Partition function zeros for the two-dimensional Ising model VI

Physica A IS4 (1989) 324-343

North-Holland, Amsterdam

PARTITION FUNCTION ZEROS FOR THE TWO-DIMENSIONAL

ISING MODEL VI*

John STEPHENSON

Physics Department, University of Alberta, Edmonton, Alberta. Canada. T6G 251

Received 6 June 1988

Special features of the complex temperature zeros of the partition function of the

two-dimensional Ising model on completely anisotropic triangular lattices with all “odd”

interactions, such as one with interaction ratios 5 : 3 : 1, are investigated. A bifurcation of one

of the boundary lines at the imaginary axis occurs. An analytical explanation of this

bifurcation is provided, and the algebraic reduction of the bifurcation eliminant to a

symmetrical form is reported. Also, the end-points of lines of pure imaginary zeros are

related to the critical point and disorder point equations.

1. Introduction

In this paper we complete our study [l ,2] (to be referred to as IV, V,

respectively) of the boundaries of the temperature zeros of the partition

function of the two-dimensional Ising model on triangular lattices. We examine

some of the special features of boundary zeros on completely anisotropic

triangular lattices with three “odd” interactions. The occurrence of a line of’

pure imaginary zeros on class (i) partially anisotropic lattices with “odd”

interactions was noticed by Stephenson and van Aalst [3] (to be referred to as

III), and has been discussed in detail in that context in V [2]. Furthermore it

was shown in V that one of the boundary lines on partially anisotropic lattices

with odd interactions bifurcates on meeting the imaginary axis. These two

features are also common to all class (i) completely anisotropic triangular

lattices with all “odd” interactions, as we explain below, where the analytical

criteria for a “boundary bifurcation point” are established. The bifurcation

discriminant can be factorized, permitting the reduction to a symmetrical form

in the three triangular lattice interactions. The results of this (heavy) algebraic

and “computer” calculation are reported.

* This work is supported in part by NSERC through grant #A6595

0378-4371/89/$03.50 0 Elsevier Science Publishers B.V.

(North-Holland Physics Publishing Division)

.I. Stephenson i Partition function zeros for Ising model VI 325

In section 2 we summarize our notation and re-state the polynomials

associated with the Ising partition function. Numerical results for the (5,3, 1)

lattice are presented in section 3, to provide a specific illustration of the

problems to be addressed in this paper. The location of the critical and other

special points, which identify features of geometrical interest in the distribution

of zeros, are tabulated for the (5,3,1) lattice, and the “interior” special points

are identified.

In section 4 we account for the positions of the end-points of the lines of

pure imaginary zeros which can occur for triangular lattices with all odd

interactions. The bifurcation of “mesh lines” of zeros at the imaginary axis is

analyzed in section 5, and the criteria for a boundary bifurcation point (on the

imaginary axis) are established in section 6. An outline of the algebraic

extraction of the bifurcation eliminant, and its reduction to a symmetrical form

in the three triangular lattice variables, is reported in section 7. [Technical

details of this calculation will be published separately.]

2. Partition function polynomials

Using the same notation as in IV [l], the polynomial factors which determine

the zeros of the partition function of the Ising model on an anisotropic

two-dimensional triangular lattice, with interactions in the ratio a : b : c, have

the form

f(z; +,, +S) = g + ih = A + B cos 4, + C cos $s + D COS(+, + 4,) , (1)

where A, B, C, D are polynomials in z = x + iy = exp(-2Jlk, T):

A = 1 + Z2(n+b) + Z2(c+a) + ZW+d = 1 + z2d + z2e + ,2f,

B = 2Zb+c(Z2a - 1) = 2(Zd+e - zf),

c = 2y+yp - 1) = qzf+d - f) )

D = 2Za+b(Z2c _ 1) = ‘4Ze+f - Zd) ,

(24

(2b)

PC)

(24

with d = a + b, e = c + a, f= b + c. Later we will need the z-derivative of (1)

in the form

f, = zf’(z; c$,, &) = A, + B, cos 4, + C, ~0s 4s + D, cos(4, + 4s) 7 (3)

where a subscript z denotes z times a derivative with respect to z.

326 J. Stephenson I Partition function zeros for Ising model VI

3. Boundaries and special points

Diagrams of interior and boundary zeros for (5,3, l), the lowest order

completely anisotropic triangular lattice with all “odd” interactions, are dis-

played in figs. la and b, and a list of special points is supplied in table I. The

interior zeros are calculated directly from (l), and (as explained in IV) the

boundary lines are constructed by solving (1) simultaneously with the “boun-

dary equation”

sin(+, + $J)[(B’D” - B”D’) sin 4, - (CID” ~ C”D’) sin 4,Y]

+ (B’C’ - B”C’) sin $, sin +* = 0 , (4)

where we have written B = B’ + iB”, etc. This equation was derived in IV and

in ref. [4].

In IV we observed that some of the complex special points which are

solutions to the critical point equations lie on the boundary, whereas others lie

inside the region containing zeros. By examining the form of the “boundary

expression ” in (4) above [or eq. (8) or (17) in IV] in the neighbourhood of

points corresponding to complex solutions of the critical point equations, we

are able to determine whether such points lie on a boundary (at pinch points),

or not. For (.5,3,1) the special points T and U are interior special points, at

which the relevant quadratic form [(4) above, or eq. (17) in IV] is (negative)

definite, whereas L, M and S are boundary pinch points at which the quadratic

form is indefinite

1.5

1.0

0.5

0.0

-c N’

1 K'

4

(5.3 11

I

1 .

