Grassmannian Complex and Second Order Tangent Complex

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1 Punjab University Journal of Mathematics (ISSN Vol. 48((016 pp Grassmannian Complex and Second Order Tangent Complex Sadaqat Hussain Department of Mathematical Sciences University of Karachi Raziuddin Siddiqui Mathematical Sciences Research Centre Federal Urdu University Gulshan-e-Iqbal, University Road, Karachi, Pakistan Received: 01 March, 016 / Accepted: 5 July, 016 / Published online 4 August, 016 Abstract. First order tangent complex is studied by Siddiqui in [11]. There he has introduced morphisms for the first order tangent complex and connected this complex to the famous Grassmannian complex. In this paper we will extend the discussion of tangent complex for the second order. To do this we will introduce second order tangent group, denoted by TB (F, and form a tangent complex of order. Then we will write morphisms in order to connect this complex with the famous Grassmannian complex. AMS (MOS Subject Classification Codes: 18G; 14L; 16G Key Words: Tangent Complex; Grassmannian Complex; Cross-ratio; Five-term Ration; Morphism. 1. INTRODUCTION In [8] Siddiui uses configurations to relate the Grassmannian complex (see. where d : (x 1,..., x m d : C m (X C m 1 (X m ( 1 m (x 1,..., ˆx i,..., x m to both infinitesimal and tangential complexes. In one hand he connected the Grassmannian complex to the variant of Cathelineau s complexes by introducing homomorphisms for both weight and weight 3. He also proved the commutativity of corresponding diagrams. On the other hand he presented the crossratio of four points and the famous Siegel s cross-ratio identity in tangential settings and also proved that the Goncharov s projected five term relation can also be defined for tangent 91

2 9 Sadaqat Hussain and Raziuddin Siddiqui group TB (F. By using these constructions he defined maps to link the Grassmannian sub complex and Cathelineau s tangential complex for both weight and 3. To connect Grassmannian sub complex to Cathelineau s tangent complex C 5 ( F[ε] d C 4 ( F[ε] d C 3 ( F[ε] TB (F ε F F for n = he gives the morphisms τ 1,ε and τ 0,ε. For n = 3 he uses morphisms τ3,ε,τ3 1,ε and τ 3 0,ε to relate F C 6 ( 3 F[ε] d C 5 ( 3 F[ε] d C 4 ( 3 F[ε] to the Cathelineau s tangent complex of weight 3 (see chapter 4 of [8]. He studied these complexes for the first order tangent group but in the appendix of [8] he discussed a little bit about the second order tangent group TB (F. There he gives the morphisms between Grassmannian complex and Cathelineau s complex by taking only pure terms occurred in the definition of maps and neglected the non pure terms. But he himself does not consider it as an authentic work because there is no logic behind ignoring the non pure terms that s why he put that work in the appendix of his dissertation. However that work gives us a guideline to move toward higher order. In this work we have tried to study the polyloagrithmic complexes in tangential settings for second order. The second section is devoted for the essential prerequisites and basic definitions as usual. In third section we have discussed second order tangent groups. Where we have tried to define the second order tangent groups of weight and 3,denoted by TB (F and TB 3 (F respectively. We also have mentioned the relations exist in these groups. Then we moved forward to describe the geometry of configurations of TB (F, which includes the construction of Grassmannian complex for tangential settings in higher order, building of the cross ratio and triple- ratio in TB (F. Then we move to write morphisms τ and τ for n = and τ 3, τ 3 and τ 3 for n = 3 in order to connect 0,ε 1,ε 0,ε 1,ε,ε the Grassmannian complex to the the Cathelineau s tangent complex. At last we write the proofs of the commutativity of resulting diagrams (see theorems (4.,(7.1and (7... NOTATIONS AND PRELIMINARIES.1. Configurations. Let C m (X be a free abelian group generated by elements (x 1,..., x m X m.for any group G acting on a non empty set X the elements of the set G/X m, consisting of the m-tuples, is called configurations of X where G is acting diagonally on X m (see[5]. A configuration (x i x 1,..., ˆx i,..., x m of vectors in V n+1 / x i (V n+1 be a vector space of dimension n + 1is said to be a projective configuration and defined as a quotient space of dimension n formed by the projection of x j V n ; j i, projected from C m+1 (n + 1 to C m (n.. Grassmannian Complex. This complex is introduced by a famous German mathematician Hermann Grassmann. The importance of this complex is because of helping in the investigation of homology of general linear groups. let X be any non empty set. Consider C m (X be a free abelian group whose generators are configurations of m points, then

3 Grassmannian Complex and Second Order Tangent Complex 93 we define a differential map d : C m (X C m 1 (X as m d : (x 1,..., x m ( 1 m (x 1,..., ˆx i,..., x m (. 1 Also if we denote C m (n to be a group which is free abelian and whose generators are the configurations of the elements of an n-dimensional vector space V n and (x i x 1,..., ˆx i,..., x m be the projective configuration of the vectors x j along the vectors x i, where i j; j = 1,..., m, then we define a projective map d : C (m+1 (n + 1 C m (n as m d : (x 1,..., x m ( 1 m (x i x 1,..., ˆx i,..., x m (. by using these differential maps and free abelian groups generated by configurations we have a bi- complex called Grassmannian bi-complex (see section of [8]. From this bi-complex we can form many sub-complexes like C m+ (n + d C m+1 (n + 1 d C m (n or C m+ (n d C m+1 (n d C m (n These sub-complexes known as Grassmannian Sub-Complexes.3. Tangential Configuration Space. For any field F with zero characteristic, we define the ring of νth truncated polynomial by polynomial F[ε] ν. We choose l = F[ε] ν := F[ε]/ε ν where ν 1. First we will define n F[ε] ν as an affine space over the the ring of truncated a 1 a. n F. \, l ε := n F and upto l ε n := a 1,ε ν 1 a,ε ν 1. a n,ε ν 1 a n n F, then one can write (see [?] l = l + l ε ε + l ε ν 1ε ν 1 = a 1 + a 1,ε ε + + a 1,ε ν 1ε ν 1 a + a,ε ε + + a,ε ν 1ε ν 1.. a n + a n,ε ε + + a n,ε ν 1ε ν 1 a 1,ε a,ε.. a n,ε n F[ε] ν Consider the free abelian group C m ( n F[ε] ν whose generators are the configurations (l 1,..., l m of m-vectors in n F[ε] ν. For these configurations, we can define the differential map d as d : C m+1 ( n F[ε] ν C m ( n F[ε] ν where d is defined in (.1. another differential map d with projection can be defined as d : C m+1 ( n F[ε] ν C m ( n 1 F[ε] ν

