The configuration space of equilateral and equiangular hexagons

The configuration s pa ce of equilateral a nd equi a ngular hexagons Jun O’Hara No v em ber 11, 2021 Abstract W e study the configuration space of equ ilateral and equiangular spatial hexagons for any b ond angle by giving explicit expressions of all the p ossible shap es. W e sho w that the chair configuration is isolated, whereas th e b oat configuration allo ws one-dimensional deformatio ns whic h form a circle in the configuration space. Key wor ds and phr ases . Configuration space, equiangular, polygon. 2000 Mathematics Subje ct Classific ation. Primary 51N20; Second ary 51H99, 55 R80 1 In tro duction Let P be a p o ly gon wit h n vertices in R 3 . W e expre ss P by its vertices, P = ( P 0 , P 1 , · · · , P n − 1 ), with suffixes mo dulo n . A polyg on P is called e quilater al if the edge length | P i +1 − P i | is cons ta n t, and e quiangular if the angle ∠ P i − 1 P i P i +1 is cons tant. T his angle betw een tw o adjacent edg es is called the b ond angle and will b e denote by θ in this pap er. An equilateral and eq uiangular po lygon ca n b e considered as a mathematical mo del o f a cyclo alk ane . W e are in terested in the set of a ll the p ossible shap es (which are called conformations in chemistry) of cycloalk anes when the n um b er n of car b ons and the b ond angle θ is fixed, i.e. in the language of mathematics, the configuratio n space of equilater al and equiangular p oly gons. Gordon Crippen studied it for n ≤ 7 ([C]). T o be precise, what he obtained is not the configuration space itself, but the s pace of the “ metric matrices ”, whic h are n × n matrices whose en tries ar e inner pro ducts of pairs of edge vectors, and then he gav e the co r resp onding conformations. When n = 4 (cyclobutanes) and n = 5 (cyclop entanes) he cons idered all the po ssible bo nd angles, but when n = 6 (cyclohexanes) and n = 7 (cyc lo p e n tanes) he fixed the bond angle to b e the ideal tetrahedral bond ang le cos − 1 ( − 1 3 ) ≈ 1 0 9 . 47 ◦ (Figure 1). He showed that if n = 6 the conformation space is a union of a circle which con tains a b o at (Figur e 2) a nd an isolated po int of a chai r (Figure 3), and that if n = 7 it consists of t wo circles, o ne for boat/twist-bo at and the other for c hair/twist-chair. In these t wo c a ses, he sho wed it by searching out all the po ssible v alues of the entries of the metric matrix thro ugh numerical exp er iment with 0.0 5 step size. Figure 1 : θ = cos − 1 ( − 1 3 ) Figure 2 : b o at Figure 3 : chair In this pap er we study the configuratio n spa ce of hexagons for a ny b ond angle. W e give explicit expressions of all the po ssible co nfigurations in terms of our pa rameters illustrated in Figure 8 . W e show that the topo logical type of the c o nfiguration spa c e dep ends on the bond a ngle θ . 