1 .

0.

0. 0.0 9.5 1.0 1.5 0.0 0.5 1.0 1.5

Fig. I. The complex z-plane for the (5,3, I) triangular lattice showing interior zeros in (a) and boundary lines in (b). The interior points are solutions of (1) at 1.5” intervals. The complete

distribution of zeros is symmetrical on reflection in both axes.

J. Stephenson I Partition function zeros for Ising model VI 327

Table I Special points for the (5,3, 1) triangular lattice: Location of the special ooints I-V for the (5.3. 1) lattice. The origin is at 0.

I

J CD’) K (‘9 K’ (C’ ) L M N N’ P Q R s T U

V(B)

exp(ini2)

(id il.150964 0.826031 (z,)

(k) iO.826031 0.474264 + i0.759431 0.662359 + iO.562280 1.324718 (z,)

(W il.324718 exp(3iTilO) exp(iPi6) exp(in/lO) 0.874985 + iO.321308 1.024504 + i0.444748 0.659334 + iO.880844

+ il.073621

line 0” 0”

-318” 180” 180” 180”

-332” line line line

0” 180”

0” -358

180” 0”

-121” 0”

180” 180”

-133”

180” 0” 0”

-177”

4. Lines of pure imaginary zeros

On examining the zero distributions for (5,3,1) in fig. 1, one observes a line

of pure imaginary zeros lying between iz, and iz,. Some of the pure imaginary

zeros for the (5,3,1) lattice, for a fixed value of one of the angles: 4, = 355”,

are graphed in fig. 2, for $, = 5”-175” (A+,$ = 5”). For the particular value of 4,

selected (3559, the upper and lower branches of zeros commence at I when

+* = 5”. These pure imaginary zeros come from the polynomials in (1) which

correspond to angles in a triangular region in the angle diagram above the line

4, + 4X = 2~ and inside the basic range 0 d 4, d HIT, 0 d c$, S rr appropriate for

class (i) lattices (III). There are also further superposed lines of pure imagi-

nary zeros, which lie (in sections) between I and D’, and are related to the

bifurcation phenomenon discussed below.

In order to determine the upper and lower end-points of a line of pure

imaginary zeros, one may regard y = Im z as a real function of the two angles

$,, $.Y, given implicitly by the polynomial (1) for the zeros. To obtain the

maximum and minimum values of y, one needs to locate the values of the

angles at which the partial derivatives of y (with respect to the angles) vanish:

ay ay -=O and -=O. w a4s

Differentiating (1) and inserting the conditions (5), we obtain two equations


1.25

Y 1.2

1 15

1.1

1.05

0.95

0.9

0 a5 I r I I , I I I 1 I , , I / I 1 -I

0 20 40 60 a0 100 120 140

P

160 18C

.s

Fig. 2. Pure imaginary zeros for the (5, 3, 1) triangular lattice, for C#J, = 355”, graphed as a function

of c$, = ~5-175”. A+, = 5”. For the particular value of 4, (355”) selected, the zeros commence at

I=(O,l) when 4,=5O. y=Imz.

t++?+ + + ++ + + +

‘t +

+

+ + ++

++ + ++

++++ +

+++++++ ++++++++++ +++

which must be solved in combination with (1) for the angles and for z = iy:

B sin 4, = C sin 4, = -D sin(4, + cb,) (6)

In practice one can eliminate the angles algebraically (between (1) and (6)) to

obtain a polynomial in z for iy:

B2C2 t B’D’ + C2D2 - 2ABCD = 0, (7)

whose roots include iz,., iz, and k zD. The complete factorization of (i’),

derived in appendix A, is

16z d+e+f(l + Zd + zp + zf)(l + 2 - ze - 2)

x (1 - zd + zc - z’)( 1 - zd - zp + z’) )

or

16uvw( 1 + u + u + w)( 1 + u ~ u - w)

x (1 - u + u - W)(l - u - u + W) )

(84

(8b)


where

u = Zd , u = zc ) w = Zf . (84

The last three factors in (8a) are just the disorder point expressions in IV (eq.

(9)). [Disorder points were first discovered on the Ising triangular lattice by

Stephenson [.5].] Then the angles are given by the (appropriate) solutions of

cos 4, = (CD - AB) lB2 ,

cos $, = (DB - AC)IC’,

(94

(9b)

cos(+, + c&) = (BC - AD)/D’ (94

[In order to satisfy (6) and to agree with the results of a direct calculation of

pure imaginary zeros from (l), the angles must be selected to lie in an

appropriate quadrant within the basic angle range 0 s 4,~ 2n, 0 G 4.s 6 rr for

all class (i) odd interaction triangular lattices.] Details of the derivation of

these formulae are presented in appendix A. For the (5,3,1) lattice one finds

at C’ $, = 317.929. . .’ and 4x = 121.187. . .O, and at N’ 4, = 332.454. . .’ and

+s = 133.449. .O. The above formulae (6) to (9) are valid for all class (i)

completely and partially anisotropic triangular lattices.

