KAMEKOS HOMOMORPHISM AND THE ALGEBRAIC TRANSFER

Abstract. Let Pk be the graded polynomial algebra F2x1, x2, . . . , xk, with the degree of each xi being 1, regarded as a module over the mod2 Steenrod algebra A, and let GLk be the general linear group over the prime field F2. We study the algebraic transfer constructed by Singer 20, using the technique of the hit problem. This transfer is a homomorphism from the homology of the mod2 Steenrod algebra, TorA k,k+n (F2, F2), to the subspace of F2⊗APk consisting of all the GLkinvariant classes of degree n. In this paper, we develop Hưng’s result in 11 on the relation between the algebraic transfer and Kameko’s homomorphism. Using this result, we show that Singer’s conjecture for the algebraic transfer is true in the case k = 5 and the degree 7.2 s − 5 with s an arbitrary positive integer.

KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC Abstract. Let Pk be the graded polynomial algebra F2 [x1 , x2 , . . . , xk ], with the degree of each xi being 1, regarded as a module over the mod-2 Steenrod algebra A, and let GLk be the general linear group over the prime field F2 . We study the algebraic transfer constructed by Singer [20], using the technique of the hit problem. This transfer is a homomorphism from the homology of the mod-2 Steenrod algebra, TorA k,k+n (F2 , F2 ), to the subspace of F2 ⊗A Pk consisting of all the GLk -invariant classes of degree n. In this paper, we develop Hưng’s result in [11] on the relation between the algebraic transfer and Kameko’s homomorphism. Using this result, we show that Singer’s conjecture for the algebraic transfer is true in the case k = 5 and the degree 7.2s − 5 with s an arbitrary positive integer. 1. Introduction Let Vk be an elementary abelian 2-group of rank k. Denote by BVk the classifying space of Vk . Then, Pk := H ∗ (BVk ) ∼ = F2 [x1 , x2 , . . . , xk ], a polynomial algebra in k variables x1 , x2 , . . . , xk , each of degree 1. Here the cohomology is taken with coefficients in the prime field F2 of two elements. Being the cohomology of a space, Pk is a module over the mod-2 Steenrod algebra, A. The action of A on Pk is determined by the elementary properties of the Steenrod squares Sq i and subject to the Cartan formula (see Steenrod and Epstein [22]). Let GLk be the general linear group over the field F2 . This group acts naturally on Pk by matrix substitution. Since the two actions of A and GLk upon Pk commute with each other, there is an inherited action of GLk on F2 ⊗A Pk . Denote by (Pk )n the subspace of Pk consisting of all the homogeneous polynomials of degree n in Pk and by (F2 ⊗A Pk )n the subspace of F2 ⊗A Pk consisting of all the classes represented by the elements in (Pk )n . In [20], Singer defined the algebraic transfer, which is a homomorphism GLk ϕk : TorA k,k+n (F2 , F2 ) −→ (F2 ⊗A Pk )n from the homology of the Steenrod algebra to the subspace of (F2 ⊗A Pk )n consisting of all the GLk -invariant classes. It is a useful tool in describing the homology groups of the Steenrod algebra, TorA k,k+n (F2 , F2 ). The Singer algebraic transfer was studied by many authors (see Boardman [2], Bruner-Hà-Hưng [3], Hà [9], Hưng [10, 11], 12000 Mathematics Subject Classification. Primary 55S10; 55S05, 55T15. 2Keywords and phrases: Steenrod squares, hit problem, algebraic transfer. 1 ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 2 Chơn-Hà [6, 7, 8], Minami [15], Nam [16], Hưng-Quỳnh [12], Quỳnh [19], the first named author [24] and others). Singer showed in [20] that ϕk is an isomorphism for k = 1, 2. Boardman showed in [2] that ϕ3 is also an isomorphism. However, for any k 4, ϕk is not a monomorphism in infinitely many degrees (see Singer [20], Hưng [11].) Singer made the following conjecture. Conjecture 1.1 (Singer [20]). The algebraic transfer ϕk is an epimorphism for any k 0. The conjecture is true for k 3. Based on the results in [23, 25], it can be verified for k = 4. We hope that it is also true in this case. In this paper, we develop Hưng’s result in [11] on the relation between the 0 algebraic transfer and Kameko’s homomorphism Sq ∗ : F2 ⊗A Pk −→ F2 ⊗A Pk . This homomorphism is an GLk -homomorphism induced by the F2 -linear map, also 0 denoted by Sq ∗ : Pk → Pk , given by 0 Sq ∗ (x) = y, if x = x1 x2 . . . xk y 2 , 0, otherwise, 0 for any monomial x ∈ Pk . Note that Sq ∗ is not an A-homomorphism. However, 0 0 0 Sq ∗ Sq 2t = Sq t Sq ∗ and Sq ∗ Sq 2t+1 = 0 for any non-negative integer t. From the result of Carlisle and Wood [4] on the boundedness conjecture, we can see that, for any non-negative integer d, there exists a non-negative integer t such that 0 (Sq ∗ )s−t : (F2 ⊗A Pk )k(2s −1)+2s d −→ (F2 ⊗A Pk )k(2t −1)+2t d is an isomorphism of GLk -modules for every s t. However, this result does not confirm how large t should be. Let n be a positive integer. Denote by α(n) the number of ones in dyadic expansion of n and by ζ(n) the greatest integer u such that n is divisible by 2u . That means n = 2ζ(n) m with m an odd integer. For any non-negative integer d, set t(k, d) = max{0, k − α(d + k) − ζ(d + k)}. The following is one of our main results. Theorem 1.2. Let d be an arbitrary non-negative integer. Then 0 (Sq ∗ )s−t : (F2 ⊗A Pk )k(2s −1)+2s d −→ (F2 ⊗A Pk )k(2t −1)+2t d is an isomorphism of GLk -modules for every s It is easy to see that t(k, d) following. t if and only if t k − 2 for every d and k t(k, d). 2. So, one gets the Corollary 1.3 (See Hưng [11]). Let d be an arbitrary non-negative integer. Then 0 (Sq ∗ )s−k+2 : (F2 ⊗A Pk )k(2s −1)+2s d −→ (F2 ⊗A Pk )k(2k−2 −1)+2k−2 d is an isomorphism of GLk -modules for every s k − 2. KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 3 Corollary 1.3 shows that the number t = k − 2 commonly serves for every degree d. In [11], Hưng predicted that t = k − 2 is the minimum number for this purpose and proved it for k = 5. It is easy to see that for d = 2k − k + 1, we have t(k, d) = k − 2. So, his prediction is true for all k 2. An application of Theorem 1.2 is the following theorem. Theorem 1.4. Singer’s conjecture is true for k = 5 and in degree 7.2s − 5 with s an arbitrary positive integer. For d = 2, we have t(5, 2) = 2 and 5(2s − 1) + 2s d = 7.2s − 5. So, by Theorem 1.2, 0 (Sq ∗ )s−2 : (F2 ⊗A P5 )7.2s −5 −→ (F2 ⊗A P5 )23 is an isomorphism of GL5 -modules for every s 2. Hence, by an explicit compu5 tation of (F2 ⊗A P5 )7.2s −5 and (F2 ⊗A P5 )GL 7.2s −5 for s = 1, 2, one gets the following. Theorem 1.5. Let m = 7.2s − 5 with s a positive integer. Then i) dim(F2 ⊗A P5 )m = 191 for s = 1 and dim(F2 ⊗A P5 )m = 1248 for any s 5 ii) (F2 ⊗A P5 )GL = 0 for any s 1. m 2. The second part of this theorem has been proved by Singer [20] for s = 1. In 5 for s 2. [11], Hưng computed the dimensions of (F2 ⊗A P5 )m and (F2 ⊗A P5 )GL m However, his result for dim(F2 ⊗A P5 )m is not true. From the results of Tangora [26], Lin [14] and Chen [5], s Ext5,7.2 (F2 , F2 ) = A P h1 , if s = 1, hs gs−1 , if s 2, s and hs gs−1 = 0, where hs denote the Adams element in Ext1,2 A (F2 , F2 ), P is the 4,2s+2 +2s+1 Adams periodicity operator [1] and gs−1 ∈ ExtA (F2 , F2 ) for s 2. By passing to the dual, one gets TorA 5,7.2s (F2 , F2 ) = (P h1 )∗ , if s = 1, ∗ (hs gs−1 ) , if s 2. Hence, by Theorem 1.5(ii), the homomorphism GL5 ϕ5 : TorA 5,7.2s (F2 , F2 ) −→ (F2 ⊗A P5 )7.2s −5 is an epimorphism. However, it is not a monomorphism for all s 1. Theorem 1.4 is proved. For d = 5, we have t(5, 5) = 2 and 5(2s − 1) + 2s d = 10.2s − 5. So, by using Theorem 1.2, we see that 0 (Sq ∗ )s−2 : (F2 ⊗A P5 )10.2s −5 −→ (F2 ⊗A P5 )35 is an isomorphism of GL5 -modules for every s 2. Hence, by computing the space (F2 ⊗A P5 )10.2s −5 for s = 1, 2, we obtain the following. Theorem 1.6. Let n = 10.2s −5 with s a positive integer. Then dim(F2 ⊗A P5 )n = 432 for s = 1 and dim(F2 ⊗A P5 )n = 1171 for any s 2. This theorem is proved in [24] for s = 1. In [11], Hưng study the dimension of this space by using computer calculation. However, his result is also not true for s 2. ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 4 From Tangora [26], Lin [14] and Chen [5], we have h40 h4 , h1 d0 , if s = 1, hs ds−1 , if s 2, s Ext5,10.2 (F2 , F2 ) = A and hs ds−1 = 0, where ds−1 ∈ Ext4,2 A dual, we obtain TorA 5,10.2s (F2 , F2 ) = s+3 +2s (F2 , F2 ) for s 1. By passing to the (h40 h4 )∗ , (h1 d0 )∗ , if s = 1, (hs ds−1 )∗ , if s 2. 5 In [24], we have proved that dim(F2 ⊗A P5 )GL = 2 and 15 GL5 ϕ5 : TorA 5,20 (F2 , F2 ) −→ (F2 ⊗A P5 )15 is an isomorphism. Combining the results of Singer [20] and Hà [9], we see that 5 ϕ5 ((hs ds−1 )∗ ) = 0. So, dim(F2 ⊗A P5 )GL 1 for any s 1. Hưng stated in [11] 10.2s −5 GL5 that dim(F2 ⊗A P5 )35 = 1. If this result is true, then the homomorphism GL5 ϕ5 : TorA 5,10.2s (F2 , F2 ) −→ (F2 ⊗A P5 )10.2s −5 is an isomorphism for s 1. Hence, Singer’s conjecture is also true for k = 5 and in degree 10.2s − 5. This paper is organized as follows. In Section 2, we recall some needed information on the admissible monomials in Pk , Singer’s criterion on the hit monomials and Kameko’s homomorphism. Theorems 1.2, 1.5 and 1.6 are respectively proved in Sections 3-5. Finally, in the appendix we list all the admissible monomials of degrees 9, 15, 16, 23, 35 in P4 and P5 . 2. Preliminaries In this section, we recall some needed information from Kameko [13] and Singer [21], which will be used in the next section. Notation 2.1. We denote Nk = {1, 2, . . . , k} and xj , J = {j1 , j2 , . . . , js } ⊂ Nk , XJ = X{j1 ,j2 ,...,js } = j∈Nk \J In particular, XNk = 1, X∅ = x1 x2 . . . xk , Xj = x1 . . . x ˆj . . . xk , 1 j k, and X := Xk ∈ Pk−1 . Let αi (a) denote the i-th coefficient in dyadic expansion of a non-negative integer a. That means a = α0 (a)20 + α1 (a)21 + α2 (a)22 + . . . , for αi (a) = 0 or 1 with i 0. Let x = xa1 1 xa2 2 . . . xakk ∈ Pk . Denote νj (x) = aj , 1 j k. Set Jt (x) = {j ∈ Nk : αt (νj (x)) = 0}, for t 0. Then, we have x = t t 0 XJ2t (x) . Definition 2.2. For a monomial x in Pk , define two sequences associated with x by ω(x) = (ω1 (x), ω2 (x), . . . , ωi (x), . . .), σ(x) = (ν1 (x), ν2 (x), . . . , νk (x)), where ωi (x) = 1 j k αi−1 (νj (x)) = deg XJi−1 (x) , i called the weight vector of x. 1. The sequence ω(x) is KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 5 Let ω = (ω1 , ω2 , . . . , ωi , . . .) be a sequence of non-negative integers. The sequence ω is called the weight vector if ωi = 0 for i 0. The sets of all the weight vectors and the sigma vectors are given the left lexicographical order. For a weight vector ω, we define deg ω = i>0 2i−1 ωi . Denote by Pk (ω) the subspace of Pk spanned by all monomials y such that deg y = deg ω, ω(y) ω, and by Pk− (ω) the subspace of Pk spanned by all monomials y ∈ Pk (ω) such that ω(y) < ω. Definition 2.3. Let ω be a weight vector and f, g two polynomials of the same degree in Pk . i) f ≡ g if and only if f − g ∈ A+ Pk . If f ≡ 0 then f is called hit. ii) f ≡ω g if and only if f − g ∈ A+ Pk + Pk− (ω). Obviously, the relations ≡ and ≡ω are equivalence ones. Denote by QPk (ω) the quotient of Pk (ω) by the equivalence relation ≡ω . Then, we have QPk (ω) = Pk (ω)/((A+ Pk ∩ Pk (ω)) + Pk− (ω)). For a polynomial f ∈ Pk , we denote by [f ] the class in F2 ⊗A Pk represented by f . If ω is a weight vector and f ∈ Pk (ω), then denote by [f ]ω the class in QPk (ω) represented by f . Denote by |S| the cardinal of a set S. It is easy to see that QPk (ω) ∼ = QPkω := {[x] ∈ QPk : x is admissible and ω(x) = ω} . So, we get QPkω ∼ = (F2 ⊗A Pk )n = deg ω=n QPk (ω). deg ω=n Hence, we can identify the vector space QPk (ω) with QPkω ⊂ QPk . We note that the weight vector of a monomial is invariant under the permutation of the generators xi , hence QPk (ω) has an action of the symmetric group Σk . For 1 i k, define the A-homomorphism gi : Pk → Pk , which is determined by gi (xi ) = xi+1 , gi (xi+1 ) = xi , gi (xj ) = xj for j = i, i + 1, 1 i < k, and gk (x1 ) = x1 + x2 , gk (xj ) = xj for j > 1. Note that the general linear group GLk is generated by the matrices associated with gi , 1 i k, and the symmetric group Σk is generated by the ones associated with gi , 1 i < k. So, a homogeneous polynomial f ∈ Pk is an GLk -invariant if and only if gi (f ) ≡ f for 1 i k. If gi (f ) ≡ f for 1 i < k, then f is an Σk -invariant. We have the following. Lemma 2.4. If x is a monomial in Pk , then gk (x) ∈ Pk (ω(x)). Proof. If ν1 (x) = 0, then x = gk (x) and ω(gk (x)) = ω(x). Suppose ν1 (x) > 0 and ν1 (x) = 2t1 + . . . + 2tb , where 0 t1 < . . . < tb , b 1. t Since x = t 0 XJ2t (x) ∈ Pk and gk is a homomorphism of algebras, b t (gk (XJt (x) ))2 = gk (x) = t 0 (x1 + x2 )XJtu (x)∪1 u=1 2tu t XJ2t (x) . t=t1 ,t2 ,...,tb ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 6 Then, gk (x) is a sum of monomials of the form c y¯ = x2 XJtu j j=1 2 tu t XJ2t (x) , (x)∪1 t=tu1 ,...,tuc where 0 c b. If c = 0, then y¯ = x and ω(¯ y ) = ω(x). Suppose c > 0. If 2 ∈ Jtuj (x) for all j, 1 j c, then ω(¯ y ) = ω(x) and y¯ ∈ Pk (ω(x)). Suppose there is an index j such that 2 ∈ Jtuj (x). Let j0 be the smallest index such that 2 ∈ Jtuj (x). Then, we have 0 ωi (¯ y) = ωi (x), if i tuj0 , ωi (x) − 2, if i = tuj0 + 1. Hence ω(¯ y ) < ω(x) and y¯ ∈ Pk (ω(x)). The lemma is proved. Lemma 2.4 implies that if ω is a weight vector and x ∈ Pk (ω), then gk (x) ∈ Pk (ω). Moreover, QPk (ω) is an GLk -module. Definition 2.5. Let x, y be monomials of the same degree in Pk . We say that x < y if and only if one of the following holds: i) ω(x) < ω(y); ii) ω(x) = ω(y) and σ(x) < σ(y). Definition 2.6. A monomial x is said to be inadmissible if there exist monomials m y1 , y2 , . . . , ym such that yt < x for t = 1, 2, . . . , m and x − t=1 yt ∈ A+ Pk . A monomial x is said to be admissible if it is not inadmissible. Obviously, the set of all the admissible monomials of degree n in Pk is a minimal set of A-generators for Pk in degree n. Theorem 2.7 (See Kameko [13]). Let x, y, w be monomials in Pk such that ωi (x) = 0 for i > r > 0, ωs (w) = 0 and ωi (w) = 0 for i > s > 0. r i) If w is inadmissible, then xw2 is also inadmissible. s ii) If w is strictly inadmissible, then wy 2 is also strictly inadmissible. Now, we recall a result of Singer [21] on the hit monomials in Pk . Definition 2.8. A monomial z in Pk is called a spike if νj (z) = 2dj − 1 for dj a non-negative integer and j = 1, 2, . . . , k. If z is a spike with d1 > d2 > . . . > dr−1 dr > 0 and dj = 0 for j > r, then it is called the minimal spike. For a positive integer n, by µ(n) one means the smallest number r for which it is possible to write n = 1 i r (2di − 1), where di > 0. In [21], Singer showed that if µ(n) k, then there exists uniquely a minimal spike of degree n in Pk . Lemma 2.9 (See [18]). All the spikes in Pk are admissible and their weight vectors are weakly decreasing. Furthermore, if a weight vector ω is weakly decreasing and ω1 k, then there is a spike z in Pk such that ω(z) = ω. The following is a criterion for the hit monomials in Pk . Theorem 2.10 (See Singer [21]). Suppose x ∈ Pk is a monomial of degree n, where µ(n) k. Let z be the minimal spike of degree n. If ω(x) < ω(z), then x is hit. This result implies a result of Wood, which original is a conjecture of Peterson [17]. KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 7 Theorem 2.11 (See Wood [27]). If µ(n) > k, then (F2 ⊗A Pk )n = 0. Denote by 0 (Sq ∗ )(k,m) : (F2 ⊗A Pk )2m+k −→ (F2 ⊗A Pk )m 0 Kameko’s homomorphism Sq ∗ : (F2 ⊗A Pk ) −→ (F2 ⊗A Pk in degree 2m + k. Theorem 2.12 (See Kameko [13]). Let m be a positive integer. If µ(2m + k) = k, then 0 (Sq ∗ )(k,m) : (F2 ⊗A Pk )2m+k −→ (F2 ⊗A Pk )m is an isomorphism of GLk -modules. Now, we recall some notations and definitions in [25], which will be used in the next sections. We set Pk0 = {x = xa1 1 xa2 2 . . . xakk : a1 a2 . . . ak = 0} , Pk+ = {x = xa1 1 xa2 2 . . . xakk : a1 a2 . . . ak > 0} . It is easy to see that Pk0 and Pk+ are the A-submodules of Pk . Furthermore, we have the following. Proposition 2.13. We have a direct summand decomposition of the F2 -vector spaces F2 ⊗A Pk = QPk0 ⊕ QPk+ . Here QPk0 = F2 ⊗A Pk0 and QPk+ = F2 ⊗A Pk+ . Denote Nk = (i; I); I = (i1 , i2 , . . . , ir ), 1 i < i1 < . . . < ir k, 0 r 2r − 1, iii) αr−t (νiu −1 (x)) = 1, ∀t, 1 t u, iv) αr−t (νit −1 (x)) = 1, ∀t, u < t r. Clearly, a monomial x can be u-compatible with a given (i; I) ∈ Nk for at most one value of u. By convention, x is 1-compatible with (i; ∅). For 1 i k, define the homomorphism fi : Pk−1 → Pk of algebras by substituting fi (xj ) = xj , xj+1 , Definition 2.15. Let (i; I) ∈ Nk , x(I,u) (I) > 0, x(∅,1) = 1. For a monomial x in in Pk by setting  r 2 −1  fi (x))/x(I,u) , (xi φ(i;I) (x) =   0, if 1 if i j < i, j < k. r−1 r−u r−t 2 = x2iu +...+2 for r = u d2 > . . . > ds−1 ds > 0 such that s n = 2d1 + 2d2 + . . . + 2ds−1 + 2ds − s = (2di − 1). i=1 (3.1) KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 9 Proof. Assume that µ(n) = s. Set β(n) = min{u ∈ N : α(n + u) u}. We prove µ(n) = β(n). Suppose β(n) = t. Then α(n + t) = r t and n = 2c1 + 2c2 + . . . + 2cr − t, where c1 > c2 > . . . > cr 0. If cr t − r then α(n + t − 1) = α(2c1 + 2c2 + . . . + 2cr−1 + 2cr − 1) = r − 1 + cr t − 1. Hence β(n) t − 1. This contradicts the fact that β(n) = t. So, cr > t − r. If r = t then ct = cr > t − r = 0. Set di = ci , i = 1, 2, . . . , t. We obtain t n = 2d1 + 2d2 + . . . + 2dt−1 + 2dt − t = (2di − 1), i=1 where d1 > d2 > . . . > dt−2 > dt−1 > dt > 0. Hence µ(n) Suppose r < t. Obviously, t = β(n). 2cr = 2cr −1 + . . . + 2cr −t+r+1 + 2cr −t+r + 2cr −t+r . Set di = ci , i = 1, 2, . . . , r − 1, dr+ = cr − − 1 > 0, = 0, 1, . . . , t − r − 2, dt−1 = dt = cr − t + r > 0. Then, we get t n = 2d1 + 2d2 + . . . + 2dt−1 + 2dt − t = (2di − 1), i=1 with d1 > d2 > . . . > dt−2 > dt−1 = dt > 0. Hence µ(n) t = β(n). s Since µ(n) = s, n = i=1 (2hi − 1) with hi positive integers. Then α(n + s) = s hi α( i=1 2 ) s. So we get µ(n) = s β(n). Hence t = β(n) = µ(n) = s and n is of the form (3.1). Now, assume that n is of the form (3.1). Then µ(n) s. We prove µ(n) = s by induction on s. If s = 1 then µ(n) = 1, since µ(n) > 0. If s = 2 then α(n+1) = α(2d1 +2d2 −1) = 1 + d2 > 1. Hence, µ(n) = β(n) 2. So, µ(n) = 2. Suppose s > 2. By the inductive hypothesis, µ(n + 1 − 2d1 ) = s − 1. It is well-known that there exists uniquely an integer d such that 2d n + 1 < d+1 2 . Since d1 > d2 > . . . > ds−1 ds > 0, we have 2d1 n + 1 < 2d1 + 2d1 −1 + . . . + 2d1 −s+2 + 2d1 −s+2 = 2d1 +1 . So, we get d1 = d. Set µ(n) = t s. There exists u1 > u2 > . . . > ut−1 ut > 0 such that n = 2u1 +2u2 +. . .+2ut −t. Then, u1 = d = d1 and α(n+1−2d +t−1) t − 1. Hence, t − 1 β(n + 1 − 2d ) = µ(n + 1 − 2d ) = s − 1. This implies t s and µ(n) = t = s. By induction on i, we get ui = di for 1 i s. The lemma is proved. From this lemma we easily obtain the following. Corollary 3.2 (See Kameko [13]). Let n, k be positive integers. Then i) µ(n) > k if and only if α(n + k) > k. n−µ(n) µ(n), µ 2n + µ(n) = µ(n). 2 ii) µ ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 10 Suppose that m is a positive integer and µ(2m + k) = s < k. By Lemma 3.1, there exists a sequence of integers d1 > d2 > . . . > ds−1 ds > 0 such that s 2d1 −1 2d2 −1 di 2ds −1 2m + k = i=1 (2 − 1). Set z = x1 x2 . . . xs ∈ (Pk )2m+k . Since z is a 0 spike and s < k, we have [z] = 0 and (Sq ∗ )(k,m) ([z]) = 0. So, one gets the following. Corollary 3.3. Let m be a positive integer. If µ(2m + k) < k, then 0 (Sq ∗ )(k,m) : (F2 ⊗A Pk )2m+k −→ (F2 ⊗A Pk )m is not a monomorphism. Proof of Theorem 1.2. Set q = α(d + k), r = ζ(d + k) and m = k(2t − 1) + 2t d. From the proof of Lemma 3.1 and Corollary 3.2, we see that if q > k, then µ(k(2s − 1) + 2s d) for any s µ(d) > k 0 = t(k, d). By Theorem 2.11, (F2 ⊗A Pk )k(2s −1)+2s d = 0, (F2 ⊗A Pk )d = 0. So, the theorem holds. Assume that q k. Using Theorem 2.12, Corollaries 3.2 and 3.3, we see that the homomorphism 0 (Sq ∗ )s−t : (F2 ⊗A Pk )k(2s −1)+2s d −→ (F2 ⊗A Pk )k(2t −1)+2t d is an isomorphism of GLk -modules for every s t if and only if µ(2m + k) = k. Since α(d + k) = q and ζ(d + k) = r, there exists a sequence of integers c1 > c2 > . . . > cq−1 > cq = r 0 such that d + k = 2c1 + 2c2 + . . . + 2cq . If q = k, then 2m + k = k(2t+1 − 1) + 2t+1 d = k(2t+1 − 1) + 2t+1 (2c1 + 2c2 + . . . + 2ck − k) = 2c1 +t+1 + 2c2 +t+1 + . . . + 2ck +t+1 − k. By Lemma 3.1, µ(2m + k) = k for any t Suppose that q < k. Then, we have 0 = t(k, d). Hence, the theorem holds. 2m + k = 2c1 +t+1 + 2c2 +t+1 + . . . + 2cq−1 +t+1 + 2r+t+1 − k = 2c1 +t+1 + 2c2 +t+1 + . . . + 2cq−1 +t+1 + 2r+t + 2r+t−1 + . . . + 2r+t−(k−q−1) + 2r+t−(k−q−1) − k. (3.2) If q + r k, then r + t − (k − q − 1) = q + r − k + 1 + t > 0 for any t 0 = t(k, d). By Lemma 3.1, µ(2m + k) = k. So, the theorem is true. If q +r < k, then from Lemma 3.1 and the relation (3.2), we see that µ(2m+k) = k if and only if t k − q − r = t(k, d). The theorem is completely proved. 4. Proof of Theorem 1.5 In this section we prove Theorem 1.5 by explicitly determining all admissible monomials of degree 7.2s − 5 in P5 . Using this results, we prove that for all s 1, 5 (F2 ⊗A P5 )GL 7.2s −5 = 0. Recall that by Theorem 1.2, 0 (Sq ∗ )s−2 : (F2 ⊗A P5 )7.2s −5 −→ (F2 ⊗A Pk )23 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER is an isomorphism of GL5 -modules for every s theorem for s = 1, 2. 11 2. So, we need only to prove the 4.1. The case s = 1. For s = 1, we have 7.2s − 5 = 9. Since Kameko’s homomorphism 0 (Sq ∗ )(5,2) : (F2 ⊗A P5 )9 −→ (F2 ⊗A Pk )2 0 is an epimorphism, we have (F2 ⊗A P5 )9 ∼ = Ker(Sq ∗ )(5,2) (F2 ⊗A P5 )2 . It is easy to see that B5 (2) = {xi xj : 1 i 0}. By Theorem 2.7, xi xj y 2 is inadmissible. Since x = xi xj y 2 is admissible, one gets x = du for suitable u. We now prove the set {[du ], 1 u 5} is linearly independent in F2 ⊗A P5 . Suppose that S = γ1 d1 + γ2 d2 + γ3 d3 + γ4 d4 + γ5 d5 ≡ 0 with γu ∈ F2 . By a simple computation using Theorem 2.10, we have p(1;2) (S) ≡ γ3 [x31 x2 x23 x44 ] + γ4 [x31 x2 x43 x2 ] + γ5 [x31 x42 x3 x24 ] ≡ 0, p(2;5) (S) ≡ (γ1 + γ3 )[x1 x2 x23 x64 ] + γ1 [x1 x22 x3 x64 ] + γ2 [x1 x22 x43 x34 ] ≡ 0. These relations imply γu = 0 for all u. The first part of the proposition is proved. ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 16 4.3. The case s = 2. For s = 2, we have 7.2s − 5 = 23. Since Kameko’s homomorphism 0 (Sq ∗ )(5,9) : (F2 ⊗A P5 )23 −→ (F2 ⊗A Pk )9 ∼ Ker(Sq 0 )(5,9) is an epimorphism, we have (F2 ⊗A P5 )23 = ∗ we need to compute (F2 ⊗A P5 )9 . Hence, 0 Ker(Sq ∗ )(5,9) . Lemma 4.3.1. If x is an admissible monomial of degree 23 in P5 and [x] ∈ 0 Ker(Sq ∗ )(5,9) , then ω(x) is one of the following sequences: (3, 2, 2, 1), (3, 2, 4), (3, 4, 1, 1), (3, 4, 3). 7 Proof. Note that z = x15 1 x2 x3 is the minimal spike of degree 23 in P5 and ω(z) = (3, 2, 2, 1). Since [x] = 0, by Theorem 2.10, either ω1 (x) = 3 or ω1 (x) = 5. If ω1 (x) = 5, then x = X∅ y 2 with y a monomial of degree 9 in P5 . Since x is 0 admissible, by Theorem 2.7, y is admissible. Hence, (Sq ∗ )(5,9) ([x]) = [y] = 0. 0 This contradicts the fact that [x] ∈ Ker(Sq ∗ )(5,9) , so ω1 (x) = 3. Then, we have x = xi xj x y12 with y1 an admissible monomial of degree 10 in P5 . Now, the lemma follows from Lemma 4.2.1. Using Lemma 4.3.1 and a result in [25], we get 0 4 QP5+ (ω(j) ) . Ker(Sq ∗ )(5,9) = (F2 ⊗A P50 )23 j=1 Here ω(1) = (3, 2, 2, 1), ω(2) = (3, 4, 1, 1), ω(3) = (3, 4, 3), ω(4) = (3, 2, 4). From a result in [25], we easily obtain dim(F2 ⊗A P50 )23 = 635. Proposition 4.3.2. B5+ (ω(1) ) = Φ(B4+ (23))∪C, where C is the set of the following monomials: 7 2 12 3 5 x1 x22 x43 x94 x75 , x1 x22 x53 x84 x75 , x1 x22 x53 x94 x65 , x1 x22 x12 3 x4 x5 , x 1 x2 x3 x4 x5 , 7 3 4 8 7 3 4 9 6 3 5 8 6 3 4 8 7 x1 x22 x12 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x31 x2 x43 x94 x65 , x31 x2 x53 x84 x65 , x31 x42 x33 x44 x95 , x31 x42 x33 x84 x55 , x31 x42 x33 x12 4 x5 , x31 x42 x73 x84 x5 , x31 x42 x93 x24 x55 , x31 x42 x93 x34 x45 , x31 x52 x3 x84 x65 , x31 x72 x43 x84 x5 , 2 5 3 12 3 4 3 12 3 4 7 3 4 8 x31 x72 x83 x44 x5 , x31 x12 2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x71 x32 x83 x44 x5 , x71 x82 x3 x24 x55 , x71 x82 x3 x34 x45 , x71 x82 x33 x4 x45 , x71 x92 x23 x44 x5 . By a direct computation, we can see that |Φ(B4+ (23)) ∪ C| = 293, hence one gets dim QP5+ (ω(1) ) = 293. The following lemma is proved by a direct computation. Lemma 4.3.3. The following monomials are strictly inadmissible: i) x2j x xt x3u , j < < t; x2j x xt xu x2v , j < < t < u; x1 x22 x23 x4 x5 . Here (j, , t, u, v) is a permutation of (1, 2, 3, 4, 5). ii) fi (¯ x), 1 i 5, where x ¯ is one of the following monomials: 7 3 12 7 3 12 3 5 3 4 9 7 x31 x12 2 x3 x4 , x 1 x2 x3 x4 , x 1 x2 x3 x4 , x 1 x2 x3 x4 , x31 x52 x93 x64 , x31 x52 x83 x74 , x71 x82 x33 x54 . KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 6 6 3 4 iii) x31 x12 x31 x12 x31 x12 x31 x42 x3 x94 x65 , 2 x3 x4 x5 , 2 x3 x4 x5 , 2 x3 x4 x5 , 3 4 9 6 3 4 8 7 3 4 8 7 3 4 8 7 x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x71 x82 x33 x44 x5 , 4 x31 x52 x83 x4 x65 , x31 x52 x83 x64 x5 , x31 x42 x83 x34 x55 , x31 x42 x11 3 x4 x5 . 17 x31 x42 x93 x4 x65 , x1 x62 x83 x34 x55 , Proof of Proposition 4.3.2. Let x be an admissible monomial in P5+ such that ω(x) = ω(1) . Then x = xj x xt y 2 with y ∈ B5 (2, 2, 1). Let z ∈ B5 (2, 2, 1) such that xj x xt z 2 ∈ P5+ . By a direct computation using the results in Subsection 4.2, we see that if xj x xt z 2 ∈ / Φ(B4+ (23)) ∪ C, then there is a u monomial w which is given in Lemma 4.3.3 such that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 and x is admissible, one gets x ∈ Φ(B4+ (23)) ∪ C. This implies B5+ (ω(1) ) ⊂ Φ(B4+ (23)) ∪ C. We now prove the set [Φ(B4+ (23)) ∪ C] is linearly independent in (F2 ⊗A P5 )23 . Suppose there is a linear relation 293 γt dt ≡ 0, S= t=1 where γt ∈ F2 and dt = d23,t ∈ Φ(B4+ (23)) ∪ C. We explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 . From the relation p(i;j) (S) ≡ 0 with 1 i < j 5, one gets γt = 0 for all t ∈ / J, where J = {1, 15, 18, 82, 98, 99, 113, 116, 121, 122, 160, 161, 168, 170, 173,199, 201, 202, 211, 214, 219, 220} and γt = γ1 , for all t ∈ J. So, S = γ1 p ≡ 0 with p = t∈J dt . Consider the homomorphism g4 : P5 → P5 . By computing g4 (p) in terms of dt ’s one gets g4 (S) ≡ γ1 (d19 + other terms) ≡ 0. Since 19 ∈ / J, we obtain γ1 = 0. Hence γt = 0 for 1 follows. t 293. The proposition Proposition 4.3.4. B5 (ω(4) ) = B5+ (ω(4) ) = ∅. That means QP5 (ω(4) ) = 0. Proof. Let x be an admissible monomial in P5+ such that ω(x) = ω(4) . Then x = xj x xt y 2 with y ∈ B5 (2, 4). By a direct computation using Theorem 2.7, Proposition 4.2.2 and Lemma 4.3.3, we see that x is a permutation of one of the monomials: x31 x42 x43 x54 x75 , x31 x42 x53 x54 x65 . A simple computation shows: 2 5 2 2 5 7 5 2 6 7 x31 x42 x43 x54 x75 = Sq 1 (x31 x2 x23 x94 x75 + x31 x2 x23 x54 x11 5 ) + Sq (x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 ) + Sq 4 (x31 x22 x23 x54 x75 + x31 x2 x23 x64 x75 ) mod(P5− (B5 (ω(4) ))). This relation implies [x31 x42 x43 x54 x75 ]ω(4) = 0. By a similar computation, we have [x31 x42 x53 x54 x65 ]ω(4) = 0. The proposition is proved. 