in Combinatorics: Proc. Oxford Conference on Combinatorial Mathematics (DJ.A. Welsh and D.R. Woodall, eds.), Institute of Mathematics and Its Applications, Oxford, 1972, pp. 356-357. 5. P.w Kasteleyn, "The statistics of dimers on a lattice," Physica 27 (1961), 1209-1225. 6. M,E. Fisher, "Statistical mechanics of dimers on a plane lattice," PhiS. Rev. 124 (1961), 1664-1672. 7. F.Y. Wu, "Number of spanning trees on a lattice," J. Phis. A: Math. Gen. 10 (1977), U13-ll5. 8. RJ. Baxter, "F model on a triangular lattice," 1. Math. PhiS. 10 (1969),1211-1216.
9. L. Lewin, Dilogarithms and Associated Functions, Macdonald, London, 1958.
214
GLASSER AND WU
10. H.N.Y. Temper]ey and E.H. Lieb, "Relations between the 'Percolation' and 'Colouring' problem and other graph-theoretical problems associated with regular planar lattices: Some exact results for the 'Percolation' problem," Proc. Roy. Soc. Lolldoll A. 322 (197]), 251-280. II. R.J. Baxter, S.B. Kelland, and F.Y. Wu, "Equivalence of the potts model or Whitney polynomial with an ice-type model," J. Phys. A: Math. Gell. 9 (1976), 397-406. ]2. F.Y. Wu, "The Potts model," Rev. Mod. Phys. 54 (1982), 235-268. 13. c.M. Fortuin and P.W. Kasteleyn, "On the random-cluster model, I. Introduction and relation to other models," Physica, 57 (1972), 536-406. 14. C. Fan and F.Y. Wu, "General lattice model of phase transitions," Phy.[ Rev. B 2 (1970),723-733. ]5. c.K. Lin, "Equivalence of the 8-vel1ex model on a kagome lattice with the 32-vertex model on a triangular lattice," 1. Phys. A: Math. Gell. 9 (1976), LI83-186. 16. R.J. Baxter, H.N,v. Temperley, and S.E. Ashley, "Triangular Potts model at its transition temperature, related models, Proc. Roy. Soc. Lolldol! A. 358 (1978), 535-559.
and