5. Bifurcation at the imaginary axis

The behaviour of zeros near bifurcation points is illustrated for a (5,3,1)

triangular lattice in figs. 3 and 4. For a fixed value of $,: 355”, the real and

imaginary parts of the zeros are graphed separately in figs. 3a and b as

functions of +s = 178”-180” (A$, = 0.1’). Bifurcations occur approximately at

$s = 179.1” and 179.4”, and between these angles there are four lines of pure

imaginary zeros. Table II contains the numerical data for these figures, and

permits one to identify the pure imaginary zeros and the corresponding real

and imaginary parts of complex zeros.

Fig. 4 shows the actual complex zeros close to the imaginary axis (for the

(5,3,1) triangular lattice, with 4, held fixed at 355”). Two “mesh lines”

(together with their mirror images in the y-axis) approach the imaginary axis,

where bifurcation occurs - approximately at y = 1.08 and 1.14.

Some of the mesh lines along which 4,. is held constant meet the imaginary

axis between I and D’. Near D’ each point on the imaginary axis is a

bifurcation point for two of the mesh lines with distinct values of 4,, and

associated values of 4s. Table III lists corresponding values of the angles for

330

0.06

0.05

X 0.04

0.03

0.02

0.01

J. Stephenson I Parliiion function zeros for Ising model VI

l-

a

+ +

t 178 178.2 i 78.4 178.6 I 78.8 179 179.2 179.4 179.6 i 79.8 180

1.2 , -- l

1.18 -

1.16 -

Y 1.14 -

1.12 -

1.1 -

1.08 - -- +

1.06 -

1.04 -

1.02 -

l-

0.98 -

0.96 -- +

+ + + + + + + +

+ + + + + + + +

++++++ + +

+ + + + + +

+ +

+ +

+ + + + +

+ +

+ +

t +

b

t- + +

t- + +

c + +

0.94;,, , , , , , , , , , , , , /, I,,

178 i 78.2 i 78.4 178.6 i 78.8 179 179.2 179.4 179.6 i 79.8 iao

Fig. 3. Behaviour of zeros near bifurcation points illustrated for the (5,3, 1) triangular lattice. For

q%, = 355”. the real and imaginary parts are graphed in (a) and (b) as functions of 4, = 178”-180” (A+, = 0.1”). Bifurcations occur approximately at I$, = 179.1” and 179.4”, and between these angles

there are four lines of pure imaginary zeros. Table II contains the numerical data for these figures.

x=Rez,y=Imz.

Table II


Zeros near bifurcation poinfs for the (5,3. 1) triangular lattice: 4, = 355”, x = Re z, y = Im z.

178.0 0.0578 1.0755 178.0 0 1.1892 178.0 0 0.9595 178.1 0.0555 1.0758 178.1 0 1.1878 178.1 0 0.9603 178.2 0.0531 1.0761 178.2 0 1.1864 178.2 0 0.9612 178.3 0.0505 1.0764 178.3 0 1.1848 178.3 0 0.9620 178.4 0.0476 1.0768 178.4 0 1.1832 178.4 0 0.9629 178.5 0.0444 1.0772 178.5 0 1.1815 178.5 0 0.9639 178.6 0.0407 1.0776 178.6 0 1.1797 178.6 0 0.9649

178.7 0.0365 1.0781 178.7 0 1.1777 178.7 0 0.9659 178.8 0.0315 1.0787 178.8 0 1.1755 178.8 0 0.9670 178.9 0.0250 1.0793 178.9 0 1.1731 178.9 0 0.9681 179.0 0.0153 1.0801 179.0 0 1.1703 179.0 0 0.9694 179.1 0 1.0951 179.1 0 1.0670 179.1 0 1.1671 179.1 0 0.9707 179.2 0 1.1084 179.2 0 1.0562 179.2 0 1.1632 179.2 0 0.9721 179.3 0 1.1196 179.3 0 1.0487 179.3 0 1.1580 179.3 0 0.9736

179.4 0 1.1337 179.4 0 1.0424 179.4 0 1.1485 179.4 0 0.9753 179.5 0.0148 1.1431 179.5 0 1.0366 179.5 0 0.9771 179.6 0.0217 1.1447 179.6 0 1.0312 179.6 0 0.9792 179.7 0.0264 1.1462 179.7 0 1.0257 179.7 0 0.9817 179.8 0.0302 1.1476 179.8 0 1.0200 179.8 0 0.9847 179.9 0.0334 1.1489 179.9 0 1.0134 179.9 0 0.9889 180.0 0.0362 1.1500 180.0 0 0.9999 180.0 0 0.9999

x Y

the appropriate range of values of y for the (.5,3,1) lattice. When a line of

complex zeros, on which 4, (say) is fixed and +S is increasing (as in table II),

meets the imaginary axis, a pair of pure imaginary zeros separate, as illustrated

in fig. 3a. One of these pure imaginary zeros descends to I, where it joins with

the independent lower branch of pure imaginary zeros. The other pure

imaginary zero ascends towards a second bifurcation point, where it meets the

upper branch of pure imaginary zeros, shown in fig. 2. Another line of complex

zeros emerges from the second bifurcation point, as $,v continues to increase

(towards 180’).