4 94 Sadaqat Hussain and Raziuddin Siddiqui where d is defined in (. Let ω detv n be the volume element in a n- dimensional vector space V n (see section 3 of [5], we define a n n determinant (x 1,..., x n = ω, x 1 x... x n ; x i V n For simplicity, we consider the the following cases.3.1. Case n = and ν = 3. In this situation, we can write (x p, x q = (x p, x q ε 0 + (x p, x q ε 1ε + (x p, x q ε ε where (x p, x q ε 0 = (x p, x q, (x p, x q ε 1ε = (x p, x q,ε + (x p,ε, x q and (x p, x q ε = (x p, x q,ε + (x p,ε, x q,ε + (x p,ε, x q.3.. Case n = 3 and ν = 3. (x p, x q, x r = (x p, x q, x r ε 0 + (x p, x q, x r ε 1ε + (x p, x q, x r ε ε where (x p, xq, xr ε 0 = (x p, x q, x r, (x p, xq, xr ε 1ε = (x p, x q, x r,ε + (x p, x q,ε, x r + (x p,ε, x q, x r and (x p, x q, x r ε = (x p, x q, x r + (x p, x q, x r,ε + (x p, x q,ε, x r + (x p,ε, x q, x r + (x p, x q, x r,ε + (x p, x q,ε, x r,ε + (x p, x q,ε, x r + (x p,ε, x q, x r,ε + (x p,ε, x q,ε, x r + (x p,ε, x q, x r.4. Cross-ratio in F[ε]. Let (x 0, x 1, x, x 3 C 4( F[ε] ν be the configuration of four points, then their cross ratio is r(x 0, x 1, x, x 3 = (x 0, x 3 (x 1, x (x 0, x (x 1, x 3 (. 3 But expansion of r(x 0, x 1, x, x 3 over the truncated polynomial ring F[ε] 3 is r(x0, x 1, x, x 3 = ( r ε 0 + r ε ε + r ε ε (x0, x 1, x, x 3 (. 4 which gives us the following values(see [8] and [9]. Put ν = 1, we get r(x 0, x 1, x, x 3 = r ε 0(x 0, x 1, x, x 3 = r(x 0, x 1, x, x 3 = (x 0, x 3 (x 1, x (x 0, x (x 1, x 3 (. 5 r ε (x 0, x 1, x, x 3 = { (x 0, x 3 (x 1, x ε (x 0, x (x 1, x 3 r(x 0, x 1, x, x 3 { (x 0, x (x 1, x 3 ε (x 0, x (x 1, x 3 (. 6 r ε (x 0, x 1, x, x 3 ={(x 0, x 3 (x 1, x ε (x 0, x (x 1, x 3 r ε (x 0, x 1, x, x 3 {(x 0, x (x 1, x 3 ε (x 0, x (x 1, x 3 r(x 0, x 1, x, x 3 {(x 0, x (x 1, x 3 ε (x 0, x (x 1, x 3 (. 7 where (ab ε = a ε b + ab ε and (ab ε = a ε b + a ε b ε + ab ε

5 Grassmannian Complex and Second Order Tangent Complex First Order Tangent Group. For any a, a F and a; a ] = [a+a ε] [a] Z[F[ε] ], we define the first order tangent Group, denoted by TB (F, is a Z-module generated by the elements of the form a; a ] [F[ε] ] an quotient by the expression where a; a ] b; b ] ( b b + a ; a a(1 b + b(1 a ; ] ( a(1 b b(1 a 1 b 1 a ; ] ( ] 1 b 1 a, a, b 0, 1, a b; (a, a F (. 8 ( b = ab a ( b 1 b ; = (1 ba (1 ab a a 1 a (1 a ( a(1 b = b(1 ba a(1 ab b(1 a (b(1 a 3. SECOND ORDER TANGENT GROUPS First of all we will give definition of TB (F and will try to construct the cross ratio, triple ratio and identities of the vectors in the configuration space C m ( n F[ε] 3 for m = 4,..., 7 ; n =, 3. We will also give the relations satisfied by TB (F and will prove the five term relation. The tangent group of order TB (F is a Z-module whose generators are the elements of the form a; a, a ] [F[ε] 3 ], where a; a, a ] = [a + a ε + a ε ] [a], (a, a, a F and quotient by the expression (see appendix of [8] where where and a; a, a ] b; b, b ] ( ( b b b + a ;, a a ( a(1 b a(1 b + b(1 a ;, b(1 a ( b = a b aba aa b + b(a ; a a 3 = ( a(1 b b(1 a ] 1 b 1 a ; ( a(1 b b(1 a ( 1 b, 1 a ( ] 1 b 1 a ], a, b 0, 1, a b (3. 9 ( 1 b A = 1 a (1 a ; 3 B a 3 (1 b 3 (3. 10 A = (1 a(1 ba (1 a b (1 aa b + (1 b(a B =(b a 3 bb a 3 + bb a (b a bb a + (b a + bab a ba b +b 3 aa b 3 (a b 3 a b aa + b (a + b a ( Functional Equations of TB (F. The tangent group TB (F satisfies the following relations (see [7] (1 Two term relation: a; b 1, b ] = 1 a; b 1, b ( Inversion relation: a; b 1, b ] = 1 a ; ] b 1 ab (b 1 a a 3 (3 The five term relation of TB (F is equation ( 3. 9 above.

6 96 Sadaqat Hussain and Raziuddin Siddiqui Lemma 3.. For (x 0, x 1, x, x 3 C 4( F[ε] 3, we have { (x 0, x 1 (x, x 3 ε = { (x 0, x (x 1, x 3 ε { (x 0, x 3 (x 1, x ε Proof. See Lemma (4.4 of [9] for proof. 4. DILOGARATHMIC COMPLEXES OF TANGENT GROUPS Cathelineau formed a complex using tangents groups known as tangent complex and we have another well known complex known as Grassmannian complex. In his work Sidddiqui connected these two complexes by defining maps τ 0,ε and τ 1,ε for n = ; ν =. In this section we also connect the complexes mentioned above for the case n = ; ν = 3. Let us denote C m ( F[ε] 3 is a free abelian group whose generators are the configuration (x0,..., x m 1 F[ε] 3 for any affine space F[ε] 3 over F[ε] 3 then we have the following Grassmannian complex Consider the following diagram... C 5 ( F[ε] 3 d C 4 ( F[ε] 3 d C 3 ( F[ε] 3 m d : (x0,..., x m ( 1 i (x0,..., ˆx i,..., x m C 5 ( d F[ε] 3 C 4 ( F[ε] 3 d C 3 ( F[ε] 3 (D where ɛ τ 1,ε TB (F ε τ 0,ε F F F : a; b 1, b ] b a b 1 (1 a + b a (1 a + b 1 a (1 a + b a b 1 b a (1 a + b 1 ; a, b (1 a 1, b F (4. 1 Now we come to define the vertical maps τ 0,ɛ and τ 1,ɛ of the diagram. The map τ 0,ɛ can be written as the sum of two maps where τ 1 (x 0,x 1, x = and ( ( 1 i (x i, x i+1 ɛ (x i, x i+1 τ (x0, x 1, x = ( 1 i (x i, x i+1 ɛ (x i, x i+1 (x i, x i+1 ɛ (x i, x i+1 τ 0,ɛ = τ 1 + τ (x i, x i+1 ɛ (x i, x i+ (x i, x i+1 (x i+1, x i+ (x i, x i+ ɛ (x i, x i+ (x i, x i+ ɛ (x i, x i+, i mod 3 (4. 13, i mod 3 (4. 14

