TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE. 1. Statement of results

Transcription

1 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE DONALD M. DAVIS Abstract. We determne, wthn 1, the topologcal complexty of many planar polygon spaces mod sometry. In all cases consdered except for those homeomorphc to real projectve space or n-tor, the upper and lower bounds gven by dmenson and cohomology consderatons dffer by 1. The spaces whch we consder are those whose genetc codes, n the sense of Hausmann and Rodrguez, have a sngle gene, and ts sze s Statement of results The topologcal complexty, TC(X, of a topologcal space X s, roughly, the number of rules requred to specfy how to move between any two ponts of X. A rule must be such that the choce of path vares contnuously wth the choce of endponts. (See [3, 4]. We study TC(X where X = M(l s the space of equvalence classes of orented n-gons n the plane wth consecutve sdes of length l 1,..., l n, dentfed under translaton, rotaton, and reflecton. (See, e.g., [4, 6]. Here l = (l 1,..., l n s an n-tuple of postve real numbers. Thus M(l = {(z 1,..., z n (S 1 n : l 1 z l n z n = 0}/O(2. We can thnk of the sdes of the polygon as lnked arms of a robot, and then TC(X s the number of rules requred to program the robot to move from any confguraton to any other. Snce permutng the l s does not affect the space up to homeomorphsm, we may assume l 1 l n. We assume that l n < l 1 + l n 1, so that the space M(l has more than one pont. We also assume that l s generc, whch means that there s Date: September 25, Key words and phrases. Topologcal complexty, planar polygon spaces Mathematcs Subject Classfcaton: 55M30, 58D29, 55R80. 1

2 2 DONALD M. DAVIS no subset M [n] = {1,..., n} wth n l = 1 l 2. When l s generc, M(l s an M =1 (n 3-manfold ([4, p.314], and hence, by [3, Cor 4.15], satsfes TC(M(l 2n 5. (1.1 It s well-understood that the homeomorphsm type of M(l s determned by whch subsets S of [n] are short, whch means that n l < 1 l 2. For generc l, a subset S =1 whch s not short s called long. Defne a partal order on the power set of [n] by S T f S = {s 1,..., s l } and T {t 1,..., t l } wth s t for all. As ntroduced n [5], the genetc code of l s the set of maxmal elements (called genes n the set of short subsets of [n] whch contan n. The homeomorphsm type of M(l s determned by the genetc code of l. Note that f l = (l 1,..., l n, then all genes have largest element n. Our man theorem s Theorem 1.2. If the genetc code of l conssts of a sngle gene of sze 2, 3, or 4, wth largest element n 5, and s not {5, 2, 1} or {6, 3, 2, 1}, then TC(M(l 2n 6. The two excluded cases are homeomorphc to tor (S 1 2 and (S 1 3. It s known ([3, (4.12] that TC((S 1 n = n + 1; ts genetc code s {n, n 3, n 4,..., 1}. The other known case n whch t s not true that TC(M(l 2n 6 s the case n whch the genetc code s {n}. Ths space M(l s homeomorphc to RP n 3, for whch the TC s often much less than 2n 7.([3, 4.8] In [1], we showed that TC(M(1 n 1, n 2k 2n 6 f 2 < 2k < n. (We use exponents for repetton n length vectors. The genetc code of ths s {n, n 1,..., n k + 1}. Thus n all cases consdered so far, except for tor and real projectve spaces, the upper bound for topologcal complexty of these planar polygon spaces gven by dmenson and the lower bound gven by our cohomologcal argument dffer by 1. Ths ncludes 12 out of the 21 6-gon spaces but only 26 out of the gon spaces.([6] Only the complcated nature of the cohomology rngs H (M(l; Z 2 hnders nvestgaton of spaces wth more complcated genetc codes. It seems not unreasonable to thnk that perhaps all spaces M(l of n-gons satsfy TC(M(l 2n 6 except for real projectve spaces and tor.

3 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE3 We wll prove n Theorems 2.3, 2.5, and 3.24 that, f X = M(l wth sngle gene of sze 2, 3, and 4, respectvely, and wth largest element n, then there are classes v 1,..., v 2n 7 n H 1 (X such that (v v 0 H (X X. (1.3 2n 7 =1 Here and throughout, all cohomology groups have coeffcents n Z 2. Theorem 1.2 then follows from [3, Cor 4.40]. 2. Proof of Theorem 1.2 for genetc codes wth a sngle gene of sze 2 and 3 The frst two of the three cases of (1.3 are handled n ths secton. We revew the determnaton of H (M(l obtaned n [4], smlar to the nterpretaton n [1, Thm 2.1]. Proposton 2.1. If l has length n, then the rng H (M(l s generated by classes R, V 1,..., V n 1 n H 1 (M(l subject to only the followng relatons: (1 All monomals of the same degree whch are dvsble by the same V s are equal. If S denotes ths set of s and d the degree, ths element s denoted by T d S. (2 If S [n 1] has S {n} long, then S V = 0. (3 If L [n 1] s long and L d + 1, then TS d = 0. (2.2 S L S {n} short Note that the d s not an exponent, and that f S =, then T d S = Rd s ncluded n the above sum. We wll often denote T d {}, T d {,j}, and T d {,j,k} by T d, T d,j, and T d,j,k, respectvely, and wll omt the superscrpt when t t clear from the context. If the LHS of (2.2 s denoted R d L, then the fact that { V l R d R d+1 L {l} L = Rd+1 L l L 0 l L and RR d L = Rd+1 L mples that the relatons (2.2 span the deal whch they generate.