1 If θ is big ( π 3 < θ < 2 π 3 ) the situation is same as that o f cyclo hexanes of idea l tetra hedral b ond angle studied b y Cr ippe n. On the other hand, if θ is s ma ll (0 < θ < π 3 ), a new configura tion (“ inw ard cr own ” illustrated in Figur e 11) app e a rs, and the one dimensiona l contin uum of deformatio n of a b oat is div ided int o tw o pieces, which implies that we cannot deform a boat into its mirr or image. W e remar k that we dis ting uish vertices in our study . Hence our configura tion s pace is not equal to the s pace of shap es. W e hav e a n exceptional case if θ = π 3 , when the inw ard cr own, which degenerates to a doubly cov ered triangle, c an b e deformed to boats via newly appear ed families o f configura tions. In any cas e, a chair is an isola ted co nfiguration, wherea s a boat allows one - dimensional deformatio ns starting fr om and ending at it. Moving pictures of deforma tion of a b oat ca n be found at http:/ /www. comp.tmu.ac.jp/knotNRG/math/configuration.html Of course, as extremal cases, we hav e a 6 times cov ered multple edge and a reg ular hexagon when θ = 0 , 2 π 3 . There hav e b een a grea t n umber of studies of the configura tion spaces of linkages . An excelle n t survey can b e fo und in [D-OR]. If we dro p the conditio n of being equiang ular, it is known tha t the space o f equilateral polygo ns has a symplectic structur e ([Kp-M]). This pap er is based on the author’s talk at “ Knots and soft-matter physics, T op ology of po lymers and related topics in physics, mathematics and biolog y”, YITP , Kyoto, 200 8 (A short announcement of the r e s ult without pro of was r ep orted in Bussei Kenky uu vol. 92 (20 09) 119 – 122). Notations . Throughout the pap er, we a gree that C = cos( θ 2 ) and S = sin( θ 2 ). The s uffixes are understo o d mo dulo n . The angle ∠ P i means ∠ P i − 1 P i P i +1 . 2 Preliminaries Definition 2.1 Put f M n ( θ ) =  P = ( P 0 , . . . , P n − 1 ) ( P i ∈ R 3 )     | P i − P i +1 | = 1 , ∠ P i − 1 P i P i +1 = θ ( ∀ i (mo d. n ))  . Let G be the group of orientation pre s erving isometries of R 3 . Put M n ( θ ) = f M n ( θ ) / G , and call it the c onfigu r ation sp ac e of θ -e quiangular unit e quilater al n -gons (0 ≤ θ < π ). Let us denote the equiv alence class o f a plolygon P by [ P ]. Remark 2.2 (1) W e allow in tersections o f edges and ov erlapping of vertices. (2) W e distinguish the vertices w he n we consider our configura tion s pace. Ther efore, tw o configurations illustrated in Figures 3 corresp ond to differ ent p oints in M 6 , a lthough their shap es ar e the same. (3) When w e ex press an eq uila teral and equia ngular p olygon we may fix the first three vertices, P 0 , P 1 , and P 2 . There are ( n − 3 ) more vertices, wher eas we hav e ( n − 2) co nditions for the lengths of the edges, and ( n − 1) conditions for the ang le s . Therefore, we ma y exp e ct that the dimension of M n ( θ ) is equa l to 3( n − 3) − ( n − 2) − ( n − 1) = n − 6 in genera l if the conditions are indep endent, which is not the c a se when n ≤ 6 as we will see. When n ≤ 5 the co nfiguration space is given as follows. Prop ositio n 2.3 ([C]) The c onfigu r ation sp ac es M n ( θ ) of e quilater al and θ -e quiangular n -gons ( n = 3 , 4 , 5) and the shap es of p olygons which c orr esp ond to t he elements ar e given by the fol lowing. M 3 ( θ ) ∼ =  { 1 p oint } ( θ = π 3 ) regular tr ia ngle ∅ otherwise M 4 ( θ ) ∼ =        { 1 p oint } ( θ = 0) 4-folded edg e (Figure 5 left) { 2 p oints } (0 < θ < π 4 ) folded rhom bus (Figure 5 center) and its mirror image { 1 p oint } ( θ = π 4 ) square (Figure 5 rig ht ) ∅ otherwise, M 5 ( θ ) ∼ =    { 1 p oint } ( θ = π 5 ) regular star shape (Figure 6) { 1 p oint } ( θ = 3 π 5 ) regular p entagon (Figure 7) ∅ otherwise. 2 Figure 4 : A chair when θ is small and its mirr or image Figure 5 : n = 4 ca se. The middle is a non-planar config ur ation. Figure 6 : Regula star shape Figure 7 : Regular p entagon Pro of. The cases when n = 3 , 4 are obvious. Suppo se n = 5. Let ( P 0 , . . . , P 4 ) ∈ f M 5 ( θ ). Then θ 6 = 0 , π . W e may assume, after a motion of R 3 , that P 0 = (0 , 0 , 0) , P 1 = (1 , 0 , 0), and P 4 = (co s θ , sin θ , 0). Then P 2 and P 3 can be expressed by P 2 =   1 − cos θ sin θ cos ϕ 2 sin θ sin ϕ 2   , P 3 =   cos θ sin θ 0 sin θ − cos θ 0 0 0 1     1 − cos θ sin θ cos ϕ 3 sin θ sin ϕ 3   for some ϕ 2 and ϕ 3 . Using sy mmetry in a plane that contains the angle bisector of ∠ P 1 P 0 P 4 and the z -axis, we can deduce from | P 2 P 4 | 2 = | P 1 P 3 | 2 =  2 sin θ 2  2 that cos ϕ 2 = co s ϕ 3 = 1 − 2 cos θ + 2 cos 2 θ 2 sin 2 θ , which implies sin ϕ 2 = ± sin ϕ 3 . Then − − − → P 1 P 2 · − − − → P 2 P 3 = − cos θ implies − 4 cos 2 θ + 2 cos θ + 1 4(1 − cos θ ) = sin 2 θ sin ϕ 2 (sin ϕ 2 − sin ϕ 3 ) . (2.1) 3 It follo ws that, whether sin ϕ 2 = sin ϕ 3 or sin ϕ 2 = − sin ϕ 3 , cos θ = 1 ± √ 5 4 , namely , θ = π 5 , 3 π 5 , and ϕ 2 = ϕ 3 = 0, which means only r egular star shap e and reg ula r pentagon can app e ar as equilater al and equiangular p entagons. ✷ 3 Equilateral and equiangular h exagons Put C = cos( θ 2 ) and S = sin( θ 2 ). First note that P 0 , P 2 , and P 4 form a regular triangle of edge length 2 S . W e ma y fix P 0 =   − S 0 0   , P 2 =   S 0 0   , P 4 =   0 √ 3 S 0   . (3.1) An y element in M 6 ( θ ) has exactly one representativ e hexagon with P 0 , P 2 , and P 4 being a s ab ove. Let us us e a “ double cone ”o r susp ension ex pr ession o f a hexagon (Figur e 8), namely , we express P 1 , P 3 , a nd P 5 by Figure 8 : Double cone or susp ension expression P 1 =    0 − C cos ϕ 1 C sin ϕ 1    , P 3 =    1 2 S + √ 3 2 C cos ϕ 3 √ 3 2 S + 1 2 C cos ϕ 3 C sin ϕ 3    , P 5 =    − 1 2 S − √ 3 2 C cos ϕ 5 √ 3 2 S + 1 2 C cos ϕ 5 C sin ϕ 5    (3.2) for some ϕ 1 , ϕ 3 , and ϕ 5 . Now the c o nditions | P j − P j +1 | = 1 ( ∀ j ) a nd ∠ P 1 = ∠ P 3 = ∠ P 5 = θ ar e satisfied. The condition ∠ P i +1 = θ ( i = 1 , 3 , 5) is equiv alent to C 2 (cos ϕ i cos ϕ i +2 − 2 sin ϕ i sin ϕ i +2 ) + √ 3 S C (cos ϕ i + cos ϕ i +2 ) = 3 − 5 C 2 , (3.3) which is equiv alen t to C  C cos ϕ i + √ 3 S  cos ϕ i +2 − 2 C 2 sin ϕ i sin ϕ i +2 = 3 − 5 C 2 − √ 3 S C cos ϕ i . (3.4) Remark that the equations ax + b y = d ( a 2 + b 2 > 0) and x 2 + y 2 = 1 hav e solutions if and only if a 2 + b 2 − d 2 ≥ 0, when we hav e x = ad ± b √ a 2 + b 2 − d 2 a 2 + b 2 , y = bd ∓ a √ a 2 + b 2 − d 2 a 2 + b 2 . (3.5) In o ur case (3.4), b y substituting a = C  C cos ϕ i + √ 3 S  , b = − 2 C 2 sin ϕ i , d = 3 − 5 C 2 − √ 3 S C cos ϕ i , (3.6) 4 we have a 2 + b 2 = 4 − ( S − √ 3 C cos ϕ i ) 2 , which is positive unless θ = π 3 and ϕ i = π , and a 2 + b 2 − d 2 = − √ 3  C cos ϕ i − √ 3 S  √ 3 C cos ϕ i − (3 − 8 C 2 ) S  . (3.7) It follows tha t when ( θ, ϕ 1 ) 6 = ( π 3 , π ) there ar e ϕ 3 and ϕ 5 so that ∠ P 0 = ∠ P 2 = θ if and only if cos ϕ 1 satisfies (3 − 8 C 2 ) S √ 3 C ≤ co s ϕ 1 ≤ √ 3 S C , which can happ en if and only if C = cos θ 2 ≥ 1 2 , i.e. 0 ≤ θ ≤ 2 π 3 (Figure ). Figure 9 : The reg ion of cos ϕ 1 so that there are ϕ 3 and ϕ 5 satisfying ∠ P 0 = ∠ P 2 = θ (1) Let us first assume that θ and ϕ 1 satisfy the conditions ab ov e men tioned and search for the case when ϕ 3 and ϕ 5 that make ∠ P 0 = ∠ P 2 = θ also s a tisfy ∠ P 4 = θ . (W e will study the cas e when ( θ, ϕ 1 ) = ( π 3 , π ) later.) W e hav e tw o ca ses, either ϕ 3 6 = ϕ 5 or ϕ 3 = ϕ 5 . Case I. Assume ϕ 3 6 = ϕ 5 . Remar k that this can o ccur if and only if ϕ 1 satisfies (3 − 8 C 2 ) S √ 3 C < co s ϕ 1 < √ 3 S C  0 < θ < 2 π 3  . Then the conditions ∠ P 0 = ∠ P 2 = θ imply that ϕ 3 and ϕ 5 are g iven by { (cos ϕ 3 , sin ϕ 3 ) , (cos ϕ 5 , sin ϕ 5 ) } = ( ad ± b √ a 2 + b 2 − d 2 a 2 + b 2 , bd ∓ a √ a 2 + b 2 − d 2 a 2 + b 2 !) , (3.8) where a, b , and d are given by (3.6). Computating the left ha nd side of (3.3), we have C 2 a 2 d 2 − b 2 ( a 2 + b 2 − d 2 ) ( a 2 + b 2 ) 2 − 2 b 2 d 2 − a 2 ( a 2 + b 2 − d 2 ) ( a 2 + b 2 ) 2 ! + √ 3 S C 2 ad a 2 + b 2 = 3 − 5 C 2 , which implies tha t the condition ∠ P 4 = θ is alwa ys satisfied in this case. Now (3 .2 ) shows tha t P 3 and 5 P 5 are g iven by P 3 =            S − √ 3 C cos ϕ 1 − (3 − 8 C 2 ) S ± √ 3 C sin ϕ 1 √ a 2 + b 2 − d 2 4 − ( S − √ 3 C cos ϕ i ) 2 2 √ 3 S − C cos ϕ 1 − 1 √ 3 (3 − 8 C 2 ) S ± C sin ϕ 1 √ a 2 + b 2 − d 2 4 − ( S − √ 3 C cos ϕ i ) 2 − 2 C  3 − 5 C 2 − √ 3 S C cos ϕ 1  sin ϕ 1 ±  C cos ϕ 1 + √ 3 S  √ a 2 + b 2 − d 2 4 − ( S − √ 