The Cycloheptane Molecule

The Cycloheptane Molecule A Challenge to Computer Algebra (Invited lecture, ISSAC 97, Maui, Hawaii, USA, (version Jan. 98)) A.H.M. Levelt University of Nijmegen, The Netherlands ahml@sci.kun.nl 1

Thanks to: G. Pfister and colleagues at Kaiserslautern for their hospitality, introduction to Singular and help with my computations. J.-C. Faugère and D. Lazard for their attention, suggestions and explanation, in particular of FGB. G.J.M. Ariaans, Organic Chemistry Department of the University of Nijmegen, for opening to me the fascinating world of molecular computation and visualization. Mark van Hoeij for our many discussions, his suggestions and criticism. 2

Overview How it started First computations Cycloheptane enters the scene New symmetric computations Pictures Gröbner basis exercises Symmetry Does it matter to the chemists? 3

How it started About 1993 some Dutch mathematicians were playing with robot-like toys made of plastic electricity tubes as I show you here (picture on slide 6). The 6-toy (resp. 7-toy ) consists of 6 (resp. 7) rods of equal length = 1, connected in cyclic order, the angle between successive rods being straight. If one rod is fixed the next one can freely rotate about the first one in a plane perpendicular to that first one (forget about the remaining rods). Interesting observation: both toys have one degree of freedom (as you may feel while handling them). Other observation: the 6-toy has a second, rigid, realization. Problem: Try to describe and understand the movements of the toys from a mathematical point of view. Easy for the 6-toy. Algebraic geometrist tried (a bit) to do the 7. No success. As an ignorant in robotics I looked at the problem as a pastime and solved the problem (that is to say...) using a computer algebra system (Maple). From time to time I returned to the problem and did some more computations. 4

Somewhat later cycloheptane appeared. Study of these concrete objects has many aspects: Algebraic geometry (curves, real and complex) Gröbner bases Symmetry and invariants Symbolic computation Numerical computation Graphics Applications (chemistry) Cycloheptane offers a good opportunity to demonstrate many well-known techniques of computer algebra. 5

First computations Cf. A lecture in honour of B. Buchberger, Nijmegen, October 1993 = P 0 7 P x y z P P P P 1 2 3 4 5 P 6 P a a a a a a 2 4 6 7 3 5 a 1 α α α α α α α The notations in this picture are used everywhere in the sequel. 6

For vectors u = [u 1, u 2, u 3 ], v = [v 1, v 2, v 3 ] in 3-space write (u, v) = u 1 v 1 + u 2 v 2 + u3v3, the inner product. Voici les relations à satisfaire: (1) (a1, a2) = (a2, a3) = = (a6, a7) = (a7, a1) = c, (a1, a1) = (a2, a2) = = (a7, a7) = 1, a1 + a2 + + a7 = 0. 7

Assume a1 = [1, 0, 0], a2 = [0, 1, 0]. Then a3 = [cos(α1), 0, sin(α1)] satifies (a2, a3) = 0, (a3, a3) = 1. Note: (a1, a3) = cos(α1). Next try a4 = (x, y, z). (a3, a4) = 0 implies z = x cos(α1)/ sin(α1). Substitute in 1 = (a4, a4) = x 2 + y 2 + z 2 and find x 2 sin(α1) 2 + y2 = 1. Put x/ sin(α1) = sin(α2), y = cos(α2). Then a4 = [sin(α1) sin(α2), cos(α2), cos(α1) sin(α2)]. Note: (a2, a4) = cos(α2). Introducing two more angles one finds: a6 = [cos(α4), sin(α3) sin(α4), cos(α3) sin(α4)], a7 = [0, cos(α3), sin(α3)]. Note that: (a2, a7) = cos(α3), (a1, a6) = cos(α4). The closing condition a1 + + a7 = 0 yields a5 = a1 a2 a3 a4 a6 a7, an expression in sines and cosines of α1,..., α4. Conditions for a5: (a5, a5) = 1, (a4, a5) = 0, (a5, a6) = 0 = 3 polynomial equations in sines and cosines of α1,..., α4. 8