Part way down the imaginary axis, fairly close to I, the mesh lines form a

boundary envelope, which itself meets the real axis at a boundary bifurcation

point B (or V in the data table), which is to be analyzed below and again in

section 6.

In order to determine bifurcation points between D’ and B, where zs = iy,,

we seek joint solutions off = 0 andf, = 0, from (1) and (3). First eliminate one

of the angles, 4, say, to obtain a relation between y = Im z and the remaining

angle, 4S. For a given value of y (in an appropriate range -between D’ and

B), one can obtain solutions of this relation for 4S, and hence determine the

332 J. Stephenson i Purtition function zeros for lsing model Vl

1.2 ,

1.18

1.16 1

y i

+ +++ t 1.14 +

1.12

1.1

1.08

1.06

1.04

1.02

1 1

+

0 0.02 0.04

X 0.06

Fig. 4. The complex z-plane for the (5, 3, 1) triangular lattice showing interior zeros close to the

imaginary axis. Here 4, is held fixed at 355”. Two “mesh lines” (together with their mirror images

in the y-axis) approach the imaginary axis, where bifucations occurs - approximately at y = 1.08

and 1.14. The corresponding angles are graphed in fig. 3, and the numerical data is in table II.

corresponding values of 4,. These solutions yield the bifurcation values

associated with the various mesh lines, where they meet the imaginary axis

between I and D’. To locate the boundary bifurcation point B, one seeks the

value of y at which the (two relevant) solutions of the above relation coincide.

Details of the algebra required to extract the bifurcation conditions are given

in appendix B. Here we provide an outline of the main steps. (There arc

several ways of proceeding.) First eliminate COS(C$, + 4,) between (1) and (3),

and express cos 4, in terms of cos +S:

cos 4, = (DA; - ADZ) + (DC, - CD,) cos 4,

(BD, - DB,) . (10)

Next substitute back in (1) to obtain

sin 4, =

(BAZ - ABJ + (BC, - CB, + DA, - AD,) cos & + (DC, - CD;)(cos &)2

(BD, - DB,) sin C&

(11)


Now use the identity cos’ + sin’ = 1, to obtain a cubic equation for cos +S = 2,

say [the quartic term has a zero coefficient]:

a’Z”+b’Z2+c’Z+d’=0, (12)

with coefficients as in eq. (B. 1) of appendix B. On giving y various values (less

than z,,), one finds that in general the above cubic has three real solutions for

Z, two of which yield real values for the angle +,C. The corresponding values of

$, are obtained from (10) and (11). For the (5,3,1) lattice fig. 5 displays

graphs of the angles as functions of y, and fig. 6 shows the corresponding lines

in the angle diagram. Table III contains the numerical data for these figures.

There are four pure imaginary zeros for angles lying inside the “boomerang”

shaped region in fig. 6, and two for angles in the remaining part of the angle

diagram above the line 4, + c$~ = 2~.

Table III

Bifurcation angles and points ( y = Im z) for the (5,3, 1) triangular lattice.

Y 4, (+) 4, (+) 4, (X) k(X)

1.0736 358.480 177.350 358.480 177.350 1.0740 357.831 178.125 358.924 176.217 1.0745 357.424 178.402 359.083 175.507 1.0750 357.104 178.561 359.174 174.948 1.0755 356.829 178.670 359.236 174.468 1.0770 356.158 178.863 359.347 173.294 1.0800 355.169 179.035 359.445 171.562 1.0850 354.035 179.140 359.505 169.567 1.0900 353.258 179.177 359.526 168.193 1.0950 352.716 179.188 359.532 167.229 1.1000 352.356 179.189 359.532 166.581 1.1050 352.149 179.187 359.531 166.199 1.1100 352.078 179.185 359.529 166.057 1.1150 352.136 179.187 359.531 166.143 1.1200 352.323 179.196 359.536 166.456 1.1250 352.645 179.216 359.547 167.008 1.1300 353.116 179.249 359.566 167.826 1.1350 353.767 179.302 359.597 168.963 1.1400 354.657 179.385 359.645 170.529 1.1450 355.935 179.518 359.721 172.786 1.1480 357.084 179.647 359.797 174.821 1.1500 358.318 179.794 359.881 177.012 1.1505 358.830 179.856 359.917 177.921 1.1507 359.117 179.891 359.937 178.430 1.1508 359.303 179.914 359.951 178.762 1.1510 360.000 180.000 360.000 180.000


360

172

170

168

166

1.07 1 .ll Y 1.15

Fig. 5. Graphs of the bifurcation angles for the (5,3, 1) triangular lattice as functions of y = Im z

along the imaginary axis, showing 4, in (a) and +s in (b). There are two values of each angle for

each value of y = Im z in the interval B to D’. Associated angles d,. c#J,, related by (10) and (11).

are denoted by the same symbol (+ or X) in each of the figures. The boundary bifurcation point is

at y, = 1.073621. , where 4, -358”, and 4, - 177”. At D’, y,, = 1.150964. Table III contains

the numerical data for these figures.