7 Grassmannian Complex and Second Order Tangent Complex 97 The map τ 1 carries the elements of C 3 ( F[ε] 3 into F F and the image of the elements of C 3 ( F[ε] 3 under the map τ is in F. τ 1,ɛ (x 0,..., x 3 = r(x 0,..., x 3 ; r ɛ (x0,..., x 3, r ɛ (x 0,..., x 3 ] (4. 15 where r(x 0,..., x 3, r ɛ (x0,..., x 3 and r ɛ (x 0,..., x 3 are the respective coefficients of the ε 0, ε 1 and ε in the cross ratio. First we have to show the maps τ and τ are both well 0,ɛ 1,ɛ defined. Since we have defined τ as the cross ratio of four points so it is not necessary 1,ɛ to check. Therefore we only check for τ. 0,ɛ Lemma 4.1. τ 0,ɛ does not depend upon the volume form ω by the vectors. Proof. According to the definition we can write τ as 0,ɛ ( 1 i (x i, x i+1 ɛ (x i, x i+1 ɛ (x i, x i+1 (x i, x i+1 (x i, x i+ (x i+1, x i+ + ( 1 i (x i, x i+1 ɛ (x i, x i+1 ɛ (x i, x i+1 (x i, x i+1 (x i, x i+ ɛ (x i, x i+ (x i, x i+ ɛ (x i, x i+ ; i mod 3 Due to the homogeneity between the factors of the terms of above expression it is not possible to get a different value by replacing the volume element ω by λω. This shows that the map τ is independent of the volume element. 0,ɛ Theorem 4.. The diagram (D commutes.i.e. τ 0,ɛ d = ɛ τ 1,ɛ Proof. We have already defined the map τ 1,ɛ (x 0,..., x 3 as τ 1,ɛ (x 0,..., x 3 = r(x 0,..., x 3 ; r ɛ (x 0,..., x 3, r ɛ (x 0,..., x 3 ] For simplicity we put a = r(x 0,..., x 3, b 1 = r ɛ (x 0,..., x 3 and b = r ɛ (x 0,..., x 3 then τ 1,ɛ (x 0,..., x 3 = a; b 1, b ] Here we evaluate value of the expression b a b 1 a. Since we have b 1 = r ε (l 0,..., l 3 = { (x 0, x 3 (x 1, x ε (x 0, x (x 1, x 3 r(x 0,..., x 3 { (x 0, x (x 1, x 3 ε (x 0, x (x 1, x 3 Dividing by a and using a shorthand (xi, x j ε = (x i x j ε, we have ( b a b 1 (x a = 0 x3 ε + (x 1 x ε (x 0 x ε (x 1 x 3 ε (x 0 x 3 (x 1 x (x 0 x (x 1 x 3 (4. 16 (x 0 x 3 ε (x 0 x 3 (x 1 x ε (x 1 x + (x 0 x ε (x 0 x + (x 1 x 3 ε (4. 17 (x 1 x 3 Similarly we can evaluate the expression as b 1 a b ( 1 (x (1 a = 0 x ε + (x 1 x 3 ε (x 0 x 1 ε (x 1 x 3 ε (x 0 x (x 1 x 3 (x 0 x 1 (x x 3 (x 0 x ε (x 0 x (x 1 x 3 ε (x 1 x 3 + (x 0 x 1 ε (x 0 x 1 + (x x 3 ε (x x 3 (4. 18

8 98 Sadaqat Hussain and Raziuddin Siddiqui Now by taking composition of the map τ with the map 1,ɛ ɛ, we get a large expression. Therefore by using homomorphic property we can split this composition into two parts to explain them separately ɛ τ 1,ɛ (x 0,..., x 3 = ( ( 1 ɛ a; b1, b ] + ɛ a; b1, b ] (4. 19 where 1 carries the elements of TB ɛ (F into F F and carries the elements of ɛ TB (F into F. Therefore we have ( ( ( 1 ɛ a; b1, b ] b = a b 1 b (1 a + a (1 a + b 1 a (1 a ɛ ( a; b1, b ] { = (x 0 x 3 ɛ + (x 1 x ɛ (x 0 x ɛ (x 0 x 3 (x 1 x (x 0 x + (x 1, x 3 ɛ (x 1, x 3 (x 0, x 3 ɛ (x 0, x 3 (x 1, x ɛ (x 1, x { + (x ox ɛ + (x 1 x 3 ɛ (x 0 x (x 1 x 3 (x 0 x 1 ɛ (x 0 x 1 (x 1 x 3 ɛ (x 1 x 3 (x ox 1 (x x 3 (x 0 x (x 1 x 3 (x x 3 ɛ (x x 3 + (x 0, x 1 ɛ (x 0, x 1 + (x, x 3 ɛ (x, x 3 (x 0, x ɛ (x 0, x (x 1, x 3 ɛ (x 1, x 3 = ( b a b 1 a ( b (1 a + b 1 (1 a { = (x ox3 ɛ + (x 1 x ɛ (x 0 x ɛ (x 0 x 3 (x 1 x (x 0 x + (x 1, x 3 ɛ (x 1, x 3 (x 0, x 3 ɛ (x 0, x 3 (x 1, x ɛ (x 1, x (x 0 x 1 ɛ (x 0 x 1 (x x 3 ɛ (x x 3 (x 1 x 3 ɛ (x 1 x 3 + (x 0, x ɛ (x 0, x (x ox 3 (x 1 x (x 0 x (x 1 x 3 + (x 0, x ɛ (x 0, x { (x ox ɛ + (x 1 x 3 ɛ (x 0 x (x 1 x 3 ( (x 0, x 1 ɛ (x 0, x 1 + (x, x 3 ɛ (x, x 3 (x 0, x ɛ (x 0, x (x 1, x 3 ɛ (x 1, x 3 (4. 1 Using ( and ( 4. 14, we split the second map τ d(x 0,ɛ 0,..., x 3 into two parts. First part can be written as τ 1 d(x 0,..., x 3 { =Ãlt (013 ( 1 i ( (x i, x i+1 ɛ (x i, x i+1 (x i, x i+1 ɛ (x i, x i+ (x i, x i+1 (x i+1, x i+ ; i mod 3 (4. where Ãlt (013 denotes the alternation sum. The expansion of inner expression give us a total of six terms. From which three will be of the form (x i,x j ε (x i,x j x y and other three will of the form (x i,x j ε (x i,x j x y. Using the property a b c = a b a c, terms will become double and form of the terms will be (x i,x j ε (x i,x j x and (x i,x j ε (x i,x j y. Then we will expand the alternation sum which gives us a total of 48 terms. After simplification we will obtain an expression identical with ( The second part of τ d(x 0,ɛ 0,..., x 3 can be written as

9 Grassmannian Complex and Second Order Tangent Complex 99 τ d(x 0,..., x 3 =Ãlt (013 { ( 1 i (x i, x i+1 ɛ (x i, x i+1 ɛ (x i, x i+1 (x i, x i+1 (x i, x i+ ɛ (x i, x i+ (x i, x i+ ɛ (x i, x i+ ; i mod 3 (4. 3 The expansion of inner sum will give us a total of 1 terms. From which six will be of the form (x i,x j ε (x i,x j x and remaining six will be of the form (x i,x j ε (x i,x j y. Applying the alternation sum number of trms will increase up to 48. After cancellation of like terms with opposite signs we obtain an expression identical to ( Since the sum of ( 4. 0 and ( 4. 1 represents the value of ɛ τ 1,ɛ which is equal to the sum of the values of τ 1 d(x 0,..., x 3 and τ d(x 0,..., x 3. Corollary 4.3. The following are complexes. (1 C 4 ( 3 F[ε] 3 d C 3 ( F[ε] 3 ( C 5 ( 3 F[ε] 3 d C 4 ( F[ε] 3 τ 0,ε F F F τ 1,ε TB (F Proof. To prove above result we have to show τ d = 0 and τ d = 0 which requires 0,ε 1,ε direct calculations. 5. TRILOGARATHMIC COMPLEXES OF TANGENT GROUPS This section is devoted to discuss about the suitable maps which connects the Grassmannian sub-complex and Cathelineau s tangential complex of second order for the case weight 3. We also show the commutativity of the resulting diagram Definition of TB 3 (F. The second order tangent group for weight 3 is denoted by TB 3 (F is a Z-module over the truncated polynomial ring F[ε] 3 whose generators are the elements of the form a; a, a ] [F[ε] 3 ], where a; b 1, b ] = [a + b 1 ε + b ε ] [a], (a, b 1, b F and quotient by the ker where where the map is defined as : [F[ε] 3 ] ( TB (F F (F B (F ( a; b 1, b ] c + u [v] = b a b 1 (1 a c b a (1 a + b 1 a c (1 a +u (1 v v + b a b 1 b a (1 a + b 1 (5. 4 (1 a 5.. Projected Five Term relations in TB (F. We already have projected five term relations in B (F (see [5] and TB (F (see [9]. Here we will prove the existence of projected five term relation in TB (F.