18 ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC Proposition 4.3.5. B5 (ω(2) ) = B5+ (ω(2) ) is the set of the following monomials: 2 2 7 11 2 2 15 3 2 3 2 15 2 3 3 14 x1 x22 x23 x34 x15 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 3 7 10 2 3 14 3 2 3 15 2 2 7 2 11 x1 x22 x33 x64 x11 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 7 10 3 2 7 11 2 2 15 2 3 2 15 3 2 x1 x22 x73 x34 x10 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 2 3 14 3 2 6 11 3 2 7 10 3 2 14 3 x1 x32 x23 x24 x15 5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 2 3 3 2 14 3 3 6 10 3 3 14 2 3 6 2 11 x1 x32 x23 x15 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 6 10 3 3 6 11 2 3 7 2 10 3 7 10 2 x1 x32 x63 x34 x10 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 3 3 14 3 2 3 15 2 2 7 2 2 11 7 2 3 10 x1 x32 x14 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 3 7 2 11 2 7 3 2 10 7 3 10 2 7 10 2 3 x1 x72 x23 x10 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 2 7 11 2 2 15 2 2 3 15 2 3 2 15 3 2 2 x1 x72 x10 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 3 2 3 14 3 2 6 11 3 2 7 10 3 2 14 3 x31 x2 x23 x24 x15 5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 2 3 3 2 14 3 3 6 10 3 3 14 2 3 6 2 11 x31 x2 x23 x15 4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 3 6 10 3 3 6 11 2 3 7 2 10 3 7 10 2 x31 x2 x63 x34 x10 5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 2 3 3 14 3 2 3 15 2 2 3 3 2 14 3 3 6 10 x31 x2 x14 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 2 3 3 5 2 10 3 3 5 10 2 3 3 13 2 2 3 5 2 2 11 x31 x32 x3 x14 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 5 2 10 3 3 5 2 11 2 3 5 3 2 10 3 5 3 10 2 x31 x52 x23 x34 x10 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 3 3 5 10 3 2 3 5 11 2 2 3 7 2 10 3 7 10 2 x31 x52 x10 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 2 3 3 13 2 3 2 3 13 3 2 2 3 15 2 2 x31 x72 x93 x24 x25 , x31 x13 2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 7 2 3 10 7 2 10 3 7 2 11 2 7 3 2 10 x71 x2 x23 x24 x11 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 7 10 2 3 7 10 3 2 7 11 2 2 7 3 2 10 x71 x2 x33 x10 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 7 3 9 2 2 7 9 2 2 3 7 9 2 3 2 7 9 3 2 2 x71 x32 x3 x10 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 2 15 2 2 3 15 2 3 2 15 3 2 2 15 3 2 2 x71 x11 2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 . Lemma 4.3.6. The following monomials are strictly inadmissible: i) x2j x x2t x3u x3v , i < j; x2j x3 x3t x3u . Here (j, , t, u, v) is a permutation of (1, 2, 3, 4, 5). ii) x1 x22 x63 x34 x35 , x1 x62 x23 x34 x35 , x1 x62 x33 x24 x35 , x1 x62 x33 x34 x25 . The proof of this lemma is straightforward. Proof of Proposition 4.3.5. Let x be an admissible monomial in P5+ such that ω(x) = ω(2) . Then x = xj x xt y 2 with y ∈ B5 (4, 1, 1). Let z ∈ B5 (4, 1, 1) such that xj x xt z 2 ∈ P5+ . Denote by D the set of all the monomials as given in this proposition. By a direct computation using Proposition 4.2.2, we see that if xj x xt z 2 ∈ / D, then there is a monomial w which is given in u Lemma 4.3.6 such that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 with y ∈ B5 (4, 1, 1) and x is admissible, one gets x ∈ D. This implies B5+ (ω(2) ) ⊂ D. We now prove the set [D] is linearly independent in (F2 ⊗A P5 )23 . Suppose there 398 is a linear relation S = t=294 γt dt ≡ 0, where γt ∈ F2 and dt = d23,t ∈ D. We KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 19 explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 . From the relation p(i;j) (S) ≡ 0 with 1 i < j 5, one gets γt = 0 for all 294 t 398. To prove [D]ω(2) is a basis of QP5 (ω(2) ), we need to show that [x]ω(2) = 0 for all x ∈ D. Observe that if x ∈ D then x is a permutation of one of the following monomials: 2 2 7 11 2 3 6 11 3 3 6 10 a1 = x1 x22 x23 x34 x15 5 , a2 = x1 x2 x3 x4 x5 , a3 = x1 x2 x3 x4 x5 , a4 = x1 x2 x3 x4 x5 , 2 3 3 14 3 3 13 2 2 3 5 2 2 11 a5 = x1 x22 x33 x74 x10 5 , a6 = x1 x2 x3 x4 x5 , a7 = x1 x2 x3 x4 x5 , a8 = x1 x2 x3 x4 x5 , a9 = x31 x72 x93 x24 x25 , a10 = x31 x32 x53 x24 x10 5 . It is easy to see that [a5 ]ω(2) = [σ1 a3 ]ω(2) , [a7 ]ω(2) = [σ2 a6 ]ω(2) , [a8 ]ω(2) = [σ3 a3 ]ω(2) , [a9 ]ω(2) = [σ4 a3 ]ω(2) , [a10 ]ω(2) = [σ5 a4 ]ω(2) , where σu are suitable permutations in Σ5 . By a similar argument as given in the proof of Proposition 4.3.2, we can prove that the set [B5+ (ω(1) ) ∪ {a1 , a2 , a3 , a4 , a6 }] is linearly independent in the space (F2 ⊗A P5 )23 . This fact shows that [av ]ω(2) = 0 for all v. So, [x]ω(2) = 0 for all x ∈ D. The proposition follows. Proposition 4.3.7. B5 (ω(3) ) = B5+ (ω(3) ) is the set of the following monomials: x1 x32 x63 x64 x75 , x1 x32 x63 x74 x65 , x1 x32 x73 x64 x65 , x1 x72 x33 x64 x65 , x31 x2 x63 x64 x75 , x31 x2 x63 x74 x65 , x31 x2 x73 x64 x65 , x31 x32 x53 x64 x65 , x31 x52 x23 x64 x75 , x31 x52 x23 x74 x65 , x31 x52 x33 x64 x65 , x31 x52 x63 x24 x75 , x31 x52 x63 x34 x65 , x31 x52 x63 x64 x35 , x31 x52 x63 x74 x25 , x31 x52 x73 x24 x65 , x31 x52 x73 x64 x25 , x31 x72 x3 x64 x65 , x31 x72 x53 x24 x65 , x31 x72 x53 x64 x25 , x71 x2 x33 x64 x65 , x71 x32 x3 x64 x65 , x71 x32 x53 x24 x65 , x71 x32 x53 x64 x25 . The following lemma is proved by a direct computation. Lemma 4.3.8. The following monomials are strictly inadmissible: xj x6 x3t x6u x7v , j < < t; xj x2 x6t x7u x7v . Here (j, , t, u, v) is a permutation of (1, 2, 3, 4, 5). Proof of Proposition 4.3.7. Let x be an admissible monomial in P5+ such that ω(x) = ω(3) . Then x = xj x xt y 2 with y ∈ B5 (4, 3). Let z ∈ B5 (4, 3) such that xj x xt z 2 ∈ P5+ . Denote by E the set of all the monomials as given in proposition 4.3.7. By a direct computation using Proposition 4.2.2, we see that if xj x xt z 2 ∈ / E, then there is a monomial w which is given in u Lemma 4.3.8 such that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 with y ∈ B5 (4, 3) and x is admissible, one gets x ∈ E. This implies B5+ (ω(3) ) ⊂ E. We now prove the set [E] is linearly independent in (F2 ⊗A P5 )23 . Suppose there 422 is a linear relation S = t=399 γt dt ≡ 0, where γt ∈ F2 and dt ∈ E. We explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 . From the relation p(i;j) (S) ≡ 0 with 1 i < j 5, one gets γt = 0 for all t. To prove [E]ω(3) is a basis of QP5 (ω(3) ), we need to show that [x]ω(3) = 0 for all x ∈ E. ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 20 It is easy to see that if x ∈ E then x is a permutation of one of the following monomials: b1 = x1 x32 x63 x64 x75 , b2 = x31 x32 x53 x64 x65 , b3 = x21 x32 x53 x64 x75 . We have [b3 ]ω(3) = [b1 ]ω(3) . By a similar argument as given in the proof of Proposition 4.3.2, we can prove that the set [B5+ (3, 2, 2, 1)∪B5 (3, 4, 1, 1)∪{b1 , b2 }] is linearly independent in the space (F2 ⊗A P5 )23 . This implies [b1 ]ω = 0 and [b2 ]ω(3) = 0, hence [x]ω(3) = 0 for all x ∈ E. The proposition follows. GL5 5 = 0, using Theorem 2.7, Now we compute (F2 ⊗A P5 )GL 23 . Since (F2 ⊗A P5 )9 0 5 5 we have (F2 ⊗A P5 )GL = Ker(Sq ∗ )GL 23 (5,9) . Recall that 0 Ker(Sq ∗ )(5,9) = QP5 (ω(1) ) QP5 (ω(2) ) QP5 (ω(3) ). By proposition 4.3.7, dim QP5 (ω(3) ) = 24 with the basis [B5 (b1 )]ω(3) ∪[B5 (b2 )]ω(3) . Proposition 4.3.9. QP5 (ω(3) )GL5 = 0. By a direct computation we easily obtain the following lemma. Lemma 4.3.10. i) The subspaces [Σ5 (b1 )]ω(3) and [Σ5 (b2 )]ω(3) are Σ5 -submodules of QP5 (ω(3) ). ii) We have a direct summand decomposition of the Σ5 -modules: QP5 (ω(3) ) = [Σ5 (b1 )]ω(3) Lemma 4.3.11. [Σ5 (b1 )]ω(3) Σ5 [Σ5 (b2 )]ω(3) . = [p(b1 )]ω(3) and [Σ5 (b2 )]ω(3) Σ5 = 0. Proof. From Proposition 4.3.7, we see that dim [Σ5 (b2 )]ω(3) = 4 with a basis consisting of the classes represented by the following monomials: u1 = x31 x32 x53 x64 x65 , u2 = x31 x52 x33 x64 x65 , u3 = x31 x52 x63 x34 x65 , u4 = x31 x52 x63 x64 x35 . 4 Suppose f = t=1 γt ut with γt ∈ F2 and [f ] ∈ [Σ5 (b2 )]ω(3) computation, we have Σ5 . By a direct g1 (f ) + f ≡ω(3) (γ2 + γ3 + γ4 )u1 ≡ω(3) 0, g2 (f ) + f ≡ω(3) (γ1 + γ2 )u1 ≡ω(3) 0, g3 (f ) + f ≡ω(3) (γ2 + γ3 )u2 ≡ω(3) 0, g4 (f ) + f ≡ω(3) (γ3 + γ4 )u3 ≡ω(3) 0. From the above relations one gets γt = 0 for t = 1, 2, 3, 4. By a similar computation we obtain [Σ5 (b1 )]ω(3) Σ5 = [p(b1 )]ω(3) . Proof of Proposition 4.3.9. Let f ∈ P5 (ω(3) ) such that [f ]ω(3) ∈ QP5 (ω(3) )GL5 . Since [f ]ω(3) ∈ QP5 (ω(3) )Σ5 , using Lemmas 4.3.10, and 4.3.11, we have f ≡ω(3) γp(b1 ) with γ ∈ F2 . By computing g5 (f ) + f in terms of the admissible monomials, we obtain g5 (f ) + f ≡ω(3) γb1 + other terms ≡ω(3) 0. This relation implies γ = 0. The proposition is proved. Proposition 4.3.12. QP5 (ω(2) )GL5 = 0. KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 21 By computing from Proposition 4.3.5, we see that dim QP5 (ω(2) ) = 105 with the 4 basis j=1 [B(aj )]ω(2) . By a direct computation, we obtain the following. Lemma 4.3.13. i) The subspaces [Σ5 (aj )]ω(2) , 1 j 4, are Σ5 -submodules of QP5 (ω(2) ). ii) We have a direct summand decomposition of the Σ5 -modules: 4 QP5 (ω(2) ) = [Σ5 (aj )]ω(2) . j=1 Lemma 4.3.14. [Σ5 (aj )]ω(2) [p4 ]ω(2) ] , where Σ5 = [p(aj )]ω(2) , j = 1, 2, 3, and [Σ5 (a4 )]ω(2) Σ5 = 2 3 14 3 2 7 3 10 2 7 10 3 3 2 3 14 p4 = x1 x22 x33 x34 x14 5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 3 3 3 2 14 3 3 14 2 3 14 2 3 3 14 3 2 + x1 x32 x23 x14 4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 7 2 10 3 7 3 2 10 7 3 10 2 7 10 2 3 + x1 x72 x23 x34 x10 5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 3 2 3 2 3 14 3 2 14 3 3 3 2 14 3 3 14 2 + x1 x72 x10 3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 2 3 3 14 3 2 3 3 2 14 3 3 14 2 3 3 13 2 2 + x31 x2 x14 3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 2 2 3 3 13 2 3 2 3 13 3 2 2 7 2 3 10 7 2 10 3 + x31 x13 2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 7 3 10 2 7 10 2 3 7 10 3 2 7 3 2 10 + x71 x2 x33 x24 x10 5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 2 7 3 9 2 2 7 9 2 2 3 7 9 2 3 2 7 9 3 2 2 + x71 x32 x3 x10 4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 + x1 x2 x3 x4 x5 . Proof of Proposition 4.3.12. Let f ∈ P5 (ω(2) ) such that [f ]ω(2) ∈ QP5 (ω(2) )GL5 . Since [f ]ω(2) ∈ QP5 (ω(2) )Σ5 , using Lemmas 4.3.13, and 4.3.14, we have f ≡ω(2) γ1 p(a1 ) + γ2 p(a2 ) + γ3 p(a3 ) + γ4 p4 , with γj ∈ F2 . By computing g5 (f ) + f in terms of the admissible monomials, we obtain g5 (f ) + f ≡ω(2) + other terms ≡ω(2) 0. This relation implies γj = 0 with j = 1, 2, 3, 4. Proposition 4.3.12 is proved. Proposition 4.3.15. QP5 (ω(1) )GL5 = 0. By using Proposition 4.3.4, we see that dim QP5 (ω(1) ) = 928. Consider the following monomials: 3 5 15 3 7 13 2 5 15 2 4 15 c1 = x1 x72 x15 3 , c2 = x1 x2 x3 , c3 = x1 x2 x3 , c4 = x1 x2 x3 x4 , c5 = x1 x2 x3 x4 x5 , 3 5 14 3 6 13 3 7 12 c6 = x1 x22 x73 x13 4 , c7 = x1 x2 x3 x4 , c8 = x1 x2 x3 x4 , c9 = x1 x2 x3 x4 , 2 7 12 3 6 12 c10 = x31 x52 x63 x94 , c11 = x1 x2 x23 x64 x13 5 , c12 = x1 x2 x3 x4 x5 , c13 = x1 x2 x3 x4 x5 , 2 3 5 12 2 5 6 9 2 5 7 8 c14 = x1 x22 x33 x44 x13 5 , c15 = x1 x2 x3 x4 x5 , c16 = x1 x2 x3 x4 x5 , c17 = x1 x2 x3 x4 x5 . Lemma 4.3.16. i) The following subspaces are Σ5 -submodules of QP5 (ω(1) ): [Σ5 (cj )] , j = 1, 2, 3, 4, 5; [Σ5 (c6 , . . . , c10 )] ; [Σ5 (c11 , . . . , c17 )] . ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 22 ii) We have a direct summand decomposition of the Σ5 -modules: 5 QP5 (ω(1) ) = [Σ5 (aj )]ω(2) [Σ5 (c6 , . . . , c10 )] [Σ5 (c11 , . . . , c17 )] . j=1 Lemma 4.3.17. i) [Σ5 (aj )]ω(2) Σ5 = [p(aj )]ω(2) , j = 1, 2, 3, 4, 5. ii) [Σ5 (c11 , . . . , c17 )] Σ5 = 0. iii) [Σ5 (c6 , . . . , c10 )] Σ5 = [p5 + p6 ], [p6 + p7 ] , where 7 14 7 14 7 14 7 14 p5 = x2 x3 x74 x14 5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x2 x3 x4 x5 7 14 7 14 7 14 7 14 + x1 x3 x74 x14 5 + x1 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 7 14 7 14 7 14 7 14 + x1 x2 x73 x14 4 + x1 x2 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 7 14 7 14 7 14 7 14 + x1 x72 x14 3 x5 + x1 x2 x3 x4 + x1 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 7 14 7 14 7 14 7 14 + x71 x2 x14 4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 6 3 13 6 3 13 6 3 13 6 3 13 6 + x2 x33 x13 4 x5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x1 x3 x4 x5 6 3 13 6 3 13 6 3 13 6 3 13 6 + x1 x32 x13 4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x3 x4 x5 + x1 x3 x4 x5 6 3 13 6 3 13 6 3 13 6 3 13 6 + x31 x13 3 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x4 x5 6 3 13 6 3 13 6 3 13 6 3 13 6 + x31 x13 2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 4 7 11 4 7 11 4 7 11 4 7 11 4 + x72 x11 3 x4 x5 + x2 x3 x4 x5 + x1 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 4 7 11 4 7 11 4 7 11 4 7 11 4 + x71 x11 2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 + x2 x63 x74 x95 + x1 x63 x74 x95 + x1 x62 x74 x95 + x1 x62 x73 x95 + x1 x62 x73 x94 + x32 x43 x74 x95 + x31 x43 x74 x95 + x31 x42 x74 x95 + x31 x42 x73 x95 + x31 x42 x73 x94 . p6 = x2 x73 x74 x85 + x72 x3 x74 x85 + x72 x73 x4 x85 + x72 x73 x84 x5 + x1 x73 x74 x85 + x1 x72 x74 x85 + x1 x72 x73 x85 + x1 x72 x73 x84 + x71 x3 x74 x85 + x71 x73 x4 x85 + x71 x73 x84 x5 + x71 x2 x74 x85 + x71 x2 x73 x85 + x71 x2 x73 x84 + x71 x72 x4 x85 + x71 x72 x84 x5 + x71 x72 x3 x85 + x71 x72 x3 x84 + x71 x72 x83 x5 + x71 x72 x83 x4 7 3 12 7 3 12 3 7 12 3 7 12 + x2 x33 x74 x12 5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 3 7 12 7 3 12 7 3 12 7 3 12 + x1 x32 x73 x12 5 + x1 x2 x3 x4 + x1 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 7 3 12 7 3 12 7 3 12 7 3 12 + x71 x32 x12 4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 + x2 x73 x64 x95 + x72 x3 x64 x95 + x1 x73 x64 x95 + x1 x72 x64 x95 + x1 x72 x63 x95 + x1 x72 x63 x94 + x71 x3 x64 x95 + x71 x2 x64 x95 + x71 x2 x63 x95 + x71 x2 x63 x94 + x32 x73 x44 x95 + x32 x73 x94 x45 + x72 x33 x44 x95 + x72 x33 x94 x45 + x31 x73 x44 x95 + x31 x73 x94 x45 + x31 x72 x44 x95 + x31 x72 x94 x45 + x31 x72 x43 x95 + x31 x72 x43 x94 + x31 x72 x93 x45 + x31 x72 x93 x44 + x71 x33 x44 x95 + x71 x33 x94 x45 + x71 x32 x44 x95 + x71 x32 x94 x45 + x71 x32 x43 x95 + x71 x32 x43 x94 + x71 x32 x93 x45 + x71 x32 x93 x44 . KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 23 3 14 5 3 5 14 3 5 14 3 5 14 p7 = x2 x33 x54 x14 5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x2 x3 x4 x5 3 14 5 3 5 14 3 14 5 3 5 14 + x1 x33 x54 x14 5 + x1 x3 x4 x5 + x1 x2 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 3 14 5 3 14 5 3 5 14 3 5 14 + x1 x32 x53 x14 4 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x3 x4 x5 + x1 x3 x4 x5 3 5 14 3 5 14 3 5 14 3 5 14 + x31 x53 x14 4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x4 x5 3 5 14 3 5 14 3 5 14 3 5 14 + x31 x52 x14 4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 6 3 13 3 6 13 6 3 13 3 6 13 + x2 x33 x64 x13 5 + x2 x3 x4 x5 + x1 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 3 6 13 6 3 13 6 3 13 6 3 13 + x1 x32 x63 x13 5 + x1 x2 x3 x4 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 3 7 12 3 7 12 3 7 12 3 7 12 + x32 x3 x74 x12 5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x1 x3 x4 x5 + x1 x3 x4 x5 3 7 12 3 7 12 3 7 12 3 7 12 + x31 x73 x12 4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x4 x5 3 7 12 3 7 12 3 7 12 3 7 12 + x31 x72 x12 4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 5 6 11 5 6 11 5 6 11 5 6 11 5 + x2 x63 x11 4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 3 3 13 4 3 4 3 13 3 3 4 13 3 3 13 4 + x32 x33 x44 x13 5 + x2 x3 x4 x5 + x2 x3 x4 x5 + x1 x3 x4 x5 + x1 x3 x4 x5 3 3 4 13 3 3 13 4 3 3 4 13 3 3 4 13 + x31 x43 x34 x13 5 + x1 x2 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 4 3 3 13 4 3 4 3 13 3 4 3 13 3 4 3 13 + x31 x32 x13 3 x5 + x1 x2 x3 x4 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 5 3 4 11 5 3 4 11 5 3 4 11 5 3 4 11 5 + x32 x43 x11 4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 3 3 12 5 3 3 5 12 3 3 12 5 3 3 5 12 + x32 x33 x54 x12 5 + x2 x3 x4 x5 + x1 x3 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5 5 3 3 5 12 3 3 5 12 3 3 12 5 3 3 12 5 + x31 x32 x12 4 x5 + x1 x2 x3 x5 + x1 x2 x3 x4 + x1 x2 x3 x5 + x1 x2 x3 x4 . The proofs of the above lemmas are straightforward. Proof of Proposition 4.3.15. Let f ∈ P5 (ω(1) ) such that [f ] ∈ QP5 (ω(1) )GL5 . Since [f ] ∈ QP5 (ω(2) )Σ5 , using Lemmas 4.3.16, and 4.3.17, we have 5 f≡ γj p(cj ) + γ6 (p5 + p6 ) + γ7 (p6 + p7 ), j=1 with γj ∈ F2 , 1 j 7. By computing g5 (f ) + f in terms of the admissible monomials using Theorem 2.10, we obtain 3 15 5 3 7 13 6 15 g5 (f ) + f ≡ γ1 x2 x73 x15 4 + γ2 x2 x3 x4 + γ3 x1 x2 x3 + γ4 x2 x3 x4 x5 2 4 7 14 3 5 14 + γ5 x1 x15 2 x3 x4 x5 + γ6 x2 x3 x4 x5 + γ7 x2 x3 x4 x5 + other terms ≡ 0. This relation implies γj = 0 with j = 1, 2, 3, 4. Proposition 4.3.15 is proved. 