The boundary bi&rcation point is now located by seeking the condition

under which the cubic has two coincident real roots, with absolute value less

than unity. For repeated roots the derivative quadratic also vanishes:

3a’Z’ + 2b’Z + c’ = 0. (13)

On subtracting Z x (13) from the cubic (12), one obtains a second quadratic


360

PI, 358

356

354

I; ,

352c 166 168 170 172 174 176 178 1

0s

Fig. 6. Lines in the angle diagram 4, versus 4s for the (5,3, 1) triangular lattice corresponding to

the bifurcation angles (+ or X) in fig. 5.

expression, which also must vanish when the cubic has repeated roots:

b’Z* + 2c’Z + 3d’ = 0. (14)

These two quadratics have an eliminant:

A = (9a’d’ - b’c’)* - 4(b” - 3a’c’)(c” - 3b’d’) (15)

(See appendix B.) This eliminant vanishes when y is chosen so the repeated

root condition is satisfied. Then cos $X is obtained from:

cos ~, _ z = (9a’d’ - b’c’) = 2(c” - 3b’d’) 5 2(bt2 - 3~2’~‘) (9a’d’ - b’c’) (16)

Numerical evaluation for the (5,3, 1) lattice reveals that the eliminant is

negative between I and D’, except for a narrow region between y = 1.071010

and 1.073621. At the boundary bifurcation point ye = Im z = 1.073621. . . , at

4, = 358.480. . .’ and c$, = 177.350. . .‘.

6. Boundary bifurcation at the imaginary axis

An alternative way of establishing the boundary bifurcation condition is to

regard eqs. (1) and (3) as yielding bifurcation solutions for y and one of the

angles, @s say, which depend on the other angle as a parameter. Then at the

boundary bifurcation point, y is an extremum (minimum) as a function of the

chosen parametric angle. Since c$~ is now a function of +,, changes in y are

336 J. Stephenson I Parliiion function zeros for Ising model VI

related by

$f dY af af d4 __+++-_2=()

3~ d4r a4, a4.7 d4,

and

af’ dy af' af' d4 ---_++++_2=~, a~ d& a4 WY d6

( 174

(17b)

As a function 4, alone, y is an extremum at the boundary bifurcation point, so

dyld$, = 0 in (17a, b). Consequently

d4S af d4r = % i (18)

whence the (third) condition required to locate the boundary bifurcation point

is

J_ a(f> f'> = af af' af af' _ __ = a(4,, 4,) a4, 84, a4, a4, O’

(19)

The Jacobian can be evaluated from (1) and (2):

J = (BC, - CB,) sin 4, sin 4,

+ sin(4, + 4,T)[(BD, - LIB,) sin 4, - (CD, - DC,) sin +.,I . (20)

J vanishes at the boundary bifurcation point, as located by our earlier criterion.

(This was first noticed by numerical evaluation!) On expressing / in terms of

cos 4,, by using eqs. (lo), (11) and (B.12) which are derived from (1) and (3),

one retrieves the quadratic equation (13) previously written down on the basis

of the repeated root condition. The three criteria: f = 0, f, = 0 and J = 0, in

(I), (3) and (20), can be put into symmetrical algebraic form, as we explain in

appendix B.

7. The boundary bifurcation discriminant

Here we wish to report that, in spite of the extensive computer algebra

entailed, we have achieved the complete elimination of the angles. Of necessity

the discriminant A in (15) was obtained by an “asymmetrical” calculation, and

so is not immediately completely symmetrical in the three triangular lattice


interaction variables. The final (reduced) eliminant should be symmetrical (in

d, e, f and U, u, w). The situation is analogous to that for partially anisotropic

lattices discussed previously in V, but much more complicated.

Also one expects the eliminant to have four critical point factors, as in IV

(eq. (S)), since the polynomial (1) has double roots when both angles are 0 as

at C, or n as at N. However an analysis of the corresponding algebraic problem

for partially anisotropic lattices reveals that this factorization occurs for every coefficient of the three (homogeneous quadratic) combinations of d, e, f which

can appear: d2, de and e2, only after one has expressed these coefficients in

terms of U, u, w. [i.e. the factorization does not occur sooner at a higher level.]

In a preliminary calculation for (5,3,1) we found that the discriminant has

the (four) critical point expressions as repeated factors. There are also other

unexpected factors, and the lowest degree polynomial obtained (of degree 33 in

z”) has extraneous real (positive) roots.

For a completely anisotropic lattice one has all combinations of d, e, f of

(homogeneous) degree 8, of which there are 45. Each combination has a

corresponding coefficient, which is a polynomial in U, u, w. We call these

combinations and their coefficients cases. The number of cases which have to

be calculated out in full could be reduced to 10 if complete symmetry under

permutation of d, e, f and CL, u, w were present, but unfortunately this is not so.

The discriminant calculated by the above “asymmetric” method is in fact not

immediately symmetrical at the d, e, f and U, u, w level, and has “irrelevant”

factors. However our procedure has retained symmetry in two (pairs) of the

variables: d and f, and u and w. Consequently there are 25 independent cases

to calculate.

Standard computer algebra methods enable one to obtain the coefficients in

the quadratic equations (13) and (14) fairly easily. Then in the discriminant (or

eliminant) (15) set

r’ = 9a’d’ - b’c’ , SI = br2 - ‘Ja’c’ , t’ = Cl2 - 3b’d’ . (21)

The full expressions for r’, s’, t’ respectively contain 1145, 954, and 1287 terms, each comprising an integer coefficient, a homogeneous combination of d, e, f of degree 8, and products of (some or all of) U, u, w to various degrees. To

expedite the calculation further, we calculated the polynomial coefficients of

each of the 25 independent cases separately. [Thus we only had to calculate

1442 559 of the possible 1145 x 1145 + 954 x 1287 = 2 538 823 terms in A.]