4 4 DONALD M. DAVIS The frst case of (1.3, and hence Theorem 1.2, follows from the followng result. We remark that one such l s (1 a, 2 n a 1, 2n a 5. Theorem 2.3. If l has genetc code {n, a} wth 1 a n 1, and X = M(l, then (V V 1 n 3 (R R n 4 0 H (X X. Proof. Snce the genetc code has no sets of length 3, for all 2-subsets S, S {n} s long, and hence all products V V j are 0. Snce {n, } s long for > a, we have V = 0 for > a. The long subsets contaned n [n 1] are the complements of the short subsets contanng n. These long subsets are exactly the (n 2-sets {n 1,..., î,..., 1}, where the omtted element satsfes a. Thus the only relatons of type (3 occur n degree n 3. Therefore, for 1 d n 4, a bass for H d (X s {R d, T d 1,..., T d a }. The subsets S {n 1,..., î,..., 1} for whch S {n} s short are just and {j} for j a and j. Thus the relatons of type (3 are a R n 3 + T n 3 j = 0. j=1 j Subtractng pars of relatons reduces ths set of relatons to T1 n 3 = = Ta n 3 and R n 3 = (a 1T1 n 3. Note that ths mples H n 3 (X = Z 2, whch must be true snce X s an (n 3-manfold. Let m = n 3. We obtan, n bdegree (m, m 1, (V V 1 m (R R m 1 = ( m ( m 1 m V 1 R m V1 m R 1 = ( ( 2m 1 m + 1T m 1 T1 m 1 + T1 m R m 1. (2.4 Here we have used that ( ( m m 1 ( m = 2m 1 m and noted that all terms n the sum are of the form T1 m T1 m 1 except the one wth = m. Snce {R m 1, T1 m 1 } s lnearly ndependent and T1 m 0, (2.4 s nonzero. Now let X = M(l wth genetc code {n, a + b, a} wth n > a + b > a > 0 and n 6. Although rrelevant to our proof, we note ths can be realzed by l =

5 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE5 (2 a, 3 b, 6 n a b 1, 6n 3b 4a 17. For a class v H 1 (X, we wll let Dv = v 1+1 v. (Thnk dagonal. Note also that we are swtchng to tensor product notaton, usng the Künneth somorphsm. The second case of (1.3 and hence Theorem 1.2 follows mmedately from the followng result, the proof of whch wll occupy the rest of ths secton. Theorem 2.5. Let X = M(l wth genetc code {n, a + b, a} wth n > a + b > a > 0 and n 6. Let m = n 3 and suppose 2 e 1 < m 2 e. Then n H (X X, a. f a s even, then (DV 1 2m 1 2e (DV a+b (DR 2e 1 0; b. f a s odd and m 2 e 1 +1, then (DV 1 m 1 (DV a+b 2 (DR m 2 0; c. f a s odd and m = 2 e 1 + 1, then (DV 1 m (DV a+b m 1 0. Smlarly to the prevous proof, Proposton 2.1 easly shows that, for 2 d n 5, a bass for H d (X s {R d } {T d : 1 a + b} {T,j d : a, < j a + b}, and all other TS d are 0. Indeed, the sets S such that S {n} s short are {j, } wth < j a + b and a, {} wth a + b, and. There are no addtonal relatons untl degree n 4 snce the smallest L [n 1] for whch L {n} s long has L = n 3 = m. All the classes V wth a play the same role n the relatons of type (3 n Proposton 2.1, as do all the classes V wth a < a+b. The way that we wll show a class z n H 2m 1 (X X s nonzero s by constructng a unform homomorphsm ψ : H m 1 (X Z 2 such that (φ ψ(z 0, where φ : H m (X Z 2 s the Poncaré dualty somorphsm. By unform homomorphsm, we mean one satsfyng ψ(t = ψ(t j f, j a or f a <, j a + b, and ψ(t,j = ψ(t,k f j, k a or f a < j, k a + b. We wll let Y 1 refer to any T wth a, and Y 2 to any T wth a < a+b. Smlarly, Y 1,1 denotes T,j wth < j a, whle Y 1,2 s T,j wth a < j a + b. Usually the superscrpt, ndcatng the gradng, wll be mplct. If θ : [a + b] {1, 2} s defned by θ( = 1 f a, and 2 otherwse, then Y θ(s = T S, where θ(s s a multset. If ψ s a unform homomorphsm, then ψ(y W s a well-defned element of Z 2 for each of the four possble W. We also let w θ( = V. Thus V 1 and V a+b n Theorem 2.5