5. Proof of Theorem 1.6 In this section we prove Theorem 1.6 by explicitly determining all admissible monomials of degree 10.2s − 5 in P5 . Recall that from Theorem 1.2, we have 0 (Sq ∗ )s−2 : (F2 ⊗A P5 )10.2s −5 −→ (F2 ⊗A Pk )35 is an isomorphism of GL5 -modules for every s theorem for s = 1, 2. 2. So, we need only to prove this ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 24 5.1. The case s = 1. 5 For s = 1, we have 10.2s − 5 = 15. The spaces (F2 ⊗A P5 )15 and (F2 ⊗A P5 )GL 15 have been computed in [24]. Theorem 5.1 (See [24]). There exist exactly 432 admissible monomials of degree 15 in P5 . Consequently dim(QP5 )15 = 432. 5 By using Theorem 5.1, we have computed (QP5 )GL 15 . 5 Theorem 5.2 (See [24]). (QP5 )GL is an F2 -vector space of dimension 2 with a 15 basis consisting of the 2 classes represented by the following polynomials: 15 15 15 15 14 14 14 14 14 p = x15 1 + x2 + x3 + x4 + x5 + x1 x2 + x1 x3 + x1 x4 + x1 x5 + x2 x3 14 14 14 14 2 12 2 12 2 12 + x2 x14 4 + x2 x5 + x3 x4 + x3 x5 + x4 x5 + x1 x2 x3 + x1 x2 x4 + x1 x2 x5 2 12 2 12 2 12 2 12 2 12 2 12 + x1 x23 x12 4 + x1 x3 x5 + x1 x4 x5 + x2 x3 x4 + x2 x3 x5 + x2 x4 x5 + x3 x4 x5 + x1 x22 x43 x84 + x1 x22 x43 x85 + x1 x22 x44 x85 + x1 x23 x44 x85 + x2 x23 x44 x85 + x1 x22 x43 x44 x45 , q = x1 x2 x3 x64 x65 + x1 x2 x63 x4 x65 + x1 x2 x63 x64 x5 + x1 x62 x3 x4 x65 + x1 x62 x3 x64 x5 + x1 x32 x63 x4 x45 + x1 x32 x63 x44 x5 + x1 x62 x33 x4 x45 + x1 x62 x33 x44 x5 + x31 x2 x3 x44 x65 + x31 x2 x3 x64 x45 + x31 x2 x43 x4 x65 + x31 x2 x43 x64 x5 + x31 x42 x3 x4 x65 + x31 x42 x3 x64 x5 + x1 x32 x33 x44 x45 + x31 x2 x33 x44 x45 + x31 x32 x3 x44 x45 + x31 x32 x43 x4 x45 + x31 x32 x43 x44 x5 + x31 x42 x33 x4 x45 + x31 x42 x33 x44 x5 . 5.2. The admissible monomials of degree 16 in P5 . To prove Theorem 1.6 for s = 2, we need to determine all the admissible monomials of degree 16 in P5 . Lemma 5.2.1. If x is an admissible monomial of degree 16 in P5 , then ω(x) is one of the following sequences: (2, 1, 1, 1), (2, 1, 3), (2, 3, 2), (4, 2, 2), (4, 4, 1). Proof. Observe that z = x15 1 x2 is the minimal spike of degree 16 in P5 and ω(z) = (2, 1, 1, 1). Since [x] = 0, by Theorem 2.10, either ω1 (x) = 2 or ω1 (x) = 2. If ω1 (x) = 2, then x = xi xj y 2 with y a monomial of degree 7 in P5 . Since x is admissible, by Theorem 2.7, y is admissible. A routine computation shows that either ω(y) = (1, 1, 1) or ω(y) = (1, 3) or ω(y) = (3, 2). If ω1 (x) = 4, then x = Xj y12 with y1 an admissible monomial of degree 6 in P5 . It is easy to see that either ω(y1 ) = (2, 2) or ω(y1 ) = (4, 1). The lemma is proved. Using this lemma, we see that (F2 ⊗A P5 )16 = (F2 ⊗A P50 )16 QP5+ (2, 3, 2) QP5+ (2, 1, 1, 1) QP5+ (4, 2, 2) QP5+ (2, 1, 3) QP5+ (4, 4, 1). From a result in [25], we have dim(F2 ⊗A P50 )16 = 255. Proposition 5.2.2. i) B5+ (2, 1, 1, 1) = {x1 x2 x23 x44 x85 , x1 x22 x3 x44 x85 , x1 x22 x43 x4 x85 , x1 x22 x43 x84 x5 }. ii) B5+ (2, 1, 3) is the set of the following monomials: x1 x22 x43 x44 x55 , x1 x22 x43 x54 x45 , x1 x22 x53 x44 x45 , x1 x32 x43 x44 x45 , x31 x2 x43 x44 x45 . KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 25 iii) B5+ (2, 3, 2) is the set of the following monomials: x1 x2 x23 x64 x65 , x1 x2 x63 x24 x65 , x1 x2 x63 x64 x25 , x1 x22 x3 x64 x65 , x1 x22 x53 x24 x65 , x1 x22 x53 x64 x25 , x1 x32 x23 x44 x65 , x1 x32 x23 x64 x45 , x1 x32 x43 x24 x65 , x1 x32 x43 x64 x25 , x1 x32 x63 x24 x45 , x1 x32 x63 x44 x25 , x31 x2 x23 x44 x65 , x31 x2 x23 x64 x45 , x31 x2 x43 x24 x65 , x31 x2 x43 x64 x25 , x31 x2 x63 x24 x45 , x31 x2 x63 x44 x25 , x31 x52 x23 x24 x45 , x31 x52 x23 x44 x25 . iv) B5+ (4, 2, 2) = Φ(B4+ (4, 2, 2)) ∪ F , where F is the set of the following monomials: x31 x42 x3 x4 x75 , x31 x42 x3 x74 x5 , x31 x42 x73 x4 x5 , x31 x72 x43 x4 x5 , x71 x32 x43 x4 x5 , x31 x42 x3 x34 x55 , x31 x42 x33 x4 x55 , x31 x42 x33 x54 x5 . v) B5+ (4, 4, 1) = Φ(B4+ (4, 4, 1)). By a simple computation, we have |B5+ (4, 2, 2)| = 110, |B5+ (4, 2, 2)| = 49, So dim(F2 ⊗A P5 )16 = 443. The following lemma is proved by a direct computation. Lemma 5.2.3. The following monomials are strictly inadmissible: i) x2j x xt , j < < t; x2j x x2t x3u , j < ; x2j x xt x2u x2v , j < ; x3j x12 xt , x3j x4 x9t , 3 5 8 xj x xt , x3j x4 xt x8u , j < < t; x2j x3 x3t . ii) x3j x4 x4t x5u , x3j x4 xt x4u x4v , j < < t. iii) xj x6 x3t x6u , xj x6 xt x2u x6v , j < < t; xj x2 x2t x5u x6v , u 4; xj x2 x3t x4u x6v , t 3; 3 4 2 6 3 4 6 2 2 2 4 7 2 6 7 xj x xt xu , xj x xt xu xv ; x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 . iv) x2j x x3t x3u x3v , j < ; x2j x xt xu x3v , j < < t < u; x31 x42 x33 x34 x35 . Here (j, , t, u, v) is a permutation of (1, 2, 3, 4, 5). Proof of Proposition 5.2.2. We prove Part (iii) of the proposition. The others are proved by a similar computation. Denote d1 = x1 x2 x23 x64 x65 , d2 = x1 x2 x63 x24 x65 , d3 = x1 x2 x63 x64 x25 , d4 = x1 x22 x3 x64 x65 , d5 = x1 x22 x53 x24 x65 , d6 = x1 x22 x53 x64 x25 , d7 = x1 x32 x23 x44 x65 , d8 = x1 x32 x23 x64 x45 , d9 = x1 x32 x43 x24 x65 , d10 = x1 x32 x43 x64 x25 , d11 = x1 x32 x63 x24 x45 , d12 = x1 x32 x63 x44 x25 , d13 = x31 x2 x23 x44 x65 , d14 = x31 x2 x23 x64 x45 , d15 = x31 x2 x43 x24 x65 , d16 = x31 x2 x43 x64 x25 , d17 = x31 x2 x63 x24 x45 , d18 = x31 x2 x63 x44 x25 , d19 = x31 x52 x23 x24 x45 , d20 = x31 x52 x23 x44 x25 . Let x be an admissible monomial in P5+ such that ω(x) = (2, 3, 2) := ω. Then x = xj x xt y 2 with y ∈ B5 (2, 3). Let z ∈ B5 (2, 3) such that xj x z 2 ∈ P5+ . By a direct computation, we see that if xj x z 2 = dt , for all t, t 20, then there is a monomial w which is given u 2 in Lemma 5.2.3 such that xj x z = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x z 2 is inadmissible. Since x = xj x y 2 with y ∈ B5 (2, 3) and x is admissible, one see that x = du for some u, u 20. Hence, QP5+ (ω) is spanned by the set {[dt ] : 1 t 20}. We now prove the set {[dt ] : 1 t 20} is linearly independent in QP5 (ω). 20 Suppose there is a linear relation S = t=1 γt dt ≡ω 0, where γt ∈ F2 . ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 26 From a result in [25], dim QP4+ (2, 3, 2) = 4, with the basis {[wu ] : 1 where u 4}, w1 = x1 x32 x63 x64 , w2 = x31 x2 x63 x64 , w3 = x31 x52 x23 x64 , w4 = x31 x52 x63 x24 . By a direct computation using Lemma 2.17, we get p(1;2) (S) ≡ω γ4 w2 + γ5 w3 + γ6 w4 ≡ω 0, p(1;3) (S) ≡ω γ1 w1 + γ9 w3 + γ10 w4 ≡ω 0. The above relations imply γt = 0 for t = 1, 2, 5, 6, 9, 10. Then, p(1;4) (S) ≡ω γ2 w1 + γ7 w2 + γ12 w4 ≡ω 0, p(1;5) (S) ≡ω γ3 w1 + γ8 w2 + γ11 w3 ≡ω 0, p(2;3) (S) ≡ω γ15 w3 + γ16 w4 ≡ω 0, p(2;4) (S) ≡ω γ2 w1 + γ13 w2 + γ18 w4 ≡ω 0, p(2;5) (S) ≡ω γ3 w1 + γ14 w2 + γ17 w3 ≡ω 0. From the last equalities, we obtain γt = 0 for t = 19, 20. Then, p(3;4) (S) ≡ω γ20 w4 ≡ω 0, p(3;5) (S) ≡ω γ19 w3 ≡ω 0. So, γt = 0 for all t, 1 t 20, completing the proof. 5.3. The case s = 2. For s = 2, we have 10.2s − 5 = 35. Since Kameko’s homomorphism 0 (Sq ∗ )(5,15) : (F2 ⊗A P5 )35 −→ (F2 ⊗A Pk )15 0 is an epimorphism, we have (F2 ⊗A P5 )35 ∼ = Ker(Sq ∗ )(5,15) we need to compute (F2 ⊗A P5 )15 . Hence, 0 Ker(Sq ∗ )(5,15) . Lemma 5.3.1. If x is an admissible monomial of degree 35 in P5 and [x] ∈ 0 Ker(Sq ∗ )(5,15) , then ω(x) is one of the following sequences: (3, 2, 1, 1, 1), (3, 2, 1, 3), (3, 2, 3, 2), (3, 4, 2, 2), (3, 4, 4, 1). 3 Proof. Note that z = x31 1 x2 x3 is the minimal spike of degree 35 in P5 and ω(z) = (3, 2, 1, 1, 1). Since [x] = 0, by Theorem 2.10, either ω1 (x) = 3 or ω1 (x) = 5. If ω1 (x) = 5, then x = X∅ y 2 with y a monomial of degree 15 in P5 . Since x is 0 admissible, by Theorem 2.7, y is admissible. Hence, (Sq ∗ )(5,15) ([x]) = [y] = 0. 0 This contradicts the fact that [x] ∈ Ker(Sq ∗ )(5,15) , so ω1 (x) = 3. Then, we have x = xi xj x y12 with y1 an admissible monomial of degree 16 in P5 . Now, the lemma follows from Lemma 5.2.1. Using Lemma 5.3.1 and a result in [25], one gets 0 5 QP5+ (¯ ω(j) ) . Ker(Sq ∗ )(5,15) = (F2 ⊗A P50 )35 j=1 Here ω ¯ (1) = (3, 2, 1, 1, 1), ω ¯ (2) = (3, 2, 1, 3), ω ¯ (3) = (3, 2, 3, 2), ω ¯ (4) = (3, 4, 2, 2), ω ¯ (5) = (3, 4, 4, 1). Using a result in [25], we can easily see that dim(F2 ⊗A P50 )23 = 460, QP5 (¯ ω(1) ) = (F2 ⊗A P50 )35 QP5+ (¯ ω(1) ) and QP5 (¯ ω(j) ) = QP5+ (¯ ω(j) ) for j = 2, 3, 4, 5. KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 27 Proposition 5.3.2. There exist exactly 169 admissible monomials in P5+ such that their weight vectors are ω ¯ (1) . Consequently dim QP5+ (¯ ω(1) ) = 169. We denote the monomials in B5 (¯ ω(1) ) by d35,t , 1 t 169 (see Appendix). Lemma 5.3.3. The following monomials are strictly inadmissible: i) fi (¯ x), 1 i 5, where x ¯ is one of the following monomials: 3 3 28 3 3 7 24 7 3 24 3 5 25 2 3 4 25 3 x31 x28 2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , 3 3 4 9 19 3 4 11 17 3 5 8 19 3 5 9 18 3 5 10 17 x31 x52 x24 3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , 16 3 7 8 17 7 3 8 17 3 7 9 16 7 3 9 16 x31 x52 x11 3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 , x1 x2 x3 x4 . ii) 2 3 28 2 3 4 25 2 3 4 25 2 3 4 25 2 x31 x28 2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 2 3 5 24 2 3 4 24 3 3 4 24 3 3 4 24 3 x31 x52 x24 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 4 8 19 3 4 8 19 3 4 9 18 3 4 9 18 x31 x42 x3 x84 x19 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 4 10 17 3 4 11 16 3 4 11 16 3 4 11 16 x31 x42 x93 x18 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 5 8 18 3 5 10 16 3 5 10 16 3 7 8 16 x31 x52 x83 x4 x18 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 7 3 8 16 7 3 8 16 3 4 9 2 17 3 4 9 17 2 x31 x72 x83 x16 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 5 8 17 2 3 5 9 2 16 3 5 9 16 2 3 4 8 3 17 x31 x52 x83 x24 x17 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 3 4 9 3 16 3 4 9 16 3 3 5 8 3 16 3 5 8 16 3 x31 x42 x83 x17 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 . Proof of Proposition 5.3.2. Let x be an admissible monomial in P5+ such that ω(x) = ω ¯ (1) . Then x = xj x xt y 2 with y ∈ B5 (2, 1, 1, 1). Let z ∈ B5 (2, 1, 1, 1) such that xj x xt z 2 ∈ P5+ . By a direct computation using the results in Subsection 5.2, we see that if xj x xt z 2 = d35,t , 1 t 169, then there is a monomial w which is given in one of Lemmas 4.3.3(i) and 5.3.3 such u that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 and x is admissible, one gets x = d35,t for some t, 1 t 169. This implies B5+ (¯ ω(1) ) ⊂ {d35,t : 1 t 169}. We now prove the set {[d35,t ] : 1 t 169} is linearly independent in (F2 ⊗A P5 )35 . Suppose there is a linear relation 169 S= γt d35,t ≡ 0, t=1 where γt ∈ F2 . For (i; I) ∈ N5 , we explicitly compute p(i;I) (S) in terms of the admissible monomials in P4 . From the relations p(i;I) (S) ≡ 0 with and (I) 2, we obtain γt = 0 for 1 t 169. The proposition follows. Proposition 5.3.4. QP5 (¯ ω(2) ) = 0. Lemma 5.3.5. All permutations of the following monomials are strictly inadmissible: 3 4 9 9 10 3 5 8 8 11 3 5 8 9 10 3 7 8 8 9 x31 x42 x83 x94 x11 5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 . 