Choose rational parametrizations of the type cos(α) = 1 s2 2s 1 + s2, sin(α) = 1 + s 2 = three polynomial equations g1 = 0, g2 = 0, g3 = 0 in s1, s2, s3, s4. Here is g 1 : 17 + s3 2 s4 2 s1 2 + s3 2 s2 2 s4 2 + s2 2 s4 2 s1 2 + s3 2 s2 2 s1 2 + 9 s2 2 + 9 s3 2 + 9 s4 2 + 9 s1 2 + 8 s1 s3 + s4 2 s1 2 + s3 2 s2 2 s4 2 s1 2 8 s3 s4 8 s2 s4 + s3 2 s2 2 + 8 s1 s2 + 9 s2 2 s4 2 16 s3 s2 + s3 2 s4 2 + 9 s3 2 s1 2 + 16 s4 s1 + s2 2 s1 2 + 8 s1 s3 2 s4 2 s2 8 s2 2 s3 s1 2 s4 8 s1 s3 s4 2 s2 2 + 8 s1 s3 s4 2 + 8 s2 s4 2 s1 + 8 s3 2 s4 s2 s1 2 + 8 s3 s4 s1 2 8 s1 s3 s2 2 + 16 s4 s1 s2 2 8 s3 2 s4 s2 + 8 s2 s3 2 s1 + 8 s2 s1 2 s4 + 8 s2 2 s4 s3 16 s3 s2 s1 2 Final computation, elinination by means of resultants: res1 = resultant s 4 (g1, g2), res2 = resultant s 4 (g1, g3). res1 and res2 have some spurious factors. There remain 2 irreducible polynomials in s1, s2, s3. Taking resultants w.r.t. s3 and s2 leads to 2 irreducible symmetric polynomials in s1, s2, resp. s1, s3 of degree 32 with 145 terms, 10 digits coefficients. 9

Cycloheptane enters the scene Early 1994 the Dutch physicist P.W. Kasteleyn draw my attention to: L.J. Oosterhoff, Restricted free rotation and cyclic molecules, Ph.D. thesis, Leiden, 1949. It contains a nice symmetric approach to cyclohexane. Cyclohexane is like our 6-toy, the straight angle is replaced by a 109 degree one (cosine = 1/3). I applied his idea to cycloheptane. 10

New symmetric computations x 7 x 1 α a 7 α a 1 x 6 a 6 α α a 2 x 2 α a 5 a 3 x 5 α a 4 α x 3 Here are the relations defining our model: x 4 (1) (a1, a2) = (a2, a3) = = (a6, a7) = (a7, a1) = c, (a1, a1) = (a2, a2) = = (a7, a7) = 1, a1 + a2 + + a7 = 0. c = 0 for the 7-toy, c = 1/3 for cycloheptane. Other values of c between -1 and 1 may be considered. The equations (1) make sense in Euclidean n-space. In our case n 3. The constraints imposed by the 3- dimensionality will be analyzed below. 11

Number of equations and unknowns For ai in 3-space the unknowns are the 21 coordinates of a1,..., a7. (1) is a system of 17 equations in the coordinates of a1,..., a7. Not all solutions lead to different configurations: congruent ones should be identified. Normalize by picking 1 out of each congruence class. This can be realized by imposing 3 more equations: a1 2 = 0, a1 3 = 0, a2 3 = 0. Now we have 20 polynomial equations in 21 unknowns. From affine algebraic geometry we know that (over an algebraically closed ground field) each irreducible component of the solution space has dimension 1. (Note: the solution space may be empty!) 12

The matrix S Define the 7x7 matrix S by S i,j = (ai, aj) (inner product of ai and aj). Easy to see: all S i,j are simple expressions in x 1 = S 2,7, x 2 = S 1,3, x 3 = S 2,4, x 4 = S 3,5, x 5 = S 4,6, x 6 = S 5,7, x 7 = S 1,6. a1,..., a7 are 7 vectors in 3-space. = any 4 of them linearly dependent. E.g. for a1, a2, a3, a4 λ1,..., λ4 not all = 0, such that λ1a1 + λ4a4 = 0. Take inner product with aj (j = 1,..., 7): λ1s 1,j + λ4s 4,j = 0. = first 4 columns of S linearly dependent. Hence necessary condition: (2) rank(s) 3. Equivalent: all 35 principal subdeterminants of S of order 4 vanish. I denotes the ideal generated by these subdeterminants. 13