6 6 DONALD M. DAVIS are replaced by w 1 and w 2, respectvely. Also, R wll sometmes be denoted by w 0. Fnally, an element of [a + b] s of type 1 f t s a, and otherwse s of type 2. The relatons of type (3 n H m 1 (X have L obtaned from [n 1] by deletng one element of type 1 and another of ether type 1 or type 2. Call these relatons R 1,1 and R 1,2. If a = 1, there are no R 1,1 relatons. Occasonally, we use a superscrpt wth R W to denote the gradng. If a > 1 and ψ s a unform homomorphsm, then ψ(r 1,1 s ψ(r m 1 + (a 2ψ(Y 1 + bψ(y 2 + ( a 2 ψ(y1,1 + (a 2bψ(Y 2 1,2 = 0. (2.6 These coeffcents count the number of relevant subsets S L. For example, n the last term, snce ths L contans a 2 numbers of type 1 and b numbers of type 2, there are (a 2b ways to choose a set S L such that S {n} s short and S has one type-1 element and one type-2 element, and ψ sends each of them to the same element of Z 2. Smlarly, abbrevatng ψ(y W as ψ W, and ψ(r as ψ 0, ψ(r 1,2 s the relaton ψ 0 + (a 1ψ 1 + (b 1ψ 2 + ( a 1 2 ψ1,1 + (a 1(b 1ψ 1,2 = 0. (2.7 Proposton 2.8. Let φ : H m (X Z 2 be the Poncaré dualty somorphsm, and let φ W = φ(y W. Then φ 1,1 = φ 1,2 = 1 φ 2 = a 1 φ 1 = a + b φ 0 = (a 1b + ( a 1 2 Proof. By symmetry, φ s unform. Usng the notaton ntroduced above, there are relatons n H m (X of the form R 1 and R 2 satsfyng, respectvely, φ 0 + (a 1φ 1 + bφ 2 + ( a 1 φ1,1 + (a 1bφ 2 1,2 = 0 φ 0 + aφ 1 + (b 1φ 2 + ( a 2 φ1,1 + a(b 1φ 1,2 = 0, as well as relatons R m 1,1 and R m 1,2 lke (2.6 and (2.7, but wth ψ replaced by φ. As one can check by row-reducton or substtuton, the nonzero soluton of these four equatons (mod 2 s the one stated n the proposton..

7 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE7 Next we expand the expressons n Theorem 2.5. The party of a s not an ssue n these expansons. Part (a expands, n bdegree (m, m 1, as (n our new notaton 2m 2 2 e =1 ( 2m 1 2 e ( 2 e 1 m 1 w 1 w 2 R m 1 w 2m 1 2e 1 R 2e m+ +w 2 R m 1 w1 2m 1 2e R 2e m + w1 2m 1 2e w 2 R 2e m R m 1 + 2m 2 2 e =1 + ( 2 e 1 m The frst lne s ( 2m 1 2 e ( 2 e 1 m R m w 2m 1 2e w 1 R m w 2m 1 2e 1 w 2 R 2e m+ 1 1 w 2 R 2e m 1 + ( 2 e 1 2 e +1 m w 2m 1 2 e 1 R 2e +1 m w 2 R m 2. 2m 2 2 e =1 ( 2m 1 2 e( 2 e 1 m 1 tmes Y1,2 Y 1. Ths sum s easly seen to be 0 mod 2. Smlarly the sum on the thrd lne s 0. We obtan that the expanson n part (a s, n bdegree (m, m 1, Y 2 Y 1 + Y 1,2 R m 1 + (1 + δ m,2 er m Y 1,2 + Y 1 Y 2. (2.9 We frequently use Lucas s Theorem for evaluaton of mod 2 bnomal coeffcents. For example, here we use that ( 2 e 1 1 for all nonnegatve 2 e 1. Part (b expands, n bdegree (m, m 1, as m 2 =1 For m 2 e 1 + 1, and m 2 =1 m 2 + =2 ( m 1 ( m 2 m 2 w 1 w2r 2 m 2 w1 m 1 R ( m 1 ( m 2 m w 1 R m w m 1 1 w 2 2R 2 +w 2 2R m 2 w m mw m 1 1 R w 2 2R m 3. m 2 =1 ( m 1 ( m 2 ( m 2 2m 3 m ( m 1 ( m 2 m 2 ( 2m 3 + m δm,2 e + m, and so, smlarly to (2.4, the expanson equals m Y 1,2 Y 1 + (δ m,2 e + my 1 Y 1,2 + Y 2 Y 1 + my 1 Y 2. (2.10