28 ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC Proof of Proposition 5.3.4. Let x be an admissible monomial in P5+ such that ω(x) = ω ¯ (2) . Then x = xj x xt y 2 with y ∈ B5 (2, 1, 3). Let z ∈ B5 (2, 1, 3) such that xj x xt z 2 ∈ P5+ . By a direct computation using the results in Subsection 5.2, we see that if xj x xt z 2 is not a permutation of one of monomials as given in Lemma 5.3.5, then there is a monomial w which is given in u Lemma 4.3.3(i) such that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 and x is admissible, x is a permutation of one of monomials as given in Lemma 5.3.5. Now the proposition follows from Lemma 5.3.5. Proposition 5.3.6. There exist exactly 26 admissible monomials in P5 such that their weight vectors are ω ¯ (3) . Consequently dim QP5 (¯ ω(3) ) = 26. We denote the monomials in B5 (¯ ω(3) ) by d35,t , 170 t 195 (see Appendix). Lemma 5.3.7. The following monomials are strictly inadmissible: i) x3j x4 x5t x7u , x3j x5 x5t x6u . ii) x3j x4 xt x4u x7v , x3j x4 xt x5u x6v , x3j x4 x2t x5u x5v , j < < t; x3j x4 x3t x4u x5v , j < , t > 3. Here (j, , t, u, v) is a permutation of (1, 2, 3, 4, 5). iii) 13 2 7 13 12 7 2 12 13 7 2 13 12 7 2 12 13 x1 x22 x73 x12 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 12 6 3 12 13 6 3 13 12 6 11 4 13 6 11 13 4 x71 x2 x23 x13 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 4 13 7 10 13 4 7 10 4 13 7 10 13 4 7 11 4 12 x1 x72 x10 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 12 4 7 11 4 12 7 11 12 4 7 11 4 12 7 11 12 4 x1 x72 x11 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 5 12 6 11 12 5 7 10 5 12 7 10 12 5 7 10 5 12 x1 x62 x11 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 12 5 3 3 4 12 13 3 3 4 13 12 3 3 12 4 13 3 3 12 13 4 x71 x2 x10 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 4 12 3 3 13 12 4 3 4 3 12 13 3 4 3 13 12 3 3 12 5 12 x31 x32 x13 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 12 5 3 4 11 4 13 3 4 11 13 4 x31 x32 x12 3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 . Proof of Proposition 5.3.6. Let x be an admissible monomial in P5+ such that ω(x) = ω ¯ (3) . Then x = xj x xt y 2 with y ∈ B5 (2, 3, 2). Let z ∈ B5 (2, 3, 2) such that xj x xt z 2 ∈ P5+ . By a direct computation using Proposition 5.2.2, we see that if xj x xt z 2 = d35,t , 170 t 195, then there is a monomial w which is given in one of Lemmas 4.3.3 and 5.3.7 such that xj x xt z 2 = u wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 with y ∈ B5 (2, 3, 2) and x is admissible, x ∈ d35,t for some t, 170 t 195. We now prove the set {[d35,t ] : 170 t 195} is linearly independent in (F2 ⊗A 195 P5 )35 . Suppose there is a linear relation S = t=170 γt d35,t ≡ 0, where γt ∈ F2 . We explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 . From the relation p(i;j) (S) ≡ 0 with 1 i < j 5, one gets γt = 0 for all t. By a direct computation, we can show that the set [B5 (¯ ω(1) )]∪{[d35,t ]} is linearly independent in (F2 ⊗A P5 )35 . Hence [d35,t ]ω¯ (3) = 0 for all t, 170 t 195. This implies that {[d35,t ]ω¯ (3) : 170 t 195} is a basis of QP5 (¯ ω(3) ). The proposition is proved. Proposition 5.3.8. There exist exactly 69 admissible monomials in P5 such that their weight vectors are ω ¯ (4) . Consequently dim QP5 (¯ ω(4) ) = 69. KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER We denote the monomials in B5 (¯ ω(4) ) by d35,t , 196 t 29 264 (see Appendix). Lemma 5.3.9. The following monomials are strictly inadmissible: i) x2j x x2t x3u x3v , j < ; x2j x3 x3t x3u . Here (j, , t, u, v) is a permutation of (1, 2, 3, 4, 5). ii) All permutations of the monomials: 15 2 7 10 15 2 7 11 14 3 6 10 15 7 7 10 10 x1 x22 x33 x14 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 15 2 2 7 9 15 2 2 7 11 13 2 3 5 10 15 2 7 7 9 10 x21 x22 x33 x13 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 . iii) 6 11 3 14 6 11 3 6 14 11 3 6 14 11 3 14 7 10 x1 x32 x14 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 3 14 3 14 7 10 7 10 3 14 3 13 2 14 3 3 13 14 2 3 x1 x72 x10 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 14 3 2 3 5 2 14 11 3 5 14 2 11 3 5 14 11 2 3 13 2 6 11 x31 x13 2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , 6 2 11 3 13 6 11 2 3 13 2 7 10 3 13 7 2 10 3 13 7 10 2 x31 x13 2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 14 3 3 5 14 3 10 3 5 14 10 3 3 13 3 6 10 3 13 6 3 10 x31 x52 x10 3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 6 10 3 3 5 6 10 11 3 5 10 6 11 3 5 10 7 10 x31 x13 2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 , x 1 x2 x3 x4 x5 . Proof of Proposition 5.3.8. Let x be an admissible monomial in P5+ such that ω(x) = ω ¯ (4) . Then x = xj x xt y 2 with y ∈ B5 (4, 2, 2). Let z ∈ B5 (4, 2, 2) such that xj x xt z 2 ∈ P5+ . By a direct computation using the results in Subsection 5.2, we see that if xj x xt z 2 = d35,t , 196 t 264, then there is a monomial w which is given in one of Lemmaa 4.3.6 and 5.3.9 such u that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 and x is admissible, one gets x = d35,t for some t, 196 t 264. This implies B5+ (¯ ω(4) ) ⊂ {d35,t : 196 t 264}. We now prove the set {[d35,t ] : 196 t 264} is linearly independent in (F2 ⊗A P5 )35 . Suppose there is a linear relation 264 S= γt d35,t ≡ 0, t=196 where γt ∈ F2 . We explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 . From the relations p(i;i) (S) ≡ 0 with 1 i < j 5, one gets γt = 0 for all t, 196 t 264. By a similar computation, we can show that the set [B5+ (¯ ω(1) ) ∪ B5+ (¯ ω(3) )] ∪ {[d35,t ]} is linearly independent in (F2 ⊗A P5 )35 for all t, 196 t 264. Hence [d35,t ]ω¯ (4) = 0 and {[d35,t ]ω¯ (4) : 196 t 264} is a basis of QP5 (¯ ω(4) ). The proposition is proved. Proposition 5.3.10. There exist exactly 15 admissible monomials in P5 such that their weight vectors are ω ¯ (5) . Consequently dim QP5 (¯ ω(5) ) = 15. We denote the monomials in B5 (¯ ω(5) ) by d35,t , 265 t 279 (see Appendix). Lemma 5.3.11. The following monomials are strictly inadmissible: 6 7 7 14 i) All permutations of the monomials: x1 x62 x63 x74 x15 5 , x 1 x2 x3 x4 x5 . ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 30 ii) 7 3 5 14 6 7 3 5 14 7 6 3 13 6 6 7 3 13 6 7 6 x31 x52 x63 x14 4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , x1 x2 x3 x4 x5 , 7 6 6 7 7 9 6 6 x31 x13 2 x3 x4 x5 , x 1 x2 x3 x4 x5 . Proof of Proposition 5.3.10. Let x be an admissible monomial in P5+ such that ω(x) = ω ¯ (4) . Then x = xj x xt y 2 with y ∈ B5 (4, 4, 1). Let z ∈ B5 (4, 4, 1) such that xj x xt z 2 ∈ P5+ . By a direct computation using the results in Subsection 5.2, we see that if xj x xt z 2 = d35,t , 265 t 279, then there is a monomial w which is given in one of Lemmas 4.3.6 and 5.3.11 such u that xj x xt z 2 = wz12 with suitable monomial z1 ∈ P5 , and u = max{j ∈ Z : ωj (w) > 0}. By Theorem 2.7, xj x xt z 2 is inadmissible. Since x = xj x xt y 2 and x is admissible, one gets x = d35,t for some t, 265 t 279. Hence, B5+ (¯ ω(4) ) ⊂ {d35,t : 265 t 279}. We now prove the set {[d35,t ] : 265 t 279} is linearly independent in (F2 ⊗A P5 )35 . Suppose there is a linear relation 279 γt d35,t ≡ 0, S= t=265 where γt ∈ F2 . We explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 . From the relations p(i;i) (S) ≡ 0 with 1 i < j 5, one gets γt = 0 for all t, 265 t 279. By a direct computation, we can show that the set [B5+ (¯ ω(1) ) ∪ B5+ (¯ ω(3) ) ∪ B5+ (¯ ω(4) )] ∪ {[d35,t ]} is linearly independent in (F2 ⊗A P5 )35 for all t, 265 t 279. Hence [d35,t ]ω¯ (5) = 0 and {[d35,t ]ω¯ (5) : 265 t 279} is a basis of QP5 (¯ ω(5) ). The proposition is proved. Acknowledgment. The first named author would like to thank Prof. Nguyễn H. V. Hưng for valuable discussions on Kameko’s homomorphism. The paper was written when the first named author was visiting to Viet Nam Institute for Advanced Study in Mathematics (VIASM) in September, 2015. He would like to thank the VIASM for support and kind hospitality. The second named author would like to thank the University of Technical Education of Ho Chi Minh city for supporting this work. References [1] J. F. Adams, A periodicity theorem in homological algebra, Math. Proc. Cambridge Phil. Soc. 62 (1966) 365-377, MR0194486. [2] J. M. Boardman, Modular representations on the homology of power of real projective space, in: M. C. Tangora (Ed.), Algebraic Topology, Oaxtepec, 1991, in: Contemp. Math., vol. 146, 1993, pp. 49-70, MR1224907. [3] R. R. Bruner, L. M. Hà and N. H. V. Hưng, On behavior of the algebraic transfer, Trans. Amer. Math. Soc. 357 (2005) 473-487, MR2095619. [4] D. P. Carlisle and R. M. W. Wood, The boundedness conjecture for the action of the Steenrod algebra on polynomials, in: N. Ray and G. Walker (ed.), Adams Memorial Symposium on Algebraic Topology 2, (Manchester, 1990), in: London Math. Soc. Lecture Notes Ser., Cambridge Univ. Press, Cambridge, vol. 176, 1992, pp. 203-216, MR1232207. KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 31 [5] T. W. Chen, Determination of Ext5,∗ A (Z/2, Z/2), Topology Appl., 158 (2011) 660-689, MR2774051. [6] P. H. Chơn and L. M. Hà, Lambda algebra and the Singer transfer, C. R. Math. Acad. Sci. Paris 349 (2011) 21-23, MR2755689. [7] P. H. Chơn and L. M. Hà, On May spectral sequence and the algebraic transfer, Manuscripta Math. 138 (2012) 141-160, MR2898751. [8] P. H. Chơn and L. M. Hà, On the May spectral sequence and the algebraic transfer II, Topology Appl. 178 (2014) 372-383, MR3276753. [9] L. M. Hà, Sub-Hopf algebras of the Steenrod algebra and the Singer transfer, “Proceedings of the International School and Conference in Algebraic Topology, Hà Nô.i 2004”, Geom. Topol. Monogr., Geom. Topol. Publ., Coventry, vol. 11 (2007), 81-105, MR2402802. [10] N. H. V. Hưng, The weak conjecture on spherical classes, Math. Zeit. 231 (1999) 727-743, MR1709493 [11] N. H. V. Hưng, The cohomology of the Steenrod algebra and representations of the general linear groups, Trans. Amer. Math. Soc. 357 (2005) 4065-4089, MR2159700. [12] N. H. V. Hưng and V. T. N. Quỳnh, The image of Singer’s fourth transfer, C. R. Math. Acad. Sci. Paris 347 (2009) 1415-1418, MR2588792. [13] M. Kameko, Products of projective spaces as Steenrod modules, PhD Thesis, The Johns Hopkins University, ProQuest LLC, Ann Arbor, MI, 1990. 29 pp, MR2638633. 5,∗ [14] W. H. Lin, Ext4,∗ A (Z/2, Z/2) and ExtA (Z/2, Z/2), Topology Appl., 155 (2008) 459-496, MR2380930. [15] N. Minami, The iterated transfer analogue of the new doomsday conjecture, Trans. Amer. Math. Soc. 351 (1999) 2325-2351, MR1443884. [16] T. N. Nam, Transfert algébrique et action du groupe linéaire sur les puissances divisées modulo 2, Ann. Inst. Fourier (Grenoble) 58 (2008) 1785-1837, MR2445834. [17] F. P. Peterson, Generators of H ∗ (RP ∞ × RP ∞ ) as a module over the Steenrod algebra, Abstracts Amer. Math. Soc. No. 833, April 1987, pp. 55-89. - . V. Phúc and N. Sum, On the generators of the polynomial algebra as a mod[18] D ule over the Steenrod algebra, C. R. Math. Acad. Sci. Paris, Ser. I (2015), DOI: 10.1016/j.crma.2015.09.002. [19] V. T. N. Quỳnh, On behavior of the fifth algebraic transfer, “Proceedings of the International School and Conference in Algebraic Topology, Hà Nô.i 2004”, Geom. Topol. Monogr., Geom. Topol. Publ., Coventry, vol. 11 (2007), 309-326, MR2402811. [20] W. M. Singer, The transfer in homological algebra, Math. Zeit. 202 (1989) 493-523, MR1022818. [21] W. M. Singer, On the action of the Steenrod squares on polynomial algebras, Proc. Amer. Math. Soc. 111 (1991) 577-583, MR1045150. [22] N. E. Steenrod and D. B. A. Epstein, Cohomology operations, Annals of Mathematics Studies 50, Princeton University Press, Princeton N.J (1962), MR0145525. [23] N. Sum, On the hit problem for the polynomial algebra, C. R. Math. Acad. Sci. Paris, Ser. I, 351 (2013) 565-568, MR3095107. [24] N. Sum, On the Peterson hit problem of five variables and its applications to the fifth Singer transfer, East-West J. of Mathematics, 16 (2014) 47-62. [25] N. Sum, On the Peterson hit problem, Adv. Math. 274 (2015) 432-489, MR3318156. [26] M. C. Tangora, On the cohomology of the Steenrod algebra, Math. Zeit. 116 (1970) 18-64, MR0266205. [27] R. M. W. Wood, Steenrod squares of polynomials and the Peterson conjecture, Math. Proc. Cambridge Phil. Soc. 105 (1989) 307-309, MR0974986. † Department of Mathematics, Quy Nhơn University, - ịnh, Viet Nam. 170 An Dương Vương, Quy Nhơn, Bình D E-mail: nguyensum@qnu.edu.vn ‡ Department of Foundation Sciences, University of Technical Education of Ho Chi Minh city, - ức district, Ho Chi Minh city, Viet Nam. 01 Võ Văn Ngân, Thu’ D Email: tinnk@hcmute.edu.vn ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 32 APPENDIX In the appendix, we list all admissible monomials of degrees 9, 15, 16, 23, 35 in P4 and P5 . We order a set B of some monomials of degree n in Pk by using the order as in Definition 2.5 and denote its elements by di = dn,i , i > 0 in such a way that dn,i < dn,j if and only if i < j. A1.1. The admissible monomials of degree 9 in P4 . B4 (9) is the set of 46 monomials: 1. x2 x3 x74 6. x72 x3 x4 11. x1 x2 x3 x64 16. x1 x22 x3 x54 21. x1 x32 x53 26. x31 x53 x4 31. x31 x42 x3 x4 36. x71 x2 x3 41. x31 x33 x34 46. x31 x32 x33 . 2. x2 x33 x54 7. x1 x3 x74 12. x1 x2 x23 x54 17. x1 x22 x53 x4 22. x1 x62 x3 x4 27. x31 x2 x54 32. x31 x52 x4 37. x32 x33 x34 42. x31 x2 x23 x34 3. x2 x73 x4 8. x1 x33 x54 13. x1 x2 x33 x44 18. x1 x32 x54 23. x1 x72 x4 28. x31 x2 x3 x44 33. x31 x52 x3 38. x1 x22 x33 x34 43. x31 x2 x33 x24 4. x32 x3 x54 9. x1 x73 x4 14. x1 x2 x63 x4 19. x1 x32 x3 x44 24. x1 x72 x3 29. x31 x2 x43 x4 34. x71 x3 x4 39. x1 x32 x23 x34 44. x31 x32 x34 5. x32 x53 x4 10. x1 x2 x74 15. x1 x2 x73 20. x1 x32 x43 x4 25. x31 x3 x54 30. x31 x2 x53 35. x71 x2 x4 40. x1 x32 x33 x24 45. x31 x32 x3 x24 A1.2. The admissible monomials of degree 9 in P5 . B5 (9) = B50 (9) ∪ B5+ (9), where B50 (9) = Φ0 (B4 (9)), |B50 (9)| = 160 and B5+ (9) is the set of 31 monomials. 1.2.1. There are 6 admissible monomials in B5+ (9) with the weight vector (3, 1, 1). 1. x1 x2 x3 x24 x45 5. x1 x22 x3 x44 x5 2. x1 x2 x23 x4 x45 3. x1 x2 x23 x44 x5 6. x1 x22 x43 x4 x5 . 4. x1 x22 x3 x4 x45 1.2.2. There are 15 admissible monomials in P5+ with the weight vector (3, 3). 7. x1 x2 x23 x24 x35 11. x1 x22 x3 x34 x25 15. x1 x22 x33 x24 x5 19. x31 x2 x3 x24 x25 8. x1 x2 x23 x34 x25 12. x1 x22 x23 x4 x35 16. x1 x32 x3 x24 x25 20. x31 x2 x23 x4 x25 9. x1 x2 x33 x24 x25 10. x1 x22 x3 x24 x35 2 2 3 13. x1 x2 x3 x4 x5 14. x1 x22 x33 x4 x25 17. x1 x32 x23 x4 x25 18. x1 x32 x23 x24 x5 21. x31 x2 x23 x24 x5 . 1.2.3. There are 10 admissible monomials with the weight vector (5, 2). 22. x1 x2 x3 x34 x35 26. x1 x32 x3 x34 x5 30. x31 x2 x33 x4 x5 23. x1 x2 x33 x4 x35 24. x1 x2 x33 x34 x5 27. x1 x32 x33 x4 x5 28. x31 x2 x3 x4 x35 31. x31 x32 x3 x4 x5 . 