10 100 Sadaqat Hussain and Raziuddin Siddiqui Lemma 5.3. Let l 0,..., l 4 F[ε] 3 be 5 points in generic position, then 4 ( 1 i r(l i l 0,..., ˆl i,..., l 4 ; r ε (li l 0,..., ˆl i,..., l 4, r ε (l i l 0,..., ˆl i,..., l 4 ] = 0 in TB (F, where l i = l i + l i ε + l i ε F[ε] 3, l i, l i, l i F and r(l i l 0,..., ˆl i,..., l 4 =r(l i l 0,..., ˆl i,..., l 4 + r ε (l i l 0,..., ˆl i,..., l 4 ε +r ε (l i l 0,..., ˆl i,..., l 4 ε (5. 5 where the LHS denotes the projected cross-ratio of any four points from l 0,..., l 4 F[ε] 3 projected from the fifth one. Proof. See [[7]] for proof Triple-ratio in F[ε] 3 : Let C 6 ( F[ε] 3 be an abelian group,preferably free abelian, whose generators are configurations of six points. Then for (l 0,..., l 5 C 6( 3 F[ε] 3 the triple-ratio for ν = 3 can be define as(see article (4.1. [8] where r 3,ε = Alt 6 { {(l 0 l 1 l 3 (l 1 l l 4 (l l 0 l 5 ε (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 r 3 (l 0,..., l 5 {(l 1 l l 4 (l 1 l l 5 (l l 0 l 3 ε (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 r 3,ε (l0, l 1, l, l 3, l 4, l 5 {(l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 ε (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (5. 6 (abc ε = a ε 0b ε 0c ε + a ε 0b ε c ε + a ε 0b ε c ε 0 + a ε b ε 0c ε + a ε b ε c ε + a ε b ε 0c ε 0 (5. 7 Using above constructions we are now able to write maps to relate Grassmannian complex with the famous Cathelineau s tangent complex for weight 3 and ν = 3. After connecting we obtain the following diagram C 6 ( 3 F[ε] 3 d C 5 ( 3 F[ε] 3 d C 4 ( 3 F[ε] 3 (A τ 3,ε TB 3 (F ε τ 3 1,ε ( TB (F F (F B (F ε τ 3 0,ε ( F F ( 3 F where 3 ( τ 3 (l 0,ε 0,..., l 3 = ( 1 i (l 0,..., ˆl i,..., l 3 ε (l 0,..., ˆl i,..., l 3 j=0 j i (l 0,..., ˆl i,..., l 3 ε (l 0,..., ˆl i,..., l 3 (l 0,..., ˆl i+1,..., l 3 (l 0,..., ˆl i+,..., l 3 (l 0,..., ˆl i+3,..., l 3 (l 0,..., ˆl i+,..., l 3 3 ( + (l 0,..., ˆl j,..., l 3 ε (l 0,..., ˆl j,..., l 3 ε (l 0,..., ˆl j,..., l 3 (l 0,..., ˆl j,..., l 3, i mod 4, (5. 8

11 Grassmannian Complex and Second Order Tangent Complex 101 τ 3 (l 1,ε 0,..., l 4 = 1 4 ( r(li ( 1 i l 0,..., ˆl i,..., l 4 ; r ε (li 3 l 0,..., lˆ i,..., l4, r ε (l i l 0,..., lˆ i,..., l4 ] 4 (l0 (ˆl i, ˆl j +,..., ˆl i,..., ˆl k,..., l 4 ε (l 0,..., ˆl i,..., ˆl k,..., l 4 [ r(l i l 0,..., ˆl i,..., l 4 ] i j 4 m=0 m i and k=0 k i (l 0,..., ˆl i,..., ˆl m,..., l 4 ε (l 0,..., ˆl i,..., ˆl m,..., l 4 τ 3,ε (l 0,..., l 5 = 45 Alt 6 [ r(l i l 0,..., ˆl i,..., l 4 ] (5. 9 r3 (l 0,..., l 5 ; r 3,ε (l 0, l 1, l, l 3, l 4, l 5, r 3,ε (l 0, l 1, l, l 3, l 4, l 5 ] 3 (5. 30 where and r 3 (l 0,..., l 5 = (l 0l 1 l 3 (l 1 l l 4 (l l 0 l 5 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (5. 31 r 3,ε (l 0, l 1, l, l 3, l 4, l 5 = {(l 0 l 1 l 3 (l 1 l l 4 (l l 0 l 5 ε (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (l 0l 1 l 3 (l 1 l l 4 (l l 0 l 5 {(l0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 ε (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (5. 3 r 3,ε (l0,..., l 5 = {(l 0 l 1 l 3 (l 1 l l 4 (l l 0 l 5 ε r 3,ε (l0 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3,..., l 5 {(l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 ε (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (l 0l 1 l 3 (l 1 l l 4 (l l 0 l 5 {(l0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 ε (5. 33 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 we define the map ε as 1 ε ( a; b 1, b ] c + x [y] = b a b 1 (1 a c b a (1 a b 1 a c (1 a +x (1 y y + b a b 1 b a (1 a + b 1 x (5. 34 (1 a and for a, b 1, b, c, c 1, c F ε ( c; c 1, c ] 3 = a; b 1, b ] a + c c c 1 [a] (5. 35 Theorem 5.5. The square right to the diagram (A is commutes, i.e. c τ 3 0,ε d = ε τ 3 1,ε

12 10 Sadaqat Hussain and Raziuddin Siddiqui Proof. The map τ 3 0,ε is defined in (5.8 which is too lengthy for calculations therefore we write the map τ 3 0,ε as sum of two maps τ(1 and τ ( then we evaluate τ (1 d and τ ( d separately. First we find τ (1 d(l0,..., l 4 = τ3 0,ε 4 ( 1 i (l0,..., ˆl i,..., l 4 ( 3 = Ãlt (0134 ( 1 i (l 0,..., ˆl i,..., l 3 ε (l 0,..., ˆl i,..., l 3, i mod 4 (l 0,..., ˆl i,..., l 3 ε (l 0,..., ˆl i,..., l 3 (l 0,..., ˆl i+1,..., l 3 (l 0,..., ˆl i+,..., l 3 (l 0,..., ˆl i+3,..., l 3 (l 0,..., ˆl i+,..., l 3 (5. 36 where Ãlt (0134 represents the alternation sum. We start our calculation from the expansion of the inner sum. The inner sum gives us four summands and each summand can further be simplified by using the relations p (q+r = p q+ p r ; p (qr = p q+ p r and p q r = p q p r in order to obtain 4 terms. And after applying the alternation sum we have a total of 10 terms containing 60 terms of the form p ε q r and remaining 60 of the form p ε q r. After that we combine the terms having common factor (l i,l j,l k (l i,l j,l k. For example the terms having common factor (l 4,l,l 3 ε (l 4,l,l 3 will be (l 4, l, l 3 ε (l 4, l, l 3 ( (l 0, l 3, l 4 (l 0, l, l 3 (l 0, l 3, l 4 (l 0, l, l 4 (l 0, l, l 4 (l 0, l, l 3 + (l 1, l 3, l 4 (l 1, l, l 4 (l 1, l 3, l 4 (l 1, l, l 3 + (l 1, l, l 4 (l 1, l, l 3 an the terms with common factor (l 1,l,l 3 ε (l 1,l,l 3 will be (l1, l, l 3 ε ( (l 0, l, l 3 (l 0, l 1, l (l 0, l, l 3 (l 0, l 1, l 3 (l 0, l 1, l 3 (l 0, l 1, l (l 1, l, l 3 + (l, l 3, l 4 (l 1, l 3, l 4 (l, l 3, l 4 (l 1, l, l 4 + (l 1, l 3, l 4 (l 1, l, l 4 and so on. This completes the calculation of first part τ (1 d. The second part τ ( d can be written as = Ãlt ( ( 1 i 4 j=0 j i ( (l 0,..., ˆl j,..., l 3 ε (l 0,..., ˆl j,..., l 3 ε (l 0,..., ˆl j,..., l 3 (l 0,..., ˆl j,..., l 3, i mod 4 (5. 37 By expanding the inner sum and wedge product for i = 0,..., 3 ; j = 0,..., 3, we get eight terms of the form a b c then pass each sum through Ãlt (0134 the terms will increase up to 40. This is the final value of τ ( d(l 0,..., l 4. Now we come to evaluate ε τ 3 1,ε.