8 8 DONALD M. DAVIS The expanson of part (c of Theorem 2.5 s easer. It equals, n bdegree (m, m 1, w 1 w 2e 1 2 w 2e w 2e w 2e 1 2 = Y 1,2 Y 1 + Y 1 Y 2. (2.11 Now we show, one-at-a-tme, that there are unform homomorphsms ψ such that φ ψ sends (2.9, (2.10, and (2.11 to 1. If ψ : H m 1 (X Z 2 s a unform homomorphsm, applyng φ ψ to (2.9 wth a even yelds, usng Proposton 2.8, ψ(y 1 + ψ(y 0 + ε 1 ψ(y 1,2 + bψ(y 2. (2.12 Here and n the followng, ε t denotes an element of Z 2 whose value turns out to be rrelevant. To have (2.12 be nonzero, we need a unform homomorphsm ψ satsfyng the system wth the followng augmented matrx. The columns represent ψ(y 0, ψ(y 1, ψ(y 2, ψ(y 1,1, and ψ(y 1,2, respectvely, and the second and thrd rows are (2.6 and ( b 0 ε b ε b 1 ε 2 b 1 0 Here we have noted that snce a s even, ( a seen to have a soluton, provng part (a of Theorem 2.5. ( a 1 mod 2. Ths system s easly Applyng φ ψ to (2.10 wth a odd and a > 1 smlarly yelds ψ(y 1 + ε 3 ψ(y 1,2 + m(b + 1ψ(Y 2. Now ψ must satsfy the followng system, whch s also easly seen to have a soluton. 0 1 m(b ε b ε 4 b b 1 ε Here we have used that f a s odd, then ( ( a 1 2 a If a = 1, then the fourth column and second row are removed, and agan there s a soluton. Fnally, applyng φ ψ to (2.11 wth a odd leads to the followng system for ψ, whch agan has a soluton. 0 1 b b ε 4 b b 1 ε Agan, the fourth column and second row are omtted f a = 1, but there s stll a soluton. Ths completes the proof of Theorem 2.5.

9 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE9 3. Proof of 1.2 for genetc codes wth a sngle gene of sze 4 In ths secton we prove (1.3 for X = M(l when the genetc code of l s {n, a + b + c, a + b, a} wth n > a + b + c > a + b > a > 0. The analyss s smlar to that of {n, a + b, a} n the prevous secton, except that there are many more cases to consder. Proposton 2.1 easly mples Proposton 3.1. For X as above, f 3 d n 6, H d (X has bass {R d } {T d : 1 a + b + c} {T d,j : 1 a + b, < j a + b + c} {T d,j,k : 1 a, < j a + b, j < k a + b + c}. For n 5 d n 3, these classes span H d (X but are subject to addtonal relatons. We adopt notaton smlar to that of the precedng proof, usng θ : [a + b + c] [3] sendng ntervals [1, a], (a, a + b], and (a + b, a + b + c] to 1, 2, and 3, respectvely. As before, we let Y θ(s = T S, w θ( = V, and m = n 3. We begn by provng, smlarly to Proposton 2.8, Theorem 3.2. Let φ : H m (X Z 2 be the Poncaré dualty somorphsm, and let φ W = φ(y W. Then φ 1,1,1 = φ 1,1,2 = φ 1,1,3 = 1 Proof. Let S = φ 1,2,2 = φ 1,2,3 = 1 φ 2,2 = φ 2,3 = a 1 φ 1,3 = a + b φ 1,1 = φ 1,2 = a + b + c 1 φ 3 = (a 1(b 1 + ( a φ 2 = (a 1(b + c + ( a 2 φ 1 = (a 1(a + b + c 1 + ( a ( + b 2 + (b 1(c 1 φ 0 = ( a 2 (a + b + c 1 + (a 1( ( b 2 + (b 1(c 1. {(1, (2, (3, (1, 1, (1, 2, (1, 3, (2, 2, (2, 3, (1, 1, 1, (1, 1, 2, (1, 1, 3, (1, 2, 2, (1, 2, 3}