25. x1 x32 x3 x4 x35 29. x31 x2 x3 x34 x5 A2.1. The admissible monomials of degree 15 in P4 . B4 (15) is the set of 75 monomials: 1. x15 4 6. x2 x14 3 11. x1 x22 x12 4 16. x2 x73 x74 21. x72 x73 x4 2. x3 x14 4 7. x15 2 12. x1 x22 x43 x84 17. x32 x53 x74 22. x1 x73 x74 3. x15 3 8. x1 x14 4 13. x1 x22 x12 3 18. x32 x73 x54 23. x1 x2 x63 x74 4. x2 x14 4 9. x1 x23 x12 4 14. x1 x14 2 19. x72 x3 x74 24. x1 x2 x73 x64 5. x2 x23 x12 4 10. x1 x14 3 15. x15 1 20. x72 x33 x54 25. x1 x22 x53 x74 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 26. 31. 36. 41. 46. 51. 56. 61. 66. 71. x1 x22 x73 x54 x1 x62 x3 x74 x1 x72 x23 x54 x31 x73 x54 x31 x32 x43 x54 x31 x52 x74 x31 x52 x73 x71 x3 x74 x71 x2 x23 x54 x71 x32 x3 x44 27. 32. 37. 42. 47. 52. 57. 62. 67. 72. x1 x32 x43 x74 x1 x62 x33 x54 x1 x72 x33 x44 x31 x2 x43 x74 x31 x32 x53 x44 x31 x52 x3 x64 x31 x72 x54 x71 x33 x54 x71 x2 x33 x44 x71 x32 x43 x4 28. 33. 38. 43. 48. 53. 58. 63. 68. 73. x1 x32 x53 x64 x1 x62 x73 x4 x1 x72 x63 x4 x31 x2 x53 x64 x31 x42 x3 x74 x31 x52 x23 x54 x31 x72 x3 x44 x71 x73 x4 x71 x2 x63 x4 x71 x32 x53 29. 34. 39. 44. 49. 54. 59. 64. 69. 74. x1 x32 x63 x54 x1 x72 x74 x1 x72 x73 x31 x2 x63 x54 x31 x42 x33 x54 x31 x52 x33 x44 x31 x72 x43 x4 x71 x2 x74 x71 x2 x73 x71 x72 x4 30. 35. 40. 45. 50. 55. 60. 65. 70. 75. 33 x1 x32 x73 x44 x1 x72 x3 x64 x31 x53 x74 x31 x2 x73 x44 x31 x42 x73 x4 x31 x52 x63 x4 x31 x72 x53 x71 x2 x3 x64 x71 x32 x54 x71 x72 x3 . A2.2. The admissible monomials of degree 15 in P5 . B5 (15) = B50 (15) ∪ B5+ (15), where B50 (15) = Φ0 (B4 (15)), |B50 (15)| = 270 and B5+ (15) = B5+ (1, 1, 3) ∪ B5+ (3, 2, 2) ∪ B5 (3, 4, 1) ∪ B5 (5, 3, 1) ∪ B5 (5, 5). 2.2.1. B5 (1, 1, 3) = {x1 x22 x43 x44 x45 }. 2.2.2. B5+ (3, 2, 2) is the set of 75 monomials: 2. x1 x2 x3 x64 x65 6. x1 x2 x23 x74 x45 10. x1 x2 x63 x24 x55 14. x1 x22 x3 x44 x75 18. x1 x22 x33 x44 x55 22. x1 x22 x43 x74 x5 26. x1 x22 x53 x64 x5 30. x1 x32 x3 x64 x45 34. x1 x32 x43 x4 x65 38. x1 x32 x53 x24 x45 42. x1 x62 x3 x24 x55 46. x1 x62 x33 x44 x5 50. x31 x2 x3 x44 x65 54. x31 x2 x33 x44 x45 58. x31 x2 x43 x64 x5 62. x31 x32 x3 x44 x45 66. x31 x42 x3 x24 x55 70. x31 x42 x33 x44 x5 74. x71 x2 x3 x24 x45 3. x1 x2 x23 x44 x75 7. x1 x2 x33 x44 x65 11. x1 x2 x63 x34 x45 15. x1 x22 x3 x54 x65 19. x1 x22 x33 x54 x45 23. x1 x22 x53 x4 x65 27. x1 x22 x73 x4 x45 31. x1 x32 x23 x44 x55 35. x1 x32 x43 x24 x55 39. x1 x32 x63 x4 x45 43. x1 x62 x3 x34 x45 47. x1 x72 x3 x24 x45 51. x31 x2 x3 x64 x45 55. x31 x2 x43 x4 x65 59. x31 x2 x53 x24 x45 63. x31 x32 x43 x4 x45 67. x31 x42 x3 x34 x45 71. x31 x52 x3 x24 x45 75. x71 x2 x23 x4 x45 4. x1 x2 x23 x54 x65 8. x1 x2 x33 x64 x45 12. x1 x2 x63 x64 x5 16. x1 x22 x3 x64 x55 20. x1 x22 x43 x4 x75 24. x1 x22 x53 x24 x55 28. x1 x22 x73 x44 x5 32. x1 x32 x23 x54 x45 36. x1 x32 x43 x34 x45 40. x1 x32 x63 x44 x5 44. x1 x62 x3 x64 x5 48. x1 x72 x23 x4 x45 52. x31 x2 x23 x44 x55 56. x31 x2 x43 x24 x55 60. x31 x2 x63 x4 x45 64. x31 x32 x43 x44 x5 68. x31 x42 x3 x64 x5 72. x31 x52 x23 x4 x45 76. x71 x2 x23 x44 x5 . 5. x1 x2 x23 x64 x55 9. x1 x2 x63 x4 x65 13. x1 x2 x73 x24 x45 17. x1 x22 x3 x74 x45 21. x1 x22 x43 x34 x55 25. x1 x22 x53 x34 x45 29. x1 x32 x3 x44 x65 33. x1 x32 x33 x44 x45 37. x1 x32 x43 x64 x5 41. x1 x62 x3 x4 x65 45. x1 x62 x33 x4 x45 49. x1 x72 x23 x44 x5 53. x31 x2 x23 x54 x45 57. x31 x2 x43 x34 x45 61. x31 x2 x63 x44 x5 65. x31 x42 x3 x4 x65 69. x31 x42 x33 x4 x45 73. x31 x52 x23 x44 x5 2.2.3. B5+ (3, 4, 1) is the set of 40 monomials: 77. x1 x22 x23 x34 x75 81. x1 x22 x33 x64 x35 85. x1 x32 x23 x24 x75 89. x1 x32 x33 x24 x65 93. x1 x32 x73 x24 x25 97. x31 x2 x23 x24 x75 101. x31 x2 x33 x24 x65 105. x31 x2 x73 x24 x25 109. x31 x52 x23 x24 x35 113. x71 x2 x23 x24 x35 78. x1 x22 x23 x74 x35 82. x1 x22 x33 x74 x25 86. x1 x32 x23 x34 x65 90. x1 x32 x33 x64 x25 94. x1 x72 x23 x24 x35 98. x31 x2 x23 x34 x65 102. x31 x2 x33 x64 x25 106. x31 x32 x3 x24 x65 110. x31 x52 x23 x34 x25 114. x71 x2 x23 x34 x25 79. x1 x22 x33 x24 x75 83. x1 x22 x73 x24 x35 87. x1 x32 x23 x64 x35 91. x1 x32 x63 x24 x35 95. x1 x72 x23 x34 x25 99. x31 x2 x23 x64 x35 103. x31 x2 x63 x24 x35 107. x31 x32 x3 x64 x25 111. x31 x52 x33 x24 x25 115. x71 x2 x33 x24 x25 80. x1 x22 x33 x34 x65 84. x1 x22 x73 x34 x25 88. x1 x32 x23 x74 x25 92. x1 x32 x63 x34 x25 96. x1 x72 x33 x24 x25 100. x31 x2 x23 x74 x25 104. x31 x2 x63 x34 x25 108. x31 x32 x53 x24 x25 112. x31 x72 x3 x24 x25 116. x71 x32 x3 x24 x25 . ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 34 2.2.4. B5+ (5, 3, 1) is the set of 45 monomials: 117. 121. 125. 129. 133. 137. 141. 145. 149. 153. 157. 161. x1 x2 x33 x34 x75 x1 x32 x3 x74 x35 x1 x32 x33 x74 x5 x1 x72 x3 x34 x35 x31 x2 x3 x74 x35 x31 x2 x33 x74 x5 x31 x32 x3 x4 x75 x31 x32 x33 x4 x55 x31 x32 x73 x4 x5 x31 x72 x3 x4 x35 x71 x2 x33 x4 x35 x71 x32 x33 x4 x5 . 118. 122. 126. 130. 134. 138. 142. 146. 150. 154. 158. x1 x2 x33 x74 x35 x1 x32 x33 x4 x75 x1 x32 x53 x34 x35 x1 x72 x33 x4 x35 x31 x2 x33 x4 x75 x31 x2 x53 x34 x35 x31 x32 x3 x34 x55 x31 x32 x33 x54 x5 x31 x52 x3 x34 x35 x31 x72 x3 x34 x5 x71 x2 x33 x34 x5 119. 123. 127. 131. 135. 139. 143. 147. 151. 155. 159. x1 x2 x73 x34 x35 x1 x32 x33 x34 x55 x1 x32 x73 x4 x35 x1 x72 x33 x34 x5 x31 x2 x33 x34 x55 x31 x2 x73 x4 x35 x31 x32 x3 x54 x35 x31 x32 x53 x4 x35 x31 x52 x33 x4 x35 x31 x72 x33 x4 x5 x71 x32 x3 x4 x35 120. 124. 128. 132. 136. 140. 144. 148. 152. 156. 160. x1 x32 x3 x34 x75 x1 x32 x33 x54 x35 x1 x32 x73 x34 x5 x31 x2 x3 x34 x75 x31 x2 x33 x54 x35 x31 x2 x73 x34 x5 x31 x32 x3 x74 x5 x31 x32 x53 x34 x5 x31 x52 x33 x34 x5 x71 x2 x3 x34 x35 x71 x32 x3 x34 x5 2.2.5. B5 (5, 5) = {x31 x32 x33 x34 x35 }. A3.1. The admissible monomials of degree 16 in P4 . B4 (16) is the set of 73 monomials: 1. x3 x15 4 6. x2 x23 x13 4 11. x32 x3 x12 4 16. x1 x3 x14 4 21. x1 x2 x14 4 26. x1 x22 x43 x94 31. x1 x32 x43 x84 36. x31 x13 4 41. x31 x2 x12 3 46. x1 x32 x63 x64 51. x1 x32 x53 x74 56. x31 x2 x53 x74 61. x31 x52 x73 x4 66. x71 x2 x73 x4 71. x31 x32 x73 x34 2. x33 x13 4 7. x2 x33 x12 4 12. x32 x13 3 17. x1 x23 x13 4 22. x1 x2 x23 x12 4 27. x1 x22 x53 x84 32. x1 x32 x12 3 37. x31 x3 x12 4 42. x31 x13 2 47. x31 x2 x63 x64 52. x1 x32 x73 x54 57. x31 x2 x73 x54 62. x31 x72 x3 x54 67. x71 x32 x3 x54 72. x31 x72 x33 x34 3. x15 3 x4 8. x2 x14 3 x4 13. x15 2 x4 18. x1 x33 x12 4 23. x1 x2 x14 3 28. x1 x22 x12 3 x4 33. x1 x14 2 x4 38. x31 x13 3 43. x15 1 x4 48. x31 x52 x23 x64 53. x1 x72 x3 x74 58. x31 x32 x53 x54 63. x31 x72 x53 x4 68. x71 x32 x53 x4 73. x71 x32 x33 x34 . 4. x2 x15 4 9. x2 x15 3 14. x15 2 x3 19. x1 x14 3 x4 24. x1 x22 x13 4 29. x1 x22 x13 3 34. x1 x14 2 x3 39. x31 x2 x12 4 44. x15 1 x3 49. x31 x52 x63 x24 54. x1 x72 x33 x54 59. x31 x52 x3 x74 64. x71 x2 x3 x74 69. x71 x72 x3 x4 5. x2 x3 x14 4 10. x32 x13 4 15. x1 x15 4 20. x1 x15 3 25. x1 x22 x3 x12 4 30. x1 x32 x12 4 35. x1 x15 2 40. x31 x2 x43 x84 45. x15 1 x2 50. x1 x2 x73 x74 55. x1 x72 x73 x4 60. x31 x52 x33 x54 65. x71 x2 x33 x54 70. x31 x32 x33 x74 A3.2. The admissible monomials of degree 16 in P5 . B5 (16) = B50 (16) ∪ B5+ (16), where B50 (156) = Φ0 (B4 (16)), |B50 (16)| = 255 and B5+ (16) = B5+ (2, 1, 1, 1) ∪ B5+ (2, 1, 3) ∪ B5+ (2, 3, 2) ∪ B5+ (4, 2, 2) ∪ B5+ (4, 4, 1). 3.2.1. B5+ (2, 1, 1, 1) is the set of 4 monomials: 1. x1 x2 x23 x44 x85 2. x1 x22 x3 x44 x85 3. x1 x22 x43 x4 x85 4. x1 x22 x43 x84 x5 . 3.2.2. B5+ (2, 1, 3) is the set of 5 monomials: 5. x1 x22 x43 x44 x55 6. x1 x22 x43 x54 x45 9. x31 x2 x43 x44 x45 . 7. x1 x22 x53 x44 x45 8. x1 x32 x43 x44 x45 3.2.3. B5+ (2, 3, 2) is the set of 20 monomials: 10. x1 x2 x23 x64 x65 14. x1 x22 x53 x24 x65 11. x1 x2 x63 x24 x65 15. x1 x22 x53 x64 x25 12. x1 x2 x63 x64 x25 16. x1 x32 x23 x44 x65 13. x1 x22 x3 x64 x65 17. x1 x32 x23 x64 x45 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 18. x1 x32 x43 x24 x65 22. x31 x2 x23 x44 x65 26. x31 x2 x63 x24 x45 19. x1 x32 x43 x64 x25 23. x31 x2 x23 x64 x45 27. x31 x2 x63 x44 x25 20. x1 x32 x63 x24 x45 24. x31 x2 x43 x24 x65 28. x31 x52 x23 x24 x45 21. x1 x32 x63 x44 x25 25. x31 x2 x43 x64 x25 29. x31 x52 x23 x44 x25 . 3.2.4. B5+ (4, 2, 2) is the set of 110 monomials: 30. x1 x2 x3 x64 x75 34. x1 x2 x73 x4 x65 38. x1 x62 x73 x4 x5 42. x71 x2 x3 x4 x65 46. x1 x2 x23 x74 x55 50. x1 x22 x53 x4 x75 54. x1 x72 x3 x24 x55 58. x71 x2 x23 x4 x55 62. x1 x2 x73 x34 x45 66. x1 x32 x43 x74 x5 70. x1 x72 x33 x4 x45 74. x31 x2 x43 x4 x75 78. x31 x42 x3 x4 x75 82. x31 x72 x3 x44 x5 86. x71 x2 x33 x44 x5 90. x1 x2 x33 x54 x65 94. x1 x32 x3 x64 x55 98. x1 x32 x63 x54 x5 102. x31 x2 x3 x54 x65 106. x31 x2 x63 x4 x55 110. x31 x52 x63 x4 x5 114. x1 x32 x53 x24 x55 118. x31 x52 x23 x4 x55 122. x1 x32 x43 x34 x55 126. x31 x2 x43 x34 x55 130. x31 x32 x43 x4 x55 134. x31 x42 x3 x34 x55 138. x31 x52 x33 x4 x45 31. x1 x2 x3 x74 x65 35. x1 x2 x73 x64 x5 39. x1 x72 x3 x4 x65 43. x71 x2 x3 x64 x5 47. x1 x2 x73 x24 x55 51. x1 x22 x53 x74 x5 55. x1 x72 x23 x4 x55 59. x71 x2 x23 x54 x5 63. x1 x32 x3 x44 x75 67. x1 x32 x73 x4 x45 71. x1 x72 x33 x44 x5 75. x31 x2 x43 x74 x5 79. x31 x42 x3 x74 x5 83. x31 x72 x43 x4 x5 87. x71 x32 x3 x4 x45 91. x1 x2 x33 x64 x55 95. x1 x32 x53 x4 x65 99. x1 x62 x3 x34 x55 103. x31 x2 x3 x64 x55 107. x31 x2 x63 x54 x5 111. x1 x22 x33 x54 x55 115. x31 x2 x23 x54 x55 119. x31 x52 x23 x54 x5 123. x1 x32 x53 x34 x45 127. x31 x2 x53 x34 x45 131. x31 x32 x43 x54 x5 135. x31 x42 x33 x4 x55 139. x31 x52 x33 x44 x5 . 32. x1 x2 x63 x4 x75 36. x1 x62 x3 x4 x75 40. x1 x72 x3 x64 x5 44. x71 x2 x63 x4 x5 48. x1 x22 x3 x54 x75 52. x1 x22 x73 x4 x55 56. x1 x72 x23 x54 x5 60. x1 x2 x33 x44 x75 64. x1 x32 x3 x74 x45 68. x1 x32 x73 x44 x5 72. x31 x2 x3 x44 x75 76. x31 x2 x73 x4 x45 80. x31 x42 x73 x4 x5 84. x71 x2 x3 x34 x45 88. x71 x32 x3 x44 x5 92. x1 x2 x63 x34 x55 96. x1 x32 x53 x64 x5 100. x1 x62 x33 x4 x55 104. x31 x2 x53 x4 x65 108. x31 x52 x3 x4 x65 112. x1 x22 x53 x34 x55 116. x31 x2 x53 x24 x55 120. x1 x32 x33 x44 x55 124. x31 x2 x33 x44 x55 128. x31 x32 x3 x44 x55 132. x31 x32 x53 x4 x45 136. x31 x42 x33 x54 x5 33. x1 x2 x63 x74 x5 37. x1 x62 x3 x74 x5 41. x1 x72 x63 x4 x5 45. x1 x2 x23 x54 x75 49. x1 x22 x3 x74 x55 53. x1 x22 x73 x54 x5 57. x71 x2 x3 x24 x55 61. x1 x2 x33 x74 x45 65. x1 x32 x43 x4 x75 69. x1 x72 x3 x34 x45 73. x31 x2 x3 x74 x45 77. x31 x2 x73 x44 x5 81. x31 x72 x3 x4 x45 85. x71 x2 x33 x4 x45 89. x71 x32 x43 x4 x5 93. x1 x32 x3 x54 x65 97. x1 x32 x63 x4 x55 101. x1 x62 x33 x54 x5 105. x31 x2 x53 x64 x5 109. x31 x52 x3 x64 x5 113. x1 x32 x23 x54 x55 117. x31 x52 x3 x24 x55 121. x1 x32 x33 x54 x45 125. x31 x2 x33 x54 x45 129. x31 x32 x3 x54 x45 133. x31 x32 x53 x44 x5 137. x31 x52 x3 x34 x45 3.2.5. B5+ (4, 4, 1) is the set of 49 monomials: 140. 144. 148. 152. 156. 160. 164. 168. 172. 176. 180. 184. 188. x1 x22 x33 x34 x75 x1 x32 x23 x74 x35 x1 x32 x73 x34 x25 x31 x2 x23 x34 x75 x31 x2 x73 x24 x35 x31 x32 x73 x4 x25 x71 x2 x23 x34 x35 x71 x32 x3 x34 x25 x1 x32 x63 x34 x35 x31 x2 x63 x34 x35 x31 x32 x33 x54 x25 x31 x52 x33 x24 x35 x31 x32 x43 x34 x35 . 141. 145. 149. 153. 157. 161. 165. 169. 173. 177. 181. 185. x1 x22 x33 x74 x35 x1 x32 x33 x24 x75 x1 x72 x23 x34 x35 x31 x2 x23 x74 x35 x31 x2 x73 x34 x25 x31 x72 x3 x24 x35 x71 x2 x33 x24 x35 x71 x32 x33 x4 x25 x1 x62 x33 x34 x35 x31 x32 x3 x34 x65 x31 x32 x53 x24 x35 x31 x52 x33 x34 x25 142. 146. 150. 154. 158. 162. 166. 170. 174. 178. 182. 186. x1 x22 x73 x34 x35 x1 x32 x33 x74 x25 x1 x72 x33 x24 x35 x31 x2 x33 x24 x75 x31 x32 x3 x24 x75 x31 x72 x3 x34 x25 x71 x2 x33 x34 x25 x1 x32 x33 x34 x65 x31 x2 x33 x34 x65 x31 x32 x3 x64 x35 x31 x32 x53 x34 x25 x31 x32 x33 x34 x45 143. 147. 151. 155. 159. 163. 167. 171. 175. 179. 183. 187. x1 x32 x23 x34 x75 x1 x32 x73 x24 x35 x1 x72 x33 x34 x25 x31 x2 x33 x74 x25 x31 x32 x3 x74 x25 x31 x72 x33 x4 x25 x71 x32 x3 x24 x35 x1 x32 x33 x64 x35 x31 x2 x33 x64 x35 x31 x32 x33 x4 x65 x31 x52 x23 x34 x35 x31 x32 x33 x44 x35 35 ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 36 A4.1. The admissible monomials of degree 23 in P4 . B4 (23) is the set of 155 monomials: 1. x1 x2 x73 x14 4 5. x1 x32 x53 x14 4 6 9. x1 x32 x13 3 x4 6 11 5 13. x1 x2 x3 x4 17. x1 x72 x63 x94 21. x1 x72 x14 3 x4 25. x31 x2 x53 x14 4 6 29. x31 x2 x13 3 x4 3 3 12 5 33. x1 x2 x3 x4 5 37. x31 x42 x11 3 x4 3 5 6 9 41. x1 x2 x3 x4 45. x31 x52 x14 3 x4 49. x31 x72 x83 x54 2 5 53. x31 x13 2 x3 x4 7 2 13 57. x1 x2 x3 x4 5 61. x71 x2 x10 3 x4 7 3 4 9 65. x1 x2 x3 x4 69. x71 x32 x12 3 x4 73. x71 x92 x33 x44 7 77. x2 x15 3 x4 3 15 5 81. x2 x3 x4 5 85. x72 x11 3 x4 15 7 89. x2 x3 x4 6 93. x1 x2 x15 3 x4 3 15 4 97. x1 x2 x3 x4 101. x1 x72 x15 3 3 4 105. x1 x15 2 x3 x4 3 7 13 109. x1 x3 x4 4 113. x31 x2 x15 3 x4 3 5 15 117. x1 x2 x3 7 121. x31 x13 2 x3 3 15 5 125. x1 x2 x3 5 129. x71 x11 3 x4 7 3 13 133. x1 x2 x4 5 137. x71 x11 2 x4 15 141. x1 x3 x74 6 145. x15 1 x2 x3 x4 15 7 149. x1 x2 x3 3 5 153. x15 1 x2 x3 7 2. x1 x2 x14 3 x4 3 6 13 6. x1 x2 x3 x4 5 10. x1 x32 x14 3 x4 7 14 14. x1 x2 x3 x4 18. x1 x72 x73 x84 7 22. x1 x14 2 x3 x4 3 6 13 26. x1 x2 x3 x4 5 30. x31 x2 x14 3 x4 3 3 13 4 34. x1 x2 x3 x4 38. x31 x52 x3 x14 4 42. x31 x52 x73 x84 46. x31 x72 x3 x12 4 50. x31 x72 x93 x44 3 4 54. x31 x13 2 x3 x4 7 3 12 58. x1 x2 x3 x4 4 62. x71 x2 x11 3 x4 7 3 5 8 66. x1 x2 x3 x4 70. x71 x72 x3 x84 4 74. x71 x11 2 x3 x4 3 5 15 78. x2 x3 x4 82. x72 x3 x15 4 86. x72 x15 3 x4 90. x1 x73 x15 4 94. x1 x22 x53 x15 4 98. x1 x62 x3 x15 4 7 102. x1 x15 2 x4 6 106. x1 x15 2 x3 x4 3 13 7 110. x1 x3 x4 114. x31 x42 x3 x15 4 118. x31 x72 x13 4 5 122. x31 x15 2 x4 7 15 126. x1 x3 x4 130. x71 x15 3 x4 134. x71 x32 x13 3 5 138. x71 x11 2 x3 15 3 5 142. x1 x3 x4 2 5 146. x15 1 x2 x3 x4 15 3 5 150. x1 x2 x4 7 154. x15 1 x2 x4 3. x1 x22 x73 x13 4 7. x1 x32 x73 x12 4 11. x1 x62 x33 x13 4 15. x1 x72 x23 x13 4 5 19. x1 x72 x10 3 x4 14 3 5 23. x1 x2 x3 x4 27. x31 x2 x73 x12 4 31. x31 x32 x43 x13 4 35. x31 x42 x33 x13 4 39. x31 x52 x23 x13 4 5 43. x31 x52 x10 3 x4 3 7 4 9 47. x1 x2 x3 x4 51. x31 x72 x12 3 x4 6 55. x31 x13 x 2 3 x4 7 6 9 59. x1 x2 x3 x4 63. x71 x2 x14 3 x4 67. x71 x32 x83 x54 71. x71 x72 x83 x4 4 75. x71 x11 2 x3 x4 3 7 13 79. x2 x3 x4 83. x72 x33 x13 4 7 87. x15 2 x3 x4 15 7 91. x1 x3 x4 5 95. x1 x22 x15 3 x4 6 15 99. x1 x2 x3 x4 6 103. x1 x15 2 x3 x4 15 7 107. x1 x2 x3 5 111. x31 x15 3 x4 3 4 15 115. x1 x2 x3 x4 119. x31 x72 x13 3 4 123. x31 x15 2 x3 x4 7 3 13 127. x1 x3 x4 131. x71 x2 x15 4 135. x71 x72 x94 139. x71 x15 2 x4 7 143. x15 1 x3 x4 15 147. x1 x2 x33 x44 3 4 151. x15 1 x2 x3 x4 15 7 155. x1 x2 x3 . 