Assume rank(s) = 3 Remark: rank(s) 2 impossible for c = 0, c = 1/3. May suppose S1 = S 1,1 S 1,2 S 1,3 S 2,1 S 2,2 S 2,3 S 3,1 S 3,2 S 3,3 is invertible. S2 def = the matrix consisting of the first 3 rows of S and U = S1 1 S2. = the j-th column of U is the coordinate vector of aj w.r.t. the basis a1, a2, a3 (assumed independent). cartesian coordinate system such that a1 = [1, 0, 0], a2 = [c, 1 c 2, 0], c(1 x2) x2 a3 = [x2, 2, + 2c 2 x2 2c 2 + 1 ]. 1 c 2 1 c 2 T def def = the 3x3 matrix with columns a1, a2, a3, C a = T U. Then a 1,..., a 7 are the successive columns (viewed as vectors) of C a. Remark: The essential point is that the coordinate expression of (ai, aj) is what you expect. 14

Explicit formulas for a1,..., a7, case c = 0 x2 2 + y2 2 = 1 (y2 = 1 x2 2 ) a1 = [1, 0, 0] a2 = [0, 1, 0] a3 = [x2, 0, y2] a4 = [ 1 x2 x3 + x6, x3, 2 1 y2 x2 ( 1 2 x2 2 x3 + 2 x6) ] 2 x2 2 1 a5 = [ x7 + x3 x6 1 2, 1 x3 + x7 x4, 2 1 y2 (2 x2 x7 2 x3 x2 + 2 x6 x2 + x2 + 2 x4) ] 2 x2 2 1 a6 = [x7, x7 1 x1 + x4, 2 1 y2 2 ( 2 x2 x7 1 2 x4 + 2 x1 2 x5) x22 1] a7 = [0, x1, 1 y2 (1 + 2 x1 + 2 x2 2 x5) 2 x2 2 ] 1 15

Explicit formulas for a1,..., a7, case c = 1/3 9(x2 1/9) 2 + 8y2 2 = (8/3) 2 (i.e. y2 = 1/4 18x2 2 + 4x2 + 14) a1 = [1, 0, 0] a2 = [ 1 3, 2 3 2, 0] a3 = [x2, 1 4 2(x2 1), y2] a4 = [ 5 6 x2 x3 + x6, 1 2(54x2 2 + 63x2 + 144x2x3 54x2x6 18x6 + 72x3 + 23), 12 9x2 + 7 1 (54x2 2 + 39x2 + 72x2x3 54x2x6 24x3 + 6x6 + 11)y2 ] 6 9x2 2 2x2 7 a7 = 16

Symmetry C: cyclic permutation C : a 1 a 2 a 7 a 1. C respects the relations (1) between a 1,..., a 7. T : permutation T : a1 a6, a2 a5, a3 a4, also respects the relations (1). C and T generate the dihedral group D 7. Note: C : x 1 x 2 x 7 x 1. T : x1 x6, x2 x5, x3 x4, D 7 operates on the polynomial ring R = Q[x1,..., x7]. All (polynomial) relations between x 1,..., x 7 following from (2) are respected by the action of D 7. Hence belong to R, the subring of invariants of R under the action of D 7. In particular I comes from an ideal Ĩ of R. The configuration space Γ of cycloheptane (resp. the c-toy ) is the zero set of I, an algebraic subspace of 7-space. Remark: Chemists say conformation instead of configuration. 17

Questions about Γ 1. Is Γ non-empty? 2. How to find points on Γ? 3. Dimension of Γ? 4. Is Γ smooth? 5. How many (real) components? 6. Find nice (e.g. small) sets of generators of I. 7. How does D 7 operate on Γ? 8. Can Γ be plotted? 9. If so, do we better understand the movements? 18