10 10 DONALD M. DAVIS denote the set of types of elements that can be deleted from [n 1] yeldng a long subset L [n 1]. For example, (1, 2 refers to a set [n 1] {x, y} wth x a and a < y a + b. For U S and 1 3, let u denote the number of s n U. For example, f U = (1, 1, 2, then u 1 = 2, u 2 = 1, and u 3 = 0. For each U S, there s a relaton R U of type (2.2, and φ(r U s φ 0 + U S ( a u1 u 1 ( b u2 u 2 ( c u3 u 3 φu = 0. (3.3 For example, (2.7 s of ths form, ncludng only a and b (not c and correspondng to U = (1, 2, and wth ψ nstead of φ. The set of all equatons (3.3 s a system of 13 homogeneous equatons over Z 2 n 14 unknowns φ U, and ts nonzero soluton s the one stated n the theorem. Ths soluton was found manually by row reducton, and then checked by Maple, notng that the system only depends on a mod 4, b mod 4, and c mod 2. The program verfed that the soluton worked n all 32 cases. There are specal consderatons when a = 1 or 2, or b = 1. For example, f b = 1, then ( b 2 2 should be 0 for us, but s 1 n most bnomal coeffcent formulas. But the relatons R U n whch such coeffcents appear are not present, and so the equatons (3.3 for these U need not be consdered. Let S = {(1, 1, (1, 2, (1, 3, (2, 2, (2, 3, (1, 1, 1, (1, 1, 2, (1, 1, 3, (1, 2, 2, (1, 2, 3}. (3.4 For ψ : H m 1 (X Z 2 to be a unform homomorphsm wth ψ W = ψ(y W, the condtons that must be satsfed are, for U S, ψ 0 + ( a u1 ( b u2 ( c u3 u U 1 u 2 u ψu = 0. (3.5 3 S The dfference between the stuaton here and n the precedng proof s that relatons n H m (X allow sets L (n (2.2 wth L = m + 1 = n 2, but relatons n H m 1 (X do not.

11 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE 11 To prove (1.3, we seek ψ : H m 1 (X Z 2 satsfyng (3.5 for all U S, and {v 1,..., v 2m 1 } such that 2m 1 (φ ψ( (Dv = 1. (3.6 =1 The classes v that we wll use depend on the mod 4 values of a and b, and c mod 2. Frst we deal wth two cases that turn out to be exceptonal. The followng easly-verfed lemma wll be useful. Lemma 3.7. ( A 2 + AB + AC + BC + ( B 2 0 mod 2 ff A + B 0 mod 4 or A + B + 2C 1 mod 4. The followng result s our frst verfcaton of (3.6. Proposton 3.8. Suppose a b 1 mod 4 and c 1 mod 2. The somorphsm φ : H m (X Z 2 sends each Y,j,k to 1 and other monomals to 0. There exsts ψ : H m 1 (X Z 2 sendng each Y,j to 1 and other monomals to 0. Let ε = 1 f m 2 s a 2-power, and ε = 2 otherwse. Then (φ ψ((dw 1 m ε (Dw 2 2 (Dw 3 3 (DR m 6+ε = 1. Proof. The φ-part s easly checked usng Theorem 3.2, and the ψ-part follows from Lemma 3.7. Indeed, (3.5 becomes ( a u1 2 + (a u1 (b u 2 + (a u 1 (c u 3 + ( b u (b u2 (c u 3, whch, by 3.7, s 0 for the prescrbed a, b, c f u 1 + u 2 2 mod 4 or u 1 + u 2 + 2u 3 3 mod 4, and ths s easly verfed to be true for the ten elements of S. In the expanson of (Dw 1 m ε (Dw 2 2 (Dw 3 3 (DR m 6+ε, there are no terms wth repeated subscrpts n ether Y factor snce t only nvolves one element of each type. Also, there are no Y 1,2,3 Y 1,2 or Y 1,2,3 Y 2,3 terms, snce (Dw 2 2 = w w 2 2. When ε = 2, the Y 1,2,3 Y 1,3 terms come from + m 3 =1 =1 ( m 2 ( m 4 m 4 w 1 w2w 2 3R 2 m 4 w1 m 2 w 3 R m 3 ( m 2 ( m 4 m 3 w 1 w2w 2 3 R m 3 w1 m 2 w3r 2 1 = ( ( 2m 6 m 4 + 1Y1,2,3 Y 1,3 = Y 1,2,3 Y 1,3