13 Grassmannian Complex and Second Order Tangent Complex 103 First we split this map into two parts 1 ε and ε then ε τ 3 1,ε = 1 ε τ 3 1,ε + ε τ 3 1,ε and then using (5.9 we can write 1 ε τ 3 1,ε ( = ε 1 4 ( r(li ( 1 i l 0,..., ˆl i,..., l 4 ; r ε (li 3 l 0,..., lˆ i,..., l4, r ε (l i l 0,..., lˆ i,..., l4 ] 4 (ˆl i, ˆl j + (l 0,..., ˆl i,..., ˆl k,..., l 4 ε (l 0,..., ˆl i,..., ˆl k,..., l 4 [ r(l i l 0,..., ˆl i,..., l 4 ] i j 4 m=0 m i k=0 k i (l 0,..., ˆl i,..., ˆl m,..., l 4 ε (l 0,..., ˆl i,..., ˆl m,..., l 4 [ r(l i l 0,..., ˆl i,..., l 4 ] = 1 4 { ( 1 i 3 r ε (l i l 0,..., ˆl i,..., l 4 r ε (li l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 ( 1 r(li l 0,..., ˆl i,..., l 4 (ˆl i, ˆl j i j r ε (l i l 0,..., ˆl i,..., l 4 1 r(l i l 0,..., ˆl i,..., l 4 r ε (li l 0,..., ˆl i,..., l 4 1 r(l i l 0,..., ˆl i,..., l 4 ( r(l i l 0,..., ˆl i,..., l 4 (ˆl i, ˆl j 4 + k=0 k i i j (l 0,..., ˆl i,..., ˆl k,..., l 4 ε (l 0,..., ˆl i,..., ˆl k,..., l 4 4 m=0 m i ( 1 r(l i l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 The second part of the map ε τ 3 1,ε will be (l 0,..., ˆl i,..., ˆl m,..., l 4 ε (l 0,..., ˆl i,..., ˆl m,..., l 4 (5. 38 ε τ 3 (l 1,ε 0,..., l 4 = 1 4 { ( 1 i 3 r ε (l i l 0,..., ˆl i,..., l 4 r ε (li l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 r ε (l i l 0,..., ˆl i,..., l 4 1 r(l i l 0,..., ˆl i,..., l 4 r ε (li l 0,..., ˆl i,..., l 4 1 r(l i l 0,..., ˆl i,..., l 4 4 (l 0,..., ˆl i,..., ˆl k,..., l 4 4 ε (l 0,..., ˆl i,..., ˆl k,..., l 4 (l0,..., ˆl i,..., ˆl m,..., l4 ε (5. 39 (l 0,..., ˆl i,..., ˆl m,..., l 4 k=0 k i First we calculate 1 τ 3 (l ε 1,ε 0,..., l 4. For this purpose we have to calculate the values ( of r ε (l i l 0,...,ˆl i,...,l 4 rε (l i l 0,...,ˆl ( i,...,l 4 r r(l i l 0,...,ˆl i,...,l 4 r(l i l 0,...,ˆl i,...,l 4 and ε (li l 0,...,ˆl i,...,l 4 rε (l. i l 0,...,ˆl i,...,l 4 1 r(l i l 0,...,ˆl i,...,l 4 1 r(l i l 0,...,ˆl i,...,l 4 Take the case i = 4 and using (.5, (.6 and (.7 we have m=0 m i

14 104 Sadaqat Hussain and Raziuddin Siddiqui ( rε (l 4 l 0, l 1, l, l 3 r(l 4 l 0, l 1, l, l 3 = r ε(l 4 l 0, l 1, l, l 3 r(l 4 l 0, l 1, l, l 3.r ε(l 4 l 0, l 1, l, l 3 r(l 4 l 0, l 1, l, l 3 = = (l 4 l 0 l 3 ε (l 4 l 0 l 3 + (l 4 l 1 l ε (l 4 l 1 l + (l 4 l 0 l ε (l 4 l 0 l + (l 4 l 1 l 3 ε (l 4 l 1 l 3 + (l 4 l 0 l 3 ε (l 4 l 0 l 3 (l 4 l 0 l 3 ε (l 4 l 0 l 3 (l4 l 0 l ε (l 4 l 0 l (l 4 l 0 l 3 ε (l 4 l 0 l 3 ( (l 4 l0 l 3 ε (l 4 l 0 l 3 + (l 4 l 1 l ε (l 4 l 1 l (l 4 l 0 l ε (l 4 l 0 l (l 4 l 1 l 3 ε (l 4 l 1 l 3 (l4 l 1 l 3 ε (l 4 l 1 l 3 (l 4 l 0 l ε (l 4 l 0 l (l4 l 1 l ε (l 4 l 1 l + (l 4 l 0 l ε (l 4 l 0 l (l4 l 1 l ε (l 4 l 1 l (l 4 l 1 l 3 ε (l 4 l 1 l 3 (l 4 l 1 l 3 ε (l 4 l 1 l 3 (l 4 l 1 l ε (l 4 l 1 l (5. 40 and = r ε (l 4 l 0, l 1, l, l 3 r(l 4 l 0, l 1, l, l 3 = (l 4 l 0 l 3 ε (l 4 l 0 l 3 + (l 4 l 1 l ε (l 4 l 1 l (l 4 l 0 l ε (l 4 l 0 l ( (l 4 l0 l 3 ε (l 4 l 0 l 3 + (l 4 l 1 l ε (l 4 l 1 l (l 4 l 0 l ε (l 4 l 0 l (l 4 l 1 l 3 ε (l 4 l 1 l 3 (l 4 l 1 l 3 ε + (l 4 l 0 l 3 ε (l4 l 1 l ε (l 4 l 1 l 3 (l 4 l 0 l 3 (l 4 l 1 l (l 4 l 0 l ε (l 4 l 0 l ( (l 4 l0 l ε (l 4 l 0 l + (l 4 l 1 l 3 ε (l 4 l 1 l 3 (l 4 l 1 l 3 ε (l 4 l 1 l 3 (5. 41 After a simple calculation we obtain r ε (l i l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 r ε (li l 0,..., ˆl i,..., l 4 r(l i l 0,..., ˆl i,..., l 4 ( (l = 4 l0 l 3 ε + (l 4 l 1 l ε (l 4 l 0 l ε (l 4 l 0 l 3 (l 4 l 1 l (l 4 l 0 l (l 4 l 1 l 3 ε (l 4 l 1 l 3 (l 4 l 0 l 3 ε (l 4 l 0 l 3 (l 4 l 1 l ε (l 4 l 1 l + (l 4 l 0 l ε (l 4 l 0 l + (l 4 l 1 l 3 ε (l 4 l 1 l 3 (5. 4 similarly r ε (l i l 0,..., ˆl i,..., l 4 1 r(l i l 0,..., ˆl i,..., l 4 r ε (li l 0,..., ˆl i,..., l 4 1 r(l i l 0,..., ˆl i,..., l 4 ( (l = 4 l0 l ε + (l 4 l 1 l 3 ε (l 4 l 0 l 1 ε (l 4 l 0 l (l 4 l 1 l 3 (l 4 l 0 l 1 (l 4 l l 3 ε (l 4 l l 3 (l 4 l 0 l ε (l 4 l 0 l (l 4 l 1 l 3 ε (l 4 l 1 l 3 + (l 4 l 0 l 1 ε (l 4 l 0 l 1 + (l 4 l l 3 ε (l 4 l l 3 (5. 43 By the substitution of ( 5. 4 and ( we can easily find the values of ( and ( for i = 4. Where ( contains 384 terms of type (l i,l j,l k ε (l i,l j,l k l m and (l i,l j,l k ε (l i,l j,l k l m. In the same way we can calculate the value of ( for i = 0, 1,, 3 and get a total of = 190 terms. Then we recombine the terms having common factor (l i,l j,l k (l i,l j,l k. e.g.