7 4. x1 x22 x13 3 x4 3 12 7 8. x1 x2 x3 x4 12. x1 x62 x73 x94 16. x1 x72 x33 x12 4 4 20. x1 x72 x11 3 x4 14 7 24. x1 x2 x3 x4 7 28. x31 x2 x12 3 x4 3 3 5 12 32. x1 x2 x3 x4 36. x31 x42 x73 x94 40. x31 x52 x33 x12 4 4 44. x31 x52 x11 3 x4 3 7 5 8 48. x1 x2 x3 x4 6 52. x31 x13 2 x3 x4 7 14 56. x1 x2 x3 x4 60. x71 x2 x73 x84 64. x71 x32 x3 x12 4 68. x71 x32 x93 x44 72. x71 x92 x23 x54 76. x2 x73 x15 4 7 80. x32 x13 3 x4 7 7 9 84. x2 x3 x4 3 5 88. x15 2 x3 x4 92. x1 x2 x63 x15 4 96. x1 x32 x43 x15 4 100. x1 x72 x15 4 2 5 104. x1 x15 2 x3 x4 3 5 15 108. x1 x3 x4 112. x31 x2 x43 x15 4 116. x31 x52 x15 4 7 120. x31 x13 2 x4 3 15 4 124. x1 x2 x3 x4 128. x71 x73 x94 132. x71 x2 x15 3 136. x71 x72 x93 140. x71 x15 2 x3 7 144. x15 1 x2 x4 15 148. x1 x2 x63 x4 3 4 152. x15 1 x2 x3 x4 A4.2. The admissible monomials of degree 23 in P5 . 0 We have B5 (23) = B50 (23)∪ψ(B5 (9))∪ B5+ (23) ∩ Ker(Sq ∗ )(5,9) , where B50 (23) = Φ0 (B4 (23)), |B50 (35)| = 635, |ψ(B5 (9))| = 191 with ψ : P5 → P5 , ψ(x) = X∅ x2 , and 0 B5+ (23) ∩ Ker(Sq ∗ )(5,9) = B5+ (3, 2, 2, 1) ∪ B5+ (3, 4, 1, 1) ∪ B5+ (3, 4, 3). KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 37 4.2.1. B5+ (3, 2, 2, 1) is the set of 293 monomials: 1. x1 x2 x3 x64 x14 5 5. x1 x2 x23 x64 x13 5 5 9. x1 x2 x23 x14 4 x5 3 12 6 13. x1 x2 x3 x4 x5 17. x1 x2 x63 x34 x12 5 4 21. x1 x2 x63 x11 4 x5 7 10 4 25. x1 x2 x3 x4 x5 6 29. x1 x2 x14 3 x4 x5 33. x1 x22 x3 x64 x13 5 5 37. x1 x22 x3 x14 4 x5 5 41. x1 x22 x33 x12 x 4 5 45. x1 x22 x43 x74 x95 49. x1 x22 x53 x4 x14 5 53. x1 x22 x53 x74 x85 4 57. x1 x22 x53 x11 4 x5 61. x1 x22 x73 x54 x85 7 65. x1 x22 x12 3 x4 x5 2 13 2 5 69. x1 x2 x3 x4 x5 4 73. x1 x22 x15 3 x4 x5 3 14 4 77. x1 x2 x3 x4 x5 4 81. x1 x32 x23 x13 4 x5 3 4 2 13 85. x1 x2 x3 x4 x5 89. x1 x32 x43 x84 x75 93. x1 x32 x43 x14 4 x5 4 97. x1 x32 x53 x10 4 x5 3 6 8 5 101. x1 x2 x3 x4 x5 105. x1 x32 x73 x84 x45 6 109. x1 x32 x12 3 x4 x5 6 113. x1 x2 x3 x4 x14 5 117. x1 x62 x3 x74 x85 121. x1 x62 x33 x4 x12 5 125. x1 x62 x33 x94 x45 129. x1 x62 x93 x24 x55 133. x1 x72 x3 x24 x12 5 137. x1 x72 x23 x44 x95 141. x1 x72 x23 x12 4 x5 145. x1 x72 x63 x84 x5 4 149. x1 x72 x10 3 x4 x5 14 3 4 153. x1 x2 x3 x4 x5 2 4 157. x1 x15 2 x3 x4 x5 3 6 12 161. x1 x2 x3 x4 x5 165. x31 x2 x23 x54 x12 5 4 169. x31 x2 x33 x12 4 x5 3 4 6 9 173. x1 x2 x3 x4 x5 5 177. x31 x2 x43 x10 4 x5 3 5 6 8 181. x1 x2 x3 x4 x5 185. x31 x2 x63 x44 x95 6 2. x1 x2 x3 x14 4 x5 2 7 12 6. x1 x2 x3 x4 x5 4 10. x1 x2 x23 x15 4 x5 3 14 4 14. x1 x2 x3 x4 x5 18. x1 x2 x63 x64 x95 22. x1 x2 x63 x14 4 x5 6 26. x1 x2 x14 3 x4 x5 15 2 4 30. x1 x2 x3 x4 x5 34. x1 x22 x3 x74 x12 5 4 38. x1 x22 x3 x15 4 x5 4 42. x1 x22 x33 x13 x 4 5 46. x1 x22 x43 x94 x75 50. x1 x22 x53 x24 x13 5 54. x1 x22 x53 x84 x75 58. x1 x22 x53 x14 4 x5 62. x1 x22 x73 x84 x55 3 5 66. x1 x22 x12 3 x4 x5 2 13 3 4 70. x1 x2 x3 x4 x5 74. x1 x32 x3 x44 x14 5 78. x1 x32 x23 x44 x13 5 82. x1 x32 x33 x44 x12 5 86. x1 x32 x43 x34 x12 5 90. x1 x32 x43 x94 x65 94. x1 x32 x53 x24 x12 5 98. x1 x32 x63 x4 x12 5 102. x1 x32 x63 x94 x45 6 106. x1 x32 x12 3 x4 x5 3 13 2 4 110. x1 x2 x3 x4 x5 114. x1 x62 x3 x24 x13 5 5 118. x1 x62 x3 x10 4 x5 6 3 4 9 122. x1 x2 x3 x4 x5 126. x1 x62 x33 x12 4 x5 130. x1 x62 x93 x34 x45 134. x1 x72 x3 x64 x85 138. x1 x72 x23 x54 x85 142. x1 x72 x33 x44 x85 146. x1 x72 x83 x24 x55 4 150. x1 x72 x10 3 x4 x5 14 154. x1 x2 x3 x64 x5 2 4 158. x1 x15 2 x3 x4 x5 3 12 6 162. x1 x2 x3 x4 x5 5 166. x31 x2 x23 x12 4 x5 3 4 14 170. x1 x2 x3 x4 x5 174. x31 x2 x43 x74 x85 4 178. x31 x2 x43 x11 4 x5 3 5 8 6 182. x1 x2 x3 x4 x5 186. x31 x2 x63 x54 x85 3. x1 x2 x23 x44 x15 5 7 7. x1 x2 x23 x12 4 x5 3 4 14 11. x1 x2 x3 x4 x5 15. x1 x2 x63 x4 x14 5 19. x1 x2 x63 x74 x85 23. x1 x2 x73 x24 x12 5 2 5 27. x1 x2 x14 3 x4 x5 31. x1 x22 x3 x44 x15 5 7 35. x1 x22 x3 x12 4 x5 39. x1 x22 x33 x44 x13 5 43. x1 x22 x43 x4 x15 5 5 47. x1 x22 x43 x11 4 x5 51. x1 x22 x53 x34 x12 5 55. x1 x22 x53 x94 x65 59. x1 x22 x73 x4 x12 5 63. x1 x22 x73 x94 x45 7 67. x1 x22 x12 3 x4 x5 2 13 6 71. x1 x2 x3 x4 x5 75. x1 x32 x3 x64 x12 5 79. x1 x32 x23 x54 x12 5 4 83. x1 x32 x33 x12 4 x5 3 4 6 9 87. x1 x2 x3 x4 x5 5 91. x1 x32 x43 x10 4 x5 3 5 6 8 95. x1 x2 x3 x4 x5 99. x1 x32 x63 x44 x95 103. x1 x32 x63 x12 4 x5 2 5 107. x1 x32 x12 x 3 4 x5 3 14 111. x1 x2 x3 x4 x45 115. x1 x62 x3 x34 x12 5 4 119. x1 x62 x3 x11 4 x5 6 3 5 8 123. x1 x2 x3 x4 x5 127. x1 x62 x73 x4 x85 4 131. x1 x62 x11 3 x4 x5 7 10 4 135. x1 x2 x3 x4 x5 139. x1 x72 x23 x84 x55 143. x1 x72 x33 x84 x45 147. x1 x72 x83 x34 x45 6 151. x1 x14 2 x3 x4 x5 14 3 155. x1 x2 x3 x4 x45 2 4 159. x1 x15 2 x3 x4 x5 3 4 163. x1 x2 x3 x14 4 x5 3 2 13 4 167. x1 x2 x3 x4 x5 171. x31 x2 x43 x24 x13 5 175. x31 x2 x43 x84 x75 179. x31 x2 x43 x14 4 x5 4 183. x31 x2 x53 x10 4 x5 3 6 8 5 187. x1 x2 x3 x4 x5 4. x1 x2 x23 x54 x14 5 6 8. x1 x2 x23 x13 4 x5 3 6 12 12. x1 x2 x3 x4 x5 16. x1 x2 x63 x24 x13 5 5 20. x1 x2 x63 x10 4 x5 7 6 8 24. x1 x2 x3 x4 x5 3 4 28. x1 x2 x14 3 x4 x5 2 5 14 32. x1 x2 x3 x4 x5 6 36. x1 x22 x3 x13 4 x5 40. x1 x22 x33 x54 x12 5 44. x1 x22 x43 x34 x13 5 48. x1 x22 x43 x15 4 x5 52. x1 x22 x53 x64 x95 5 56. x1 x22 x53 x10 4 x5 60. x1 x22 x73 x44 x95 64. x1 x22 x73 x12 4 x5 6 68. x1 x22 x13 x 3 4 x5 2 15 72. x1 x2 x3 x4 x45 6 76. x1 x32 x3 x12 4 x5 3 2 12 5 80. x1 x2 x3 x4 x5 84. x1 x32 x43 x4 x14 5 88. x1 x32 x43 x74 x85 4 92. x1 x32 x43 x11 4 x5 3 5 8 6 96. x1 x2 x3 x4 x5 100. x1 x32 x63 x54 x85 104. x1 x32 x73 x44 x85 3 4 108. x1 x32 x12 3 x4 x5 3 14 4 112. x1 x2 x3 x4 x5 116. x1 x62 x3 x64 x95 120. x1 x62 x3 x14 4 x5 124. x1 x62 x33 x84 x55 128. x1 x62 x73 x84 x5 4 132. x1 x62 x11 3 x4 x5 7 2 136. x1 x2 x3 x4 x12 5 140. x1 x72 x23 x94 x45 144. x1 x72 x63 x4 x85 148. x1 x72 x93 x24 x45 2 5 152. x1 x14 2 x3 x4 x5 14 3 4 156. x1 x2 x3 x4 x5 160. x31 x2 x3 x44 x14 5 164. x31 x2 x23 x44 x13 5 168. x31 x2 x33 x44 x12 5 172. x31 x2 x43 x34 x12 5 176. x31 x2 x43 x94 x65 180. x31 x2 x53 x24 x12 5 184. x31 x2 x63 x4 x12 5 188. x31 x2 x63 x94 x45 ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 38 189. 193. 197. 201. 205. 209. 213. 217. 221. 225. 229. 233. 237. 241. 245. 249. 253. 257. 261. 265. 269. 273. 277. 281. 285. 289. 293. x31 x2 x63 x12 4 x5 2 5 x31 x2 x12 3 x4 x5 3 14 x1 x2 x3 x4 x45 x31 x32 x43 x4 x12 5 x31 x32 x43 x94 x45 4 x31 x32 x12 3 x4 x5 3 4 3 12 x1 x2 x3 x4 x5 4 x31 x42 x3 x11 4 x5 3 4 3 5 8 x1 x2 x3 x4 x5 x31 x42 x73 x4 x85 4 x31 x42 x11 3 x4 x5 4 x x31 x52 x3 x10 4 5 x31 x52 x23 x84 x55 x31 x52 x33 x84 x45 x31 x52 x83 x34 x45 x31 x72 x3 x44 x85 x31 x72 x83 x4 x45 3 4 x31 x12 2 x3 x4 x5 x71 x2 x3 x24 x12 5 x71 x2 x23 x44 x95 x71 x2 x23 x12 4 x5 x71 x2 x63 x84 x5 4 x71 x2 x10 3 x4 x5 7 3 4 8 x1 x2 x3 x4 x5 x71 x82 x3 x24 x55 x71 x92 x23 x4 x45 2 4 x15 1 x2 x3 x4 x5 . 190. 194. 198. 202. 206. 210. 214. 218. 222. 226. 230. 234. 238. 242. 246. 250. 254. 258. 262. 266. 270. 274. 278. 282. 286. 290. x31 x2 x73 x44 x85 3 4 x31 x2 x12 3 x4 x5 3 14 4 x1 x2 x3 x4 x5 x31 x32 x43 x44 x95 x31 x32 x43 x12 4 x5 4 x31 x32 x12 3 x4 x5 3 4 6 9 x1 x2 x3 x4 x5 x31 x42 x3 x14 4 x5 x31 x42 x33 x84 x55 x31 x42 x73 x84 x5 x31 x52 x3 x24 x12 5 x31 x52 x23 x4 x12 5 x31 x52 x23 x94 x45 x31 x52 x63 x4 x85 x31 x52 x93 x24 x45 x31 x72 x3 x84 x45 x31 x72 x83 x44 x5 2 4 x31 x13 2 x3 x4 x5 x71 x2 x3 x64 x85 x71 x2 x23 x54 x85 x71 x2 x33 x44 x85 x71 x2 x83 x24 x55 4 x71 x2 x10 3 x4 x5 7 3 4 8 x1 x2 x3 x4 x5 x71 x82 x3 x34 x45 x71 x92 x23 x44 x5 191. 195. 199. 203. 207. 211. 215. 219. 223. 227. 231. 235. 239. 243. 247. 251. 255. 259. 263. 267. 271. 275. 279. 283. 287. 291. x31 x2 x73 x84 x45 6 x31 x2 x12 3 x4 x5 3 3 4 12 x1 x2 x3 x4 x5 x31 x32 x43 x54 x85 x31 x32 x53 x44 x85 x31 x42 x3 x4 x14 5 x31 x42 x3 x74 x85 x31 x42 x33 x4 x12 5 x31 x42 x33 x94 x45 x31 x42 x93 x24 x55 x31 x52 x3 x64 x85 x31 x52 x23 x44 x95 x31 x52 x23 x12 4 x5 x31 x52 x63 x84 x5 4 x31 x52 x10 3 x4 x5 x31 x72 x43 x4 x85 2 5 x31 x12 2 x3 x4 x5 2 4 x31 x13 x x 2 3 4 x5 4 x x71 x2 x3 x10 4 5 7 2 8 5 x1 x2 x3 x4 x5 x71 x2 x33 x84 x45 x71 x2 x83 x34 x45 x71 x32 x3 x44 x85 x71 x32 x83 x4 x45 x71 x82 x33 x4 x45 2 4 x15 1 x2 x3 x4 x5 192. 196. 200. 204. 208. 212. 216. 220. 224. 228. 232. 236. 240. 244. 248. 252. 256. 260. 264. 268. 272. 276. 280. 284. 288. 292. 6 x31 x2 x12 3 x4 x5 3 13 2 4 x1 x2 x3 x4 x5 4 x31 x32 x3 x12 4 x5 3 3 4 8 5 x1 x2 x3 x4 x5 x31 x32 x53 x84 x45 x31 x42 x3 x24 x13 5 5 x31 x42 x3 x10 4 x5 3 4 3 4 9 x1 x2 x3 x4 x5 x31 x42 x33 x12 4 x5 x31 x42 x93 x34 x45 x31 x52 x3 x84 x65 x31 x52 x23 x54 x85 x31 x52 x33 x44 x85 x31 x52 x83 x24 x55 4 x31 x52 x10 3 x4 x5 x31 x72 x43 x84 x5 3 4 x31 x12 2 x3 x4 x5 2 4 x31 x13 x x 2 3 4 x5 x71 x2 x23 x4 x12 5 x71 x2 x23 x94 x45 x71 x2 x63 x4 x85 x71 x2 x93 x24 x45 x71 x32 x3 x84 x45 x71 x32 x83 x44 x5 x71 x92 x3 x24 x45 2 4 x15 1 x2 x3 x4 x5 297. 301. 305. 309. 313. 317. 321. 325. 329. 333. 337. 341. 345. 349. 353. 357. 361. 365. 369. 373. x1 x22 x33 x24 x15 5 3 x1 x22 x33 x14 4 x5 2 7 10 3 x1 x2 x3 x4 x5 x1 x32 x23 x24 x15 5 3 x1 x32 x23 x14 4 x5 3 3 14 2 x1 x2 x3 x4 x5 2 x1 x32 x63 x11 4 x5 3 14 3 2 x1 x2 x3 x4 x5 3 x1 x72 x23 x10 4 x5 7 10 2 3 x1 x2 x3 x4 x5 2 3 2 x1 x15 2 x3 x4 x5 3 2 6 11 x1 x2 x3 x4 x5 x31 x2 x33 x24 x14 5 x31 x2 x63 x34 x10 5 2 x31 x2 x73 x10 4 x5 2 14 3 3 x1 x2 x3 x4 x5 2 x31 x32 x53 x10 4 x5 3 5 2 10 3 x1 x2 x3 x4 x5 2 3 x31 x52 x10 3 x4 x5 3 7 10 2 x1 x2 x3 x4 x5 4.2.2. B5+ (3, 4, 1, 1) is the set of 105 monomials: 294. 298. 302. 306. 310. 314. 318. 322. 326. 330. 334. 338. 342. 346. 350. 354. 358. 362. 366. 370. x1 x22 x23 x34 x15 5 x1 x22 x33 x34 x14 5 2 x1 x22 x33 x15 4 x5 2 7 11 2 x1 x2 x3 x4 x5 x1 x32 x23 x34 x14 5 2 x1 x32 x23 x15 4 x5 3 6 2 11 x1 x2 x3 x4 x5 x1 x32 x73 x24 x10 5 2 2 x1 x32 x15 3 x4 x5 2 x1 x72 x23 x11 4 x5 7 10 3 2 x1 x2 x3 x4 x5 3 2 2 x1 x15 2 x3 x4 x5 3 2 7 10 x1 x2 x3 x4 x5 x31 x2 x33 x64 x10 5 3 x31 x2 x63 x10 4 x5 3 14 2 3 x1 x2 x3 x4 x5 x31 x32 x3 x64 x10 5 2 2 x31 x32 x13 3 x4 x5 2 x31 x52 x23 x11 4 x5 3 5 10 3 2 x1 x2 x3 x4 x5 295. 299. 303. 307. 311. 315. 319. 323. 327. 331. 335. 339. 343. 347. 351. 355. 359. 363. 367. 371. x1 x22 x23 x74 x11 5 x1 x22 x33 x64 x11 5 x1 x22 x73 x24 x11 5 2 3 x1 x22 x15 3 x4 x5 x1 x32 x23 x64 x11 5 x1 x32 x33 x24 x14 5 x1 x32 x63 x34 x10 5 2 x1 x32 x73 x10 4 x5 7 2 2 11 x1 x2 x3 x4 x5 x1 x72 x33 x24 x10 5 2 2 x1 x72 x11 3 x4 x5 x31 x2 x23 x24 x15 5 3 x31 x2 x23 x14 4 x5 3 3 14 2 x1 x2 x3 x4 x5 2 x31 x2 x63 x11 4 x5 3 14 3 2 x1 x2 x3 x4 x5 2 x31 x32 x3 x14 4 x5 3 5 2 2 11 x1 x2 x3 x4 x5 x31 x52 x33 x24 x10 5 2 2 x31 x52 x11 3 x4 x5 296. 300. 304. 308. 312. 316. 320. 324. 328. 332. 336. 340. 344. 348. 352. 356. 360. 364. 368. 372. 3 x1 x22 x23 x15 4 x5 2 3 7 10 x1 x2 x3 x4 x5 x1 x22 x73 x34 x10 5 3 2 x1 x22 x15 3 x4 x5 x1 x32 x23 x74 x10 5 x1 x32 x33 x64 x10 5 3 x1 x32 x63 x10 4 x5 3 14 2 3 x1 x2 x3 x4 x5 x1 x72 x23 x34 x10 5 2 x1 x72 x33 x10 4 x5 15 2 2 3 x1 x2 x3 x4 x5 x31 x2 x23 x34 x14 5 2 x31 x2 x23 x15 4 x5 3 6 2 11 x1 x2 x3 x4 x5 x31 x2 x73 x24 x10 5 2 2 x31 x2 x15 3 x4 x5 x31 x32 x53 x24 x10 5 x31 x52 x23 x34 x10 5 2 x31 x52 x33 x10 4 x5 3 7 2 10 x1 x2 x3 x4 x5 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 374. 378. 382. 386. 390. 394. 398. x31 x72 x93 x24 x25 2 2 x31 x15 2 x3 x4 x5 7 2 11 2 x1 x2 x3 x4 x5 3 2 x71 x2 x10 3 x4 x5 7 3 9 2 2 x1 x2 x3 x4 x5 2 2 x71 x11 2 x3 x4 x5 15 3 x1 x2 x3 x24 x25 . 375. 379. 383. 387. 391. 395. 2 2 3 x31 x13 2 x3 x4 x5 7 2 2 11 x1 x2 x3 x4 x5 x71 x2 x33 x24 x10 5 2 2 x71 x2 x11 3 x4 x5 x71 x92 x23 x24 x35 2 2 3 x15 1 x2 x3 x4 x5 376. 380. 384. 388. 392. 396. 2 3 2 x31 x13 2 x3 x4 x5 7 2 3 10 x1 x2 x3 x4 x5 2 x71 x2 x33 x10 4 x5 7 3 2 10 x1 x2 x3 x4 x5 x71 x92 x23 x34 x25 2 3 2 x15 1 x2 x3 x4 x5 377. 381. 385. 389. 393. 397. 3 2 2 x31 x13 2 x3 x4 x5 7 2 10 3 x1 x2 x3 x4 x5 2 3 x71 x2 x10 3 x4 x5 7 3 10 2 x1 x2 x3 x4 x5 x71 x92 x33 x24 x25 3 2 2 x15 1 x2 x3 x4 x5 4.2.2. B5+ (3, 4, 3) is the set of 24 monomials: 399. 403. 407. 411. 415. 419. x1 x32 x63 x64 x75 x31 x2 x63 x64 x75 x31 x52 x23 x64 x75 x31 x52 x63 x34 x65 x31 x52 x73 x64 x25 x71 x2 x33 x64 x65 400. 404. 408. 412. 416. 420. x1 x32 x63 x74 x65 x31 x2 x63 x74 x65 x31 x52 x23 x74 x65 x31 x52 x63 x64 x35 x31 x72 x3 x64 x65 x71 x32 x3 x64 x65 401. 405. 409. 413. 417. 421. x1 x32 x73 x64 x65 x31 x2 x73 x64 x65 x31 x52 x33 x64 x65 x31 x52 x63 x74 x25 x31 x72 x53 x24 x65 x71 x32 x53 x24 x65 402. 406. 410. 414. 418. 422. x1 x72 x33 x64 x65 x31 x32 x53 x64 x65 x31 x52 x63 x24 x75 x31 x52 x73 x24 x65 x31 x72 x53 x64 x25 x71 x32 x53 x64 x25 . A5.1. The admissible monomials of degree 35 in P4 . B4 (35) is the set of 120 monomials: 1. x2 x33 x31 4 5. x32 x33 x29 4 9. x32 x31 3 x4 3 13. x31 2 x3 x4 31 3 17. x1 x3 x4 21. x1 x2 x73 x26 4 25. x1 x22 x33 x29 4 29. x1 x22 x31 3 x4 33. x1 x32 x33 x28 4 37. x1 x32 x73 x24 4 41. x1 x32 x31 3 45. x1 x72 x27 4 49. x1 x72 x26 3 x4 3 53. x1 x31 x 2 4 3 31 57. x1 x3 x4 3 61. x31 x29 3 x4 3 2 29 65. x1 x2 x3 x4 69. x31 x2 x63 x25 4 73. x31 x2 x30 3 x4 77. x31 x32 x43 x25 4 81. x31 x42 x3 x27 4 85. x31 x52 x3 x26 4 89. x31 x52 x27 3 3 93. x31 x29 2 x4 3 31 97. x1 x2 x4 101. x71 x27 3 x4 105. x71 x2 x33 x24 4 109. x71 x32 x3 x24 4 3 113. x31 1 x3 x4 2 117. x31 1 x2 x3 x4 2. x2 x73 x27 4 6. x32 x53 x27 4 10. x72 x3 x27 4 3 14. x31 2 x3 x4 18. x1 x2 x23 x31 4 3 22. x1 x2 x30 3 x4 2 5 27 26. x1 x2 x3 x4 30. x1 x32 x31 4 34. x1 x32 x43 x27 4 3 38. x1 x32 x28 3 x4 6 27 42. x1 x2 x3 x4 46. x1 x72 x3 x26 4 50. x1 x72 x27 3 2 54. x1 x31 2 x3 x4 3 3 29 58. x1 x3 x4 62. x31 x31 3 x4 66. x31 x2 x33 x28 4 70. x31 x2 x73 x24 4 74. x31 x2 x31 3 78. x31 x32 x53 x24 4 82. x31 x42 x33 x25 4 86. x31 x52 x23 x25 4 90. x31 x72 x25 4 2 94. x31 x29 2 x3 x4 3 31 98. x1 x2 x3 102. x71 x2 x27 4 106. x71 x2 x26 3 x4 110. x71 x32 x25 3 3 114. x31 1 x3 x4 3 118. x31 1 x2 x3 3 3. x2 x31 3 x4 3 7 25 7. x2 x3 x4 11. x72 x33 x25 4 15. x1 x33 x31 4 19. x1 x2 x33 x30 4 2 23. x1 x2 x31 3 x4 2 7 25 27. x1 x2 x3 x4 31. x1 x32 x3 x30 4 35. x1 x32 x53 x26 4 2 39. x1 x32 x29 3 x4 6 3 25 43. x1 x2 x3 x4 47. x1 x72 x23 x25 4 3 51. x1 x30 2 x3 x4 31 2 55. x1 x2 x3 x4 59. x31 x53 x27 4 63. x31 x2 x31 4 67. x31 x2 x43 x27 4 3 71. x31 x2 x28 3 x4 3 3 29 75. x1 x2 x4 79. x31 x32 x28 3 x4 83. x31 x42 x27 3 x4 87. x31 x52 x33 x24 4 91. x31 x72 x3 x24 4 2 95. x31 x29 2 x3 x4 99. x71 x3 x27 4 103. x71 x2 x3 x26 4 107. x71 x2 x27 3 111. x71 x27 2 x4 3 115. x31 x 1 2 x4 31 3 119. x1 x2 x4 39 4. x32 x3 x31 4 3 8. x32 x29 3 x4 7 27 12. x2 x3 x4 16. x1 x73 x27 4 20. x1 x2 x63 x27 4 24. x1 x22 x3 x31 4 3 28. x1 x22 x29 3 x4 3 2 29 32. x1 x2 x3 x4 36. x1 x32 x63 x25 4 40. x1 x32 x30 3 x4 44. x1 x62 x27 3 x4 48. x1 x72 x33 x24 4 3 52. x1 x30 2 x3 x4 3 56. x1 x31 2 x3 3 7 25 60. x1 x3 x4 64. x31 x2 x3 x30 4 68. x31 x2 x53 x26 4 2 72. x31 x2 x29 3 x4 3 3 28 76. x1 x2 x3 x4 80. x31 x32 x29 3 84. x31 x52 x27 4 88. x31 x52 x26 3 x4 92. x31 x72 x25 3 3 96. x31 x29 2 x3 7 3 25 100. x1 x3 x4 104. x71 x2 x23 x25 4 108. x71 x32 x25 4 112. x71 x27 2 x3 2 116. x31 x 1 2 x3 x4 31 3 120. x1 x2 x3 . ´ TÍN‡ NGUYỄN SUM† AND NGUYỄN KHĂC 40 A5.2. The admissible monomials of degree 35 in P5 . 