Elimination (Will lead to partial answers to the questions.) Computations in R = Q[x 1,..., x 7 ], c = 0, resp. c = 1/3. Symbols like f 125 denote an element of R only depending on x 1, x 2, x 5. Input for the next computation: 2 simple principal subdeterminants of order 3 of S: f 125, f 12356. The D 7 -symmetry is applied several times. E.g. C : f 125 f 236, C 3 : f 125 f 256. Try to find relations between as few x s as possible. Method: elimination using resultants. Cf. scheme on next slide. f 12 has total degree 32, 289 terms and 30-digits coefficients (12-digits for 7 ). Cpu time for computing f 23, f 36, f 35 375 secs (MapleV4, Pentium 200 MHz) 19

Elimination scheme f 125 x 6 g 235 x 1 f 2356 x 2 f 35 f 12356 x 5 g 236 f 256 x 6 f, f 23 36 x 2 x 6 f 235 f 236 Input Elimination scheme Output The arrows connecting f 125, f 12356 with f 2356 and the symbol x 1 mean: f 2356 = resultant x1 (f 125, f 12356 ). 20

Computing points on Γ Recent result: For all ξ 1, ξ 2 C with at most finitely many exceptions there exist precisely one sequence ξ 3,..., ξ 7 C such that ξ 1,..., ξ 7 is a point of Γ. Consequence: The projection mapping: Γ f 12 = 0 is a birational morphism of algebraic curves. Algorithm for computing ξ 3,..., ξ 7. Idea: Assume Γ. Let ξ 1,..., ξ 7 Γ. Then f 12 (ξ 1, ξ 2 ) = 0, f 15 (ξ 1, ξ 5 ) = 0, f 125 (ξ 1, ξ 2, ξ 5 ) = 0. f 15 and f 125 are known. f 125 has degree 2 in x 5. r 125 = prem(f 15, f 125, x 5 ). r 125 has degree 1 in x 5 and r 125 (ξ 1, ξ 2, ξ 5 ) = 0. Hence ξ 5 is a rational function expression in ξ 1, ξ 2. Consequence: with a finite number of exceptions ξ 1, ξ 2 determine ξ 5 uniquely. In a similar way one finds ξ 4, ξ 3, ξ 6, ξ 7. (x,x,x,x,x ) 1 2 4 5 6 (x,x,x,x,x ) 1 2 4 5 7 f 13457 (x,x,x,x,x 1 3 ) 4 5 7 f 12456 f f12457 f (x,x,x,x ) 1245 (x,x,x ) 1 2 4 5 1 2 5 f 125 Solution scheme (x 1,x ) 2 Algorithm works for floating point numbers, points with values in a finite field, resp. algebraic numbers. A point on Γ with values in algebraic numbers can be computed. 21

Pictures Example of an old (1993) plot: f13 for 7, plotted by Maple s implicitplot : c=0, x1, x3 0.8 0.6 0.4 x3 0.2 x1-1 -0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 0-0.2-0.4-0.6-0.8-1 and f 12 for cycloheptane: c=1/3, x1, x2 0.2 x1-1 -0.8-0.6-0.4-0.2 0.2 0-0.2-0.4 x2-0.6-0.8-1 22

Shapes of projections Three different shapes for the projections on the 2- dimensional coordinate planes. Quite similar for c=0 and c=1/3. But look at the loose part in the cycloheptane plot. Two problems: Problem 1. The projections of Γ are highly singular. Is Γ itself singular? What do the plots say about the move- Problem 2. ments? The algorithm for computing points on Γ leads to a (visual) solution to Problem 1. Projections of Γ onto 3- dimensional coordinate spaces can be computed. These projections clearly show that (the real part) of Γ is smooth and has 2 components. 23