12 12 DONALD M. DAVIS snce m 2 s not a 2-power. If m 2 s a 2-power, the smlar calculaton, nvolvng ( m 1 ( m 5 ( m 4 and m 1 ( m 5 ( m 3, gves just 2m 6 m 4 = 1. The other exceptonal case verfyng (3.6 s smlar. Proposton 3.9. Suppose a 2 mod 4, b 4 mod 4, and c 1 mod 2. The somorphsm φ : H m (X Z 2 sends Y,j,k, Y 2,2, and Y 2,3 to 1, and other monomals to 0. There exsts ψ : H m 1 (X Z 2 sendng each Y,j to 1 and other monomals to 0. Let ε = 1 f m 2 s a 2-power, and ε = 2 otherwse. Then (φ ψ((dw 1 2 (Dw 2 2 (Dw 3 m ε (DR m 5+ε = 1. Proof. The only term n the expanson whch s mapped nontrvally s Y 2,3 Y 1,3. It appears as m ε 1 =1 ( m ε wth coeffcent ( 2m 5 m 2 ( m 5+ε m 2 + ( m 5+ε w 2 2 w 3R m 2 w 2 1w m ε 3 R +ε 3 m 2 ( + m 5+ε ε 2 = 1. Let a (resp. b denote the mod 4 value of a (resp. b, and c the mod 2 value of c. For the other 30 cases of a, b, and c (or 90 f you consder devatons regardng whether m or m 1 s a 2-power, we use Maple to tell that an approprate ψ can be found. To accomplsh ths n all cases, several choces for the exponents of Dw 1, Dw 2, and Dw 3 are requred. Possbly some choce of exponents mght work n all cases, but we dd not fnd one. Most of our results wll be obtaned usng the followng result.

13 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE 13 Proposton If nether m nor m 1 s a 2-power, then the component of (Dw 1 α (Dw 2 2 (Dw 3 (DR 2m 4 α n bdegree (m, m 1 equals ( 2m 4 α (Y0 Y 1,2,3 + Y 1 Y 1,2,3 m + ( 2m 4 α (Y1,2,3 Y m Y 1,2,3 Y 1 + Y 3 Y 1,2 + Y 1,3 Y 1,2 + ( 2m 4 α (Y2 Y m 2 1,3 + Y 1,2 Y 3 + ( 2m 4 α (Y2,3 Y m Y 1,2,3 Y 1 + Y 1,3 Y 2 + Y 1,3 Y 1,2 + ( 2m 4 α (Y1 Y m 4 2,3 + Y 1 Y 1,2,3. If m s a 2-power, there s an addtonal term Y 1 Y 1,2,3. If m 1 s a 2-power, Y 1,2,3 Y 1 + Y 1,3 Y 1,2 must be added to the above expanson. Proof. The desred expresson expands as α ( α 2m 4 α ( m 3 w 1 w2w 2 3 R m 3 w1 α R m 1 α =0 α =0 α =0 α =0 ( α 2m 4 α ( m w 1 R m w1 α w2w 2 3 R m 4 α ( α 2m 4 α ( m 2 w 1 w2r 2 m 2 w1 α w 3 R m 2 α+ ( α 2m 4 α ( m 1 w 1 w 3 R m 1 w1 α w2r 2 m 3 α+. If nether m nor m 1 s a 2-power, the coeffcents ( 2m 4 m t for t = 3, 0, 2, 1, whch occur as the sum of all coeffcents on a lne, are 0. Thus the four lnes equal, respectvely, ( 2m 4 α (Y2,3 Y m Y 1,2,3 Y 1 + ( 2m 4 α (Y1,2,3 Y m α Y 1,2,3 Y 1, ( 2m 4 α (Y0 Y 1,2,3 + Y 1 Y 1,2,3 + ( 2m 4 α (Y1 Y 2,3 + Y 1 Y 1,2,3, m ( 2m 4 α m 2 ( 2m 4 α m 1 m α (Y2 Y 1,3 + Y 1,2 Y 1,3 + ( 2m 4 α (Y1,2 Y m α Y 1,2 Y 1,3, (Y3 Y 1,2 + Y 1,3 Y 1,2 + ( 2m 4 α (Y1,3 Y m 1 α 2 + Y 1,3 Y 1,2. The sum of these s easly manpulated nto the clamed form. If m s a 2-power, then ( 2m 4 ( m s odd, whle f m 1 s a 2-power, 2m 4 ( m 1 and 2m 4 m 3 are odd, yeldng the addtonal terms n the clam. The followng result follows mmedately from Proposton 3.10 and Theorem 3.2.