0 We have B5 (35) = B50 (35) ∪ ψ(B5 (15)) ∪ B5+ (35) ∩ Ker(Sq ∗ )(5,15) , where B50 (35) = Φ0 (B4 (35)), |B50 (35)| = 460, |ψ(B5 (15))| = 432 and 0 B5+ (15) ∩ Ker(Sq ∗ )(5,15) = B5+ (3, 2, 1, 1, 1) ∪ B5+ (3, 2, 3, 2) ∪ B5+ (3, 4, 2, 2) ∪ B5+ (3, 4, 4, 1). 5.2.1. B5+ (3, 2, 1, 1, 1) is the set of 169 monomials: 1. x1 x2 x3 x24 x30 5 5. x1 x2 x23 x24 x29 5 9. x1 x2 x23 x64 x25 5 13. x1 x2 x23 x30 4 x5 2 17. x1 x2 x33 x28 4 x5 21. x1 x2 x63 x26 x 4 5 25. x1 x22 x3 x4 x30 5 29. x1 x22 x3 x54 x26 5 2 33. x1 x22 x3 x29 4 x5 2 3 5 24 37. x1 x2 x3 x4 x5 41. x1 x22 x43 x94 x19 5 45. x1 x22 x53 x4 x26 5 49. x1 x22 x53 x94 x18 5 2 53. x1 x22 x53 x25 4 x5 2 7 9 16 57. x1 x2 x3 x4 x5 2 61. x1 x22 x29 3 x4 x5 3 6 24 65. x1 x2 x3 x4 x5 69. x1 x32 x23 x54 x24 5 73. x1 x32 x43 x24 x25 5 17 77. x1 x32 x43 x10 4 x5 3 4 26 81. x1 x2 x3 x4 x5 2 85. x1 x32 x53 x24 4 x5 3 6 24 89. x1 x2 x3 x4 x5 93. x1 x62 x3 x4 x26 5 97. x1 x62 x33 x4 x24 5 101. x1 x72 x3 x24 x24 5 105. x1 x72 x23 x24 4 x5 109. x31 x2 x3 x24 x28 5 113. x31 x2 x23 x4 x28 5 117. x31 x2 x33 x44 x24 5 121. x31 x2 x43 x84 x19 5 3 125. x31 x2 x43 x24 4 x5 3 5 8 18 129. x1 x2 x3 x4 x5 133. x31 x2 x63 x84 x17 5 2 137. x31 x2 x28 3 x4 x5 3 3 4 8 17 141. x1 x2 x3 x4 x5 145. x31 x42 x3 x4 x26 5 149. x31 x42 x33 x4 x24 5 2. x1 x2 x3 x64 x26 5 6. x1 x2 x23 x34 x28 5 10. x1 x2 x23 x74 x24 5 14. x1 x2 x33 x24 x28 5 18. x1 x2 x63 x4 x26 5 22. x1 x2 x73 x24 x24 5 26. x1 x22 x3 x24 x29 5 30. x1 x22 x3 x64 x25 5 34. x1 x22 x3 x30 4 x5 38. x1 x22 x33 x28 4 x5 17 42. x1 x22 x43 x11 4 x5 2 5 2 25 46. x1 x2 x3 x4 x5 17 50. x1 x22 x53 x10 4 x5 2 5 26 54. x1 x2 x3 x4 x5 58. x1 x22 x73 x24 4 x5 2 62. x1 x22 x29 x 3 4 x5 3 28 2 66. x1 x2 x3 x4 x5 70. x1 x32 x23 x28 4 x5 74. x1 x32 x43 x34 x24 5 16 78. x1 x32 x43 x11 4 x5 3 5 2 24 82. x1 x2 x3 x4 x5 86. x1 x32 x63 x4 x24 5 90. x1 x32 x73 x84 x16 5 94. x1 x62 x3 x24 x25 5 98. x1 x62 x33 x84 x17 5 102. x1 x72 x23 x4 x24 5 106. x1 x72 x33 x84 x16 5 110. x31 x2 x3 x44 x26 5 114. x31 x2 x23 x44 x25 5 118. x31 x2 x43 x4 x26 5 122. x31 x2 x43 x94 x18 5 2 126. x31 x2 x43 x25 4 x5 3 5 10 16 130. x1 x2 x3 x4 x5 134. x31 x2 x63 x94 x16 5 2 138. x31 x2 x28 3 x4 x5 142. x31 x32 x43 x94 x16 5 146. x31 x42 x3 x24 x25 5 150. x31 x42 x33 x84 x17 5 2 3. x1 x2 x3 x30 4 x5 7. x1 x2 x23 x44 x27 5 3 11. x1 x2 x23 x28 4 x5 15. x1 x2 x33 x44 x26 5 19. x1 x2 x63 x24 x25 5 2 23. x1 x2 x30 3 x4 x5 27. x1 x22 x3 x34 x28 5 31. x1 x22 x3 x74 x24 5 35. x1 x22 x33 x4 x28 5 39. x1 x22 x43 x4 x27 5 3 43. x1 x22 x43 x25 4 x5 2 5 3 24 47. x1 x2 x3 x4 x5 16 51. x1 x22 x53 x11 4 x5 2 7 24 55. x1 x2 x3 x4 x5 3 59. x1 x22 x28 3 x4 x5 3 2 28 63. x1 x2 x3 x4 x5 67. x1 x32 x23 x4 x28 5 71. x1 x32 x33 x44 x24 5 75. x1 x32 x43 x84 x19 5 3 79. x1 x32 x43 x24 4 x5 3 5 8 18 83. x1 x2 x3 x4 x5 87. x1 x32 x63 x84 x17 5 2 91. x1 x32 x28 3 x4 x5 6 3 24 95. x1 x2 x3 x4 x5 99. x1 x62 x33 x94 x16 5 103. x1 x72 x23 x84 x17 5 2 107. x1 x30 2 x3 x4 x5 3 6 24 111. x1 x2 x3 x4 x5 115. x31 x2 x23 x54 x24 5 119. x31 x2 x43 x24 x25 5 17 123. x31 x2 x43 x10 4 x5 4 26 3 127. x1 x2 x3 x4 x5 2 131. x31 x2 x53 x24 4 x5 3 6 24 135. x1 x2 x3 x4 x5 139. x31 x32 x3 x44 x24 5 143. x31 x32 x43 x24 4 x5 147. x31 x42 x3 x34 x24 5 151. x31 x42 x33 x94 x16 5 4. x1 x2 x23 x4 x30 5 8. x1 x2 x23 x54 x26 5 2 12. x1 x2 x23 x29 4 x5 16. x1 x2 x33 x64 x24 5 20. x1 x2 x63 x34 x24 5 2 24. x1 x2 x30 3 x4 x5 28. x1 x22 x3 x44 x27 5 3 32. x1 x22 x3 x28 4 x5 2 3 4 25 36. x1 x2 x3 x4 x5 40. x1 x22 x43 x34 x25 5 44. x1 x22 x43 x27 4 x5 48. x1 x22 x53 x84 x19 5 3 52. x1 x22 x53 x24 4 x5 2 7 8 17 56. x1 x2 x3 x4 x5 3 60. x1 x22 x28 3 x4 x5 3 4 26 64. x1 x2 x3 x4 x5 68. x1 x32 x23 x44 x25 5 72. x1 x32 x43 x4 x26 5 76. x1 x32 x43 x94 x18 5 2 80. x1 x32 x43 x25 4 x5 3 5 10 16 84. x1 x2 x3 x4 x5 88. x1 x32 x63 x94 x16 5 2 92. x1 x32 x28 3 x4 x5 96. x1 x62 x3 x26 4 x5 100. x1 x62 x33 x24 4 x5 104. x1 x72 x23 x94 x16 5 2 108. x1 x30 2 x3 x4 x5 2 112. x31 x2 x3 x28 4 x5 3 2 28 116. x1 x2 x3 x4 x5 120. x31 x2 x43 x34 x24 5 16 124. x31 x2 x43 x11 4 x5 5 2 24 3 128. x1 x2 x3 x4 x5 132. x31 x2 x63 x4 x24 5 136. x31 x2 x73 x84 x16 5 140. x31 x32 x43 x4 x24 5 144. x31 x32 x53 x84 x16 5 148. x31 x42 x3 x26 4 x5 152. x31 x42 x33 x24 4 x5 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER x31 x52 x3 x24 x24 5 x31 x52 x23 x4 x24 5 x31 x52 x33 x84 x16 5 x71 x2 x23 x84 x17 5 x71 x32 x3 x84 x16 5 . 153. 157. 161. 165. 169. 154. 158. 162. 166. x31 x52 x3 x84 x18 5 x31 x52 x23 x84 x17 5 x31 x72 x3 x84 x16 5 x71 x2 x23 x94 x16 5 155. 159. 163. 167. 16 x31 x52 x3 x10 4 x5 3 5 2 9 16 x1 x2 x3 x4 x5 x71 x2 x3 x24 x24 5 x71 x2 x23 x24 4 x5 41 156. 160. 164. 168. 2 x31 x52 x3 x24 4 x5 3 5 2 24 x1 x2 x3 x4 x5 x71 x2 x23 x4 x24 5 x71 x2 x33 x84 x16 5 173. 177. 181. 185. 189. 193. 12 x1 x72 x33 x12 4 x5 3 3 5 12 12 x1 x2 x3 x4 x5 12 x31 x52 x23 x13 4 x5 3 5 10 12 5 x1 x2 x3 x4 x5 12 x31 x72 x3 x12 4 x5 7 3 12 12 x1 x2 x3 x4 x5 199. 203. 207. 211. 215. 219. 223. 227. 231. 235. 239. 243. 247. 251. 255. 259. 263. 14 x1 x32 x33 x14 4 x5 3 14 3 14 x1 x2 x3 x4 x5 6 10 x1 x72 x11 3 x4 x5 3 3 14 14 x1 x2 x3 x4 x5 3 14 x31 x2 x14 3 x4 x5 3 3 5 14 10 x1 x2 x3 x4 x5 14 x31 x52 x23 x11 4 x5 3 5 7 10 10 x1 x2 x3 x4 x5 14 2 x31 x52 x11 3 x4 x5 3 7 9 2 14 x1 x2 x3 x4 x5 10 2 x31 x72 x13 3 x4 x5 7 3 10 14 x1 x2 x3 x4 x5 10 x71 x32 x3 x14 4 x5 7 3 9 14 2 x1 x2 x3 x4 x5 x71 x92 x33 x24 x14 5 5 2 10 x71 x11 2 x3 x4 x5 15 2 2 x15 1 x2 x3 x4 x5 5.2.2. B5+ (3, 2, 3, 2) is the set of 26 monomials: 13 x1 x32 x63 x12 4 x5 3 6 12 13 x1 x2 x3 x4 x5 5 12 x31 x42 x11 3 x4 x5 3 5 3 12 12 x1 x2 x3 x4 x5 13 4 x31 x52 x10 3 x4 x5 3 7 9 4 12 x1 x2 x3 x4 x5 x71 x32 x93 x44 x12 5 170. 174. 178. 182. 186. 190. 194. 171. 175. 179. 183. 187. 191. 195. 12 x1 x32 x63 x13 4 x5 3 6 13 12 x1 x2 x3 x4 x5 12 5 x31 x42 x11 3 x4 x5 3 5 10 4 13 x1 x2 x3 x4 x5 4 12 x31 x52 x11 3 x4 x5 3 7 9 12 4 x1 x2 x3 x4 x5 4 x71 x32 x93 x12 4 x5 . 172. 176. 180. 184. 188. 192. 12 x1 x32 x73 x12 4 x5 3 7 12 12 x1 x2 x3 x4 x5 13 x31 x52 x23 x12 4 x5 3 5 10 5 12 x1 x2 x3 x4 x5 12 4 x31 x52 x11 3 x4 x5 7 3 12 12 x1 x2 x3 x4 x5 5.2.3. B5+ (3, 4, 2, 2) is the set of 69 monomials: 196. 200. 204. 208. 212. 216. 220. 224. 228. 232. 236. 240. 244. 248. 252. 256. 260. 264. 15 x1 x22 x23 x15 4 x5 3 6 11 14 x1 x2 x3 x4 x5 14 3 x1 x32 x14 3 x4 x5 15 2 2 15 x1 x2 x3 x4 x5 14 x31 x2 x63 x11 4 x5 3 14 14 3 x1 x2 x3 x4 x5 2 14 x31 x32 x13 3 x4 x5 3 5 3 10 14 x1 x2 x3 x4 x5 3 14 x31 x52 x10 3 x4 x5 3 7 10 14 x1 x2 x3 x4 x5 x31 x72 x93 x64 x10 5 2 3 14 x31 x13 2 x3 x4 x5 10 x71 x2 x33 x14 4 x5 7 3 5 10 10 x1 x2 x3 x4 x5 2 10 x71 x32 x13 3 x4 x5 7 9 3 6 10 x1 x2 x3 x4 x5 5 10 2 x71 x11 2 x3 x4 x5 15 15 x1 x2 x3 x24 x25 . 197. 201. 205. 209. 213. 217. 221. 225. 229. 233. 237. 241. 245. 249. 253. 257. 261. 2 15 x1 x22 x15 3 x4 x5 3 7 10 14 x1 x2 x3 x4 x5 14 x1 x72 x33 x10 4 x5 15 2 15 2 x1 x2 x3 x4 x5 14 x31 x2 x73 x10 4 x5 3 3 14 14 x1 x2 x3 x4 x5 6 10 x31 x32 x13 3 x4 x5 3 5 3 14 10 x1 x2 x3 x4 x5 2 14 x31 x52 x11 3 x4 x5 3 7 14 10 x1 x2 x3 x4 x5 2 x31 x72 x93 x14 4 x5 3 13 3 2 14 x1 x2 x3 x4 x5 6 10 x71 x2 x11 3 x4 x5 7 3 9 2 14 x1 x2 x3 x4 x5 10 2 x71 x32 x13 3 x4 x5 7 9 3 14 2 x1 x2 x3 x4 x5 2 2 15 x15 1 x2 x3 x4 x5 198. 202. 206. 210. 214. 218. 222. 226. 230. 234. 238. 242. 246. 250. 254. 258. 262. 15 2 x1 x22 x15 3 x4 x5 3 7 14 10 x1 x2 x3 x4 x5 10 x1 x72 x33 x14 4 x5 15 15 2 2 x1 x2 x3 x4 x5 10 x31 x2 x73 x14 4 x5 3 3 5 10 14 x1 x2 x3 x4 x5 14 2 x31 x32 x13 3 x4 x5 3 5 6 11 10 x1 x2 x3 x4 x5 6 10 x31 x52 x11 3 x4 x5 3 7 5 10 10 x1 x2 x3 x4 x5 2 10 x31 x72 x13 3 x4 x5 3 13 3 14 2 x1 x2 x3 x4 x5 14 x71 x32 x3 x10 4 x5 7 3 9 6 10 x1 x2 x3 x4 x5 x71 x92 x23 x34 x14 5 6 10 x71 x11 2 x3 x4 x5 2 15 2 x15 1 x2 x3 x4 x5 5.2.4. B5+ (3, 4, 4, 1) is the set of 15 monomials: 265. 269. 273. 277. x31 x52 x63 x64 x15 5 6 x31 x52 x73 x14 4 x5 6 6 x x x31 x72 x13 3 4 5 6 6 x71 x32 x13 3 x4 x5 266. 270. 274. 278. x31 x52 x63 x74 x14 5 6 6 x31 x52 x15 3 x4 x5 5 6 6 x31 x15 2 x3 x4 x5 5 6 6 x x71 x11 2 3 x4 x5 267. 271. 275. 279. 6 268. x31 x52 x73 x64 x14 x31 x52 x63 x15 5 4 x5 6 272. x31 x72 x53 x14 x31 x72 x53 x64 x14 4 x5 5 7 3 5 14 6 276. x x x x x x71 x32 x53 x64 x14 5 1 2 3 4 5 3 5 6 6 x15 1 x2 x3 x4 x5 . [...]... the Singer transfer, C R Math Acad Sci Paris 349 (2011) 21-23, MR2755689 [7] P H Chơn and L M Hà, On May spectral sequence and the algebraic transfer, Manuscripta Math 138 (2012) 141-160, MR2898751 [8] P H Chơn and L M Hà, On the May spectral sequence and the algebraic transfer II, Topology Appl 178 (2014) 372-383, MR3276753 [9] L M Hà, Sub-Hopf algebras of the Steenrod algebra and the Singer transfer, ... x31 x2 x43 x4 + x31 x42 x4 x5 + x31 x42 x3 x5 + x31 x42 x3 x4 Proof We prove the Part (ii) of the lemma The others can be proved by a similar computation From Propositions 4.1.2 and 4.1.3, we see that dim [Σ5 (u4 )] = 21 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 13 with a basis consisting of the classes represented by the following monomials: v1 = x1 x2 x3 x24 x45 , v2 = x1 x2 x23 x4 x45 , v3... admissible, one gets x ∈ D This implies B5+ (ω(2) ) ⊂ D We now prove the set [D] is linearly independent in (F2 ⊗A P5 )23 Suppose there 398 is a linear relation S = t=294 γt dt ≡ 0, where γt ∈ F2 and dt = d23,t ∈ D We KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 19 explicitly compute p(i;j) (S) in terms of the admissible monomials in P4 From the relation p(i;j) (S) ≡ 0 with 1 i < j 5, one gets γt = 0... we obtain g5 (f ) + f ≡ω(3) γb1 + other terms ≡ω(3) 0 This relation implies γ = 0 The proposition is proved Proposition 4.3.12 QP5 (ω(2) )GL5 = 0 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 21 By computing from Proposition 4.3.5, we see that dim QP5 (ω(2) ) = 105 with the 4 basis j=1 [B(aj )]ω(2) By a direct computation, we obtain the following Lemma 4.3.13 i) The subspaces [Σ5 (aj )]ω(2) , 1... 4, 5 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 27 Proposition 5.3.2 There exist exactly 169 admissible monomials in P5+ such that their weight vectors are ω ¯ (1) Consequently dim QP5+ (¯ ω(1) ) = 169 We denote the monomials in B5 (¯ ω(1) ) by d35,t , 1 t 169 (see Appendix) Lemma 5.3.3 The following monomials are strictly inadmissible: i) fi (¯ x), 1 i 5, where x ¯ is one of the following monomials:... representations on the homology of power of real projective space, in: M C Tangora (Ed.), Algebraic Topology, Oaxtepec, 1991, in: Contemp Math., vol 146, 1993, pp 49-70, MR1224907 [3] R R Bruner, L M Hà and N H V Hưng, On behavior of the algebraic transfer, Trans Amer Math Soc 357 (2005) 473-487, MR2095619 [4] D P Carlisle and R M W Wood, The boundedness conjecture for the action of the Steenrod algebra... Ray and G Walker (ed.), Adams Memorial Symposium on Algebraic Topology 2, (Manchester, 1990), in: London Math Soc Lecture Notes Ser., Cambridge Univ Press, Cambridge, vol 176, 1992, pp 203-216, MR1232207 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER 31 [5] T W Chen, Determination of Ext5,∗ A (Z/2, Z/2), Topology Appl., 158 (2011) 660-689, MR2774051 [6] P H Chơn and L M Hà, Lambda algebra and the. .. Observe that z = x71 x32 is the minimal spike of degree 10 in P5 and ω(z) = (2, 2, 1) Since [x] = 0, by Theorem 2.10, either ω1 (x) = 2 or ω1 (x) = 4 If ω1 (x) = 2, then x = xi xj y 2 with y a monomial of degree 4 in P5 and i < j Since x is admissible, by Theorem 2.7, y is admissible and y ∈ P50 Using a result in [25], one gets either ω(y) = (2, 1) or ω(y) = (4, 0) If ω1 (x) = 4, then x = Xj y12 with y1... Amer Math Soc No 833, April 1987, pp 55-89 - V Phúc and N Sum, On the generators of the polynomial algebra as a mod[18] D ule over the Steenrod algebra, C R Math Acad Sci Paris, Ser I (2015), DOI: 10.1016/j.crma.2015.09.002 [19] V T N Quỳnh, On behavior of the fifth algebraic transfer, “Proceedings of the International School and Conference in Algebraic Topology, Hà Nô.i 2004”, Geom Topol Monogr.,... [24] N Sum, On the Peterson hit problem of five variables and its applications to the fifth Singer transfer, East-West J of Mathematics, 16 (2014) 47-62 [25] N Sum, On the Peterson hit problem, Adv Math 274 (2015) 432-489, MR3318156 [26] M C Tangora, On the cohomology of the Steenrod algebra, Math Zeit 116 (1970) 18-64, MR0266205 [27] R M W Wood, Steenrod squares of polynomials and the Peterson conjecture, ... prove the Part (ii) of the lemma The others can be proved by a similar computation From Propositions 4.1.2 and 4.1.3, we see that dim [Σ5 (u4 )] = 21 KAMEKO’S HOMOMORPHISM AND THE ALGEBRAIC TRANSFER. .. Chơn and L M Hà, On the May spectral sequence and the algebraic transfer II, Topology Appl 178 (2014) 372-383, MR3276753 [9] L M Hà, Sub-Hopf algebras of the Steenrod algebra and the Singer transfer, ... Hưng’s result in [11] on the relation between the algebraic transfer and Kameko’s homomorphism Sq ∗ : F2 ⊗A Pk −→ F2 ⊗A Pk This homomorphism is an GLk -homomorphism induced by the F2 -linear map,