Parametrization of cycloheptane y-axis L Q P C E -1-7/9 α 0 1 x-axis Needed: parametrization of the algebraic function y = 1/4 14 + 4x 18x 2, i.e. x and y related by E : 9(x 1/9) 2 + 8y 2 = (8/3) 2. P = (x, y) is on the ellips E. L line through (1, 0) and P. L cuts circle C in Q. t: slope of L. Then: ( ) 8t 2 7 P = 8t 2 + 9, 16t. 8t 2 + 9 Conversely: t = y/(x 1). Also: ( ) 1 + 15 cos(a) P = 17 cos(a), 16 sin(a). 17 cos(a) Conversely: cos(a) = (17x 1)/(x + 15). 24

Pictures of relations between angles cycloheptane 280 260 240 220 200 angle2 180 160 140 120 100 80 80 100 120 140 160 180 200 220 240 260 280 angle1 cycloheptane 280 260 240 220 200 angle3 180 160 140 120 100 80 280 260 240 220 200 180 160 angle2 140 120 100 80 280 260 240 220 200 180 160 angle1 140 80 100 120 Notice the two components 25

Gröbner basis exercises Some results: For 7 one can write down a system of 12 equations in 13 cartesian coordinates. G. Pfister (1995) solved this system (i.e. he computed f 12) using Singular. Computing time: 10 hrs on a HP workstation. (Cf. elimination scheme ) G. Pfister using the eliminate command of Singular computed f 23 in about 1800 secs. The harder problem of computing f35 took too much time. For the same problem J.-C. Faugère used FGb to compute f 35. Computing time 350 secs! (Comparable to the resultant method.) Why use Gröbner bases for this kind of problems? With fast Gröbner basis algorithms, the cycloheptane computations become automatic and trivial. For cycloheptane they may become as easy as the use of today s GB-packages for (the much simpler) cyclohexane. The cyclohexane computations will be shown now. 26

GB-computation for cyclohexane Setup similar to cycloheptane case. Here S is 6x6 matrix and there are 3 unknowns: x1, x2, x3. The ideal I of the principal 3x3 subdeterminants is generated bij 9 of them. The zero set of I is the configuration curve Γ 3 in 3-space. The symmetry group is the full S3. I is invariant under S 3. The Gröbner basis G ( plex ordering) is computed in a few seconds. G is generated by g1, g2, g3, g4 where g1 = f13 = 9x1 2 x3 2 + 6x1 2 x3 + 6x1 x3 2 23x1 2 20x1 x3 23x3 2 34x1 34x3 15, g2 = 9x1 x3 3 9x2 x3 3 9x1 x3 3 9x2 x3 2 6x3 3 + 20x1 x2 +41x1 x3 + 41x2 x3 + 18x3 2 + 21x1 + 21x2 + 54x3 + 18, g3 = (x3 + 3)(3x3 + 1)(9x2 x3 2 + 6x3 2 + 9x1 x3 2 + 6x1 x3 +6x2 x3 20x3 23x1 23x2 34 94 = f23 (= f13(x2, x3)). f13 is symmetric and transformed into f23, f12 by S 3. Note that (f 12, f 13, f 23) I. (Look at x1 = x2 = x3 = 1/3.) Similar to cycloheptane, but much simpler. 27

Primary decomposition of I for cyclohexane Primary decomposition of the configuration ideal I, computed by Singular: I = Q1 Q2 Q3, where Q1, Q2, Q3 are the prime ideals: Q1 = (9x2 x3 2 + 6x3 2 + 9x1 x3 2 + 6x1 x3 + 6x2 x3 20x3 23x1 23x2 34, f23, 3x1 x2 + 3x1 x3 + 3x2 x3 + 2x2 + 2x2 + 2x3 + 1), Q2 = (x1 + 3, x2 + 3, x3 + 3), Q3 = (3x1 + 1, 3x2 + 1, 3x3 + 1). Q2 belongs to the point [ 3, 3, 3] (no meaning for our model). Q3 belongs to [ 1/3, 1/3, 1/3] and corresponds to the (rigid) chair form of cyclohexane. Q1 corresponds to the 1 degree of freedom configuration of cyclohexane. Easy to verify: Q defines a (singular) plane curve of genus 1. 28