3 C cos ϕ i ) 2            P 5 =            − S + √ 3 C cos ϕ 1 − (3 − 8 C 2 ) S ∓ √ 3 C sin ϕ 1 √ a 2 + b 2 − d 2 4 − ( S − √ 3 C cos ϕ i ) 2 2 √ 3 S − C cos ϕ 1 − 1 √ 3 (3 − 8 C 2 ) S ∓ C sin ϕ 1 √ a 2 + b 2 − d 2 4 − ( S − √ 3 C cos ϕ i ) 2 − 2 C  3 − 5 C 2 − √ 3 S C cos ϕ 1  sin ϕ 1 ∓  C cos ϕ 1 + √ 3 S  √ a 2 + b 2 − d 2 4 − ( S − √ 3 C cos ϕ i ) 2            , (3.9) where a 2 + b 2 − d 2 is g iven by (3.7). Case I I. Assume ϕ 3 = ϕ 5 . Let us first s tudy the condition for ∠ P 4 = θ without assuming ∠ P 0 = ∠ P 2 = θ . If ϕ 3 = ϕ 5 , which we deno te by ϕ , a nd ∠ P 4 = θ , then (3.3) implies that ϕ must satisfy cos ϕ = − √ 3 S C or S √ 3 C . Case I I-1 . Assume (cos ϕ 3 , sin ϕ 3 ) = (cos ϕ 5 , sin ϕ 5 ) = S √ 3 C , √ 4 C 2 − 1 √ 3 C ! . Then (3.3) implies that ∠ P 0 = ∠ P 2 = θ if and only if (cos ϕ 1 , sin ϕ 1 ) = S √ 3 C , √ 4 C 2 − 1 √ 3 C ! or (3 − 8 C 2 ) S √ 3 C , (4 C 2 − 3) √ 4 C 2 − 1 √ 3 C ! . Note that we have − − − → P 0 P 5 = − − − → P 2 P 3 by (3.2). The p o ints P 1 and P 4 are in the opp osite (or same) side of the pla ne containing P 0 , P 2 , P 3 , and P 5 if co s ϕ 1 = S √ 3 C (or r esp ectively cos ϕ 1 = (3 − 8 C 2 ) S √ 3 C ). Namely , the hexagon is a chair if ϕ 1 = ϕ 3 = ϕ 5 and a boa t if ϕ 1 6 = ϕ 3 = ϕ 5 in this case. Both coincide if and only if θ = 0 o r 2 π 3 , when P is a 6-times co vered multiple edge or a regular hexa gon. The c hair is giv en by P 1 =     0 − S √ 3 √ 4 C 2 − 1 √ 3     , P 3 =     S 2 S √ 3 √ 4 C 2 − 1 √ 3     , P 5 =     − S 2 S √ 3 √ 4 C 2 − 1 √ 3     , whereas the b oat is giv en by substituting (cos ϕ 1 , sin ϕ 1 ) =  (3 − 8 C 2 ) S √ 3 C , (4 C 2 − 3) √ 4 C 2 − 1 √ 3 C  to (3.9), P 1 =     0 − (3 − 8 C 2 ) S √ 3 (4 C 2 − 3) √ 4 C 2 − 1 √ 3     , P 3 =     S 2 S √ 3 √ 4 C 2 − 1 √ 3     , P 5 =     − S 2 S √ 3 √ 4 C 2 − 1 √ 3     . Case I I-2 . Assume (cos ϕ 3 , sin ϕ 3 ) = (cos ϕ 5 , sin ϕ 5 ) = − √ 3 S C , √ 4 C 2 − 3 C ! , 6 which can o ccur if and only if √ 3 2 ≤ C ≤ 1, namely , 0 ≤ θ ≤ π 3 . Then (3.3) implies that ∠ P 0 = ∠ P 2 = θ if and only if 2 C √ 4 C 2 − 3 sin ϕ 1 = 8 C 2 − 6 . Therefore, when θ 6 = π 3 (w e will study the case when θ = π 3 and ϕ 3 = ϕ 5 = π later ) then ∠ P 0 = ∠ P 2 = θ if and only if (cos ϕ 1 , sin ϕ 1 ) = − √ 3 S C , √ 4 C 2 − 3 C ! or √ 3 S C , √ 4 C 2 − 3 C ! . Note that P 3 and P 5 are above P 0 and P 2 resp ectively . When ϕ 1 = ϕ 3 = ϕ 5 the hexagon is an “inw ard crown” (Figur e 11) giv en by P 1 =    0 √ 3 S √ 4 C 2 − 3    , P 3 =    − S 0 √ 4 C 2 − 3    , P 5 =    S 0 √ 4 C 2 − 3    , whereas the other is given by substituting (co s ϕ 1 , sin ϕ 1 ) =  √ 3 S C , √ 4 C 2 − 3 C  to (3 .9), P 1 =    0 − √ 3 S √ 4 C 2 − 3    , P 3 =    − S 0 √ 4 C 2 − 3    , P 5 =    S 0 √ 4 C 2 − 3    . Let us summarize the argument ab ov e when θ 6 = π 3 . Theorem 3.1 Supp ose a θ - e qu iangular unit