Moreover, the four repeated critical point factors were extracted individually

for each stage. The resulting (partially factorized) discriminant contained 3972

terms, and was partially symmetrical: i.e. under d ++ f, and u * w simulta-

neously. It was now necessary to obtain, and remove, an extraneous partially


symmetrical factor of unknown homogeneous degree in d, e, f, with polynomial

coefficients in ~1, u. W. This was achieved. The resulting quotient eliminant is

completely symmetrical in d, e, f and U, u, W, and comprises 841 terms. It

reduces correctly, after further factorization, for partially anisotropic lattices

(Stephenson [2]). The extraneous factor. which is a perfect square, and the 7

independent component cases, from which the entire 841 term symmetrical

eliminant can be constructed, are displayed in table IV. [Technical details of

the calculation will be published separately.]

Explicit evaluation of the final reduced symmetrical eliminant confirms all

our trial calculations for (5, 3, 1) described above.

8. Concluding remarks

In this series of papers [6, 7, 3, 1, 21 (to be referred to as I, 11, III. IV, V.

respectively) and the present paper, we have endeavoured to describe and

account for the geometrical, numerical and analytical properties of interior and

boundary temperature zeros on the various classes of anisotropic triangular and

quadratic lattices, and we have derived expressions for the density of zeros

valid near critical points. It remains to investigate the density of zeros

throughout the regions containing zeros.

Table IV Reduced symmetrical bifurcation discriminant factor and extraneous partially symmetrical factor. The seven independent cases from which the bifurcation discriminant can be constructed are tabulated according to the degrees of d, e. f, which are indicated in the form [lmn]. The total homogeneous degree is six, in d, e. f. In each case the terms in the polynomial comprise an integer coefficient, and products of IL. u, w to various degrees, as listed below the corresponding variable. The complete symmetrical discriminant contains all the (symmetrical) combinations of the basic cases. [222] is already symmetrical. [600] must be combined with [060] and [006], etc. Remember that when d, e, f are permuted, U, U. w must also be permuted in the same manner. An overall factor of -3uuw has been removed. The extraneous factor is of homogeneous degree two in d. e, f. and is a perfect square. It is symmetrical under interchange of d, Sand ~1. w.

(a) Extraneous partially symmetrical factor

d (’ f U U H’ d c f U U M’

I 2 0 0 2 4 2 I 0 2 0 4 2 0 -2 2 0 0 2 2 2 -2 (1 2 0 2 2 2

1 2 0 0 2 0 2 I 0 2 0 0 2 4 -2 I 1 0 3 3 1 2 0 I I 3 3 1

2 I I 0 3 1 I -2 0 I I 3 I 1 2 1 1 0 I 3 3 -2 0 I 1 1 3 3

-2 I I 0 1 I 3 2 0 1 I I I 3 -2 I 0 I 2 4 2 I 0 0 2 2 4 2

4 1 0 1 2 2 2 -2 0 0 2 2 2 2 -2 1 0 1 2 0 2 I 0 0 2 2 0 2

J. Stephenson I Partition function zeros for Ising model VI

Table IV (continued)

339

(b) Reduced symmetrical bifurcation discriminant factor

16001 u L’ w ~510] u u !+ [222] u ” w [330] u ” w

108 6 2 2 144 5 3 1

27 5 1 1 144 5 3 3 27 S I 5 76 7 3 1 27 5 5 1 36 5 1 3 4 8 2 0 16 7 I 3 4 8 0 2 16 7 1 1 4 8 2 2 12 6 2 4

-1 9 1 I 12 6 2 0 -18 7 I 3 2Yl I -18 7 1 1 -12 6 4 2 -18 7 3 1 -12 8 2 0 -54 5 3 1 -12 8 2 2 -54 5 I 3 -12 6 4 0 -54 5 3 3 -1X 5 1 5

-1X 5 1 1 -126 5 5 I -216 6 2 2

[420] u v w

255 5 5 1 X4 6 2 2 36 3 5 1 36 3 5 3 36 3 3 3 36 6 4 0 36 b 4 2 12 8 2 2 12 4 2 0 12 4 2 6 12 8 2 0 12 4 6 0 12 4 6 2 2711 2 7 I 3 2 5 I 3

-I 5 1 1 -1 Y 1 1 -1 5 1 5

-12 4 2 2 -12 4 2 4 -1x 3 7 1 -18 3 3 5 -18 3 3 1 -24 b 2 0 -24 6 2 4 -24 4 4 4 -24 4 4 0 -30 5 3 I -30 5 3 3

-116 7 3 1 -228 4 4 2

[411] Id ” w

456 6 2 2 22X 4 4 2 22x 3 2 4 60 5 1 5 60 5 5 I 54 3 1 3 54 3 3 1 48 4 4 4 36 8 2 2 32 5 1 1 18 3 7 1 18 3 I 7