Symmetry For the notions and results cf. B. Sturmfels, Algorithms in Invarant Theory, Springer Verlag, 1993. In particular pp 58-64. D 7 operates on Γ = Γ 7. Challenge: Compute the relative orbit variety Γ/D 7. Little hope that Sturmfels s method works for Γ 7 : Gröbner bases computations are too big. Computations possible for Γ 3 with group S 3. Gives good idea of method as will be shown now. S 3 operates on 3-space and on R = Q[x1, x2, x3]. R, the subring of invariants, is Q[y1, y2, y3], where y1 = x1 + x2 + x3, y2 = x1 x2 + x2 x3 + x3 x1, y3 = x1 x2 x3. 29

Now view x1,..., y3 as indeterminates. T = Q[x1, x2, x3, y1, y2, y3] and J = (y1 (x1 + x2 + x3), y2 (x1 x2 + x2 x3 + x3 x1), y3 x1 x2 x3). J is the ideal belonging to the graph of the inclusion map of Q[y1, y2, y3] Q[x1, x2, x3]. The ideal H of T generated by J and I (= (g1, g2, g3, g4)) corresponds to graph of Γ 3 Γ 3 /S 3. Hence Γ 3 /S 3 corresponds to H Q[y1, y2, y3]. Generators can be found by standard methods of Gröbner basis theory. Fix an admissible ordering on T where x1, x2, x3 precede y1, y2, y3. The ideal H of T is generated by and g1, g2, g3, g4 y1 (x1+x2+x3), y2 (x1 x2+x2 x3+x1 x3), y3 x1 x2 x3. Compute Gröbner basis G H of H. It has 8 members, and G H Q[y1, y2, y3] is the set 17472y1 + 621y2 y3 + 26231y2 162y3 2 411y3 + 8721, 2093y2 2 + 1524y2 y3 1986y2 540y3 2 1188y3 + 405, 621y2 y3 2 + 16790y2 y3 + 621y2 162y3 3 4785y3 2 11112y3 405. 30

The zero set is the relative orbit space Γ 3 /S 3. It consists of 3 components, two points Y 1 = [ 9, 27, 27], Y 2 = [ 1, 1/3, 1/27] and the (parametrized) straight line s [ 9/23 s 34/23, 6/23, s + 15/23, s]. Over Y 1 (resp. Y 2) lies the point [ 3, 3, 3] (resp. [ 1/3, 1/3, 1/3]) that we have met before. Let X3 be the part of Γ lying over Y 3. One can prove: X3 Y 3 is a ramified covering of degree 6. On X3 there are 12 ramification points with ramification index 2, lying over the points y3 = 7 81, 9, 19 3 + 23 9 6, 19 3 23 9 Agreement with Hurwitz formula (g = genus) 2g(X3) 2 = deg(x3/y 3)(2g(Y 3) 2) + 6. P X3 (e P 1) where e P is the ramification index of the mapping at P. (Fill in: g(x3) = 1, g(y 3) = 0, deg(x3/y 3) = 6, e P = 2 at the 12 ramification points.) The ramification points correspond to special configurations, e.g. [ 1/3, 1/3, 7/9] X3 over y3 = 7/81. But what is special here? 31

Does it matter to the chemists? The answer is: NO!. In our model the geometry rules: the building blocks are rigid; in chemistry energy rules and nothing is rigid. Distances between carbon molecules may vary, just as angles between bonds. The molecule is viewed as a conglomerate of atoms kept together by the various forces between the constituents. The geometry of the molecule is the result of the balance of forces. The chemists flexible model is the opposite of the rigid one in this paper. Also in another respect the chemists model is more realistic. It takes into account the other (non-carbon) atoms. In their cycloheptane the 14 hydrogene atoms play a role. There exist powerful software for molecular modeling and visualization. A package like CS Chem3D Pro enables you draw organic chemical structures you have in mind. After the command minimize energy a computation starts that finds the nearest (local) minimum of energy by repeated small changes of the geometry. In this way one finds for cycloheptane the two characteristic shapes ( boat and chair ) depending on the starting position. Our Γ vanishes completely in this approach. For more complicate molecules there is a danger of overlooking some potential wells. 32