e qu ilater al hexagon ( θ 6 = π 3 ) is p ar ametrize d by the agnles ϕ 1 , ϕ 3 , and ϕ 5 by (3.1) , (3.2 ) . L et C = cos  θ 2  and S = sin  θ 2  as b efor e. (1) When θ = 2 π 3 i.e. C = 1 2 we have ϕ 1 = ϕ 3 = ϕ 5 = 0 , whi ch c orr esp onds to a r e gular hexagon. (2) When π 3 < θ < 2 π 3 i.e. 1 2 < C < √ 3 2 we have − arccos  (3 − 8 C 2 ) S √ 3 C  ≤ ϕ 1 ≤ a rccos  (3 − 8 C 2 ) S √ 3 C  . (i) When ϕ 1 = ± arccos  (3 − 8 C 2 ) S √ 3 C  we have ϕ 3 = ϕ 5 = ∓ arccos  S √ 3 C  , which c orr esp onds to a b o at (Figur e 2). (ii) When − ar ccos  (3 − 8 C 2 ) S √ 3 C  < ϕ 1 < a rccos  (3 − 8 C 2 ) S √ 3 C  we have either * ϕ 3 6 = ϕ 5 , which ar e given by (3.8) , * ϕ 1 = ϕ 3 = ϕ 5 = ± ar ccos  S √ 3 C  , which c orr esp onds t o a chair (Figur e 3). (3) When 0 < θ < π 3 i.e. √ 3 2 < C < 1 we have a rccos  √ 3 S C  ≤ | ϕ 1 | ≤ arcc o s  (3 − 8 C 2 ) S √ 3 C  . (i) When ϕ 1 = ± arccos  (3 − 8 C 2 ) S √ 3 C  we have ϕ 3 = ϕ 5 = ± arccos  S √ 3 C  , which c orr esp onds to a b o at (Figur e 10). (ii) When ϕ 1 = ± arccos  √ 3 S C  we have ϕ 3 = ϕ 5 = ± ar ccos  − √ 3 S C  = ±  π − arccos  √ 3 S C  . (iii) When arc c o s  √ 3 S C  < | ϕ 1 | < a rccos  (3 − 8 C 2 ) S √ 3 C  we have either * ϕ 3 6 = ϕ 5 , which ar e given by (3.8) , * ϕ 1 = ϕ 3 = ϕ 5 = ± ar ccos  S √ 3 C  , which c orr esp onds t o a chair (Figur e 4). * ϕ 1 = ϕ 3 = ϕ 5 = ± ar ccos  − √ 3 S C  , which c orr esp onds to an “inwar d cr own ” (Figur e 11). (4) When θ = 0 the hexagon de gener ates t o a 6 times c over e d mult iple e dge. 7 Figure 1 0 : A boat w ith a s mall b ond angle Figure 11: “ inw ard crown ” in a prism. Two edges intersect each other in a side face of the prism. Corollary 3.2 The c onfigur ation sp ac e M 6 ( θ ) of θ -e quiangular unit e quilater al hexagons ( θ 6 = π 3 ) is home omorphic to a p oint if θ = 0 , 2 π 3 , the un ion of a cir cle and a p air of p oints if π 3 < θ < 2 π 3 , the un ion of two cir cles and four p oints if 0 < θ < π 3 , and the empty set if θ < 0 or θ > 2 π 3 . Bo at c onfigur ations ar e include d in cir cles ab ove mentione d, and chairs ar e isolate d. W e will see that a b o a t degenera tes to a planar co nfiguration when θ = π 3 . Corollary 3.3 (1) A b o at and its mirr or image c an b e joine d by a p ath in the c onfigur ation sp ac e, i.e. they c an b e c ontinu ously defo rme d fr om one to t he other, if and only if the b ond angle satisfies π 3 < θ < 2 π 3 . (2) A b o at and a chair c annot b e joine d by a p ath in the c onfigur ation s p ac e, i.e. they c annot b e c ontinu ously defo rme d fr om one to the other. (2) Finally we study the exceptional case θ = π 3 , when the cas es w he n ϕ j = π ( j = 1 , 3 , 5) hav e no t bee n co ns idered yet. When θ = π 3 the equa tion (3.3) beco mes (cos ϕ i + 1)(cos ϕ i +2 + 1) − 2 sin ϕ i sin ϕ i +2 = 0 . If ∠ P 2 = π 