-4 9 1 1 -10 7 1 1 -12 6 2 4 -12 b 4 2 -18 3 1 1 -18 3 5 3 -1x 3 3 5 -24 4 2 6 -24 4 b 2 -36 3 3 3 -54 3 5 1 -54 3 1 5 -74 7 1 3 -74 7 3 I -92 5 I 3 -92 5 3 1

-204 4 2 2 -56X 5 3 3

480 2 4 4 480 4 4 2 4x0 4 2 4 144 4 4 4 X4 3 3 3 X4 2 2 2 36 2 2 8 36 8 2 2 36 2 8 2 12 2 6 2 12 6 2 2 12 2 2 6 8 3 1 5 8 1 3 5 x 5 3 1 8 1 5 3 8 3 5 1 8 5 1 3 6 5 1 5 6 5 5 1 6 1 5 5 4 1 1 7 4 1 7 1 4 7 I 1 4 1 3 I 4 1 1 3 4 3 1 1

-1 9 1 1 -1 1 Y 1 -1 1 1 9 -1 1 1 1 -2 1 3 7 -2 7 3 I -2 3 7 1 -2 1 7 3 -2 7 1 3 -2 3 1 7 -6 5 1 1 -6 1 1 5 -6 1 5 1

-10 3 1 3 -10 I 3 3 -10 3 3 1 -36 6 4 2 -36 2 6 4 -36 6 2 4 -36 2 4 b -36 4 6 2 -36 4 2 6

-132 2 2 4 -132 2 4 2 -132 4 2 2 -490 5 3 3 -490 3 5 3

490 3 3 5

480 4 4 2 76 3 7 1 76 7 3 1 4X 4 4 4 4x 4 4 0 24 2 2 4 20 3 3 5 20 3 3 1 12 2 6 4

12 6 2 0 12 6 2 4 12 2 4 4 12 4 2 4 12 4 2 2 12 2 4 2 12 2 b 0 8 6 2 2 8 2 6 2 4 2 2 0 4 2 2 8

-4 8 2 0 -4 8 2 2 -4 2 x 2 -4 2 8 0

-12 4 2 0 -12 2 4 6 -12 2 4 0 -12 4 2 6 -16 2 2 2

-16 2 2 6 -36 6 4 0 -36 6 4 2 -36 4 6 2 -36 4 6 0 -40 3 3 3

-Y6 3 5 3 -96 3 5 1 -96 5 3 1 -96 5 3 3

-312 5 5 1

[321] u u us

452 5 3 3 204 4 2 2 154 3 5 1 118 7 3 1 86 3 3 5 60 4 6 2 60 6 2 4 36 2 2 6 36 2 4 6

12 2 8 2 12 2 2 2 6 3 1 3 651 I 2 9 1 1 2 3 1 7

-2 7 1 3 -2 5 1 5 -2 3 1 1 -4 5 I 3 -6 3 1 5 -6 7 I 1

-10 3 5 3 -12 2 6 2 -12 2 2 8 -12 3 3 3 -12 2 4 2 -12 4 2 6 -24 2 4 4 -36 6 4 2

-36 2 6 4 -36 8 2 2 -36 2 2 4 -42 5 5 1 -44 5 3 1 -4X 4 4 4 -74 3 3 1 -7x 3 7 1

-180 6 2 2 -192 4 2 4 -264 4 4 2

340 .I. Stepherwon I Partition function zeros fbr Ising model VI

Acknowledgements

I am very grateful to R. Teshima for assistance with the computer algebra,

and his algebraic and numerical analysis of the (5, 3,1) discriminant polyno-

mial. Without the benefit of his advice and insights, much of this work would

not have been attempted, nor would it have succeeded.

Appendix A

Here we provide details of the algebra leading to (7)-(9). The problem is to

eliminate the angles between

f(~;~,.~,~)-g+ih=A+Bcos~,+CcOs~,+~cos(~,+~,) (A.1)

and

B sin 4, = C sin 4, = -D sin(4, + 4,,) , (A.21

where A, B, C, D are the polynomials in (2), which we write in the form

A = 1 + u2 + r,? + w2, B = 2(uu - w) ,

C=2(wu-u), D = 2(uw - U) , (A.3)

with

ll = Zd , v = zr ) w = zJ . (A.4)

To derive (9a), first multiply the A, B, C, D terms in (A.l) respectively by the

-D, C, B, and -D terms in (A.2), and cancel the factor of sin(+, + 4,) which

now appears to obtain the equation

AD - BC + D ’ cos( 4, + &) = 0 , (A.9

which yields (SC). The expressions for the other cosines can be derived in the

same way, or written down directly by symmetry in B, C, D. When these

cosine formulae are substituted in the polynomial (A.l), the final algebraic

expression is, as in (7),

B2C2 + B2D2 + C2D’ - 2ABCD = 0. (A.6)


Using (A.6) one obtains corresponding expressions for the sines:

sin2+, = (B2 + C* + D* - A2)lB2 (A.7)

and

sin2+,y = (B* + C2 + D2 - A2)/C2 . 64.8)

Thus all the terms in (A.2) have the common value

B sin +, = c sin +s = -D sin(+, + 4,) = *[B* + C2 + D2 - A2]“2 )

(A.9)

where the 2 sign is chosen for numerical agreement. (A.6) is a polynomial of

seventh degree in u, u, and W. Numerical evaluation shows that +iz,, *Liz,

and +z, are roots of (A.6) for all permutations of (d, e, f), which fact yields

three disorder point (equation) type factors, as in IV (eq. (9)). The complete

factorization is

16uuw(l+ u + u + w)(l+ u - u - w)(l - ~1+ u - w)(l- u - u + w).