14 14 DONALD M. DAVIS Corollary Let q t = ( 2m 4 α m t for 0 t 4, and ψw = ψ(y W, where ψ : H m 1 (X Z 2 s a unform homomorphsm. Then (φ ψ((dw 1 α (Dw 2 2 (Dw 3 (DR 2m 4 α (3.12 = q 1 ψ 0 + (q 1 + q 3 aψ 1 + q 3 (a + bψ 2 + q 2 (a + b + c 1ψ 3 +(q 1 (ab ( a 2 + q3 (a + bψ 1,2 + q 2 ((a 1(b + c + ( a 2 ψ1,3 +q 4 ((a 1(a + b + c 1 + ( ( a b 2 + (b 1(c 1(ψ2,3 + ψ 1,2,3 +q 0 ((a + ( ( a 2 1(a + b + c 1 + a 1 ( 2 + a b 2 + a(b 1(c 1ψ1,2,3. Lemma If m = 2 e + m wth 2 m 2 e 1 and α = 2m 3 and 0 t 4, then ( 2m 4 α m t 1 mod 2. Proof. The top of the bnomal coeffcent s 2 e+1 1, whle the bottom s 2 e+1 1. The frst of several verfcatons of (1.3 for multple values of (a, b, c appears n the followng result. Theorem If m = 2 e + m wth 2 m 2 e 1, then (Dw 1 2m 3 (Dw 2 2 (Dw 3 (DR 2e H 2m 1 (X X for the values of a, b, and c whch have an n the 3.14 column of Table Proof. Ths s the case descrbed n Lemma 3.13, so that q 0 = = q 4 = 1 n (3.12. We need values of ψ W such that the RHS of (3.12 equals 1, and (3.5 s satsfed for all U S. (Recall that the relatonshp of U to (3.5 s that u s the number of occurrences of n U. Altogether ths s 11 equatons over Z 2 n 14 unknowns. The coeffcents of the equatons depend only on a, b, and c. It s a smple matter to run Maple on these 32 cases, and t tells us that there s a soluton n exactly the clamed cases. The Maple program, nput and output, can be seen at [2]. The two cases, (a, b, c = (1, 1, 1 and (2, 4, 1, consdered n Propostons 3.8 and 3.9 are not ncluded n Table 3.23 because they dd not yeld a soluton n any of the stuatons whose results appear as a column of that table. The specal stuaton when a = 1 or 2 or b = 1 s not a problem, exactly as n the proof of Theorem 3.2.

15 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE 15 The next result s very smlar. The only dfference s a small change n the exponent of Dw 1 (and hence also of DR. Ths changes the values of q t. Theorem If m = 2 e + m wth 2 m 2 e 1, then (Dw 1 2m 2 (Dw 2 2 (Dw 3 (DR 2e H 2m 1 (X X for the values of a, b, and c whch have an n the 3.15 column of Table Proof. In ths case, q t = m t 1. That s the only change from the proof of Theorem Here we requre that soluton must exst both when q = (1, 0, 1, 0, 1 and (0, 1, 0, 1, 0, coverng ether party of m. Here and later q = (q 0, q 1, q 2, q 3, q 4. The thrd result also just nvolves a change n the exponent of Dw 1. Ths tme q t = ( m t+2 2, so we requre a soluton for all four of the vectors q, correspondng to mod 4 values of m. Theorem If m = 2 e + m wth 2 m 2 e 1, then (Dw 1 2m 1 (Dw 2 2 (Dw 3 (DR 2e H 2m 1 (X X for the values of a, b, and c whch have an n the 3.16 column of Table We can fll n the mssng cases by changng the exponent of Dw 2. We begn wth the followng analogue of Proposton 3.10, whose proof s totally analogous. Proposton If nether m nor m 1 s a 2-power, then the component of (Dw 1 α (Dw 2 (Dw 3 (DR 2m 3 α n bdegree (m, m 1 equals ( 2m 3 α (Y0 Y 1,2,3 + Y 1 Y 1,2,3 m + ( 2m 3 α (Y1,2,3 Y m Y 1,2,3 Y 1 + Y 3 Y 1,2 + Y 1,3 Y 1,2 + Y 2 Y 1,3 + Y 1,2 Y 1,3 + ( 2m 3 α (Y2,3 Y m Y 1,2,3 Y 1 + Y 1,3 Y 2 + Y 1,3 Y 1,2 + Y 1,2 Y 3 + Y 1,2 Y 1,3 + ( 2m 3 α (Y1 Y m 3 2,3 + Y 1 Y 1,2,3. If m s a 2-power, there s an addtonal Y 1 Y 1,2,3. We do not need to use ths proposton when m 1 s a 2-power.