3 then ϕ 3 is deter mined by ϕ 1 as follows; • if ϕ 1 = π then ϕ 3 is a rbitrary , • if ϕ 1 = 0 then ϕ 3 = π , • if ϕ 1 6 = 0 , π then ϕ 3 = π or f ( ϕ 1 ) ( f ( ϕ 1 ) 6 = π ), where f ( ϕ ) ( ϕ 6 = ± π ) is given b y (cos f ( ϕ ) , sin f ( ϕ )) =  − (cos ϕ + 1) 2 + 4 sin 2 ϕ (cos ϕ + 1) 2 + 4 sin 2 ϕ , 4 sin ϕ (cos ϕ + 1 ) (cos ϕ + 1) 2 + 4 sin 2 ϕ  . (3.10) Remark that f (0) = π and that f ( ϕ ) = ϕ if and o nly if ϕ = ± ar ccos( 1 3 ). Put f ( π ) = 0 as lim ϕ → π f ( ϕ ) = 0. Theorem 3.4 Supp ose a π 3 -e qu iangular u nit e quilater al hexagon is p ar ametrize d by the agnles ϕ 1 , ϕ 3 , and ϕ 5 by (3.1) , (3.2) . Then we have { ϕ 1 , ϕ 3 , ϕ 5 } = { π , ϕ, f ( ϕ ) } , { π , π , ϕ } , or  ± arcco s  1 3  , ± arcco s  1 3  , ± arcco s  1 3  , wher e ϕ is arbitr ary. The first c ase c ontains a b o at when { ϕ 1 , ϕ 3 , ϕ 5 } = { π , ± ar ccos  1 3  , ± arcco s  1 3  } , and the last triples c orr esp ond to a cha ir. Corollary 3.5 The c onfigur ation sp ac e M 6 ( π 3 ) of e quilater al and π 3 -e qu iangular hexagons is home omor- phic to the union of a p air of p oints and the sp ac e X il lustr ate d in Figur e 12 which is a 1 -skelton of a tetr ahe dr on with e dges b eing double d. The author would like to clo se the article with an op en problem: find a new inv arian which can show that a boa t canno t b e defo rmed contin uously into a chair. 8 Figure 12: The co nfiguration s pace X = M 6 ( π 3 ) \ { c hairs } . The num b e rs 0 , 2, and 4 in the figure indicate the vertices P 0 , P 2 , and P 4 resp ectively . The figur es of six hexag ons a round X are seen from a b ove. The non-planar co nfigurations left below is a b o at, when P i o ccupy five vertices of a regular o ctahedro n. The four vertices A, B , C , and D of X corres po nd to planar config urations parametrized by ( ϕ 1 , ϕ 3 , ϕ 5 ) = ( π , π , π ) , ( π , π , 0) , (0 , π, π ), and ( π, 0 , π ) r esp ectively . The cir cle throug h A and D consists of the config- urations parametrized b y ( ϕ 1 , ϕ 3 , ϕ 5 ) = ( π , ϕ, π ) ( − π ≤ ϕ ≤ π ). The circle hro ugh B and C consists of the co nfigurations par ametrized by ( ϕ 1 , ϕ 3 , ϕ 5 ) = ( ϕ, π , f ( ϕ )) ( − π ≤ ϕ ≤ π ). 9 References [C] Gordon M. Cripp en, Explo ring the c o nformation s pace of cycloalk anes by linear ized embedding. J. Comput. Chem., 1 3(3) (19 9 2), 351 – 361. [D-OR] E.D.Demaine and J. O’Ro urke, Geometric F olding Algorithms: Link a ges, Origami, Polyhedra. Cambridge Universit y P ress, (2007). [Kp-M] M. Ka povich and J. Millson, The symplectic ge o metry of po lygons in Euclidea n space. J. Different ial Geom. 44 (1996), 479 – 513. Department of Mathematics and Information Sciences, T o kyo Metrop olita n Universit y , 1-1 Mina mi-Ohsaw a, Ha chiouji-Shi, T okyo 1 92-03 97, JAP AN. E-mail: o hara@tmu.ac.jp F ax : 81 -42-6 77-24 81 10