(A.lO)

Using the expansion of (A.lO) one finds that

B’ + C2 + D2 _ A2 = -l(juuw = -l(jZd+‘+f= 1(jy2(a+b+c) (A.ll)

since a + b + c is odd. This result can be used to simplify the RHS of (A.7),

(A.8) and (A.9).

Appendix B

Here we provide details of the algebra leading to the resolution of the

boundary bifurcation point problem, as in (12)-(16). If one expands cos(+, +

+s) in (1) and (3), by the trigonometric addition formula, then one has two

expressions involving cos 4, and sin c$, in a “symmetrical” manner, from which

one obtains (10) and (11). Then elimination of c$, using cos2 + sin2 = 1 leads to

the cubic (12) with coefficients:

342 J. Stephensorz I Partition function zero5 for Ising model VI

a’ = 2(DCz - CD;)(BC; - CB;) ,

b’ = (DC, - CD,)’ + (BC, ~ CB,)’ + (BD, ~ DB,)’

+ 2(DA, - AD,)(BC, - CB,) + 2(DCz - CD;)(BA. - AB,) ,

(B.1) c’ = 2[(DCz - CD,)(DAz - AD,) + (BA; - ABJ(DA, - ADZ)

+ (BA, ~ AB,)(BCz - CB:)] >

d’ = (DA; - AD,)’ + (AB; - BAZ)’ - (BDz - DBJ’

These coefficients are to be substituted in the eliminant, A in (1.5). Any two

simultaneous quadratic forms, such as those in (13) and (14),

3a’Z2 + 2b’Z

and

b’Z2 + 2c’Z i

have an eliminant

tc'=O or aZ’+pZ+r=O (B.2)

3d’=O or cu’Z’+p’Z+r’=O,

(or discriminant):

(B.3)

A = (cq’ - y’y’)’ - (a@’ ~ @‘)( Py’ - yp’) , 03.4)

which yields (15). Then

(F’ - T’) (Pr’ - rP’) cm-3 4, = z= (@, _ pa’) = (rat _ cyyl) ’ (B.5)

from which one derives (16). Numerical evaluation for the (.5,3, 1) lattice

reveals that the discriminant is negative (and large!) between I and D’, except

for a rather narrow region between 1’ = 1.071010 and 1.073621, at B. Con-

sequently it is better to compare the ratios in eq. (BS), or (16), for Z, in

locating zeros of A.

In order to exhibit the algebraic criteria f = 0, f; = 0, and J = 0, in (l), (3)

and (20), for the boundary bifurcation point in a useful and symmetrical form,

it is convenient to introduce a compact notation for the skew-symmetrical

combinations of A, B, C, D and their z-derivatives, which frequently appear,

by setting

(BC) - BCz - CB, , etc.

[These bracket expressions are not all independent. For example:

(CA)(BD) + (BA)(DC) - (BC)(DA) = 0,

(B.6)

(B.7)


which can be used to express (CA) in terms of the other five (independent)

bracket symbols.] Also set

R=cos d,, S=COS+,~ and T=cos(~~ + $,s). (B.8)

Then one can rearrange the Jacobian in terms of cosines as

J = (BC)(RS - T) + (DC)(R - ST) + (BD)(S - RT) , (B.9)

and the two polynomials (1) and (3) as

A+BR+CS+DT=O, A, + B,R + C,S + D,T = 0. (B.lO)

Here R, S, T are identically inter-related by

2RST-R2-S2-T2=l. (B.ll)

The elimination can start from these compact symmetrical forms, (B.9)-

(B. ll), for example by solving (B. 10) as two simultaneous equations for R and

S to obtain:

R = [(CA) - (DC)T] I(BC) , S = [(BA) + (BD) T] I(CB) , (B.12)

which are consistent with (10). On substituting in the Jacobian in (B.9) one

obtains a quadratic in T, whereas the identity (B.ll) yields a cubic equation in

T, from which T is to be eliminated. A second quadratic is obtained by

eliminating the cubic term between the quadratic and cubic equations. The

resulting expressions are equivalent to (12), (13) and (14) under a cyclic

permutation of variables. The final eliminant is constructed as before. [Alter-

natively, as in the text, we could have kept S as the final variable to be

eliminated.]

References

l] J. Stephenson, Physica A 148 (1988) 88-106, to be referred to as IV. 21 J. Stephenson, Physica A 148 (1988) 107-123, to be referred to as V.

31 J. Stephenson and J. van Aalst, Physica A 136 (1986) 160-175, to be referred to as III.

41 J. Stephenson, J. Phys. A: Math. Gen. 20 (1987) L331-L335.

51 J. Stephenson, J. Math. Phys. 11 (1970) 420-431.

61 J. Stephenson, and R. Couzens, Physica A 129 (1984) 201-210, to be referred to as I. 71 J. Stephenson, Physica A 136 (1986) 147-159, to be referred to as II.

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Partition function zeros for the two-dimensional Ising model VI