15 Grassmannian Complex and Second Order Tangent Complex 105 The terms having common factor (l 4,l,l 3 ε (l 4,l,l 3 will be of the form 3 (l 4, l, l 3 ε (l 4, l, l 3 ( (l 0, l 3, l 4 (l 0, l, l 3 (l 0, l 3, l 4 (l 0, l, l 4 (l 0, l, l 4 (l 0, l, l 3 + (l 1, l 3, l 4 (l 1, l, l 4 (l 1, l 3, l 4 (l 1, l, l 3 + (l 1, l, l 4 (l 1, l, l 3 the coefficient -3 can be canceled with the first factor 1 3 of ( and it becomes identical with that of ( Similarly the value of (5.39 can be calculated by expanding the sum for i = 0,..., 4 which gives us terms of the form a b c. After direct calculation we get an expression equal to (5.37. Theorem 5.6. The left part of the diagram (A commutes, i.e. τ 3,ε ε = d τ 3 1,ε Proof. The map τ 3,ε gives us a large number of terms but most of them are identical due to symmetry an remains only 10 different terms. Due to the long calculations we will use the technique of combinatorics. From the definition ( 5. 30, we have τ 3 (l,ε 0,..., l 5 = 45 Alt 6 r3 (l 0,..., l 5 ; r 3,ε (l0,..., l 5, r 3,ε (l 0,..., l 5 ] 3 (5. 44 From now we use a short hand,that is,we write r 3 (l i l j l k l l l m l n instead of r 3 (l i, l j, l k, l l, l m, l n and (li l j l k ε in place of (l i l j l k ε. In other words we ignore and commas for simplicity. Now ε τ 3,ε (l 0... l 5 = 45 Alt 6 + { r3 (l 0... l 5 ; r 3,ε(l0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] r 3(l 0... l 5 r 3,ε (l 0... l 5 r 3 (l 0... l 5 r 3,ε (l 0... l 5 r 3 (l 0... l 5 [r 3(l 0... l 5 ] (5. 45 We are going to evaluate the value of r 3,ε (l 0...l 5 r 3 (l 0...l 5 r 3,ε (l 0...l 5 r 3 (l 0...l 5. Using ( 5. 31, ( 5. 3 and ( we get r 3,ε (l 0... l 5 r 3 (l 0... l 5 r 3,ε (l 0... l 5 r3 (l 0... l 5 ( (l = 0 l1 l 3 ε + (l 1 l l 4 ε (l 0 l 1 l 3 (l 1 l l 4 + (l l 0 l 5 ε (l l 0 l 5 (l 0 l 1 l 4 ε (l 0 l 1 l 4 (l 1 l l 5 ε (l 1 l l 5 (l l 0 l 3 ε (l l 0 l 3 + (l 0 l 1 l 4 ε (l 0 l 1 l 4 + (l 1 l l 5 ε (l 1 l l 5 + (l l 0 l 3 ε (l l 0 l 3 (l 0 l 1 l 3 ε (l 0 l 1 l 3 (l 1 l l 4 ε (l 1 l l 4 (l l 0 l 5 ε (l l 0 l 5 (5. 46

16 106 Sadaqat Hussain and Raziuddin Siddiqui Then ( becomes = { r3 45 Alt 6 (l 0... l 5 ; r 3,ε(l0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3 (l 1 l l 4 (l l 0 l 5 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 ( + (l 0 l 1 l 3 ε + (l 1 l l 4 ε + (l l 0 l 5 ε (l 0 l 1 l 4 ε (l 1 l l 5 ε (l l 0 l 3 ε (l 0 l 1 l 3 (l 1 l l 4 (l l 0 l 5 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 + (l 0 l 1 l 4 ε (l 0 l 1 l 4 + (l 1 l l 5 ε (l 1 l l 5 + (l l 0 l 3 ε (l l 0 l 3 (l 0 l 1 l 3 ε (l 0 l 1 l 3 (l 1 l l 4 ε (l 1 l l 4 (l l 0 l 5 ε [r 3 (l 0... l 5 ] (l l 0 l 5 (5. 47 Above expression consist of two summands of the form a b and c [b].to discuss both separately, consider the first summand = 45 Alt 6 { r3 (l 0... l 5 ; r 3,ε(l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3 (l 1 l l 4 (l l 0 l 5 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (5. 48 Which can further be written as =Alt 6 { r3 (l 0... l 5 ; r 3,ε (l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3 +Alt 6 { r3 (l 0... l 5 ; r 3,ε (l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 1l l 4 +Alt 6 { r3 (l 0... l 5 ; r 3,ε (l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l l 0 l 5 Alt 6 { r3 (l 0... l 5 ; r 3,ε (l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 4 Alt 6 { r3 (l 0... l 5 ; r 3,ε (l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 1l l 5 Alt 6 { r3 (l 0... l 5 ; r 3,ε (l 0... l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l l 0 l 3 (5. 49 According to the technique used in ([8] and ([9] we have Alt 6 { r3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l 0 l 1 l l 3 l 4 l 5,, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3 =Alt 6 { r3 (l 1 l l 0 l 4 l 5 l 3 ; r 3,ε (l 1 l l 0 l 4 l 5 l 3,, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 1l l 4 and r3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l 0 l 1 l l 3 l 4 l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] = r 3 (l 1 l l 0 l 4 l 5 l 3 ; r 3,ε (l 1 l l 0 l 4 l 5 l 3, r 3,ε (l 1 l l 0 l 4 l 5 l 3 ] (5. 50 And this symmetry is true for all other ratios obtained by going through the alternation. Therefore ( becomes = 15 Alt 6{ r3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l 0 l 1 l l 3 l 4 l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3 r 3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l 0 l 1 l l 3 l 4 l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 4 Applying the odd permutation (l 3 l 4 (or (l 3 l 4 = 4 15 Alt { 6 r3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l0 l 1 l l 3 l 4 l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3