16 16 DONALD M. DAVIS Theorem If m = 2 e + m wth 2 m 2 e 1, then (Dw 1 2m 1 (Dw 2 (Dw 3 (DR 2e H 2m 1 (X X for the values of a, b, and c whch have an n the 3.18 column of Table Proof. Let q t = ( 2m 3 α m t for 0 t 3. Usng Proposton 3.17 and Theorem 3.2, we fnd that (φ ψ((dw 1 α (Dw 2 (Dw 3 (DR 2m 3 α (3.19 = q 1ψ 0 + (q 1 + q 2aψ 1 + q 2(a + bψ 2 + q 2(a + b + c 1ψ 3 +(q 1(ab ( a 2 + q 2 (a + bψ 1,2 +(q 1(a(b + c + a 1 + ( a 2 + q 2 (a + b + c 1ψ 1,3 +q 3((a 1(a + b + c 1 + ( ( a b 2 + (b 1(c 1(ψ2,3 + ψ 1,2,3 +q 0((a + ( ( a 2 1(a + b + c 1 + a 1 ( 2 + a b 2 + a(b 1(c 1ψ1,2,3. Smlarly to Lemma 3.13, wth α = 2m 1, we have ( 2m 3 α m t m t 1 mod 2, and the result follows smlarly to the three prevous ones, havng Maple check whether there s a soluton to the system of 11 equatons n 14 unknowns, whose frst equaton s that the RHS of (3.19 equals 1 and others are, as before, (3.5 for each U S. Ths tme a soluton s requred for both q = (0, 1, 0, 1 and (1, 0, 1, 0. Referrng to Table 3.23 and Theorems 3.14, 3.15, 3.16, and 3.18, accompaned by Propostons 3.8 and 3.9, we fnd that, when nether m nor m 1 s a 2-power, (1.3 s satsfed for all (a, b, c, establshng Theorem 1.2 when nether m nor m 1 s a 2-power. Next we handle the case when m = 2 e. Theorem If m = 2 e, then, for 1 ε 3, (Dw 1 2e ε (Dw 2 2 (Dw 3 (DR 2e +ε 4 0 H 2m 1 (X X for the values of a, b, and c whch have an n the 3.20(ε column of Table Proof. Because of the change due to m = 2 e noted n Proposton 3.10, the expresson n Corollary 3.11 has an extra ((a 1(a +b+c 1+ ( ( a b 2 +(b 1(c 1ψ1,2,3 added. The vectors q are (0, 1, 1, 1, 1, (0, 0, 1, 0, 1, and (0, 0, 0, 1, 1 for ε = 3, 2, 1,

17 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE 17 respectvely. The other 10 equatons for the ψ W s are as before. Maple tells us when the system has a soluton. The next result s the 2 e -analogue of Theorem As shown n Table 3.23, ths, Theorem 3.20, and Propostons 3.8 and 3.9 mply Theorem 1.2 when m = 2 e. Theorem If m = 2 e, then (Dw 1 2e 1 (Dw 2 (Dw 3 (DR 2e 2 0 H 2m 1 (X X for the values of a, b, and c whch have an n the 3.21 column of Table Proof. Because of the change due to m = 2 e noted n Proposton 3.17, the expresson n (3.19 has an extra ((a 1(a + b + c 1 + ( ( a b 2 + (b 1(c 1ψ1,2,3 added. The vector q s (0, 0, 1, 0, and the other 10 equatons for the ψ W s are as before. Maple tells us when the system has a soluton. Fnally, we handle the case m = 2 e + 1. Theorem If m = 2 e + 1, then, for ε = ±1, (Dw 1 2e +ε (Dw 2 2 (Dw 3 (DR 2e ε 2 0 H 2m 1 (X X for the values of a, b, and c whch have an n the 3.22(ε column of Table Proof. Because of the change due to m = 2 e + 1 noted n Proposton 3.10, the expresson n Corollary 3.11 has an extra ψ 1 + (a + bψ 1,2 added. The vectors q are (0, 0, 0, 0, 1 and (0, 0, 1, 1, 1 for ε = 1 and 1, respectvely. The other 10 equatons for the ψ W s are as before. Maple tells us when the system has a soluton.

18 18 DONALD M. DAVIS Table m 2 e, 2 e + 1 m = 2 e m = 2 e + 1 a b c (3 3.20(2 3.20( (1 3.22(

19 TOPOLOGICAL COMPLEXITY OF PLANAR POLYGON SPACES WITH SMALL GENETIC CODE 19 The Maple program that performed all these verfcatons can be vewed at [2]. We conclude that Theorem If X = M(l wth genetc code {n, a + b + c, a + b, a}, then (3.6 holds for some set of 2m 1 classes v. Proof. Table 3.23 shows that for all (a, b, c except (1, 1, 1 and (2, 4, 1 one of the tabulated theorems apples, whle the two exceptonal cases are covered n Propostons 3.8 and 3.9. References [1] D.M.Davs, Topologcal complexty of some planar polygon spaces, on arxv. [2], dmd1/mapleresults.pdf. [3] M.Farber, Invtaton to topologcal robotcs, European Math Socety (2008. [4] J.-C.Hausmann and A.Knutson, The cohomology rngs of polygon spaces, Ann Inst Fourer (Grenoble 48 ( [5] J.-C.Hausmann and E.Rodrguez, The space of clouds n Eucldean space, Experment Math 13 ( [6], Department of Mathematcs, Lehgh Unversty, Bethlehem, PA 18015, USA E-mal address: dmd1@lehgh.edu