17 Grassmannian Complex and Second Order Tangent Complex 107 Once again using the permutation (l 0 l 3 ( or (l0 l 3 we get = 15 Alt { 6 r3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l0 l 1 l l 3 l 4 l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] (l 0l 1 l 3 r 3 (l 3 l 1 l l 0 l 4 l 5 ; r 3,ε (l3 l 1 l l 0 l 4 l 5, r 3,ε (l 3 l 1 l l 0 l 4 l 5 ] (l 3l 1 l 0 we have another symmetry (l 0 l 1 l 3 = (l 3 l 1 l 0 = (l 0 l 3 l 1 up to -torsion, then = 15 Alt 6{( r3 (l 0 l 1 l l 3 l 4 l 5 ; r 3,ε (l0 l 1 l l 3 l 4 l 5, r 3,ε (l 0 l 1 l l 3 l 4 l 5 ] r 3 (l 3 l 1 l l 0 l 4 l 5 ; r 3,ε (l3 l 1 l l 0 l 4 l 5, r 3,ε (l 3 l 1 l l 0 l 4 l 5 ] (l0 l 1 l 3 (5. 51 Here again we follow the method used in ([8],[9] that is we can express the triple ratio as the ratio of two cross ratios. In ([8],[9] it has been done for r 3 (l 3 l 1 l l 0 l 4 l 5 and r 3,ε (l3 l 1 l l 0 l 4 l 5, so we will do it only for r 3,ε (l 0 l 1 l l 3 l 4 l 5. Consider Since r 3 (l 0 l 1 l l 3 l 4 l 5 = (l 0l 1 l 3 (l 1 l l 4 (l l 0 l 5 (l 0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 r 3 (l 0 l 1 l l 3 l 4 l 5 = r 3(l 0 l 1 l l 3 l 4 l 5 + ( r 3 (l 0 l 1 l l 3 l 4 l 5 ε ε + ( r 3 (l 0 l 1 l l 3 l 4 l 5 ε ε so we can write ( (l r 3,ε (l0 l 1 l l 3 l 4 l 5 = 0 l1 l 3 (l 1 l l 4 (l l 0 l 5 (l0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 ε Using the projection here by l 1 and l ( (l r 3,ε (l0 l 1 l l 3 l 4 l 5 = 1 l l 3 (l 0 l 1 l 3 (l 1 l l 4 (l l 0 l 5 (l0 l 1 l 4 (l 1 l l 5 (l l 0 l 3 (l 1 l l 3 = ε Therefore ( can be written as = { ( 15 Alt r(l l 1 l 0 l 5 l 3 r(l 6 r(l 1 l 0 l l 3 l 4 ; l1 l 0 l 5 l 3 r(l1 l 0 l l 3 l 4 ( r(l l 1 l 3 l 5 l 0 r(l r(l 1 l 3 l l 0 l 4 ; l1 l 3 l 5 l 0 r(l1 l 3 l l 0 l 4, ε, ε ( r(l l1 l 0 l 5 l 3 r(l1 l 0 l l 3 l 4 ( r(l l1 l 3 l 5 l 0 r(l 1 l 3 l l 0 l 4 Applying the five term relation in TB (F = ( r(l 15 Alt 6{ l 1 l 0 l 5 l 3 ; r ε (l l 1 l 0 l 5 l 3, r ε (l l 1 l 0 l 5 l 3 ] r(l 1 l 0 l l 3 l 4 ; r ε (l1 l 0 l l 3 l 4, r ε (l 1 l 0 l l 3 l 4 ] ( r(l l 1 l 5 l 3 l 0 r(l r(l 1 l 0 l 3 l 4 l ; l1 l 5 l 3 l 0 ( r(l l1 l 5 l 3 l 0 r(l1 l 0 l 3 l 4 l r(l1 l 0 l 3 l 4 l, ε ε ( r(l l1 l 0 l 5 l 3 ] r(l 1 l 0 l l 3 l 4 ] ε ] ε ε (l 0 l 1 l 3 (l 0 l 1 l 3 (l 0 l 1 l 3 (5. 5 From above three terms of individual determinant (l 0 l 1 l 3,Consider the last term. Which is ( 15 Alt r(l l 1 l 5 l 3 l 0 r(l 6 r(l 1 l 0 l 3 l 4 l ; l1 l 5 l 3 l 0 ( r(l l1 l 5 l 3 l 0 ] r(l1 l 0 l 3 l 4 l r(l1 l 0 l 3 l 4 l (l 0 l 1 l 3, ε ε

18 108 Sadaqat Hussain and Raziuddin Siddiqui Which can further be written as = ( 15 Alt Alt r(l l 1 l 5 l 3 l 0 r(l (l 0 l 1 l 3 (l l 4 l 5 r(l 1 l 0 l 3 l 4 l ; l1 l 5 l 3 l 0 r(l1 l 0 l 3 l 4 l, ε ( r(l l1 l 5 l 3 l 0 r(l 1 l 0 l 3 l 4 l ε ] (l 0 l 1 l 3 Consider a permutation subgroup S 3 S 3 in S 6, Where S 3 permuting {l 0, l 1, l 3 and {l, l 4, l 5, which fixes the determinant (l 0 l 1 l 3 as (l 0 l 1 l 3 (l 3 l 1 l 0 (l 3 l 0 l 1. Take ( r(l l 1 l 5 l 3 l 0 r(l Alt (l0 l 1 l 3 (l l 4 l 5 r(l 1 l 0 l 3 l 4 l ; l1 l 5 l 3 l 0 ( r(l r(l1 l 0 l 3 l 4 l, l1 l 5 l 3 l 0 ] r(l ε 1 l 0 l 3 l 4 l (l 0 l 1 l 3 ε ( (l l 5 l 3 (l 1 l 0 l 4 (l =Alt (l0 l 1 l 3 (l l 4 l 5 (l l 5 l 0 (l 1 l 3 l 4 ; l 5 l 3 (l 1 l 0 l 4 ( (l l 5 l 3 (l 1 l 0 l 4 (l l 5 l 0 (l 1 l 3 l 4 (l l 5 l 0 (l 1 l 3 l 4 (l 0 l 1 l 3 This becomes zero if we use odd permutation (l l 5 and ( 5. 5 will be of the form, ε ε = {( r(l 15 Alt 6 l 1 l 0 l 5 l 3 ; r ε (l l 1 l 0 l 5 l 3, r ε (l l 1 l 0 l 5 l 3 ] r(l 1 l 0 l l 3 l 4 ; r ε (l1 l 0 l l 3 l 4, r ε (l 1 l 0 l l 3 l 4 ] (l 0 l 1 l 3 (5. 53 Take first term of above { 15 Alt 6 r(l l 1 l 0 l 5 l 3 ; r ε (l l 1 l 0 l 5 l 3, r ε (l l 1 l 0 l 5 l 3 ] (l 0l 1 l 3 = { 1 { 15 Alt 6 36 Alt (l 0 l 1 l 3 (l l 4 l 5 r(l l 1 l 0 l 5 l 3 ; r ε (l l 1 l 0 l 5 l 3, r ε (l l 1 l 0 l 5 l 3 ] (l 0l 1 l 3 Since the ratio is projected by so, the the permutation (l 0 l 1 l 3 will have no influence in above ratio. Therefore we can write above as = 15 Alt 6 { 1 { 6 Alt (l l 4 l 5 r(l l 1 l 0 l 5 l 3 ; r ε (l l 1 l 0 l 5 l 3, r ε (l l 1 l 0 l 5 l 3 ] (l 0l 1 l 3 By the expansion of inner alternation sum we get = 1 {( r(l4 45 Alt 6 l 1 l 0 l l 3 ; r ε (l4 l 1 l 0 l l 3, r ε (l 4 l 1 l 0 l l 3 ] r(l l 1 l 0 l 4 l 3 ; r ε (l l 1 l 0 l 4 l 3, r ε (l l 1 l 0 l 4 l 3 ] + r(l 5 l 1 l 0 l 4 l 3 ; r ε (l 5 l 1 l 0 l 4 l 3, r ε (l 5 l 1 l 0 l 4 l 3 ] r(l 4 l 1 l 0 l 5 l 3 ; r ε (l4 l 1 l 0 l 5 l 3, r ε (l 4 l 1 l 0 l 5 l 3 ] + r(l l 1 l 0 l 5 l 3 ; r ε (l l 1 l 0 l 5 l 3, r ε (l l 1 l 0 l 5 l 3 ] r(l 5 l 1 l 0 l l 3 ; r ε (l 5 l 1 l 0 l l 3, r ε (l 5 l 1 l 0 l l 3 ] (l 0 l 1 l 3

19 Grassmannian Complex and Second Order Tangent Complex 109 Using projected five-term relation for TB (F = 1 {( r(l0 45 Alt 6 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] r(l 1 l 0 l l 3 l 4 ; r ε (l 1 l 0 l l 3 l 4, r ε (l 1 l 0 l l 3 l 4 ] r(l 3 l 0 l 1 l l 4 ; r ε (l3 l 0 l 1 l l 4, r ε (l 3 l 0 l 1 l l 4 ] + r(l 0 l 1 l 4 l 3 l 5 ; r ε (l 0 l 1 l 4 l 3 l 5, r ε (l 0 l 1 l 4 l 3 l 5 ] r(l 1 l 0 l 4 l 3 l 5 ; r ε (l1 l 0 l 4 l 3 l 5, r ε (l 1 l 0 l 4 l 3 l 5 ] r(l 3 l 0 l 1 l 4 l 5 ; r ε (l 3 l 0 l 1 l 4 l 5, r ε (l 3 l 0 l 1 l 4 l 5 ] + r(l 0 l 1 l 5 l 3 l ; r ε (l0 l 1 l 5 l 3 l, r ε (l 0 l 1 l 5 l 3 l ] r(l 1 l 0 l 5 l 3 l ; r ε (l 1 l 0 l 5 l 3 l, r ε (l 1 l 0 l 5 l 3 l ] r(l 3 l 0 l 1 l 5 l ; r ε (l3 l 0 l 1 l 5 l, r ε (l 3 l 0 l 1 l 5 l ] (l 0 l 1 l 3 Use the cyclic permutation (l 0 l 1 l 3 (l l 4 l 5 we get = 1 { 45 9Alt 6 r(l0 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] (l 0l 1 l 3 Similarly we will write the second term of ( as Alt 6 { r(l1 l 0 l l 3 l 4 ; r ε (l 1 l 0 l l 3 l 4, r ε (l 1 l 0 l l 3 l 4 ] (l 0l 1 l 3 Using ( and ( we can write ( as = 1 {( 45 Alt 6 9 r(l 0 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] 6 r(l 1 l 0 l l 3 l 4 ; r ε (l1 l 0 l l 3 l 4, r ε (l 1 l 0 l l 3 l 4 ] (l 0 l 1 l 3 The permutation (l 0 l 1 l 3 (l l 4 l 5 will give us = 1 { 3 Alt 6 r(l0 l 1 l l 3 l 4 ; r ε (l 0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] (l 0l 1 l 3 (5. 54 (5. 55 (5. 56 (5. 57 (5. 58 This is the final value of the first summand of ( Now we proceed for the second summand of ( Since B (F satisfies five-term relation therefore second summand can be written as = 1 3 Alt 6 (l 0 l 1 l 3 ε (l 0 l 1 l 3 + (l 0 l 1 l 3 ε (l 0 l 1 l 3 (l 0 l 1 l 3 ε [r(l (l 0 l 1 l 3 0 l 1 l l 3 l 4 ] (5. 59 Use ( and ( to write ( as = 1 { r(l0 3 Alt 6 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] (l 0l 1 l 3 + (l 0 l 1 l 3 ε (l 0 l 1 l 3 ε [r(l (l 0 l 1 l 3 (l 0 l 1 l 3 0 l 1 l l 3 l 4 ] (5. 60 This completes the calculations of one side of the proof. Now to compute other side τ 3 d(l 1,ε 0... l 5 we will write the map τ3 1,ε in the form of

20 110 Sadaqat Hussain and Raziuddin Siddiqui alternation sum. That is { r(l0 τ 3 (l 1,ε 0... l 4 =1 3 Alt 6 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] (l 0l 1 l + (l 0 l 1 l ε (l 0 l 1 l ε [r(l (l 0 l 1 l (l 0 l 1 l 0 l 1 l l 3 l 4 ] (5. 61 Applying the cycle (l 0 l 1 l l 3 l 4 l 5 for the map d and the expansion through the alternation Alt 5 we get { r(l0 τ 3 1ε d(l 0... l 5 =1 3 Alt 6 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] (l 0l 1 l + (l 0 l 1 l ε (l 0 l 1 l ε [r(l (l 0 l 1 l (l 0 l 1 l 0 l 1 l l 3 l 4 ] (5. 6 The odd cycle (l l 3 will make ( 5. 6 like { r(l0 τ 3 1ε d(l 0... l 5 =1 3 Alt 6 l 1 l l 3 l 4 ; r ε (l0 l 1 l l 3 l 4, r ε (l 0 l 1 l l 3 l 4 ] (l 0l 1 l 3 + (l 0 l 1 l 3 ε (l 0 l 1 l 3 ε [r(l (l 0 l 1 l 3 (l 0 l 1 l 3 0 l 1 l l 3 l 4 ] (5. 63 At last two-term relation in TB (F and B (F will give us the correct sign. The final result obtained is same as ( Corollary 5.7. The maps defined below are zero (1 C 5 ( 4 F[ε] 3 d C 4 ( 3 F[ε] 3 ( C 6 ( 4 F[ε] 3 d C 5 ( 3 F[ε] 3 τ 3 0,ε F F 3 F τ 3 1,ε ( TB (F F (F B (F Proof. Simply we have to show τ 3 0,ε d = 0 and τ 3 1,ε d = 0 6. ACKNOWLEDGMENTS My heartfelt thanks and deep gratitude to my thesis supervisor Dr.Raziuddin Siddiqui. He exerted his utmost in encouraging and in offering me their Valuable advice without which such accomplishment would never have been possible. REFERENCES [1] J. L. Cathelineau, Infinitesimal Polylogarithms, multiplicative Presentations of Kähler Differentials and Goncharov complexes, talk at the workshop on polylogarthms, Essen, May 1-4 (1997. [] J. L. Cathelineau, The tangent complex to the Bloch-Suslin complex, Bull. Soc. Math. France 135, ( [3] J. L. Cathelineau, Projective Configurations, Homology of Orthogonal Groups, and Milnor K-theory, Duke Mathematical Journal, No. 11 (004. [4] Ph. Elbaz-Vincent and H. Gangl, On Poly(analogs I, Compositio Mathematica, 130, ( [5] A. B. Goncharov, Geometry of Configurations, Polylogarithms and Motivic Cohomology, Adv. Math., 114, ( [6] A. B. Goncharov and J. Zhao, Grassmannian Trilogarithms, Compositio Mathematica, 17, ( [7] S. Hussain and R. Siddiqui, Projected Five Term Relation in TB (F,International Journal of Algebra, 6, No. 8 (

21 Grassmannian Complex and Second Order Tangent Complex 111 [8] R. Siddiqui, Configuration Complexes and Tangential and Infinitesimal versions of Polylogarithmic Complexes, Doctoral thesis, Durham University. Available at Durham E-Theses Online: (010. [9] R. Siddiqui, Tangent to Bloch-Suslin and Grassmannian Complexes over the dual numbers, (01 arxiv: v [math.nt].